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\input acphdr % Answer pages (double-check position of figures)
\open0=v2ans3.inx
\chpar19←1 % this will prepare a ".JST" file containing justification statistics
\runninglefthead{ANSWERS TO EXERCISES}
\titlepage\setcount00
\null
\vfill
\tenpoint
\ctrline{ANSWER PAGES for THE ART OF COMPUTER PROGRAMMING}
\ctrline{(Volume 2)}
\ctrline{(third half of the answers)}
\ctrline{$\copyright$ 1980 Addison--Wesley Publishing Company, Inc.}
\vfill
\ninepoint
\runningrighthead{ANSWERS TO EXERCISES}
\section{4.5.3}
\eject
\setcount0 608
% Page 608
% moved over from V2ANS2 because of page break:
\def\sim{\mathrel{\char'430}}
\ansno 42. Suppose that $\|qX\|=|qX-p|$. We can always
find integers $u$ and $v$ such that
$q=uq↓{n-1}+vq↓n$ and $p=up↓{n-1}+vp↓n$, where $p↓n=Q↓{n-1}(A↓2,\ldotss,A↓n)$,
since $q↓np↓{n-1}-q↓{n-1}p↓n=\pm1$. We must have $uv<0$, hence $u(q↓{n-1}X-p↓{n-1})$
has the same sign as $v(q↓nX-p↓n)$, and $|qX-p|=|u|\,|q↓{n-1}X-p↓{n-1}|+
|v|\,|q↓nX-p↓n|$. This completes the proof, since $u≠0$. See Theorem↔6.4S for
a generalization.

\ansno 43. If $x$ is representable, so is the father of $x$ in the \α{Stern--Peirce
tree} of exercise 40; thus the representable numbers form a subtree of that binary
tree. Let $(u/u↑\prime)$ and $(v/v↑\prime)$ be adjacent representable numbers. Then
one is an ancestor of the other; say $(u/u↑\prime)$ is an ancestor of $(v/v↑\prime)$,
since the other case is similar. Then $(u/u↑\prime)$ is the nearest left
ancestor of $(v/v↑\prime)$, so all numbers between $u/u↑\prime$ and $v/v↑\prime$
are left descendants of $(v/v↑\prime)$ and the mediant
$\biglp(u+v)/(u↑\prime+v↑\prime)\bigrp$ is its left son. According to the
relation between regular continued fractions and the binary tree, the mediant
and all of its left descendants will have $(u/u↑\prime)$ as their last
representable $p↓i/q↓i$, while all of the mediant's right descendants will
have $(v/v↑\prime)$ as one of the $p↓i/q↓i$.\xskip(The numbers $p↓i/q↓i$ label
the {\sl fathers} of the `turning-point' nodes on the path to↔$x$.)

\ansno 44. A counterexample for $M=N=100$ is $(u/u↑\prime)={1\over3}$, $(v/v↑\prime)=
{67\over99}$. However, the identity is almost always true, because of (12);
it fails only when $u/u↑\prime+v/v↑\prime$ is very nearly equal to a fraction
that is simpler than $(u/u↑\prime)$.

\ansno 45. See M. S. \α{Waterman}, {\sl BIT \bf17} (1977), 465--478.
% Vicinity of page 608
% folio 790 galley 10 	*
\ansbegin{4.5.4}

\ansno 1. If $d↓k$ isn't prime, its
prime factors are cast out before $d↓k$ is tried.

\ansno 2. No; the algorithm would fail if $p↓{t-1} = p↓t$, giving
``1'' as a spurious prime factor.

\ansno 3. Let $P$ be the product of the first 168 primes.\xskip [{\sl
Note:} Although $P=19590\ldotsm5910$ is a 416-digit number, such a gcd can
be computed in much less time than it would take to do 
168 divisions, if we just want to test whether or not $n$ is prime.]

\ansno 4. In the notation of exercise 3.1--11,$$\sum↓{\mu,\,λ}2↑{\lceil\lg
\max(\mu+1,λ)\rceil}P(\mu,λ)={1\over m}\sum↓{l≥1}f(l)\prod↓{1≤k<l}
\bigglp1-{k\over m}\biggrp,$$
where $f(l)=\sum↓{1≤λ≤l}2↑{\lceil\lg\max(l-λ,λ)\rceil}$.
If $l=2↑{k+\theta}$, where $0<\theta≤1$, we have $f(l)=l↑2(3\cdot2↑{-\theta}-2\cdot
2↑{-2\theta})$, where the function $3\cdot2↑{-\theta}-2\cdot2↑{-2\theta}$
reaches a maximum of $9\over8$ at $\theta=\lg(4/3)$ and has a minimum of 1 at
$\theta=0$ and 1. Therefore the average value of $2↑{\lceil\lg\max(\mu+1,λ)\rceil}$
lies between 1.0 and 1.125 times the average value of $\mu+λ$, and the result
follows.

[Algorithm↔B is a refinement of \α{Pollard}'s original algorithm, which was based
on exercise 3.1--6(b) instead of the (yet undiscovered) result in exercise 3.1--7.
He showed that the least $n$ such that $X↓{2n}=X↓n$ has average value $\sim (π↑2/12)
Q(m)$; this constant $π↑2/12$ is explained by Eq.\ 4.5.3--21. Hence the average
value of $3n$ in his original method is $\sim (π/2)↑{5/2}\sqrt m=3.092\sqrt m$.\xskip
Richard \α{Brent} has observed that, as $m→∞$, the density $\prod↓{1≤k<l}\biglp1-k/m
\bigrp=\exp\biglp-l(l-1)/2m+O(l↑3/m↑2)\bigrp$ approaches a normal distribution,
and we may assume that $\theta$ is uniformly distributed. Then ${3{\cdot}2↑{-\theta}
-2{\cdot}2↑{-2\theta}}$ takes the average value $3/(4\ln2)$, and the average
number of iterations needed by Algorithm↔B comes to $\sim \biglp3/(4\ln2)+{1\over2}
\bigrp\sqrt{πm/2}=1.983\sqrt m$. A similar analysis of the more general method
in the answer to exercise 3.1--7 gives $\sim 1.926\sqrt m$, when $p=2.4771$ is
chosen ``optimally'' as the root of $(p↑2-1)\ln p=p↑2-p+1$.]

\ansno 5. $x\mod 3 = 0$; $x \mod 5 = 0$,
1, 4; $x \mod 7 = 0$, 1, 6; $x \mod 8 = 1$, 3, 5, 7; $x > 103$.
The first try is $x = 105$; and $(105)↑2 - 10541 = 484 = 22↑2$. This
would also have been found by Algorithm↔C in a relatively short
time. Thus $10541 = 83 \cdot 127$.

\ansno 6. Let us count the number of solutions
$(x, y)$ of the congruence $N ≡ (x - y)(x + y)\modulo p$,
where $0 ≤ x, y < p$. Since $N \neqv 0$ and $p$ is prime,
$x + y \neqv 0$. For each $v \neqv 0$ there is a unique
$u$ (modulo $p$) such that $N ≡ uv$. The congruences $x - y
≡ u$, $x + y ≡ v$ now uniquely determine $x\mod p$ and $y\mod
p$, since $p$ is odd. Thus the stated congruence has exactly
$p - 1$ solutions $(x, y)$. If $(x, y)$ is a solution, so is
$(x, p - y)$ if $y ≠ 0$, since $(p - y)↑2 ≡ y↑2$; and if $(x,
y↓1)$ and $(x, y↓2)$ are solutions with $y↓1 ≠ y↓2$, we have
$y↑{2}↓{1} ≡ y↑{2}↓{2}$; hence $y↓1 = p - y↓2$. Thus the number
of different $x$ values among the solutions $(x, y)$ is $(p
- 1)/2$ if $N ≡ x↑2$ has no solutions, or $(p + 1)/2$ if $N
≡ x↑2$ has solutions.

\ansno 7. One procedure is to keep two indices for each modulus,
one for the current word position and one for the current bit
position; loading two words of the table and doing an indexed
shift command will bring the table entries into proper alignment.\xskip
(Many computers have special facilities for such bit manipulation.)

\ansno 8. (We may assume that $N = 2M$
is even.)\xskip The following algorithm uses an auxiliary table $X[1]$,
$X[2]$, $\ldotss$, $X[M]$, where $X[k]$ represents the primality
of $2k + 1$.

\def\\#1. {\yskip\noindent\hangindent 38pt\hbox to 38pt{\hfill\bf#1. }}
\\S1. Set $X[k] ← 1$
for $1 ≤ k ≤ M$. Also set $j ← 1$, $p ← 1$, $p ← 3$, $q ← 4$.\xskip (During
this algorithm $p = 2j + 1$, $q = 2j + 2j↑2$; the integer variables
$j$, $k$, $p$, $q$ may readily be manipulated in index registers.)

\\S2. If $X[j] = 0$, go to S4.
Otherwise output $p$, which is prime, and set $k ← q$.

\\S3. If $k ≤ M$, then set $X[k] ← 0$, $k ← k + p$, and repeat this step.

\\S4. Set $j ← j + 1$, $p ← p + 2$, $q
← q + 2p - 2$. If $j ≤ M$, return to S2.\quad\blackslug

\yskip\noindent A major part of this calculation
could be made noticeably faster if $q$ (instead of $j$) were
tested against $M$ in step S4, and if a new loop were appended
that outputs $2j + 1$ for all remaining $X[j]$ that equal
1, suppressing the manipulation of $p$ and $q$.

Improvements in the efficiency of sieve methods for generating primes are discussed in
exercise 5.2.3--15 and in Section 7.1.

\ansno 9. If $p↑2$ is a divisor of $n$ for some
prime $p$, then $p$ is a divisor of $λ(n)$, but not of $n -
1$. If $n = p↓1p↓2$, where $p↓1 < p↓2$ are primes, then $p↓2
- 1$ is a divisor of $λ(n)$ and therefore $p↓1p↓2 - 1 ≡ 0
\modulo{p↓2 - 1}$. Since $p↓2 ≡ 1$, this means $p↓1 - 1$ is a multiple
of $p↓2 - 1$, contradicting the assumption $p↓1 < p↓2$.\xskip [Values
of $n$ for which $λ(n)$ properly divides $n - 1$ are called
``\α{Carmichael numbers}.'' For example, here are some small Carmichael numbers with
up to six prime factors: $3\cdot11\cdot17$, $5\cdot13\cdot17$, $7\cdot11\cdot13
\cdot41$, $5\cdot7\cdot17\cdot19\cdot73$, $5\cdot7\cdot17\cdot73\cdot89\cdot107
$.]

 \ansno 10. Let $k↓p$ be the order of $x↓p$ modulo $n$, and let
$λ$ be the least common multiple of all the $k↓p$'s. Then $λ$
is a divisor of $n - 1$ but not of any $(n - 1)/p$, so $λ =
n - 1$. Since $x↑{\varphi (n)}↓{p} \mod n = 1$, $\varphi (n)$
is a multiple of $k↓p$ for all $p$, so $\varphi (n) ≥ λ$. But
$\varphi (n) < n - 1$ when $n$ is not prime.\xskip (Another way to
carry out the proof is to construct an element $x$ of order
$n - 1$ from the $x↓p$'s, by the method of exercise↔\hbox{3.2.1.2--15}.)

\anskip\null\penalty1000\vskip-6pt
\vbox{\halign to size{\hbox to 19pt{\hfill\bf#}\tabskip0pt plus100pt⊗
\rt{#}\tabskip10pt⊗\rt{#}⊗\rt{#}⊗\rt{#}⊗\rt{#}⊗\rt{#}⊗$\rt{#}$\tabskip 0pt plus100pt
\cr
11. ⊗$U$\hfill⊗$V$\hfill⊗$A$\hfill⊗$P$\hfill⊗$S$⊗$T$⊗\hbox{Output}\cr
\noalign{\vskip2pt}
⊗1984⊗1⊗0⊗992⊗0⊗---\cr
⊗1981⊗1981⊗1⊗992⊗1⊗1981\cr
⊗1983⊗4⊗495⊗993⊗0⊗1⊗993↑2 ≡+2↑2 \cr
⊗1983⊗991⊗2⊗98109⊗1⊗991\cr
⊗1981⊗4⊗495⊗2⊗0⊗1⊗2↑2 ≡+2↑2 \cr
⊗1984⊗1981⊗1⊗99099⊗1⊗1981\cr
⊗1984⊗1⊗1984⊗99101⊗0⊗1⊗99101↑2 ≡+2↑0 \cr}}
\yskip\noindent The factorization $199 \cdot 991$ is evident from
the first or last outputs. The shortness of the cycle, and the
appearance of the well-known number 1984, are probably just
coincidences.

\ansno 12. The following algorithm makes use of an auxiliary
$(m + 1) \times (m + 1)$ matrix of single-precision integers
$E↓{jk}$, $0 ≤ j, k ≤ m$; a single-precision vector $(b↓0, b↓1,
\ldotss , b↓m)$; and a multiple-precision vector $(x↓0, x↓1,
\ldotss , x↓m)$ with entries in the range $0 ≤ x↓k < N$.

\algstep F1. [Initialize.] Set
$b↓i ← -1$ for $0 ≤ i ≤ m$; then set $j ← 0$.

\algstep F2. [Next solution.] Get the next output
$(x, e↓0, e↓1, \ldotss , e↓m)$ produced by Algorithm↔E\null.\xskip (It is
convenient to regard Algorithms E and↔F as coroutines.)\xskip Set
$k ← 0$.

\algstep F3. [Search for odd.] If $k > m$ go to step
F5. Otherwise if $e↓k$ is even, set $k ← k + 1$ and repeat this
step.

\algstep F4. [Linear dependence?] If $b↓k ≥ 0$, then
set $i ← b↓k$, $x ← (x↓ix)\mod N$, $e↓r ← e↓r + E↓{ir}$ for $0
≤ r ≤ m$; set $k ← k + 1$ and return to F3. Otherwise set $b↓k
← j$, $x↓j ← x$, $E↓{jr} ← e↓r$ for $0 ≤ r ≤ m$; set $j ← j + 1$
and return to F2.\xskip (In the latter case we have a new linearly
independent solution, modulo 2, whose first odd component is
$e↓k$.)

\algstep F5. [Try to factor.] (Now $e↓0$, $e↓1$, $\ldotss$,
$e↓m$ are even.)\xskip Set
$$y ← \biglp (-1) ↑{e↓0/2}p↑{e↓1/2}↓{1} \ldotss p↑{e↓m/2}↓{m}\bigrp
\mod N.$$
If $x = y$ or if $x + y = N$, return
to F2. Otherwise compute $\gcd(x - y, N)$, which is a proper
factor of $N$, and terminate the algorithm.\quad\blackslug

\yyskip\noindent It can be shown that this
algorithm finds a factor, whenever it is possible to deduce one from the
given outputs of Algorithm↔E.\xskip[{\sl Proof:} Let the outputs of
Algorithm↔E be $(X↓i,E↓{i0},\ldotss,E↓{im})$ for $1≤i≤t$, and suppose that we
could find a factorization $N=N↓1N↓2$ when $x≡X↓1↑{a↓1}\ldotss X↓t↑{a↓t}$ and
$y≡(-1)↑{e↓0/2}p↓1↑{e↓1/2}\ldotss p↓m↑{e↓m/2}\modulo N$, where $e↓j=
a↓1E↓{1j}+\cdots+a↓tE↓{tj}$ is even for all $j$. Then $x≡\pm y\modulo{N↓1}$
and $x≡\mp y\modulo{N↓2}$. It is not difficult to see that this solution can
be transformed into a pair $(x,y)$ that appears in step↔F5, by a series of
steps that systematically replace $(x,y)$ by $(xx↑\prime,yy↑\prime)$ where $x↑\prime≡
\pm y↑\prime\modulo N$.]

\ansno 13. There are $2↑d$ values of $x$ having the same exponents $(e↓0,\ldotss,
e↓m)$, since we can choose the sign of $x$ modulo↔$q↓i↑{f↓i}$ arbitrarily when
$N=q↓1↑{f↓1}\ldotss q↓d↑{f↓d}$. Exactly two of these $2↑d$ values will fail to
yield a factor.

\ansno 14. Since $P↑2 ≡ kNQ↑2\modulo p$ for any prime divisor
$p$ of $V$, we have $1 ≡ P↑{2(p-1)/2} ≡ (kNQ↑2)↑{(p-1)/2} ≡
(kN)↑{(p-1)/2}\modulo p$, if $P \neqv 0$.
% Vicinity of page 610
% folio 794 galley 11a 	*
\ansno 15. $U↓n = (a↑n - b↑n)/\sqrt{D}$, where $a = {1\over
2}(P + \sqrt{D})$, $b = {1\over 2}(P - \sqrt{D})$, $D = P↑2 - 4Q$.
Then $2↑{n-1}U↓n = \sum ↓k{n\choose 2k+1}P↑{n-2k-1}D↑k$; so
$U↓p ≡ D↑{(p-1)/2}\modulo p$ if $p$ is an odd prime. Similarly,
if $V↓n = a↑n + b↑n = U↓{n+1} - QU↓{n-1}$, then $2↑{n-1}V↓n
= \sum ↓k {n\choose 2k}P↑{n-2k}D↑k$, and $V↓p ≡ P↑n ≡ P$. Thus
if $U↓p ≡ -1$, we find that $U↓{p+1}\mod p = 0$. If $U↓p ≡
1$, we find that $(QU↓{p-1})\mod p = 0$; here if $Q$ is a multiple
of $p$, $U↓n ≡ P↑{n-1}\modulo p$ for $n > 0$, so $U↓n$ is
never a multiple of $p$; if $Q$ is not a multiple of $p$, $U↓{p-1}
\mod p = 0$. Therefore as in Theorem↔L\null, $U↓t\mod N = 0$ if
$N = p↑{e↓1}↓{1} \ldotss p↑{e↓r}↓{r}$, $\gcd(N, Q) = 1$, and $t
=\lcm↓{1≤j≤r}\biglp p↑{e↓j-1}↓{j}(p↓j + ε↓j)\bigrp $. Under
the assumptions of this exercise, the rank of apparition of
$N$ is $N + 1$; hence $N$ is prime to $Q$ and $t$ is a multiple
of $N + 1$. Also, the assumptions of this exercise imply that
each $p↓j$ is odd and each $ε↓j$ is $\pm 1$, so $t ≤ 2↑{1-r}\prod p↑{e↓j-1}↓{j}(p↓j
+ {1\over 3}p↓j) = 2({2\over 3})↑rN$; hence $r = 1$ and $t =
p↑{e↓1}↓{1} + ε↓1p↑{e↓1-1}↓{1}$. Finally, therefore, $e↓1 = 1$
and $ε↓1 = 1$.

{\sl Note:} If this test for primality
is to be any good, we must choose $P$ and $Q$ in such a way that
the test will probably work. Lehmer suggests taking
$P = 1$ so that $D = 1 - 4Q$, and choosing $Q$ so that $\gcd(N,
QD) = 1$.\xskip (If the latter condition fails, we know already that
$N$ is not prime, unless $|QD| ≥ N$.)\xskip Furthermore, the derivation
above shows that we will want $ε↓1 = 1$, that is, $D↑{(N-1)/2}
≡ -1\modulo N$. This is another condition that determines
the choice of $Q$. Furthermore, if $D$ satisfies this condition,
and if $U↓{N+1}\mod N ≠ 0$, we know that $N$ is {\sl not}
prime.

{\sl Example:} If $P = 1$ and $Q = -1$, we have
the \α{Fibonacci sequence}, with $D = 5$. Since $5↑{11} ≡ -1 \modulo
{23}$, we might attempt to prove that 23 is prime by using the
Fibonacci sequence:
$$\langle F↓n\mod 23\rangle = 0, 1, 1, 2, 3, 5, 8, 13, 21,
11, 9, 20, 6, 3, 9, 12, 21, 10, 8, 18, 3, 21, 1, 22, 0,\ldotss,$$
so 24 is the rank of apparition of 23
and the test works. However, the Fibonacci sequence cannot be
used in this way to prove the primality of 13 or 17, since $F↓7
\mod 13 = 0$ and $F↓9\mod 17 = 0$. When $p ≡ \pm 1\modulo{10}$,
we have $5↑{(p-1)/2}\mod p = 1$, so $F↓{p-1}$ (not $F↓{p+1}$)
is divisible by $p$.

