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\begin_body

\begin_layout Standard
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\backslash
rexerc3[20]
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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?
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answer 
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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

.
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\backslash
rexerc5[21]
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Design an algorithm that takes a circular list such as (1) and reverses the direction of all the arrows.
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answer 
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This is similar to exercise 2.2.3.7.
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\begin_layout Enumerate
Set 
\begin_inset Formula $\mathtt{P}\gets\Lambda$
\end_inset

 and 
\begin_inset Formula $\mathtt{C}\gets\mathtt{PTR}$
\end_inset

.
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\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.
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\backslash
exerc7[10]
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Why is it useful to assume that the 
\family typewriter
ABC
\family default
 fields of a polynomial list appear in decreasing order?
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answer 
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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.
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\backslash
rexerc8[10]
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Why is it useful to have 
\family typewriter
Q1
\family default
 trailing one step behind 
\family typewriter
Q
\family default
 in Algorithm A?
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answer 
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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
.
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\backslash
rexerc9[23]
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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

?
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answer 
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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.
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\backslash
rexerc10[20]
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\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?
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answer 
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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.
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\backslash
rexerc17[22]
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\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?
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answer 
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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.
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\backslash
rexerc18[25]
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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.
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answer 
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\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.
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