16 files changed, 15889 insertions, 0 deletions
diff --git a/vol2/3.1.lyx b/vol2/3.1.lyx
new file mode 100644
index 0000000..81d7d26
--- /dev/null
+++ b/vol2/3.1.lyx
@@ -0,0 +1,1204 @@
+#LyX 2.4 created this file. For more info see https://www.lyx.org/
+\lyxformat 620
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\begin_preamble
+\input defs
+\end_preamble
+\use_default_options true
+\maintain_unincluded_children no
+\language english
+\language_package default
+\inputencoding utf8
+\fontencoding auto
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_roman_osf false
+\font_sans_osf false
+\font_typewriter_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\float_placement class
+\float_alignment class
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_formatted_ref 0
+\use_minted 0
+\use_lineno 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style english
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tablestyle default
+\tracking_changes false
+\output_changes false
+\change_bars false
+\postpone_fragile_content false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\docbook_table_output 0
+\docbook_mathml_prefix 1
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc1[20]
+\end_layout
+
+\end_inset
+
+Suppose that you wish to obtain a decimal digit at random,
+ not using a computer.
+ Which of the following methods would be suitable?
+\end_layout
+
+\begin_layout Enumerate
+Open a telephone directory to a random place by sticking your finger in it somewhere,
+ and use the units digit of the first number found on the selected page.
+\end_layout
+
+\begin_layout Enumerate
+Same as (a),
+ but use the units digit of the 
+\emph on
+page
+\emph default
+ number.
+\end_layout
+
+\begin_layout Enumerate
+Roll a die that is in the shape of a regular icosahedron,
+ whose twenty faces have been labeled with the digits 0,
+ 0,
+ 1,
+ 1,
+ ...,
+ 9,
+ 9.
+ Use the digit that appears on the top,
+ when the die comes to rest.
+ (A felt-covered table with a hard surface is recommended for rolling dice.)
+\end_layout
+
+\begin_layout Enumerate
+Expose a geiger counter to a source of radioactivity for one minute (shielding yourself) and use the units digit of the resulting count.
+ Assume that the geiger counter displays the number of counts in decimal notation,
+ and that the count is initially zero.
+\end_layout
+
+\begin_layout Enumerate
+Glance at your wristwatch;
+ and if the position of the second-hand is between 
+\begin_inset Formula $6n$
+\end_inset
+
+ and 
+\begin_inset Formula $6(n+1)$
+\end_inset
+
+ seconds,
+ choose the digit 
+\begin_inset Formula $n$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Ask a friend to think of a random digit,
+ and use the digit he names.
+\end_layout
+
+\begin_layout Enumerate
+Ask an enemy to think of a random digit,
+ and use the digit he names.
+\end_layout
+
+\begin_layout Enumerate
+Assume that 10 horses are entered in a race and that you know nothing whatever about their qualifications.
+ Assign to these horses the digits 0 to 9,
+ in arbitrary fashion,
+ and after the race use the winner's digit.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+This would work for a typical phone book.
+ It wouldn't work if the book has,
+ say,
+ entries ordered by number with a fixed number of entries per page that is a multiple of 2 or 5.
+ And,
+ according to the official solution,
+ in some places people can choose phone numbers (presumably in the US where Knuth lives) so it wouldn't work there as people would tend to choose round numbers.
+\end_layout
+
+\begin_layout Enumerate
+This would be suitable,
+ although some adjustment would have to be made to account for the fact that left pages would have even numbers and right pages would have round numbers (say,
+ we could take the even number and divide it by two) and to avoid the last few pages if the total number is not a multiple of 20.
+\end_layout
+
+\begin_layout Enumerate
+This would work very well.
+\end_layout
+
+\begin_layout Enumerate
+This would work well assuming there are several digits between the most significant one and the units.
+\end_layout
+
+\begin_layout Enumerate
+This would work assuming you only have to do it once at a time and not at a specific time.
+\end_layout
+
+\begin_layout Enumerate
+This wouldn't work,
+ as humans are not good at mentally generating random numbers.
+\end_layout
+
+\begin_layout Enumerate
+This definitely wouldn't work,
+ as the enemy would probably be compelled to choose the number strategically instead of randomly.
+\end_layout
+
+\begin_layout Enumerate
+This could work,
+ although it would be time consuming.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc3[10]
+\end_layout
+
+\end_inset
+
+What number follows 1010101010 in the middle-square method?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula $1010101010^{2}=1020304050403020100$
+\end_inset
+
+,
+ so the next number would be 3040504030.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc6[M21]
+\end_layout
+
+\end_inset
+
+Suppose that we want to generate a sequence of integers 
+\begin_inset Formula $X_{0},X_{1},X_{2},\dots$
+\end_inset
+
+,
+ in the range 
+\begin_inset Formula $0\leq X_{n}<m$
+\end_inset
+
+.
+ Let 
+\begin_inset Formula $f(x)$
+\end_inset
+
+ be any function such that 
+\begin_inset Formula $0\leq x<m$
+\end_inset
+
+ implies 
+\begin_inset Formula $0\leq f(x)<m$
+\end_inset
+
+.
+ Consider a sequence formed by the rule 
+\begin_inset Formula $X_{n+1}=f(X_{n})$
+\end_inset
+
+.
+ (Examples are the middle-square method and Algorithm K.)
+\end_layout
+
+\begin_layout Enumerate
+Show that the sequence is ultimately periodic,
+ in the sense that there exist numbers 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ and 
+\begin_inset Formula $\mu$
+\end_inset
+
+ for which the values
+\begin_inset Formula 
+\[
+X_{0},X_{1},\dots,X_{\mu},\dots,X_{\mu+\lambda-1}
+\]
+
+\end_inset
+
+are distinct,
+ but 
+\begin_inset Formula $X_{n+\lambda}=X_{n}$
+\end_inset
+
+ when 
+\begin_inset Formula $n\geq\mu$
+\end_inset
+
+.
+ Find the maximum and minimum possible values of 
+\begin_inset Formula $\mu$
+\end_inset
+
+ and 
+\begin_inset Formula $\lambda$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:e316b"
+
+\end_inset
+
+(R.
+ W.
+ Floyd.) Show that there exists an 
+\begin_inset Formula $n>0$
+\end_inset
+
+ such that 
+\begin_inset Formula $X_{n}=X_{2n}$
+\end_inset
+
+;
+ and the smallest such value of 
+\begin_inset Formula $n$
+\end_inset
+
+ lies in the range 
+\begin_inset Formula $\mu\le n\leq\mu+\lambda$
+\end_inset
+
+.
+ Furthermore the value of 
+\begin_inset Formula $X_{n}$
+\end_inset
+
+ is unique in the sense that if 
+\begin_inset Formula $X_{n}=X_{2n}$
+\end_inset
+
+ and 
+\begin_inset Formula $X_{r}=X_{2r}$
+\end_inset
+
+,
+ then 
+\begin_inset Formula $X_{r}=X_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Use the idea of part 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "enu:e316b"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ to design an algorithm that calculates 
+\begin_inset Formula $\mu$
+\end_inset
+
+ and 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ for any given function 
+\begin_inset Formula $f$
+\end_inset
+
+ and any given 
+\begin_inset Formula $X_{0}$
+\end_inset
+
+,
+ using only 
+\begin_inset Formula $O(\mu+\lambda)$
+\end_inset
+
+ steps and only a bounded number of memory locations.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Assume 
+\begin_inset Formula $f:\mathbb{Z}_{m}\to\mathbb{Z}_{m}$
+\end_inset
+
+,
+ as otherwise the statements below are not necessarily true (e.g.
+ if 
+\begin_inset Formula $f:\mathbb{R}\to\mathbb{R}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f(x)\coloneqq\frac{x}{2}$
+\end_inset
+
+,
+ and 
+\begin_inset Formula $X_{0}=\frac{m}{2}$
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Enumerate
+Require that 
+\begin_inset Formula $\mu\in\mathbb{N}$
+\end_inset
+
+ and 
+\begin_inset Formula $\lambda\in\mathbb{N}^{*}$
+\end_inset
+
+,
+ as otherwise it is unclear what is meant and,
+ in particular,
+ the statement is obviously true when 
+\begin_inset Formula $\mu\in\mathbb{N}$
+\end_inset
+
+ and 
+\begin_inset Formula $\lambda=0$
+\end_inset
+
+.
+ Now,
+ since 
+\begin_inset Formula $\{X_{0},\dots,X_{m}\}\subseteq\{0,\dots,m-1\}$
+\end_inset
+
+,
+ then 
+\begin_inset Formula $|\{X_{0},\dots,X_{m}\}|\leq m$
+\end_inset
+
+ and there must be integers 
+\begin_inset Formula $i$
+\end_inset
+
+ and 
+\begin_inset Formula $j$
+\end_inset
+
+,
+ 
+\begin_inset Formula $0\leq i<j\leq m$
+\end_inset
+
+,
+ such that 
+\begin_inset Formula $X_{i}=X_{j}$
+\end_inset
+
+.
+ We choose 
+\begin_inset Formula $i$
+\end_inset
+
+ and 
+\begin_inset Formula $j$
+\end_inset
+
+ such that 
+\begin_inset Formula $j$
+\end_inset
+
+ is minimum,
+ which ensures that 
+\begin_inset Formula $X_{0},\dots,X_{j-1}$
+\end_inset
+
+ are distinct and uniquely determines 
+\begin_inset Formula $i$
+\end_inset
+
+,
+ and we set 
+\begin_inset Formula $\mu\coloneqq i$
+\end_inset
+
+ and 
+\begin_inset Formula $\lambda\coloneqq j-i$
+\end_inset
+
+;
+ then 
+\begin_inset Formula $X_{0},\dots,X_{\mu+\lambda-1=j-1}$
+\end_inset
+
+ are all distinct and,
+ for 
+\begin_inset Formula $n\geq\mu$
+\end_inset
+
+,
+\begin_inset Formula 
+\[
+X_{n+\lambda}=X_{n-i+j}=f^{n-i}(X_{j})=f^{n-i}(X_{i})=X_{n}.
+\]
+
+\end_inset
+
+It's easy to check that these values are unique:
+ 
+\begin_inset Formula $\mu+\lambda$
+\end_inset
+
+ must be the first value such that 
+\begin_inset Formula $X_{\mu+\lambda}$
+\end_inset
+
+ equals some other value earlier in the sequence,
+ and 
+\begin_inset Formula $\mu$
+\end_inset
+
+ must be the only value less than 
+\begin_inset Formula $\mu+\lambda$
+\end_inset
+
+ such that 
+\begin_inset Formula $X_{\mu}=X_{\mu+\lambda}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+As for the maximum and minimum,
+ since 
+\begin_inset Formula $0\leq i<j\leq m$
+\end_inset
+
+,
+ we have 
+\begin_inset Formula $0\leq\mu\leq m-1$
+\end_inset
+
+ and 
+\begin_inset Formula $1\leq\lambda\leq m$
+\end_inset
+
+.
+ These bounds can be reached:
+ when 
+\begin_inset Formula $f$
+\end_inset
+
+ is the identity,
+ 
+\begin_inset Formula $\mu=0$
+\end_inset
+
+ and 
+\begin_inset Formula $\lambda=1$
+\end_inset
+
+;
+ when 
+\begin_inset Formula $f(x)=(x+1)\bmod m$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\lambda=m$
+\end_inset
+
+,
+ and when 
+\begin_inset Formula $f(x)=\max\{0,x-1\}$
+\end_inset
+
+ and 
+\begin_inset Formula $X_{0}=m-1$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\mu=m-1$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Let 
+\begin_inset Formula $k\coloneqq\lceil\frac{\mu}{\lambda}\rceil$
+\end_inset
+
+ and 
+\begin_inset Formula $n\coloneqq k\lambda$
+\end_inset
+
+,
+ then 
+\begin_inset Formula $\frac{\mu}{\lambda}\leq k\leq\frac{\mu}{\lambda}+1$
+\end_inset
+
+ and therefore 
+\begin_inset Formula $\mu\leq n\leq\mu+\lambda$
+\end_inset
+
+,
+ and 
+\begin_inset Formula $X_{n}=X_{n+\lambda}=X_{n+2\lambda}=\dots=X_{n+k\lambda=2n}$
+\end_inset
+
+.
+ For the uniqueness,
+ let 
+\begin_inset Formula $r>0$
+\end_inset
+
+ be such that 
+\begin_inset Formula $X_{r}=X_{2r}$
+\end_inset
+
+,
+ so necessarily 
+\begin_inset Formula $2r\geq\mu+\lambda$
+\end_inset
+
+.
+ We note that,
+ for any 
+\begin_inset Formula $s\geq\mu$
+\end_inset
+
+,
+ by induction,
+ 
+\begin_inset Formula $X_{s}=X_{s-\lfloor\frac{s-\mu}{\lambda}\rfloor\lambda}=X_{\mu+((s-\mu)\bmod\lambda)}$
+\end_inset
+
+,
+ and we call 
+\begin_inset Formula $R(s)\coloneqq\mu+((s-\mu)\bmod\lambda)\in\{\mu,\dots,\mu+\lambda-1\}$
+\end_inset
+
+.
+ With this,
+ 
+\begin_inset Formula $X_{R(2r)}=X_{2r}=X_{r}$
+\end_inset
+
+,
+ so either 
+\begin_inset Formula $r=R(2r)$
+\end_inset
+
+ or 
+\begin_inset Formula $r\geq\mu+\lambda$
+\end_inset
+
+;
+ in any case 
+\begin_inset Formula $r\geq\mu$
+\end_inset
+
+ and 
+\begin_inset Formula $X_{R(2r)}=X_{R(r)}$
+\end_inset
+
+.
+ This means 
+\begin_inset Formula $R(2r)=R(r)$
+\end_inset
+
+,
+ so 
+\begin_inset Formula $r\equiv0\mod{\lambda}$
+\end_inset
+
+ and 
+\begin_inset Formula $R(r)=\mu+(-\mu\bmod\lambda)\equiv0\bmod\lambda$
+\end_inset
+
+,
+ so 
+\begin_inset Formula $R(r)$
+\end_inset
+
+ is unique and 
+\begin_inset Formula $X_{r}=X_{R(r)}=X_{R(n)}=X_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+First we set 
+\begin_inset Formula $a,b\gets X_{0}$
+\end_inset
+
+ and then repeatedly set 
+\begin_inset Formula $a\gets f(a)$
+\end_inset
+
+ and 
+\begin_inset Formula $b\gets f^{2}(b)$
+\end_inset
+
+ until 
+\begin_inset Formula $a=b$
+\end_inset
+
+;
+ the number of times we do this operation is 
+\begin_inset Formula $n$
+\end_inset
+
+,
+ and in the end 
+\begin_inset Formula $a=X_{n}$
+\end_inset
+
+.
+ Then we set 
+\begin_inset Formula $b\gets a$
+\end_inset
+
+ and repeat 
+\begin_inset Formula $b\gets f(b)$
+\end_inset
+
+ until 
+\begin_inset Formula $b=a$
+\end_inset
+
+;
+ the number of times we do this operation is 
+\begin_inset Formula $\lambda$
+\end_inset
+
+.
+ Finally,
+ since 
+\begin_inset Formula $a$
+\end_inset
+
+ is a multiple of 
+\begin_inset Formula $\lambda$
+\end_inset
+
+,
+ 
+\begin_inset Formula $X_{\mu}=X_{\mu+n}$
+\end_inset
+
+,
+ so we set 
+\begin_inset Formula $b\gets X_{0}$
+\end_inset
+
+ and repeat 
+\begin_inset Formula $b\gets f(b)$
+\end_inset
+
+ and 
+\begin_inset Formula $a\gets f(a)$
+\end_inset
+
+ until 
+\begin_inset Formula $a=b$
+\end_inset
+
+;
+ the number of times we do this operation is 
+\begin_inset Formula $\mu$
+\end_inset
+
+.
+ In total this uses two slots of memory and runs in time 
+\begin_inset Formula $O(n)+O(\lambda)+O(\mu)=O(\mu+\lambda)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc7[M21]
+\end_layout
+
+\end_inset
+
+(R.
+ P.
+ Brent,
+ 1977.) Let 
+\begin_inset Formula $\ell(n)$
+\end_inset
+
+ be the greatest power of 2 that is less than or equal to 
+\begin_inset Formula $n$
+\end_inset
+
+;
+ thus,
+ for example,
+ 
+\begin_inset Formula $\ell(15)=8$
+\end_inset
+
+ and 
+\begin_inset Formula $\ell(\ell(n))=\ell(n)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Show that,
+ in terms of the notation in exercise 6,
+ there exists an 
+\begin_inset Formula $n>0$
+\end_inset
+
+ such that 
+\begin_inset Formula $X_{n}=X_{\ell(n)-1}$
+\end_inset
+
+.
+ Find a formula that expresses the least such 
+\begin_inset Formula $n$
+\end_inset
+
+ in terms of the periodicity numbers 
+\begin_inset Formula $\mu$
+\end_inset
+
+ and 
+\begin_inset Formula $\lambda$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Apply this result to design an algorithm that can be used in conjunction with any random number generator of the type 
+\begin_inset Formula $X_{n+1}=f(X_{n})$
+\end_inset
+
+,
+ to prevent it from cycling indefinitely.
+ Your algorithm should calculate the period length 
+\begin_inset Formula $\lambda$
+\end_inset
+
+,
+ and it should only use a small amount of memory space—
+you must not simply store all the computed sequence values!
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+For sufficiently large 
+\begin_inset Formula $r$
+\end_inset
+
+,
+ 
+\begin_inset Formula $n\coloneqq2^{r}-1+\lambda<2^{r+1}$
+\end_inset
+
+ and then 
+\begin_inset Formula $X_{\ell(n)-1}=X_{2^{r}-1}=X_{n}$
+\end_inset
+
+,
+ and it's easy to convince ourselves that all such 
+\begin_inset Formula $n$
+\end_inset
+
+ are of the form 
+\begin_inset Formula $2^{r}-1+k\lambda$
+\end_inset
+
+ with 
+\begin_inset Formula $r\in\mathbb{N}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $k\in\mathbb{N}^{*}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $k\lambda-1<2^{r}$
+\end_inset
+
+,
+ and 
+\begin_inset Formula $2^{r}-1\geq\mu$
+\end_inset
+
+.
+ The minimum value,
+ then,
+ happens with 
+\begin_inset Formula $k=1$
+\end_inset
+
+ and 
+\begin_inset Formula $2^{r}$
+\end_inset
+
+ is the minimum such that 
+\begin_inset Formula $\lambda,\mu+1\leq2^{r}$
+\end_inset
+
+,
+ so if 
+\begin_inset Formula $u(n)$
+\end_inset
+
+ is the lowest power of 2 that is greater than or equal to 
+\begin_inset Formula $n$
+\end_inset
+
+,
+ then
+\begin_inset Formula 
+\[
+n=u(\max\{\lambda,\mu+1\})+\lambda-1.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+We set 
+\begin_inset Formula $a\gets b\gets X_{0}$
+\end_inset
+
+ and 
+\begin_inset Formula $r\gets0$
+\end_inset
+
+.
+ Then repeat 
+\begin_inset Formula $b\gets f(b)$
+\end_inset
+
+ up to 
+\begin_inset Formula $2^{r}$
+\end_inset
+
+ times;
+ if at any point 
+\begin_inset Formula $b=a$
+\end_inset
+
+,
+ terminate with 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ equal to the number of times that 
+\begin_inset Formula $b\gets f(b)$
+\end_inset
+
+ has been run in this iteration;
+ otherwise set 
+\begin_inset Formula $a\gets b$
+\end_inset
+
+,
+ 
+\begin_inset Formula $r\gets r+1$
+\end_inset
+
+,
+ and repeat this step.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc16[15]
+\end_layout
+
+\end_inset
+
+A sequence generated as in exercise 6 must begin to repeat after at most 
+\begin_inset Formula $m$
+\end_inset
+
+ values have been generated.
+ Suppose we generalize the method so that 
+\begin_inset Formula $X_{n+1}$
+\end_inset
+
+ depends on 
+\begin_inset Formula $X_{n-1}$
+\end_inset
+
+ as well as on 
+\begin_inset Formula $X_{n}$
+\end_inset
+
+;
+ formally,
+ let 
+\begin_inset Formula $f(x,y)$
+\end_inset
+
+ be a function such that 
+\begin_inset Formula $0\leq x,y<m$
+\end_inset
+
+ implies 
+\begin_inset Formula $0\leq f(x,y)<m$
+\end_inset
+
+.
+ The sequence is constructed by selecting 
+\begin_inset Formula $X_{0}$
+\end_inset
+
+ and 
+\begin_inset Formula $X_{1}$
+\end_inset
+
+ arbitrarily,
+ and then letting
+\begin_inset Formula 
+\begin{align*}
+X_{n+1} & =f(X_{n},X_{n-1}), & \text{for }n & >0.
+\end{align*}
+
+\end_inset
+
+What is the maximum period conceivably attainable in this case?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+See exercise 17.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc17[10]
+\end_layout
+
+\end_inset
+
+Generalize the situation in the previous exercise so that 
+\begin_inset Formula $X_{n+1}$
+\end_inset
+
+ depends on the preceding 
+\begin_inset Formula $k$
+\end_inset
+
+ values of the sequence.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+The maximum conceivable would be 
+\begin_inset Formula $m^{k}$
+\end_inset
+
+.
+ Why this is always attainable is not at all obvious.
+\end_layout
+
+\end_body
+\end_document
diff --git a/vol2/3.2.1.1.lyx b/vol2/3.2.1.1.lyx
new file mode 100644
index 0000000..7eec653
--- /dev/null
+++ b/vol2/3.2.1.1.lyx
@@ -0,0 +1,719 @@
+#LyX 2.4 created this file. For more info see https://www.lyx.org/
+\lyxformat 620
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\begin_preamble
+\input defs
+\end_preamble
+\use_default_options true
+\maintain_unincluded_children no
+\language english
+\language_package default
+\inputencoding utf8
+\fontencoding auto
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_roman_osf false
+\font_sans_osf false
+\font_typewriter_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\float_placement class
+\float_alignment class
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_formatted_ref 0
+\use_minted 0
+\use_lineno 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style english
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tablestyle default
+\tracking_changes false
+\output_changes false
+\change_bars false
+\postpone_fragile_content false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\docbook_table_output 0
+\docbook_mathml_prefix 1
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc3[M25]
+\end_layout
+
+\end_inset
+
+Many computers do not provide the ability to divide a two-word number by a one-word number;
+ they provide only operations on single-word numbers,
+ such as 
+\begin_inset Formula $\text{himult}(x,y)=\lfloor xy/w\rfloor$
+\end_inset
+
+ and 
+\begin_inset Formula $\text{lomult}(x,y)=xy\bmod w$
+\end_inset
+
+,
+ when 
+\begin_inset Formula $x$
+\end_inset
+
+ and 
+\begin_inset Formula $y$
+\end_inset
+
+ are nonnegative integers less than the word size 
+\begin_inset Formula $w$
+\end_inset
+
+.
+ Explain how to evaluate 
+\begin_inset Formula $ax\bmod m$
+\end_inset
+
+ in terms of 
+\begin_inset Formula $\text{himult}$
+\end_inset
+
+ and 
+\begin_inset Formula $\text{lomult}$
+\end_inset
+
+,
+ assuming that 
+\begin_inset Formula $0\leq a,x<m<w$
+\end_inset
+
+ and that 
+\begin_inset Formula $m\bot w$
+\end_inset
+
+.
+ You may use precomputed constants that depend on 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ 
+\begin_inset Formula $m$
+\end_inset
+
+,
+ and 
+\begin_inset Formula $w$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+
+\begin_inset Note Greyedout
+status open
+
+\begin_layout Plain Layout
+(I had to look up the answer,
+ I would have never guessed it.)
+\end_layout
+
+\end_inset
+
+Let 
+\begin_inset Formula $b\coloneqq aw\bmod m$
+\end_inset
+
+ and 
+\begin_inset Formula $c\in\{0,\dots,w-1\}$
+\end_inset
+
+ such that 
+\begin_inset Formula $mc\equiv1\pmod w$
+\end_inset
+
+,
+ we compute 
+\begin_inset Formula $y\gets\text{lomult}(b,x)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $z\gets\text{himult}(b,x)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $t\gets\text{lomult}(c,y)$
+\end_inset
+
+,
+ and 
+\begin_inset Formula $u\gets\text{himult}(m,t)$
+\end_inset
+
+,
+ and return 
+\begin_inset Formula $z-u+[z<u]m$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+To see that this works,
+ we note that 
+\begin_inset Formula 
+\[
+mt=m(cbx\bmod w)\equiv mcbx\equiv bx\pmod w,
+\]
+
+\end_inset
+
+so
+\begin_inset Formula 
+\[
+bx-mt=\left(\left\lfloor \frac{bx}{w}\right\rfloor -\left\lfloor \frac{mt}{w}\right\rfloor \right)w=(z-u)w
+\]
+
+\end_inset
+
+and,
+ taking modulo 
+\begin_inset Formula $m$
+\end_inset
+
+ on both sides,
+ 
+\begin_inset Formula $awx\bmod m=(z-u)w\bmod m$
+\end_inset
+
+ and 
+\begin_inset Formula $ax\bmod m=(z-u)\bmod m$
+\end_inset
+
+ by canceling 
+\begin_inset Formula $w$
+\end_inset
+
+.
+ Finally,
+ 
+\begin_inset Formula $0\leq z=\lfloor\frac{(aw\bmod m)x}{w}\rfloor\leq\lfloor\frac{mx}{w}\rfloor<m$
+\end_inset
+
+ and 
+\begin_inset Formula $0\leq u=\lfloor\frac{m(cy\bmod w)}{w}\rfloor<m$
+\end_inset
+
+,
+ so 
+\begin_inset Formula $-m<z-u<m$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc4[21]
+\end_layout
+
+\end_inset
+
+Discuss the calculation of linear congruential sequences with 
+\begin_inset Formula $m=2^{32}$
+\end_inset
+
+ on two's-complement machines such as the System/370 series.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+When multiplying 
+\begin_inset Formula $a$
+\end_inset
+
+ times 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ one or two of them could instead be 
+\begin_inset Formula $a-m$
+\end_inset
+
+ or 
+\begin_inset Formula $x-m$
+\end_inset
+
+;
+ however,
+ for 
+\begin_inset Formula $i,j\in\{0,1\}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $(a-im)(x-jm)\equiv am\pmod m$
+\end_inset
+
+,
+ so no further action is necessary given that taking modulo 
+\begin_inset Formula $m=2^{32}$
+\end_inset
+
+ is trivial.
+ Of course,
+ the resulting number could be negative,
+ but the bit pattern would be the same.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc6[20]
+\end_layout
+
+\end_inset
+
+The previous exercise suggests that subtraction mod 
+\begin_inset Formula $m$
+\end_inset
+
+ is easier to perform than addition mod 
+\begin_inset Formula $m$
+\end_inset
+
+.
+ Discuss sequences generated by the rule
+\begin_inset Formula 
+\[
+X_{n+1}=(aX_{n}-c)\bmod m.
+\]
+
+\end_inset
+
+Are these sequences essentially different from linear congruential sequences as defined in the text?
+ Are they more suited to efficient computer calculation?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+These are the same,
+ as subtracting 
+\begin_inset Formula $c$
+\end_inset
+
+ is the same modulo 
+\begin_inset Formula $m$
+\end_inset
+
+ as adding 
+\begin_inset Formula $m-c$
+\end_inset
+
+.
+ Although they are equivalent,
+ computing them this way could be slightly more efficient.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc8[20]
+\end_layout
+
+\end_inset
+
+Write a 
+\family typewriter
+MIX
+\family default
+ program analogous to (2) that computes 
+\begin_inset Formula $(aX)\bmod(w-1)$
+\end_inset
+
+.
+ The values 0 and 
+\begin_inset Formula $w-1$
+\end_inset
+
+ are to be treated as equivalent in the input and output of your program.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Let 
+\begin_inset Formula $c\coloneqq\text{himult}(a,X)$
+\end_inset
+
+ and 
+\begin_inset Formula $d\coloneqq\text{lomult}(a,X)$
+\end_inset
+
+,
+ then
+\begin_inset Formula 
+\[
+aX=cw+d\equiv(cw+d)-c(w-1)=c+d\pmod m,
+\]
+
+\end_inset
+
+so assuming the overflow bit was off,
+ we could write:
+\begin_inset listings
+inline false
+status open
+
+\begin_layout Plain Layout
+
+LDA  X
+\end_layout
+
+\begin_layout Plain Layout
+
+MUL  A
+\end_layout
+
+\begin_layout Plain Layout
+
+STX  TEMP
+\end_layout
+
+\begin_layout Plain Layout
+
+ADD  TEMP
+\end_layout
+
+\begin_layout Plain Layout
+
+JNOV *+2
+\end_layout
+
+\begin_layout Plain Layout
+
+INCA 1
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+The last instruction is valid because the value of 
+\begin_inset Formula $c+d$
+\end_inset
+
+ can never reach 
+\begin_inset Formula $2w-1$
+\end_inset
+
+,
+ so 
+\family typewriter
+INCA
+\family default
+ cannot produce a second overflow.
+ However 
+\begin_inset Formula $c+d$
+\end_inset
+
+ could be 
+\begin_inset Formula $w-1$
+\end_inset
+
+ or 0 and no effort is made to make rA be equal in these two cases.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc9[M25]
+\end_layout
+
+\end_inset
+
+Most high-level programming languages do not provide a good way to divide a two-word integer by a one-word integer,
+ nor do they provide the 
+\begin_inset Formula $\text{himult}$
+\end_inset
+
+ operation of exercise 3.
+ The purpose of this exercise is to find a reasonable way to cope with such limitations when we wish to evaluate 
+\begin_inset Formula $ax\bmod m$
+\end_inset
+
+ for variable 
+\begin_inset Formula $x$
+\end_inset
+
+ and for constants 
+\begin_inset Formula $0<a<m$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Prove that if 
+\begin_inset Formula $q=\lfloor m/a\rfloor$
+\end_inset
+
+,
+ we have 
+\begin_inset Formula $a(x-(x\bmod q))=\lfloor x/q\rfloor(m-(m\bmod a))$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Use the identity of part 1 to evaluate 
+\begin_inset Formula $ax\bmod m$
+\end_inset
+
+ without computing any numbers that exceed 
+\begin_inset Formula $m$
+\end_inset
+
+ in absolute value,
+ assuming that 
+\begin_inset Formula $a^{2}\leq m$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+First I challenge the idea that most high-level languages do not provide these features.
+ While this might have been true when the book was written,
+ it's not true today with languages like Python,
+ Ruby,
+ or JavaScript dominating the industry,
+ you can do that but it's slow.
+ (I guess the the text said 
+\begin_inset Quotes eld
+\end_inset
+
+most
+\begin_inset Quotes erd
+\end_inset
+
+ was about the Lisp family of languages.) Anyway,
+ a lot of them like Go are still stuck in the 70s so...
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $a(x-(x\bmod q))=a\lfloor\frac{x}{q}\rfloor q=\lfloor\frac{x}{q}\rfloor a\lfloor\frac{m}{a}\rfloor=\lfloor\frac{x}{q}\rfloor(m-(m\bmod a))$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+If we operate modulo 
+\begin_inset Formula $m$
+\end_inset
+
+,
+\begin_inset Formula 
+\[
+ax=a(x\bmod q)+\left\lfloor \frac{x}{q}\right\rfloor (m-(m\bmod a))\equiv a(x\bmod q)-\left\lfloor \frac{x}{q}\right\rfloor (m\bmod a)\mod m,
+\]
+
+\end_inset
+
+so 
+\begin_inset Formula $ax\bmod m=\left(a(x\bmod q)-\lfloor\frac{x}{q}\rfloor(m\bmod a)\right)\bmod m$
+\end_inset
+
+.
+ Since 
+\begin_inset Formula $a^{2}\leq m$
+\end_inset
+
+,
+ 
+\begin_inset Formula $a\leq\frac{m}{a}$
+\end_inset
+
+ and therefore 
+\begin_inset Formula $a\leq q$
+\end_inset
+
+,
+ so 
+\begin_inset Formula $\lfloor\frac{x}{q}\rfloor\leq\lfloor\frac{x}{a}\rfloor\leq\frac{x}{a}$
+\end_inset
+
+ and,
+ multiplied by 
+\begin_inset Formula $m\bmod a$
+\end_inset
+
+ (a constant),
+ the product is no greater than 
+\begin_inset Formula $x$
+\end_inset
+
+.
+ In the first factor,
+ however,
+ 
+\begin_inset Formula $x\bmod q\leq q\leq\frac{m}{a}$
+\end_inset
+
+,
+ so again the product by 
+\begin_inset Formula $a$
+\end_inset
+
+ is no greater than 
+\begin_inset Formula $m$
+\end_inset
+
+.
+\end_layout
+
+\end_body
+\end_document
diff --git a/vol2/3.2.1.2.lyx b/vol2/3.2.1.2.lyx
new file mode 100644
index 0000000..ad58b78
--- /dev/null
+++ b/vol2/3.2.1.2.lyx
@@ -0,0 +1,1542 @@
+#LyX 2.4 created this file. For more info see https://www.lyx.org/
+\lyxformat 620
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\begin_preamble
+\input defs
+\end_preamble
+\use_default_options true
+\maintain_unincluded_children no
+\language english
+\language_package default
+\inputencoding utf8
+\fontencoding auto
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_roman_osf false
+\font_sans_osf false
+\font_typewriter_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\float_placement class
+\float_alignment class
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_formatted_ref 0
+\use_minted 0
+\use_lineno 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style english
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tablestyle default
+\tracking_changes false
+\output_changes false
+\change_bars false
+\postpone_fragile_content false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\docbook_table_output 0
+\docbook_mathml_prefix 1
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc1[10]
+\end_layout
+
+\end_inset
+
+What is the length of the period of the linear congruential sequence with 
+\begin_inset Formula $X_{0}=5772156648$
+\end_inset
+
+,
+ 
+\begin_inset Formula $a=3141592621$
+\end_inset
+
+,
+ 
+\begin_inset Formula $c=2718281829$
+\end_inset
+
+,
+ and 
+\begin_inset Formula $m=10000000000$
+\end_inset
+
+?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Obviously 
+\begin_inset Formula $c$
+\end_inset
+
+ is relatively prime to 
+\begin_inset Formula $m$
+\end_inset
+
+.
+ Furthermore,
+ the prime divisors of 
+\begin_inset Formula $m$
+\end_inset
+
+ are 2 and 5 and 
+\begin_inset Formula $a-1$
+\end_inset
+
+ is a multiple of both 4 and 5.
+ Therefore the length is 
+\begin_inset Formula $m$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc2[10]
+\end_layout
+
+\end_inset
+
+Are the following two conditions sufficient to guarantee the maximum length period,
+ when 
+\begin_inset Formula $m$
+\end_inset
+
+ is a power of 2?