\ansno 17. Let $f(q) = 2\lg q - 1$. When $q = 2$ or 3, the
tree has at most $f(q)$ nodes. When $q > 3$ is prime, let $q
= 1 + q↓1 \ldotsm q↓t$ where $t ≥ 2$ and $q↓1$, $\ldotss$, $q↓t$
are prime. The size of the tree is $≤1 + \sum f(q↓k) = 2 + f(q
- 1) - t < f(q)$.\xskip [{\sl SIAM J. Computing \bf 7} (1975), \hbox{214--220}.]

\ansno 18. $x\biglp G(α) - F(α)\bigrp$ is the number of $n ≤
x$ whose second-largest prime factor is $≤x↑α$ and whose largest
prime factor is $>x↑α$. Hence$$xG↑\prime (t)\,dt = \biglp π(x↑{t+dt})
- π(x↑t)\bigrp\cdot x↑{1-t}\biglp G\biglp t/(1 - t)\bigrp - F\biglp t/(1
- t)\bigrp\bigrp.$$ The probability that $p↓{t-1} ≤ \sqrt{\chop to 0pt{p↓t}}$ is
$\int ↑{1}↓{0} F\biglp t/2(1 - t)\bigrp t↑{-1}\,dt$.\xskip
[Curiously, it can be shown that this
also equals $\int ↑{1}↓{0} F\biglp t/(1 - t)\bigrp\,dt$, the
average value of $\log p↓t/\!\log x$, and it also equals \α{Golomb's constant} $λ$
of exercises 1.3.3--23 and 3.1--13. The derivative $G↑\prime
(0)$ can be shown to equal $\int ↑{1}↓{0} F\biglp t/(1 - t)\bigrp
t↑{-2}\,dt = F(1) + 2F({1\over 2}) + 3F({1\over 3}) +\cdots
= e↑\gamma$. The third-largest prime factor has $H(α
) = \int ↑{α}↓{0} \biglp H\biglp t/(1 - t)\bigrp - G\biglp t/(1 - t)\bigrp\bigrp
t↑{-1}\,dt$ and $H↑\prime(0) = ∞$. See P. \α{Billingsley}, {\sl Period.\ Math.\
Hungar.\ \bf2} (1972), 283--289; J. \α{Galambos}, {\sl Acta Arith.\ \bf31} (1976),
213--218; D. E. \α{Knuth} and L. \α{Trabb Pardo},
{\sl Theoretical Comp.\ Sci.\ \bf3} (1976), 321--348.]

\ansno 19. $M = 2↑D - 1$ is a multiple of all $p$ for which
the order of 2 modulo $p$ divides $D$. To extend this idea, let $a↓1 = 2$
and $a↓{j+1}
= a↑{q↓j}↓{j}\mod N$, where $q↓j = p↑{e↓j}↓{j}$, $p↓j$ is the
$j$th prime, and $e↓j = \lfloor \log 1000/\!\log p↓j\rfloor $;
let $A = a↓{169}$. Now compute $b↓q =\gcd(A↑q - 1, N)$ for
all primes $q$ between $10↑3$ and $10↑5$. One way to do this is
to start with $A↑{1009}\mod N$ and then to multiply alternately
\inx{MIX (actually 1009)}
by $A↑4\mod N$ and $A↑2\mod N$.\xskip (A similar method was used
in the 1920s by D. N. Lehmer, but he didn't publish it.)\xskip As
\inx{Lehmer, D. N.}
with Algorithm↔B we can avoid most of the gcd's by batching;
e.g., since $b↓{30r-k} =\gcd(A↑{30r} - A↑k, N)$, we might try
batches of\penalty1000\ 8, computing $c↓r = (A↑{30r} - A↑{29})(A↑{30r} -
A↑{23}) \ldotsm (A↑{30r} - A)\mod N$, then $\gcd(c↓r, N)$ for
$33 < r ≤ 3334$.

\ansno 22. Algorithm P fails only when the random
number $x$ does not reveal the fact that $n$ is nonprime. 
Say $x$ is {\sl bad\/} if $x↑q\mod n=1$ or
if one of the numbers $x↑{2↑jq}$ is $≡-1\modulo n$ for $0≤j<k$. Since 1 is bad,
we have $p↓n=(b↓n-1)/(n-2)<b↓n/(n-1)$, where $b↓n$ is the number of bad $x$
such that $1≤x<n$, when $n$ is not prime.

Every bad $x$ satisfies $x↑{n-1}≡1\modulo n$. When $p$ is prime, the number of
solutions to the congruence $x↑q≡1\modulo{p↑e}$ for $1≤x≤p↑e$ is the number of
solutions of $qy≡0$ $\biglp$\hbox{modulo}\penalty1000\ $p↑{e-1}(p-1)\bigrp$
for $0≤y<p↑{e-1}(p-1)$, namely $\gcd\biglp q,p↑{e-1}(p-1)\bigrp$, since we
may replace $x$ by $a↑y$ where $a$ is a primitive root.

Let $n=n↓1↑{e↓1}\ldotss n↓r↑{e↓r}$, where the $n↓i$ are distinct primes. According
to the
Chinese remainder theorem, the number of solutions to the congruence
$x↑{n-1}≡1\modulo n$ is
$\prod↓{1≤i≤r}\gcd\biglp n-1,n↓i↑{e↓i-1}(n↓i-1)\bigrp$, and this is at most
$\prod↓{1≤i≤r}(n↓i-1)$ since $n↓i$ is relatively prime to $n-1$. If
some $e↓i>1$, we have $n↓i-1≤{2\over9}n↓i↑{e↓i}$, hence the number of solutions
is at most ${2\over9}n$; in this case $b↓n≤{2\over9}n≤{1\over4}(n-1)$, since
$n≥9$.

Therefore we may assume that $n$ is the product $n↓1\ldotsm n↓r$ of distinct
primes. Let $n↓i=1+2↑{k↓i}q↓i$, where $k↓1≤\cdots≤k↓r$. Then $\gcd(n-1,n↓i-1)
=2↑{k↑\prime↓i}q↑\prime↓i$, where $k↑\prime↓i=\min(k,k↓i)$ and $q↑\prime↓i=
\gcd(q,q↓i)$. Modulo $n↓i$, the number of $x$ such that $x↑q≡1$ is $q↑\prime↓i$;
and the number of $x$ such that $x↑{2↑jq}≡-1$ is $2↑jq↓i↑\prime$ for
$0≤j<k↓i↑\prime$, otherwise 0. Since $k≥k↓1$, we have $b↓n=q↓1↑\prime\ldotsm
q↓r↑\prime\,\biglp1+\sum↓{0≤j<k↓1}2↑{jr}\bigrp$.

To complete the proof, it suffices to show that $b↓n≤{1\over4}q↓1\ldotsm q↓r
2↑{k↓1+\cdots+k↓r}={1\over4}\varphi(n)$, since $\varphi(n)<n-1$. We have
$$\eqalign{\textstyle\biglp1+\sum↓{0≤j<k↓1}2↑{jr}\bigrp\hbox{\:a/}2↑{k↓1+\cdots+k↓r}
⊗\textstyle≤
\biglp1+\sum↓{0≤j<k↓1}2↑{jr}\bigrp\hbox{\:a/}2↑{k↓1r}\cr⊗=1/(2↑r-1)+
(2↑r-2)\hbox{\:a/}\biglp2↑{k↓1r}(2↑r-1)\bigrp≤1/2↑{r-1},\cr}$$
so the result follows
unless $r=2$ and $k↓1=k↓2$. If $r=2$, exercise↔9 shows that $n-1$ is not a
multiple of both $n↓1-1$ and $n↓2-1$. Thus if $k↓1=k↓2$ we cannot have both
$q↓1↑\prime=q↓1$ and $q↓2↑\prime=q↓2$; it follows that $q↓1↑\prime q↓2↑\prime
≤{1\over3}q↓1q↓2$ and $b↓n≤{1\over6}\varphi(n)$ in this case.

\yyskip [{\sl Reference: J. Number Theory} (1980), to appear.
The above proof shows that $p↓n$ is near $1\over 4$ in only two cases, when $n=
(1+2q↓1)(1+4q↓1)$ or $(1+2q↓1)(1+2q↓2)(1+2q↓3)$. For example, when $n=
49939\cdot99877$ we have $b↓n={1\over4}(49938\cdot99876)$ and $p↓n\approx
.2499925$. See the next answer for further remarks.]

\ansno 23. (a)\9 The proofs are simple except perhaps for the reciprocity law.
Let $p=p↓1\ldotsm p↓s$ and $q=q↓1\ldotsm q↓r$, where the $p↓i$ and $q↓j$ are
prime. Then
$$\bigglp{p\over q}\biggrp=\prod↓{i,j}\bigglp{p↓i\over q↓j}\biggrp
=\prod↓{i,j}(-1)↑{(p↓i-1)(q↓j-1)/4}\,\bigglp{q↓j\over p↓i}\biggrp=
(-1)↑{\vcenter{\hbox{\:b\char6}}↓{i,j}
(p↓i-1)(q↓j-1)/4}\,\bigglp{q\over p}\biggrp,$$
so we need only verify that $\sum↓{i,j\,}(p↓i-1)(q↓j-1)/4≡(p-1)(q-1)/4\modulo 2$.
But $\sum↓{i,j\,}(p↓i-1)(q↓j-1)/4=\biglp\sum↓i(p↓i-1)/2\bigrp\biglp\sum↓j
(q↓j-1)/2\bigrp$ is odd iff an odd number of the $p↓i$ and an odd number of the
$q↓j$ are $≡3\modulo 4$, and this holds iff $(p-1)(q-1)/4$ is odd.

\def\\#1{\raise 2pt\hbox{$\scriptstyle#1$}}
(b) As in exercise↔22, we may assume that $n=n↓1\ldotsm n↓r$ where the $n↓i=1+
2↑{k↓i}q↓i$ are distinct primes, and $k↓1≤\cdots≤k↓r$;
we let $\gcd(n-1,n↓i-1)=2↑{k↑\prime↓i}q↑\prime↓i$ and we call
$x$ {\sl bad} if it falsely makes $n$ look prime. Let $\Pi↓n=\prod↓{1≤i≤r}
q↓i↑\prime\,2↑{\min(k↓i,k-1)}$ be the number of solutions of $x↑{(n-1)/2}≡1$.
The number of bad $x$ with $({\\x\over n})=1$ is $\Pi↓n$, times an extra factor
of $1\over2$ if $k↓1<k$.\xskip (This factor $1\over2$ is needed to
ensure that $({\\x\over n↓i})=-1$ 
for an even number of the $n↓i$ with $k↓i<k$.)\xskip
The number of bad $x$ with $({\\x\over n})=-1$ is $\Pi↓n$ if $k↓1
=k$, otherwise 0.\xskip[If $x↑{(n-1)/2}≡-1\modulo{n↓i}$, we have $({\\x\over
n↓i})=-1$ if $k↓i=k$, $({\\x\over n↓i})=+1$ if $k↓i>k$, and a contradiction if
$k↓i<k$. If $k↓1=k$, there are an odd number of $k↓i$ equal to
$k$.]

{\sl Notes:} The probability of a 
bad guess is $>{1\over4}$ only if $n$ is a \α{Carmichael
number} with $k↓r<k$; for example, $n=7\cdot13\cdot19=1729$,
a number made famous by \α{R\=am\=anujan} in another context. 
Louis \α{Monier} has extended the above analyses to obtain the following closed
formulas for the number of bad $x$ in general:
$$\eqalign{b↓n⊗=\bigglp1+{2↑{rk↓1}-1\over2↑r-1}\biggrp\prod↓{1≤i≤r}q↓i↑\prime;\cr
b↓n↑\prime⊗=\delta↓n\prod↓{1≤i≤r}\gcd\left({n-1\over2},\;n↓i-1\right).\cr}$$
Here $b↓n↑\prime$ is the number of bad $x$ in this exercise, and $\delta↓n$ is
either 2 (if $k↓1=k$), or $1\over2$ (if $k↓i<k$ and $e↓i$ is odd for some $i$), or
1 (otherwise).

(c) If $x↑q\mod n=1$, then $1=({\\x↑q\over n})=({\\x\over n})↑q=({\\x\over n})$.
If $x↑{2↑jq}≡-1\modulo n$, then the order of $x$ modulo $n↓i$ must be an odd multiple
of $2↑{j+1}$ for all prime divisors $n↓i$ of↔$n$. Let $n=n↓1↑{e↓1}\ldotss n↓r↑{e↓r}$
and $n↓i=1+2↑{j+1}q↓i↑{\prime\prime}$; then $({\\x\over n↓i})=(-1)↑{q↓i↑{\prime
\prime}}$, so $({\\x\over n})=+1$ or $-1$ according as $\sum e↓iq↓i↑{\prime\prime}$
is even or odd. Since $n≡\biglp1+2↑{j+1}\sum e↓iq↓i↑{\prime\prime}\bigrp
\modulo{2↑{j+2}}$, the sum $\sum e↓iq↓i↑{\prime\prime}$ is odd iff $j+1=k$.\xskip
[Univ.\ de Paris-Sud, Lab.\ Rech.\ Informatique, Rapport 20 (1978).]

\ansno 24. Let $M↓1$ be a matrix having one row for each nonprime odd number in the
range $1≤n≤N$
and having $N-1$ columns numbered from 2 to $N$; 
the entry in row↔$n$ column $x$ is 1 if $n$ passes the $x$ test of
Algorithm↔P\null, otherwise it is zero. When $N=qn+r$ and $0≤r<n$, we 
know that row $n$ contains at most ${1\over4}qn+\min\biglp{1\over4}n,r\bigrp=
{1\over4}N+\min\biglp{1\over4}n-{1\over4}r,{3\over4}r\bigrp≤{1\over4}N+{3\over16}n
<{1\over2}N$ entries equal to 0, so at least
half of the entries in the matrix are 1. Thus, some column $x↓1$ of $M↓1$ has
at least half of its entries equal to 1. Removing column $x↓1$ and all rows in
which this column contains 1 leaves a matrix $M↓2$ having similar properties; a
repetition of this construction produces matrix $M↓r$ with $N-r$ columns and
fewer than $N↓{\null}/2↑r$ rows, and with
at least ${1\over2}(N-1)$ entries per row equal to 1.\xskip
[Cf.\ {\sl Proc.\ IEEE Symp.\
Foundations of Comp.\ Sci.\ \bf19} (1978), 78.]

[A similar proof implies the existence of a {\sl single} infinite
sequence $x↓1<x↓2<\cdots$ such that the number $n>1$ is prime if and only if it
passes the $x$ test of Algorithm↔P for $x=x↓1$, $\ldotss$, $x=x↓m$, where $m=
{1\over2}\lfloor\lg n\rfloor\biglp\lfloor\lg n\rfloor-1\bigrp$. Does there exist a
sequence $x↓1<x↓2<\cdots$ having this property but with $m=O(\log n)$?]

\ansno 25. This theorem was first proved rigorously by \α{von Mangoldt}
[{\sl J. f\"ur die reine und angew.\ Math.\ \bf114} (1895), 255--305], who
showed in fact that the $O(1)$ term is equal to ${1\over2}\hbox{($x$ is prime)}+
\hbox{constant}+\int↓x↑∞ dt/(t↑2-1)t\ln t$. The constant is 
\inx{logarithmic integral}
$\mathop{\hbox{li}}2-\ln2=\gamma+\ln\ln2+\sum↓{n≥2}(\ln2)↑n/nn!=
0.35201$ 65995 57547 47542 73567 67736 43656 84471$+$.

\ansno 26. If $N$ is not prime, it has a prime factor $q≤\sqrt N$. By hypothesis,
every prime divisor $p$ of $f$ has an integer $x↓p$ such that the order of
$x↓p$ modulo $q$ is a divisor of $N-1$ but not of $(N-1)/p$. Therefore if
$p↑k$ divides $f$, the order of $x↓p$ modulo $q$ is a multiple of↔$p↑k$.
Exercise 3.2.1.2--15 now tells us that there is an element $x$ of order
$f$ modulo↔$q$. But this is impossible, since it implies that $q↑2≥(f+1)↑2≥
(f+1)\,r≥N$, and equality cannot hold.\xskip[{\sl AMM \bf64} (1957), 703--710.]

\ansno 27. If $k$ is not divisible by 3 and if $k≤2↑n+1$, the number $k{\cdot}2↑n
+1$ is prime iff $3↑{2↑{n-1}k}≡-1\modulo{k{\cdot}2↑n+1}$. For if this condition
holds, $k{\cdot}2↑n+1$ is prime by exercise↔26; and if $k{\cdot}2↑n+1$ is prime,
the number 3 is a quadratic nonresidue mod $k{\cdot}2↑n+1$ by the law of
\α{quadratic reciprocity}, since $k{\cdot}2↑n+1\mod12=5$.\xskip
[This test was stated without proof by \α{Proth} in {\sl Comptes Rendus \bf87}
(Paris, 1878), 926.]

To implement Proth's test with the necessary efficiency, we need to be able to
\inx{multiplication modulo m}
compute $x↑2\mod(k{\cdot}2↑n+1)$ with about the same speed as we can compute
the quantity
$x↑2\mod(2↑n-1)$. Let $x↑2=A{\cdot}2↑n+B$ and let $A=qk+r$ where $0≤r<k$;
we can determine $q$ and $r$ rapidly, when $k$ is a single-precision number.
Then $x↑2≡{B-q+r{\cdot}2↑n}\modulo{k{\cdot}2↑n+1}$, so the remainder is
easily obtained.

[To test numbers of the form $3{\cdot}2↑n+1$ for primality, the job is only
slightly more difficult; we first try random single-precision numbers until
finding one that is a quadratic nonresidue mod $3{\cdot}2↑n+1$ by the law of
quadratic reciprocity, then use this number in place of ``3'' in the above test.
If $n\mod4≠0$, the number 5 can be used. It turns out that $3{\cdot}2↑n+1$ is
prime when $n=1$, 2, 5, 6, 8, 12, 18, 30, 36, 41, 66, 189, 201, 209, 276, 353,
408, 438, 534, 2208, 2816, 3168, 3189, 3912;
and $5{\cdot}2↑n+1$ is prime when $n=1$, 3, 7, 13, 15, 25, 39,
55, 75, 85, 127, 1947. See R.↔M. \α{Robinson}, {\sl Proc.\ Amer.\ Math.\ Soc.\
\bf9} (1958), 673--681; the additional numbers listed here
were found by M.↔F. \α{Plass}.]

\ansno 28. $f(p, p↑2d) = 2/(p + 1) + f(p, d)/p$, since $1/(p
+ 1)$ is the probability that $A$ is a multiple of $p$.\xskip $f(p,
pd) = 1/(p + 1)$ when $d\mod p ≠ 0$.\xskip $f(2, 4k + 3) = {1\over
3}$ since $A↑2 - (4k + 3)B↑2$ cannot be a multiple of 4; $f(2,
8k + 5) = {2\over 3}$ since $A↑2 - (8k + 5)B↑2$ cannot be a
multiple of 8; $f(2, 8k + 1) = {1\over 3} + {1\over 3} + {1\over
3} + {1\over 6} + {1\over 12} +\cdots = {4\over 3}$.\xskip
$f(p, d) = \biglp2p/(p↑2 - 1), 0\bigrp$ if $d↑{(p-1)/2}\mod p =
(1, p - 1)$, respectively, for odd $p$.

\ansno 29. The number of solutions to the equation $x↓1+\cdots+x↓m≤r$ in
nonnegative integers $x↓i$ is ${m+r\choose r}≥m↑r/r!$, and each of these
corresponds to a unique integer ${p↓1↑{x↓1}\ldotss p↓m↑{x↓m}≤n}$.\xskip[For sharper
estimates, in the special case that $p↓j$ is the $j$th prime for all $j$, see N. G.
\α{de Bruijn}, {\sl Indag.\ Math.\ \bf28} (1966), 240--247; H. \α{Halberstam},
{\sl Proc. London Math.\ Soc.\ \rm(3) \bf21} (1970), 102--107.]

