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\begin_body

\begin_layout Standard
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\backslash
rexerc[21]
\end_layout

\end_inset

Formulas (5) and (6) have been derived from the assumption that 
\begin_inset Formula $0\leq\mathtt{I}_{r}\leq d_{r}$
\end_inset

 for 
\begin_inset Formula $1\leq r\leq k$
\end_inset

.
 Give a general formula that applies to the case 
\begin_inset Formula $l_{r}\leq\mathtt{I}_{r}\leq u_{r}$
\end_inset

,
 where 
\begin_inset Formula $l_{r}$
\end_inset

 and 
\begin_inset Formula $u_{r}$
\end_inset

 are any lower and upper bounds on the dimensionality.
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\backslash
answer 
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\end_inset


\begin_inset Formula 
\begin{align*}
\mathtt{LOC(A[}\mathtt{I}_{1},\mathtt{I}_{2},\dots,\mathtt{I}_{k}\mathtt{])} & =\mathtt{LOC(A[}l_{1},l_{2},\dots,l_{k}\mathtt{])}+\sum_{1\leq r\leq k}a_{r}(\mathtt{I}_{r}-l_{r})=\\
 & =\left(\mathtt{LOC(A[}l_{1},l_{2},\dots,l_{k}\mathtt{])}-\sum_{1\leq r\leq k}a_{r}l_{r}\right)+\sum_{1\leq r\leq k}a_{r}\mathtt{I}_{r},
\end{align*}

\end_inset

where
\begin_inset Formula 
\[
a_{r}=c\prod_{r<s\leq k}(u_{s}-l_{s}+1).
\]

\end_inset


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\backslash
rexerc6[24]
\end_layout

\end_inset

Consider the 
\begin_inset Quotes eld
\end_inset

tetrahedral arrays
\begin_inset Quotes erd
\end_inset

 
\begin_inset Formula $\mathtt{A[}i\mathtt{,}j\mathtt{,}k\mathtt{]},$
\end_inset


\begin_inset Formula $\mathtt{B[}i\mathtt{,}j\mathtt{,}k\mathtt{]}$
\end_inset

,
 where 
\begin_inset Formula $0\leq k\leq j\leq i\leq n$
\end_inset

 in 
\family typewriter
A
\family default
,
 and 
\begin_inset Formula $0\leq i\leq j\leq k\leq n$
\end_inset

 in 
\family typewriter
B
\family default
.
 Suppose that both of these arrays are stored in consecutive memory locations in lexicographic order of the indices;
 show that 
\begin_inset Formula $\mathtt{LOC(A[I,J,K])}=a_{0}+f_{1}\mathtt{(I)}+f_{2}\mathtt{(J)}+f_{3}\mathtt{(K)}$
\end_inset

 for certain functions 
\begin_inset Formula $f_{1}$
\end_inset

,
 
\begin_inset Formula $f_{2}$
\end_inset

,
 
\begin_inset Formula $f_{3}$
\end_inset

.
 Can 
\begin_inset Formula $\mathtt{LOC(B[I,J,K])}$
\end_inset

 be expressed in a similar manner?
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\backslash
answer 
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\end_inset

Let 
\begin_inset Formula $c$
\end_inset

 be the size of an entry,
 
\begin_inset Formula $a_{0}\coloneqq\mathtt{LOC(A[}0\mathtt{,}0\mathtt{,}0\mathtt{])}$
\end_inset

,
 and 
\begin_inset Formula $x(i,j,k)$
\end_inset

 be the number of positions from 
\begin_inset Formula $a_{0}$
\end_inset

 to 
\begin_inset Formula $\mathtt{LOC(A[}i\mathtt{,}j\mathtt{,}k\mathtt{])}$
\end_inset

,
 so that 
\begin_inset Formula $\mathtt{LOC(A[}i\mathtt{,}j\mathtt{,}k\mathtt{])}=a_{0}+cx(i,j,k)$
\end_inset

,
 then
\begin_inset Formula 
\[
x(i,j,k)=x(i,0,0)+(x(i,j,0)-x(i,0,0))+(x(i,j,k)-x(i,j,0)),
\]

\end_inset

but
\begin_inset Formula 
\begin{align*}
x(i,j,k)-x(i,j,0) & =k,\\
x(i,j,0)-x(i,0,0) & =\sum_{p=0}^{j-1}(x(i,p+1,0)-x(i,p,0))=\sum_{p=0}^{j-1}(p+1)=\sum_{p=1}^{j}p=\frac{j(j+1)}{2},
\end{align*}