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $c$
+\end_inset
+
+ is odd;
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $a\bmod4=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Yes,
+ by theorem A.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc6[20]
+\end_layout
+
+\end_inset
+
+Find all multipliers 
+\begin_inset Formula $a$
+\end_inset
+
+ that satisfy the conditions of Theorem A when 
+\begin_inset Formula $m=10^{6}-1$
+\end_inset
+
+.
+ (See Table 3.2.1.1–1.)
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+By the table,
+ 
+\begin_inset Formula $m=3^{3}\cdot7\cdot11\cdot13\cdot37$
+\end_inset
+
+.
+ Multipliers are numbers 
+\begin_inset Formula $a$
+\end_inset
+
+ such that 
+\begin_inset Formula 
+\[
+a\bmod3=a\bmod7=a\bmod11=a\bmod13=a\bmod37=1.
+\]
+
+\end_inset
+
+These identities tell us that 
+\begin_inset Formula $a\equiv1\pmod{3\cdot7\cdot11\cdot13\cdot37=m/3^{2}=111111}$
+\end_inset
+
+,
+ so the possible values are 
+\begin_inset Formula $1,111112,222223,333334,444445,555556,666667,777778,888889$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc7[M23]
+\end_layout
+
+\end_inset
+
+The period of a congruential sequence need not start with 
+\begin_inset Formula $X_{0}$
+\end_inset
+
+,
+ but we can always find indices 
+\begin_inset Formula $\mu\geq0$
+\end_inset
+
+ and 
+\begin_inset Formula $\lambda>0$
+\end_inset
+
+ such that 
+\begin_inset Formula $X_{n+\lambda}=X_{n}$
+\end_inset
+
+ whenever 
+\begin_inset Formula $n\geq\mu$
+\end_inset
+
+,
+ and for which 
+\begin_inset Formula $\mu$
+\end_inset
+
+ and 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ are the smallest possible values with this property.
+ (See exercises 3.1–6 and 3.2.1–1.) If 
+\begin_inset Formula $\mu_{j}$
+\end_inset
+
+ and 
+\begin_inset Formula $\lambda_{j}$
+\end_inset
+
+ are the indices corresponding to the sequences
+\begin_inset Formula 
+\[
+(X_{0}\bmod p_{j}^{e_{j}},a\bmod p_{j}^{e_{j}},c\bmod p_{j}^{e_{j}},p_{j}^{e_{j}}),
+\]
+
+\end_inset
+
+and if 
+\begin_inset Formula $\mu$
+\end_inset
+
+ and 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ correspond to the composite sequence 
+\begin_inset Formula $(X_{0},a,c,p_{1}^{e_{1}}\cdots p_{t}^{e_{t}})$
+\end_inset
+
+,
+ Lemma Q states that 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ is the least common multiple of 
+\begin_inset Formula $\lambda_{1},\dots,\lambda_{t}$
+\end_inset
+
+.
+ What is the value of 
+\begin_inset Formula $\mu$
+\end_inset
+
+ in terms of the values of 
+\begin_inset Formula $\mu_{1},\dots,\mu_{t}$
+\end_inset
+
+?
+ What is the maximum possible value of 
+\begin_inset Formula $\mu$
+\end_inset
+
+ obtainable by varying 
+\begin_inset Formula $X_{0}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ and 
+\begin_inset Formula $c$
+\end_inset
+
+,
+ when 
+\begin_inset Formula $m=p_{1}^{e_{1}}\cdots p_{t}^{e_{t}}$
+\end_inset
+
+ is fixed?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Since the 
+\begin_inset Formula $j$
+\end_inset
+
+th sequence has values 
+\begin_inset Formula $X_{n}\bmod p_{j}^{e_{j}}$
+\end_inset
+
+,
+ and the 
+\begin_inset Formula $X_{n}$
+\end_inset
+
+ are uniquely determined by such values for all 
+\begin_inset Formula $j$
+\end_inset
+
+,
+ 
+\begin_inset Formula $(X_{n})_{n}$
+\end_inset
+
+ enters into a loop whenever all the 
+\begin_inset Formula $X_{n}\bmod p_{j}^{e_{j}}$
+\end_inset
+
+ have entered into a loop,
+ that is,
+ 
+\begin_inset Formula $\mu=\max_{j}\mu_{j}$
+\end_inset
+
+.
+ We are left to see what is the maximum 
+\begin_inset Formula $\mu$
+\end_inset
+
+ when 
+\begin_inset Formula $m=p^{e}$
+\end_inset
+
+ for some prime number 
+\begin_inset Formula $p$
+\end_inset
+
+ and positive integer 
+\begin_inset Formula $e$
+\end_inset
+
+ (for 
+\begin_inset Formula $m=1$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\mu=0$
+\end_inset
+
+).
+ In this case,
+ 
+\begin_inset Formula $\mu$
+\end_inset
+
+ is the lowest number such that
+\begin_inset Formula 
+\[
+X_{\mu}=X_{\mu+\lambda}\equiv X_{\mu}a^{\lambda}+\frac{a^{\lambda}-1}{a-1}c\pmod{p^{e}},
+\]
+
+\end_inset
+
+if and only if 
+\begin_inset Formula $X_{\mu}(1-a^{\lambda})\equiv\frac{a^{\lambda}-1}{a-1}c\pmod{p^{e}}$
+\end_inset
+
+,
+ so for 
+\begin_inset Formula $\mu>0$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\mu-1$
+\end_inset
+
+ must be a number such that
+\begin_inset Formula 
+\begin{align*}
+X_{\mu-1}(1-a^{\lambda}) & \not\equiv\frac{a^{\lambda}-1}{a-1}c, & aX_{\mu-1}(1-a^{\lambda}) & \equiv a\frac{a^{\lambda}-1}{a-1}c &  & \pmod{p^{e}},
+\end{align*}
+
+\end_inset
+
+where the first equation comes from changing 
+\begin_inset Formula $\mu$
+\end_inset
+
+ to 
+\begin_inset Formula $\mu-1$
+\end_inset
+
+ above and the second one from changing 
+\begin_inset Formula $X_{\mu}$
+\end_inset
+
+ to 
+\begin_inset Formula $aX_{\mu-1}+c$
+\end_inset
+
+ and rearranging terms.
+ This means that 
+\begin_inset Formula $a$
+\end_inset
+
+ must be a multiple of 
+\begin_inset Formula $p$
+\end_inset
+
+.
+ Then,
+\begin_inset Formula 
+\begin{align*}
+X_{e} & \equiv X_{0}a^{e}+\frac{a^{e}-1}{a-1}c\equiv(a^{e-1}+a^{e-2}+\dots+a+1)c, & \pmod{p^{e}},\\
+X_{e+\lambda} & \equiv X_{0}a^{e+\lambda}+\frac{a^{e+\lambda}-1}{a-1}c\equiv(a^{e-1}+a^{e-2}+\dots+a+1)c, & \pmod{p^{e}},
+\end{align*}
+
+\end_inset
+
+since 
+\begin_inset Formula $a^{e}\equiv0$
+\end_inset
+
+,
+ so 
+\begin_inset Formula $\mu\leq e$
+\end_inset
+
+.
+ The value 
+\begin_inset Formula $\mu=e$
+\end_inset
+
+ is reached when 
+\begin_inset Formula $(X_{0},a,c)=(1,p,0)$
+\end_inset
+
+.
+ Summing up,
+ the maximum for 
+\begin_inset Formula $m=p_{1}^{e_{1}}\cdots p_{t}^{e_{t}}$
+\end_inset
+
+ is 
+\begin_inset Formula $\max\{e_{1},\dots,e_{t}\}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc9[M22]
+\end_layout
+
+\end_inset
+
+(W.
+ E.
+ Thompson.) When 
+\begin_inset Formula $c=0$
+\end_inset
+
+ and 
+\begin_inset Formula $m=2^{e}\geq16$
+\end_inset
+
+,
+ Theorems B and C say that the period has length 
+\begin_inset Formula $2^{e-2}$
+\end_inset
+
+ if and only if the multiplier 
+\begin_inset Formula $a$
+\end_inset
+
+ satisfies 
+\begin_inset Formula $a\bmod8=3$
+\end_inset
+
+ or 
+\begin_inset Formula $a\bmod8=5$
+\end_inset
+
+.
+ Show that every such sequence is essentially a linear congruential sequence with 
+\begin_inset Formula $m=2^{e-2}$
+\end_inset
+
+,
+ having 
+\emph on
+full
+\emph default
+ period,
+ in the following sense:
+\end_layout
+
+\begin_layout Enumerate
+If 
+\begin_inset Formula $X_{n+1}=(4c+1)X_{n}\bmod2^{e}$
+\end_inset
+
+,
+ and 
+\begin_inset Formula $X_{n}=4Y_{n}+1$
+\end_inset
+
+,
+ then
+\begin_inset Formula 
+\[
+Y_{n+1}=((4c+1)Y_{n}+c)\bmod2^{e-2}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+If 
+\begin_inset Formula $X_{n+1}=(4c-1)X_{n}\bmod2^{e}$
+\end_inset
+
+,
+ and 
+\begin_inset Formula $X_{n}=((-1)^{n}(4Y_{n}+1))\bmod2^{e}$
+\end_inset
+
+,
+ then
+\begin_inset Formula 
+\[
+Y_{n+1}=((1-4c)Y_{n}-c)\bmod2^{e-2}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+We start with 
+\begin_inset Formula $(Y_{n})_{n}$
+\end_inset
+
+ as defined and see that 
+\begin_inset Formula $(X_{n})_{n}$
+\end_inset
+
+ calculated from there matches the initial definition of 
+\begin_inset Formula $(X_{n})_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+We have 
+\begin_inset Formula $Y_{n}=\left((4c+1)^{n}Y_{0}+\frac{(4c+1)^{n}-1}{4\cancel{c}}\cancel{c}\right)\bmod2^{e-2}$
+\end_inset
+
+,
+ so
+\begin_inset Formula 
+\[
+4Y_{n}+1\equiv(4Y_{0}+1)(4c+1)^{n}\equiv(4c+1)^{n}X_{0}\equiv X_{n}\pmod{2^{e}}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+We have 
+\begin_inset Formula $Y_{n}=\left((1-4c)^{n}Y_{0}-\frac{(1-4c)^{n}-1}{-4c}c\right)\bmod2^{e-2}$
+\end_inset
+
+,
+ so
+\begin_inset Formula 
+\begin{multline*}
+(-1)^{n}(4Y_{n}+1)\equiv(-1)^{n}\left(4(1-4c)^{n}Y_{0}+(1-4c)^{n}\right)=(4Y_{0}+1)(4c-1)^{n}=\\
+=(4c-1)^{n}X_{0}\equiv X_{n+1}\pmod{2^{e}}.
+\end{multline*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc16[M24]
+\end_layout
+
+\end_inset
+
+
+\emph on
+(Existence of primitive roots.)
+\emph default
+ Let 
+\begin_inset Formula $p$
+\end_inset
+
+ be a prime number.
+\end_layout
+
+\begin_layout Enumerate
+Consider the polynomial 
+\begin_inset Formula $f(x)=x^{n}+c_{1}x^{n-1}+\dots+c_{n}$
+\end_inset
+
+,
+ where the 
+\begin_inset Formula $c$
+\end_inset
+
+'s are integers.
+ Given that 
+\begin_inset Formula $a$
+\end_inset
+
+ is an integer for which 
+\begin_inset Formula $f(a)\equiv0\pmod p$
+\end_inset
+
+,
+ show that there exists a polynomial
+\begin_inset Formula 
+\[
+q(x)=x^{n-1}+q_{1}x^{n-2}+\dots+q_{n-1}
+\]
+
+\end_inset
+
+with integer coefficients such that 
+\begin_inset Formula $f(x)\equiv(x-a)q(x)\pmod p$
+\end_inset
+
+ for all integers 
+\begin_inset Formula $x$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Let 
+\begin_inset Formula $f(x)$
+\end_inset
+
+ be a polynomial as in (1).
+ Show that 
+\begin_inset Formula $f(x)$
+\end_inset
+
+ has at most 
+\begin_inset Formula $n$
+\end_inset
+
+ distinct 
+\begin_inset Quotes eld
+\end_inset
+
+roots
+\begin_inset Quotes erd
+\end_inset
+
+ modulo 
+\begin_inset Formula $p$
+\end_inset
+
+;
+ that is,
+ there are at most 
+\begin_inset Formula $n$
+\end_inset
+
+ integers 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ with 
+\begin_inset Formula $0\leq a<p$
+\end_inset
+
+,
+ such that 
+\begin_inset Formula $f(a)\equiv0\pmod p$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Because of exercise 15(2),
+ the polynomial 
+\begin_inset Formula $f(x)=x^{\lambda(p)}-1$
+\end_inset
+
+ has 
+\begin_inset Formula $p-1$
+\end_inset
+
+ distinct roots;
+ hence there is an integer 
+\begin_inset Formula $a$
+\end_inset
+
+ with order 
+\begin_inset Formula $p-1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+It suffices to prove that,
+ for 
+\begin_inset Formula $f\in\mathbb{Z}_{p}[X]$
+\end_inset
+
+ and 
+\begin_inset Formula $a\in\mathbb{Z}_{p}$
+\end_inset
+
+ with 
+\begin_inset Formula $f(a)=0$
+\end_inset
+
+,
+ there exists 
+\begin_inset Formula $q\in\mathbb{Z}_{p}[X]$
+\end_inset
+
+ such that 
+\begin_inset Formula $f(x)\equiv(x-a)q(x)$
+\end_inset
+
+,
+ since then,
+ for 
+\begin_inset Formula $f$
+\end_inset
+
+ to have degree 
+\begin_inset Formula $n$
+\end_inset
+
+ and leading coefficient 1,
+ 
+\begin_inset Formula $q$
+\end_inset
+
+ must have degree 
+\begin_inset Formula $n-1$
+\end_inset
+
+ and leading coefficient 
+\begin_inset Formula $q$
+\end_inset
+
+.
+ This is immediate for 
+\begin_inset Formula $f=0$
+\end_inset
+
+,
+ so we prove it by induction on the degree 
+\begin_inset Formula $n\geq0$
+\end_inset
+
+ of 
+\begin_inset Formula $f$
+\end_inset
+
+.
+ If 
+\begin_inset Formula $n=0$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f(a)=c_{0}\neq0\#$
+\end_inset
+
+,
+ so this follows trivially.
+ If 
+\begin_inset Formula $n>0$
+\end_inset
+
+,
+ let 
+\begin_inset Formula $g(x)\coloneqq f(x)-c_{0}(x-a)x^{n-1}$
+\end_inset
+
+ is a polynomial of degree 
+\begin_inset Formula $n-1$
+\end_inset
+
+ such that 
+\begin_inset Formula $g(a)=f(a)=0$
+\end_inset
+
+,
+ so by induction there exists 
+\begin_inset Formula $r\in\mathbb{Z}_{p}(x)$
+\end_inset
+
+ with 
+\begin_inset Formula $g(x)=(x-a)r(x)$
+\end_inset
+
+.
+ But then 
+\begin_inset Formula $f(x)=g(x)+c_{0}(x-a)x^{n-1}=(x-a)(r(x)+c_{0}x^{n-1})$
+\end_inset
+
+.
+ Note that this proof is valid even when changing 
+\begin_inset Formula $\mathbb{Z}_{p}$
+\end_inset
+
+ to any other domain.
+\end_layout
+
+\begin_layout Enumerate
+If 
+\begin_inset Formula $a$
+\end_inset
+
+ and 
+\begin_inset Formula $b$
+\end_inset
+
+ are different roots of 
+\begin_inset Formula $f$
+\end_inset
+
+,
+ then 
+\begin_inset Formula $f(x)=(x-a)q(x)$
+\end_inset
+
+ for some 
+\begin_inset Formula $q\in\mathbb{Z}_{p}[X]$
+\end_inset
+
+,
+ and 
+\begin_inset Formula $b$
+\end_inset
+
+ is still a root of 
+\begin_inset Formula $q$
+\end_inset
+
+ as,
+ if it wasn't,
+ it wouldn't be a root of 
+\begin_inset Formula $f$
+\end_inset
+
+ either.
+ Therefore,
+ by induction,
+ a polynomial 
+\begin_inset Formula $f\in\mathbb{Z}_{p}[X]\setminus0$
+\end_inset
+
+ with 
+\begin_inset Formula $n$
+\end_inset
+
+ different roots has at least degree 
+\begin_inset Formula $n$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Yes.
+ Trivial.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc20[M24]
+\end_layout
+
+\end_inset
+
+(G.
+ Marsaglia.) The purpose of this exercise is to study the period length of an 
+\emph on
+arbitrary
+\emph default
+ linear congruential sequence.
+ Let 
+\begin_inset Formula $Y_{n}=1+a+\dots+a^{n-1}$
+\end_inset
+
+,
+ so that 
+\begin_inset Formula $X_{n}=(AY_{n}+X_{0})\bmod m$
+\end_inset
+
+ for some constant 
+\begin_inset Formula $A$
+\end_inset
+
+ by Eq.
+ 3.2.1–(8).
+\end_layout
+
+\begin_layout Enumerate
+Prove that the period length of 
+\begin_inset Formula $\langle X_{n}\rangle$
+\end_inset
+
+ is the period length of 
+\begin_inset Formula $\langle Y_{n}\bmod m'\rangle$
+\end_inset
+
+,
+ where 
+\begin_inset Formula $m'=m/\gcd(A,m)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Prove that the period length of 
+\begin_inset Formula $\langle Y_{n}\bmod p^{e}\rangle$
+\end_inset
+
+ satisfies the following when 
+\begin_inset Formula $p$
+\end_inset
+
+ is prime:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+If 
+\begin_inset Formula $a\bmod p=0$
+\end_inset
+
+,
+ it is 1.
+\end_layout
+
+\begin_layout Enumerate
+If 
+\begin_inset Formula $a\bmod p=1$
+\end_inset
+
+,
+ it is 
+\begin_inset Formula $p^{e}$
+\end_inset
+
+,
+ except when 
+\begin_inset Formula $p=2$
+\end_inset
+
+ and 
+\begin_inset Formula $e\geq2$
+\end_inset
+
+ and 
+\begin_inset Formula $a\bmod4=3$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+If 
+\begin_inset Formula $p=2$
+\end_inset
+
+,
+ 
+\begin_inset Formula $e\geq2$
+\end_inset
+
+,
+ and 
+\begin_inset Formula $a\bmod4=3$
+\end_inset
+
+,
+ it is twice the order of 
+\begin_inset Formula $a$
+\end_inset
+
+ modulo 
+\begin_inset Formula $p^{e}$
+\end_inset
+
+ (see exercise 11),
+ unless 
+\begin_inset Formula $a\equiv-1\pmod{2^{e}}$
+\end_inset
+
+ when it is 2.
+\end_layout
+
+\begin_layout Enumerate
+If 
+\begin_inset Formula $a\bmod p>1$
+\end_inset
+
+,
+ it is the order of 
+\begin_inset Formula $a$
+\end_inset
+
+ modulo 
+\begin_inset Formula $p^{e}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula 
+\[
+X_{n}=X_{n+m}\iff AY_{n}\equiv AY_{n+m}\pmod m\iff Y_{n}\equiv Y_{n+m}\pmod{m'}.
+\]
+
+\end_inset
+
+Note that 
+\begin_inset Formula $\langle Y_{n}\bmod t\rangle$
+\end_inset
+
+,
+ 
+\begin_inset Formula $t\in\mathbb{Z}^{+}$
+\end_inset
+
+,
+ is the linear congruential sequence with 
+\begin_inset Formula $Y_{0}=0$
+\end_inset
+
+,
+ 
+\begin_inset Formula $c=1$
+\end_inset
+
+,
+ 
+\begin_inset Formula $a=a$
+\end_inset
+
+,
+ and 
+\begin_inset Formula $m=t$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+For 
+\begin_inset Formula $n\geq e-1$
+\end_inset
+
+,
+ 
+\begin_inset Formula $Y_{n}=(1+a+a^{2}+\dots+a^{e-1})\bmod p$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Then we're in the conditions of Theorem A.
+\end_layout
+
+\begin_layout Enumerate
+If 
+\begin_inset Formula $e=2$
+\end_inset
+
+,
+ then 
+\begin_inset Formula $a\equiv-1\pmod 4$
+\end_inset
+
+ and this part doesn't apply,
+ so we assume 
+\begin_inset Formula $e>2$
+\end_inset
+
+.
+ Let 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ be the order of 
+\begin_inset Formula $a$
+\end_inset
+
+ modulo 
+\begin_inset Formula $2^{e}$
+\end_inset
+
+,
+ obviously 
+\begin_inset Formula $\lambda>1$
+\end_inset
+
+,
+ so
+\begin_inset Formula 
+\begin{multline*}
+Y_{k}=1+a+\dots+a^{k-1}=\frac{a^{k}-1}{a-1}\equiv Y_{0}=0\pmod{2^{e}}\iff\\
+\iff a^{k}-1\equiv0\pmod{2^{e}(a-1)}\iff2^{e}(a-1)\mid a^{k}-1.
+\end{multline*}
+
+\end_inset
+
+The order 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ of 
+\begin_inset Formula $a$
+\end_inset
+
+ modulo 
+\begin_inset Formula $2^{e}$
+\end_inset
+
+ is the smallest number such that 
+\begin_inset Formula $2^{e}\mid a^{k}-1$
+\end_inset
+
+,
+ and in fact
+\begin_inset Formula 
+\[
+2^{e}\mid a^{k}-1\iff a^{k}\equiv1\pmod{2^{e}}\iff\lambda\mid k.
+\]
+
+\end_inset
+
+Exercise 11 gives us an unique 
+\begin_inset Formula $f>1$
+\end_inset
+
+ such that 
+\begin_inset Formula $a\bmod2^{f+1}=2^{f}\pm1$
+\end_inset
+
+,
+ and then says that 
+\begin_inset Formula $\lambda=2^{e-f}$
+\end_inset
+
+.
+ This can be used to prove that 
+\begin_inset Formula $a^{\lambda}-1$
+\end_inset
+
+ is not a multiple of 
+\begin_inset Formula $2^{e}(a-1)$
+\end_inset
+
+,
+ as if this were the case,
+ then 
+\begin_inset Formula $a^{\lambda-1}$
+\end_inset
+
+ would be a multiple of 
+\begin_inset Formula $2^{e+1}$
+\end_inset
+
+ because 
+\begin_inset Formula $a-1$
+\end_inset
+
+ is even,
+ but by Exercise 11 the order of 
+\begin_inset Formula $a$
+\end_inset
+
+ modulo 
+\begin_inset Formula $e+1$
+\end_inset
+
+ is 
+\begin_inset Formula $2^{e+1-f}>\lambda\#$
+\end_inset
+
+.
+ However,
+ since 
+\begin_inset Formula $a^{\lambda}-1$
+\end_inset
+
+ is a multiple of both 
+\begin_inset Formula $2^{e}$
+\end_inset
+
+ and 
+\begin_inset Formula $a^{\lambda}-1$
+\end_inset
+
+,
+ it is a multiple of 
+\begin_inset Formula $\text{lcm}\{2^{e},a^{\lambda}-1\}=\frac{2^{e}(a^{\lambda}-1)}{2}$
+\end_inset
+
+ and therefore 
+\begin_inset Formula $a^{2\lambda}-1=(a^{\lambda}-1)(a^{\lambda}+1)=(a^{\lambda}-1)^{2}+2(a^{\lambda}-1)$
+\end_inset
+
+ is a multiple of 
+\begin_inset Formula $2^{e}(a^{\lambda}-1)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Like before,
+ 
+\begin_inset Formula $Y_{k}\equiv0\pmod{2^{e}}\iff p^{e}(a-1)\mid a^{k}-1$
+\end_inset
+
+.
+ This time,
+ however,
+ 
+\begin_inset Formula $\gcd\{a-1,p^{e}\}=1$
+\end_inset
+
+,
+ so 
+\begin_inset Formula $p^{e}(a-1)\mid a^{k}-1\iff p^{e},a-1\mid a^{k}-1\iff p^{e}\mid a^{k}-1$
+\end_inset
+
+,
+ and the lowest 
+\begin_inset Formula $k$
+\end_inset
+
+ for which this happens is precisely the order of 
+\begin_inset Formula $a$
+\end_inset
+
+ modulo 
+\begin_inset Formula $p^{e}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc22[M25]
+\end_layout
+
+\end_inset
+
+Discuss the problem of finding moduli 
+\begin_inset Formula $m=b^{k}\pm b^{l}\pm1$
+\end_inset
+
+ so that the subtract-with-borrow and add-with-carry generators of exercise 3.2.1.1–14 will have very long periods.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+First we would need the prime factorization of 
+\begin_inset Formula $m$
+\end_inset
+
+,
+ which is not easily characterized from 
+\begin_inset Formula $b$
+\end_inset
+
+,
+ 
+\begin_inset Formula $k$
+\end_inset
+
+,
+ and 
+\begin_inset Formula $l$
+\end_inset
+
+.
+ Once we have that we may apply Theorems A–C.
+ The solution in the book suggests looking for 
+\begin_inset Formula $m$
+\end_inset
+
+ to be prime.
+\end_layout
+
+\end_body
+\end_document
diff --git a/vol2/3.2.1.3.lyx b/vol2/3.2.1.3.lyx
new file mode 100644
index 0000000..e94ce2c
--- /dev/null
+++ b/vol2/3.2.1.3.lyx
@@ -0,0 +1,273 @@
+#LyX 2.4 created this file. For more info see https://www.lyx.org/
+\lyxformat 620
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\begin_preamble
+\input defs
+\end_preamble
+\use_default_options true
+\maintain_unincluded_children no
+\language english
+\language_package default
+\inputencoding utf8
+\fontencoding auto
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_roman_osf false
+\font_sans_osf false
+\font_typewriter_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\float_placement class
+\float_alignment class
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_formatted_ref 0
+\use_minted 0
+\use_lineno 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style english
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tablestyle default
+\tracking_changes false
+\output_changes false
+\change_bars false
+\postpone_fragile_content false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\docbook_table_output 0
+\docbook_mathml_prefix 1
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc1[M10]
+\end_layout
+
+\end_inset
+
+Show that,
+ no matter what the byte size 
+\begin_inset Formula $B$
+\end_inset
+
+ of 
+\family typewriter
+MIX
+\family default
+ happens to be,
+ the code (3) yields a random number generator of maximum period.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+This generator has 
+\begin_inset Formula $a=B^{2}+1$
+\end_inset
+
+,
+ 
+\begin_inset Formula $c=1$
+\end_inset
+
+,
+ and 
+\begin_inset Formula $m=B^{5}$
+\end_inset
+
+.
+ Since the prime divisors of 
+\begin_inset Formula $B^{5}$
+\end_inset
+
+ and 
+\begin_inset Formula $B^{2}$
+\end_inset
+
+ are the same,
+ the conditions of Theorem 3.2.1.2A are satisfied.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc2[10]
+\end_layout
+
+\end_inset
+
+What is the potency of the generator represented by the 
+\family typewriter
+MIX
+\family default
+ code (3)?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+3.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc6[20]
+\end_layout
+
+\end_inset
+
+Which of the values of 
+\begin_inset Formula $m=w\pm1$
+\end_inset
+
+ in Table 3.2.1–1 can be used in a linear congruential sequence of maximum period whose potency is 4 or more?
+ (Use the result of exercise 5.)
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+By exercise 5,
+ for a modulus 
+\begin_inset Formula $m=p_{1}^{e_{1}}\cdots p_{t}^{e_{t}}$
+\end_inset
+
+ with 
+\begin_inset Formula $e_{1}\geq\dots\geq e_{t}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $a=p_{1}+1$
+\end_inset
+
+ has the maximum potency,
+ which is 
+\begin_inset Formula $e_{1}$
+\end_inset
+
+,
+ except that if 
+\begin_inset Formula $m=p_{1}$
+\end_inset
+
+,
+ then the maximum potency is 0,
+ corresponding to 
+\begin_inset Formula $a=1$
+\end_inset
+
+.
+ Thus the values from the table that can be used are 
+\begin_inset Formula $10^{9}-1=3^{4}\cdot37\cdot333667$
+\end_inset
+
+ and 
+\begin_inset Formula $2^{27}+1=3^{4}\cdot19\cdot87211$
+\end_inset
+
+.
+\end_layout
+
+\end_body
+\end_document
diff --git a/vol2/3.2.1.lyx b/vol2/3.2.1.lyx
new file mode 100644
index 0000000..f78e3ee
--- /dev/null
+++ b/vol2/3.2.1.lyx
@@ -0,0 +1,320 @@
+#LyX 2.4 created this file. For more info see https://www.lyx.org/
+\lyxformat 620
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\begin_preamble
+\input defs
+\end_preamble
+\use_default_options true
+\maintain_unincluded_children no
+\language english
+\language_package default
+\inputencoding utf8
+\fontencoding auto
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_roman_osf false
+\font_sans_osf false
+\font_typewriter_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\float_placement class
+\float_alignment class
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_formatted_ref 0
+\use_minted 0
+\use_lineno 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style english
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tablestyle default
+\tracking_changes false
+\output_changes false
+\change_bars false
+\postpone_fragile_content false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\docbook_table_output 0
+\docbook_mathml_prefix 1
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc1[10]
+\end_layout
+
+\end_inset
+
+Example (3) shows a situation in which 
+\begin_inset Formula $X_{4}=X_{0}$
+\end_inset
+
+,
+ so the sequence begins again from the beginning.
+ Give an example of a linear congruential sequence with 
+\begin_inset Formula $m=10$
+\end_inset
+
+ for which 
+\begin_inset Formula $X_{0}$
+\end_inset
+
+ never appears again in the sequence.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+This happens if 
+\begin_inset Formula $a=c=0$
+\end_inset
+
+ and 
+\begin_inset Formula $0<X_{0}<m$
+\end_inset
+
+;
+ then the sequence starts with 
+\begin_inset Formula $X_{0}\neq0$
+\end_inset
+
+ and then it's constant at 0.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc2[M20]
+\end_layout
+
+\end_inset
+
+Show that if 
+\begin_inset Formula $a$
+\end_inset
+
+ and 
+\begin_inset Formula $m$
+\end_inset
+
+ are relatively prime,
+ the number 
+\begin_inset Formula $X_{0}$
+\end_inset
+
+ will always appear in the period.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+For 
+\begin_inset Formula $a\geq2$
+\end_inset
+
+,
+ by Euler's theorem,
+ 
+\begin_inset Formula $a^{\varphi(m)}\equiv1\mod m$
+\end_inset
+
+,
+ so 
+\begin_inset Formula $X_{\varphi(m)}=X_{0}$
+\end_inset
+
+ by Equation (6).
+ For 
+\begin_inset Formula $a=1$
+\end_inset
+
+,
+ 
+\begin_inset Formula $X_{m}=(X_{0}+mc)\bmod m=X_{0}$
+\end_inset
+
+,
+ and for 
+\begin_inset Formula $a=0$
+\end_inset
+
+,
+ 
+\begin_inset Formula $a$
+\end_inset
+
+ and 
+\begin_inset Formula $m$
+\end_inset
+
+ are not relatively prime.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc3[M10]
+\end_layout
+
+\end_inset
+
+If 
+\begin_inset Formula $a$
+\end_inset
+
+ and 
+\begin_inset Formula $m$
+\end_inset
+
+ are not relatively prime,
+ explain why the sequence will be somewhat handicapped and probably not very random;
+ hence we will generally want the multiplier 
+\begin_inset Formula $a$
+\end_inset
+
+ to be relatively prime to the modulus 
+\begin_inset Formula $m$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Let 
+\begin_inset Formula $p$
+\end_inset
+
+ be a common prime between 
+\begin_inset Formula $a$
+\end_inset
+
+ and 
+\begin_inset Formula $c$
+\end_inset
+
+,
+\begin_inset Formula 
+\[
+X_{n}=(a^{n}X_{0}+(a^{n-1}+a^{n-2}+\dots+1)c)\bmod m,
+\]
+
+\end_inset
+
+so 
+\begin_inset Formula 
+\[
+X_{n}\bmod p=c\bmod p
+\]
+
+\end_inset
+
+ for any 
+\begin_inset Formula $n\geq1$
+\end_inset
+
+.
+\end_layout
+
+\end_body
+\end_document
diff --git a/vol2/3.2.2.lyx b/vol2/3.2.2.lyx
new file mode 100644
index 0000000..173f721
--- /dev/null
+++ b/vol2/3.2.2.lyx
@@ -0,0 +1,834 @@
+#LyX 2.4 created this file. For more info see https://www.lyx.org/
+\lyxformat 620
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\begin_preamble
+\input defs
+\end_preamble
+\use_default_options true
+\maintain_unincluded_children no
+\language english
+\language_package default
+\inputencoding utf8
+\fontencoding auto
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_roman_osf false
+\font_sans_osf false
+\font_typewriter_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\float_placement class
+\float_alignment class
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_formatted_ref 0
+\use_minted 0
+\use_lineno 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style english
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tablestyle default
+\tracking_changes false
+\output_changes false
+\change_bars false
+\postpone_fragile_content false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\docbook_table_output 0
+\docbook_mathml_prefix 1
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc1[12]
+\end_layout
+
+\end_inset
+
+In practice,
+ we form random numbers using 
+\begin_inset Formula $X_{n+1}=(aX_{n}+c)\bmod m$
+\end_inset
+
+,
+ where the 
+\begin_inset Formula $X$
+\end_inset
+
+'s are 
+\emph on
+integers
+\emph default
+,
+ afterwards treating them as the 
+\emph on
+fractions
+\emph default
+ 
+\begin_inset Formula $U_{n}=X_{n}/m$
+\end_inset
+
+.
+ The recurrence relation for 
+\begin_inset Formula $U_{n}$
+\end_inset
+
+ is actually
+\begin_inset Formula 
+\[
+U_{n+1}=(aU_{n}+c/m)\bmod1.
+\]
+
+\end_inset
+
+Discuss the generation of random sequences using this relation 
+\emph on
+directly
+\emph default
+,
+ by making use of floating point arithmetic on the computer.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Despite the appeal of the idea for simplicity,
+ there is the problem of precision,
+ as the computer needs to be able to represent numbers as high as 
+\begin_inset Formula $a$
+\end_inset
+
+ (almost) with a precision of 
+\begin_inset Formula $1/m$
+\end_inset
+
+;
+ otherwise some numbers would be truncated and the period would be smaller than expected.
+ Also,
+ there might be issues due to rounding if 
+\begin_inset Formula $m$
+\end_inset
+
+ is not a divisor of a power of the computer's base (i.e.