\ansno 30. If $p↓1↑{e↓1}\ldotss p↓m↑{e↓m}≡x↓i↑2\modulo{q↓i}$, we can find $y↓i$
such that $p↓1↑{e↓1}\ldotss p↓m↑{e↓m}≡(\pm y↓i)↑2\modulo{q↓i↑{d↓i}}$, hence
by the Chinese remainder theorem we obtain $2↑d$ values of $X$ such that $X↑2≡
p↓1↑{e↓1}\ldotss p↓m↑{e↓m}\modulo N$. Such $(e↓1,\ldotss,e↓m)$ correspond to at
most $r\choose r/2$ pairs $(e↓1↑\prime,\ldotss,e↓m↑\prime;
e↓1↑{\prime\prime},\ldotss,e↓m↑{\prime\prime})$ having the hinted properties.
Now for each of the $2↑d$ binary numbers $a=(a↓1\ldotsm a↓d)↓2$, let $n↓a$ be
the number of exponents $(e↓1↑\prime,\ldotss,e↓m↑\prime)$ such that
$(p↓1↑{e↓1↑\prime}\ldotss p↓m↑{e↓m↑\prime})↑{(q↓i-1)/2}≡(-1)↑{a↓i}\modulo{q↓i}$;
we have proved that the required number of integers $X$ is $≥2↑d\biglp\sum↓a
n↓a↑2\bigrp\vcenter{\hbox{\:@\char'16}}{r\choose r/2}$. Since $\sum↓a n↓a$ is
the number of ways to choose at most $r/2$ objects from a set of $m$ objects
\inx{combinations with repetitions}
with repetitions permitted, namely $m+r/2\choose r/2$, we have $\sum↓an↓a↑2≥
{m+r/2\choose r/2}↑2/2↑d≥m↑r\vcenter{\hbox{\:@\char'16}}\biglp2↑d(r/2)!\bigrp$.\xskip
[Cf.\ {\sl J. Number Theory}, to appear.]

\ansno 31. Set $r=M$, $qM=4m$, $εM=2m$.

\def\\{\spose{\raise4.5pt\hbox{\hskip2.3pt$\scriptscriptstyle3$}}\sqrt N}
\ansno 32. Let $M=\lfloor\\\rfloor$, and let the places $x↓i$ of each message
be restricted to the range $0≤x<M↑3-M↑2$. If $x≥M$, encode it as $x↑3\mod N$ as
before, but if $x<M$ change the encoding to $(x+yM)↑3\mod N$, where $y$ is a
random number in the range $M↑2-M≤y≤M↑2$. To decode, first take the cube root;
\inx{random numbers, using}
and if the result is $M↑3-M↑2$ or more, take the remainder mod $M$.

\ansno 34. Let $P$ be the probability that $x↑m\mod p=1$ and let $Q$ be
the probability
that $x↑m\mod q=1$. The probability that $\gcd(x↑{m\!}-1,N)=p$ or $q$ is ${P(1\!-\!Q)+
Q(1\!-\!P)}=P+Q-2PQ$. If $P≤{1\over2}$ or $Q≤{1\over2}$, this probability is
$≥{1\over2}\max(P,Q)≥{1\over2}10↑{-6}$, so we have a good chance of finding a
factor after about $10↑6\log m$ arithmetic operations modulo↔$N$. On the other
hand if $P>{1\over2}$ and $Q>{1\over2}$ then $P\approx Q\approx1$, since we have
the general formula $P=\gcd(m,p-1)/p$; thus $m$ is a multiple of lcm$(p-1,q-1)$
in this case. Let $m=2↑kr$ where $r$ is odd, and form the sequence $x↑r\mod N$,
${x↑{2r}\mod N}$, $\ldotss$, $x↑{2↑kr}\mod N$; with high probability we find
that the first appearance of 1 is preceded 
by a value $y$ other than $N-1$, as in
Algorithm↔P, hence $\gcd(y-1,N)=p$ or↔$q$.

\ansno 35. Let $f=(p↑{q-1}-q↑{p-1})\mod N$. Since $p\mod4=q\mod4=3$,
we have $({-1\over p})=({-1\over q})=({f\over p})=-({f\over q})=-1$, and we also
have $({2\over p})=-({2\over q})=1$. Given a message $x$ in the range
$0≤x≤{1\over8}(N-5)$, let $\=x=4x+2$ or $8x+4$, whichever satisfies
$({\=x\over N})=1$; then transmit the message $\=x↑2\mod N$.

 To decode this message, we first use a SQRT box to find the unique number $y$
such that $y↑2≡\=x↑2\mod N$ and $({y\over N})=1$ and $y$ is even. Then $y=\=x$,
since the other three square roots of $\=x↑2$ are $N-\=x$ and $(\pm f\=x)\mod N$;
the first of these is odd, and the other two have negative Jacobi symbols. The
decoding is now completed by setting $x=\lfloor y/4\rfloor$ if $y\mod4=2$,
otherwise $x←\lfloor y/8\rfloor$.

Anybody who can decode such encodings can also find the factors of $N$, because
the decoding of a false message $\=x↑2\mod N$ when $({\=x\over N})=-1$ reveals
$(\pm f)\mod N$, a number that has a nontrivial gcd with $N$.
% To appear

\ansno 36. The $m$th \α{prime} equals
$$m\ln m+m\ln\ln m-m+m\ln\ln m/\!\ln m-2m/\!\ln m+O\biglp
m(\log\log m)↑2(\log m)↑{-2}\bigrp,$$by (4), although for this problem we
need only the weaker estimate $p↓m=m\ln m+O(m\log\log m)$.\xskip(We will assume that
\inx{prime number theorem}
$p↓m$ is the $m$th prime, since this corresponds to the assumption that
$V$ is uniformly distributed.)\xskip If we choose $\ln m={1\over2}c\sqrt{
\,\ln N\ln\ln N}$, where $c=O(1)$, we find that $r=c↑{-1}\sqrt{\,\ln N/\!\ln\ln N}
-c↑{-2}-c↑{-2}(\ln\ln\ln N/\!\ln\ln N)-2c↑{-2}(\ln{1\over2}c)/\!\ln\ln N+
O(\sqrt{\,\ln\ln N/\!\ln N}\,)$. The estimated running time (21) now
simp\-lifies somewhat surprisingly to the formula $\exp\biglp f(c,N)\sqrt{\,
\ln N\ln\ln N}+O(\log\log N)\bigrp$, where we have
$f(c,N)=c+\biglp1-(1+\ln2)/\!\ln\ln N\bigrp
c↑{-1}$. The value of $c$ that minimizes $f(c,N)$ is $\sqrt{1-(1+\ln2)/\!\ln\ln N}$,
so we obtain the estimate$$\textstyle
\exp\biglp2\sqrt{\,\ln N\ln\ln N}\sqrt{1-(1+\ln2)/\!
\ln\ln N}+O(\log\log N)\bigrp.$$When $N=10↑{50}$ this gives $ε(N)
\approx.33$, which is still much larger than the observed behavior.

{\sl Note:} The partial quotients of $\sqrt D$ seem to behave according to the
distribution obtained for random real numbers in Section↔4.5.3. For example, the
first million partial quotients of the number $10↑{18}+314159$ include exactly
(415236, 169719, 93180, 58606) cases where $A↓n$ is respectively $(1,2,3,4)$. Since
$V↓n$ lies between $(\sqrt D+1)/(A↓n+1)$ and $2\sqrt D/A↓n$, it is likely that
$V↓n≤2\sqrt D y$ with probability about $\lg(1+y)$. This is not much different
from a uniform distribution, so something besides the size of $V↓n$ must account
for the unreasonable effectiveness of Algorithm↔E.

\ansno 37. Apply exercise 4.5.3--12 to the number $\sqrt D+R$, to see that the
periodic part begins immediately, and run the period backwards to verify the
palindromic property.\xskip[It follows that the second half of the period gives
the same $V$'s as the first, and Algorithm↔E could be shut down earlier by
terminating it when $U=U↑\prime$ or $V=V↑\prime$ in step↔E5. However, the period
is generally so long, we never even get close to halfway through it, so there is
no point in making the algorithm more complicated.]

\chpar7←1000 % inhibit break after relations, especially "←"
\ansno 38. [{\sl Inf.\ Proc.\ Letters \bf8} (1979), 28--31.]\xskip
Note that $x\mod y=x-y\,\lfloor x/y\rfloor$ can be computed easily on
such a machine, and we can get simple constants like $0=x-x$, $1=\lfloor
x/x\rfloor$, $2=1+1$; we can test $x>0$ by testing whether $x=1$ or $\lfloor
x/(x-1)\rfloor≠0$.

(a)\9 First compute $l=\lfloor\lg n\rfloor$ in $O(\log n)$ steps, by repeatedly
dividing by 2; at the same time compute $k=2↑l$ and $A←2↑{2↑{l+1}}$ in
$O(\log n)$ steps by repeatedly setting $k←2k$, $A←A↑2$. For the main
computation, suppose we know that $t=A↑m$, $u=(A+1)↑m$, and $v=m!$; then
we can increase the value of $m$ by 1 by setting $m←m+1$, $t←At$, $u←(A+1)u$,
$v←vm$; and we can {\sl double} the value of $m$ by setting $m←2m$, $u←u↑2$,
$v←\biglp\lfloor u/t\rfloor\mod A\bigrp v↑2$, $t←t↑2$, provided that $A$ is
sufficiently large.\xskip$\biglp$Consider the number $u$ in radix-$A$ notation;
$A$ must be greater than $2m\choose m$.$\bigrp$\xskip Now if $n=(a↓l\ldotsm a↓0)
↓2$, let $n↓j=(a↓l\ldotsm a↓j)↓2$; if $m=n↓j$ and $k=2↑j$ and $j>0$ we can
decrease $j$ by 1 by setting $k←\lfloor k/2\rfloor$, $m←2m+\biglp\lfloor n/k
\rfloor\mod2\bigrp$. Hence we can compute $n↓j!$ for $j=l$, $l-1$, $\ldotss$,
0 in $O(\log n)$ steps.\xskip[Another solution, due to Julia Robinson, is to
\inx{Robinson, Julia}
compute $n!=\lfloor B↑n/{B\choose n}\rfloor$ when $B>(2n)↑{n+1}$; cf.\
{\sl AMM \bf80} (1973), 250--251, 266.]

(b)\9 First compute $A=2↑{2↑{l+2}}$ as in (a), then find the least $k≥0$ such that
$2↑{k+1}!\mod n=0$. If $\gcd(n,2↑k!)≠1$, let $f(n)$ be this value; note that this
gcd can be computed in $O(\log n)$ steps by Euclid's algorithm. Otherwise we will
find the least integer $m$ such that ${m\choose\lfloor m/2\rfloor}\mod n=0$, and
let $f(n)=\gcd(m,n)$.\xskip$\biglp$Note that in this case $2↑k<m≤2↑{k+1}$,
hence $\lceil m/2\rceil≤2↑k$ and $\lceil m/2\rceil!$ is relatively prime to
$n$; therefore ${m\choose\lfloor m/2\rfloor}\mod n=0$ iff $m!\mod n=0$.
Furthermore $n≠4$.$\bigrp$

To compute $m$ with a bounded number of registers, we can use \α{Fibonacci numbers}
(cf.\ Algorithm 6.2.1F\null). Suppose we know that
$s=F↓j$, $s↑\prime=F↓{j+1}$, $t=A↑{F↓j}$, $t↑\prime=
A↑{F↓{j+1}}$, $u=(A+1)↑{2F↓j}$, $u↑\prime=(A+1)↑{2F↓{j+1}}$, $v=A↑m$,
$w=(A+1)↑{2m}$, ${2m\choose m}\mod n≠0$, and ${2(m+s)\choose m+s}=0$.
It is easy to reach this state of affairs with $m=F↓{j+1}$, for suitably large
$j$, in $O(\log n)$ steps; furthermore $A$ will be larger than $2↑{2(m+s)}$.
If $s=1$, we set $f(n)=\gcd(2m+1,n)$ or $\gcd(2m+2,n)$, whichever is $≠1$,
and terminate the algorithm. Otherwise we reduce $j$ by 1 as follows: Set
$r←s$, $s←s↑\prime-s$, $s←r$, $r←t$, $t←\lfloor t↑\prime/t\rfloor$, $t↑\prime←r$,
$r←u$, $u←\lfloor u↑\prime/u\rfloor$, $u↑\prime←r$; then
if $\biglp\lfloor wu/vt\rfloor \mod A\bigrp\mod n≠0$, set $m←m+s$, $w←wu$, $v←vt$.

[Can this problem be solved with fewer than $O(\log n)$ operations? Can the
smallest, or the largest, prime factor of $n$ be computed in $O(\log n)$
operations?]

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\ansbegin{4.6}

\ansno 1. $9x↑2 + 7x + 9$;\xskip $5x↑3 + 7x↑2
+ 2x + 6$.

\ansno 2. (a)\9 True.\xskip (b) False if the algebraic
system $S$ contains ``zero divisors,'' nonzero numbers whose
product is zero, as in exercise↔1; otherwise true.\xskip (c) True when $m≠n$, but
false in general when $m=n$, since the leading coefficients might cancel.

\ansno 3. Assume that $r ≤ s$. For $0
≤ k ≤ r$ the maximum is $m↓1m↓2(k + 1)$; for $r ≤ k ≤ s$ it
is $m↓1m↓2(r + 1)$; for $s ≤ k ≤ r + s$ it is $m↓1m↓2(r + s
+ 1 - k)$. The least upper bound valid for all $k$ is $m↓1m↓2(r
+ 1)$.\xskip (The solver of this exercise will know how to factor
the polynomial $x↑7 + 2x↑6 + 3x↑5 + 3x↑4 + 3x↑3 + 3x↑2 + 2x
+ 1$.)

\ansno 4. If one of the polynomials has fewer than
$2↑t$ nonzero coefficients, the product can be formed by putting
exactly $t - 1$ zeros between each of the coefficients, then
multiplying in the binary number system, and finally using a
\α{logical} \.{\α{AND}} instruction (present on most binary computers, cf.\
Section 4.5.4) to zero out the extra bits. For example, if $t
= 3$, the multiplication in the text would become $(1001000001)↓2
\times (1000001001)↓2 = (1001001011001001001)↓2$; if we \.{AND}
this result with the constant $(1001001 \ldotsm 1001)↓2$, the desired
answer is obtained. A similar technique can be used to multiply
polynomials with nonnegative coefficients, when it is known that
the coefficients will not be too large.

\ansno 5. Polynomials of degree $≤2n$ can be represented
as $U↓1(x)x↑n + U↓0(x)$ where deg$(U↓1)$ and deg$(U↓0) ≤ n$;
and $\biglp U↓1(x)x↑n + U↓0(x)\bigrp\biglp V↓1(x)x↑n
+ V↓0(x)\bigrp = U↓1(x)V↓1(x)(x↑{2n} + x↑n) + \biglp U↓1(x)
+ U↓0(x)\bigrp\biglp V↓1(x) + V↓0(x)\bigrp x↑n + U↓0(x)V↓0(x)(x↑n
+ 1)$.\xskip (This equation assumes that arithmetic is being done
modulo 2.)\xskip Thus Eqs.\ 4.3.3--3, 4, 5 hold.

{\sl Notes:} S. A. \α{Cook} has shown that Algorithm
4.3.3C can be extended in a similar way, and exercise 4.6.4--57
describes a method requiring even fewer operations for large↔$n$.
But these ideas are not useful for ``sparse'' polynomials
(having mostly zero coefficients).

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\ansbegin{4.6.1}

\ansno 1. $q(x) = 1 \cdot 2↑3x↑3 + 0
\cdot 2↑2x↑2 - 2 \cdot 2x + 8 = 8x↑3 - 4x + 8$;\xskip$r(x) = 28x↑2
+ 4x + 8$.

\ansno 2. The monic sequence of polynomials produced
during Euclid's algorithm has the coefficients $(1, 5, 6, 6,
1, 6, 3)$, $(1, 2, 5, 2, 2, 4, 5)$, $(1, 5, 6, 2, 3, 4)$, $(1, 3,
4, 6)$, 0. Hence the greatest common divisor is $x↑3 + 3x↑2 +
4x + 6$.\xskip (The greatest common divisor of a polynomial and its
reverse is always symmetric, in the sense that it is a unit
multiple of its own reverse.)

\ansno 3. The procedure of Algorithm↔4.5.2X is
valid, with polynomials over $S$ substituted for integers. When
the algorithm terminates, we have $U(x) = u↓2(x)$, $V(x) = u↓1(x)$.
Let $m =\\deg(u)$, $n =\\deg(v)$. It is easy to prove by induction
that $\\deg(u↓3) +\\deg(v↓1) = n$, $\\deg(u↓3) +\\deg(v↓2) = m$,
after step X3, throughout the execution of the algorithm, provided
that $m ≥ n$. Hence if $m$ and $n$ are greater than $d =\+deg\biglp
\gcd(u, v)\bigrp$ we have $\\deg(U) < m - d$, $\\deg(V) < n - d$;
the exact degrees are $m - d↓1$ and $n - d↓1$, where $d↓1$ is
the degree of the second-last nonzero remainder. If $d = \min(m,
n)$, say $d = n$, we have $U(x) = 0$ and $V(x) = 1$.

When $u(x) = x↑m - 1$ and $v(x) = x↑n - 1$, the
identity $(x↑m - 1)\mod(x↑n - 1) = x↑{m\mod n} - 1$ shows that
all polynomials occurring during the calculation are monic, with
integer coefficients. When $u(x) = x↑{21} - 1$ and $v(x) = x↑{13}
- 1$, we have $V(x) = x↑{11} + x↑8 + x↑6 + x↑3 + 1$ and $U(x) =
-(x↑{19} + x↑{16} + x↑{14} + x↑{11} + x↑8 + x↑6 + x↑3 + x)$.\xskip
[See also Eq.\ 3.3.3--29, which gives an alternative formula
for $U(x)$ and $V(x)$.]

\ansno 4. Since the quotient $q(x)$ depends only on $v(x)$
and the first $m - n$ coefficients of $u(x)$, the remainder
$r(x) = u(x) - q(x)v(x)$ is uniformly distributed and independent
of $v(x)$. Hence each step of the algorithm may be regarded
as independent of the others; this algorithm is much more well-behaved
than Euclid's algorithm over the integers.

The probability that $n↓1 = n - k$ is $p↑{1-k}(1
- 1/p)$, and $t = 0$ with probability $p↑{-n}$. Each succeeding
step has essentially the same behavior; hence we can see that
any given sequence of degrees $n$, $n↓1$, $\ldotss$, $n↓t$, $-∞$ occurs
with probability $(p - 1)↑t/p↑n$. To find the average value
of $f(n↓1, \ldotss , n↓t)$, let $S↓t$ be the sum of $f(n↓1, \ldotss
, n↓t)$ over all sequences $n > n↓1 >\cdots > n↓t
≥ 0$ having a given value of $t$; then the average is $\sum
↓t S↓t(p - 1)↑t/p↑n$.

Let $f(n↓1, \ldotss , n↓t) = t$; then
$S↓t = {n\choose t}(t + 1)$, so the average is $n(1 - 1/p)$. Similarly,
if $f(n↓1, \ldotss , n↓t) = n↓1 +\cdots + n↓t$, then
$S↓t = {n\choose2}{n-1\choose t-1}$, and the average is ${n\choose2}(1
- 1/p)$. Finally, if $f(n↓1, \ldotss , n↓t) = (n - n↓1)n↓1 +\cdots
+ (n↓{t-1} - n↓t)n↓t$, then $S↓t = {{n+2\choose t+2}
- (n + 1){n+1\choose t+1}}+ {n+1\choose2}{n\choose t}$, and the
average is ${{n+1\choose2} - (n + 1)p/(p - 1)} + {\biglp p/(p -
1)\bigrp ↑2(1 - 1/p↑{n+1})}$.

As a consequence we can see that if $p$ is large
there is very high probability that $n↓{j+1} = n↓j - 1$ for
all $j$.\xskip (If this condition fails over the rational numbers,
it fails for all $p$, so we have further evidence for the text's
claim that Algorithm↔C almost always finds $\delta↓2 =\cdots
= 1$.)

\ansno 5. Using the formulas developed
in exercise↔4, with $f(n↓1, \ldotss , n↓t) = \delta ↓{n↓t0}$,
we find that the probability is $1 - 1/p$ if $n > 0$, 1 if
$n = 0$.

\ansno 6. Assuming that the constant terms $u(0)$
and $v(0)$ are nonzero, imagine a ``right-to-left'' division
algorithm, $u(x) = v(x)q(x) + x↑{m-n}r(x)$, where $\\deg(r) <\\deg(v)$.
We obtain a gcd algorithm analogous to Algorithm↔4.5.2B\null, which
is essentially Euclid's algorithm applied to the ``\α{reverse}''
of the original inputs (cf.\ exercise↔2), afterwards reversing
the answer and multiplying by an appropriate power of $x$.

