Changed layout for more manageable volumes

1 files changed, 0 insertions, 644 deletions

diff --git a/2.2.4.lyx b/2.2.4.lyx
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-#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
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-\index Index
-\shortcut idx
-\color #008000
-\end_index
-\secnumdepth 3
-\tocdepth 3
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-\end_header
-
-\begin_body
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc3[20]
-\end_layout
-
-\end_inset
-
-What does operation (3) do if 
-\begin_inset Formula $\mathtt{PTR}_{1}$
-\end_inset
-
- and 
-\begin_inset Formula $\mathtt{PTR}_{2}$
-\end_inset
-
- are both pointing to nodes in the 
-\emph on
-same
-\emph default
- circular list?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-It splits the list into two parts;
- the one that goes from the right of 
-\begin_inset Formula $\mathtt{PTR}_{1}$
-\end_inset
-
- to 
-\begin_inset Formula $\mathtt{PTR}_{2}$
-\end_inset
-
- is preserved but the other one becomes a memory leak.
- If the two pointers are equal,
- nothing happens,
- just 
-\begin_inset Formula $\mathtt{PTR}_{2}$
-\end_inset
-
- is set to 
-\begin_inset Formula $\Lambda$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc5[21]
-\end_layout
-
-\end_inset
-
-Design an algorithm that takes a circular list such as (1) and reverses the direction of all the arrows.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-This is similar to exercise 2.2.3.7.
-\end_layout
-
-\begin_layout Enumerate
-Set 
-\begin_inset Formula $\mathtt{P}\gets\Lambda$
-\end_inset
-
- and 
-\begin_inset Formula $\mathtt{C}\gets\mathtt{PTR}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Enumerate
-If 
-\begin_inset Formula $\mathtt{C}\neq\Lambda$
-\end_inset
-
-,
- set 
-\begin_inset Formula $\mathtt{N}\gets\mathtt{LINK(C)}$
-\end_inset
-
-,
- 
-\begin_inset Formula $\mathtt{LINK(C)}\gets\mathtt{P}$
-\end_inset
-
-,
- 
-\begin_inset Formula $\mathtt{P}\gets\mathtt{C}$
-\end_inset
-
-,
- 
-\begin_inset Formula $\mathtt{C}\gets\mathtt{N}$
-\end_inset
-
-,
- and repeat.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc7[10]
-\end_layout
-
-\end_inset
-
-Why is it useful to assume that the 
-\family typewriter
-ABC
-\family default
- fields of a polynomial list appear in decreasing order?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Many operations on polynomials require us to match monomials with equal exponents of the variables.
- By having a total order on them it's easy to match these in linear time.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc8[10]
-\end_layout
-
-\end_inset
-
-Why is it useful to have 
-\family typewriter
-Q1
-\family default
- trailing one step behind 
-\family typewriter
-Q
-\family default
- in Algorithm A?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-We need 
-\family typewriter
-Q
-\family default
- to advance in order to check if the next monomial has greater,
- lower,
- or equal coefficient than the one pointed to by 
-\family typewriter
-P
-\family default
-,
- and if it's lower,
- we need 
-\family typewriter
-Q1
-\family default
- to insert the one pointed to by 
-\family typewriter
-P
-\family default
- right before the one pointed to by 
-\family typewriter
-Q
-\family default
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc9[23]
-\end_layout
-
-\end_inset
-
-Would Algorithm A work properly if 
-\begin_inset Formula $\mathtt{P}=\mathtt{Q}$
-\end_inset
-
- (i.e.,
- both pointer variables point at the same polynomial)?
- Would Algorithm M work properly if 
-\begin_inset Formula $\mathtt{P}=\mathtt{M}$
-\end_inset
-
-,
- if 
-\begin_inset Formula $\mathtt{P}=\mathtt{Q}$
-\end_inset
-
-,
- or if 
-\begin_inset Formula $\mathtt{M}=\mathtt{Q}$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Algorithm A would work properly,
- doubling the polynomial being pointed to (this is assuming no coefficient is 0).
- As for Algorithm M,
- it would work for 
-\begin_inset Formula $\mathtt{P}=\mathtt{M}$
-\end_inset
-
- since they're not mutated.