+ 2,
+ or in some old computers 10),
+ but if it is,
+ then it's just faster to use integer arithmetic.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc2[M20]
+\end_layout
+
+\end_inset
+
+A good source of random numbers will have 
+\begin_inset Formula $X_{n-1}<X_{n+1}<X_{n}$
+\end_inset
+
+ about one-sixth of the time,
+ since each of the six possible relative orders of 
+\begin_inset Formula $X_{n-1}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $X_{n}$
+\end_inset
+
+,
+ and 
+\begin_inset Formula $X_{n+1}$
+\end_inset
+
+ should be equally probable.
+ However,
+ show that the ordering above 
+\emph on
+never
+\emph default
+ occurs if the Fibonacci sequence (5) is used.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+If 
+\begin_inset Formula $X_{n+1}<X_{n}$
+\end_inset
+
+,
+ then 
+\begin_inset Formula $X_{n-1}=(X_{n+1}-X_{n})\bmod m=X_{n+1}-X_{n}+m>X_{n+1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc4[00]
+\end_layout
+
+\end_inset
+
+Why is the most significant byte used in the first line of program (14),
+ instead of some other byte?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+It follows Algorithm B,
+ and it's the byte that's the most random.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc5[20]
+\end_layout
+
+\end_inset
+
+Discuss using 
+\begin_inset Formula $X_{n}=Y_{n}$
+\end_inset
+
+ in Algorithm M,
+ in order to improve the speed of generation.
+ Is the result analogous to Algorithm B?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+This would use the next element in the sequence to choose which of the previous elements to use.
+ By the discussion about Exercise 15,
+ the period of the sequence would be the same as the original one,
+ and the resulting might actually be less random.
+ This is different than Algorithm B where the element used to choose is the one that was chosen from the table in the previous iteration,
+ so the index 
+\begin_inset Formula $j$
+\end_inset
+
+ itself is shuffled.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc6[10]
+\end_layout
+
+\end_inset
+
+In the binary method (10),
+ the text states that the low-order bit of 
+\family typewriter
+X
+\family default
+ is random,
+ if the code is performed repeatedly.
+ Why isn't the entire 
+\emph on
+word
+\emph default
+ 
+\family typewriter
+X
+\family default
+ random?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Because there is a high correlation between one element and the next:
+ if the higher order bit is 0,
+ the next word is simply twice the previous one.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc9[M24]
+\end_layout
+
+\end_inset
+
+(R.
+ R.
+ Coveyou.) Use the result of exercise 8 to prove that the modified middle-square method (4) has a period of length 
+\begin_inset Formula $2^{e-2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+After several trials of the middle-square method,
+ we hypothesize that,
+ when 
+\begin_inset Formula $e\geq2$
+\end_inset
+
+,
+ 
+\begin_inset Formula $X_{n}=4Y_{n}+2$
+\end_inset
+
+ for each 
+\begin_inset Formula $n\geq0$
+\end_inset
+
+,
+ where 
+\begin_inset Formula $(Y_{n})_{n}$
+\end_inset
+
+ is the quadratic congruential sequence given by
+\begin_inset Formula 
+\begin{align*}
+X_{0} & =4Y_{0}+2, & Y_{n+1} & =(4Y_{n}+1)(Y_{n}+1)\bmod2^{e-2}.
+\end{align*}
+
+\end_inset
+
+This sequence has 
+\begin_inset Formula $d=4$
+\end_inset
+
+,
+ 
+\begin_inset Formula $a=5$
+\end_inset
+
+,
+ and 
+\begin_inset Formula $c=1$
+\end_inset
+
+,
+ so it meets the conditions from exercise 8 and therefore has period 
+\begin_inset Formula $2^{e-2}$
+\end_inset
+
+.
+ We can prove this identity by induction,
+ as if 
+\begin_inset Formula $X_{n}=4Y_{n}+2$
+\end_inset
+
+,
+ then
+\begin_inset Formula 
+\begin{multline*}
+X_{n+1}=X_{n}(X_{n}+1)\bmod2^{e}=(4Y_{n}+2)(4Y_{n}+3)\bmod2^{e}=\\
+=16Y_{n}^{2}+20Y_{n}+6\bmod2^{e}=4(4Y_{n}+1)(Y_{n}+1)+2\bmod2^{e}=4Y_{n+1}+2.
+\end{multline*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc22[M24]
+\end_layout
+
+\end_inset
+
+The text restricts discussion of the extended linear sequences (8) to the case that 
+\begin_inset Formula $m$
+\end_inset
+
+ is prime.
+ Prove that reasonably long periods can also be obtained when 
+\begin_inset Formula $m$
+\end_inset
+
+ is 
+\begin_inset Quotes eld
+\end_inset
+
+squarefree,
+\begin_inset Quotes erd
+\end_inset
+
+ that is,
+ the product of distinct primes.
+ (Examinations of Table 3.2.1.1–1 shows that 
+\begin_inset Formula $m=w\pm1$
+\end_inset
+
+ often satisfies this hypothesis;
+ many of the results of the text can therefore be carried over to that case,
+ which is somewhat more convenient for calculation.)
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+If 
+\begin_inset Formula $m=p_{1}\cdots p_{t}$
+\end_inset
+
+ where the 
+\begin_inset Formula $p_{k}$
+\end_inset
+
+ are distinct primes,
+ then integers 
+\begin_inset Formula $a\in\{0,\dots,m-1\}$
+\end_inset
+
+ can be associated with the tuples 
+\begin_inset Formula $(a\bmod p_{1},\dots,a\bmod p_{t})$
+\end_inset
+
+ because of the Chinese remainder theorem.
+ Furthermore,
+ equation (8) respects this association,
+ in the sense that 
+\begin_inset Formula $X_{n}\bmod p_{i}=((a_{1}\bmod p_{i})(X_{n-1}\bmod p_{i})+\dots+(a_{k}\bmod p_{i})(X_{n-k}\bmod p_{i}))\bmod p_{i}$
+\end_inset
+
+ and therefore we could as well calculate the sequence using these 
+\begin_inset Quotes eld
+\end_inset
+
+coordinates
+\begin_inset Quotes erd
+\end_inset
+
+.
+ We also know that the 
+\begin_inset Formula $a_{j}\bmod p_{i}$
+\end_inset
+
+ can be chosen to make 
+\begin_inset Formula $(X_{n}\bmod p_{i})_{n}$
+\end_inset
+
+ have period 
+\begin_inset Formula $p_{i}^{k}-1$
+\end_inset
+
+,
+ so the 
+\begin_inset Formula $a_{j}$
+\end_inset
+
+ can be chosen to make 
+\begin_inset Formula $(X_{n})_{n}$
+\end_inset
+
+ have period 
+\begin_inset Formula $\text{lcm}\{p_{1}^{k}-1,\dots,p_{t}^{k}-1\}$
+\end_inset
+
+,
+ which we deem reasonably long.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc23[20]
+\end_layout
+
+\end_inset
+
+Discuss the sequence defined by 
+\begin_inset Formula $X_{n}=(X_{n-55}-X_{n-24})\bmod m$
+\end_inset
+
+ as an alternative to (7).
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+The period is still at least 
+\begin_inset Formula $2^{55}-1$
+\end_inset
+
+,
+ since the sequence defined by the lowest bit in (7) has this period and the one defined by the lowest bit in this formula is exactly the same.
+ We don't know if exercise 30 still applies as we haven't done it.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+begin{samepage}
+\end_layout
+
+\end_inset
+
+
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc33[M23]
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+Let 
+\begin_inset Formula $g_{n}(z)=X_{n+30}+X_{n+29}z+\dots+X_{n}z^{30}+X_{n+54}z^{31}+\dots+X_{n+31}z^{54}$
+\end_inset
+
+,
+ where the 
+\begin_inset Formula $X$
+\end_inset
+
+'s satisfy the lagged Fibonacci recurrence (7).
+ Find a simple relation between 
+\begin_inset Formula $g_{n}(z)$
+\end_inset
+
+ and 
+\begin_inset Formula $g_{n+t}(z)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Express 
+\begin_inset Formula $X_{500}$
+\end_inset
+
+ in terms of 
+\begin_inset Formula $X_{0},\dots,X_{54}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+end{samepage}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer
+\end_layout
+
+\end_inset
+
+
+\begin_inset Note Greyedout
+status open
+
+\begin_layout Plain Layout
+(I obviously had to look at the answers.)
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula 
+\begin{align*}
+g_{n+1}(z) & =X_{n+31}+X_{n+30}z+\dots+X_{n+1}z^{30}+X_{n+55}z^{31}+\dots+X_{n+32}z^{54}\\
+ & =zg_{n}(z)+X_{n+31}-X_{n}z^{31}+X_{n+55}z^{31}-X_{n+31}z^{55}\\
+ & =zg_{n}(z)+X_{n+31}(z^{31}-z^{55}+1)+km,
+\end{align*}
+
+\end_inset
+
+for some small integer 
+\begin_inset Formula $k$
+\end_inset
+
+.
+ If we see the 
+\begin_inset Formula $g_{n}$
+\end_inset
+
+ as polynomials on 
+\begin_inset Formula $z$
+\end_inset
+
+,
+ in the domain 
+\begin_inset Formula $\mathbb{Z}[X]$
+\end_inset
+
+,
+ then 
+\begin_inset Formula $g_{n+t}(z)\equiv z^{t}g_{n}(z)$
+\end_inset
+
+ in 
+\begin_inset Formula $A\coloneqq\mathbb{Z}[X]/(m\mathbb{Z}[X]+(z^{55}-z^{31}-1)\mathbb{Z}[X])\cong\mathbb{Z}_{m}[X]/(z^{55}-z^{31}-1)\mathbb{Z}_{m}[X]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+We prove by strong induction that,
+ if 
+\begin_inset Formula $a_{0}+\dots+a_{54}z^{54}=z^{k}\bmod(z^{55}-z^{31}-1)$
+\end_inset
+
+ in 
+\begin_inset Formula $\mathbb{Z}_{m}[X]$
+\end_inset
+
+,
+ then 
+\begin_inset Formula $X_{k}=a_{0}X_{0}+\dots+a_{54}X^{54}\bmod m$
+\end_inset
+
+.
+ If 
+\begin_inset Formula $k<55$
+\end_inset
+
+,
+ this is obvious;
+ otherwise 
+\begin_inset Formula $X_{k}=X_{k-55}+X_{k-24}$
+\end_inset
+
+ and 
+\begin_inset Formula $z^{k}\bmod(z^{55}-z^{31}-1)=(z^{k-55}+z^{k-24})\bmod(z^{55}-z^{31}-1)$
+\end_inset
+
+ by the division algorithm.
+ With this,
+ we run 
+\family typewriter
+divide(z^500,
+ z^55-z^31-1,
+ z)
+\family default
+ in Maxima and we get that 
+\begin_inset Formula $X_{500}=120X_{54}+462X_{50}+455X_{57}+X_{44}+120X_{43}+1001X_{40}+18X_{37}+9X_{36}+1287X_{33}+16X_{30}+462X_{26}+105X_{23}+210X_{19}+364X_{16}+X_{13}+36X_{12}+715X_{9}+17X_{6}+X_{5}+792X_{2}\bmod m$
+\end_inset
+
+.
+\end_layout
+
+\end_body
+\end_document
diff --git a/vol2/3.3.1.lyx b/vol2/3.3.1.lyx
new file mode 100644
index 0000000..f2bfc5c
--- /dev/null
+++ b/vol2/3.3.1.lyx
@@ -0,0 +1,607 @@
+#LyX 2.4 created this file. For more info see https://www.lyx.org/
+\lyxformat 620
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\begin_preamble
+\input defs
+\end_preamble
+\use_default_options true
+\maintain_unincluded_children no
+\language english
+\language_package default
+\inputencoding utf8
+\fontencoding auto
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_roman_osf false
+\font_sans_osf false
+\font_typewriter_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\float_placement class
+\float_alignment class
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_formatted_ref 0
+\use_minted 0
+\use_lineno 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style english
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tablestyle default
+\tracking_changes false
+\output_changes false
+\change_bars false
+\postpone_fragile_content false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\docbook_table_output 0
+\docbook_mathml_prefix 1
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+\begin_inset FormulaMacro
+\newcommand{\stirla}[2]{\genfrac[]{0pt}{}{#1}{#2}}
+{\begin{bmatrix}{\textstyle #1}\\
+{\textstyle #2}
+\end{bmatrix}}
+\end_inset
+
+
+\begin_inset FormulaMacro
+\newcommand{\stirlb}[2]{\genfrac\{\}{0pt}{}{#1}{#2}}
+{\begin{Bmatrix}{\textstyle #1}\\
+{\textstyle #2}
+\end{Bmatrix}}
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc1[00]
+\end_layout
+
+\end_inset
+
+What line of the chi-square table should be used to check whether or not the value 
+\begin_inset Formula $V=7\frac{7}{48}$
+\end_inset
+
+ of Eq.
+ (5) is improbably high?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Row 
+\begin_inset Formula $\nu=k-1=11-1=10$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc3[23]
+\end_layout
+
+\end_inset
+
+Some dice that were loaded as described in the previous exercise were rolled 144 times,
+ and the following values were observed:
+\begin_inset Formula 
+\[
+\begin{array}{rrrrrrrrrrrr}
+\text{value of }s= & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12\\
+\text{observed number, }Y_{s}= & 2 & 6 & 10 & 16 & 18 & 32 & 20 & 13 & 16 & 9 & 2
+\end{array}
+\]
+
+\end_inset
+
+Apply the chi-square test to 
+\emph on
+these
+\emph default
+ values,
+ using the probabilities in (1),
+ pretending that the dice are not in fact known to be faulty.
+ Does the chi-square test detect the bad dice?
+ If not,
+ explain why not.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+We take the values 
+\begin_inset Formula $np_{s}$
+\end_inset
+
+ from (2),
+ to get:
+\begin_inset Formula 
+\begin{align*}
+V & =\sum_{s=2}^{12}\frac{(Y_{s}-np_{s})^{2}}{np_{s}}=\frac{4}{4}+\frac{4}{8}+\frac{4}{12}+\frac{0}{16}+\frac{4}{20}+\frac{64}{24}+\frac{0}{20}+\frac{9}{16}+\frac{16}{12}+\frac{1}{8}+\frac{4}{4}\\
+ & =1+\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{8}{3}+\frac{9}{16}+\frac{4}{3}+\frac{1}{8}+1=2+\frac{13}{3}+\frac{19}{16}+\frac{1}{5}\\
+ & =7+\frac{80+45+48}{240}=7+\frac{173}{240}.
+\end{align*}
+
+\end_inset
+
+Using 
+\begin_inset Formula $n=10$
+\end_inset
+
+ we get a probability between 
+\begin_inset Formula $.25$
+\end_inset
+
+ and 
+\begin_inset Formula $.5$
+\end_inset
+
+,
+ which is not suspect.
+ This seems to be because the bias of one die compensates that of the other,
+ smoothing out the probability differences.
+ The difference could be discovered with a large enough value of 
+\begin_inset Formula $n$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc4[23]
+\end_layout
+
+\end_inset
+
+The author actually obtained the data in experiment 1 of (9) by simulating dice in which one was normal,
+ the other was loaded so that it always turned up 1 or 6.
+ (The latter two possibilities were equally probable.) Compute the probabilities that replace (1) in this case,
+ and by using a chi-square test decide if the results of that experiment are consistent with the dice being loaded in this way.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+We compute the table with the sum of the two dice:
+\begin_inset Formula 
+\[
+\begin{array}{r|rrrrrr}
+ & 1 & 2 & 3 & 4 & 5 & 6\\
+\hline 1 & 2 & 3 & 4 & 5 & 6 & 7\\
+6 & 7 & 8 & 9 & 10 & 11 & 12
+\end{array}
+\]
+
+\end_inset
+
+This gives us the following table of probabilities:
+\begin_inset Formula 
+\[
+\begin{array}{rrrrrrrrrrrr}
+s= & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12\\
+144p_{s}= & 12 & 12 & 12 & 12 & 12 & 24 & 12 & 12 & 12 & 12 & 12
+\end{array}
+\]
+
+\end_inset
+
+Thus,
+\begin_inset Formula 
+\begin{align*}
+V & =\frac{1}{12}\left(8^{2}+2^{2}+2^{2}+1^{2}+8^{2}+\frac{6^{2}}{2}+6^{2}+1^{2}+1^{2}+2^{2}+1^{2}\right)\\
+ & =\frac{1}{12}(64+4+4+1+64+18+26+1+1+2+1)=\frac{186}{12}=15+\frac{1}{2}.
+\end{align*}
+
+\end_inset
+
+With 
+\begin_inset Formula $n=10$
+\end_inset
+
+,
+ this is somewhat in the middle of 
+\begin_inset Formula $p=.75$
+\end_inset
+
+ and 
+\begin_inset Formula $p=.95$
+\end_inset
+
+,
+ which is consistent.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc8[00]
+\end_layout
+
+\end_inset
+
+The text describes an experiment in which 20 values of the statistic 
+\begin_inset Formula $K_{10}^{+}$
+\end_inset
+
+ were obtained in the study of a random sequence.
+ These values were plotted,
+ to obtain Fig.
+ 4,
+ and a KS statistic was computed from the resulting graph.
+ Why were the table entries for 
+\begin_inset Formula $n=20$
+\end_inset
+
+ used to study the resulting statistic,
+ instead of the table entries for 
+\begin_inset Formula $n=10$
+\end_inset
+
+?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Because the value of 
+\begin_inset Formula $n$
+\end_inset
+
+ to use is not about the underlying probability distribution (which can be an arbitrary real-valued one,
+ not just 
+\begin_inset Formula $K_{n}^{+}$
+\end_inset
+
+ or 
+\begin_inset Formula $K_{n}^{-}$
+\end_inset
+
+),
+ but rather it is the number of observations we make for this distribution,
+ which is 20.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc9[20]
+\end_layout
+
+\end_inset
+
+The experiment described in the text consisted of plotting 20 values of 
+\begin_inset Formula $K_{10}^{+}$
+\end_inset
+
+,
+ computed from the maximum-of-5 test applied to different parts of a random sequence.
+ We could have computed also the corresponding 20 values of 
+\begin_inset Formula $K_{10}^{-}$
+\end_inset
+
+;
+ since 
+\begin_inset Formula $K_{10}^{-}$
+\end_inset
+
+ has the same distribution as 
+\begin_inset Formula $K_{10}^{+}$
+\end_inset
+
+,
+ we could lump together the 40 values thus obtained (that is,
+ 20 of the 
+\begin_inset Formula $K_{10}^{+}$
+\end_inset
+
+'s and 20 of the 
+\begin_inset Formula $K_{10}^{-}$
+\end_inset
+
+'s),
+ and a KS test could be applied so that we would get new values 
+\begin_inset Formula $K_{40}^{+}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $K_{40}^{-}$
+\end_inset
+
+.
+ Discuss the merits of this idea.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+The issue here is that the 40 points would not be independent;
+ if the maximum of 5 is low,
+ the minimum of 5 must be necessarily lower,
+ the probability of it being higher is 0.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc10[20]
+\end_layout
+
+\end_inset
+
+Suppose a chi-square test is done by making 
+\begin_inset Formula $n$
+\end_inset
+
+ observations,
+ and the value 
+\begin_inset Formula $V$
+\end_inset
+
+ is obtained.
+ Now we repeat the test on these same 
+\begin_inset Formula $n$
+\end_inset
+
+ observations over again (getting,
+ of course,
+ the same results),
+ and we put together the data from both tests,
+ regarding it as a single chi-square test with 
+\begin_inset Formula $2n$
+\end_inset
+
+ observations.
+ (This procedure violates the text's stipulation that all of the observations must be independent of one another.) How is the second value of 
+\begin_inset Formula $V$
+\end_inset
+
+ related to the first one?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Let 
+\begin_inset Formula $Y'_{s}=2Y_{s}$
+\end_inset
+
+ be the number of observations of category 
+\begin_inset Formula $s$
+\end_inset
+
+ in the second test,
+ the second value of 
+\begin_inset Formula $V$
+\end_inset
+
+ is
+\begin_inset Formula 
+\[
+V'=\sum_{s=1}^{k}\frac{(Y'_{s}-2np_{s})^{2}}{2np_{s}}=\sum_{s=1}^{k}\frac{(2Y_{s}-2np_{s})^{2}}{2np_{s}}=2V.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc11[10]
+\end_layout
+
+\end_inset
+
+Solve exercise 10 substituting the KS test for the chi-square test.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+This time,
+ after sorting the 
+\begin_inset Formula $2n$
+\end_inset
+
+ observations 
+\begin_inset Formula $X'_{1},\dots,X'_{2n}$
+\end_inset
+
+,
+ we have 
+\begin_inset Formula $X_{j}=X'_{2j-1}=X'_{2j}$
+\end_inset
+
+,
+ so
+\begin_inset Formula 
+\[
+K_{2n}^{+}=\sqrt{2n}\max_{1\leq j\leq2n}\left(\frac{j}{2n}-F(X'_{j})\right)=\sqrt{2n}\max_{1\leq j\leq n}\left(\frac{2j}{2n}-F(X_{j})\right)=\sqrt{2}K_{n}^{+},
+\]
+
+\end_inset
+
+and similarly,
+\begin_inset Formula 
+\[
+K_{2n}^{-}=\sqrt{2n}\max_{1\leq j\leq2n}\left(F(X'_{j})-\frac{j-1}{2n}\right)=\sqrt{2n}\max_{1\leq j\leq n}\left(F(X_{j})-\frac{2j-2}{2n}\right)=\sqrt{2}K_{n}^{-}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_body
+\end_document
diff --git a/vol2/3.3.2.lyx b/vol2/3.3.2.lyx
new file mode 100644
index 0000000..03a3027
--- /dev/null
+++ b/vol2/3.3.2.lyx
@@ -0,0 +1,1009 @@
+#LyX 2.4 created this file. For more info see https://www.lyx.org/
+\lyxformat 620
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\begin_preamble
+\input defs
+\end_preamble
+\use_default_options true
+\maintain_unincluded_children no
+\language english
+\language_package default
+\inputencoding utf8
+\fontencoding auto
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_roman_osf false
+\font_sans_osf false
+\font_typewriter_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\float_placement class
+\float_alignment class
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_formatted_ref 0
+\use_minted 0
+\use_lineno 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style english
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tablestyle default
+\tracking_changes false
+\output_changes false
+\change_bars false
+\postpone_fragile_content false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\docbook_table_output 0
+\docbook_mathml_prefix 1
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc1[10]
+\end_layout
+
+\end_inset
+
+Why should the serial test described in part B be applied to 
+\begin_inset Formula 
+\[
+(Y_{0},Y_{1}),(Y_{2},Y_{3}),\dots,(Y_{2n-2},Y_{2n-1})
+\]
+
+\end_inset
+
+instead of to 
+\begin_inset Formula $(Y_{0},Y_{1}),(Y_{1},Y_{2}),\dots,(Y_{n-1},Y_{n})$
+\end_inset
+
+?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Otherwise the points wouldn't be independent.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc2[10]
+\end_layout
+
+\end_inset
+
+State an appropriate way to generalize the serial test to triples,
+ quadruples,
+ etc.,
+ instead of pairs.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+For 
+\begin_inset Formula $k$
+\end_inset
+
+-tuples,
+ use points 
+\begin_inset Formula $(Y_{0},\dots,Y_{k-1}),(Y_{k},\dots,Y_{2k-1}),\dots,(Y_{k(n-1)},\dots,Y_{kn-1})$
+\end_inset
+
+.
+ The chi-square method is applied to the 
+\begin_inset Formula $d^{k}$
+\end_inset
+
+ possible categories and at least 
+\begin_inset Formula $5d^{k}$
+\end_inset
+
+ values of 
+\begin_inset Formula $U$
+\end_inset
+
+ should be taken.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc3[M20]
+\end_layout
+
+\end_inset
+
+How many 
+\begin_inset Formula $U$
+\end_inset
+
+'s need to be examined in the gap test (Algorithm G) before 
+\begin_inset Formula $n$
+\end_inset
+
+ gaps have been found,
+ on average,
+ assuming that the sequence is random?
+ What is the standard deviation of this quantity?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+The probability of a given value being the end of a gap is 
+\begin_inset Formula $p\coloneqq\frac{1}{\beta-\alpha}$
+\end_inset
+
+,
+ so the gap lengths have approximately exponential distribution and therefore we should take about 
+\begin_inset Formula $np=\frac{n}{\beta-\alpha}$
+\end_inset
+
+ 
+\begin_inset Formula $U$
+\end_inset
+
+'s.
+ More precisely,
+ let 
+\begin_inset Formula $q\coloneqq1-p$
+\end_inset
+
+,
+ the probability of a gap with length 
+\begin_inset Formula $k$
+\end_inset
+
+ is 
+\begin_inset Formula $q^{k-1}p$
+\end_inset
+
+,
+ so the average is
+\begin_inset Formula 
+\[
+\sum_{k=1}^{\infty}q^{k-1}pk=p\sum_{k=1}^{\infty}kq^{k-1}=p\sum_{j=1}^{\infty}\sum_{k=j}^{\infty}q^{k-1}=p\sum_{j=1}^{\infty}\frac{q^{j-1}}{1-q}=\sum_{j=0}^{\infty}q^{j}=\frac{1}{1-q}=\frac{1}{p},
+\]
+
+\end_inset
+
+so the approximation above is actually the exact value of the mean gap length and the average is exactly 
+\begin_inset Formula $\frac{n}{\beta-\alpha}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc7[08]
+\end_layout
+
+\end_inset
+
+Apply the coupon collector's test procedure (Algorithm C),
+ with 
+\begin_inset Formula $d=3$
+\end_inset
+
+ and 
+\begin_inset Formula $n=7$
+\end_inset
+
+,
+ to the sequence 1101221022120202001212201010201121.
+ What length do the seven subsequences have?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula $5,3,5,6,5,5,4$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc8[M22]
+\end_layout
+
+\end_inset
+
+How many 
+\begin_inset Formula $U$
+\end_inset
+
+'s need to be examined in the coupon collector's test,
+ on the average,
+ before 
+\begin_inset Formula $n$
+\end_inset
+
+ complete sets have been found by Algorithm C,
+ assuming that the sequence is random?
+ What is the standard deviation?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Given that 
+\begin_inset Formula $\stirlb rd=0$
+\end_inset
+
+ for 
+\begin_inset Formula $r<d$
+\end_inset
+
+,
+ we can write the generating function corresponding to the coupon collector's probability distribution as
+\begin_inset Formula 
+\begin{align*}
+F(z) & \coloneqq\sum_{r\geq1}\frac{d!}{d^{r}}\stirlb{r-1}{d-1}z^{r}=d!\frac{z}{d}\sum_{r\geq0}\stirlb r{d-1}\left(\frac{z}{d}\right)^{r}=\frac{(d-1)!z\left(\frac{z}{d}\right)^{d-1}}{(1-\tfrac{1}{d}z)(1-\tfrac{2}{d}z)\cdots(1-\tfrac{d-1}{d}z)}\\
+ & =\frac{(d-1)!z^{d}}{(d-z)(d-2z)\cdots(d-(d-1)z)}=\frac{z^{d}}{(d-z)(\frac{d}{2}-z)\cdots(\frac{d}{d-1}-z)}.
+\end{align*}
+
+\end_inset
+
+using Eq.
+ 1.2.9(28).
+ We derive this to get
+\begin_inset Formula 
+\begin{align*}
+F'(z) & =\frac{dz^{d-1}}{(d-z)\cdots(\frac{d}{d-1}-z)}-\frac{z^{d}(d-z)\cdots(\frac{d}{d-1}-z)\left(-\frac{1}{d-z}-\dots-\frac{1}{\frac{d}{d-1}-z}\right)}{(d-z)^{2}\cdots(\tfrac{d}{d-1}-z)^{2}}\\
+ & =\frac{dz^{d-1}+z^{d}\left(\frac{1}{d-z}+\dots+\frac{1}{\frac{d}{d-1}-z}\right)}{(d-z)\cdots(\frac{d}{d-1}-z)};\\
+F''(z) & =\frac{d(d-1)z^{d-2}+dz^{d-1}\left(\frac{1}{d-z}+\dots+\frac{1}{\frac{d}{d-1}-z}\right)+z^{d}\left(\frac{1}{(d-z)^{2}}+\dots+\frac{1}{(\frac{d}{d-1}-z)^{2}}\right)}{(d-z)\cdots(\frac{d}{d-1}-z)}+\\
+ & +\left(\frac{1}{d-z}+\dots+\frac{1}{\frac{d}{d-1}-z}\right)\frac{dz^{d-1}+z^{d}\left(\frac{1}{d-z}+\dots+\frac{1}{\frac{d}{d-1}-z}\right)}{(d-z)\cdots(\frac{d}{d-1}-z)}.
+\end{align*}
+
+\end_inset
+
+We are interested in 
+\begin_inset Formula $F'(1)$
+\end_inset
+
+ and 
+\begin_inset Formula $F''(1)$
+\end_inset
+
+,
+ for which we should note that
+\begin_inset Formula 
+\begin{align*}
+\frac{1}{\frac{d}{k}-1} & =\frac{k}{d-k}, & (d-1)(\tfrac{d}{2}-1)\cdots(\tfrac{d}{d-2}-1)(\tfrac{d}{d-1}-1) & =(d-1)\tfrac{d-2}{2}\cdots\tfrac{2}{d-2}\tfrac{1}{d-1}=1.
+\end{align*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Thus,
+ if 
+\begin_inset Formula $D_{1}\coloneqq\sum_{k=1}^{d}\frac{k}{d-k}$
+\end_inset
+
+ and 
+\begin_inset Formula $D_{2}\coloneqq\sum_{k=1}^{d}\left(\frac{k}{d-k}\right)^{2}$
+\end_inset
+
+,
+\begin_inset Formula 
+\begin{align*}
+F'(1)= & D_{1}+d; & F''(1)= & d(d-1)+dD_{1}+dD_{1}+D_{1}^{2}+D_{2}=(d+D_{1})^{2}-d+D_{2}.
+\end{align*}
+
+\end_inset
+
+With this,
+\begin_inset Formula 
+\begin{align*}
+\mu & =F'(1)=D_{1}+d;\\
+\sigma & =\sqrt{F''(1)+F'(1)-F'(1)^{2}}=\sqrt{D_{2}+D_{1}}.
+\end{align*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc11[00]
+\end_layout
+
+\end_inset
+
+The 
+\begin_inset Quotes eld
+\end_inset
+
+runs up
+\begin_inset Quotes erd
+\end_inset
+
+ in a particular permutation are displayed in (9);
+ what are the 
+\begin_inset Quotes eld
+\end_inset
+
+runs down
+\begin_inset Quotes erd
+\end_inset
+
+ in that permutation?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\begin{array}{|c|c|cccc|c|cc|c|}
+1 & 2 & 9 & 8 & 5 & 3 & 6 & 7 & 0 & 4\end{array}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc14[M15]
+\end_layout
+
+\end_inset
+
+If we 
+\begin_inset Quotes eld
+\end_inset
+
+throw away
+\begin_inset Quotes erd
+\end_inset
+
+ the element that immediately follows a run,
+ so that when 
+\begin_inset Formula $X_{j}$
+\end_inset
+
+ is greater than 
+\begin_inset Formula $X_{j+1}$
+\end_inset
+
+ we start the next run with 
+\begin_inset Formula $X_{j+2}$
+\end_inset
+
+,
+ the run lengths are independent,
+ and a simple chi-square test may be used (instead of the horribly complicated method derived in the text).
+ What are the appropriate run-length probabilities for this simple run test?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+When the 
+\begin_inset Formula $X_{j}$
+\end_inset
+
+ are real numbers,
+ the probability of a repeated element in a finite sample is 0,
+ so we can unambiguously take a permutation of the elements in such a way that all permutations are equally likely.
+ Let 
+\begin_inset Formula $p_{r}$
+\end_inset
+
+ be the probability that 
+\begin_inset Formula $U_{1},U_{2},U_{3},\dots$
+\end_inset
+
+ starts with a run of length 
+\begin_inset Formula $r$
+\end_inset
+
+ for 
+\begin_inset Formula $1\leq r<t$
+\end_inset
+
+,
+ and let 
+\begin_inset Formula $p_{t}$
+\end_inset
+
+ be the probability that it starts with a run of length 
+\begin_inset Formula $t$
+\end_inset
+
+ or more,
+ we have 
+\begin_inset Formula $p_{r}=\frac{r}{(r+1)!}$
+\end_inset
+
+,
+ since 
+\begin_inset Formula $U_{1},\dots,U_{r}$
+\end_inset
+
+ must be ordered like 
+\begin_inset Formula $U_{1}<\dots<U_{r}$
+\end_inset
+
+ and 
+\begin_inset Formula $U_{r+1}$
+\end_inset
+
+ must be inserted in this permutation in any of the 
+\begin_inset Formula $r$
+\end_inset
+
+ places that is not the last one (that is,
+ 
+\begin_inset Formula $U_{r+1}<U_{r}$
+\end_inset
+
+).
+ Similarly,
+ 
+\begin_inset Formula $p_{t}=\frac{1}{t!}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc15[M10]
+\end_layout
+
+\end_inset
+
+In the maximum-of-
+\begin_inset Formula $t$
+\end_inset
+
+ test,
+ why are 
+\begin_inset Formula $V_{0}^{t},V_{1}^{t},\dots,V_{n-1}^{t}$
+\end_inset
+
+ supposed to be uniformly distributed between zero and one?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula $P(V_{0}^{t}\leq x)=P(V_{0}\leq\sqrt[t]{x})=\sqrt[t]{x}^{t}=x$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc6[15]
+\end_layout
+
+\end_inset
+
+Mr.
+ J.
+ H.
+ Quick (a student) wanted to perform the maximum-of-
+\begin_inset Formula $t$
+\end_inset
+
+ test for several different values of 
+\begin_inset Formula $t$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Letting 
+\begin_inset Formula $Z_{jt}=\max(U_{j},U_{j+1},\dots,U_{j+t-1})$
+\end_inset
+
+ he found a clever way to go from the sequence 
+\begin_inset Formula $Z_{0(t-1)},Z_{1(t-1)},\dots$
+\end_inset
+
+,
+ to the sequence 
+\begin_inset Formula $Z_{0t},Z_{1t},\dots$
+\end_inset
+
+,
+ using very little time and space.
+ What was his bright idea?
+\end_layout
+
+\begin_layout Enumerate
+He decided to modify the maximum-of-
+\begin_inset Formula $t$
+\end_inset
+
+ method so that the 
+\begin_inset Formula $j$
+\end_inset
+
+th observation would be 
+\begin_inset Formula $\max(U_{j},\dots,U_{j+t-1})$
+\end_inset
+
+;
+ in other words,
+ he took 
+\begin_inset Formula $V_{j}=Z_{jt}$
+\end_inset
+
+ instead of 
+\begin_inset Formula $V_{j}=Z_{(tj)t}$
+\end_inset
+
+ as the text says.