\ansno 7. The units of $S$ (as polynomials of degree
zero).
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\ansno 8. If $u(x) = v(x)w(x)$, where $u(x)$ has
integer coefficients while $v(x)$ and $w(x)$ have rational coefficients,
there are integers $m$ and $n$ such that $m\cdot v(x)$ and $n \cdot
w(x)$ have integer coefficients. Now $u(x)$ is primitive, so
we have
$$u(x) = \pm\,\+pp\biglp m \cdot v(x)\bigrp\+pp\biglp n \cdot w(x)\bigrp.$$

\ansno 9. We can extend Algorithm↔E as follows:
Let $\biglp u↓{1\hskip-.35pt}(x), u↓2(x), u↓3, u↓4(x)\bigrp$ and $\biglp v↓
{1\hskip-.35pt}(x),\hskip-1pt
v↓2(x)$, $v↓3, v↓4(x)\bigrp$ be quadruples that satisfy the relations
$u↓1(x)u(x) + u↓2(x)v(x) = u↓3u↓4(x)$, \ $v↓1(x)u(x) + v↓2(x)v(x)
= v↓3v↓4(x)$. The extended algorithm starts with the quadruples $\biglp 1, 0,
\\cont(u), \\pp(u(x))\bigrp$ and $\biglp 0, 1, \\cont(v), \\pp(v(x))\bigrp$
and manipulates them in such a way as to preserve
the above conditions, where $u↓4(x)$ and $v↓4(x)$ run through the
same sequence as $u(x)$ and $v(x)$ do in Algorithm↔E\null. If $au↓4(x)
= q(x)v↓4(x) + br(x)$, we have $av↓3\biglp u↓1(x), u↓2(x)\bigrp
- q(x)u↓3\biglp v↓1(x), v↓2(x)\bigrp = \biglp r↓1(x), r↓2(x)\bigrp
$, where $r↓1(x)u(x) + r↓2(x)v(x) = bu↓3v↓3r(x)$, so the extended
algorithm can preserve the desired relations. If $u(x)$ and
$v(x)$ are relatively prime, the extended algorithm eventually
finds $r(x)$ of degree zero, and we obtain $U(x) = r↓2(x)$, $V(x)
= r↓1(x)$ as desired.\xskip$\biglp$In practice we would divide $r↓1(x)$,
$r↓2(x)$, and $bu↓3v↓3$ by $\gcd\biglp\\cont(r↓1),\\ cont(r↓2)\bigrp$.$\bigrp
$\xskip Conversely, if such $U(x)$ and $V(x)$ exist, then $u(x)$ and
$v(x)$ have no common prime divisors, since they are primitive
and have no common divisors of positive degree.

\ansno 10. By successively factoring reducible polynomials
into polynomials of smaller degree, we must
obtain a finite factorization of any polynomial into irreducibles.
The factorization of the {\sl content} is unique. To show that
there is at most one factorization of the primitive part, the
key result is to prove that if $u(x)$ is an irreducible factor
of $v(x)w(x)$, but not a unit multiple of the irreducible polynomial
$v(x)$, then $u(x)$ is a factor of $w(x)$. This can be proved
by observing that $u(x)$ is a factor of $v(x)w(x)U(x) = rw(x)
- w(x)u(x)V(x)$ by the result of exercise↔9, where $r$ is a
nonzero constant.

\ansno 11. The only row names needed
would be $A↓1$, $A↓0$, $B↓4$,
$B↓3$, $B↓2$, $B↓1$, $B↓0$, $C↓1$, $C↓0$, $D↓0$. In general, let $u↓{j+2}(x)
= 0$; then the rows needed for the proof are $A↓{n↓2-n↓j}$
through $A↓0$, $B↓{n↓1-n↓j}$ through $B↓0$, $C↓{n↓2-n↓j}$ through $C↓0$, 
$D↓{n↓3-n↓j}$ through $D↓0$, etc.

\ansno 12. If $n↓k = 0$, the text's proof of (24) shows that the value of
the determinant is $\pm h↓k$, and this equals $\pm\lscr↓k↑{n↓{k-1}}/
\prod↓{1<j<k}\lscr↓{\!j}↑{\delta↓{j-1}(\delta↓j-1)}$.
If the polynomials have a factor
of positive degree, we can artificially assume that the polynomial
zero has degree zero and use the same formula with $\lscr↓k=
0$.

{\sl Notes:} The value $R(u, v)$ of Sylvester's determinant
is called the {\sl \α{resultant}} of $u$ and↔$v$, and the quantity
$(-1)↑{\hbox{\:e deg}(u)(\hbox{\:e deg}(u)-1)/2}\lscr(u)↑{-1}R(u, u↑\prime )$ is
called the {\sl \α{discriminant}} of↔$u$, where $u↑\prime$ is the derivative of $u$. 
If $u(x)$ has the factored form ${a(x - α↓1)
\ldotsm (x - α↓m)}$, and if $v(x) = b(x - β↓1) \ldotsm (x - β↓n)$,
the resultant $R(u, v)$ is $a↑nv(α↓1) \ldotsm v(α↓m) = (-1)↑{mn}b↑mu(β↓1)
\ldotsm u(β↓n) = a↑nb↑m \prod↓{1≤i≤m,1≤j≤n}(α↓i - β↓j)$. It follows
that the polynomials of degree $mn$ in $y$ defined as the respective
resultants with $v(x)$ of $u(y - x)$, $u(y + x)$, $x↑mu(y/x)$, and $u(yx)$
have as respective roots the sums $α↓i + β↓j$, differences
$α↓i - β↓j$, products $α↓iβ↓j$, and quotients $α↓i/β↓j$ $\biglp$when
$v(0) ≠ 0\bigrp $. This idea has been used by R. G. K.
\α{Loos} to construct algorithms for arithmetic on algebraic numbers
[{\sl SIAM J. Computing}, to appear].

If we replace each row $A↓i$ in Sylvester's matrix by
$$(b↓0A↓i + b↓1A↓{i+1} +\cdots + b↓{n↓2-1-i}A↓{n↓2-1})
- (a↓0B↓i + a↓1B↓{i+1} + \cdots + a↓{n↓2-1-i}B↓{n↓2-1}),$$
and then delete rows $B↓{n↓2-1}$
through $B↓0$ and the last $n↓2$ columns, we obtain an $n↓1
\times n↓1$ determinant for the resultant instead of the original
$(n↓1 + n↓2) \times (n↓1 + n↓2)$ determinant. In some cases
the resultant can be evaluated efficiently by means of this
determinant; see {\sl CACM \bf 12} (1969), 23--30, 302--303.

J. T. \α{Schwartz} has shown that it is possible to evaluate
resultants and \α{Sturm} sequences for
polynomials of degree $n$ with a total of $O\biglp n(\log n)↑2\bigrp$ arithmetic
operations as $n→∞$.\xskip [{\sl JACM}, to appear.]

\ansno 13. One can show by induction on $j$ that the values of $\biglp
u↓{j+1}(x),g↓{j+1},h↓j\bigrp$ are replaced respectively by $\biglp\lscr↑{1+p↓{j\,}}
w(x)u↓j(x),\lscr↑{2+p↓{j\,}}g↓j,\lscr↑{p↓{j\,}}h↓j\bigrp$ for $j≥2$, 
where $p↓j=n↓1+n↓2-2n↓j$.\xskip
[In spite of this growth, the bound (26) remains valid.]

\ansno 14. Let $p$ be a prime of the domain, and let $j,k$ be
maximum such that $p↑k\rslash v↓n = \lscr(v)$, $p↑j\rslash v↓{n-1}$.
Let $P = p↑k$. By Algorithm↔R we may write $q(x) = a↓0 + Pa↓1x
+\cdots + P↑sa↓sx↑s$, where $s = m - n ≥ 2$. Let
us look at the coefficients of $x↑{n+1}$, $x↑n$, and $x↑{n-1}$
in $v(x)q(x)$, namely $Pa↓1v↓n + P↑2a↓2v↓{n-1} +\cdotss$, $a↓0v↓n
+ Pa↓1v↓{n-1} +\cdotss$, and $a↓0v↓{n-1} + Pa↓1v↓{n-2} +\cdotss$,
each of which is a multiple of $P↑3$. We conclude from the first
that $p↑j\rslash a↓1$, from the second that $p↑{\min(k,2j)}\rslash
a↓0$, then from the third that $P\rslash a↓0$. Hence $P\rslash
r(x)$.\xskip [If $m$ were only $n + 1$, the best we could prove would
be that $p↑{\lceil k/2\rceil }$ divides $r(x)$; e.g., consider
$u(x) = x↑3 + 1$, $v(x) = 4x↑2 + 2x + 1$, $r(x) = 18$. On the other hand, an
argument based on determinants of matrices like (21) and (22) can be used to
show that \def\\{\hbox{\:e deg}}$\lscr(r)↑{\\(v)-\\(r)-1}r(x)$ is always a
multiple of $\lscr(v)↑{(\\(u)-\\(v))(\\(v)-\\(r)-1)}$.]
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\ansno 15. Let $c↓{ij} = a↓{i1}a↓{j1} +\cdots +
a↓{in}a↓{jn}$; we may assume that $c↓{ii} > 0$ for all $i$.
If $c↓{ij} ≠ 0$ for some $i ≠ j$, we can replace row $i$ by
$(a↓{i1} - ca↓{j1}, \ldotss , a↓{in} - ca↓{jn})$, where $c =
c↓{ij}/c↓{jj}$; this does not change the value of the determinant,
and it decreases the value of the upper bound we wish to prove,
since $c↓{ii}$ is replaced by $c↓{ii} - c↑{2}↓{ij}/c↓{ij}$.
These replacements can be done in a systematic way for increasing
$i$ and for $j < i$, until $c↓{ij} = 0$ for all $i ≠ j$.\xskip [The
latter algorithm is called ``\α{Schmidt's orthogonalization process}'';
see {\sl Math.\ Annalen \bf 63} (1907), 442.]\xskip Then $\det(A)↑2
= \det(AA↑T) = c↓{11} \ldotsm c↓{nn}$.

\ansno 16. Let $f(x↓1, \ldotss , x↓n) =
g↓m(x↓2, \ldotss , x↓n)x↑{m}↓{1} +\cdots + g↓0(x↓2,
\ldotss , x↓n)$ and $g(x↓2, \ldotss , x↓n) = g↓m(x↓2, \ldotss
, x↓n)↑2 +\cdots + g↓0(x↓2, \ldotss , x↓n)↑2$; the
latter is not identically zero. We have $$a↓N ≤ m(2N + 1)↑{n-1}
+ (2N + 1)b↓N,$$where $b↓N$ counts the integer solutions
of $g(x↓2, \ldotss , x↓n) = 0$ with variables bounded by↔$N$.
Hence $\lim↓{N→∞} a↓N/(2N + 1)↑n = \lim↓{N→∞} b↓N/(2N + 1)↑{n-1}$,
and this is zero by induction.

\ansno 17. (a)\9 For convenience, let us describe the algorithm
only for $A = \{a, b\}$. The hy\-poth\-eses imply that $\\deg(Q↓1U)
=\\deg(Q↓2V) ≥ 0$, $\\deg(Q↓1) ≤\\deg(Q↓2)$. If $\\deg(Q↓1)
= 0$, then $Q↓1$ is just a nonzero rational number, so we set
$Q = Q↓2/Q↓1$. Otherwise we let $Q↓1 = aQ↓{11} + bQ↓{12} + r↓1$,
$Q↓2 = aQ↓{21} + bQ↓{22} + r↓2$, where $r↓1$ and $r↓2$ are rational
numbers; it follows that
$$Q↓1U - Q↓2V = a(Q↓{11}U - Q↓{21}V) + b(Q↓{12}U - Q↓{22}V)
+ r↓1U - r↓2V.$$
We must have either $\\deg(Q↓{11}) =\\deg(Q↓1) -
1$ or $\\deg(Q↓{12}) =\\deg(Q↓1) - 1$. In the former case, $\\deg(Q↓{11}U
- Q↓{21}V) <\\deg(Q↓{11}U)$, by considering the terms of highest
degree that start with $a$; so we may replace 
$Q↓1$ by $Q↓{11}$, $Q↓2$ by $Q↓{21}$, and repeat the process. Similarly in
the latter case, we may replace $(Q↓1, Q↓2)$
by $(Q↓{12}, Q↓{22})$ and repeat the process.

(b)\9 We may assume that $\\deg(U) ≥\\deg(V)$. If
$\\deg(R) ≥\\deg(V)$, note that $Q↓1U - Q↓2V = Q↓1R - {(Q↓2 -
Q↓1Q)}V$ has degree less than $\\deg(V) ≤\\deg(Q↓1R)$, so we can
repeat the process with $U$ replaced by $R$; we obtain $R =
Q↑\prime V + R↑\prime$, $U = (Q + Q↑\prime )V + R↑\prime $, where
$\\deg(R↑\prime ) <\\deg(R)$, so eventually a solution will be
obtained.

(c)\9 The algorithm of (b) gives $V↓1 = UV↓2 + R$,
$\\deg(R) <\\deg(V↓2)$; by homogeneity, $R = 0$ and $U$ is homogeneous.

(d)\9 We may assume that $\\deg(V) ≤\\deg(U)$. If
$\\deg(V) = 0$, set $W ← U$; otherwise use (c) to find $U = QV$,
so that $QVV = VQV$, $(QV - VQ)V = 0$. This implies that $QV =
VQ$, so we can set $U ← V$, $V ← Q$ and repeat the process.

For further details about the subject of this
exercise, see P. M. Cohn, {\sl Proc.\ Cambridge Phil.\ Soc.\ \bf 57}
(1961), 18--30. The considerably more difficult problem of characterizing
{\sl all} string polynomials such that $UV = VU$ has been solved
by G. M. \α{Bergman} [Ph.D. thesis, Harvard University, 1967].

\ansno 18. [P. M. \α{Cohn,} {\sl Transactions Amer.\ Math.\ Soc.\ \bf 109}
(1963), 332--356.]

\yskip\hang\textindent{\bf C1.} Set $u↓1 ← U↓1$, $u↓2
← U↓2$, $v↓1 ← V↓1$, $v↓2 ← V↓2$, $z↓1 ← z↑\prime↓{2} ← w↓1 ← w↑\prime↓{2}
← 1$, $z↑\prime↓{1} ← z↓2 ← w↑\prime↓{1} ← w↓2 ← 0$, $n
← 0$.

\yskip\hang\textindent{\bf C2.} (At this point the identities
given in the exercise hold, and $u↓1v↓1 = u↓1v↓2$; $v↓2 =
0$ if and only if $u↓1 = 0$.)\xskip If $v↓2 = 0$, the algorithm terminates
with gcrd$(V↓1, V↓2) = v↓1$, lclm$(V↓1, V↓2) = z↑\prime↓{1}V↓1
= -z↑\prime↓{2}V↓2$.\xskip (Also, by symmetry, we have gcld$(U↓1, U↓2) = u↓2$ and
lcrm$(U↓1, U↓2) = U↓1w↓1 = -U↓2w↓2$.)

\yskip\hang\textindent{\bf C3.} Find $Q$ and $R$ such that $v↓1
= Qv↓2 + R$, where $\\deg(R) <\\deg(v↓2)$.\xskip$\biglp$We have $u↓1(Qv↓2
+ R) = u↓2v↓2$, so $u↓1R = (u↓2 - u↓1Q)v↓2 = R↑\prime v↓2.\bigrp$

\yskip\hang\textindent{\bf C4.} Set $(w↓1$, $w↓2$, $w↑\prime↓{1}$,
$w↑\prime↓{2}$, $z↓1$, $z↓2$, $z↑\prime↓{1}$, $z↑\prime↓{2}$,
$u↓1$, $u↓2$, $v↓1$, $v↓2) ← (w↑\prime↓{1} - w↓1Q$, $w↑\prime↓{2}
- w↓2Q$, $w↓1$, $w↓2$, $z↑\prime↓{1}$, $z↑\prime↓{2}$, $z↓1 - Qz↑\prime↓{1}$,
$z↓2 - Qz↑\prime↓{2}$, $u↓2 - u↓1Q$, $u↓1$, $v↓2$, $v↓1 - Qv↓2)$ and
$n ← n + 1$. Go back to C2.\quad\blackslug
% Vicinity of page 620
% folio 802 galley 1a 	*
\def\\#1({\mathop{\hbox{#1}}(}\def\+#1\biglp{\mathop{\hbox{#1}}\biglp}
\yyskip This extension of Euclid's algorithm includes most
of the features we have seen in previous extensions, all at
the same time, so it provides new insight into the special cases
already considered. To prove that it is valid, note first that
deg$(v↓2)$ decreases in step C4, so the algorithm certainly
terminates. At the conclusion of the algorithm, $v↓1$ is a common
right divisor of $V↓1$ and $V↓2$, since $w↓1v↓1 = (-1)↑nV↓1$
and $-w↓2v↓1 = (-1)↑nV↓2$; also if $d$ is any common right divisor
of $V↓1$ and $V↓2$, it is a right divisor of $z↓1V↓1 + z↓2V↓2
= v↓1$. Hence $v↓1 =\\gcrd(V↓1, V↓2)$. Also if $m$ is any common
left multiple of $V↓1$ and $V↓2$, we may assume without loss
of generality that $m = U↓1V↓1 = U↓2V↓2$, since the sequence
of values of $Q$ does not depend on $U↓1$ and $U↓2$. Hence $m
= (-1)↑n(-u↓2z↑\prime↓{1})V↓1 = (-1)↑n(u↓2z↑\prime↓{2})V↓2$
is a multiple of $z↑\prime↓{1}V↓1$.

In practice, if we just want to calculate gcrd$(V↓1,
V↓2)$, we may suppress the computation of $n$, $w↓1$, $w↓2$, $w↑\prime↓{1}$,
$w↑\prime↓{2}$, $z↓1$, $z↓2$, $z↑\prime↓{1}$, $z↑\prime↓{2}$.
These additional quantities were added to the algorithm primarily
to make its validity more readily established.

{\sl Note:} Nontrivial factorizations of string
polynomials, such as the example given with this exercise, can
be found from matrix identities such as
\def\choptop#1{\vbox to 0pt{\vss\hbox{$#1$}}}
$$\left({a \atop 1}\hskip 1.5em{1\atop 0}\right)\left({\choptop b\atop 1}\hskip 1.5em{1\atop
0}\right)\left({c\atop 1}\hskip 1.5em{1\atop0}\right)\left({0\atop 1}\9
{\quad1\atop -c}\right)\left({0\atop 1}\9{\quad1\atop -b}\right)\left({0\atop
1}\9{\quad1\atop -a}\right) = \left({1\hskip 1.5em 0\atop 0\hskip 1.5em 1}\right),$$
since these identities hold even when multiplication is not commutative.
For example,
$$(abc + a + c)(1 + ba) = (ab + 1)(cba + a + c).$$
(Compare this with the ``\α{continuant polynomials}'' of Section 4.5.3.)

\ansno 19. [Cf.\ Eug\`ene \α{Cahen,} {\sl Th\'eorie des Nombres \bf1} (Paris: A.
Hermann & fils, 1914), \hbox{336--338}.]\xskip
If such an algorithm
exists, $D$ is a gcrd by the argument in exercise↔18. Let us
regard $A$ and $B$ as a single $2n \times n$ matrix $C$ whose
first $n$ rows are those of $A$, and whose second $n$ rows are
those of $B$. Similarly, $P$ and $Q$ can be combined into a
$2n \times n$ matrix $R$; $X$ and $Y$ can be combined into an
$n \times 2n$ matrix $Z$. The desired conditions now reduce
to two equations $C = RD$, $D = ZC$. If we can find a $2n \times
2n$ integer matrix $U$ with determinant $\pm 1$ such that the last
$n$ rows of $U↑{-1}C$ are all zero, then $R = ($first $n$ columns
of $U)$, $D = ($first $n$ rows of $U↑{-1}C)$, $Z = ($first $n$ rows
of↔$U↑{-1})$ solves the desired conditions. Hence, for example,
the following algorithm may be used (with $m=2n$):

\algbegin Algorithm T (\α{Triangularization}). Let $C$ be an $m \times n$ matrix
\inx{matrix triangularization}
of integers. This algorithm finds $m \times m$ integer matrices
$U$ and $V$ such that $UV = I$ and $VC$ is ``upper triangular.''\xskip
(The entry in row $i$ and column $j$ of $VC$ is zero if $i >
j$.)

\algstep T1. [Initialize.] Set $U ← V ← I$, the
$m \times m$ identity matrix; and set $T ← C$.\xskip (Throughout the
algorithm we will have $T = VC$ and $UV = 1$.)

\algstep T2. [Iterate on $j$.] Do step T3 for $j = 1$, 2, $\ldotss
$, $\min(m, n)$, and terminate the algorithm.