\end_inset

and 
\begin_inset Formula $x(i,0,0)$
\end_inset

 is already a function of 
\begin_inset Formula $i$
\end_inset

,
 so in fact locations can be put in this form.
\end_layout

\begin_layout Standard
Now let 
\begin_inset Formula $b_{0}\coloneqq\mathtt{LOC(B[}0\mathtt{,}0\mathtt{,}0\mathtt{])}$
\end_inset

 and 
\begin_inset Formula $y(i,j,k)\coloneqq\frac{\mathtt{LOC(B[}0\mathtt{,}0\mathtt{,}0\mathtt{])}-\mathtt{LOC(B[}i\mathtt{,}j\mathtt{,}k\mathtt{])}}{c}$
\end_inset

,
 the functions,
 if they exist,
 would depend on 
\begin_inset Formula $n$
\end_inset

,
 since 
\begin_inset Formula $y(0,1,1)=y(0,0,n)+1=n+1$
\end_inset

,
 but still
\begin_inset Formula 
\begin{align*}
y(i,j,k) & =y(i,n,n)+(y(i,j,n)-y(i,n,n))+(y(i,j,k)-y(i,j,n))
\end{align*}

\end_inset

with
\begin_inset Formula 
\begin{align*}
y(i,j,k)-y(i,j,n) & =n-k,\\
y(i,j,n)-y(i,n,n) & =\sum_{p=j+1}^{n}(y(i,p-1,n)-y(i,p,n))=\sum_{p=j+1}^{n}(1+y(i,p,p)-y(i,p,n))\\
 & =\sum_{p=j+1}^{n}(n-p+1)=\sum_{p=1}^{n-p}p,
\end{align*}

\end_inset

and 
\begin_inset Formula $y(i,n,n)$
\end_inset

 is already a function of 
\begin_inset Formula $i$
\end_inset

 (and 
\begin_inset Formula $n$
\end_inset

).
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\begin_layout Standard
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\backslash
rexerc12[20]
\end_layout

\end_inset

What are 
\family typewriter
VAL(Q0)
\family default
,
 
\family typewriter
VAL(P0)
\family default
,
 and 
\family typewriter
VAL(P1)
\family default
 at the beginning of step S7,
 in terms of the notation 
\begin_inset Formula $a$
\end_inset

,
 
\begin_inset Formula $b$
\end_inset

,
 
\begin_inset Formula $c$
\end_inset

,
 
\begin_inset Formula $d$
\end_inset

 used in (13)?
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answer 
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\begin_inset Formula $\mathtt{VAL(P1)}=d$
\end_inset

,
 
\begin_inset Formula $\mathtt{VAL(Q0)}=c$
\end_inset

,
 
\begin_inset Formula $\mathtt{VAL(P0)}=b/a$
\end_inset

.
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\backslash
rexerc13[22]
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\end_inset

Why were circular lists used in Fig.
 14 instead of straight linear lists?
 Could Algorithm S be rewritten so that it does not make use of the circular linkage?
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\backslash
answer 
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\end_inset

Circular lists are used as an optimization and simplification:
 since many lists have to be traversed many times,
 it is advantageous to not have to reset the pointer to the list every time as the pointer is already situated at the beginning of the list by the time the previous iteration ends.
\end_layout

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\backslash
rexerc24[25]
\end_layout

\end_inset

(
\emph on
The sparse array trick.
\emph default
) Suppose you want to use a large array for random access,
 although you won't actually be referring to very many of its entries.
 You want 
\begin_inset Formula $\mathtt{A[}k\mathtt{]}$
\end_inset

 to be zero the first time you access it,
 yet you don't want to spend the time to set every location to zero.
 Explain how it is possible to read and write any desired elements 
\begin_inset Formula $\mathtt{A[}k\mathtt{]}$
\end_inset

 reliably,
 given 
\begin_inset Formula $k$
\end_inset

,
 without assuming anything about the actual initial memory contents,
 by doing only a small fixed number of additional operations per array access.
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answer 
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\end_inset

One could have an auxiliary array 
\family typewriter
L
\family default
 containing pairs of index and value,
 and have 
\family typewriter
A
\family default
 contain pointers to 
\family typewriter
L
\family default
.
 If the pointer is within bounds and 
\family typewriter
L
\family default
 contains the index 
\begin_inset Formula $k$
\end_inset

 in that position,
 then the value contained is the value of 
\begin_inset Formula $\mathtt{A[}k\mathtt{]}$
\end_inset

,
 otherwise the value is 0.
 
\family typewriter
L
\family default
 could actually be the size of 
\family typewriter
A
\family default
 but have an upper limit 
\begin_inset Formula $n$
\end_inset

 with the number of elements actually allocated,
 which is considered the upper bound and which increases whenever a value that has not been yet allocated is written to.
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