+ He reasoned that 
+\emph on
+all
+\emph default
+ of the 
+\begin_inset Formula $Z$
+\end_inset
+
+'s should have the same distribution,
+ so the test is even stronger if each 
+\begin_inset Formula $Z_{jt}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $0\leq j<n$
+\end_inset
+
+,
+ is used instead of just every 
+\begin_inset Formula $t$
+\end_inset
+
+th one.
+ But when he tried a chi-square equidistribution test on the values of 
+\begin_inset Formula $V_{j}^{t}$
+\end_inset
+
+,
+ he got extremely high values of the statistic 
+\begin_inset Formula $V$
+\end_inset
+
+,
+ which got even higher as 
+\begin_inset Formula $t$
+\end_inset
+
+ increased.
+ Why did this happen?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula 
+\begin{multline*}
+Z_{jt}=\max\{U_{j},\dots,U_{j+t-1}\}=\\
+=\max\{\max\{U_{j},\dots,U_{j+t-2}\},\max\{U_{j+1},\dots,U_{j+t-1}\}\}=\max\{Z_{j,t-1},Z_{j+1,t-1}\}.
+\end{multline*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+The values of the 
+\begin_inset Formula $Z_{jt}$
+\end_inset
+
+ for fixed 
+\begin_inset Formula $t$
+\end_inset
+
+ and for all 
+\begin_inset Formula $j$
+\end_inset
+
+ are not independent,
+ as they represent overlapping ranges of elements in 
+\begin_inset Formula $(U_{j})_{j}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc31[M21]
+\end_layout
+
+\end_inset
+
+The recurrence 
+\begin_inset Formula $Y_{n}=(Y_{n-24}+Y_{n-55})\bmod2$
+\end_inset
+
+,
+ which describe the least significant bits of the lagged Fibonacci generator 3.2.2–(7) as well as the second-least significant bits of 3.2.2–(7'),
+ is known to have period length 
+\begin_inset Formula $2^{55}-1$
+\end_inset
+
+;
+ hence every possible nonzero pattern of bits 
+\begin_inset Formula $(Y_{n},Y_{n+1},\dots,Y_{n+54})$
+\end_inset
+
+ occurs equally often.
+ Nevertheless,
+ prove that if we generate 79 consecutive random bits 
+\begin_inset Formula $Y_{n},\dots,Y_{n+78}$
+\end_inset
+
+ starting at a random point in the period,
+ the probability is more than 
+\begin_inset Formula $\unit[51]{\%}$
+\end_inset
+
+ that there are more 1s than 0s.
+ If we use such bits to define a 
+\begin_inset Quotes eld
+\end_inset
+
+random walk
+\begin_inset Quotes erd
+\end_inset
+
+ that moves to the right when the bit is 1 and to the left when the bit is 0,
+ we'll finish to the right of our starting point significantly more than half of the time.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+The values 
+\begin_inset Formula $(Y_{n},\dots,Y_{n+54})$
+\end_inset
+
+ can be any nonzero pattern of bits with equal probability.
+ It is enough to prove this when also include the zero pattern,
+ as the probability of finding more 1s than 0s is even higher when we don't.
+ 
+\end_layout
+
+\begin_layout Standard
+In this case,
+ each 
+\begin_inset Formula $Y_{n+j}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $0\leq j\leq54$
+\end_inset
+
+,
+ can be 0 or 1 with equal probability,
+ each independent of the other,
+ and 
+\begin_inset Formula $Y_{n+j}=(Y_{n+j-55}+Y_{n+j-24})\bmod2$
+\end_inset
+
+ for 
+\begin_inset Formula $55\leq j\leq78$
+\end_inset
+
+.
+ Then,
+ for 
+\begin_inset Formula $0\leq j\leq23$
+\end_inset
+
+,
+ 
+\begin_inset Formula $Z_{n+j}\coloneqq Y_{n+j}+Y_{n+31+j}+Y_{n+55+j}$
+\end_inset
+
+ is 2 with probability 
+\begin_inset Formula $\frac{3}{4}$
+\end_inset
+
+ and 0 with probability 
+\begin_inset Formula $\frac{1}{4}$
+\end_inset
+
+,
+ and for 
+\begin_inset Formula $24\leq j\leq30$
+\end_inset
+
+,
+ 
+\begin_inset Formula $Y_{n+j}$
+\end_inset
+
+ is 1 or 0 with equal probability 
+\begin_inset Formula $\frac{1}{2}$
+\end_inset
+
+.
+ Furthermore,
+ these probabilities are now independent.
+\end_layout
+
+\begin_layout Standard
+With this in mind,
+\begin_inset Formula 
+\begin{align*}
+F(z) & \coloneqq\sum_{k}P(Y_{n}+\dots+Y_{n+78}=k)z^{k}\\
+ & =\sum_{k_{0},\dots,k_{30}}P(Z_{n}=k_{0})\cdots P(Z_{n+23}=k_{23})P(Y_{24}=k_{24})\cdots P(Y_{30}=k_{30})z^{k_{1}+\dots+k_{30}}\\
+ & =\left(\frac{3}{4}z^{2}+\frac{1}{4}\right)^{24}\left(\frac{1}{2}(z+1)\right)^{7}.
+\end{align*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+In Maxima,
+ we can type
+\begin_inset listings
+inline false
+status open
+
+\begin_layout Plain Layout
+
+p:''(expand((3/4*z^2+1/4)^24*(1/2*(z+1))^7))$
+\end_layout
+
+\begin_layout Plain Layout
+
+sum(coeff(p,z,j),j,40,79),numer;
+\end_layout
+
+\end_inset
+
+giving us a probability of having 1s than 0s of 
+\begin_inset Formula $0.5137...$
+\end_inset
+
+
+\end_layout
+
+\end_body
+\end_document
diff --git a/vol2/3.3.3.lyx b/vol2/3.3.3.lyx
new file mode 100644
index 0000000..d03c773
--- /dev/null
+++ b/vol2/3.3.3.lyx
@@ -0,0 +1,1383 @@
+#LyX 2.4 created this file. For more info see https://www.lyx.org/
+\lyxformat 620
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\begin_preamble
+\input defs
+\end_preamble
+\use_default_options true
+\maintain_unincluded_children no
+\language english
+\language_package default
+\inputencoding utf8
+\fontencoding auto
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_roman_osf false
+\font_sans_osf false
+\font_typewriter_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\float_placement class
+\float_alignment class
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_formatted_ref 0
+\use_minted 0
+\use_lineno 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style english
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tablestyle default
+\tracking_changes false
+\output_changes false
+\change_bars false
+\postpone_fragile_content false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\docbook_table_output 0
+\docbook_mathml_prefix 1
+\end_header
+
+\begin_body
+
+\begin_layout Subsubsection
+First Set
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc1[M10]
+\end_layout
+
+\end_inset
+
+Express 
+\begin_inset Formula $x\bmod y$
+\end_inset
+
+ in terms of the sawtooth and 
+\begin_inset Formula $\delta$
+\end_inset
+
+ functions.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+We have 
+\begin_inset Formula $((x))-\frac{1}{2}\delta(x)=x-\lfloor x\rfloor-\frac{1}{2}$
+\end_inset
+
+,
+ so 
+\begin_inset Formula $\lfloor x\rfloor=x-((x))+\tfrac{1}{2}(\delta(x)-1)$
+\end_inset
+
+.
+ Therefore
+\begin_inset Formula 
+\begin{multline*}
+x\bmod y=x-y\left\lfloor \frac{x}{y}\right\rfloor =x-y\left(\frac{x}{y}-\left(\left(\frac{x}{y}\right)\right)+\frac{1}{2}\left(\delta(\tfrac{x}{y})-1\right)\right)=\\
+=\left(\left(\left(\frac{x}{y}\right)\right)+\frac{1-\delta(\frac{x}{y})}{2}\right)y.
+\end{multline*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc4[M19]
+\end_layout
+
+\end_inset
+
+If 
+\begin_inset Formula $m=10^{10}$
+\end_inset
+
+,
+ what is the highest possible value of 
+\begin_inset Formula $d$
+\end_inset
+
+ (in the notation of Theorem P),
+ given that the potency of the generator is 10?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+It's 
+\begin_inset Formula $d=2\cdot5^{10}$
+\end_inset
+
+.
+ First,
+ we note that 
+\begin_inset Formula 
+\[
+(2\cdot5^{10})^{9}\bmod10^{10}=2^{9}5^{90}\bmod2^{10}5^{10}=2^{9}5^{10}(5^{80}\bmod2)=m/2\neq0,
+\]
+
+\end_inset
+
+ and that 
+\begin_inset Formula $(2\cdot5^{10})^{10}\bmod10^{10}=2^{10}5^{100}\bmod2^{10}5^{10}=0$
+\end_inset
+
+,
+ so if 
+\begin_inset Formula $b=d$
+\end_inset
+
+ we have potency 10.
+ Second,
+ we note that,
+ since 
+\begin_inset Formula $d\mid m$
+\end_inset
+
+,
+ and any divisor of 
+\begin_inset Formula $m$
+\end_inset
+
+ greater that 
+\begin_inset Formula $2\cdot5^{10}$
+\end_inset
+
+ has to be a multiple at least of 
+\begin_inset Formula $2^{2}$
+\end_inset
+
+ and of 
+\begin_inset Formula $5^{2}$
+\end_inset
+
+,
+ and therefore of 100,
+ then 
+\begin_inset Formula $b$
+\end_inset
+
+ would have to be a multiple of 100 and the potency would be at most 5.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc7[M24]
+\end_layout
+
+\end_inset
+
+Give a proof of the reciprocity law (19),
+ when 
+\begin_inset Formula $c=0$
+\end_inset
+
+,
+ by using the general reciprocity law of exercise 1.2.4–45.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+
+\begin_inset Note Greyedout
+status open
+
+\begin_layout Plain Layout
+(I had to look up the solution,
+ obviously.)
+\end_layout
+
+\end_inset
+
+In this case,
+ the law reduces to
+\begin_inset Formula 
+\begin{multline*}
+\sigma(h,k,0)+\sigma(k,h,0)=\\
+=12\sum_{0\leq j<k}\left(\left(\frac{j}{k}\right)\right)\left(\left(\frac{hj}{k}\right)\right)+12\sum_{0\leq j<h}\left(\left(\frac{j}{h}\right)\right)\left(\left(\frac{kj}{h}\right)\right)=\frac{h}{k}+\frac{k}{h}+\frac{1}{hk}-3.
+\end{multline*}
+
+\end_inset
+
+where 
+\begin_inset Formula $0<h\leq k$
+\end_inset
+
+ are coprime integers.
+ The last equation from exercise 1.2.4–45 with 
+\begin_inset Formula $k=2$
+\end_inset
+
+ tells us that
+\begin_inset Formula 
+\[
+\sum_{1\leq j<n}\left\lfloor \frac{mj}{n}\right\rfloor \left(\left\lfloor \frac{mj}{n}\right\rfloor +1\right)+2\sum_{1\leq j<m}j\left\lceil \frac{jn}{m}\right\rceil =nm(m-1)
+\]
+
+\end_inset
+
+for 
+\begin_inset Formula $m,n\in\mathbb{N}$
+\end_inset
+
+,
+ where we multiply by 2 and then set the lower bound of the sums to 1 because terms with 
+\begin_inset Formula $j=0$
+\end_inset
+
+ evaluate to 0.
+ Now,
+ since 
+\begin_inset Formula $h$
+\end_inset
+
+ and 
+\begin_inset Formula $k$
+\end_inset
+
+ are coprime,
+ for 
+\begin_inset Formula $j\in\{1,\dots,k-1\}$
+\end_inset
+
+,
+\begin_inset Formula 
+\begin{align*}
+\left(\left(\frac{hj}{k}\right)\right) & =\frac{hj}{k}-\left\lfloor \frac{hj}{k}\right\rfloor -\frac{1}{2}=\frac{hj}{k}-\left\lceil \frac{hj}{k}\right\rceil +\frac{1}{2},
+\end{align*}
+
+\end_inset
+
+so substituting above,
+\begin_inset Formula 
+\begin{multline*}
+S\coloneqq kh(h-1)=\\
+=\sum_{j=1}^{k-1}\left(\frac{hj}{k}-\left(\left(\frac{hj}{k}\right)\right)-\frac{1}{2}\right)\left(\frac{hj}{k}-\left(\left(\frac{hj}{k}\right)\right)+\frac{1}{2}\right)+\\
++2\sum_{j=1}^{h-1}j\left(\frac{kj}{h}-\left(\left(\frac{kj}{h}\right)\right)+\frac{1}{2}\right)=\\
+=\frac{h^{2}}{k^{2}}\sum_{j=1}^{k-1}j^{2}-\frac{2h}{k}\sum_{j=1}^{k-1}j\left(\left(\frac{hj}{k}\right)\right)+\sum_{j=1}^{k-1}\left(\left(\frac{hj}{k}\right)\right)^{2}-\frac{k-1}{4}+\frac{2k}{h}\sum_{j=1}^{h-1}j^{2}-\\
+-2\sum_{j=1}^{h-1}j\left(\left(\frac{kj}{h}\right)\right)+\frac{h^{2}-h}{2}.
+\end{multline*}
+
+\end_inset
+
+Now,
+\begin_inset Formula 
+\begin{align*}
+\sum_{j=1}^{k-1}\left(\left(\frac{hj}{k}\right)\right) & =\sum_{j=1}^{k-1}\left(\left(\frac{j}{k}\right)\right)=\sum_{j=1}^{k-1}\left(\frac{j}{k}-\frac{1}{2}\right)=\frac{k(k-1)}{2k}-\frac{k-1}{2}=0,
+\end{align*}
+
+\end_inset
+
+so
+\begin_inset Formula 
+\begin{multline*}
+\sigma(h,k,0)=12\sum_{j=0}^{k-1}\left(\left(\frac{j}{k}\right)\right)\left(\left(\frac{hj}{k}\right)\right)=12\sum_{j=1}^{k-1}\left(\frac{j}{k}-\frac{1}{2}\right)\left(\left(\frac{hj}{k}\right)\right)=\\
+=12\sum_{j=1}^{k-1}\frac{j}{k}\left(\left(\frac{hj}{k}\right)\right).
+\end{multline*}
+
+\end_inset
+
+In addition,
+\begin_inset Formula 
+\[
+\sum_{j=1}^{k-1}j^{2}=\frac{(k-1)k(2k-1)}{6}=\frac{k}{6}(2k^{2}-3k+1)=\frac{k^{3}}{3}-\frac{k^{2}}{2}+\frac{k}{6},
+\]
+
+\end_inset
+
+and in particular
+\begin_inset Formula 
+\begin{align*}
+\sum_{j=1}^{k-1}\left(\left(\frac{hj}{k}\right)\right)^{2} & =\sum_{j=1}^{k-1}\left(\left(\frac{j}{k}\right)\right)^{2}\\
+ & =\sum_{j=1}^{k-1}\frac{j^{2}}{k^{2}}-\sum_{j=1}^{k-1}\frac{j}{k}+\frac{k-1}{4}=\frac{k}{3}-\frac{1}{2}+\frac{1}{6k}-\frac{k-1}{2}+\frac{k-1}{4}\\
+ & =\frac{k}{3}-\frac{1}{2}+\frac{1}{6k}-\frac{k-1}{4},
+\end{align*}
+
+\end_inset
+
+so finally
+\begin_inset Formula 
+\begin{align*}
+S= & \frac{kh^{2}}{3}-\frac{h^{2}}{2}+\frac{h^{2}}{6k}-\frac{h}{6}\sigma(h,k,0)+\frac{k}{3}-\frac{1}{2}+\frac{1}{6k}-\frac{k-1}{2}+\frac{2kh^{2}}{3}-kh+\\
+ & +\frac{k}{3}-\frac{h}{6}\sigma(k,h,0)+\frac{h^{2}}{2}-\frac{h}{2}\\
+= & kh(h-1)-\frac{h}{2}+\frac{h^{2}}{6k}+\frac{k}{6}+\frac{1}{6k}-\frac{h}{6}\sigma(h,k,0)-\frac{h}{6}\sigma(k,h,0).
+\end{align*}
+
+\end_inset
+
+With this,
+\begin_inset Formula 
+\[
+\sigma(h,k,0)+\sigma(k,h,0)=-3+\frac{h}{k}+\frac{k}{h}+\frac{1}{hk}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc14[M20]
+\end_layout
+
+\end_inset
+
+The linear congruential generator that has 
+\begin_inset Formula $m=2^{35}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $a=2^{18}+1$
+\end_inset
+
+,
+ 
+\begin_inset Formula $c=1$
+\end_inset
+
+,
+ was given the serial correlation test on three batches of 1000 consecutive numbers,
+ and the result was a very high correlation,
+ between 
+\begin_inset Formula $0.2$
+\end_inset
+
+ and 
+\begin_inset Formula $0.3$
+\end_inset
+
+,
+ in each case.
+ What is the serial correlation of this generator,
+ taken over all 
+\begin_inset Formula $2^{35}$
+\end_inset
+
+ numbers of the period?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+The generator has full period by 3.2.1.2–A,
+ and 
+\begin_inset Formula $x'\coloneqq2^{18}-1$
+\end_inset
+
+ gives us 
+\begin_inset Formula $S(x')=0$
+\end_inset
+
+.
+ Thus,
+ by (17),
+ 
+\begin_inset Formula $C=(2^{35}\sigma(2^{18}+1,2^{35},1)-3+6(2^{35}-2^{18}))/(2^{70}-1)$
+\end_inset
+
+.
+ Now we calculate the Dedekind coefficient by Theorem D.
+\begin_inset Formula 
+\begin{align*}
+2^{35} & =(2^{17}-1)(2^{18}+1)+(2^{17}+1), & 1 & =0(2^{18}+1)+1;\\
+2^{18}+1 & =1(2^{17}+1)+2^{17}, & 1 & =0(2^{17}+1)+1;\\
+2^{17}+1 & =1\cdot2^{17}+1 & 1 & =0(2^{17})+1;\\
+2^{17} & =2^{17}\cdot1+0, & 1 & =1\cdot1+0.
+\end{align*}
+
+\end_inset
+
+The number 
+\begin_inset Formula $h'\coloneqq2^{35}-2^{18}+1$
+\end_inset
+
+ gives us 
+\begin_inset Formula $(2^{18}+1)h'\equiv1\pmod{2^{35}}$
+\end_inset
+
+,
+ so
+\begin_inset Formula 
+\begin{multline*}
+\sigma(2^{18}+1,2^{35},1)=\frac{(\cancel{2^{18}}+1)+(2^{35}\cancel{-2^{18}}+1)}{2^{35}}+\left((2^{17}-1)-6\cdot0+6\frac{1^{2}}{2^{35}(2^{18}+1)}\right)-\\
+-\left(1-6\cdot0+\frac{6\cdot1^{2}}{(2^{18}+1)(2^{17}+1)}\right)+\left(1-6\cdot0+6\frac{1^{2}}{(2^{17}+1)2^{17}}\right)-\left(2^{17}-6\cdot1+6\frac{1^{2}}{2^{17}}\right)\\
+\\-3-2+1=\\
+=\cancel{1}+\frac{2}{2^{35}}\cancel{+2^{17}}\cancel{-1}+\frac{6}{2^{53}+2^{35}}\cancel{-1}-\frac{6}{2^{35}+2^{18}+2^{17}+1}\cancel{+1}+\frac{6}{2^{34}+2^{17}}\cancel{-2^{17}}+6-\frac{6}{2^{17}}-4=\\
+=2+\frac{1}{2^{34}}+6\left(\frac{1}{2^{53}+2^{35}}-\frac{1}{2^{35}+3\cdot2^{17}+1}+\frac{1}{2^{34}+2^{17}}-\frac{1}{2^{17}}\right)=\\
+=2+\frac{1}{2^{34}}+6\frac{2^{17}+1-2^{35}\cancel{+2^{36}+2^{18}}-2^{53}\cancel{-2^{36}}-2^{35}\cancel{-2^{18}}}{2^{35}(2^{18}+1)(2^{17}+1)}=\\
+=2+\frac{1}{2^{34}}+3\frac{(-2^{36}+1)\cancel{(2^{17}+1)}}{2^{34}(2^{18}+1)\cancel{(2^{17}+1)}}=2+\frac{1}{2^{34}}-3\frac{2^{18}-1}{2^{34}}=2-\frac{3\cdot2^{18}-4}{2^{34}}=\\
+=2-\frac{3\cdot2^{16}-1}{2^{32}}=\frac{2^{33}-2^{17}-2^{16}+1}{2^{32}}=\frac{(2^{17}-1)(2^{16}-1)}{2^{32}}.
+\end{multline*}
+
+\end_inset
+
+Thus,
+\begin_inset Formula 
+\begin{multline*}
+C=\frac{8(2^{17}-1)(2^{16}-1)-3+6\cdot2^{18}(2^{17}-1)}{2^{70}-1}=\frac{91624920407}{393530540239137101141}\cong\\
+\cong2.33\cdot10^{-10}.
+\end{multline*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc18[M23]
+\end_layout
+
+\end_inset
+
+(U.
+ Dieter.) Given positive integers 
+\begin_inset Formula $h$
+\end_inset
+
+,
+ 
+\begin_inset Formula $k$
+\end_inset
+
+,
+ 
+\begin_inset Formula $z$
+\end_inset
+
+,
+ let
+\begin_inset Formula 
+\[
+S(h,k,c,z)=\sum_{0\leq j<z}\left(\left(\frac{hj+c}{k}\right)\right).
+\]
+
+\end_inset
+
+Show that this sum can be evaluated in closed form,
+ in terms of the generalized Dedekind sums and the sawtooth function.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Let 
+\begin_inset Formula $S_{j}\coloneqq\left(\left(\frac{hj+c}{k}\right)\right)$
+\end_inset
+
+ and note that 
+\begin_inset Formula $S_{j}=S_{j+k}$
+\end_inset
+
+ for all 
+\begin_inset Formula $j$
+\end_inset
+
+.
+ Thus,
+ if 
+\begin_inset Formula $z>k$
+\end_inset
+
+,
+\begin_inset Formula 
+\[
+\sum_{0\leq j<z}\left(\left(\frac{hj+c}{k}\right)\right)=\left\lfloor \frac{z}{k}\right\rfloor \sum_{0\leq j<k}\left(\left(\frac{hj+c}{k}\right)\right)+\sum_{0\leq j<z\bmod k}\left(\left(\frac{hj+c}{k}\right)\right),
+\]
+
+\end_inset
+
+so we may assume 
+\begin_inset Formula $z\leq k$
+\end_inset
+
+ from now on.
+ Now,
+ 
+\begin_inset Formula $\left\lfloor \frac{j}{k}\right\rfloor -\left\lfloor \frac{j-z}{k}\right\rfloor $
+\end_inset
+
+ is 1 when 
+\begin_inset Formula $0\leq j<z$
+\end_inset
+
+ and 0 when 
+\begin_inset Formula $z\leq j<k$
+\end_inset
+
+,
+ so
+\begin_inset Formula 
+\begin{multline*}
+S(h,k,c,z)=\sum_{0\leq j<k}\left(\left\lfloor \frac{j}{k}\right\rfloor -\left\lfloor \frac{j-z}{k}\right\rfloor \right)\left(\left(\frac{hj+c}{k}\right)\right)=\\
+=\sum_{0\leq j<k}\left(\cancel{\frac{j}{k}}-\left(\left(\frac{j}{k}\right)\right)\cancel{-\frac{1}{2}}-\frac{\cancel{j}-z}{k}+\left(\left(\frac{j-z}{k}\right)\right)\cancel{+\frac{1}{2}}\right)\left(\left(\frac{hj+c}{k}\right)\right)+\\
++\frac{1}{2}\left(\left(\frac{c}{k}\right)\right)-\frac{1}{2}\left(\left(\frac{hz+c}{k}\right)\right)=\\
+=\sum_{0\leq j<k}\left(\left(\left(\frac{j-z}{k}\right)\right)+\frac{z}{k}-\left(\left(\frac{j}{k}\right)\right)\right)\left(\left(\frac{hj+c}{k}\right)\right)+\frac{1}{2}\left(\left(\frac{c}{k}\right)\right)-\frac{1}{2}\left(\left(\frac{hz+c}{k}\right)\right).
+\end{multline*}
+
+\end_inset
+
+Let's evaluate the sum term by term.
+ Clearly the term 
+\begin_inset Formula $-\left(\left(\frac{j}{k}\right)\right)$
+\end_inset
+
+ sums to 
+\begin_inset Formula $-\frac{1}{12}\sigma(h,k,c)$
+\end_inset
+
+.
+ For the term with 
+\begin_inset Formula $\frac{z}{k}$
+\end_inset
+
+,
+ which is constant,
+ we use the argument used to derive Eq.
+ (13) in the text with 
+\begin_inset Formula $d\coloneqq\gcd\{h,k\}$
+\end_inset
+
+.
+ Finally,
+\begin_inset Formula 
+\begin{multline*}
+\sum_{0\leq j<k}\left(\left(\frac{j-z}{k}\right)\right)\left(\left(\frac{hj+c}{k}\right)\right)=\sum_{-z\leq j<k-z}\left(\left(\frac{j}{k}\right)\right)\left(\left(\frac{hj+c+hz}{k}\right)\right)=\\
+=\frac{1}{12}\sigma(h,k,c+hz).
+\end{multline*}
+
+\end_inset
+
+Putting it all together,
+\begin_inset Formula 
+\[
+S(h,k,c,z)=\frac{1}{12}\sigma(h,k,c+hz)-\frac{1}{12}\sigma(h,k,c)+\frac{zd}{k}\left(\left(\frac{c}{d}\right)\right)+\frac{1}{2}\left(\left(\frac{c}{k}\right)\right)-\frac{1}{2}\left(\left(\frac{hz+c}{k}\right)\right).
+\]
+
+\end_inset
+
+For 
+\begin_inset Formula $z=k$
+\end_inset
+
+,
+ this simplifies to 
+\begin_inset Formula $d\left(\left(\frac{c}{d}\right)\right)$
+\end_inset
+
+,
+ and since 
+\begin_inset Formula $\left\lfloor \frac{z}{k}\right\rfloor +\frac{z\bmod k}{k}=\frac{z}{k}$
+\end_inset
+
+,
+ it's easy to check that this formula still applies when 
+\begin_inset Formula $z>k$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc19[M23]
+\end_layout
+
+\end_inset
+
+Show that the 
+\emph on
+serial test
+\emph default
+ can be analyzed over the full period,
+ in terms of generalized Dedekind sums,
+ by finding a formula for the probability that 
+\begin_inset Formula $\alpha\leq X_{n}<\beta$
+\end_inset
+
+ and 
+\begin_inset Formula $\alpha'\leq X_{n+1}<\beta'$
+\end_inset
+
+,
+ when 
+\begin_inset Formula $\alpha$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\beta$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\alpha'$
+\end_inset
+
+,
+ and 
+\begin_inset Formula $\beta'$
+\end_inset
+
+ are given integers with 
+\begin_inset Formula $0\leq\alpha<\beta\leq m$
+\end_inset
+
+ and 
+\begin_inset Formula $0\leq\alpha'<\beta'\leq m$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula $P(x)\coloneqq\left\lfloor \frac{x-\alpha}{m}\right\rfloor -\left\lfloor \frac{x-\beta}{m}\right\rfloor $
+\end_inset
+
+ is 1 precisely when 
+\begin_inset Formula $x\in[\alpha,\beta)$
+\end_inset
+
+ and 0 for any other 
+\begin_inset Formula $x\in[0,1)$
+\end_inset
+
+.
+ 
+\begin_inset Formula $Q(x)\coloneqq\left\lfloor \frac{x-\alpha'}{m}\right\rfloor -\left\lfloor \frac{x-\beta'}{m}\right\rfloor $
+\end_inset
+
+ works in an analogous manner.
+ For a linear congruential sequence given by 
+\begin_inset Formula $S(x)\coloneqq(ax+c)\bmod m$
+\end_inset
+
+ that has maximum period,
+ the probability that 
+\begin_inset Formula $x_{n}\in[\alpha,\beta)\land x_{n+1}\in[\alpha',\beta')$
+\end_inset
+
+ is
+\begin_inset Formula 
+\begin{multline*}
+\frac{1}{m}\sum_{0\leq x<m}P(x)Q(ax+c)=\\
+=\frac{1}{m}\sum_{0\leq x<m}\left(\left\lfloor \frac{x-\alpha}{m}\right\rfloor -\left\lfloor \frac{x-\beta}{m}\right\rfloor \right)\left(\left\lfloor \frac{S(x)-\alpha'}{m}\right\rfloor -\left\lfloor \frac{S(x)-\beta'}{m}\right\rfloor \right)
+\end{multline*}
+
+\end_inset
+
+Now,
+\begin_inset Formula 
+\begin{align*}
+\left\lfloor \frac{x-\alpha}{m}\right\rfloor  & =\frac{x}{m}-\frac{\alpha}{m}-\frac{1}{2}-\left(\left(\frac{x-\alpha}{m}\right)\right)+\frac{1}{2}\delta_{x,\alpha},\\
+\left\lfloor \frac{S(x)-\alpha'}{m}\right\rfloor  & =\frac{(ax+c)\bmod m-\alpha'}{m}-\left(\left(\frac{ax+c-\alpha'}{m}\right)\right)-\frac{1}{2}+\frac{1}{2}\delta_{S(x)\alpha'}=\\
+ & =\left(\left(\frac{ax+c}{m}\right)\right)-\left(\left(\frac{ax+c-\alpha'}{m}\right)\right)-\frac{\alpha'}{m}+\frac{1}{2}(\delta_{S(x)\alpha'}-\delta_{S(x)0}).
+\end{align*}
+
+\end_inset
+
+Thus,
+ 
+\begin_inset Formula 
+\begin{multline*}
+\sum_{0\leq x<m}\left\lfloor \frac{x-\alpha}{m}\right\rfloor \left\lfloor \frac{S(x)-\alpha'}{m}\right\rfloor =\\
+=\sum_{0\leq x<m}\left(\left(\frac{x}{m}-\frac{1}{2}\right)-\frac{\alpha}{m}-\left(\left(\frac{x-\alpha}{m}\right)\right)\right)\\
+\left(\left(\left(\frac{ax+c}{m}\right)\right)-\left(\left(\frac{ax+c-\alpha'}{m}\right)\right)-\frac{\alpha'}{m}\right)+\\
++\frac{1}{2}\left(\left\lfloor \frac{S(\alpha)-\alpha'}{m}\right\rfloor +\left\lfloor \frac{S^{-1}(\alpha')-\alpha}{m}\right\rfloor -\left\lfloor \frac{S^{-1}(0)-\alpha}{m}\right\rfloor \right)+\frac{1}{4}([S(\alpha)=\alpha']-[S(\alpha)=0]).
+\end{multline*}
+
+\end_inset
+
+The terms outside this last sum can be calculated directly.
+ Inside the sum,
+ we have a product of two sums with three terms each,
+ which we may expand into 9 terms.
+ For these,
+ note that 
+\begin_inset Formula $\frac{x}{m}-\frac{1}{2}=\left(\left(\frac{x}{m}\right)\right)-\frac{1}{2}[x=0]$
+\end_inset
+
+,
+ so for example
+\begin_inset Formula 
+\[
+\sum_{0\leq x<m}\left(\frac{x}{m}-\frac{1}{2}\right)\left(\left(\frac{ax+c-\alpha'}{m}\right)\right)=\frac{1}{12}\delta(a,m,c-\alpha')-\left(\left(\frac{c-\alpha'}{m}\right)\right),
+\]
+
+\end_inset
+
+and similarly,
+\begin_inset Formula 
+\begin{multline*}
+\sum_{0\leq x<m}\left(\left(\frac{x-\alpha}{m}\right)\right)\left(\left(\frac{ax+c-\alpha'}{m}\right)\right)=\\
+=\sum_{-\alpha\leq x<m-\alpha}\left(\left(\frac{x}{m}\right)\right)\left(\left(\frac{ax+c-\alpha'+\alpha}{m}\right)\right)=\frac{1}{12}\sigma(a,m,c-\alpha'+\alpha).
+\end{multline*}
+
+\end_inset
+
+The other terms do not include sawtooth functions and can be expanded mechanically using that 
+\begin_inset Formula $\sum_{0\leq x<m}x=\frac{m(m-1)}{2}$
+\end_inset
+
+.
+ Then we can compute the initial sum as the sum of 4 of these sums.
+\end_layout
+
+\begin_layout Subsubsection
+Second Set
+\end_layout
+
+\begin_layout Standard
+In many cases,
+ exact computations with integers are quite difficult to carry out,
+ but we can attempt to study the probabilities that arise when we take the average real values of 
+\begin_inset Formula $x$
+\end_inset
+
+ instead of restricting the calculation to integer values.
+ Although these results are only approximate,
+ they shed some light on the subject.
+\end_layout
+
+\begin_layout Standard
+It is convenient to deal with numbers 
+\begin_inset Formula $U_{n}$
+\end_inset
+
+ between zero and one;
+ for linear congruential sequences,
+ 
+\begin_inset Formula $U_{n}=X_{n}/m$
+\end_inset
+
+,
+ and we have 
+\begin_inset Formula $U_{n+1}=\{aU_{n}+\theta\}$
+\end_inset
+
+,
+ where 
+\begin_inset Formula $\theta=c/m$
+\end_inset
+
+ and 
+\begin_inset Formula $\{x\}$
+\end_inset
+
+ denotes 
+\begin_inset Formula $x\bmod1$
+\end_inset
+
+.
+ For example,
+ the formula for serial correlation now becomes
+\begin_inset Formula 
+\[
+C=\left(\int_{0}^{1}x\{ax+\theta\}\text{d}x-\left(\int_{0}^{1}x\text{d}x\right)^{2}\right)\biggg/\left(\int_{0}^{1}x^{2}\text{d}x-\left(\int_{0}^{1}x\text{d}x\right)^{2}\right).
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc21[HM23]
+\end_layout
+
+\end_inset
+
+(R.
+ R.
+ Coveyou.) What is the value of 
+\begin_inset Formula $C$
+\end_inset
+
+ in the formula just given?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+We have
+\begin_inset Formula 
+\begin{align*}
+\int_{0}^{1}x\text{d}x & =\left.\frac{x^{2}}{2}\right|_{x=0}^{1}=\frac{1}{2}, & \int_{0}^{1}x^{2}\text{d}x & =\left.\frac{x^{3}}{3}\right|_{x=0}^{1}=\frac{1}{3},
+\end{align*}
+
+\end_inset
+
+and we just need to calculate the more complex integral.