\algstep T3. [Zero out column $j$.] Perform the following transformation
zero or more times until $T↓{ij}$ is zero for all $i > j$: Let
$T↓{kj}$ be a nonzero element of $\{T↓{ij}, T↓{(j+1)j}, \ldotss,
T↓{mj}\}$ having the smallest absolute value. Interchange
rows $k$ and $j$ of $T$ and of $V$; interchange columns $k$
and $j$ of $U$. Then subtract $\lfloor T↓{ij}/T↓{jj}\rfloor$
times row $j$ from row $i$, in matrices $T$ and $V$, and add
the same multiple of column $i$ to column $j$ in matrix $U$,
for $j < i ≤ m$.\quad\blackslug

\yyskip\noindent For the stated example, the
algorithm yields \def\\#1#2#3#4{({#1\atop#3}\>{#2\atop#4})}
$\\1234=\\1032\\1{\quad2}0{-1}$, $\\4321=\\4523\\1{\quad2}0{-1}$,
$\\1{\quad2}0{-1}=\\1{\quad0}2{-2}\\1234
+\\0010\\4321$.\xskip (Actually
{\sl any} matrix with determinant $\pm 1$ would be a gcrd in this particular case.)

\ansno 20. It may be helpful to consider the construction of exercise 4.6.2--22,
with $p↑m$ replaced by a small number $ε$.

\ansno 21. Note that Algorithm R is used only when $m - n ≤
1$; furthermore, the coefficients are bounded by (25) with $m =
n$.\xskip [The stated formula is, in fact, the execution time observed
in practice, not merely an upper bound. For more detailed information
see G. E. \α{Collins,} {\sl Proc. 1968 Summer Inst.\ on Symbolic
Math.\ Comp.}, Robert G. \α{Tobey,} ed.\ (IBM Federal Systems Center:
June 1969), 195--231.]

\ansno 22. A sequence of signs cannot contain two consecutive
zeros, since $u↓{k+1}(x)$ is a nonzero constant in (29). Moreover
we cannot have ``+, 0, +'' or ``$-$, 0, $-$'' as subsequences. The formula
$V(u, a) - V(u, b)$ is clearly valid when $b = a$, so we must
only verify it as $b$ increases. The polynomials $u↓j(x)$ have
finitely many roots, and $V(u, b)$ changes only when $b$ encounters
or passes such roots. Let $x$ be a root of some (possibly several)
$u↓j$. When $b$ increases from $x - ε$ to $x$, the sign sequence
near $j$ goes from ``+, $\pm$, $-$'' to ``+, 0, $-$'' or from ``$-$, $\pm$, +''
to ``$-$, 0, +'' if $j > 0$; and from ``+, $-$'' to ``0, $-$'' or from ``$-$, +'' to
``0, +'' if $j = 0$.\xskip (Since $u↑\prime (x)$ is the derivative, $u↑\prime
(x)$ is negative when $u(x)$ is decreasing.)\xskip Thus the net change
in $V$ is $-\delta ↓{j0}$. When $b$ increases from $x$ to $x
+ ε$, a similar argument shows that $V$ remains unchanged.

[L. E. \α{Heindel,} {\sl JACM} {\bf 18} (1971), 533--548,
has applied these ideas to construct algorithms for isolating
the real zeros of a given polynomial $u(x)$, in time bounded
by a polynomial in deg$(u)$ and log $N$, where all coefficients
$y↓j$ are integers with $|u↓j| ≤ N$, and all operations are
guaranteed to be exact.]

\ansno 23. If $v$ has $n - 1$ real roots occurring between the
$n$ real roots of $u$, then (by considering sign changes) $u(x)\mod
v(x)$ has $n - 2$ real roots lying between the $n - 1$ roots of $v$.

\ansno 24. First show that $h↓j=g↓{\!j}↑{\delta↓{j-1}}g↓{\!j-1}↑{\delta↓{j-2}
(1-\delta↓{j-1})}\ldotss g↓2↑{\delta↓1(1-\delta↓2)\ldotsm(1-\delta↓{j-1})}$.
Then show that the exponent of $g↓2$ on the left-hand side of (18) has the form
$\delta↓2+\delta↓1x$, where $x=\delta↓2+\cdots+\delta↓{j-1}+1-\delta↓2(\delta↓3+
\cdots+\delta↓{j-1}+1)-\delta↓3(1-\delta↓2)(\delta↓4+\cdots+\delta↓{j-1}+1)-\cdots
-{\delta↓{j-1}(1-\delta↓2)\ldotsm(1-\delta↓{j-2})(1)}$. But $x=1$, since
it is seen to be independent of $\delta↓{j-1}$ and we can set $\delta↓{j-1}=0$,
etc. A similar derivation works for $g↓3$, $g↓4$, $\ldotss$, and a simpler
derivation works for (23).

\ansno 25. Each coefficient of $u↓j(x)$ can be expressed as a determinant in
which one column contains only $\lscr(u)$, $\lscr(v)$, and zeros. To use
this fact, modify Algorithm↔C as follows: In step C1, set $g←\gcd\biglp\lscr(u),
\lscr(v)\bigrp$ and $h←0$.
In step C3, if $h=0$, set $u(x)←v(x)$, $v(x)←r(x)/g$, $h←\lscr(u)↑\delta/g$,
$g←\lscr(u)$, and return to C2; otherwise proceed as in the unmodified
algorithm. The effect of this new initialization is simply to replace
$u↓j(x)$ by $u↓j(x)/\!\gcd\biglp\lscr(u),\lscr(v)\bigrp$ for all $j≥3$;
thus, $\lscr↑{2j-4}$ will become $\lscr↑{2j-5}$ in (28).

\def\dg{\mathop{\char d\char e\char g}}
\ansno 26. In fact, even more is true. Note that the algorithm in exercise↔3
computes $\pm p↓n(x)$ and $\mp q↓n(x)$ for $n≥-1$. Let $e↓n=\dg
(q↓n)$ and $d↓n=\dg(p↓nu-q↓nv)$; we observed in exercise↔3 that $d↓{n-1}+
e↓n=\dg(u)$ for $n≥0$. We shall prove that the conditions $\dg(q)<e↓n$ and
$\dg(pu-qv)<d↓{n-2}$ imply that $p(x)=c(x)p↓{n-1}(x)$ and $q(x)=c(x)q↓{n-1}(x)$:
Given such $p$ and $q$, we can find $c(x)$ and $d(x)$ such that $p(x)=c(x)p↓{n-1}(x)
+d(x)p↓n(x)$ and $q(x)=c(x)q↓{n-1}(x)+d(x)q↓n(x)$, since $p↓{n-1}(x)q↓n(x)-
p↓n(x)q↓{n-1}(x)=\pm1$. Hence $pu-qv=c(p↓{n-1}u-q↓{n-1}v)+d(p↓nu-q↓nv)$. If
${d(x)≠0}$, we must have $\dg(c)+e↓{n-1}=\dg(d)+e↓n$, since $\dg(q)<\dg(q↓n)$;
it follows that $\dg(c)+d↓{n-1}>\dg(d)+d↓n$, since this is surely true if $d↓n=
-∞$ and otherwise we have $d↓{n-1}+e↓n=d↓n+e↓{n+1}>d↓n+e↓{n-1}$. Therefore
$\dg(pu-qv)=\dg(c)+d↓{n-1}$; but we have assumed that $\dg(pu-qv)<d↓{n-2}=
d↓{n-1}+e↓n-e↓{n-1}$, so $\dg(c)<e↓n-e↓{n-1}$ and $\dg(d)<0$, a contradiction.

[This result is essentially due to L. \α{Kronecker},
{\sl Monatsberichte K\"onigl.\ Preu\ss.\ Akad.\ Wiss.\ }(Berlin: 1881),
535--600.
It implies the following theorem: ``Let $u(x)$ and↔$v(x)$ be relatively prime
polynomials over a field and let $d≤\dg(v)<\dg(u)$. If $q(x)$ is a polynomial
of least degree such that there exist polynomials $p(x)$ and↔$r(x)$ with
$p(x)u(x)-q(x)v(x)=r(x)$ and $\dg(r)=d$, then $p(x)/q(x)=p↓n(x)/q↓n(x)$ for
some↔$n$.'' For if $d↓{n-2}>d≥d↓{n-1}$, there are solutions $q(x)$
with $\dg(q)={e↓{n-1}+d-d↓{n-1}<e↓n}$, and we have proved that all solutions
of such low degree have the stated property.]
% Vicinity of page 622
% folio 804 galley 1b 	*
\ansbegin{4.6.2}

\ansno 1. By the principle of \α{inclusion
and exclusion} (Section 1.3.3), the number of polynomials without
linear factors is $\sum ↓{k≤n} {p\choose k}p↑{n-k}(-1)↑k = p↑{n-p}(p
- 1)↑{\mskip 1mu p}$. The stated probability when $n≥p$ is therefore $1 - (1 - 1/p)↑{\mskip 1mu p}$,
which is greater than ${1\over 2}$.\xskip [In fact, the stated probability
is greater than ${1\over 2}$ for all $n ≥ 1$.]\xskip The average number of
linear factors is $p$ times the average number of times $x$ is a factor,
so it is ${p\over p-1}(1-p↑{-n})=1+p↑{-1}+\cdots+p↑{1-n}$.\xskip
[See answer 38 for further comments on the average number of linear factors.]

\ansno 2. (a)\9 We know that $u(x)$ has a representation
as a product of irreducible polynomials; and the leading coefficients
of these polynomials must be units, since they divide the leading
coefficient of $u(x)$. Therefore we may assume that $u(x)$ has
a representation as a product of monic irreducible polynomials
$p↓1(x)↑{e↓1}\ldotsm p↓r(x)↑{e↓r}$, where $p↓1(x)$,
$\ldotss$, $p↓r(x)$ are distinct. This representation is unique,
except for the order of the factors, so the conditions on $u(x)$,
$v(x)$, $w(x)$ are satisfied if and only if
$$v(x) = p↓1(x)↑{\lfloor e↓1/2\rfloor} \ldotss p↓r(x)↑{\lfloor e↓r/2\rfloor},
\qquad w(x) = p↓1(x)↑{e↓1\mod2} \ldotss p↓r(x)↑{e↓r\mod2}.$$

(b)\9 The generating function for the number
of monic polynomials of degree $n$ is $1 + pz + p↑2z↑2 + \cdots
= 1/(1 - pz)$. The generating function for the number of polynomials
of degree $n$ having the form $v(x)↑2$, where $v(x)$ is monic,
is $1 + pz↑2 + p↑2z↑4 + \cdots = 1/(1 - pz↑2)$. If the generating
function for the number of monic squarefree polynomials of degree
$n$ is $g(z)$, then by part (a)  we must have $1/(1 - pz) = g(z)/(1 - pz↑2)$.
Hence $g(z) = (1 - pz↑2)/(1 - pz) = 1 + pz + (p↑2 - p)z↑2 +
(p↑3 - p↑2)z↑3 + \cdotss$. The answer is $p↑n - p↑{n-1}$ for
$n ≥ 2$.\xskip [Curiously, this proves that $\gcd\biglp u(x), u↑\prime
(x)\bigrp = 1$ with probability $1 - 1/p$; it is the same as
the probability that $\gcd\biglp u(x), v(x)\bigrp = 1$ when $u(x)$
and $v(x)$ are {\sl independent}, by exercise 4.6.1--5.]

{\sl Note:} By a similar argument, every $u(x)$ has a unique representation
$v(x)w(x)↑r$, where $v(x)$ is not divisible by the $r$th power of any
irreducible; the number of such monic polynomials $v(x)$ is $p↑n-p↑{n-r+1}$
for $n≥r$.

\ansno 3. Let $u(x) = u↓1(x) \ldotsm u↓r(x)$. There is {\sl at
most} one such $v(x)$, by the argument of Theorem↔4.3.2C\null. There
is {\sl at least} one if, for each $j$, we can solve the system
with $w↓j(x) = 1$ and $w↓k(x) = 0$ for $k ≠ j$. A solution to
the latter is $v↓1(x)\prod↓{k≠j}u↓k(x)$, where $v↓1(x)$
and $v↓2(x)$ can be found satisfying
$$\textstyle v↓1(x)\prod↓{k≠j}u↓k(x)\,+\,v↓2(x)u↓j(x) = 1,\qquad
\hbox{deg}(v↓1) <\hbox{deg}(u↓j),$$
by the extension of Euclid's algorithm (exercise↔4.6.1--3).
% Vicinity of page 623
% folio 805 galley 2 	*
\ansno 4. By unique factorization, we have
$(1 - pz)↑{-1} =\prod↓{n≥1} (1 - z↑n)↑{-a↓{np}}$;
after taking logarithms, this can be rewritten$$\textstyle\sum ↓{j≥1} G↓p(z↑j)/j
= \sum ↓{k,j≥1} a↓{kp}z↑{kj}/j = \ln\biglp 1/(1 - pz)\bigrp.$$
The stated identity now yields the answer $G↓p(z) = \sum ↓{m≥1}
\mu (m)m↑{-1}\ln\biglp 1/(1 - pz↑m)\bigrp $, from which we
obtain $a↓{np} = \sum ↓{d\rslash n} \mu (n/d)n↑{-1}p↑d$; thus $\lim↓{p→∞}
a↓{np}/p↑n = 1/n$. To prove the stated identity, note that
$$\textstyle\sum
↓{n,j≥1}\mu (n)g(z↑{nj})n↑{-t}j↑{-t} = \sum ↓{m≥1} g(z↑m)m↑{-t}
\sum ↓{n\rslash m} \mu (n) = g(z).$$

\ansno 5. Let $a↓{npr}$ be the number of monic
polynomials of degree $n$ modulo $p$ having exactly $r$ irreducible
factors. Then $\Gscr↓p(z, w) = \sum ↓{n,r≥0} a↓{npr}z↑nw↑r =
\exp\biglp\lower6pt\null
 \sum ↓{k≥1} G↓p(z↑k)w↑k/k\bigrp$, where $G↓p$ is the generating
function in exercise↔4; cf.\ Eq.\ 1.2.9--34.
We have $$\baselineskip15pt\eqalign{\textstyle\sum ↓{n≥0} A↓{np}z↑n
= d\Gscr↓p(z/p, w)/d↓{\null}w\,|↓{w=1}⊗=\textstyle
\biglp \sum ↓{k≥1}G↓p(z↑k/p↑k)\bigrp \exp\biglp\ln(1/(1
- z))\bigrp\cr⊗\textstyle=\biglp\sum↓{n≥1}\ln\biglp 1/(1-p↑{1-n}z↑n)\bigrp\varphi
(n)/n\bigrp/(1-z),\cr}$$ hence $A↓{np} = H↓n + 1/2p + O(p↑{-2})$ for
$n ≥ 2$. The average value of $2↑r$ is the coefficient of $z↑n$
in $\Gscr↓p(z/p, 2)$, namely $n + 1 + (n - 1)/p + O(p↑{-2})$.\xskip (The
variance is of order $n↑3$, however: set $w = 4$.)

\ansno 6. For $0 ≤ s < p$, $x - s$ is a factor of
$x↑{\mskip 1mu p} - x$ (modulo $p$) by Fermat's theorem. So $x↑{\mskip 1mu p} - x$ is
a multiple of $\lcm\biglp x - 0, x - 1, \ldotss , x - (p - 1)\bigrp=x↑{\underline p}
$.\xskip [{\sl Note:} Therefore the \α{Stirling numbers}
$p\comb[]k$ are multiples of $p$ except when $k = 1$, $k =
p$. Equation 1.2.6--41 shows that the same statement is valid
for Stirling numbers $p\comb\{\} k$ of the other kind.]

\ansno 7. The factors on the right are relatively
prime, and each is a divisor of $u(x)$, so their product divides
$u(x)$. On the other hand, $u(x)$ divides
$$\textstyle v(x)↑{\mskip 1mu p} - v(x) = \prod ↓{0≤s<p}
\biglp v(x) - s\bigrp ,$$
so it divides the right-hand side by exercise↔4.5.2--2.

\ansno 8. The vector (18) is the only output whose $k$th component is nonzero.

\ansno 9. For example, start with $x ← 1$ and $y ← 1$;
then repeatedly set $R[x] ← y$, ${x ←2x \mod 101}$, $y ← 51y
\mod 101$, one hundred times.

\ansno 10. The matrix $Q - I$ below has a null space generated
by the two vectors $v↑{[1]} =(1, 0, 0, 0, 0, 0, 0, 0)$, $v↑{[2]}
= (0, 1, 1, 0, 0, 1, 1, 1)$. The factorization is
$$(x↑6 + x↑5 + x↑4 + x + 1)(x↑2 + x + 1).$$

\vskip-4pt
\hbox to size{\qquad$\dispstyle{p=2\lower5pt\null\atop
\left(\,\vcenter{\halign{\hfill#⊗\hbox to 15pt{\hfill#}⊗\!
\hbox to 15pt{\hfill#}⊗\hbox to 15pt{\hfill#}⊗\hbox to 15pt{\hfill#}⊗\!
\hbox to 15pt{\hfill#}⊗\hbox to 15pt{\hfill#}⊗\hbox to 15pt{\hfill#}\cr
0⊗0⊗0⊗0⊗0⊗0⊗0⊗0\cr
0⊗1⊗1⊗0⊗0⊗0⊗0⊗0\cr
0⊗0⊗1⊗0⊗1⊗0⊗0⊗0\cr
0⊗0⊗0⊗1⊗0⊗0⊗1⊗0\cr
1⊗0⊗0⊗1⊗0⊗0⊗1⊗0\cr
1⊗0⊗1⊗1⊗1⊗0⊗0⊗0\cr
0⊗0⊗1⊗0⊗1⊗1⊗0⊗1\cr
1⊗1⊗0⊗1⊗1⊗1⊗0⊗1\cr}}\,\right)}\hfill{p=5\lower5pt\null\atop
\left(\,\vcenter{\halign{\hfill#⊗\hbox to 15pt{\hfill#}⊗\!
\hbox to 15pt{\hfill#}⊗\hbox to 15pt{\hfill#}⊗\!
\hbox to 15pt{\hfill#}⊗\hbox to 15pt{\hfill#}⊗\hbox to 15pt{\hfill#}\cr
0⊗0⊗0⊗0⊗0⊗0⊗0\cr
0⊗4⊗0⊗0⊗0⊗1⊗0\cr
0⊗2⊗2⊗0⊗4⊗3⊗4\cr
0⊗1⊗4⊗4⊗4⊗2⊗1\cr
2⊗2⊗2⊗3⊗4⊗3⊗2\cr
0⊗0⊗4⊗0⊗1⊗3⊗2\cr
3⊗0⊗2⊗1⊗4⊗2⊗1\cr}}\,\right)}$\qquad}

\ansno 11. Removing the trivial factor % WATCH PAGING HERE IN FUTURE EDITIONS
$x$, the matrix $Q - I$ on the previous page has a null space generated by
$(1, 0, 0, 0, 0, 0, 0)$ and $(0, 3, 1, 4, 1, 2, 1)$. The factorization
is
$$x(x↑2 + 3x + 4)(x↑5 + 2x↑4 + x↑3 + 4x↑2 + x + 3).$$

\ansno 12. If $p = 2$, $(x + 1)↑4 = x↑4 + 1$. If $p
= 8k + 1$, $Q - I$ is the zero matrix, so there are four factors.
For other values of $p$ we have
$$\lower18pt\hbox{$Q-I=\null$}
{p=8k+3\lower5pt\null\atop
\left(\,\vcenter{\halign{\hfill#⊗\hbox to 20pt{\hfill$#$}⊗\!
\hbox to 20pt{\hfill$#$}⊗\hbox to 20pt{\hfill$#$}⊗\hbox to 20pt{\hfill$#$}\cr
0⊗0⊗0⊗0\cr0⊗-1⊗0⊗1\cr0⊗0⊗-2⊗0\cr0⊗1⊗0⊗-1\cr}}\,\right)}
{p=8k+5\lower5pt\null\atop
\left(\,\vcenter{\halign{\hfill#⊗\hbox to 20pt{\hfill$#$}⊗\!
\hbox to 20pt{\hfill$#$}⊗\hbox to 20pt{\hfill$#$}⊗\hbox to 20pt{\hfill$#$}\cr
0⊗0⊗0⊗0\cr0⊗-2⊗0⊗0\cr0⊗0⊗0⊗0\cr0⊗0⊗0⊗-2\cr}}\,\right)}
{p=8k+7\lower5pt\null\atop
\left(\,\vcenter{\halign{\hfill#⊗\hbox to 20pt{\hfill$#$}⊗\!
\hbox to 20pt{\hfill$#$}⊗\hbox to 20pt{\hfill$#$}⊗\hbox to 20pt{\hfill$#$}\cr
0⊗0⊗0⊗0\cr0⊗-1⊗0⊗-1\cr0⊗0⊗-2⊗0\cr0⊗-1⊗0⊗-1\cr}}\,\right)}$$
Here $Q - I$ has rank 2, so there are $4
- 2 = 2$ factors.\xskip [But it is easy to prove that $x↑4 + 1$ is
irreducible over the integers, since it has no linear factors
and the coefficient of $x$ in any factor of degree two must
be less than or equal to 2 in absolute value by exercise↔20.
For all $k ≥ 2$, H. P. F. \α{Swinnerton-Dyer} has exhibited polynomials
of degree $2↑k$ that are irreducible over the integers, but
they split completely into linear and quadratic factors modulo
every prime. For degree 8, his example is $x↑8 - 16x↑6 + 88x↑4
+192x↑2 + 144$, having roots $\pm\sqrt 2\pm\sqrt3\pm i$ [see {\sl Math.\
Comp.\ \bf24} (1970), 733--734]. According to the theorem of \α{Frobenius} cited
in exercise↔37, any irreducible polynomial of degree $n$
whose \α{Galois group} contains no $n$-cycles will have factors modulo almost
all primes.]