- If 
-\begin_inset Formula $\mathtt{P}=\mathtt{Q}$
-\end_inset
-
- it wouldn't work,
- because if 
-\family typewriter
-M
-\family default
- has more than one monomial then the addition of 
-\begin_inset Formula $\mathtt{P}$
-\end_inset
-
- times the second monomial or later (the second iteration of step M2) would add the wrong value to 
-\family typewriter
-Q
-\family default
-.
- If 
-\begin_inset Formula $\mathtt{M}=\mathtt{Q}$
-\end_inset
-
- it wouldn't work either as,
- for example,
- if initially 
-\begin_inset Formula $\mathtt{M}=\mathtt{Q}=1$
-\end_inset
-
- and 
-\begin_inset Formula $\mathtt{P}=-1$
-\end_inset
-
-,
- then addition in the first iteration of M2 would free the current monomial pointed to by 
-\begin_inset Formula $\mathtt{M}$
-\end_inset
-
- and in the next iteration of M1,
- 
-\begin_inset Formula $\mathtt{M}$
-\end_inset
-
- would point to free memory with unpredictable consequences.
- In cases where the constant coefficient of 
-\family typewriter
-P
-\family default
- is not 
-\begin_inset Formula $-1$
-\end_inset
-
-,
- it amazingly works,
- as writes to 
-\family typewriter
-Q
-\family default
- are always behind where we read from 
-\family typewriter
-M
-\family default
- when considering monomials of 
-\family typewriter
-P
-\family default
- with positive degree and,
- for the constant coefficient,
- it writes to the same position where 
-\family typewriter
-M
-\family default
- is but only to the 
-\family typewriter
-COEF
-\family default
- field—
-unless 
-\family typewriter
-P
-\family default
- is 
-\begin_inset Formula $-1$
-\end_inset
-
- in which case the node pointed to by 
-\family typewriter
-M
-\family default
- is freed.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc10[20]
-\end_layout
-
-\end_inset
-
-The algorithms in this section assume that we are using three variables 
-\begin_inset Formula $x$
-\end_inset
-
-,
- 
-\begin_inset Formula $y$
-\end_inset
-
-,
- and 
-\begin_inset Formula $z$
-\end_inset
-
- in the polynomials,
- and that their exponents individually never exceed 
-\begin_inset Formula $b-1$
-\end_inset
-
- (where 
-\begin_inset Formula $b$
-\end_inset
-
- is the byte size in 
-\family typewriter
-MIX
-\family default
-'s case).
- Suppose instead that we want to do addition and multiplication of polynomials in only one variable,
- 
-\begin_inset Formula $x$
-\end_inset
-
-,
- and to let its exponent take on values up to 
-\begin_inset Formula $b^{3}-1$
-\end_inset
-
-.
- What changes should be made to Algorithms A and M?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-None,
- as they work out of the box if we consider the 
-\family typewriter
-ABC
-\family default
- field to hold a single coefficient.
- The sums and comparisons made on this field in the algorithms are equally valid independently of which of the two interpretations we want to give to this field,
- assuming no overflow.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc17[22]
-\end_layout
-
-\end_inset
-
-What advantage is there in representing polynomials with a circular list as in this section,
- instead of with a straight linear linked list terminated by 
-\begin_inset Formula $\Lambda$
-\end_inset
-
- as in the previous section?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Steps A2 and A5 don't need special cases to handle the end of a list,
- only A3 does.
- Other than that there's not much of a difference.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc18[25]
-\end_layout
-
-\end_inset
-
-Devise a way to represent circular lists inside a computer in such a way that the list can be traversed efficiently in both directions,
- yet only one link field is used per node.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-If the link field combines the location of the next and previous fields with some monoidal operation like XOR,
- and we have an item 
-\begin_inset Formula $x_{i}$
-\end_inset
-
-,
- then knowing the location of 
-\begin_inset Formula $x_{i+1}$
-\end_inset
-
- allows us to get the location of 
-\begin_inset Formula $x_{i-1}$
-\end_inset
-
- and vice versa,
- so knowing the location of two consecutive items allows us to traverse the list in both directions.
-\end_layout
-
-\end_body
-\end_document