+ Assume 
+\begin_inset Formula $a>0$
+\end_inset
+
+.
+ Then the graph for 
+\begin_inset Formula $\{ax+\theta\}$
+\end_inset
+
+ is a sequence of lines.
+ The first goes from 
+\begin_inset Formula $(0,\theta)$
+\end_inset
+
+ to 
+\begin_inset Formula $(\frac{1-\theta}{a},1)$
+\end_inset
+
+,
+ the next one from 
+\begin_inset Formula $(\frac{1-\theta}{a},0)$
+\end_inset
+
+ to 
+\begin_inset Formula $(\frac{2-\theta}{a},1)$
+\end_inset
+
+,
+ etc.,
+ and the last one goes from 
+\begin_inset Formula $(1-\tfrac{\theta}{a},0)$
+\end_inset
+
+ to 
+\begin_inset Formula $(1,\theta)$
+\end_inset
+
+ (we used that 
+\begin_inset Formula $a$
+\end_inset
+
+ is an integer to calculate this).
+ Thus
+\begin_inset Formula 
+\[
+\int_{0}^{1}x\{ax+\theta\}\text{d}x=\sum_{k=1}^{a-1}\int_{\frac{k-\theta}{a}}^{\frac{k+1-\theta}{a}}x(ax+\theta-k)\text{d}x+\int_{0}^{\frac{1-\theta}{a}}x(ax+\theta)\text{d}x+\int_{1-\frac{\theta}{a}}^{1}x(ax+\theta-a)\text{d}x.
+\]
+
+\end_inset
+
+Now,
+\begin_inset Formula 
+\[
+\int x(ax+\theta-k)\text{d}x=\frac{a}{3}x^{3}+\frac{\theta-k}{2}x^{2}+C,
+\]
+
+\end_inset
+
+so if we call 
+\begin_inset Formula $x_{k}\coloneqq\frac{k-\theta}{a}$
+\end_inset
+
+ except that 
+\begin_inset Formula $x_{0}\coloneqq0$
+\end_inset
+
+ and 
+\begin_inset Formula $x_{a+1}\coloneqq1$
+\end_inset
+
+,
+ we have
+\begin_inset Formula 
+\begin{multline*}
+\int_{0}^{1}x\{ax+\theta\}\text{d}x=\sum_{0\leq k\leq a}\int_{x_{k}}^{x_{k+1}}x(ax+\theta-k)\text{d}x=\\
+=\sum_{0\leq k\leq a}\left(\frac{a}{3}x_{k+1}^{3}+\frac{\theta-k}{2}x_{k+1}^{2}-\frac{a}{3}x_{k}^{3}-\frac{\theta-k}{2}x_{k}^{2}\right)=\frac{a}{3}+\frac{\theta-a}{2}+\sum_{0<k\leq a}\frac{1}{2}x_{k}^{2}=\\
+=\frac{a}{3}+\frac{\theta-a}{2}+\frac{1}{2a^{2}}\sum_{k=1}^{a}(k^{2}-2k\theta+\theta^{2})=\frac{\theta}{2}-\frac{a}{6}+\frac{(a+1)(2a+1)}{12a}-\frac{(a+1)\theta}{2a}+\frac{\theta^{2}}{2a}=\\
+=\frac{\theta(\theta-1)}{2a}+\frac{1}{12a}+\frac{1}{4}.
+\end{multline*}
+
+\end_inset
+
+Putting it all together,
+\begin_inset Formula 
+\[
+C=\left(\frac{\theta(\theta-1)}{2a}+\frac{1}{12a}+\frac{1}{4}-\frac{1}{4}\right)\biggg/\left(\frac{1}{3}-\frac{1}{4}\right)=\frac{6\theta(\theta-1)+1}{a}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc22[M22]
+\end_layout
+
+\end_inset
+
+Let 
+\begin_inset Formula $a$
+\end_inset
+
+ be an integer,
+ and let 
+\begin_inset Formula $0\leq\theta<1$
+\end_inset
+
+.
+ If 
+\begin_inset Formula $x$
+\end_inset
+
+ is a random real number,
+ uniformly distributed between 0 and 1,
+ and if 
+\begin_inset Formula $s(x)=\{ax+\theta\}$
+\end_inset
+
+,
+ what is the probability that 
+\begin_inset Formula $s(x)<x$
+\end_inset
+
+?
+ (This is the 
+\begin_inset Quotes eld
+\end_inset
+
+real number
+\begin_inset Quotes erd
+\end_inset
+
+ analog of Theorem P.)
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+As in the previous exercise,
+ let 
+\begin_inset Formula $x_{0}\coloneqq0$
+\end_inset
+
+,
+ 
+\begin_inset Formula $x_{k}\coloneqq\frac{k-\theta}{a}$
+\end_inset
+
+ for 
+\begin_inset Formula $k\in\{1,\dots,a\}$
+\end_inset
+
+,
+ and 
+\begin_inset Formula $x_{a+1}\coloneqq1$
+\end_inset
+
+,
+ for 
+\begin_inset Formula $k\in\{0,\dots,a\}$
+\end_inset
+
+ and 
+\begin_inset Formula $x\in[x_{k},x_{k+1})$
+\end_inset
+
+ we have 
+\begin_inset Formula $\lfloor ax+\theta\rfloor=k$
+\end_inset
+
+ and
+\begin_inset Formula 
+\[
+s(x)=ax+\theta-\lfloor ax+\theta\rfloor<x\iff(a-1)x<\lfloor ax+\theta\rfloor-\theta=k-\theta\iff x<\frac{k-\theta}{a-1},
+\]
+
+\end_inset
+
+so in particular 
+\begin_inset Formula $s(x)\geq x$
+\end_inset
+
+ for 
+\begin_inset Formula $x<x_{1}$
+\end_inset
+
+ and the probability is
+\begin_inset Formula 
+\begin{multline*}
+\int_{0}^{1}[s(x)<x]\text{d}x=\sum_{k=1}^{a}\int_{x_{k}}^{x_{k+1}}[x<\tfrac{k-\theta}{a-1}]\text{d}x=\sum_{k=1}^{a-1}\left(\frac{k-\theta}{a-1}-\frac{k-\theta}{a}\right)+1-\frac{a-\theta}{a}=\\
+=\sum_{k=1}^{a-1}\frac{k-\theta}{a(a-1)}+\frac{\theta}{a}=\left(\frac{1}{2}-\frac{\theta}{a}\right)+\frac{\theta}{a}=\frac{1}{2}.
+\end{multline*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc25[M25]
+\end_layout
+
+\end_inset
+
+Let 
+\begin_inset Formula $\alpha$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\beta$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\alpha'$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\beta'$
+\end_inset
+
+ be real numbers with 
+\begin_inset Formula $0\leq\alpha<\beta\leq1$
+\end_inset
+
+,
+ 
+\begin_inset Formula $0\leq\alpha'<\beta'\leq1$
+\end_inset
+
+.
+ Under the assumptions of exercise 22,
+ what is the probability that 
+\begin_inset Formula $\alpha\leq x<\beta$
+\end_inset
+
+ and 
+\begin_inset Formula $\alpha'\leq s(x)<\beta'$
+\end_inset
+
+?
+ (This is the 
+\begin_inset Quotes eld
+\end_inset
+
+real number
+\begin_inset Quotes erd
+\end_inset
+
+ analog of exercise 19.)
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+In the notation of the answer to exercise 22,
+ assume that 
+\begin_inset Formula $\alpha\in[x_{p},x_{p+1})$
+\end_inset
+
+ and 
+\begin_inset Formula $\beta\in[x_{q},x_{q+1})$
+\end_inset
+
+,
+ with 
+\begin_inset Formula $0\leq p\leq q\leq a$
+\end_inset
+
+.
+ For 
+\begin_inset Formula $x\in[x_{k},x_{k+1})$
+\end_inset
+
+,
+\begin_inset Formula 
+\[
+\alpha'\leq s(x)<\beta'\iff\alpha'\leq ax+\theta-k<\beta'\iff x\in\left[\frac{\alpha'+k-\theta}{a},\frac{\beta'+k-\theta}{a}\right).
+\]
+
+\end_inset
+
+Note that,
+ since 
+\begin_inset Formula $s(x)$
+\end_inset
+
+ goes from 0 to 1 when 
+\begin_inset Formula $x$
+\end_inset
+
+ goes from 
+\begin_inset Formula $x_{k}$
+\end_inset
+
+ to 
+\begin_inset Formula $x_{k+1}$
+\end_inset
+
+,
+ each of these intervals has 
+\begin_inset Formula $s(x)$
+\end_inset
+
+ enter and exit 
+\begin_inset Formula $[\alpha',\beta')$
+\end_inset
+
+ and fully contains the interval above.
+ Let 
+\begin_inset Formula $s_{k}^{-1}(y)\coloneqq\frac{y+k-\theta}{a}$
+\end_inset
+
+.
+ If 
+\begin_inset Formula $p=q$
+\end_inset
+
+,
+ the probability is
+\begin_inset Formula 
+\[
+\int_{\alpha}^{\beta}[\alpha'\leq s(x)<\beta']\text{d}x=\max\left\{ 0,\min\{\beta,s_{p}^{-1}(\beta')\}-\max\{\alpha,s_{p}^{-1}(\alpha')\}\right\} .
+\]
+
+\end_inset
+
+If 
+\begin_inset Formula $p\neq q$
+\end_inset
+
+,
+ we have to consider the two extremes and the 
+\begin_inset Formula $q-p-1$
+\end_inset
+
+ intervals in the middle,
+ each of which contributes 
+\begin_inset Formula $\frac{\beta'-\alpha'}{a}$
+\end_inset
+
+,
+ so the probability in this case is
+\begin_inset Formula 
+\begin{multline*}
+\max\left\{ 0,s_{p}^{-1}(\beta')-\max\{\alpha,s_{p}^{-1}(\alpha')\}\right\} +(q-p-1)\frac{\beta'-\alpha'}{a}+\\
++\max\left\{ 0,\min\{\beta,s_{q}^{-1}(\beta')\}-s_{q}^{-1}(\alpha')\right\} .
+\end{multline*}
+
+\end_inset
+
+Note that this last formula is still valid when 
+\begin_inset Formula $p=q$
+\end_inset
+
+.
+\end_layout
+
+\end_body
+\end_document
diff --git a/vol2/3.3.4.lyx b/vol2/3.3.4.lyx
new file mode 100644
index 0000000..1c06b60
--- /dev/null
+++ b/vol2/3.3.4.lyx
@@ -0,0 +1,781 @@
+#LyX 2.4 created this file. For more info see https://www.lyx.org/
+\lyxformat 620
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\begin_preamble
+\input defs
+\end_preamble
+\use_default_options true
+\maintain_unincluded_children no
+\language english
+\language_package default
+\inputencoding utf8
+\fontencoding auto
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_roman_osf false
+\font_sans_osf false
+\font_typewriter_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\float_placement class
+\float_alignment class
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_formatted_ref 0
+\use_minted 0
+\use_lineno 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style english
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tablestyle default
+\tracking_changes false
+\output_changes false
+\change_bars false
+\postpone_fragile_content false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\docbook_table_output 0
+\docbook_mathml_prefix 1
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc1[M10]
+\end_layout
+
+\end_inset
+
+To what does the spectral test reduce in 
+\emph on
+one
+\emph default
+ dimension?
+ (In other words,
+ what happens when 
+\begin_inset Formula $t=1$
+\end_inset
+
+?)
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+In this case 
+\begin_inset Formula $\nu_{1}^{-1}$
+\end_inset
+
+ is the maximum distance between points in 
+\begin_inset Formula $\{x/m\}_{x=0}^{m-1}$
+\end_inset
+
+,
+ which is 
+\begin_inset Formula $m^{-1}$
+\end_inset
+
+,
+ so 
+\begin_inset Formula $\nu_{1}=m$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc4[M23]
+\end_layout
+
+\end_inset
+
+Let 
+\begin_inset Formula $u_{11}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $u_{12}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $u_{21}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $u_{22}$
+\end_inset
+
+ be elements of a 
+\begin_inset Formula $2\times2$
+\end_inset
+
+ integer matrix such that 
+\begin_inset Formula $u_{11}+au_{12}\equiv u_{21}+au_{22}\equiv0\pmod m$
+\end_inset
+
+ and 
+\begin_inset Formula $u_{11}u_{22}-u_{21}u_{12}=m$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Prove that all integer solutions 
+\begin_inset Formula $(y_{1},y_{2})$
+\end_inset
+
+ to the congruence 
+\begin_inset Formula $y_{1}+ay_{2}\equiv0\pmod m$
+\end_inset
+
+ have the form 
+\begin_inset Formula $(y_{1},y_{2})=(x_{1}u_{11}+x_{2}u_{21},x_{1}u_{12}+x_{2}u_{22})$
+\end_inset
+
+ for integer 
+\begin_inset Formula $x_{1}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $x_{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+If,
+ in addition,
+ 
+\begin_inset Formula $2|u_{11}u_{21}+u_{12}u_{22}|\leq u_{11}^{2}+u_{12}^{2}\leq u_{21}^{2}+u_{22}^{2}$
+\end_inset
+
+,
+ prove that 
+\begin_inset Formula $(y_{1},y_{2})=(u_{11},u_{12})$
+\end_inset
+
+ minimizes 
+\begin_inset Formula $y_{1}^{2}+y_{2}^{2}$
+\end_inset
+
+ over all nonzero solutions to the congruence.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Assume 
+\begin_inset Formula $a\in\mathbb{Z}$
+\end_inset
+
+ and 
+\begin_inset Formula $m\in\mathbb{N}^{*}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Clearly all pairs of integers 
+\begin_inset Formula $(p,q)$
+\end_inset
+
+ can be written as 
+\begin_inset Formula $(p,q)=z_{1}(m,0)+z_{2}(-a,1)+z_{3}(1,0)$
+\end_inset
+
+ for some 
+\begin_inset Formula $z_{1},z_{2},z_{3}\in\mathbb{Z}$
+\end_inset
+
+ with 
+\begin_inset Formula $0\leq z_{3}<m$
+\end_inset
+
+.
+ Moreover,
+ solutions to the congruence are precisely those pairs with 
+\begin_inset Formula $z_{3}=0$
+\end_inset
+
+,
+ and we just have to prove that 
+\begin_inset Formula $(m,0)$
+\end_inset
+
+ and 
+\begin_inset Formula $(-a,1)$
+\end_inset
+
+ can be expressed as an integer linear combination of 
+\begin_inset Formula $(u_{11},u_{12})$
+\end_inset
+
+ and 
+\begin_inset Formula $(u_{21},u_{22})$
+\end_inset
+
+.
+ 
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+For 
+\begin_inset Formula $(m,0)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $u_{22}(u_{11},u_{12})-u_{12}(u_{21},u_{22})=(m,0)$
+\end_inset
+
+.
+ For 
+\begin_inset Formula $(a,-1)$
+\end_inset
+
+,
+ if 
+\begin_inset Formula $j,k\in\mathbb{Z}$
+\end_inset
+
+ are such that 
+\begin_inset Formula $u_{11}=jm-au_{12}$
+\end_inset
+
+ and 
+\begin_inset Formula $u_{21}=km-au_{22}$
+\end_inset
+
+,
+ we can expand to get 
+\begin_inset Formula $m=u_{11}u_{22}-u_{21}u_{12}=ju_{22}m-ku_{12}m$
+\end_inset
+
+,
+ so 
+\begin_inset Formula $ju_{22}-ku_{12}=1$
+\end_inset
+
+,
+ and then 
+\begin_inset Formula $ju_{21}-ku_{11}=-aju_{22}+aku_{12}=-a(ju_{22}-ku_{12})=-a$
+\end_inset
+
+,
+ so 
+\begin_inset Formula $(-a,1)=-k(u_{11},u_{12})+j(u_{21},u_{22})$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+If 
+\begin_inset Formula $x_{1},x_{2}\in\mathbb{Z}$
+\end_inset
+
+ are not both 0,
+ then
+\begin_inset Formula 
+\begin{multline*}
+(x_{1}u_{11}+x_{2}u_{21})^{2}+(x_{1}u_{12}+x_{2}u_{22})^{2}=\\
+=x_{1}^{2}(u_{11}^{2}+u_{12}^{2})+x_{2}^{2}(u_{21}^{2}+u_{22}^{2})+2x_{1}x_{2}(u_{11}u_{21}+u_{21}u_{22}).
+\end{multline*}
+
+\end_inset
+
+If 
+\begin_inset Formula $x_{1}x_{2}(u_{11}u_{21}+u_{21}u_{22})\geq0$
+\end_inset
+
+,
+ then this is greater or equal to
+\begin_inset Formula 
+\[
+x_{1}^{2}(u_{11}^{2}+u_{12}^{2})+x_{2}^{2}(u_{21}^{2}+u_{22}^{2})\geq(x_{1}^{2}+x_{2}^{2})(u_{11}^{2}+u_{12}^{2})\geq u_{11}^{2}+u_{12}^{2}.
+\]
+
+\end_inset
+
+Otherwise 
+\begin_inset Formula $x_{1},x_{2}\neq0$
+\end_inset
+
+ and,
+ if 
+\begin_inset Formula $|x_{1}|\leq|x_{2}|$
+\end_inset
+
+,
+ then the above is greater than or equal to
+\begin_inset Formula 
+\[
+x_{1}^{2}(u_{11}^{2}+u_{12}^{2}-2(u_{11}u_{21}+u_{21}u_{22}))+x_{2}^{2}(u_{21}^{2}+u_{22}^{2})\geq x_{2}^{2}(u_{21}^{2}+u_{22}^{2})\geq u_{11}^{2}+u_{12}^{2},
+\]
+
+\end_inset
+
+whereas the case with 
+\begin_inset Formula $|x_{1}|\geq|x_{2}|$
+\end_inset
+
+ is analogous.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc15[M20]
+\end_layout
+
+\end_inset
+
+Let 
+\begin_inset Formula $U$
+\end_inset
+
+ be an integer vector satisfying (15).
+ How many of the 
+\begin_inset Formula $(t-1)$
+\end_inset
+
+-dimensional hyperplanes defined by 
+\begin_inset Formula $U$
+\end_inset
+
+ intersect the unit hypercube 
+\begin_inset Formula $\{(x_{1},\dots,x_{t})\mid0\leq x_{j}<1\text{ for }1\leq j\leq t\}$
+\end_inset
+
+?
+ (This is approximately the number of hyperplanes in the family that will suffice to cover 
+\begin_inset Formula $L_{0}$
+\end_inset
+
+.)
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+The hyperplanes are defined by 
+\begin_inset Formula $\{\{X\mid X\cdot U=q\}\}_{q\in\mathbb{Z}}$
+\end_inset
+
+,
+ so we need to find the maximum and minimum integer values for 
+\begin_inset Formula $X\cdot U$
+\end_inset
+
+ when 
+\begin_inset Formula $X\in[0,1)^{n}$
+\end_inset
+
+,
+ which exist because 
+\begin_inset Formula $0\cdot U=0\in\mathbb{Z}$
+\end_inset
+
+.
+ The maximum and minimum real values when 
+\begin_inset Formula $X\in[0,1]^{n}$
+\end_inset
+
+ are,
+ respectively,
+ 
+\begin_inset Formula $M\coloneqq u_{1}\frac{1+\text{sgn}u_{1}}{2}+\dots+u_{t}\frac{1+\text{sgn}u_{t}}{2}$
+\end_inset
+
+ and 
+\begin_inset Formula $m\coloneqq u_{1}\frac{1-\text{sgn}u_{1}}{2}+\dots+u_{t}\frac{1-\text{sgn}u_{t}}{2}$
+\end_inset
+
+,
+ which happen to be integers,
+ so we have 
+\begin_inset Formula 
+\[
+M-m+1=u_{1}\text{sgn}u_{1}+\dots+u_{t}\text{sgn}u_{t}+1=|u_{1}|+\dots+|u_{t}|+1
+\]
+
+\end_inset
+
+hyperplanes.
+\end_layout
+
+\begin_layout Standard
+However,
+ one of these hyperplanes might only cover points in 
+\begin_inset Formula $[0,1]^{n}\setminus[0,1)^{n}$
+\end_inset
+
+.
+ This happens precisely when 
+\begin_inset Formula $(1,\dots,1)\cdot U=u_{1}+\dots+u_{t}$
+\end_inset
+
+ is either 
+\begin_inset Formula $M$
+\end_inset
+
+ or 
+\begin_inset Formula $m$
+\end_inset
+
+,
+ that is,
+ when all of the 
+\begin_inset Formula $u_{i}$
+\end_inset
+
+ are nonnegative or nonpositive.
+ Thus,
+ the actual number of hyperplanes is
+\begin_inset Formula 
+\[
+|u_{1}|+\dots+|u_{t}|+1-[u_{1},\dots,u_{t}\leq0]-[u_{1},\dots,u_{t}\geq0].
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc19[HM25]
+\end_layout
+
+\end_inset
+
+Suppose step S5 were changed slightly,
+ so that a transformation with 
+\begin_inset Formula $q=1$
+\end_inset
+
+ would be performed when 
+\begin_inset Formula $2V_{i}\cdot V_{j}=V_{j}\cdot V_{j}$
+\end_inset
+
+.
+ (Thus,
+ 
+\begin_inset Formula $q=\lfloor(V_{i}\cdot V_{j}/V_{j}\cdot V_{j})+\frac{1}{2}\rfloor$
+\end_inset
+
+ whenever 
+\begin_inset Formula $i\neq j$
+\end_inset
+
+.) Would it still be possible for Algorithm S to get into an infinite loop?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+No.
+ If 
+\begin_inset Formula $2|V_{i}\cdot V_{j}|>V_{j}\cdot V_{j}$
+\end_inset
+
+ in some step,
+ then
+\begin_inset Formula 
+\[
+(V_{i}-qV_{j})\cdot(V_{i}-qV_{j})=V_{i}\cdot V_{i}-2qV_{i}\cdot V_{j}+V_{j}\cdot V_{j}<V_{i}\cdot V_{i},
+\]
+
+\end_inset
+
+because 
+\begin_inset Formula $q$
+\end_inset
+
+ has the same sign as 
+\begin_inset Formula $V_{i}\cdot V_{j}$
+\end_inset
+
+ and therefore 
+\begin_inset Formula $V_{j}\cdot V_{j}<2|V_{i}\cdot V_{j}|\leq2qV_{i}\cdot V_{j}$
+\end_inset
+
+,
+ so 
+\begin_inset Formula $V_{i}\cdot V_{i}$
+\end_inset
+
+ decreases and,
+ since it is an integer,
+ it cannot decrease for infinitely many steps.
+ Thus,
+ an infinite loop would eventually only contain steps where 
+\begin_inset Formula $2V_{i}\cdot V_{j}=V_{j}\cdot V_{j}$
+\end_inset
+
+,
+ which are the ones we allow now,
+ and since there are only finitely many integer vectors with a given norm,
+ 
+\begin_inset Formula $V$
+\end_inset
+
+ would have to repeat at some point.
+ However,
+ in these cases 
+\begin_inset Formula $q=1$
+\end_inset
+
+,
+ so the steps are equivalent to multiplying 
+\begin_inset Formula $V$
+\end_inset
+
+ by an elementary matrix with 1s at the diagonal and at some other value and 0s everywhere else.
+ These matrices cannot result in an identity matrix when multiplying them because they don't have negative entries.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc32[M21]
+\end_layout
+
+\end_inset
+
+Let 
+\begin_inset Formula $m_{1}=2^{31}-1$
+\end_inset
+
+ and 
+\begin_inset Formula $m_{2}=2^{31}-249$
+\end_inset
+
+ be the moduli of generator (38).
+\end_layout
+
+\begin_layout Enumerate
+Show that if 
+\begin_inset Formula $U_{n}=(X_{n}/m_{1}-Y_{n}/m_{2})\bmod1$
+\end_inset
+
+,
+ we have 
+\begin_inset Formula $U_{n}\approx Z_{n}/m_{1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Let 
+\begin_inset Formula $W_{0}=(X_{0}m_{2}-Y_{0}m_{1})\bmod m$
+\end_inset
+
+ and 
+\begin_inset Formula $W_{n+1}=aW_{n}\bmod m$
+\end_inset
+
+,
+ where 
+\begin_inset Formula $a$
+\end_inset
+
+ and 
+\begin_inset Formula $m$
+\end_inset
+
+ have the values stated in the text following (38).
+ Prove that there is a simple relation between 
+\begin_inset Formula $W_{n}$
+\end_inset
+
+ and 
+\begin_inset Formula $U_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $Z_{n}/m_{1}=(X_{n}/m_{1}-Y_{n}/m_{1})\bmod1\approx(X_{n}/m_{1}-Y_{n}/m_{2})\bmod1=U_{n}$
+\end_inset
+
+.
+ The difference is at most 
+\begin_inset Formula $|Y_{n}/m_{1}-Y_{n}/m_{2}|=Y_{n}\left|\frac{1}{2^{31}-1}-\frac{1}{2^{31}-249}\right|=Y_{n}\frac{248}{(2^{31}-1)(2^{31}-249)}<\frac{248}{2^{31}-1}<2^{-23}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+We have 
+\begin_inset Formula $mU_{0}=(X_{0}m/m_{1}-Y_{0}m/m_{2})\bmod m=(X_{0}m_{2}-Y_{0}m_{1})\bmod m=W_{0}$
+\end_inset
+
+,
+ and also
+\begin_inset Formula 
+\begin{multline*}
+U_{n+1}=(aX_{n}\bmod m_{1}/m_{1}-aY_{n}\bmod m_{2}/m_{2})\bmod1=\\
+=a(X_{n}/m_{1}-Y_{n}/m_{2})\bmod1=aU_{n}\bmod1,
+\end{multline*}
+
+\end_inset
+
+so by induction 
+\begin_inset Formula $W_{n}\equiv mU_{n}$
+\end_inset
+
+.
+\end_layout
+
+\end_body
+\end_document
diff --git a/vol2/3.4.1.lyx b/vol2/3.4.1.lyx
new file mode 100644
index 0000000..2cd7261
--- /dev/null
+++ b/vol2/3.4.1.lyx
@@ -0,0 +1,1677 @@
+#LyX 2.4 created this file. For more info see https://www.lyx.org/
+\lyxformat 620
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\begin_preamble
+\input defs
+\end_preamble
+\use_default_options true
+\maintain_unincluded_children no
+\language english
+\language_package default
+\inputencoding utf8
+\fontencoding auto
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_roman_osf false
+\font_sans_osf false
+\font_typewriter_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\float_placement class
+\float_alignment class
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_formatted_ref 0
+\use_minted 0
+\use_lineno 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style english
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tablestyle default
+\tracking_changes false
+\output_changes false
+\change_bars false
+\postpone_fragile_content false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\docbook_table_output 0
+\docbook_mathml_prefix 1
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc1[10]
+\end_layout
+
+\end_inset
+
+If 
+\begin_inset Formula $\alpha$
+\end_inset
+
+ and 
+\begin_inset Formula $\beta$
+\end_inset
+
+ are real numbers with 
+\begin_inset Formula $\alpha<\beta$
+\end_inset
+
+,
+ how would you generate a random real number uniformly distributed between 
+\begin_inset Formula $\alpha$
+\end_inset
+
+ and 
+\begin_inset Formula $\beta$
+\end_inset
+
+?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+By taking a uniformly distributed real number 
+\begin_inset Formula $U$
+\end_inset
+
+ between 0 and 1 and returning 
+\begin_inset Formula $\alpha+(\beta-\alpha)U$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc3[14]
+\end_layout
+
+\end_inset
+
+Discuss treating 
+\begin_inset Formula $U$
+\end_inset
+
+ as an integer and computing its 
+\emph on
+remainder
+\emph default
+ 
+\begin_inset Formula $\bmod k$
+\end_inset
+
+ to get a random integer between 0 and 
+\begin_inset Formula $k-1$
+\end_inset
+
+,
+ instead of multiplying as suggested in the text.
+ Thus (1) would be changed to
+\begin_inset Formula 
+\begin{align*}
+ & \mathtt{ENTA\ 0}; &  & \mathtt{LDX\ U}; &  & \mathtt{DIV\ K},
+\end{align*}
+
+\end_inset
+
+with the result appearing in register X.
+ Is this a good method?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+While in theory it's as good as the other one,
+ less significant bits tend to be less random in many random number generation methods,
+ so we prefer the method in the text.
+ This is specially important if we use a linear congruential method where 
+\begin_inset Formula $k$
+\end_inset
+
+ is not relatively prime to 
+\begin_inset Formula $m$
+\end_inset
+
+.
+ In addition,
+ if 
+\begin_inset Formula $k$
+\end_inset
+
+ is not much smaller than 
+\begin_inset Formula $m$
+\end_inset
+
+,
+ numbers lower than 
+\begin_inset Formula $m\bmod k$
+\end_inset
+
+ will have a larger probability of appearing,
+ whereas in the method of the text a similar effect is observed but the numbers with higher probability are more evenly distributed.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc5[21]
+\end_layout
+
+\end_inset
+
+Suggest an efficient way to compute a random variable with the distribution 
+\begin_inset Formula $F(x)=px+qx^{2}+rx^{3}$
+\end_inset
+
+,
+ where 
+\begin_inset Formula $p\geq0$
+\end_inset
+
+,
+ 
+\begin_inset Formula $q\geq0$
+\end_inset
+
+,
+ 
+\begin_inset Formula $r\geq0$
+\end_inset
+
+,
+ and 
+\begin_inset Formula $p+q+r=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Assume this formula for the distribution is valid for 
+\begin_inset Formula $x\in[0,1]$
+\end_inset
+
+,
+ as 
+\begin_inset Formula $F(0)=0$
+\end_inset
+
+ and 
+\begin_inset Formula $F(1)=1$
+\end_inset
+
+.
+ Now,
+ 
+\begin_inset Formula $F(x)=x(rx^{2}+qx+p)$
+\end_inset
+
+,
+ or more precisely,
+\begin_inset Formula 
+\[
+F(x)=\max\{0,\min\{1,x\}\}\cdot\begin{cases}
+0, & x\leq-\tfrac{q}{2r};\\
+\min\{1,rx^{2}+qx+p\}, & x\geq-\tfrac{q}{2r},
+\end{cases}
+\]
+
+\end_inset
+
+so we may take 
+\begin_inset Formula $U_{1}$
+\end_inset
+
+ and 
+\begin_inset Formula $U_{2}$
+\end_inset
+
+ uniformly distributed between 0 and 1 and return
+\begin_inset Formula 
+\[
+\max\left\{ U_{1},-\frac{q}{2r}+\sqrt{\left(\frac{q}{2r}\right)^{2}-\frac{p}{r}+\frac{U_{2}}{r}}\right\} .
+\]
+
+\end_inset
+
+Note that we take the 
+\begin_inset Quotes eld
+\end_inset
+
+
+\begin_inset Formula $+$
+\end_inset
+
+
+\begin_inset Quotes erd
+\end_inset
+
+ sign in the solution to the quadratic equation to choose the solution in the right side of the parabola.
+ The method in the book is more efficient.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc7[20]
+\end_layout
+
+\end_inset
+
+(A.
+ J.
+ Walker) Suppose we have a bunch of cubes of 
+\begin_inset Formula $k$
+\end_inset
+
+ different colors,
+ say 
+\begin_inset Formula $n_{j}$
+\end_inset
+
+ cubes of color 
+\begin_inset Formula $C_{j}$
+\end_inset
+
+ for 
+\begin_inset Formula $1\leq j\leq k$
+\end_inset
+
+,
+ and we also have 
+\begin_inset Formula $k$
+\end_inset
+
+ boxes 
+\begin_inset Formula $\{B_{1},\dots,B_{k}\}$
+\end_inset
+
+ each of which can hold exactly 
+\begin_inset Formula $n$
+\end_inset
+
+ cubes.
+ Furthermore 
+\begin_inset Formula $n_{1}+\dots+n_{k}=kn$
+\end_inset
+
+,
+ so the cubes will just fit into the boxes.
+ Prove (constructively) that there is always a way to put the cubes into the boxes so that each box contains at most two different colors of cubes;
+ in fact,
+ there is a way to do it so that,
+ wherever box 
+\begin_inset Formula $B_{j}$
+\end_inset
+
+ contains two colors,
+ one of those colors is 
+\begin_inset Formula $C_{j}$
+\end_inset
+
+.
+ Show how to use this principle to compute the 
+\begin_inset Formula $P$
+\end_inset
+
+ and 
+\begin_inset Formula $Y$
+\end_inset
+
+ tables required in (3),
+ given a probability distribution 
+\begin_inset Formula $(p_{1},\dots,p_{k})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+We prove this by induction on 
+\begin_inset Formula $k$
+\end_inset
+
+.
+ For 
+\begin_inset Formula $k=1$
+\end_inset
+
+ this is trivial.
+ For 
+\begin_inset Formula $k>1$
+\end_inset
+
+,
+ if there is a 
+\begin_inset Formula $j$
+\end_inset
+
+ such that 
+\begin_inset Formula $n=n_{j}$
+\end_inset
+
+,
+ we pack the 
+\begin_inset Formula $n_{j}$
+\end_inset
+
+ cubes of color 
+\begin_inset Formula $C_{j}$
+\end_inset
+
+ into 
+\begin_inset Formula $B_{j}$
+\end_inset
+
+;
+ otherwise there exists 
+\begin_inset Formula $i$
+\end_inset
+
+ and 
+\begin_inset Formula $j$
+\end_inset
+
+ such that 
+\begin_inset Formula $n_{i}<n<n_{j}$
+\end_inset
+
+,
+ so we pack the 
+\begin_inset Formula $n_{i}$
+\end_inset
+
+ cubes of color 
+\begin_inset Formula $C_{i}$
+\end_inset
+
+ and also 
+\begin_inset Formula $n-n_{i}$
+\end_inset
+
+ cubes of color 
+\begin_inset Formula $C_{j}$
+\end_inset
+
+ into 
+\begin_inset Formula $B_{i}$
+\end_inset
+
+.
+ Either way we are left with 
+\begin_inset Formula $k-1$
+\end_inset
+
+ colors and 
+\begin_inset Formula $k-1$
+\end_inset
+
+ boxes.
+\end_layout
+
+\begin_layout Standard
+To construct the tables,
+ we note that this proof doesn't require 
+\begin_inset Formula $n$
+\end_inset
+
+ and the 
+\begin_inset Formula $n_{j}$
+\end_inset
+
+s to be integers,
+ they can be arbitrary nonnegative nonnegative reals as long as the conditions are met,
+ so we can let 
+\begin_inset Formula $n=\frac{1}{k}$
+\end_inset
+
+ and 
+\begin_inset Formula $n_{j}=p_{j}$
+\end_inset
+
+ (the probability of event 
+\begin_inset Formula $x_{j}$
+\end_inset
+
+) for each 
+\begin_inset Formula $j$
+\end_inset
+
+.