\ansno 13. $p=8k+1$:
${\biglp x+(1+\sqrt{-1})/\sqrt2\bigrp}\*
{\biglp x+(1-\sqrt{-1})/\sqrt2\bigrp}\*
{\biglp x-(1+\sqrt{-1})/\sqrt2\bigrp}\*
{\biglp x-(1-\sqrt{-1})/\sqrt2\bigrp}$.\xskip
$p=8k+3$: ${\biglp x↑2-\sqrt{-2}x-1\bigrp}\*
{\biglp x↑2-\sqrt{-2}x-1\bigrp}$.\xskip
$p=8k+5$: ${\biglp x↑2-\sqrt{-1}\bigrp}\*
{\biglp x↑2-\sqrt{-1}\bigrp}$.\xskip
$p=8k+7$: ${\biglp x↑2-\sqrt2x+1\bigrp}\*
{\biglp x↑2+\sqrt2x+1\bigrp}$. The latter factorization also holds over the
field of real numbers.

\ansno 14. Algorithm N can be adapted to find the coefficients
\inx{null space}
of $w$: Let $A$ be the $(r + 1) \times n$ matrix whose $k$th
row contains the coefficients of $v(x)↑k \mod
u(x)$, for $0 ≤ k ≤ r$. Apply the method of Algorithm↔N until
the first dependence is found in step N3; then the algorithm
terminates with $w(x) = v↓0 + v↓1x +\cdots + v↓kx↑k$,
where $v↓j$ is defined in (18).
At this point $2 ≤ k ≤ r$; it is not necessary to know $r$ in advance, since
we can check for dependency after generating each row of $A$.


\ansno 15. We may assume that $u≠0$ and that $p$ is odd. \α{Berlekamp}'s method applied
to the polynomial $x↑2-u$ tells us that a square root exists if and only if
$u↑{(p-1)/2}\mod p=1$; let us assume that this condition holds.

Let $p-1=2↑e\cdot q$, where $q$ is odd. \α{Zassenhaus}'s factoring procedure suggests
the following square-root extraction algorithm: Set $t←0$. Evaluate
$$\twoline{\gcd\biglp(x+t)↑q-1,x↑2-u\bigrp,\;\gcd\biglp(x+t)↑{2q}-1,x↑2
-u\bigrp,}{2pt}{\gcd\biglp(x+t)↑{4q}-1,x↑2-u\bigrp,\;\gcd\biglp(x+t)↑{8q}-1,
x↑2-u\bigrp,\;\ldotss,}$$
until finding the first case where the gcd is not 1 (modulo $p$). If the
gcd is $x-v$, then $\sqrt u = \pm v$. If the gcd is $x↑2-u$, set $t←t+1$ and
repeat the calculation.

{\sl Notes:} If $(x+t)↑k\mod(x↑2-u)=ax+b$, then we have $(x+t)↑{k+1}\mod(x↑2-u)=
(b+at)x+(bt+au)$, and $(x+t)↑{2k}\mod(x↑2-u)=2abx+(b↑2+a↑2u)$; hence
$(x+t)↑q$, $(x+t)↑{2q}$, $\ldots$ are easy to evaluate efficiently, and the
calculation for fixed $t$ takes $O\biglp(\log p)↑3\bigrp$ units of time.
The square root will be found when $t=0$ with probability $1/2↑{e-1}$;
thus it will always be found immediately when $p\mod4=3$. If we choose
$t$ at random instead of increasing it sequentially, exercise↔29 gives a
rigorous proof that each↔$t$ gives success at least about half of the time;
but for practical purposes this random choice isn't needed.

Another square-root method has been suggested by D. \α{Shanks}. When $e>1$ it
requires an auxiliary constant $z$ (depending only on $p$) such that
$z↑{2↑{e-1}}≡-1\modulo p$. The value $z=n↑q\mod p$ will work for almost
one half of all integers $n$; once $z$ is known, the following algorithm
requires no more probabilistic calculation:

\yskip\hang\textindent{\bf S1.}Set $y←z$, $r←e$, $v←u↑{(q+1)/2}\mod p$,
$w←u↑q\mod p$.

\yskip\hang\textindent{\bf S2.}If $w=1$, stop; $v$ is the answer. Otherwise
find the smallest $k$ such that $w↑{2↑k}\mod p$ is equal to 1.
If $k=r$, stop (there is
no answer); otherwise set$$(y,r,v,w)←(y↑{2↑{r-k}},k,vy↑{2↑{r-k-1}},wy↑{2↑{r-k}})$$
and repeat step S2.\quad\blackslug

\yyskip The validity of this algorithm follows from the invariant congruences
$uw≡v↑2$, $y↑{2↑{r-1}}≡-1$, $w↑{2↑{r-1}}≡1\modulo p$. On the average, step S2
will require about ${1\over4}e↑2$ multiplications mod $p$.\xskip
Reference: {\sl Proc.\ Second Manitoba Conf.\ Numer.\ Math.\ }(1972),
\hbox{58--62.} A related method was published by A. \α{Tonelli,} {\sl G\"ottinger
Nachrichten} (1891), 344--346.
% Vicinity of page 626
% folio 810 galley 3 	*
\ansno 16. (a)\9 Substitute polynomials modulo $p$ for integers,
in the proof for $n = 1$.\xskip (b) The proof for $n=1$
carries over to any finite field.\xskip (c) Since $x = \xi ↑k$ for
some $k$, $x↑{p↑{\vbox to 3pt{}n}} = x$ in the field defined by $f(x)$.
Furthermore, the elements $y$ that satisfy the equation $y↑{p↑m} = y$ in the field
are closed under addition, and closed under multiplication;
so if $x↑{p↑m}= x$, then $\xi$ (being a polynomial in
$x$ with integer coefficients) satisfies $\xi↑{p↑m} =
\xi $.

\ansno 17. If $\xi$ is a primitive root, each nonzero element
is some power of $\xi $. Hence the order must be a divisor of
$13↑2 - 1 = 2↑3 \cdot 3 \cdot 7$, and $\varphi (f)$ elements have
order $f$.
\def\\#1{\hbox{$\hfill\hskip-10pt#1\hskip-10pt\hfill$}\hfill}
$$\vbox{\halign{\hfill#⊗\qquad\hfill#⊗\qquad\qquad\hfill#⊗\qquad\hfill#⊗\!
\qquad\qquad\hfill#⊗\qquad\hfill#⊗\qquad\qquad\hfill#⊗\qquad\hfill#\cr
\\f⊗\\{\varphi(f)}⊗\\f⊗\\{\varphi(f)}⊗\\f⊗\\{\varphi(f)}⊗\\f⊗\\{\varphi(f)}\cr
\noalign{\vskip3pt}
1⊗1⊗3⊗2⊗7⊗6⊗21⊗12\cr
2⊗1⊗6⊗2⊗14⊗6⊗42⊗12\cr
4⊗2⊗12⊗4⊗28⊗12⊗84⊗24\cr
8⊗4⊗24⊗8⊗56⊗24⊗168⊗48\cr}}$$

\ansno 18. (a)\9 pp$\biglp p↓1(u↓nx)\bigrp \ldotsm\hbox{pp}
\biglp p↓r(u↓nx)\bigrp$, by \α{Gauss's lemma}. For example, let
$$u(x) = 6x↑3 - 3x↑2 + 2x - 1,\qquad v(x) = x↑3 - 3x↑2 + 12x
- 36 = (x↑2 + 12)(x - 3);$$
then pp$(36x↑2 + 12) = 3x↑2 + 1$, pp$(6x - 3) =
2x - 1$.\xskip (This is a modern version of a fourteenth-century trick
used for many years to help solve algebraic equations.)

(b)\9 Let pp$\biglp w(u↓nx)\bigrp = \=w↓mx↑m
+ \cdots + \=w↓0 = w(u↓nx)/c$, where $c$ is the content
of $w(u↓nx)$ as a polynomial in $x$. Then $w(x) = (c\=w/u↑{m}↓{n})x↑m
+ \cdots + c\=w↓0$, hence $c\=w↓m = u↑{m}↓{n}$;
since $\=w↓m$ is a divisor of $u↓n$, $c$ is a multiple of $u↑{m-1}↓{n}$.

\ansno 19. If $u(x) = v(x)w(x)$ with deg$(v)$deg$(w) ≥
1$, then $u↓nx↑n ≡ v(x)w(x) \modulo p$. By unique factorization
modulo $p$, all but the leading coefficients of $v$ and $w$ are multiples of $p$,
and $p↑2$ divides $v↓0w↓0=u↓0$.

\ansno 20. (a)\9 $\sum (αu↓j - u↓{j-1})(\=α
{\=u}↓j - {\=u}↓{j-1}) = \sum (u↓j - \=αu↓{j-1})({\=u}↓j
- α{\=u}↓{j-1})$.\xskip (b) We may assume that $u↓0 ≠ 0$. Let
$m(u) = \prod↓{1≤j≤n} \min(1, |α↓j|) = u↓0/M(u)u↓n$. Whenever
$|α↓j| < 1$, change the factor $x - α↓j$ to ${\=α}↓jx -
1$ in $u(x)$; this doesn't affect $|u|$. Now looking at the
leading and trailing coefficients only, we have $|u|↑2 ≥ |u↓n|↑2m(u)↑2
+ |u↓n|↑2M(u)↑2$; hence we obtain the slightly stronger result
$M(u)↑2 ≤ \biglp |u|↑2 + (|u|↑4 - 4|u↓0u↓n|↑2)↑{1/2}\bigrp
/2|u↓n|↑2$.\xskip
[A further improvement in the estimate of $M(u)$ can often be obtained by
replacing $u(x)$ by the polynomial $\A u(x)=\biglp x↑k-s↓k/|u|↑2\bigrp u(x)$,
where $s↓k=\sum↓j\=u↓ju↓{j-k}$; since $M(\A u)=M(u)$ and $|\A u|↑2=|u|↑2-
|s↓k|↑2\!/|u|↑2$, we obtain the inequality $|u|↑2-|s↓k|↑2\!/|u|↑2≥|u↓0|↑2|s↓k|↑2
\!/\biglp|u|↑4M(u)↑2\bigrp+|u↓n|↑2M(u)↑2$ for $1≤k<n$. In the case of polynomial↔
(22), we have $s↓2=-72$ and we obtain the bound $M(u)<8.1837$.]\xskip
(c)↔$u↓j = u↓m \sum α↓{i↓1}\ldotsm α↓{i↓{m-j}}$,
an elementary symmetric function, hence
$|u↓j| ≤ |u↓m| \sum β↓{i↓1} \ldotsm β↓{i↓{m-j}}$
where $β↓i = \max(1, |α↓i|)$. We complete the proof by showing
that when $x↓1≥1$, $\ldotss$, $x↓n≥1$, and $x↓1 \ldotsm x↓n = M$,
the elementary symmetric function $\sigma ↓{nk} = \sum x↓{i↓1}
\ldotsm x↓{i↓k}$ is $≤{n-1\choose k-1}M + {n-1\choose k}$,
the value assumed when $x↓1 = \cdots = x↓{n-1} = 1$ and $x↓n
= M$.\xskip$\biglp$For if $x↓1 ≤ \cdots ≤ x↓n < M$, the transformation $x↓n
← x↓{n-1}x↓n$, $x↓{n-1} ← 1$ increases $\sigma ↓{nk}$ by $\sigma
↓{(n-2)(k-1)}(x↓n - 1)(x↓{n-1} - 1)$, which is positive.$\bigrp$\xskip
(d)↔$|v↓j| ≤ |v↓m|
\mathopen{\vcenter{\hbox{\:@\char'0}}}{m-1\choose j}M(v)+
{m-1\choose j-1}\mathclose{\vcenter{\hbox{\:@\char'1}}}≤|u↓n|
\mathopen{\vcenter{\hbox{\:@\char'0}}}{m-1\choose j}M(u)+
{m-1\choose j-1}\mathclose{\vcenter{\hbox{\:@\char'1}}}$
since $|v↓m| ≤ |u↓n|$ and
$M(v) ≤ M(u)$.\xskip [See
J. \α{Vicente Gon\c calves}, {\sl Revista de Faculdade de Ci\A encias} (2) A
{\bf1} (Univ. Lisbon, 1950), 167--171;
M. \α{Mignotte,} {\sl Math.\ Comp.\ \bf 28} (1974), 1153--1157;
Mignotte and \α{Payafar}, {\sl RAIRO Analyse num\'er.\ \bf13} (1979), 181--192.]

\ansno 21. (a)\9 $\int↑{1}↓{0}\biglp u↓ne(n\theta ) + \cdots + u↓0)({\=u}↓n
e(-n\theta ) + \cdots + {\=u}↓0)\,d\theta = |u↓n|↑2 +
\cdots + |u↓0|↑2$ since $\int ↑{1}↓{0} e(j\theta )e(-k\theta
)\,d\theta = \delta↓{jk}$; now use induction on $t$.\xskip (b) Since
$|v↓j| ≤ {m\choose j}M(v)|v↓m|$ we conclude that $|v|↑2 ≤ {2m\choose
m}M(v)↑2|v↓m|↑2$. Hence $|v|↑2|w|↑2 ≤ {2m\choose m}{2k\choose
k}M(v)↑2M(w)↑2|v↓mw↓k|↑2 = f(m, k)M(u)↑2|u↓n|↑2 ≤ f(m, k)|u|↑2$.\xskip
[Slightly better values for $f(m, k)$ are possible based on
the more detailed information in exercise↔20.]\xskip (c) The case
$t = 3$ suffices to show how to get from $t - 1$ to $t$. When
$t = 2$ we have shown that, for all $\theta ↓1$, $$\twoline{\textstyle\int ↑{1}↓{0}
\int ↑{1}↓{0} \int ↑{1}↓{0} \int ↑{1}↓{0}\left|v\biglp e(\theta
↓1), e(\phi ↓2), e(\phi ↓3)\bigrp\right|↑2\left|w\biglp e(\theta ↓1),
e(\psi ↓2), e(\psi ↓3)\bigrp \right|↑2\,d\phi ↓2\,d\phi ↓3\,d\psi ↓2\,
d\psi ↓3}{3pt}{\textstyle≤ f(m↓2, k↓2)f(m↓3, k↓3) \int ↑{1}↓{0} \int ↑{1}↓{0}
\left|v\biglp e(\theta ↓1), e(\theta ↓2), e(\theta ↓3)\bigrp \right|↑2
\left|w\biglp e(\theta ↓1), e(\theta ↓2), e(\theta ↓3)\bigrp \right|↑2
\,d\theta ↓2\,d\theta ↓3.}$$ For all $\phi ↓2$, $\phi ↓3$, $\psi ↓2$,
$\psi ↓3$ we have also shown that$$\twoline{\textstyle\int ↑{1}↓{0} \int ↑{1}↓{0}
\left|v\biglp e(\phi ↓1), e(\phi ↓2), e(\phi ↓3)\bigrp \right|↑2\left|w\biglp
e(\psi ↓1), e(\psi ↓2), e(\psi ↓3)\bigrp \right|↑2\,d\phi ↓1\,d\psi
↓1}{3pt}{\textstyle≤ f(m↓1, k↓1) \int ↑{1}↓{0}\left|v\biglp e(\theta ↓1), e(\phi
↓2), e(\phi ↓3)\bigrp \right|↑2\left|w\biglp e(\theta ↓1), e(\psi ↓2),
e(\psi ↓3)\bigrp \right|↑2\, d\theta ↓1.}$$Integrate the former inequality
with respect to $\theta ↓1$ and the latter with respect to $\phi
↓2$, $\phi ↓3$, $\psi ↓2$, $\psi ↓3$.\xskip [This method was used by A.
O. \α{Gel'fond} in {\sl Transcendental and Algebraic Numbers} (New
York: Dover, 1960), Section 3.4, to derive a slightly different
result.]

\def\\{\hbox{deg}}
\ansno 22. More generally, assume that $u(x) ≡ v(x)w(x)\modulo
q$, $a(x)v(x) + b(x)w(x) ≡ 1 \modulo p$, and $c\lscr(v) ≡ 1
\modulo r$, $\\(a) < \\(w)$, $\\(b) < \\(v)$, $\\(u)
= \\(v) + \\(w)$, where $r = \gcd(p, q)$ and $p, q$ needn't
be prime. We shall construct polynomials $V(x) ≡ v(x)$ and $W(x)
≡ w(x) \modulo q$ such that $u(x) ≡ V(x)W(x) \modulo {qr}$,
$\lscr(V) = \lscr(v)$, $\\(V) = \\(v)$, $\\(W) =\\(w)$; furthermore,
if $r$ is prime, the results will be unique modulo $qr$.

The problem asks us to find $\=v(x)$ and
$\=w(x)$ with $V(x) = v(x) + q\=v(x)$, $W(x)
= w(x) + q\=w(x)$, $\\(\=v) < \\(v)$, $\\(\=w)≤
\\(w)$; and the other condition$$\biglp v(x) + q\=v
(x)\bigrp\biglp w(x) + q\=w(x)\bigrp ≡ u(x)\;\modulo
{qr}$$is equivalent to $\=w(x)v(x) + \=v(x)w(x)
≡ f(x)\modulo r$, where $f(x)$ satisfies $u(x) ≡ v(x)w(x) + qf(x) \modulo
{qr}$. We have$$\biglp a(x)f(x) + t(x)w(x)\bigrp v(x) + \biglp
b(x)f(x) - t(x)v(x)\bigrp w(x) ≡ f(x)\;\modulo r$$for all
$t(x)$. Since $\lscr(v)$ has an inverse modulo $r$, we can find
a quotient $t(x)$ by Algorithm↔4.6.1D such that $\\(bf - tv)
< \\(v)$; for this $t(x)$, $\\(af + tw) ≤ \\(w)$, since we have
$\\(f) ≤ \\(u) = \\(v) + \\(w)$. Thus the desired
solution is $\=v(x) = b(x)f(x) - t(x)v(x) = b(x)f(x)\mod
v(x)$, $\=w(x) = a(x)f(x) + t(x)w(x)$. If $\biglp \={\=v}
(x), \={\=w}(x)\bigrp$ is another
solution, we have $\biglp \=w(x) - \={\=w}(x)\bigrp
v(x) ≡ \biglp \={\=v}(x) - \=v(x)\bigrp w(x)
\modulo r$. Thus if $r$ is prime, $v(x)$ must divide $\={\=v}
(x) - \=v(x)$; but $\\(\={\=v}
- \=v) < \\(v)$, so $\={\=v}(x) = 
\=v(x)$ and $\={\=w}(x) = \=w(x)$.

For $p = 2$, the factorization proceeds as follows
(writing only the coefficients, and using bars for negative
digits): Exercise↔10 says that $v↓1(x) = (\overline1\,\overline
1\,\overline1)$, $w↓1(x) = (\overline1\,\overline1\,\overline1\,
0\,0\,\overline1\,\overline1)$ in one-bit two's complement notation.
Euclid's extended algorithm yields $a(x) = (1\,0\,0\,0\,0\,1), b(x)
= (1\,0)$. The factor $v(x) = x↑2 + c↓1x + c↓0$ must have $|c↓1|
≤ \lfloor 1 + \sqrt{113}\rfloor = 11$, $|c↓0| ≤ 10$, by exercise↔20.
Three applications of Hensel's lemma yield $v↓4(x) = (1\,
3\,\overline1)$, $w↓4(x) = (1\,\overline3\,\overline5\,\overline4\,
4\,\overline3\,5)$. Thus $c↓1 ≡ 3$ and $c↓0 ≡ -1\modulo {16}$;
the only possible quadratic factor of $u(x)$ is $x↑2 + 3x -
1$. Division fails, so $u(x)$ is irreducible.\xskip$\biglp$Since we have
now proved the irreducibility of this beloved polynomial by
four separate methods, it is unlikely that it has any factors.$\bigrp$

Hans \α{Zassenhaus} has observed that we can often
speed up such calculations by increasing $p$ as well as $q$:
In the above notation, we can find $A(x)$, $B(x)$ such that $A(x)V(x)
+ B(x)W(x) ≡ 1 \modulo {pr}$, namely by taking $A(x) = a(x)
+ p\=a(x)$, $B(x) = b(x) + p\=b(x)$, where $\=a
(x)V(x) + \=b(x)W(x) ≡ g(x)\modulo r$, $a(x)V(x)
+ b(x)W(x) ≡ 1 - pg(x) \modulo {pr}$. We can also find $C$
with $\lscr(V)C ≡ 1\modulo {pr}$. In this way we can lift a
squarefree factorization $u(x) ≡ v(x)w(x) \modulo p$ to its
unique extensions modulo $p↑2$, $p↑4$, $p↑8$, $p↑{16}$, etc. However,
this ``accelerated'' procedure reaches a point of diminishing
returns in practice, as soon as we get to double-precision moduli,
since the time for multiplying multiprecision numbers in practical
ranges outweighs the advantage of squaring the modulus directly.
From a computational standpoint it seems best to work with the
successive moduli $p$, $p↑2$, $p↑4$, $p↑8$, $\ldotss$, $p↑E$, $p↑{E+e}$,
$p↑{E+2e}$, $p↑{E+3e}$, $\ldotss $, where $E$ is the smallest power
of 2 with $p↑E$ greater than single precision and $e$ is the
largest integer such that $p↑e$ has single precision.

Hensel's lemma, which K. \α{Hensel} introduced in order to
demonstrate the factorization of polynomials over the field
of \α{$p$-adic numbers} (see exercise 4.1--31), can be generalized
in several ways. First, if there are more factors, say $u(x)
≡ v↓1(x)v↓2(x)v↓3(x) \modulo p$, we can find $a↓1(x)$, $a↓2(x)$,
$a↓3(x)$ such that $a↓1(x)v↓2(x)v↓3(x) + a↓2(x)v↓1(x)v↓3(x) +
a↓3(x)v↓1(x)v↓2(x) ≡ 1 \modulo p$ and $\\(a↓i) < \\(v↓i)$.\xskip
$\biglp$In essence, $1/u(x)$ is expanded in partial fractions as $\sum
a↓i(x)/v↓i(x)$.$\bigrp$\xskip An exactly analogous construction now allows
us to lift the factorization without changing the leading coefficients
of $v↓1$ and $v↓2$; we take $\=v↓1(x) = a↓1(x)f(x)\mod
v↓1(x)$, $\=v↓2(x) = a↓2(x)f(x)\mod v↓2(x)$, etc. Another
important generalization is to several simultaneous moduli,
of the respective forms $p↑e$, $(x↓2 - a↓2)↑{n↓2}$, $\ldotss
$, $(x↓t - a↓t)↑{n↓t}$, when performing multivariate gcds
and factorizations. Cf.\ D. Y. Y. \α{Yun,} Ph.D. Thesis (M.I.T.,
1974).

\ansno 23. The \α{discriminant} of pp$\biglp u(x)\bigrp$ is a nonzero integer (cf.\
exercise 4.6.1--12), and there are multiple factors modulo $p$
iff $p$ divides the discriminant.\xskip[The factorization of (22)
modulo 3 is $(x + 1)(x↑2 - x - 1)↑2(x↑3 + x↑2 - x + 1)$; squared
factors for this polynomial occur only for $p = 3$, 23, 233,
and 121702457. It is not difficult to prove that the smallest
prime that is not unlucky is at most $O(n\log Nn)$, if $n
= \\(u)$ and $N$ bounds the coefficients of $u(x)$.]

% Vicinity of page 628
% folio 818 galley 4a 	*
\ansno 24. Multiply a monic polynomial with rational coefficients
by a suitable nonzero integer, to get a primitive polynomial
over the integers. Factor this polynomial over the integers,
and then convert the factors back to monic.\xskip (No factorizations
are lost in this way; see exercise↔4.6.1--8.)

\ansno 25. Consideration of the constant term shows there are
no factors of degree 1, so if the polynomial is reducible, it
must have one factor of degree 2 and one of degree 3. Modulo
2 the factors are $x(x + 1)↑2(x↑2+x+1)$; this is not much help.
Modulo 3 the factors are $(x + 2)↑2(x↑3+2x+2)$. Modulo 5 they
are $(x↑2 + x + 1)(x↑3 + 4x + 2)$. So we see that the answer
is $(x↑2 + x + 1)(x↑3 - x + 2)$.

\ansno 26. Begin with $D ← (0 \ldotsm 0 1)$, representing the
set $\{0\}$. Then for $1 ≤ j ≤ r$, set $D ← D ∨ (D\lsh d↓j)$,
where $∨$ denotes \α{logical ``or''} and $D\lsh d$ denotes $D$ shifted left
$d$ bit positions.\xskip$\biglp$Actually we need only work with a bit
vector of length $\lceil(n + 1)/2\rceil$, since $n - m$ is in the set iff
$m$ is.$\bigrp$

\ansno 27. Exercise 4 says that a random polynomial of degree
$n$ is irreducible modulo $p$ with rather low probability, about
$1/n$. But the Chinese remainder theorem implies that a random
monic polynomial of degree $n$ over the integers will be reducible
with respect to each of $k$ distinct primes with probability
about $(1 - 1/n)↑k$, and this approaches zero as $k → ∞$. Hence
almost all polynomials over the integers are irreducible with
respect to infinitely many primes; and almost all primitive
polynomials over the integers are irreducible.\xskip [Another proof
has been given by W. S. \α{Brown,} {\sl AMM \bf 70} (1963), 965--969. See
also the generalization cited in the answer to exercise↔36.]

\def\\{\hbox{deg}}
\ansno 28. Cf.\ exercise↔4; the probability is the coefficient
of $z↑n$ in ${(1\!+\!a↓{1p}z/p)}\*
{(1\!+\!a↓{2p}z↑2\!/p↑2)}\*{(1\!+\!a↓{3p}z↑3/p↑3)}\ldotsm $, which has the
limiting value $g(z) = {(1 + z)}\*{(1 + {1\over 2}z↑2)}\*
{(1 + {1\over 3}z↑3)}\ldotsm\,$. For $1 ≤ n ≤ 10$ the answers are 1, ${1\over
2}$, ${5\over 6}$, ${7\over 12}$, ${37\over 60}$, ${79\over
120}$, ${173\over 280}$, ${101\over 168}$, ${127\over 210}$, ${1033\over
1680}$.\xskip{\:a[}Let
$f(y) = \ln(1 + y) - y = O(y↑2)$. We have $$\textstyle g(z) = \exp\biglp \sum
↓{n≥1} z↑n/n + \sum ↓{n≥1} f(z↑n/n)\bigrp =
h(z)/(1 - z),$$ and it can be shown that the limiting probability
is $h(1) =
\exp\biglp \sum ↓{n≥1} f(1/n)\bigrp =e↑{-\gamma}\approx .56146$
as $n → ∞$; cf.\ D. H. \α{Lehmer,} {\sl Acta Arith.\ \bf 21} (1972),
379--388. Indeed, N.↔G. \α{de Bruijn} has established the asymptotic
formula $\lim↓{p→∞} a↓{np} =
e↑{-\gamma}+ e↑{-\gamma}/n + O(n↑{-2}\log n)$.{\:a]}

\ansno 29. Let $q↓1(x)$ and $q↓2(x)$ be any two of the irreducible divisors of
$g(x)$. By the Chinese remainder theorem (exercise↔3), choosing a random polynomial
$t(x)$ of degree $<2d$ is equivalent to choosing two random polynomials
$t↓1(x)$ and $t↓2(x)$ of degrees $<d$, where $t↓i(x)=t(x)\mod q↓i(x)$. The gcd
will be a proper factor if $t↓1(x)↑{(p↑d-1)/2}\mod q↓1(x)=1$ and
$t↓2(x)↑{(p↑d-1)/2}\mod q↓1(x)≠1$, or vice versa, and this condition holds for
exactly $2\biglp(p↑d-1)/2\bigrp\biglp(p↑d+1)/2\bigrp=(p↑{2d}-1)/2$ choices of
$t↓1(x)$ and $t↓2(x)$.

{\sl Notes:} We are considering here only the behavior with respect to two
irreducible factors, but the true behavior is probably much better. Suppose that each
irreducible factor $q↓i(x)$ has probability $1\over2$ of dividing $t(x)↑{(p↑d-1)/2}-1$
for each $t(x)$, independent of the behavior for other $q↓j(x)$ and $t(x)$; and
assume that $g(x)$ has $r$ irreducible factors in all.
Then if we encode each $q↓i(x)$ by a sequence of 0's and 1's
according as $q↓i(x)$ does or doesn't divide $t(x)↑{(p↑d-1)/2}-1$ for the successive
$t$'s tried, we obtain a random binary \α{trie} with $r$ lieves (cf.\ Section 6.3).
The cost associated with an internal node of this trie, having $m$ lieves as
descendants, is $O\biglp m↑2(\log p)\bigrp$; and the solution to
the recurrence $A↓n={n\choose2}+2↑{1-n}\sum{n\choose k}A↓k$ is $A↓n=2{n\choose2}$,
by exercise 5.2.2--36. Hence the sum of costs
in the given random trie---representing the expected time to factor $g(x)$ {\sl
completely}---is $O\biglp r↑2(\log p)↑3\bigrp$ under this
plausible assumption. The plausible assumption becomes rigorously true if we choose
$t(x)$ at random of degree $<rd$ instead of restricting it to degree $<2d$.

\ansno 30. Let $T(x)=x+x↑{\mskip 1mu p}+\cdots+x↑{p↑{d-1}}$ and $v(x)=T\biglp t(x)\bigrp
\mod q(x)$. Since $t(x)↑{p↑d}=t(x)$ in the field of polynomial remainders
modulo $q(x)$, we have $v(x)↑{\mskip 1mu p}=v(x)$ in that field; in other words, $v(x)$
is one of the $p$ roots of the equation $y↑{\mskip 1mu p}-y=0$. Hence $v(x)$ is an integer.

It follows that $\prod↓{0≤s<p}\gcd\biglp g↓d(x),T\biglp t(x)\bigrp-s\bigrp=
g↓d(x)$. In particular, when $p=2$ we can argue as in exercise 29 that
$\gcd\biglp g↓d(x),T\biglp t(x)\bigrp\bigrp$ will be a proper factor of $g↓d(x)$
with probability $≥{1\over2}$ when $g↓d(x)$ has at least two irreducible
factors and $t(x)$ is a random binary polynomial of degree $<2d$.

[Note that $T\biglp t(x)\bigrp\mod g(x)$ can be computed by starting with
$u(x)←t(x)$ and setting $u(x)←\biglp u(x)+u(x)↑{\mskip 1mu p}\bigrp\mod g(x)$ repeatedly,
$d-1$ times. The method of this exercise is based on the polynomial
factorization $x↑{p↑d}-x=\prod↓{0≤s<p}\biglp T(x)-s\bigrp$, which holds for
any↔$p$, while formula
(21) is based on the polynomial factorization $x↑{p↑d}-x=x(x↑{(p↑d-1)/2}+1)
(x↑{(p↑d-1)/2}-1)$ for odd↔$p$.]

\ansno 31. If $α$ is an element of the field of $p↑d$ elements, let $d(α)$ be
the ``degree'' of $α$, namely the smallest exponent $e$ such that $α↑{p↑e}=α$.
Then$$P↓α(x)=(x-α)(x-α↑{\mskip 1mu p})\ldotsm(x-α↑{p↑{d-1}})=q↓α(x)↑{d/d(α)},$$where
$q↓α(x)$ is an irreducible polynomial of degree $d(α)$. As $α$ runs through
all elements of the field, the corresponding
$q↓α(x)$ runs through every irreducible polynomial
of degree $e$ dividing↔$d$, where every such irreducible occurs exactly $e$
times. We have $(x+t)↑{(p↑d-1)/2}\mod q↓α(x)=1$ if and only if $(α+t)↑
{(p↑d-1)/2}=1$ in the field. If $t$ is an integer, we have $d(α+t)=d(α)$,
hence $n(p,d)$ is $d↑{-1}$ times the number of elements $α$ of degree $d$
such that $α↑{(p↑d-1)/2}=1$. Similarly, if $t↓1≠t↓2$ we want to count the
number of elements of degree $d$ such that $(α+t↓1)↑{(p↑d-1)/2}=
(α+t↓2)↑{(p↑d-1)/2}$, i.e., $\biglp(α+t↓1)/(a+t↓2)\bigrp↑{(p↑d-1)/2}=1$.
As $α$ runs through all the elements of degree $d$, so does the quantity
$(α+t↓1)/(α+t↓2)=1+(t↓1-t↓2)/(α+t↓2)$.

[We have $n(p,d)={1\over4}d↑{-1}\sum↓{c\rslash d}\biglp3+(-1)↑c\bigrp
\mu(c)(p↑{d/c}-1)$, which is about half the total number of irreducibles---exactly
half, in fact, when $d$ is odd.
This proves that $\gcd\biglp g↓d(x),(x+t)↑{(p↑d-1)/2}-1\bigrp$
has a good chance of finding factors of $g↓d(x)$ when $t$ is fixed and $g↓d(x)$ is
chosen at random; but a \α{probabilistic algorithm} is supposed to work with
guaranteed probability for fixed $g↓d(x)$ and random $t$, as in exercise↔29.]

\def\+{\hbox{$\Psi$\hskip-1.5pt}}
\ansno 32. (a)\9 Clearly $x↑n-1=\prod↓{d\rslash n}\+↓d(x)$, since every complex
$n$th root of unity is a primitive $d$th root for some unique $d\rslash n$.
The second identity follows from the first; and $\+↓n(x)$ has integer
coefficients since it is expressed in terms of products and quotients of monic
polynomials with integer coefficients.\xskip(b) The condition in the hint
suffices to prove that $f(x)=\+↓n(x)$, so we shall take the hint. When
$p$ does not divide $n$, we have $\gcd(x↑n-1,nx↑{n-1})=1$ modulo $p$, hence
$x↑n-1$ is squarefree modulo $p$. Given $f(x)$ and $\zeta$ as in the hint,
let $g(x)$ be the irreducible factor of $\+↓n(x)$ such that $g(\zeta↑{\mskip 1mu p})=0$.
If $g(x)≠f(x)$ then both $f(x)$ and $g(x)$ are distinct factors of $\+↓n(x)$,
hence they are distinct factors of $x↑n-1$, hence they have no irreducible
factors in common modulo↔$p$. However, $\zeta↑{\mskip 1mu p}$ is a root of $f(x↑{\mskip 1mu p})$,
so $\gcd\biglp g(x),f(x↑{\mskip 1mu p})\bigrp≠1$ over the integers, hence $g(x)$ is
a divisor of $f(x↑{\mskip 1mu p})$. By (5), $g(x)$ is a divisor of $f(x)↑{\mskip 1mu p}$, modulo $p$,
contradicting the assumption that $f(x)$ and $g(x)$ have no irreducible
factors in common. Therefore $f(x)=g(x)$.\xskip[The irreducibility of
$\+↓n(x)$ was first proved for prime $n$ by K. F. \α{Gauss} in {\sl Disquisitiones
Arithmetic\ae\ }(Leipzig, 1801), Art.\ 341, and for general $n$ by L.
\α{Kronecker,} {\sl J. de Math. Pures et Appliqu\'ees \bf19} (1854), 177--192.]

(c)\9 $\+↓1(x)=x-1$; and when $p$ is prime, $\+↓p(x)=1+x+\cdots+x↑{p-1}$.
If $n>1$ is odd, it is not difficult to prove that $\+↓{2n}(x)=\+↓n(-x)$.
If $p$ divides $n$, the second identity in (a) shows that $\+↓{pn}(x)=
\+↓n(x↑{\mskip 1mu p})$. If $p$ does not divide $n$, we have $\+↓{pn}(x)=\+↓n(x↑{\mskip 1mu p})/
\+↓n(x)$. For nonprime $n≤15$ we have $\+↓4(x)=x↑2+1$, $\+↓6(x)=
x↑2-x+1$, $\+↓8(x)=x↑4+1$, $\+↓9(x)=x↑6+x↑3+1$, $\+↓{10}(x)=
x↑4-x↑3+x↑2-x+1$, $\+↓{12}(x)=x↑4-x↑2+1$,
$\+↓{14}(x)=x↑6-x↑5+x↑4-x↑3+x↑2-x+1$, $\+↓{15}(x)=
x↑8-x↑7+x↑5-x↑4+x↑3-x+1$.\xskip[The formula $\+↓{pq}(x)=(1+x↑{\mskip 1mu p}+\cdots+x↑{(q-1)p})
(x-1)/(x↑q-1)$ can be used to show that $\+↓{pq}(x)$ has all coefficients
$\pm1$ or 0 when $p$ and $q$ are prime; but the coefficients of $\+↓{pqr}(x)$
can be arbitrarily large.]

\ansno 33. False; we lose all $p↓j$ with $e↓j$ divisible by
$p$. True if $p ≥ \\(u)$.\xskip[See exercise↔36.]

\ansno 34. [D. Y. Y. \α{Yun}, {\sl Proc.\ ACM Symp.\ Symbolic and Algebraic
Comp.\ }(1976), 26--35.]\xskip Set $\biglp t(x),v↓1(x),w↓1(x)\bigrp←\hbox{GCD}\biglp
u(x),u↑\prime(x)\bigrp$. If $t(x)=1$, set $e←1$; otherwise set $\biglp u↓i(x),
v↓{i+1}(x),w↓{i+1}(x)\bigrp←\hbox{GCD}\biglp v↓i(x),w↓i(x)-v↓{\!i}↑\prime(x)\bigrp$
for $i=1$, 2, $\ldotss$, $e-1$, until finding $w↓e(x)-v↓{\!e}↑\prime(x)=0$.
Finally set $u↓e(x)←v↓e(x)$.

To prove the validity of this algorithm, we observe that it computes the polynomials
$t(x)=u↓2(x)u↓3(x)↑2u↓4(x)↑3
\ldotsm$, $v↓i(x)=u↓i(x)u↓{i+1}(x)u↓{i+2}(x)\ldotsm$, and
$$\twoline{w↓i(x)=u↓{\!i}↑\prime(x)u↓{i+1}(x)u↓{i+2}(x)\ldotsm+
2u↓i(x)u↓{\!i+1}↑\prime(x)u↓{i+2}(x)\ldotsm}{3pt}{\null+
3u↓i(x)u↓{i+1}(x)u↓{\!i+2}↑\prime(x)\ldotsm+\cdotss.}$$
We have $\gcd\biglp t(x),w↓1(x)\bigrp=1$, since an irreducible factor of $u↓i(x)$
divides all but the $i$th term of $w↓1(x)$, and it is relatively prime to that
term. Fur\-ther\-more $\gcd\biglp u↓i(x),v↓{i+1}(x)\bigrp$ is clearly↔1.

[Exercise 2(b) indicates that comparatively few polynomials are squarefree, but
non-squarefree polynomials actually occur often in practice; hence this method
turns out to be quite important. See Paul S. \α{Wang} and Barry M. \α{Trager},
{\sl SIAM J. Computing \bf8} (1979), 300--305, for suggestions on how to
improve the efficiency when the given polynomial is already squarefree.]