+ Then,
+ if after proceeding as in this proof,
+ we find that 
+\begin_inset Formula $B_{i}$
+\end_inset
+
+ has 
+\begin_inset Formula $p$
+\end_inset
+
+ cubes of color 
+\begin_inset Formula $C_{i}$
+\end_inset
+
+ and 
+\begin_inset Formula $n-p$
+\end_inset
+
+ cubes of some other color 
+\begin_inset Formula $C_{j}$
+\end_inset
+
+ (where 
+\begin_inset Formula $p\in[0,n]$
+\end_inset
+
+),
+ we set 
+\begin_inset Formula $P=pk$
+\end_inset
+
+ and 
+\begin_inset Formula $Y=j$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc10[HM24]
+\end_layout
+
+\end_inset
+
+Explain how to calculate auxiliary constants 
+\begin_inset Formula $P_{j}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $Q_{j}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $Y_{j}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $Z_{j}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $S_{j}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $D_{j}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $E_{j}$
+\end_inset
+
+ so that Algorithm M delivers answers with the correct distribution.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Let 
+\begin_inset Formula $p_{1},\dots,p_{31}$
+\end_inset
+
+ be the probabilities as in the text and 
+\begin_inset Formula $p_{0}=0$
+\end_inset
+
+.
+ We may calculate probabilities 
+\begin_inset Formula $p_{16},\dots,p_{30}$
+\end_inset
+
+ by a numeral method like Simpson's rule,
+ and this gives us values that are all lower than 
+\begin_inset Formula $\frac{1}{32}$
+\end_inset
+
+.
+ 
+\end_layout
+
+\begin_layout Standard
+First,
+ calculate 
+\begin_inset Formula $p_{1},\dots,p_{31}$
+\end_inset
+
+ as defined in the text,
+ obtaining that 
+\begin_inset Formula $p_{16},\dots,p_{31}<\frac{1}{32}$
+\end_inset
+
+ (we may use some numerical method like the Simpson's rule),
+ and set 
+\begin_inset Formula $p_{0}=0$
+\end_inset
+
+.
+ Then,
+ operate like in Exercise 7 for these probabilities to obtain tables 
+\begin_inset Formula $(P_{j},Y'_{j})_{j=0}^{31}$
+\end_inset
+
+,
+ ensuring that 
+\begin_inset Formula $p_{15},\dots,p_{31}$
+\end_inset
+
+ are considered first so that all the 
+\begin_inset Formula $Y'_{j}\in\{1,\dots,15\}$
+\end_inset
+
+.
+ This gives us the 
+\begin_inset Formula $P_{j}$
+\end_inset
+
+,
+ and then we set 
+\begin_inset Formula $Y_{j}=\frac{Y'_{j}-1}{5}$
+\end_inset
+
+ and 
+\begin_inset Formula $Z_{j}=\frac{1}{5(1-P_{j})}$
+\end_inset
+
+ for 
+\begin_inset Formula $0\leq j\leq31$
+\end_inset
+
+.
+ We also set 
+\begin_inset Formula $S_{j}=\frac{j-1}{5}$
+\end_inset
+
+ for 
+\begin_inset Formula $1\leq j\leq16$
+\end_inset
+
+ and 
+\begin_inset Formula $Q_{j}=\frac{1}{5P_{j}}$
+\end_inset
+
+ for 
+\begin_inset Formula $1\leq j\leq15$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Now,
+ the wedge functions are
+\begin_inset Formula 
+\[
+f_{j+15}(x)\coloneqq\sqrt{\frac{2}{\pi}}\left(\text{e}^{-x^{2}/2}-\text{e}^{-j^{2}/50}\right)\bigg/p_{j+15}
+\]
+
+\end_inset
+
+for 
+\begin_inset Formula $1\leq j\leq15$
+\end_inset
+
+.
+ For 
+\begin_inset Formula $1\leq j\leq5$
+\end_inset
+
+,
+ the curve is concave,
+ so we set
+\begin_inset Formula 
+\begin{align*}
+b_{j} & =-f'_{j+15}(S_{j+1})=\frac{j}{5p_{j+15}}\sqrt{\frac{2}{\pi}}\text{e}^{-j^{2}/50}, & a_{j} & =f_{j+15}(S_{j}).
+\end{align*}
+
+\end_inset
+
+For 
+\begin_inset Formula $6\leq j\leq15$
+\end_inset
+
+ the curve is convex and we set 
+\begin_inset Formula $b_{j}=5f_{j+15}(S_{j})$
+\end_inset
+
+ (the slope) and 
+\begin_inset Formula $a_{j}=f_{j+15}(x)+(x-S_{j})b$
+\end_inset
+
+,
+ where 
+\begin_inset Formula $x\in[S_{j},S_{j+1}]$
+\end_inset
+
+ is such that 
+\begin_inset Formula $f'_{j+15}(x)=-b_{j}$
+\end_inset
+
+.
+ Now,
+\begin_inset Formula 
+\begin{multline*}
+f'_{j+15}(x)=-\sqrt{\frac{2}{\pi}}\frac{x\text{e}^{-x^{2}/2}}{p_{j+15}}=-\frac{5}{p_{j+15}}\sqrt{\frac{2}{\pi}}(\text{e}^{-(j-1)^{2}/50}-\text{e}^{-j^{2}/50})=-b_{j}\iff\\
+x\text{e}^{-x^{2}/2}=5(\text{e}^{-(j-1)^{2}/50}-\text{e}^{-j^{2}/50}).
+\end{multline*}
+
+\end_inset
+
+We know this value exists by Lagrange's mean value theorem,
+ although we need to calculate it numerically (for example,
+ by Newton's method,
+ using that 
+\begin_inset Formula $\frac{\text{d}}{\text{d}x}(x\text{e}^{-x^{2}/2})=(1-x^{2})\text{e}^{-x^{2}/2}$
+\end_inset
+
+).
+ Then,
+ for for 
+\begin_inset Formula $1\leq j\leq15$
+\end_inset
+
+,
+ we set 
+\begin_inset Formula $D_{j+15}=a_{j}/b_{j}$
+\end_inset
+
+ and 
+\begin_inset Formula $E_{j+15}=\sqrt{\frac{2}{\pi}}\frac{e^{-j^{2}/50}}{b_{j}p_{j+15}}$
+\end_inset
+
+,
+ so
+\begin_inset Formula 
+\[
+E_{j+15}=\begin{cases}
+\frac{5}{j}, & 1\leq j\leq5;\\
+\frac{5}{\text{e}^{(2j-1)/50}-1}, & 6\leq j\leq15.
+\end{cases}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+\begin_inset Note Comment
+status open
+
+\begin_layout Plain Layout
+\begin_inset listings
+inline false
+status open
+
+\begin_layout Plain Layout
+
+use POSIX;
+\end_layout
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\begin_layout Plain Layout
+
+my (@D,
+ @E,
+ @P,
+ @Q,
+ @S,
+ @Y,
+ @Z);
+\end_layout
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\begin_layout Plain Layout
+
+sub normal {
+\end_layout
+
+\begin_layout Plain Layout
+
+    my $rand  = rand;
+\end_layout
+
+\begin_layout Plain Layout
+
+    my $sign  = POSIX::floor( $rand * 2 );
+\end_layout
+
+\begin_layout Plain Layout
+
+    my $piece = POSIX::floor( $rand * 64 ) % 32;
+\end_layout
+
+\begin_layout Plain Layout
+
+    my $dev   = ( $rand * 64 ) % 1;
+\end_layout
+
+\begin_layout Plain Layout
+
+    my $abs;
+\end_layout
+
+\begin_layout Plain Layout
+
+    if ( $dev >= $P[$piece] ) {
+\end_layout
+
+\begin_layout Plain Layout
+
+        $abs = $Y[$piece] + $dev * $Z[$piece];
+\end_layout
+
+\begin_layout Plain Layout
+
+    }
+\end_layout
+
+\begin_layout Plain Layout
+
+    elsif ( $piece <= 15 ) {
+\end_layout
+
+\begin_layout Plain Layout
+
+        $abs = $S[$piece] + $dev * $Q[$piece];
+\end_layout
+
+\begin_layout Plain Layout
+
+    }
+\end_layout
+
+\begin_layout Plain Layout
+
+    elsif ( $piece < 31 ) {
+\end_layout
+
+\begin_layout Plain Layout
+
+        $piece -= 16;
+\end_layout
+
+\begin_layout Plain Layout
+
+        my ( $u,
+ $v );
+\end_layout
+
+\begin_layout Plain Layout
+
+        do {
+\end_layout
+
+\begin_layout Plain Layout
+
+            ( $u,
+ $v ) = ( rand,
+ rand );
+\end_layout
+
+\begin_layout Plain Layout
+
+            ( $u,
+ $v ) = ( $v,
+ $u ) if $u > $v;
+\end_layout
+
+\begin_layout Plain Layout
+
+            $abs = $S[$piece] + $u / 5;
+\end_layout
+
+\begin_layout Plain Layout
+
+        } while ( $v < $D[$piece]
+\end_layout
+
+\begin_layout Plain Layout
+
+            || $v <= $u +
+\end_layout
+
+\begin_layout Plain Layout
+
+            $E[$piece] * ( exp( ( $S[ $piece + 1 ]**2 - $abs**2 ) / 2 ) - 1 ) );
+\end_layout
+
+\begin_layout Plain Layout
+
+    }
+\end_layout
+
+\begin_layout Plain Layout
+
+    else {
+\end_layout
+
+\begin_layout Plain Layout
+
+        do { $abs = sqrt( 9 - 2 * log rand ) } while ( $abs * rand ) >= 3;
+\end_layout
+
+\begin_layout Plain Layout
+
+    }
+\end_layout
+
+\begin_layout Plain Layout
+
+    return $sign ?
+ -$abs :
+ $abs;
+\end_layout
+
+\begin_layout Plain Layout
+
+}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset listings
+inline false
+status open
+
+\begin_layout Plain Layout
+
+local D = ...
+\end_layout
+
+\begin_layout Plain Layout
+
+local E = ...
+\end_layout
+
+\begin_layout Plain Layout
+
+local P = ...
+\end_layout
+
+\begin_layout Plain Layout
+
+local Q = ...
+\end_layout
+
+\begin_layout Plain Layout
+
+local S = ...
+\end_layout
+
+\begin_layout Plain Layout
+
+local Y = ...
+\end_layout
+
+\begin_layout Plain Layout
+
+local Z = ...
+\end_layout
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\begin_layout Plain Layout
+
+local function normal ()
+\end_layout
+
+\begin_layout Plain Layout
+
+    local rand = math.random()
+\end_layout
+
+\begin_layout Plain Layout
+
+    local sign = math.floor(rand * 2)
+\end_layout
+
+\begin_layout Plain Layout
+
+    local piece = math.floor(rand * 64) % 32
+\end_layout
+
+\begin_layout Plain Layout
+
+    local dev = (rand * 64) % 1
+\end_layout
+
+\begin_layout Plain Layout
+
+    local abs
+\end_layout
+
+\begin_layout Plain Layout
+
+    if dev >= P[piece] then
+\end_layout
+
+\begin_layout Plain Layout
+
+        abs = Y[piece] + dev * Z[piece]
+\end_layout
+
+\begin_layout Plain Layout
+
+    elseif piece <= 15 then
+\end_layout
+
+\begin_layout Plain Layout
+
+        abs = S[piece] + dev * Q[piece]
+\end_layout
+
+\begin_layout Plain Layout
+
+    elseif piece < 31 then
+\end_layout
+
+\begin_layout Plain Layout
+
+        local u,
+ v
+\end_layout
+
+\begin_layout Plain Layout
+
+        repeat
+\end_layout
+
+\begin_layout Plain Layout
+
+            u,
+ v = math.random(),
+ math.random()
+\end_layout
+
+\begin_layout Plain Layout
+
+            if u > v then u,
+ v = v,
+ u end
+\end_layout
+
+\begin_layout Plain Layout
+
+            abs = S[piece-15] + u/5
+\end_layout
+
+\begin_layout Plain Layout
+
+        until v >= D[piece-15] or
+\end_layout
+
+\begin_layout Plain Layout
+
+            v > u + E[piece-15] * (math.exp((S[piece-14]^2 - abs^2) / 2) - 1)
+\end_layout
+
+\begin_layout Plain Layout
+
+    else
+\end_layout
+
+\begin_layout Plain Layout
+
+        repeat
+\end_layout
+
+\begin_layout Plain Layout
+
+            abs = math.sqrt(9 - 2 * math.log(math.random()))
+\end_layout
+
+\begin_layout Plain Layout
+
+        until abs * rand >= 3
+\end_layout
+
+\begin_layout Plain Layout
+
+    end
+\end_layout
+
+\begin_layout Plain Layout
+
+    if sign == 0 then return abs else return -abs end
+\end_layout
+
+\begin_layout Plain Layout
+
+end
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset listings
+inline false
+status open
+
+\begin_layout Plain Layout
+
+from math import exp,
+ floor,
+ log,
+ sqrt
+\end_layout
+
+\begin_layout Plain Layout
+
+from random import random
+\end_layout
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\begin_layout Plain Layout
+
+D,
+ E,
+ P,
+ Q,
+ S,
+ Y,
+ Z = [],
+ [],
+ [],
+ [],
+ [],
+ [],
+ []
+\end_layout
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\begin_layout Plain Layout
+
+def normal():
+\end_layout
+
+\begin_layout Plain Layout
+
+    rand = random()
+\end_layout
+
+\begin_layout Plain Layout
+
+    sign = floor(rand * 2)
+\end_layout
+
+\begin_layout Plain Layout
+
+    piece = floor(rand * 64) % 32
+\end_layout
+
+\begin_layout Plain Layout
+
+    dev = (rand * 64) % 1
+\end_layout
+
+\begin_layout Plain Layout
+
+    if dev >= P[piece]:
+\end_layout
+
+\begin_layout Plain Layout
+
+        result = Y[piece] + dev * Z[piece]
+\end_layout
+
+\begin_layout Plain Layout
+
+    elif piece <= 15:
+\end_layout
+
+\begin_layout Plain Layout
+
+        result = S[piece] + dev * Q[piece]
+\end_layout
+
+\begin_layout Plain Layout
+
+    elif piece < 31:
+\end_layout
+
+\begin_layout Plain Layout
+
+        piece = piece - 16
+\end_layout
+
+\begin_layout Plain Layout
+
+        while True:
+\end_layout
+
+\begin_layout Plain Layout
+
+            u,
+ v = random(),
+ random()
+\end_layout
+
+\begin_layout Plain Layout
+
+            if u > v:
+\end_layout
+
+\begin_layout Plain Layout
+
+                u,
+ v = v,
+ u
+\end_layout
+
+\begin_layout Plain Layout
+
+            result = S[piece] + u/5
+\end_layout
+
+\begin_layout Plain Layout
+
+            if v >= D[piece]:
+\end_layout
+
+\begin_layout Plain Layout
+
+                break
+\end_layout
+
+\begin_layout Plain Layout
+
+            if v < u + E[piece] * (exp((S[piece+1]**2 - abs**2)/2) - 1):
+\end_layout
+
+\begin_layout Plain Layout
+
+                break
+\end_layout
+
+\begin_layout Plain Layout
+
+    else:
+\end_layout
+
+\begin_layout Plain Layout
+
+        while True:
+\end_layout
+
+\begin_layout Plain Layout
+
+            result = sqrt(9 - 2*log(random()))
+\end_layout
+
+\begin_layout Plain Layout
+
+            if result * random() < 3:
+\end_layout
+
+\begin_layout Plain Layout
+
+                break
+\end_layout
+
+\begin_layout Plain Layout
+
+    if sign:
+\end_layout
+
+\begin_layout Plain Layout
+
+        result = -result
+\end_layout
+
+\begin_layout Plain Layout
+
+    return result
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc16[HM22]
+\end_layout
+
+\end_inset
+
+(J.
+ H.
+ Ahrens.) Develop an algorithm for gamma deviates of order 
+\begin_inset Formula $a$
+\end_inset
+
+ when 
+\begin_inset Formula $0<a<1$
+\end_inset
+
+,
+ using the rejection method with 
+\begin_inset Formula $cg(t)=t^{a-1}/\Gamma(a)$
+\end_inset
+
+ for 
+\begin_inset Formula $0<t<1$
+\end_inset
+
+,
+ and with 
+\begin_inset Formula $cg(t)=\text{e}^{-t}/\Gamma(a)$
+\end_inset
+
+ for 
+\begin_inset Formula $t\geq1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+The density function for the gamma distribution is
+\begin_inset Formula 
+\begin{align*}
+f(x) & \coloneqq\frac{1}{\Gamma(a)}x^{a-1}\text{e}^{-x}, & x & \geq0.
+\end{align*}
+
+\end_inset
+
+Let
+\begin_inset Formula 
+\[
+g(t)\coloneqq\begin{cases}
+t^{a-1}, & 0<t<1;\\
+\text{e}^{-t}, & t\geq1.
+\end{cases}
+\]
+
+\end_inset
+
+Then,
+\begin_inset Formula 
+\begin{align*}
+G_{0}(x)\coloneqq\int_{0}^{x}g(t)\text{d}t & =\begin{cases}
+\int_{0}^{x}t^{a-1}\text{d}t=\left[\frac{t^{a}}{a}\right]_{t=0}^{x}=\frac{x^{a}}{a}, & 0<t<1;\\
+a^{-1}+\int_{1}^{x}\text{e}^{-t}\text{d}t=\frac{1}{a}-\left[\text{e}^{-t}\right]_{t=1}^{x}=a^{-1}+\text{e}^{-1}-\text{e}^{-x}, & t\geq1.
+\end{cases}
+\end{align*}
+
+\end_inset
+
+If 
+\begin_inset Formula $M\coloneqq\lim_{x\to\infty}G(x)=a^{-1}+\text{e}^{-1}$
+\end_inset
+
+,
+ the real probability distribution function is 
+\begin_inset Formula $G(x)\coloneqq G_{0}(x)/M$
+\end_inset
+
+.
+ Thus,
+ a possible method could be the following:
+ Generate to independent deviates 
+\begin_inset Formula $U$
+\end_inset
+
+ and 
+\begin_inset Formula $V$
+\end_inset
+
+ uniformly distributed between 0 and 1.
+ If 
+\begin_inset Formula $U<a^{-1}/M$
+\end_inset
+
+,
+ set 
+\begin_inset Formula $X\coloneqq\sqrt[a]{aMU}=\sqrt[a]{1+\frac{a}{\text{e}}}\sqrt[a]{U}$
+\end_inset
+
+ and 
+\begin_inset Formula $Y\coloneqq\Gamma(a)f(X)/g(X)=\text{e}^{-X}$
+\end_inset
+
+.
+ Otherwise set 
+\begin_inset Formula $X\coloneqq-\ln M\cdot\ln(1-U)$
+\end_inset
+
+ and 
+\begin_inset Formula $Y\coloneqq\Gamma(a)f(X)/g(X)=X^{a-1}$
+\end_inset
+
+.
+ If 
+\begin_inset Formula $V\geq Y$
+\end_inset
+
+,
+ reject 
+\begin_inset Formula $X$
+\end_inset
+
+ and repeat all the steps.
+ Otherwise return 
+\begin_inset Formula $X$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc17[M24]
+\end_layout
+
+\end_inset
+
+What is the 
+\emph on
+distribution function
+\emph default
+ 
+\begin_inset Formula $F(x)$
+\end_inset
+
+ for the geometric distribution with probability 
+\begin_inset Formula $p$
+\end_inset
+
+?
+ What is the 
+\emph on
+generating function
+\emph default
+ 
+\begin_inset Formula $G(z)$
+\end_inset
+
+?
+ What are the mean and standard deviation of this distribution?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+F(x)=P(X\leq x)=1-P(X>x)=1-P(X>\lfloor x\rfloor)=1-(1-p)^{\max\{\lfloor x\rfloor,0\}}.
+\]
+
+\end_inset
+
+The generating function is
+\begin_inset Formula 
+\[
+G(z)\coloneqq\sum_{k\geq1}(1-p)^{k-1}pz^{k}=zp\sum_{k\geq0}(1-p)^{k}z^{k}=\frac{zp}{1-(1-p)z}=\frac{zp}{1-qz},
+\]
+
+\end_inset
+
+where 
+\begin_inset Formula $q\coloneqq1-p$
+\end_inset
+
+.
+ Then
+\begin_inset Formula 
+\begin{align*}
+G'(z) & =\frac{p-pqz+pqz}{(1-qz)^{2}}=\frac{p}{(1-qz)^{2}}, & G''(z) & =-\frac{2pq}{(1-qz)^{3}},
+\end{align*}
+
+\end_inset
+
+so the mean is
+\begin_inset Formula 
+\[
+G'(1)=\frac{p}{(1-q)^{2}}=\frac{p}{p^{2}}=\frac{1}{p},
+\]
+
+\end_inset
+
+and the variance is
+\begin_inset Formula 
+\[
+G''(1)+G'(1)-G'(1)^{2}=-\frac{2pq}{(1-q)^{3}}+\frac{1}{p}-\frac{1}{p^{2}}=\frac{2q}{p^{2}}+\frac{1}{p}-\frac{1}{p^{2}}=\frac{2q+p-1}{p^{2}}=\frac{q}{p^{2}},
+\]
+
+\end_inset
+
+so the standard deviation is 
+\begin_inset Formula $\frac{\sqrt{q}}{p}$
+\end_inset
+
+.
+\end_layout
+
+\end_body
+\end_document
diff --git a/vol2/3.4.2.lyx b/vol2/3.4.2.lyx
new file mode 100644
index 0000000..8d59cc5
--- /dev/null
+++ b/vol2/3.4.2.lyx
@@ -0,0 +1,537 @@
+#LyX 2.4 created this file. For more info see https://www.lyx.org/
+\lyxformat 620
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\begin_preamble
+\input defs
+\end_preamble
+\use_default_options true
+\maintain_unincluded_children no
+\language english
+\language_package default
+\inputencoding utf8
+\fontencoding auto
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_roman_osf false
+\font_sans_osf false
+\font_typewriter_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\float_placement class
+\float_alignment class
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_formatted_ref 0
+\use_minted 0
+\use_lineno 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style english
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tablestyle default
+\tracking_changes false
+\output_changes false
+\change_bars false
+\postpone_fragile_content false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\docbook_table_output 0
+\docbook_mathml_prefix 1
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc3[22]
+\end_layout
+
+\end_inset
+
+The 
+\begin_inset Formula $(t+1)$
+\end_inset
+
+st item in Algorithm S is selected with probability 
+\begin_inset Formula $(n-m)/(N-t)$
+\end_inset
+
+,
+ not 
+\begin_inset Formula $n/N$
+\end_inset
+
+,
+ yet the text claims that the sample is unbiased;
+ thus each item should be selected with the 
+\emph on
+same
+\emph default
+ probability.
+ How can both of these statements be true?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+This is because 
+\begin_inset Formula $\frac{n-m}{N-t}$
+\end_inset
+
+ appears as a 
+\emph on
+conditioned
+\emph default
+ probability,
+ where the condition depends on 
+\begin_inset Formula $t$
+\end_inset
+
+.
+ The actual probability of choosing the 
+\begin_inset Formula $(t+1)$
+\end_inset
+
+st element is
+\begin_inset Formula 
+\[
+\sum_{m=0}^{t}P(\{m\text{ elements have been chosen between }1\text{ and }t\})\frac{n-m}{N-t}.
+\]
+
+\end_inset
+
+Let 
+\begin_inset Formula $P_{mt}$
+\end_inset
+
+ be the probability described with words inside this formula.
+ For 
+\begin_inset Formula $m>t$
+\end_inset
+
+ or 
+\begin_inset Formula $m<0$
+\end_inset
+
+,
+ this probability is obviously 0,
+ and for 
+\begin_inset Formula $t=0$
+\end_inset
+
+,
+ 
+\begin_inset Formula $P_{00}=1$
+\end_inset
+
+.
+ For 
+\begin_inset Formula $t\geq1$
+\end_inset
+
+ and 
+\begin_inset Formula $0\leq m\leq t$
+\end_inset
+
+,
+ 
+\begin_inset Formula 
+\[
+P_{mt}=P_{m(t-1)}\left(1-\frac{n-m}{N-t+1}\right)+P_{(m-1)(t-1)}\frac{n-m+1}{N-t+1}.
+\]
+
+\end_inset
+
+Note that,
+ by this formula,
+ 
+\begin_inset Formula $P_{(n-1)t}=0$
+\end_inset
+
+ always,
+ since the second term is 0 and the first term is a multiple of 
+\begin_inset Formula $P_{(n-1)(t-1)}$
+\end_inset
+
+,
+ which is 0 by induction since 
+\begin_inset Formula $P_{(n-1)(-1)}=0$
+\end_inset
+
+.
+ Then,
+ by induction in 
+\begin_inset Formula $m$
+\end_inset
+
+,
+ 
+\begin_inset Formula $P_{mt}=0$
+\end_inset
+
+ for any 
+\begin_inset Formula $m>n$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Since the algorithm is said to be unbiased,
+ we would expect
+\begin_inset Formula 
+\[
+P_{mt}=\frac{\binom{t}{m}\binom{N-t}{n-m}}{\binom{N}{n}},
+\]
+
+\end_inset
+
+that is,
+ the number of ways to choose 
+\begin_inset Formula $m$
+\end_inset
+
+ elements among the first 
+\begin_inset Formula $t$
+\end_inset
+
+ ones and 
+\begin_inset Formula $n-m$
+\end_inset
+
+ elements among the rest,
+ divided by the total number of ways of choosing 
+\begin_inset Formula $n$
+\end_inset
+
+ elements among 
+\begin_inset Formula $N$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+We prove this by induction.
+ For 
+\begin_inset Formula $t=0$
+\end_inset
+
+ we have seen it already.
+ For 
+\begin_inset Formula $t>0$
+\end_inset
+
+,
+ if 
+\begin_inset Formula $m=0$
+\end_inset
+
+,
+\begin_inset Formula 
+\[
+P_{0t}=P_{0(t-1)}\left(1-\frac{n}{N-t+1}\right)=\frac{\binom{N-t+1}{n}}{\binom{N}{n}}\frac{N-t-n+1}{N-t+1}=\frac{\binom{N-t}{n}}{\binom{N}{n}},
+\]
+
+\end_inset
+
+and if 
+\begin_inset Formula $m>0$
+\end_inset
+
+,
+\begin_inset Formula 
+\begin{align*}
+P_{mt} & =\frac{\binom{t-1}{m}\binom{N-t+1}{n-m}}{\binom{N}{n}}\frac{N-t-n+m+1}{N-t+1}+\frac{\binom{t-1}{m-1}\binom{N-t+1}{n-m+1}}{\binom{N}{n}}\frac{n-m+1}{N-t+1}=\\
+ & =\frac{\binom{t-1}{m}\binom{N-t}{n-m}}{\binom{N}{n}}+\frac{\binom{t-1}{m-1}\binom{N-t}{n-m}}{\binom{N}{n}}=\frac{\binom{t}{m}\binom{N-t}{n-m}}{\binom{N}{n}},
+\end{align*}
+
+\end_inset
+
+where we make extensive use of Eq.
+ 1.2.6–(8).
+ Finally,
+ by Eq.
+ 1.2.6–(21),
+ the probability of choosing any given element is
+\begin_inset Formula 
+\[
+\sum_{m=0}^{t}P_{mt}\frac{n-m}{N-t}=\frac{1}{\binom{N}{n}}\sum_{m=0}^{t}\binom{t}{m}\binom{N-1-t}{n-1-m}=\frac{\binom{N-1}{n-1}}{\binom{N}{n}}=\frac{\binom{N-1}{N-n}}{\binom{N}{N-n}}=\frac{n}{N}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc11[M25]
+\end_layout
+
+\end_inset
+
+Let 
+\begin_inset Formula $p_{m}$
+\end_inset
+
+ be the probability that exactly 
+\begin_inset Formula $m$
+\end_inset
+
+ elements are put into the reservoir during the first pass of Algorithm R.
+ Determine the generating function 
+\begin_inset Formula $G(z)=\sum_{m}p_{m}z^{m}$
+\end_inset
+
+,
+ and find the mean and standard deviation.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Let 
+\begin_inset Formula $s_{k}\coloneqq P(\text{the }k\text{th element is copied to the reservoir})$
+\end_inset
+
+,
+ then 
+\begin_inset Formula $p_{m}$
+\end_inset
+
+ is the sum across all subsets 
+\begin_inset Formula $\{x_{1},\dots,x_{m}\}\in\{1,\dots,N\}$
+\end_inset
+
+ of 
+\begin_inset Formula $m$
+\end_inset
+
+ elements,
+ 
+\begin_inset Formula $x_{1}<\dots<x_{m}$
+\end_inset
+
+,
+ of the product of the 
+\begin_inset Formula $s_{x_{i}}$
+\end_inset
+
+ times the 
+\begin_inset Formula $(1-s_{j})$
+\end_inset
+
+ for 
+\begin_inset Formula $j\in\{1,\dots N\}\setminus\{x_{1},\dots,x_{m}\}$
+\end_inset
+
+.
+ Rearranging terms we can see that
+\begin_inset Formula 
+\[
+G(z)=\prod_{k=1}^{N}(s_{k}z+1-s_{k})=z^{n}\prod_{k=n+1}^{N}\left(\frac{nz}{k}+1-\frac{n}{k}\right)=z^{n}\prod_{k=n+1}^{N}\frac{n(z-1)+k}{k}.
+\]
+
+\end_inset
+
+Let 
+\begin_inset Formula $F(z)\coloneqq\prod_{k=n+1}^{N}\frac{n(z-1)+k}{k}$
+\end_inset
+
+,
+ we have 
+\begin_inset Formula $F(1)=1$
+\end_inset
+
+ and
+\begin_inset Formula 
+\begin{align*}
+\dot{F}(z) & =F(z)\sum_{k=n+1}^{N}\frac{n}{n(z-1)+k}, & \dot{F}(1) & =\sum_{k=n+1}^{N}\frac{n}{k}=n(H_{N}-H_{n});\\
+\ddot{F}(z) & =\frac{\dot{F}(z)^{2}}{F(z)}-F(z)\sum_{k=n+1}^{N}\frac{n^{2}}{(n(z-1)+k)^{2}}, & \ddot{F}(1) & =n^{2}\left((H_{N}-H_{n})^{2}-(H_{N}^{(2)}-H_{n}^{(2)})\right).
+\end{align*}
+
+\end_inset
+
+Using this,
+\begin_inset Formula 
+\begin{align*}
+\dot{G}(z) & =nz^{n-1}F(z)+z^{n}\dot{F}(z),\\
+\ddot{G}(z) & =n(n-1)z^{n-2}F(z)+2nz^{n-1}\dot{F}(z)+z^{n}\ddot{F}(z),
+\end{align*}
+
+\end_inset
+
+so
+\begin_inset Formula 
+\begin{align*}
+\text{mean}(G) & =\dot{G}(1)=n(1+H_{N}-H_{n});\\
+\text{var}(G) & =\ddot{G}(1)+\dot{G}(1)-\dot{G}(1)^{2}\\
+ & =\cancel{n(n-1)}\cancel{+2n\dot{F}(1)}+\ddot{F}(1)\cancel{+n}+\dot{F}(1)\cancel{-n^{2}}-\dot{F}(1)^{2}\cancel{-2n\dot{F}(1)}\\
+ & =n(H_{N}-H_{n})-n^{2}(H_{N}^{(2)}-H_{n}^{(2)}),\\
+\sigma(G) & =\sqrt{\text{var}(G)}=\sqrt{n(H_{N}-H_{n})-n^{2}(H_{N}^{(2)}-H_{n}^{(2)})}.
+\end{align*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc16[M25]
+\end_layout
+
+\end_inset
+
+Devise a way to compute a random sample of 
+\begin_inset Formula $n$
+\end_inset
+
+ records from 
+\begin_inset Formula $N$
+\end_inset
+
+,
+ given 
+\begin_inset Formula $N$
+\end_inset
+
+ and 
+\begin_inset Formula $n$
+\end_inset
+
+,
+ based on the idea of hashing (Section 6.4).
+ Your method should use 
+\begin_inset Formula $O(n)$
+\end_inset
+
+ storage locations and an average of 
+\begin_inset Formula $O(n)$
+\end_inset
+
+ units of time,
+ and it should present the sample as a sorted set of integers 
+\begin_inset Formula $1\leq X_{1}<X_{2}<\dots<X_{n}\leq N$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+TODO Depends on Section 6.4.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_body
+\end_document
diff --git a/vol2/3.5.lyx b/vol2/3.5.lyx
new file mode 100644
index 0000000..0268000
--- /dev/null
+++ b/vol2/3.5.lyx
@@ -0,0 +1,1191 @@
+#LyX 2.4 created this file. For more info see https://www.lyx.org/
+\lyxformat 620
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\begin_preamble
+\input defs
+\end_preamble
+\use_default_options true
+\maintain_unincluded_children no
+\language english
+\language_package default
+\inputencoding utf8
+\fontencoding auto
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_roman_osf false
+\font_sans_osf false
+\font_typewriter_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\float_placement class
+\float_alignment class
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_formatted_ref 0
+\use_minted 0
+\use_lineno 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style english
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tablestyle default
+\tracking_changes false
+\output_changes false
+\change_bars false
+\postpone_fragile_content false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\docbook_table_output 0
+\docbook_mathml_prefix 1
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc1[10]
+\end_layout
+
+\end_inset
+
+Can a periodic sequence be equidistributed?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Not if it's a real sequence,
+ because the period must be finite and,
+ if 
+\begin_inset Formula $0\leq x_{1}<\dots<x_{t}\leq1$
+\end_inset
+
+ are the numbers that appear in the period,
+ then 
+\begin_inset Formula $\text{Pr}(\frac{1}{3}(2x_{1}+x_{2})\leq x<\frac{1}{3}(x_{1}+2x_{2}))=0\neq\frac{1}{3}(x_{2}-x_{1})$
+\end_inset
+
+ (a similar proof can be made for 
+\begin_inset Formula $t=1$
+\end_inset
+
+).
+ If it's an integer sequence it can happen;
+ for example for the 
+\begin_inset Formula $b$
+\end_inset
+
+-ary sequence with period 
+\begin_inset Formula $0,1,2,\dots,b-1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc2[10]
+\end_layout
+
+\end_inset
+
+Consider the periodic binary sequence 0,
+ 0,
+ 1,
+ 1,
+ 0,
+ 0,
+ 1,
+ 1,
+ 
+\begin_inset Formula $\dots$
+\end_inset
+
+.
+ Is it 1-distributed?
+ Is it 2-distributed?
+ Is it 3-distributed?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+It is clearly 1-distributed and 2-distributed,
+ but not 3-distributed because 
+\begin_inset Quotes eld
+\end_inset
+
+111
+\begin_inset Quotes erd
+\end_inset
+
+ never appears.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc5[HM22]
+\end_layout
+
+\end_inset
+
+Let 
+\begin_inset Formula $U_{n}=(2^{\lfloor\lg(n+1)\rfloor}/3)\bmod1$
+\end_inset
+
+.
+ What is 
+\begin_inset Formula $\text{Pr}(U_{n}<\frac{1}{2})$
+\end_inset
+
+?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+We have 
+\begin_inset Formula $\lfloor\lg(n+1)\rfloor=0$
+\end_inset
+
+ for 
+\begin_inset Formula $n=0$
+\end_inset
+
+;
+ 1 for 
+\begin_inset Formula $n=1,2$
+\end_inset
+
+;
+ 2 for 
+\begin_inset Formula $n=3,4,5,6$
+\end_inset
+
+,
+ 3 for 
+\begin_inset Formula $n=7,\dots,14$
+\end_inset
+
+,
+ etc.,
+ so the sequence 
+\begin_inset Formula $(2^{\lfloor\lg(n+1)\rfloor})_{n}$
+\end_inset
+
+ has 1 1's,
+ followed by 2 2's,
+ 4 4's,
+ 8 8's,
+ etc.
+ It is easy to prove by induction that,
+ when 
+\begin_inset Formula $k\in\mathbb{N}$
+\end_inset
+
+ is even,
+ 
+\begin_inset Formula $2^{k}\equiv1\bmod3$
+\end_inset
+
+,
+ and when it's odd,
+ 
+\begin_inset Formula $2^{k}\equiv2\bmod3$
+\end_inset
+
+,
+ and so 
+\begin_inset Formula $U_{n}<\frac{1}{2}$
+\end_inset
+
+ precisely when 
+\begin_inset Formula $\lfloor\lg(n+1)\rfloor$
+\end_inset
+
+ is even,
+ which is when 
+\begin_inset Formula $2^{k}\equiv1\bmod3$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+If 
+\begin_inset Formula $\nu(n)=|\{m\leq n\mid U_{n}<\frac{1}{2}\}|$
+\end_inset
+
+,
+ then 
+\begin_inset Formula $\nu(n)/n$
+\end_inset
+
+ clearly increases when 
+\begin_inset Formula $n$
+\end_inset
+
+ is between 
+\begin_inset Formula $2^{2k}-1$
+\end_inset
+
+ and 
+\begin_inset Formula $2^{2k+1}-1$
+\end_inset
+
+,
+ and it decreases between 
+\begin_inset Formula $2^{2k-1}-1$
+\end_inset
+
+ and 
+\begin_inset Formula $2^{2k}-1$
+\end_inset
+
+,
+ for 
+\begin_inset Formula $k\in\mathbb{N}^{*}$
+\end_inset
+
+.
+ The limit exists if the subsequence made from these infinite local minima and the one made from these infinite local maxima both have a limit and these limits match.
+\end_layout
+
+\begin_layout Standard
+For the maxima,
+ 
+\begin_inset Formula $\nu(1)=1$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\nu(7)=5$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\nu(31)=21$
+\end_inset
+
+,
+ etc.
+ In general,
+ 
+\begin_inset Formula 
+\[
+\nu(2^{2k+1}-1)=\sum_{i=0}^{k}2^{2k}=\frac{1-4^{k+1}}{1-4}=\frac{4^{k+1}-1}{3},
+\]
+
+\end_inset
+
+so
+\begin_inset Formula 
+\[
+\lim_{k}\frac{\nu(2^{2k+1}-1)}{2^{2k+1}-1}=\frac{\frac{4^{k+1}-1}{3}}{2\cdot4^{k}-1}=\frac{1}{3}\frac{4\cdot4^{k}-1}{2\cdot4^{k}-1}=\frac{2}{3}.
+\]
+
+\end_inset
+
+For the minima,
+ 
+\begin_inset Formula $\nu(3)=1$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\nu(15)=5$
+\end_inset
+
+,
+ etc.,
+ and in general 
+\begin_inset Formula $\nu(2^{2k}-1)=\nu(2^{2k-1}-1)=\frac{4^{k}-1}{3}$
+\end_inset
+
+,
+ so
+\begin_inset Formula 
+\[
+\lim_{k}\frac{\nu(2^{2k}-1)}{2^{2k}-1}=\frac{1}{3}\frac{4^{k}-1}{4^{k}-1}=\frac{1}{3}.
+\]
+
+\end_inset
+
+Since 
+\begin_inset Formula $\frac{1}{3}\neq\frac{2}{3}$
+\end_inset
+
+,
+ this probability is undefined.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc10[HM22]
+\end_layout
+
+\end_inset
+
+Where was the fact that 
+\begin_inset Formula $m$
+\end_inset
+
+ divides 
+\begin_inset Formula $q$
+\end_inset
+
+ used in the proof of Theorem C?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+This is used for the sums in the second page of proof,
+ when telling the range of 
+\begin_inset Formula $t$
+\end_inset
+
+.
+ In particular,
+ it is needed when evaluating the sum over 
+\begin_inset Formula $t$
+\end_inset
+
+ in Equation (22).
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc11[M10]
+\end_layout
+
+\end_inset
+
+Use Theorem C to prove that if a sequence 
+\begin_inset Formula $\langle U_{n}\rangle$
+\end_inset
+
+ is 
+\begin_inset Formula $\infty$
+\end_inset
+
+-distributed,
+ so is the subsequence 
+\begin_inset Formula $\langle U_{2n}\rangle$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Since it's 
+\begin_inset Formula $\infty$
+\end_inset
+
+-distributed,
+ it's also 
+\begin_inset Formula $(2,2k)$
+\end_inset
+
+-distributed for all 
+\begin_inset Formula $k\in\mathbb{N}^{*}$
+\end_inset
+
+,
+ so 
+\begin_inset Formula $\text{Pr}(u_{1}\leq U_{2n}<v_{1},\dots,u_{2k}\leq U_{2n+2k-1}<v_{2k})=(v_{1}-u_{1})\cdots(v_{k}-u_{k})$
+\end_inset
+
+ for any 
+\begin_inset Formula $u_{1},v_{1},\dots,u_{k},v_{k}\in[0,1)$
+\end_inset
+
+ with each 
+\begin_inset Formula $u_{i}<v_{i}$
+\end_inset
+
+,
+ and in particular,
+ if we let 
+\begin_inset Formula $u_{2},u_{4},\dots,u_{2k}=0$
+\end_inset
+
+ and 
+\begin_inset Formula $v_{2},v_{4},\dots,v_{2k}=1$
+\end_inset
+
+ we get the formula that shows that 
+\begin_inset Formula $\langle U_{2n}\rangle$
+\end_inset
+
+ is 
+\begin_inset Formula $k$
+\end_inset
+
+-distributed.
+ And since this 
+\begin_inset Formula $k$
+\end_inset
+
+ is arbitrary,
+ 
+\begin_inset Formula $\langle U_{2n}\rangle$
+\end_inset
+
+ is 
+\begin_inset Formula $\infty$
+\end_inset
+
+-distributed.
+ Note that this argument applies to any 
+\begin_inset Formula $\langle U_{mn+j}\rangle$
+\end_inset
+
+ with 
+\begin_inset Formula $m\in\mathbb{N}^{*}$
+\end_inset
+
+ and 
+\begin_inset Formula $j\in\mathbb{N}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc18[HM22]
+\end_layout
+
+\end_inset
+
+Prove that if 
+\begin_inset Formula $U_{0},U_{1},\dots$
+\end_inset
+
+ is 
+\begin_inset Formula $k$
+\end_inset
+
+-distributed,
+ so is the sequence 
+\begin_inset Formula $V_{0},V_{1},\dots$
+\end_inset
+
+ where 
+\begin_inset Formula $V_{n}=\lfloor nU_{n}\rfloor/n$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Take any 
+\begin_inset Formula $u_{1},v_{1},\dots,u_{k},v_{k}\in[0,1)$
+\end_inset
+
+ such that each 
+\begin_inset Formula $u_{i}<v_{i}$
+\end_inset
+
+.
+ If 
+\begin_inset Formula $u_{i}\leq V_{n}<v_{i}$
+\end_inset
+
+,
+ then 
+\begin_inset Formula $u_{i}-\frac{1}{n}\leq U_{n}<v_{i}+\frac{1}{n}$
+\end_inset
+
+,
+ so if 
+\begin_inset Formula $S(n)\coloneqq\{\forall i,u_{i}\leq V_{n+i}<v_{i}\}$
+\end_inset
+
+,
+ then 
+\begin_inset Formula 
+\begin{align*}
+\overline{\text{Pr}}(S(n)) & \leq\text{Pr}\left(\forall i,u_{i}-\frac{1}{n+i}\leq U_{n+i}<v_{i}+\frac{1}{n+i}\right)\\
+ & \leq\text{Pr}\left(\forall i,u_{i}-\frac{1}{n}\leq U_{n+i}<v_{i}+\frac{1}{n}\right)\\
+ & \leq\text{Pr}\left(\forall i,u_{i}-\frac{1}{n_{0}}\leq U_{n+i}<v_{i}+\frac{1}{n_{0}}\right)
+\end{align*}
+
+\end_inset
+
+for any 
+\begin_inset Formula $n_{0}\in\mathbb{N}$
+\end_inset
+
+,
+ since the first finitely many terms of the sequence 
+\begin_inset Quotes eld
+\end_inset
+
+don't matter,
+\begin_inset Quotes erd
+\end_inset
+
+ and since 
+\begin_inset Formula $n_{0}$
+\end_inset
+
+ is arbitrary,
+ taking limits on it we see that 
+\begin_inset Formula $\overline{\text{Pr}}(S(n))\leq\text{Pr}(\forall i,u_{i}\leq U_{n+i}<v_{i})=\prod_{i}(v_{i}-u_{i})$
+\end_inset
+
+.
+ Similarly,
+ if 
+\begin_inset Formula $u_{i}+\frac{1}{n}\leq U_{n}<v_{i}-\frac{1}{n}$
+\end_inset
+
+,
+ then 
+\begin_inset Formula $u_{i}\leq V_{n}<v_{i}$
+\end_inset
+
+,
+ so
+\begin_inset Formula 
+\[
+\underline{\text{Pr}}(S(n))\geq\text{Pr}\left(\forall i,u_{i}+\frac{1}{n}\leq U_{n+i}<v_{i}+\frac{1}{n}\right)\geq\text{Pr}\left(\forall i,u_{i}+\frac{1}{n_{0}}\leq U_{n+i}<v_{i}-\frac{1}{n_{0}}\right),
+\]
+
+\end_inset
+
+this time taking 
+\begin_inset Formula $n_{0}$
+\end_inset
+
+ such that 
+\begin_inset Formula $\frac{1}{n_{0}}\leq v_{i}-u_{i}$
+\end_inset
+
+ for every 
+\begin_inset Formula $i$
+\end_inset
+
+.
+ Again we reach the conclusion that 
+\begin_inset Formula $\underline{\text{Pr}}(S(n))\geq\prod_{i}(v_{i}-u_{i})$
+\end_inset
+
+.
+ We get the result by the same argument used at the end of Theorem A.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc28[HM21]
+\end_layout
+
+\end_inset
+
+Use the sequence (11) to construct a 
+\begin_inset Formula $[0..1)$
+\end_inset
+
+ sequence that is 3-distributed,
+ for which 
+\begin_inset Formula $\text{Pr}(U_{2n}\geq\frac{1}{2})=\frac{3}{4}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Let 
+\begin_inset Formula $(W_{n})_{n}$
+\end_inset
+
+ be an 
+\begin_inset Formula $\infty$
+\end_inset
+
+-distributed real-valued sequence,
+ and let 
+\begin_inset Formula $(X_{n})_{n}$
+\end_inset
+
+ be the 3-distributed binary sequence from (11),
+ then 
+\begin_inset Formula $(U_{n}\coloneqq\frac{1}{2}(W_{n}+1-X_{n}))_{n}$
+\end_inset
+
+ satisfies the properties.
+ For if 
+\begin_inset Formula $0\leq u_{i}<v_{i}<1$
+\end_inset
+
+,
+ 
+\begin_inset Formula $i\in\{1,2,3\}$
+\end_inset
+
+,
+ and if we assume that,
+ for each 
+\begin_inset Formula $i$
+\end_inset
+
+,
+ 
+\begin_inset Formula $v_{i}\geq\frac{1}{2}$
+\end_inset
+
+ implies 
+\begin_inset Formula $u_{i}\geq\frac{1}{2}$
+\end_inset
+
+ (so the 
+\begin_inset Quotes eld
+\end_inset
+
+rectangle
+\begin_inset Quotes erd
+\end_inset
+
+ is contained in one quadrant),
+ then 
+\begin_inset Formula $u_{i}\leq U_{n}<v_{i}$
+\end_inset
+
+ if,
+ and only if,
+ 
+\begin_inset Formula $\lfloor2u_{i}\rfloor=\lfloor2U_{i}\rfloor=1-X_{n}$
+\end_inset
+
+ and 
+\begin_inset Formula $2u_{i}\bmod1\leq2U_{n}\bmod1=W_{n}\leq2v_{i}$
+\end_inset
+
+.
+ Since 
+\begin_inset Formula $W_{n}$
+\end_inset
+
+ is 
+\begin_inset Formula $(16,3)$
+\end_inset
+
+-distributed,
+ the triplets 
+\begin_inset Formula $(W_{n},W_{n+1},W_{n+2})$
+\end_inset
+
+ starting at positions where 
+\begin_inset Formula $(X_{n},X_{n+1},X_{n+2})$
+\end_inset
+
+ has a given value have the same density as those starting at positions where it has any other value,
+ so 
+\begin_inset Formula 
+\begin{multline*}
+\text{Pr}(\forall i,u_{i}\leq U_{n}<v_{i})=\text{Pr}(\forall i,\lfloor2u_{i}\rfloor=1-X_{n})\text{Pr}(\forall i,2u_{i}\bmod1\leq W_{n}\leq2v_{i}\bmod1)=\\
+=\frac{1}{8}\prod_{i}(2v_{i}-2u_{i})=\prod_{i}(v_{i}-u_{i})
+\end{multline*}
+
+\end_inset
+
+and the sequence is 3-distributed (the cases where some 
+\begin_inset Formula $\lfloor2u_{i}\rfloor\neq\lfloor2v_{i}\rfloor$
+\end_inset
+
+ can be split into cases where this is not the case).
+ In addition,
+ 
+\begin_inset Formula $\text{Pr}(U_{2n}\geq\frac{1}{2})=\text{Pr}(X_{2n}=0)=\frac{3}{4}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc34[M25]
+\end_layout
+
+\end_inset
+
+Define subsequence rules 
+\begin_inset Formula ${\cal R}_{1}$
+\end_inset
+
+,
+ 
+\begin_inset Formula ${\cal R}_{2}$
+\end_inset
+
+,
+ 
+\begin_inset Formula ${\cal R}_{3}$
+\end_inset
+
+,
+ ...
+ such that Algorithm W can be used with these rules to give an effective algorithm to construct a 
+\begin_inset Formula $[0..1)$
+\end_inset
+
+ sequence satisfying Definition R1.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+
+\begin_inset Note Greyedout
+status open
+
+\begin_layout Plain Layout
+(I had to look up the solution.)
+\end_layout
+
+\end_inset
+
+The 
+\begin_inset Quotes eld
+\end_inset
+
+algorithm
+\begin_inset Quotes erd
+\end_inset
+
+ gives us a potentially infinite amount of sequences 
+\begin_inset Formula $\langle U_{n}\rangle{\cal R}_{k}$
+\end_inset
+
+ that are 1-distributed,
+ so we may encode the properties that we want to check for in the value 
+\begin_inset Formula $k$
+\end_inset
+
+.
+ Specifically,
+ we want to check that,
+ for an increasing sequence of bases 
+\begin_inset Formula $(b_{n})_{n}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $k\in\mathbb{N}^{*}$
+\end_inset
+
+,
+ and 
+\begin_inset Formula $a_{1},\dots,a_{k}\in\{0,\dots,b-1\}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $U_{n-k}=a_{k},\dots,U_{n-1}=a_{1}$
+\end_inset
+
+,
+ so if,
+ for example,
+ 
+\begin_inset Formula $k=10^{b}10^{a_{1}}10^{a_{2}}1\cdots10a^{j}$
+\end_inset
+
+ with each 
+\begin_inset Formula $a_{i}<b$
+\end_inset
+
+,
+ we may set 
+\begin_inset Formula ${\cal R}_{k}(x_{0},\dots,x_{n-1})=1$
+\end_inset
+
+ if,
+ and only if,
+ 
+\begin_inset Formula $\lfloor bU_{n-1}\rfloor=a_{1}\land\dots\land\lfloor bU_{n-k}\rfloor=a_{k}$
+\end_inset
+
+.
+ For every other value of 
+\begin_inset Formula $k$
+\end_inset
+
+,
+ we may as well set 
+\begin_inset Formula ${\cal R}_{k}\equiv1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+TODO 44 (
+\begin_inset Formula $<$
+\end_inset
+
+4pp.,
+ 
+\begin_inset Formula $<$
+\end_inset
+
+1:53)
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc44[16]
+\end_layout
+
+\end_inset
+
+(I.
+ J.
+ Good.) Can a valid table of random digits contain just one misprint?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Yes;
+ for example,
+ both 23456782019372837458 and 23456782019372837459 are random.
+ Of course,
+ this is not true for any misprint,
+ as then all numbers would be random.
+ For example,
+ 23456782019372828221 is random but 23456782019372828222 isn't,
+ as it contains too many 2's.
+ This has been calculated with the following (terrible) code:
+\end_layout
+
+\begin_layout Standard
+\begin_inset listings
+lstparams "language=Python,numbers=left,basicstyle={\footnotesize\sffamily},breaklines=true"
+inline false
+status open
+
+\begin_layout Plain Layout
+
+import math
+\end_layout
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\begin_layout Plain Layout
+
+def israndom(digs):
+\end_layout
+
+\begin_layout Plain Layout
+
+    N = len(digs)
+\end_layout
+
+\begin_layout Plain Layout
+
+    dev = math.sqrt(N)
+\end_layout
+
+\begin_layout Plain Layout
+
+    for k in range(0,
+ math.floor(math.log10(N)) + 1):
+\end_layout
+
+\begin_layout Plain Layout
+
+        pos = 10**k
+\end_layout
+
+\begin_layout Plain Layout
+
+        expect = N/pos
+\end_layout
+
+\begin_layout Plain Layout
+
+        for ss in range(pos,
+ 2*pos):
+\end_layout
+
+\begin_layout Plain Layout
+
+          sb = str(ss)[1:]
+\end_layout
+
+\begin_layout Plain Layout
+
+          amt = len([n for n in range(len(digs)) if digs[n:n+k] == sb])
+\end_layout
+
+\begin_layout Plain Layout
+
+          if abs(amt-expect) > dev:
+\end_layout
+
+\begin_layout Plain Layout
+
+              print(
+\begin_inset Quotes eld
+\end_inset
+
+FAIL
+\begin_inset Quotes erd
+\end_inset
+
+,
+ digs,
+ sb)
+\end_layout
+
+\begin_layout Plain Layout
+
+              return False
+\end_layout
+
+\begin_layout Plain Layout
+
+    return True
+\end_layout
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\begin_layout Plain Layout
+
+for x in range(123456782019372800000,
+ 123456790000000000000):
+\end_layout
+
+\begin_layout Plain Layout
+
+    if israndom(str(x)[1:]):
+\end_layout
+
+\begin_layout Plain Layout
+
+        print(str(x)[1:])
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_body
+\end_document
diff --git a/vol2/3.6.lyx b/vol2/3.6.lyx
new file mode 100644
index 0000000..8eaaa14
--- /dev/null
+++ b/vol2/3.6.lyx
@@ -0,0 +1,496 @@
+#LyX 2.4 created this file. For more info see https://www.lyx.org/
+\lyxformat 620
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\begin_preamble
+\input defs
+\end_preamble
+\use_default_options true
+\maintain_unincluded_children no
+\language english
+\language_package default
+\inputencoding utf8
+\fontencoding auto
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_roman_osf false
+\font_sans_osf false
+\font_typewriter_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\float_placement class
+\float_alignment class
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_formatted_ref 0
+\use_minted 0
+\use_lineno 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style english
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tablestyle default
+\tracking_changes false
+\output_changes false
+\change_bars false
+\postpone_fragile_content false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\docbook_table_output 0
+\docbook_mathml_prefix 1
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc2[15]
+\end_layout
+
+\end_inset
+
+some people have been afraid that computers will someday take over the world;
+ but they are reassured by the statement that a machine cannot do anything really new,
+ since it is only obeying the commands of its master,
+ the programmer.
+ Lady Lovelace wrote in 1844,
+ 
+\begin_inset Quotes eld
+\end_inset
+
+The Analytical Engine has no pretensions to 
+\emph on
+originate
+\emph default
+ anything.
+ It can do 
+\emph on
+whatever we know how to order it
+\emph default
+ to perform.
+\begin_inset Quotes erd
+\end_inset
+
+ Her statement has been elaborated further by many philosophers.
+ Discuss this topic,
+ with random number generators in mind.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+While it is true that a machine cannot do anything that a programmer doesn't tell it to do,
+ there are a number of caveats here.
+ First is that the programmer,
+ or their boss,
+ may not have other people's best interests in mind.
+ Another one is that what the programmer tells the machine to do is not the same thing as what they 
+\emph on
+expects
+\emph default
+ them to do;
+ the software may have bugs,
+ although the bugs are unlikely to make the computer take over the world.
+ Finally,
+ if a random number generator is being used,
+ the programmer is telling the machine to do one of a number of actions,
+ chosen with some given distribution,
+ without telling it which one to choose.
+ These actions are typically limited to a given scope,
+ and the programmer still controls the family of actions that the machine may take.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc11[M25]
+\end_layout
+
+\end_inset
+
+Assuming that floating point arithmetic on numbers of type 
+\family typewriter
+double
+\family default
+ is properly rounded in the sense of Section 4.2.2 (hence exact when the values are suitably restricted),
+ convert the C routines 
+\emph on
+ran_array
+\emph default
+ and 
+\emph on
+ran_start
+\emph default
+ to similar programs that deliver double-precision random fractions in the range 
+\begin_inset Formula $[0..1)$
+\end_inset
+
+,
+ instead of 30-bit integers.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Technically this does the work:
+\begin_inset listings
+lstparams "language=C,numbers=left,basicstyle={\small\sffamily},breaklines=true"
+inline false
+status open
+
+\begin_layout Plain Layout
+
+#include <math.h>
+\end_layout
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\begin_layout Plain Layout
+
+static double ran_u[KK];
+\end_layout
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\begin_layout Plain Layout
+
+void ranf_array(double out[],
+ size_t n)
+\end_layout
+
+\begin_layout Plain Layout
+
+{
+\end_layout
+
+\begin_layout Plain Layout
+
+	size_t i,
+ j;
+\end_layout
+
+\begin_layout Plain Layout
+
+	assert(n >= KK);
+\end_layout
+
+\begin_layout Plain Layout
+
+	for (j = 0;
+ j < KK;
+ j++)
+\end_layout
+
+\begin_layout Plain Layout
+
+		out[j] = ran_u[j];
+\end_layout
+
+\begin_layout Plain Layout
+
+	for (;
+ j < n;
+ j++)
+\end_layout
+
+\begin_layout Plain Layout
+
+		out[j] = fmod(1.
+ + out[j-KK] - out[j-LL],
+ 1.);
+\end_layout
+
+\begin_layout Plain Layout
+
+	for (i = 0;
+ i < LL;
+ i++,
+ j++)
+\end_layout
+
+\begin_layout Plain Layout
+
+		ran_u[i] = fmod(1.
+ + out[j-KK] - out[j-LL],
+ 1.);
+\end_layout
+
+\begin_layout Plain Layout
+
+	for (;
+ i < KK;
+ i++,
+ j++)
+\end_layout
+
+\begin_layout Plain Layout
+
+		ran_u[i] = fmod(1.
+ + out[j-KK] - ran_x[j-LL],
+ 1.);
+\end_layout
+
+\begin_layout Plain Layout
+
+}
+\end_layout
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\begin_layout Plain Layout
+
+void ranf_start(long seed)
+\end_layout
+
+\begin_layout Plain Layout
+
+{
+\end_layout
+
+\begin_layout Plain Layout
+
+	ranf_start(seed);
+\end_layout
+
+\begin_layout Plain Layout
+
+	for (size_t i = 0;
+ i < KK;
+ i++)
+\end_layout
+
+\begin_layout Plain Layout
+
+		ran_u[i] = (double)ran_x[i] / MM;
+\end_layout
+
+\begin_layout Plain Layout
+
+}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc12[M21]
+\end_layout
+
+\end_inset
+
+What random number generator would be suitable for a minicomputer that does arithmetic only on integers in the range 
+\begin_inset Formula $[-32768..32767]$
+\end_inset
+
+?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+We could use one of the generators with very large moduli from exercise 3.2.1.1–14.
+ We could also use running generators like the one in the text 
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc15[25]
+\end_layout
+
+\end_inset
+
+Write C code that makes it convenient to generate the random integers obtained from 
+\emph on
+ran_array
+\emph default
+ by discarding all but the first 100 of every 1009 elements,
+ as recommended in the text.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset listings
+lstparams "language=C,numbers=left,basicstyle={\normalsize\sffamily},breaklines=true"
+inline false
+status open
+
+\begin_layout Plain Layout
+
+long next()
+\end_layout
+
+\begin_layout Plain Layout
+
+{
+\end_layout
+
+\begin_layout Plain Layout
+
+	static long buf[1009],
+ next = 100;
+\end_layout
+
+\begin_layout Plain Layout
+
+	if (next == 100) {
+\end_layout
+
+\begin_layout Plain Layout
+
+		ran_array(buf,
+ sizeof(buf) / sizeof(buf[0]));
+\end_layout
+
+\begin_layout Plain Layout
+
+		next = 0;
+\end_layout
+
+\begin_layout Plain Layout
+
+	}
+\end_layout
+
+\begin_layout Plain Layout
+
+	return buf[next++];
+\end_layout
+
+\begin_layout Plain Layout
+
+}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_body
+\end_document
diff --git a/vol2/4.1.lyx b/vol2/4.1.lyx
new file mode 100644
index 0000000..b68ec0b
--- /dev/null
+++ b/vol2/4.1.lyx
@@ -0,0 +1,1922 @@
+#LyX 2.4 created this file. For more info see https://www.lyx.org/
+\lyxformat 620
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\begin_preamble
+\input defs
+\end_preamble
+\use_default_options true
+\maintain_unincluded_children no
+\language english
+\language_package default
+\inputencoding utf8
+\fontencoding auto
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_roman_osf false
+\font_sans_osf false
+\font_typewriter_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\float_placement class
+\float_alignment class
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_formatted_ref 0
+\use_minted 0
+\use_lineno 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style english
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tablestyle default
+\tracking_changes false
+\output_changes false
+\change_bars false
+\postpone_fragile_content false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\docbook_table_output 0
+\docbook_mathml_prefix 1
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc2[24]
+\end_layout
+
+\end_inset
+
+Consider the following four number systems:
+\end_layout
+
+\begin_layout Enumerate
+binary (signed magnitude);
+\end_layout
+
+\begin_layout Enumerate
+negabinary (radix 
+\begin_inset Formula $-2$
+\end_inset
+
+);
+\end_layout
+
+\begin_layout Enumerate
+balanced ternary;
+ and
+\end_layout
+
+\begin_layout Enumerate
+radix 
+\begin_inset Formula $b=\frac{1}{10}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Use each of these four number systems to express each of the following three numbers:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $-49$
+\end_inset
+
+;
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $-3\frac{1}{7}$
+\end_inset
+
+ (show the repeating cycle);
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\pi$
+\end_inset
+
+ (to a few significant figures).
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $-49\to-110001$
+\end_inset
+
+.
+ Multiplying by 2,
+ 
+\begin_inset Formula $\frac{1}{7}\to\frac{2}{7}\to\frac{4}{7}\to1+\frac{1}{7}$
+\end_inset
+
+,
+ so 
+\begin_inset Formula $-3\frac{1}{7}\to-11.\overbrace{001}$
+\end_inset
+
+.
+ Finally,
+ 
+\begin_inset Formula $\pi\to11.00100100001111...$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+The trick is to express the number in binary and,
+ for the odd positions (the ones where 
+\begin_inset Formula $(-2)^{k}$
+\end_inset
+
+ is negative),
+ add 1 to the part of the number on the right,
+ because we are subtracting 
+\begin_inset Formula $2^{k}$
+\end_inset
+
+ instead of adding so we compensate for that.
+ For negative numbers,
+ we represent them in two's complement before this change,
+ effectively adding 
+\begin_inset Formula $2^{M}$
+\end_inset
+
+ for some large 
+\begin_inset Formula $M$
+\end_inset
+
+,
+ and after it we drop that upper bit to subtract 
+\begin_inset Formula $2^{M}$
+\end_inset
+
+.
+ Thus 
+\begin_inset Formula $-49\to...111001111\to11010011$
+\end_inset
+
+,
+ 
+\begin_inset Formula $-3\frac{1}{7}\to...1100.\overbrace{110110}\to1101.\overbrace{001011}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\pi\to111.01100100010000...$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+We proceed as in the text:
+ 
+\begin_inset Formula 
+\begin{align*}
+-49 & \to-1211\to...11102200\to\overline{1}11\overline{1}\overline{1};\\
+-3\frac{1}{7} & \to-10.\overbrace{010212}\to\dots1101.\overbrace{100122}\to\overline{1}0.\overbrace{0\overline{1}\overline{1}011};\\
+\pi & \to10.010211012\dots\to...1121.122022201\dots\to10.011\overline{1}111\overline{1}0\dots.
+\end{align*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+This is like the decimal system but backwards,
+ so 
+\begin_inset Formula $-49\to-9.4$
+\end_inset
+
+,
+ 
+\begin_inset Formula $-3\frac{1}{7}\to-\overbrace{758241}3$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\pi\to...51413$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc5[00]
+\end_layout
+
+\end_inset
+
+Explain why a negative integer in nines' complement notation has a representation in ten's complement notation that is always one greater,
+ if the representations are regarded as positive.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Ten's complement of a negative number 
+\begin_inset Formula $s$
+\end_inset
+
+ is 
+\begin_inset Formula $10^{M}+s$
+\end_inset
+
+,
+ for a suitably large integer 
+\begin_inset Formula $M$
+\end_inset
+
+,
+ while nines' complement is 
+\begin_inset Formula $(10^{M}-1)+s$
+\end_inset
+
+,
+ so the ten's complement is one greater.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc8[M10]
+\end_layout
+
+\end_inset
+
+Prove Eq.
+ (5).
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+The part at the left is 
+\begin_inset Formula $\sum_{j}a_{j}b^{j}$
+\end_inset
+
+,
+ while the part at the right is 
+\begin_inset Formula $\sum_{j}A_{j}b^{kj}$
+\end_inset
+
+,
+ but 
+\begin_inset Formula $A_{j}=\sum_{i=0}^{k-1}a_{kj+i}b^{i}$
+\end_inset
+
+,
+ so this is
+\begin_inset Formula 
+\[
+\sum_{j}A_{j}b^{kj}=\sum_{j}\sum_{i=0}^{k-1}(a_{kj+i}b^{i})b^{kj}=\sum_{j}\sum_{i=0}^{k-1}(a_{kj+i}b^{kj+i})=\sum_{j}a_{j}b^{j}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc9[15]
+\end_layout
+
+\end_inset
+
+Change the following 
+\emph on
+octal
+\emph default
+ numbers to 
+\emph on
+hexadecimal
+\emph default
+ notation,
+ using the hexadecimal digits 
+\family typewriter
+0
+\family default
+,
+ 
+\family typewriter
+1
+\family default
+,
+ ...,
+ 
+\family typewriter
+9
+\family default
+,
+ 
+\family typewriter
+A
+\family default
+,
+ 
+\family typewriter
+B
+\family default
+,
+ 
+\family typewriter
+C
+\family default
+,
+ 
+\family typewriter
+D
+\family default
+,
+ 
+\family typewriter
+E
+\family default
+,
+ 
+\family typewriter
+F
+\family default
+:
+ 12;
+ 5655;
+ 2550276;
+ 76545336;
+ 3726755.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+We use Eq.
+ (5) with base 2.
+\end_layout
+
+\begin_layout Standard
+\align center
+\begin_inset Tabular
+<lyxtabular version="3" rows="6" columns="3">
+<features tabularvalignment="middle">
+<column alignment="right" valignment="top">
+<column alignment="right" valignment="top">
+<column alignment="right" valignment="top">
+<row>
+<cell alignment="right" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Octal
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="right" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Binary
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="right" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Hexadecimal
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="right" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+12
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="right" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+1 010
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="right" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+
+\family typewriter
+A
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="right" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+5655
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="right" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+101 110 101 101
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="right" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+
+\family typewriter
+BAD
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="right" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+2550276
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="right" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+10 101 101 000 010 111 110
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="right" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+
+\family typewriter
+AD0BE
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="right" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+76545336
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="right" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+111 110 101 100 101 011 011 110
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="right" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+
+\family typewriter
+FACADE
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="right" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+3726755
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="right" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+11 111 010 110 111 101 101
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="right" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+
+\family typewriter
+FADED
+\end_layout
+
+\end_inset
+</cell>
+</row>
+</lyxtabular>
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Note Comment
+status open
+
+\begin_layout Plain Layout
+Is this what academics do when they get bored?
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc13[M21]
+\end_layout
+
+\end_inset
+
+In the decimal system there are some numbers with two infinite decimal expansions;
+ for example,
+ 
+\begin_inset Formula $2.3599999...=2.3600000...$
+\end_inset
+
+.
+ Does the 
+\emph on
+negadecimal
+\emph default
+ (base 
+\begin_inset Formula $-10$
+\end_inset
+
+) system have unique expansions,
+ or are there real numbers with two different infinite expansions in this base also?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+There are real numbers with two different infinite expansions:
+ 
+\begin_inset Formula 
+\[
+\tfrac{1}{11}=0.090909...=(0.090909...)_{-10}=(1.909090...)_{-10}.
+\]
+
+\end_inset
+
+Effectively,
+ 
+\begin_inset Formula $(0.090909...)_{-10}=\sum_{k\geq1}9\cdot(-10)^{-2k}=9\sum_{k\geq1}100^{-k}=9\frac{\frac{1}{100}}{\frac{99}{100}}=\frac{1}{11}$
+\end_inset
+
+,
+ but 
+\begin_inset Formula $(1.909090...)_{-10}=1+\sum_{k\geq1}9\cdot(-10)^{-2k+1}=1-90\sum_{k\geq1}100^{-k}=1-90\frac{1}{99}=1-\frac{10}{11}=\frac{1}{11}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc19[23]
+\end_layout
+
+\end_inset
+
+(David W.
+ Matula.) Let 
+\begin_inset Formula $D$
+\end_inset
+
+ be a set of 
+\begin_inset Formula $b$
+\end_inset
+
+ integers,
+ containing exactly one solution to the congruence 
+\begin_inset Formula $x\equiv j\pmod b$
+\end_inset
+
+ for 
+\begin_inset Formula $0\leq j<b$
+\end_inset
+
+.
+ Prove that all integers (positive,
+ negative,
+ or zero) can be represented in the form 
+\begin_inset Formula $m=(a_{n}\dots a_{0})_{b}$
+\end_inset
+
+,
+ where all the 
+\begin_inset Formula $a_{j}$
+\end_inset
+
+ are in 
+\begin_inset Formula $D$
+\end_inset
+
+,
+ if and only if all integers in the range 
+\begin_inset Formula $l\leq m\leq u$
+\end_inset
+
+ can be so represented,
+ where 
+\begin_inset Formula $l=-\max\{a\mid a\in D\}/(b-1)$
+\end_inset
+
+ and 
+\begin_inset Formula $u=-\min\{a\mid a\in D\}/(b-1)$
+\end_inset
+
+.
+ For example,
+ 
+\begin_inset Formula $D=\{-1,0,\dots,b-2\}$
+\end_inset
+
+ satisfies the conditions for all 
+\begin_inset Formula $b\geq3$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+For the statement to make sense,
+ we need 
+\begin_inset Formula $b\geq2$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Obvious.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Take an arbitrary 
+\begin_inset Formula $m\in\mathbb{Z}$
+\end_inset
+
+.
+ If 
+\begin_inset Formula $l\leq m\leq u$
+\end_inset
+
+,
+ we are done.
+ Otherwise,
+ let 
+\begin_inset Formula $q$
+\end_inset
+
+ and 
+\begin_inset Formula $r$
+\end_inset
+
+ be the quotient and remainder of dividing 
+\begin_inset Formula $m$
+\end_inset
+
+ by 
+\begin_inset Formula $b$
+\end_inset
+
+,
+ and let 
+\begin_inset Formula $x\in D$
+\end_inset
+
+ be such that 
+\begin_inset Formula $x\equiv r\pmod b$
+\end_inset
+
+.
+ Then write 
+\begin_inset Formula $x$
+\end_inset
+
+ and,
+ if 
+\begin_inset Formula $k\in\mathbb{Z}$
+\end_inset
+
+ is such that 
+\begin_inset Formula $x=kb+r$
+\end_inset
+
+,
+ write the number 
+\begin_inset Formula $q-k$
+\end_inset
+
+ to the right of the 
+\begin_inset Formula $x$
+\end_inset
+
+ we just wrote by applying this same process.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+We have to prove that this algorithm terminates.
+ First we note that there exists at least an integer between 
+\begin_inset Formula $l$
+\end_inset
+
+ and 
+\begin_inset Formula $u$
+\end_inset
+
+,
+ because there are 
+\begin_inset Formula $b$
+\end_inset
+
+ integers in 
+\begin_inset Formula $D$
+\end_inset
+
+ and so the maximum and minimum must differ by at least 
+\begin_inset Formula $b-1$
+\end_inset
+
+,
+ and so 
+\begin_inset Formula $u-l\geq1$
+\end_inset
+
+.
+ This means that necessarily 
+\begin_inset Formula $l\leq0$
+\end_inset
+
+ and 
+\begin_inset Formula $u\geq0$
+\end_inset
+
+,
+ for if 
+\begin_inset Formula $l>0$
+\end_inset
+
+,
+ then all numbers in 
+\begin_inset Formula $D$
+\end_inset
+
+ are negative,
+ but the integers between 
+\begin_inset Formula $l$
+\end_inset
+
+ and 
+\begin_inset Formula $u$
+\end_inset
+
+ are positive so they couldn't be represented in the given form,
+\begin_inset Formula $\#$
+\end_inset
+
+ and a similar thing would happen if 
+\begin_inset Formula $u<0$
+\end_inset
+
+.
+ Now,
+ in the algorithm above,
+ 
+\begin_inset Formula $q-k=\lfloor\frac{m}{b}\rfloor-k=\frac{m-r}{b}-\frac{x-r}{b}=\frac{m-x}{b}$
+\end_inset
+
+.
+ With this,
+ if 
+\begin_inset Formula $m>u$
+\end_inset
+
+,
+ then 
+\begin_inset Formula $m(b-1)>-\min D$
+\end_inset
+
+,
+ but and therefore
+\begin_inset Formula 
+\[
+m-(q-k)=m-\frac{m-x}{b}=\frac{m(b-1)+x}{b}>\frac{-\min D+\min D}{b}=0,
+\]
+
+\end_inset
+
+so 
+\begin_inset Formula $q-k<m$
+\end_inset
+
+,
+ and 
+\begin_inset Formula 
+\[
+m+(q-k)=\frac{m(b+1)-x}{b}>m-\frac{\max D}{b}\geq m+l,
+\]
+
+\end_inset
+
+so 
+\begin_inset Formula $q-k\geq l$
+\end_inset
+
+.
+ This means that,
+ on each step,
+ 
+\begin_inset Formula $m$
+\end_inset
+
+ is smaller by at least 1,
+ but it doesn't become smaller than 
+\begin_inset Formula $l$
+\end_inset
+
+,
+ so it eventually reaches the interval 
+\begin_inset Formula $[l,u]$
+\end_inset
+
+ in at most 
+\begin_inset Formula $|m-u|$
+\end_inset
+
+ steps.
+ And if 
+\begin_inset Formula $m<l$
+\end_inset
+
+,
+ then 
+\begin_inset Formula $m(b-1)<-\max D$
+\end_inset
+
+,
+ so
+\begin_inset Formula 
+\[
+m-(q-k)=\frac{m(b-1)+x}{b}<\frac{-\max D+\max D}{b}=0,
+\]
+
+\end_inset
+
+so 
+\begin_inset Formula $q-k>m$
+\end_inset
+
+,
+ and
+\begin_inset Formula 
+\[
+m+(q-k)=\frac{m(b+1)-x}{b}<m-\frac{\min D}{b}\leq m+u,
+\]
+
+\end_inset
+
+ so 
+\begin_inset Formula $q-k\leq u$
+\end_inset
+
+,
+ and a similar reasoning applies.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc21[M22]
+\end_layout
+
+\end_inset
+
+(C.
+ E.
+ Shannon.) Can every real number (positive,
+ negative,
+ or zero) be expressed in a 
+\begin_inset Quotes eld
+\end_inset
+
+balanced decimal
+\begin_inset Quotes erd
+\end_inset
+
+ system,
+ that is,
+ in the form 
+\begin_inset Formula $\sum_{k\leq n}a_{k}10^{k}$
+\end_inset
+
+,
+ for some integer 
+\begin_inset Formula $n$
+\end_inset
+
+ and some sequence 
+\begin_inset Formula $a_{n},a_{n-1},a_{n-2},\dots$
+\end_inset
+
+,
+ where each 
+\begin_inset Formula $a_{k}$
+\end_inset
+
+ is one of the ten numbers 
+\begin_inset Formula $\{-4\frac{1}{2},-3\frac{1}{2},-2\frac{1}{2},-1\frac{1}{2},-\frac{1}{2},\frac{1}{2},1\frac{1}{2},2\frac{1}{2},3\frac{1}{2},4\frac{1}{2}\}$
+\end_inset
+
+?
+ (Although zero is not one of the allowed digits,
+ we implicitly assume that 
+\begin_inset Formula $a_{n+1},a_{n+2},\dots$
+\end_inset
+
+ are zero.) Find all representations of zero in this number system,
+ and find all representations of unity.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Yes,
+ we can do something similar to the construction of balanced ternary.
+ First,
+ we add 
+\begin_inset Formula $5\cdot10^{n+1}$
+\end_inset
+
+ to the number,
+ and then we subtract 
+\begin_inset Formula $4\frac{1}{2}$
+\end_inset
+
+ from each digit from the 
+\begin_inset Formula $(n+1)$
+\end_inset
+
+st to the right.
+\end_layout
+
+\begin_layout Standard
+For a sum in the given form to be 0,
+ 
+\begin_inset Formula $a_{n}10^{n}$
+\end_inset
+
+ must equal 
+\begin_inset Formula $-\sum_{k<n}a_{k}10^{k}$
+\end_inset
+
+.
+ The biggest value we can write as 
+\begin_inset Formula $\sum_{k<n}a_{k}10^{k}$
+\end_inset
+
+ is 
+\begin_inset Formula $5\cdot10^{n-1}=10^{n}/2$
+\end_inset
+
+,
+ which we can only reach if all those 
+\begin_inset Formula $a_{k}=4\frac{1}{2}$
+\end_inset
+
+,
+ and the smallest is 
+\begin_inset Formula $-10^{n}/2$
+\end_inset
+
+,
+ where all the 
+\begin_inset Formula $a_{k}=-4\frac{1}{2}$
+\end_inset
+
+,
+ so the representations are 
+\begin_inset Quotes eld
+\end_inset
+
+
+\begin_inset Formula $-\frac{1}{2},4\frac{1}{2},4\frac{1}{2},4\frac{1}{2},\dots$
+\end_inset
+
+
+\begin_inset Quotes erd
+\end_inset
+
+ and 
+\begin_inset Quotes eld
+\end_inset
+
+
+\begin_inset Formula $\frac{1}{2},-4\frac{1}{2},-4\frac{1}{2},-4\frac{1}{2},\dots$
+\end_inset
+
+
+\begin_inset Quotes erd
+\end_inset
+
+,
+ with the decimal point in any arbitrary position.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+begin{sloppypar}
+\end_layout
+
+\end_inset
+
+To get one,
+ we need 
+\begin_inset Formula $\frac{1}{2}+\frac{1}{2}$
+\end_inset
+
+ or 
+\begin_inset Formula $1\frac{1}{2}-\frac{1}{2}$
+\end_inset
+
+,
+ so either 
+\begin_inset Quotes eld
+\end_inset
+
+
+\begin_inset Formula $\frac{1}{2};4\frac{1}{2},\dots,4\frac{1}{2},\dots$
+\end_inset
+
+
+\begin_inset Quotes erd
+\end_inset
+
+ or 
+\begin_inset Quotes eld
+\end_inset
+
+
+\begin_inset Formula $1\frac{1}{2};-4\frac{1}{2},\dots,-4\frac{1}{2},\dots$
+\end_inset
+
+
+\begin_inset Quotes erd
+\end_inset
+
+,
+ or 
+\begin_inset Quotes eld
+\end_inset
+
+
+\begin_inset Formula $\frac{1}{2},-4\frac{1}{2},\dots,-4\frac{1}{2},-3\frac{1}{2},-4\frac{1}{2},\dots,-4\frac{1}{2},\dots$
+\end_inset
+
+
+\begin_inset Quotes erd
+\end_inset
+
+ or 
+\begin_inset Quotes eld
+\end_inset
+
+
+\begin_inset Formula $\frac{1}{2},-4\frac{1}{2},\dots,-4\frac{1}{2};4\frac{1}{2},\dots,4\frac{1}{2},\dots$
+\end_inset
+
+
+\begin_inset Quotes erd
+\end_inset
+
+,
+ which stem from representing 
+\begin_inset Formula $\frac{1}{2}$
+\end_inset
+
+ and 
+\begin_inset Formula $1\frac{1}{2}$
+\end_inset
+
+ in a way similar to the one used to represent the zero above.
+ Here the semicolon represents the radix point.
+ The most significant digit must be positive and to the left of the radix point,
+ and we can use this fact to show that these are all the ways that the number 1 can be represented in this system.
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+end{sloppypar}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc28[M24]
+\end_layout
+
+\end_inset
+
+Show that every nonzero complex number of the form 
+\begin_inset Formula $a+b\text{i}$
+\end_inset
+
+ where 
+\begin_inset Formula $a$
+\end_inset
+
+ and 
+\begin_inset Formula $b$
+\end_inset
+
+ are integers has a unique 
+\begin_inset Quotes eld
+\end_inset
+
+revolving binary representation
+\begin_inset Quotes erd
+\end_inset
+
+
+\begin_inset Formula 
+\[
+(1+\text{i})^{e_{0}}+\text{i}(1+\text{i})^{e_{1}}-(1+\text{i})^{e_{2}}-\text{i}(1+\text{i})^{e_{3}}+\dots+\text{i}^{t}(1+\text{i})^{e_{t}},
+\]
+
+\end_inset
+
+where 
+\begin_inset Formula $e_{0}<e_{1}<\dots<e_{t}$
+\end_inset
+
+.
+ (Compare with exercise 27.)
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+First we show that such representation exists by providing an algorithm for getting it from a number 
+\begin_inset Formula $c\in\mathbb{Z}+\text{i}\mathbb{Z}$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+[Initialize.] Set 
+\begin_inset Formula $s\gets0$
+\end_inset
+
+ and 
+\begin_inset Formula $e\gets0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:e4128b"
+
+\end_inset
+
+[Case 
+\begin_inset Formula $(1+\text{i})^{0}$
+\end_inset
+
+.] If 
+\begin_inset Formula $\text{Re}c$
+\end_inset
+
+ is odd and 
+\begin_inset Formula $\text{Im}c$
+\end_inset
+
+ is even,
+ or vice versa,
+ write down 
+\begin_inset Formula $\text{i}^{s}(1+\text{i})^{e}$
+\end_inset
+
+ and set 
+\begin_inset Formula $s\gets s+1$
+\end_inset
+
+ and 
+\begin_inset Formula $c\gets-\text{i}(c-1)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:e4128c"
+
+\end_inset
+
+[Case 
+\begin_inset Formula $(1+\text{i})^{1}$
+\end_inset
+
+.] If 
+\begin_inset Formula $\text{Re}c$
+\end_inset
+
+ is odd [if,
+ and only if,
+ 
+\begin_inset Formula $\text{Im}c$
+\end_inset
+
+ is odd,] write down 
+\begin_inset Formula $\text{i}^{s}(1+\text{i})^{e+1}$
+\end_inset
+
+ and set 
+\begin_inset Formula $s\gets s+1$
+\end_inset
+
+ and 
+\begin_inset Formula $c\gets-\text{i}(c-1-\text{i})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:e4128d"
+
+\end_inset
+
+[Iterate.] If 
+\begin_inset Formula $c\neq0$
+\end_inset
+
+,
+ set 
+\begin_inset Formula $e\gets e+2$
+\end_inset
+
+ and 
+\begin_inset Formula $c\gets\frac{c}{2\text{i}}$
+\end_inset
+
+ and go back to step 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "enu:e4128b"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+The algorithm always maintains the invariant that 
+\begin_inset Formula $c_{0}=w+\text{i}^{s}(1+\text{i})^{e}c$
+\end_inset
+
+,
+ where 
+\begin_inset Formula $c_{0}$
+\end_inset
+
+ is the original value of 
+\begin_inset Formula $c$
+\end_inset
+
+ and 
+\begin_inset Formula $w$
+\end_inset
+
+ is the sum of the terms that have been written,
+ so when it finishes,
+ 
+\begin_inset Formula $w=c_{0}$
+\end_inset
+
+,
+ and it's trivial to see that 
+\begin_inset Formula $w$
+\end_inset
+
+ is a revolving binary representation.
+ To see that the algorithm terminates,
+ we see that,
+ after each iteration,
+ 
+\begin_inset Formula $c$
+\end_inset
+
+ has a smaller magnitude except if originally 
+\begin_inset Formula $c=-1,-\text{i},-1-\text{i}$
+\end_inset
+
+,
+ but in these cases the iteration after that will lower the magnitude of 
+\begin_inset Formula $c$
+\end_inset
+
+,
+ preventing an infinite loop:
+\begin_inset Formula 
+\begin{align*}
+-1 & \overset{\eqref{enu:e4128b}}{\to}2\text{i}\overset{\eqref{enu:e4128d}}{\to}1, & -\text{i} & \overset{\eqref{enu:e4128b}}{\to}\text{i}-1\overset{\eqref{enu:e4128c}}{\to}2\text{i}\overset{\eqref{enu:e4128d}}{\to}1, & -1-\text{i} & \overset{\eqref{enu:e4128c}}{\to}2\text{i}-2\overset{\eqref{enu:e4128d}}{\to}\text{i}+1.
+\end{align*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+For the uniqueness,
+ let 
+\begin_inset Formula $\sum_{k=0}^{n}\text{i}^{k}(1+\text{i})^{e_{k}}=\sum_{j=0}^{m}\text{i}^{j}(1+\text{i})^{f_{j}}$
+\end_inset
+
+ (
+\begin_inset Formula $n,m,e_{k},f_{j}\in\mathbb{N}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $e_{0}<\dots<e_{n}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f_{0}<\dots<f_{m}$
+\end_inset
+
+) be two different representations of the same number,
+ and let 
+\begin_inset Formula $s\leq m,n$
+\end_inset
+
+ be the first index where they differ (let's say 
+\begin_inset Formula $e_{s}<f_{s}$
+\end_inset
+
+).
+ Then we may remove all terms before 
+\begin_inset Formula $s$
+\end_inset
+
+ in both representations and divide each remaining term by 
+\begin_inset Formula $\text{i}^{s}(1+\text{i})^{e_{s}}$
+\end_inset
+
+.
+ But then,
+ the second representation is a multiple of 
+\begin_inset Formula $(1+\text{i})^{\min\{e_{s}+1,f_{s}\}}$
+\end_inset
+
+ in 
+\begin_inset Formula $\mathbb{Z}+\text{i}\mathbb{Z}$
+\end_inset
+
+ but the first one isn't.
+\begin_inset Formula $\#$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc34[22]
+\end_layout
+
+\end_inset
+
+(G.
+ W.
+ Reitweisner,
+ 1960.) Explain how to represent a given integer 
+\begin_inset Formula $n$
+\end_inset
+
+ in the form 
+\begin_inset Formula $(\dots a_{2}a_{1}a_{0})_{2}$
+\end_inset
+
+,
+ where each 
+\begin_inset Formula $a_{j}$
+\end_inset
+
+ is 
+\begin_inset Formula $-1$
+\end_inset
+
+,
+ 0,
+ or 1,
+ using the fewest nonzero digits.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+If 
+\begin_inset Formula $n=0$
+\end_inset
+
+,
+ we just write 0.
+ If 
+\begin_inset Formula $n>0$
+\end_inset
+
+,
+ we take 
+\begin_inset Formula $k\in\mathbb{N}$
+\end_inset
+
+ such that 
+\begin_inset Formula $|2^{k}-n|$
+\end_inset
+
+ is minimum,
+ write a 1 at position 
+\begin_inset Formula $k$
+\end_inset
+
+,
+ and then write 
+\begin_inset Formula $n-2^{k}$
+\end_inset
+
+ on the 
+\begin_inset Formula $k$
+\end_inset
+
+ positions to the right.
+ If 
+\begin_inset Formula $n<0$
+\end_inset
+
+,
+ we do the same except that first we write 
+\begin_inset Formula $\overline{1}$
+\end_inset
+
+ instead of 1.
+\end_layout
+
+\begin_layout Standard
+Now we prove that this in fact results in the representation fewest number of nonzero digits.
+ We only need to prove it for positive 
+\begin_inset Formula $n$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Now,
+ if 
+\begin_inset Formula $n=2^{k}$
+\end_inset
+
+ or 
+\begin_inset Formula $3\cdot2^{k}$
+\end_inset
+
+ for some 
+\begin_inset Formula $k\in\mathbb{N}$
+\end_inset
+
+,
+ the statement is trivial.
+ Otherwise,
+ if 
+\begin_inset Formula $2^{k}<n<2^{k+1}$
+\end_inset
+
+ for some 
+\begin_inset Formula $k\in\mathbb{N}$
+\end_inset
+
+,
+ we operate by induction.
+ First note that,
+ for such 
+\begin_inset Formula $n$
+\end_inset
+
+,
+ 
+\begin_inset Formula $k\geq2$
+\end_inset
+
+,
+ and that all possible representations start with a 1 at position 
+\begin_inset Formula $k$
+\end_inset
+
+ or 
+\begin_inset Formula $k+1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+If 
+\begin_inset Formula $n<3\cdot2^{k-1}$
+\end_inset
+
+,
+ let 
+\begin_inset Formula $\sigma(s)$
+\end_inset
+
+ be the minimum number of nonzero digits in a representation of 
+\begin_inset Formula $s$
+\end_inset
+
+,
+ then if we start with a 1 at position 
+\begin_inset Formula $k$
+\end_inset
+
+ we would have a minimum of 
+\begin_inset Formula $1+\sigma(n-2^{k})$
+\end_inset
+
+ nonzero digits,
+ whereas if we start at position 
+\begin_inset Formula $k+1$
+\end_inset
+
+ we would have a minimum of 
+\begin_inset Formula $1+\sigma(2^{k+1}-n)$
+\end_inset
+
+.
+ Let 
+\begin_inset Formula $t\coloneqq2^{k+1}-n>2^{k-1}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $t$
+\end_inset
+
+ would need to have its most significant digit in position 
+\begin_inset Formula $k$
+\end_inset
+
+ or 
+\begin_inset Formula $k-1$
+\end_inset
+
+,
+ but any representation of 
+\begin_inset Formula $t$
+\end_inset
+
+ that starts at position 
+\begin_inset Formula $k$
+\end_inset
+
+ can be converted into one of 
+\begin_inset Formula $2^{k}-n$
+\end_inset
+
+ by removing the 1 at that position,
+ and any one starting at position 
+\begin_inset Formula $k-1$
+\end_inset
+
+ can be converted into one of 
+\begin_inset Formula $2^{k}-n$
+\end_inset
+
+ by swapping the sign in that position,
+ so in general 
+\begin_inset Formula $\sigma(t)=\sigma(2^{k+1}-n)\geq\sigma(2^{k}-n)=\sigma(n-2^{k})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+A similar argument shows that,
+ if 
+\begin_inset Formula $n>3\cdot2^{k-1}$
+\end_inset
+
+,
+ then 
+\begin_inset Formula $\sigma(n-2^{k})\geq\sigma(2^{k+1}-n)$
+\end_inset
+
+.
+\end_layout
+
+\end_body
+\end_document
diff --git a/vol2/index.lyx b/vol2/index.lyx
new file mode 100644
index 0000000..816a76b
--- /dev/null
+++ b/vol2/index.lyx
@@ -0,0 +1,1394 @@
+#LyX 2.4 created this file. For more info see https://www.lyx.org/
+\lyxformat 620
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\begin_preamble
+\input{../defs}
+
+\makeatletter
+\newcommand{\biggg}{\bBigg@\thr@@}
+\newcommand{\Biggg}{\bBigg@{3.5}}
+\makeatother
+\end_preamble
+\use_default_options true
+\maintain_unincluded_children no
+\language english
+\language_package default
+\inputencoding utf8
+\fontencoding auto
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_roman_osf false
+\font_sans_osf false
+\font_typewriter_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype true
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\float_placement class
+\float_alignment class
+\paperfontsize 10
+\spacing single
+\use_hyperref true
+\pdf_bookmarks true
+\pdf_bookmarksnumbered false
+\pdf_bookmarksopen false
+\pdf_bookmarksopenlevel 1
+\pdf_breaklinks false
+\pdf_pdfborder false
+\pdf_colorlinks false
+\pdf_backref false
+\pdf_pdfusetitle true
+\papersize custom
+\use_geometry true
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_formatted_ref 0
+\use_minted 0
+\use_lineno 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\paperwidth 198mm
+\paperheight 297mm
+\leftmargin 23mm
+\topmargin 33mm
+\rightmargin 43mm
+\bottommargin 66mm
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style english
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 2
+\paperpagestyle fancy
+\tablestyle default
+\listings_params "basicstyle={\ttfamily}"
+\tracking_changes false
+\output_changes false
+\change_bars false
+\postpone_fragile_content false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\docbook_table_output 0
+\docbook_mathml_prefix 1
+\end_header
+
+\begin_body
+
+\begin_layout Title
+Exercises on 
+\emph on
+The Art Of Computer Programming
+\emph default
+
+\begin_inset Newline newline
+\end_inset
+
+
+\size larger
+Volume 2:
+ Seminumerical Algorithms
+\begin_inset Newline newline
+\end_inset
+
+Third Edition
+\end_layout
+
+\begin_layout Author
+Juan Marín Noguera
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset toc
+LatexCommand tableofcontents
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+Equivalent page size can be obtained with the following layouts (in mm):
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Tabular
+<lyxtabular version="3" rows="3" columns="7">
+<features tabularvalignment="middle">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<row>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Height
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Width
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Top
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Bottom
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Inner
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Outer
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Recommended headers
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+297
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+198
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+33
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+66
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+23
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+43
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Fancy
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+210
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+140
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+4
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+8
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+3
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+5
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Empty
+\end_layout
+
+\end_inset
+</cell>
+</row>
+</lyxtabular>
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+The first one follows Tschichold's canon except for 
+\begin_inset Formula $\unit[1]{mm}$
+\end_inset
+
+ given to the inner margin from the outer one to account for a minimal folding.
+ The golden ratio resulted in way too narrow pages on A5 paper.
+\end_layout
+
+\begin_layout Plain Layout
+Proposed notation for in-progress:
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset listings
+inline false
+status open
+
+\begin_layout Plain Layout
+
+atom = (
+\begin_inset Quotes eld
+\end_inset
+
+A
+\begin_inset Quotes erd
+\end_inset
+
+|
+\begin_inset Quotes erd
+\end_inset
+
+R
+\begin_inset Quotes erd
+\end_inset
+
+) ?(
+\begin_inset Quotes eld
+\end_inset
+
+B
+\begin_inset Quotes erd
+\end_inset
+
+|
+\begin_inset Quotes erd
+\end_inset
+
+M
+\begin_inset Quotes erd
+\end_inset
+
+) ((
+\begin_inset Quotes eld
+\end_inset
+
+0
+\begin_inset Quotes erd
+\end_inset
+
+..
+\begin_inset Quotes erd
+\end_inset
+
+4
+\begin_inset Quotes erd
+\end_inset
+
+) (
+\begin_inset Quotes eld
+\end_inset
+
+0
+\begin_inset Quotes erd
+\end_inset
+
+..
+\begin_inset Quotes erd
+\end_inset
+
+9
+\begin_inset Quotes erd
+\end_inset
+
+) / 
+\begin_inset Quotes eld
+\end_inset
+
+@
+\begin_inset Quotes erd
+\end_inset
+
+) / 1DIGIT *DIGIT / *1(1DIGIT *DIGIT) 
+\begin_inset Quotes eld
+\end_inset
+
+..
+\begin_inset Quotes erd
+\end_inset
+
+ *1(1DIGIT *DIGIT)
+\end_layout
+
+\begin_layout Plain Layout
+
+base = atom / 
+\begin_inset Quotes eld
+\end_inset
+
+(
+\begin_inset Quotes eld
+\end_inset
+
+ progress 
+\begin_inset Quotes eld
+\end_inset
+
+)
+\begin_inset Quotes erd
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+
+intersection = atom / intersection 
+\begin_inset Quotes eld
+\end_inset
+
+*
+\begin_inset Quotes erd
+\end_inset
+
+ atom
+\end_layout
+
+\begin_layout Plain Layout
+
+progress = intersection / progress (
+\begin_inset Quotes eld
+\end_inset
+
++
+\begin_inset Quotes erd
+\end_inset
+
+/
+\begin_inset Quotes erd
+\end_inset
+
+-
+\begin_inset Quotes erd
+\end_inset
+
+) intersection
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+An 
+\family typewriter
+atom
+\family default
+ represents all/recommended exercises,
+ optionally excluding M or HM/excluding HM,
+ and up to the given difficulty level/all difficulty levels,
+ or else a single exercise with the given number,
+ or a range of exercises (both ends inclusive).
+ Then 
+\family typewriter
+*
+\family default
+ is used for intersection,
+ 
+\family typewriter
++
+\family default
+ for union,
+ and 
+\family typewriter
+-
+\family default
+ for set difference.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+setcounter{chapter}{2}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Chapter
+Random Numbers
+\end_layout
+
+\begin_layout Section
+Introduction
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "3.1.lyx"
+literal "false"
+
+\end_inset
+
+
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+
+\family typewriter
+A10+R25+16
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Generating Uniform Random Numbers
+\end_layout
+
+\begin_layout Subsection
+The Linear Congruential Method
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "3.2.1.lyx"
+literal "false"
+
+\end_inset
+
+
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+
+\family typewriter
+A10+R25
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Choice of modulus
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "3.2.1.1.lyx"
+literal "false"
+
+\end_inset
+
+
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+
+\family typewriter
+A10+R25
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Choice of multiplier
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "3.2.1.2.lyx"
+literal "false"
+
+\end_inset
+
+
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+
+\family typewriter
+A10+R25
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Potency
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "3.2.1.3.lyx"
+literal "false"
+
+\end_inset
+
+
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+
+\family typewriter
+A10+R25
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Other Methods
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "3.2.2.lyx"
+literal "false"
+
+\end_inset
+
+
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+
+\family typewriter
+A10+R25
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Statistical Tests
+\end_layout
+
+\begin_layout Subsection
+General Test Procedures for Studying Random Data
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "3.3.1.lyx"
+literal "false"
+
+\end_inset
+
+
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+
+\family typewriter
+A10+R25
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Empirical Tests
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "3.3.2.lyx"
+literal "false"
+
+\end_inset
+
+
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+
+\family typewriter
+A10+R25
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Theoretical Tests
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "3.3.3.lyx"
+literal "false"
+
+\end_inset
+
+
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+
+\family typewriter
+A10+R25
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+The Spectral Test
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "3.3.4.lyx"
+literal "false"
+
+\end_inset
+
+
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+
+\family typewriter
+A10+R25
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Other Types of Random Quantities
+\end_layout
+
+\begin_layout Subsection
+Numerical Distributions
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "3.4.1.lyx"
+literal "false"
+
+\end_inset
+
+
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+
+\family typewriter
+A10+R25
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Random Sampling and Shuffling
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "3.4.2.lyx"
+literal "false"
+
+\end_inset
+
+
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+
+\family typewriter
+A10+R25-16
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+What Is a Random Sequence?
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "3.5.lyx"
+literal "false"
+
+\end_inset
+
+
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+
+\family typewriter
+A10+R25
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Summary
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "3.6.lyx"
+literal "false"
+
+\end_inset
+
+
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+
+\family typewriter
+A10+R25
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Chapter
+Arithmetic
+\end_layout
+
+\begin_layout Section
+Positional Number Systems
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "4.1.lyx"
+literal "false"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Floating Point Arithmetic
+\end_layout
+
+\begin_layout Subsection
+Single-Precision Calculations
+\end_layout
+
+\begin_layout Subsection
+Accuracy of Floating Point Arithmetic
+\end_layout
+
+\begin_layout Subsection
+Double-Precision Calculations
+\end_layout
+
+\begin_layout Standard
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+7+0;
+ 5 (0:28) -> 3d,
+ -1/3
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Distribution of Floating Point Numbers
+\end_layout
+
+\begin_layout Standard
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+10+2;
+ 5,
+ 13,
+ 17 (0:58) -> 5d
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Multiple-Precision Arithmetic
+\end_layout
+
+\begin_layout Subsection
+The Classical Algorithms
+\end_layout
+
+\begin_layout Standard
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+16+4;
+ 6,
+ 9,
+ 11,
+ 14,
+ 16,
+ 19,
+ 21,
+ 22,
+ 30,
+ 37,
+ 43 (2:58) -> 12d,
+ -2/3
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Modular Arithmetic
+\end_layout
+
+\begin_layout Standard
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+8+1;
+ 5,
+ 7,
+ 12,
+ 13 (1:13) -> 4d
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+How Fast Can We Multiply?
+\end_layout
+
+\begin_layout Standard
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+22+3;
+ 16,
+ 19 (0:56) -> 10d,
+ -2/3
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Radix Conversion
+\end_layout
+
+\begin_layout Standard
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+9+2;
+ 1,
+ 3,
+ 12,
+ 13,
+ 19 (2:22) -> 8d,
+ -2/3
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Rational Arithmetic
+\end_layout
+
+\begin_layout Subsection
+Fractions
+\end_layout
+
+\begin_layout Standard
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+3+1;
+ 5,
+ 6,
+ 8 (0:42) -> 2d,
+ -2/3
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+The Greatest Common Divisor
+\end_layout
+
+\begin_layout Standard
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+19+4;
+ 8,
+ 10,
+ 14,
+ 16,
+ 17,
+ 18,
+ 23,
+ 40 (3:19) -> 13d,
+ -1/3
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Analysis of Euclid's Algorithm
+\end_layout
+
+\begin_layout Standard
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+17+6;
+ 1,
+ 17,
+ 39,
+ 50 (1:42) -> 9d,
+ -2/3
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Factoring into Primes
+\end_layout
+
+\begin_layout Standard
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+32+6;
+ 1,
+ 8,
+ 18,
+ 19,
+ 24,
+ 26,
+ 32,
+ 35 (3:17) -> 17d
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Polynomial Arithmetic
+\end_layout
+
+\begin_layout Standard
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+2+0;
+ 1,
+ 4,
+ 5 (0:31) -> 2d,
+ -1/3
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Division of Polynomials
+\end_layout
+
+\begin_layout Standard
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+15+4;
+ 1,
+ 3,
+ 7,
+ 8,
+ 12,
+ 16,
+ 18 (2:08) -> 9d,
+ -1/3
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Factorization of Polynomials
+\end_layout
+
+\begin_layout Standard
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+17+5;
+ 1,
+ 2,
+ 10,
+ 12,
+ 18,
+ 22,
+ 34,
+ 40 (3:22) -> 13d,
+ -2/3
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Evaluation of Powers
+\end_layout
+
+\begin_layout Standard
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+21+4;
+ 3,
+ 5,
+ 9,
+ 10,
+ 12,
+ 24,
+ 26,
+ 36,
+ 39,
+ 40 (3:37) -> 14d,
+ -2/3
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Evaluation of Polynomials
+\end_layout
+
+\begin_layout Standard
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+29+10;
+ 2,
+ 19,
+ 20,
+ 24,
+ 26,
+ 29,
+ 33,
+ 35,
+ 44,
+ 45,
+ 49,
+ 51,
+ 70 (5:24) -> 20d
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Manipulation of Power Series
+\end_layout
+
+\begin_layout Standard
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+8+5;
+ 1,
+ 4,
+ 5,
+ 6,
+ 8,
+ 11,
+ 17 (1:59) -> 7d,
+ -1/3
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_body
+\end_document