\ansno 35. We have $w↓j(x)=\gcd\biglp u↓j(x),v↓{\!j}↑{\ast}(x)\bigrp\cdot
\gcd\biglp u↓{\!j+1}↑\ast(x),v↓j(x)\bigrp$, where$$u↓{\!j}↑\ast(x)=u↓j(x)
u↓{j+1}(x)\ldotsm\qquad\hbox{and}\qquad
v↓{\!j}↑\ast(x)=v↓j(x)v↓{j+1}(x)\ldotsm\,.$$
[\α{Yun} notes that the running time for squarefree factorization by the method
of exercise↔34 is at most about twice the running time to calculate $\gcd\biglp
u(x),u↑\prime(x)\bigrp$. Furthermore if we are given an arbitrary method for
discovering squarefree factorization, the method of this exercise leads to a
gcd procedure.\xskip(When $u(x)$ and $v(x)$ are squarefree, their gcd is
simply $w↓2(x)$ where $w(x)=u(x)v(x)=w↓1(x)w↓2(x)↑2$; the polynomials
$u↓j(x)$, $v↓j(x)$, $u↓{\!j}↑\ast(x)$, and $v↓{\!j}↑\ast(x)$ are all
squarefree.)\xskip Hence the problem of converting a primitive polynomial
of degree $n$ to its squarefree representation is computationally {\sl equivalent\/}
to the problem of calculating the gcd of two $n$th degree polynomials, in the
sense of asymptotic worst-case running time.]

\ansno 36. Let $U↓j(x)$ be the value computed for ``$u↓j(x)$'' by the procedure
of exercise↔34. If $\\(U↓1)+2\\(U↓2)+\cdots=\\(u)$, then $u↓j(x)=U↓j(x)$ for
all $j$. But in general we will have $e<p$ and $U↓j(x)=\prod↓{k≥0}u↓{j+pk}(x)$
for $1≤j<p$. To separate these factors further, we can calculate $t(x)/\biglp
U↓2(x)U↓3(x)↑2\ldotss U↓{p-1}(x)↑{p-2}\bigrp=\prod↓{j≥p}u↓j(x)↑{p\lfloor j/p\rfloor}
=z(x↑{\mskip 1mu p})$. After \α{recursive}ly finding the squarefree representation of
$z(x)=\biglp z↓1(x),z↓2(x),\ldotss\bigrp$, we will have $z↓k(x)=\prod↓{0≤j<p}
u↓{j+pk}(x)$, so we can calculate the individual $u↓i(x)$ by the formula
$\gcd\biglp U↓j(x),z↓k(x)\bigrp=u↓{j+pk}(x)$ for $1≤j<p$. The polynomial
$u↓{pk}(x)$ will be left when the other factors of $z↓k(x)$ have been removed.

{\sl Note:} This procedure is fairly simple but the program is lengthy.
If one's goal is to have a short program for complete factorization modulo↔$p$,
rather than an extremely efficient one, it is probably easiest to
modify the \α{distinct-degree factorization} routine so that it casts out
$\gcd\biglp x↑{p↑d}-x,u(x)\bigrp$ several times for the same value of $d$
until the gcd is↔1. In this case you needn't begin by calculating $\gcd\biglp
u(x),u↑\prime(x)\bigrp$ and removing multiple factors as suggested in the text,
since the polynomial $x↑{p↑d}-x$ is squarefree.

\ansno 37. The exact probability is $\prod↓{j≥1}(a↓{jp}/p↑j)↑{k↓j}/k↓j!$,
where $k↓j$ is the number of $d↓i$ that are
equal to $j$. Since $a↓{jp}/p↑j\approx 1/j$
by exercise↔4, we get the formula of exercise 1.3.3--21.

{\sl Notes:} This exercise says that if we fix the prime $p$ and let the
polynomial $u(x)$ be random, it will have certain probability of splitting
in a given way modulo $p$. A much harder problem is to fix the polynomial $u(x)$
and to let $p$ be ``random''; it turns out that the same asymptotic result holds
for almost all $u(x)$.\xskip G. \β{Frobenius} proved in 1880 that the integer
polynomial $u(x)$ splits modulo $p$ into factors of degrees $d↓1$, $\ldotss$, $d↓r$,
when $p$ is a large prime chosen at random, with probability equal to the
number of permutations in the \β{Galois group} $G$ of $u(x)$ having cycle lengths
$\{d↓1,\ldotss,d↓r\}$ divided by the total number of permutations in $G$.\xskip
$\biglp$If $u(x)$ has rational coefficients and distinct roots $\xi↓1$,
$\ldotss$,↔$\xi↓n$ over the complex numbers, its Galois group is the (unique) group $G$
of permutations such that the polynomial $\prod↓{p(1)\ldotsm p(n)\in G}
\biglp z+\xi↓{p(1)}y↓1+\cdots+\xi↓{p(n)}y↓n\bigrp=U(z,y↓1,\ldotss,y↓n)$ has rational
coefficients and is irreducible over the rationals.$\bigrp$\xskip Furthermore
B.↔L. \α{van↔der↔Waerden} proved in 1934 that almost all polynomials of degree $n$
have the set of all $n!$ permutations as their Galois group. Therefore
almost all fixed irreducible polynomials $u(x)$ will factor as we might expect them
to, with respect to randomly chosen large primes $p$.\xskip References:
{\sl Sitzungsberichte K\"onigl.\ Preu\ss.\ Akad.\ Wiss.\ }(Berlin: 1896),
\hbox{689--703}; {\sl Math.\ Annalen \bf109} (1934), 13--16. See also
N. \α{Chebotarev}, {\sl Math.\ Annalen \bf95} (1926), for a generalization of
Frobenius's theorem to conjugacy classes of the Galois group.

\ansno 38. (Partial solution by Peter \α{Weinberger}.)\xskip
The average number of 1-cycles in a randomly chosen element of any transitive
\α{permutation} group $G$ on $n$ objects is↔1, since the probability is $1/n$
that any given object is fixed. Since Galois groups are always transitive, the
remarks in the previous answer show that a fixed irreducible polynomial has
exactly one linear factor modulo↔$p$, on the average, as $p→∞$. Thus,
{\sl the average number of linear factors of $u(x)$ modulo↔$p$ is the
number of irreducible factors of $u(x)$ over the integers.}

Weinberger [{\sl Proc.\ Symp.\ Pure Math.\ \bf24} (Amer.\ Math.\ Soc., 1973),
321--332] has proved that if the \α{generalized Riemann hypothesis} (\α{GRH})
holds---this is a conjecture about the zeros of a generalized \α{zeta
function}---then there is an absolute constant $A↓1$ with the following property:
The number of prime ideals of norm $≤x$, in any \α{algebraic number field} of
degree↔$n$, differs from $\int↓2↑x dt/\!\ln t$ by at most $A↓1nx↑{1/2}\ln(x\Delta)$,
where $\Delta$ is the absolute value of the \α{discriminant} of the irreducible
polynomial $u$ that defines the field. The number of prime ideals of norm $≤x$
is at most $nx↑{1/2}$ different from the total number of linear factors of $u$
modulo primes $≤x$, since such linear factors correspond to ideals of norm↔$p$,
while other prime ideals have norms $≥p↑2$. Consequently if we let $N(x)$ be the
total number of linear factors of a given primitive polynomial $u$, modulo all
primes $≤x$, then the GRH implies that there is an absolute constant $A↓2$
such that $|N(x)/π(x)-r|<A↓2nx↑{-1/2}(\ln x)(\ln x\Delta)$. A proof of GRH
would yield a ``short'' proof of the number of irreducible factors of any primitive
polynomial $u$ over the integers, since we could evaluate $N(x)$ for a value of
$x$ sufficiently large to make this error bound less than ${1\over2}$. Unfortunately
$A↓2$ is quite large.
% Vicinity of page 632
% folio 819 galley 4b 	*

\ansbegin{4.6.3}

\ansno 1. $x↑m$, where $m=2↑{λ(n)}$, the highest
power of 2 less than or equal to $n$.

\ansno 2. Assume that $x$ is input in register↔A\null,
and $n$ in location↔\.{NN}; the output is in register↔X.

{\yyskip\tabskip30pt \mixfive{\!
01⊗A1⊗ENTX⊗1⊗1⊗\understep{A1. Initialize.}\cr
02⊗⊗STX⊗Y⊗1⊗$Y←1$.\cr
03⊗⊗STA⊗Z⊗1⊗$Z←x$.\cr
\\04⊗⊗LDA⊗NN⊗1⊗$N←n$.\cr
05⊗⊗JMP⊗2F⊗1⊗To A2.\cr
\\06⊗5H⊗SRB⊗1⊗L+1-K\cr
07⊗⊗STA⊗N⊗L+1-K⊗$N←\lfloor N↓{\null}/2\rfloor$.\cr
\\08⊗A5⊗LDA⊗Z⊗L⊗\understep{A5. S\hskip2.5pt}{\sl\hskip-2.5pt
q}\understep{uare $Z$.}\cr
09⊗⊗MUL⊗Z⊗L⊗$Z\times Z\mod w$\cr
10⊗⊗STX⊗Z⊗L⊗\qquad$→Z$.\cr
\\11⊗A2⊗LDA⊗N⊗L⊗\understep{A2. Halve $N$.}\cr
12⊗2H⊗JAE⊗5B⊗L+1⊗To A5 if $N$ is even.\cr
13⊗⊗SRB⊗1⊗K\cr
14⊗A4⊗JAZ⊗4F⊗K⊗Jump if $N=1$.\cr
15⊗⊗STA⊗N⊗K-1⊗$N←\lfloor N↓{\null}/2\rfloor$.\cr
\\16⊗A3⊗LDA⊗Z⊗K-1⊗\understep{A3. Multi}{\sl p\hskip-3pt}\understep{\hskip
3pt l\hskip1pt}{\sl\hskip-1pt y\hskip-2pt}\understep{\hskip2pt\ $Y$ b\hskip1pt
}{\sl\hskip-1pt y\hskip-2pt}\understep{\hskip2pt\ $Z$.}\cr
17⊗⊗MUL⊗Y⊗K-1⊗$Z\times Y\mod w$\cr
18⊗⊗STX⊗Y⊗K-1⊗\qquad$→Y$.\cr
19⊗⊗JMP⊗A5⊗K-1⊗To A5.\cr
\\20⊗4H⊗LDA⊗Z⊗1\cr
21⊗⊗MUL⊗Y⊗1⊗Do the final multiplication.\quad\blackslug\cr}}

\yyskip\noindent [It would be better programming
practice to change the instruction in line 05 to ``\.{JAP}'', followed
by an error indication. The running time is $21L + 16K + 8$,
where $L = λ(n)$ is one less than the number of bits in the
binary representation of $n$, and $K = \nu (n)$ is the number
of 1 bits in that representation.]

For the serial program, we may assume that $n$
is small enough to fit in an index register; otherwise serial
exponentiation is out of the question. The following program
leaves the output in register↔A:

{\yyskip\tabskip30pt\mixfive{\!
01⊗S1⊗LD1⊗NN⊗1⊗$\rI1←n$.\cr
02⊗⊗STA⊗X⊗1⊗$X←x$.\cr
03⊗⊗JMP⊗2F⊗1\cr
\\04⊗1H⊗MUL⊗X⊗N-1⊗$\rA\times X\mod w$\cr
05⊗⊗SLAX⊗5⊗N-1⊗\qquad$→\rA$.\cr
\\06⊗2H⊗DEC1⊗1⊗N⊗$\rI1←\rI1-1$.\cr
07⊗⊗J1P⊗1B⊗N⊗Multiply again if $\rI1>0$.\quad\blackslug\cr}}

\yyskip\noindent The running time for this program is $14N-7$; it is faster than
the previous program when $n≤7$, slower when $n≥8$.


\ansno 3. The sequences of exponents are:\xskip (a) 1, 2, 3,
6, 7, 14, 15, 30, 60, 120, 121, 242, 243, 486, 487, 974, 975
[16 multiplications];\xskip (b) 1, 2, 3, 4, 8, 12, 24, 36, 72, 108,
216, 324, 325, 650, 975 [14 multiplications];\xskip (c) 1, 2, 3, 6,
12, 15, 30, 60, 120, 240, 243, 486, 972, 975 [13 multiplications];\xskip
(d) 1, 2, 3, 6, 12, 15, 30, 60, 75, 150, 300, 600, 900, 975
[13 multiplications].\xskip [The smallest possible number of multiplications
is 12; this is obtainable by combining the factor method with
the binary method, since $975 =15 \cdot (2↑6 + 1)$.]

\ansno 4. $(\!\hbox{\it 777777\/})↓8 = 2↑{18} - 1$.

\def\ansalgstep #1 #2. {\anskip\noindent\hbox to 40pt{\!
\hbox to 19pt{\hss\bf#1\ }\hfil\bf#2. }\hangindent 40pt}
\ansalgstep 5. T1. [Initialize.]\xskip Set $\.{LINKU}[j]
← 0$ for $1 ≤ j ≤ 2↑r$, and set $k ← 0$, $\.{LINKR}[0] ← 1$, $\.{LINKR}[1]
← 0$.

\ansalgstep{} T2. [Change level.]\xskip(Now level $k$
of the tree has been linked together from left to right, starting
at $\.{LINKR}[0]$.) If $k = r$, the algorithm terminates. Otherwise
set $n ← \.{LINKR}[0]$, $m ← 0$.

\ansalgstep{} T3. [Prepare for $n$.]\xskip(Now $n$
is a node on level $k$, and $m$ points to the rightmost node
currently on level $k + 1$.)\xskip Set $q ← 0$, $s ← n$.

\ansalgstep{} T4. [Already in tree?]\xskip(Now $s$ is
a node in the path from the root to $n$.)\xskip If $\.{LINKU}[n + s]
≠ 0$, go to T6 (the value $n + s$ is already in the tree).

\ansalgstep{} T5. [Insert below $n$.]\xskip If $q
= 0$, set $m↑\prime ← n + s$. Then set $\.{LINKR}[n + s] ← q$, $\.{LINKU}[n
+ s] ← n$, $q ← n + s$.

\ansalgstep{} T6. [Move up.]\xskip Set $s ←\.{LINKU}[s]$.
If $s ≠ 0$, return to T4.

\ansalgstep{} T7. [Attach group.]\xskip If $q ≠ 0$, set
$\.{LINKR}[m] ← q$, $m ← m↑\prime $.

\ansalgstep{} T8. [Move $n$.]\xskip Set $n ←\.{LINKR}[n]$.
If $n ≠ 0$, return to T3.

\ansalgstep{} T9. [End of level.]\xskip Set $\.{LINKR}[m]
← 0$, $k ← k + 1$, and return to T2.\quad\blackslug

\ansno 6. Prove by induction that
the path to the number $2↑{e↓0} + 2↑{e↓1} + \cdots
+ 2↑{e↓t}$, if $e↓0 > e↓1 > \cdots >e↓t ≥ 0$, is 1, 2,
$2↑2$, $\ldotss$, $2↑{e↓0}$, $2↑{e↓0} + 2↑{e↓1}$, $\ldotss$,
$2↑{e↓0}+2↑{e↓1}+ \cdots + 2↑{e↓t}$; furthermore,
the sequences of exponents on each level are in decreasing lexicographic
order.

\ansno 7. The binary and factor methods require
one more step to compute $x↑{2n}$ than $x↑n$; the power tree
method requires at most one more step. Hence (a) $15 \cdot 2↑k$;
(b) $33 \cdot 2↑k$; (c)↔$23 \cdot 2↑k$; $k = 0$, 1, 2, 3,
$\ldotss\,$.

\ansno 8. The power tree always includes the node
$2m$ at one level below $m$, unless it occurs at the same level
or an earlier level; and it always includes the node $2m + 1$
at one level below $2m$, unless it occurs at the same level or an earlier level.\xskip
[It is not true that $2m$ is a son of $m$ in the power tree for all↔$m$; the
smallest example where this fails is $m=2138$, which appears on level↔15, while
4276 appears elsewhere on level↔16. In fact, $2m$ sometimes occurs on the same level
as $m$; the smallest example is $m=6029$.]

\ansno 9. Start with $N←n$, $Z←x$, and $Y↓k←1$ for $1≤k<m$; in general we will have
$x↑n=Y↓1Y↓{\!2}↑2\ldotsm Y↓{\!m-1}↑{m-1}Z↑N$. If $N>0$, set $k←N\mod m$, and if
$k≠0$ set $Y↓k←Y↓k\cdot Z$. Then set $Z←Z↑m$, $N←\lfloor N\!/m\rfloor$, and repeat. Finally
set $Y↓k←Y↓k\cdot Y↓{k+1}$ for $k=m-2$, $m-3$, $\ldotss$,↔1; the answer is
$Y↓1\ldotsm Y↓{m-1}$.

\ansno 10. By using the ``\.{FATHER}'' representation discussed
in Section 2.3.3: Make use of a table $f[j]$, $1 ≤ j ≤ 100$,
such that $f[1] = 0$ and $f[j]$ is the number of the node
just above $j$ for $j ≥ 2$.\xskip (The fact that each node of this
tree has degree at most two has no effect on the efficiency
of this representation; it just makes the tree look prettier
as an illustration.)

\ansno 11. 1, 2, 3, 5, 10, 20, (23 or 40), 43; 1, 2, 4, 8, 9,
17, (26 or 34), 43; 1, 2, 4, 8, 9, 17, 34, (43 or 68), 77; 1,
2, 4, 5, 9, 18, 36, (41 or 72), 77. If either of the latter
two paths were in the tree we would have no possibility for
$n = 43$, since the tree must contain either 1, 2, 3, 5 or 1,
2, 4, 8, 9.

\ansno 12. No such infinite tree can exist, since $l(n) ≠ l↑\ast(n)$
for some $n$.
% Vicinity of page 635
% folio 821 galley 5 	*
\ansno 13. For Case 1, use a Type-1 chain followed by $2↑{A+C}
+ 2↑{B+C} + 2↑A + 2↑B$; or use the factor method. For Case 2,
use a Type-2 chain followed by $2↑{A+C+1} + 2↑{B+C} + 2↑A +
2↑B$. For Case 3, use a Type-5 chain followed by addition of
$2↑A + 2↑{A-1}$, or use the factor method. For Case 4, $n =
135 \cdot 2↑D$, so we may use the factor method.

\ansno 14. (a)\9 It is easy to verify that steps $r - 1$ and $r
- 2$ are not both small, so let us assume that step $r - 1$
is small and step $r - 2$ is not. If $c = 1$, then $λ(a↓{r-1})
= λ(a↓{r-k})$, so $k = 2$; and since $4 ≤ \nu (a↓r) = \nu (a↓{r-1})
+ \nu (a↓{r-k}) - 1 ≤ \nu (a↓{r-1}) + 1$, we have $\nu (a↓{r-1})
≥ 3$, making $r - 1$ a star step (lest $a↓0$, $a↓1$, $\ldotss$, $a↓{r-3}$,
$a↓{r-1}$ include only one small step). Then $a↓{r-1} = a↓{r-2}
+ a↓{r-q}$ for some $q$, and if we replace $a↓{r-2}$, $a↓{r-1}$,
$a↓r$ by $a↓{r-2}$, $2a↓{r-2}$, $2a↓{r-2} + a↓{r-q} = a↓r$, we obtain
another counterexample chain in which step $r$ is small; but
this is impossible. On the other hand, if $c ≥ 2$, then $4 ≤
\nu(a↓r) ≤ \nu (a↓{r-1}) + \nu (a↓{r-k}) - 2 ≤ \nu (a↓{r-1})$;
hence $\nu (a↓{r-1}) = 4$, $\nu (a↓{r-k}) = 2$, and $c = 2$. This
leads readily to an impossible situation by a consideration
of the six types in the proof of Theorem↔B.

(b)\9 If $λ(a↓{r-k}) < m - 1$, we have $c ≥ 3$, so
$\nu (a↓{r-k}) + \nu (a↓{r-1}) ≥ 7$ by (22); therefore both
$\nu (a↓{r-k})$ and $\nu (a↓{r-1})$ are $≥3$. All small steps
must be $≤r - k$, and $λ(a↓{r-k}) = m - k + 1$. If $k ≥ 4$,
we must have $c = 4$, $k = 4$, $\nu (a↓{r-1}) = \nu (a↓{r-4}) =
4$; thus $a↓{r-1} ≥ 2↑m + 2↑{m-1} + 2↑{m-2}$, and $a↓{r-1}$
must equal $2↑m + 2↑{m-1} + 2↑{m-2} + 2↑{m-3}$; but $a↓{r-4}
≥ {1\over 8}a↓{r-1}$ now implies that $a↓{r-1} = 8a↓{r-4}$.
Thus $k = 3$ and $a↓{r-1} > 2↑m + 2↑{m-1}$. Since $a↓{r-2} <
2↑m$ and $a↓{r-3} < 2↑{m-1}$, step $r - 1$ must be a doubling;
but step $r - 2$ is a nondoubling, since $a↓{r-1} ≠ 4a↓{r-3}$.
Furthermore, since $\nu (a↓{r-3}) ≥ 3$, $r - 3$ is a star step;
and $a↓{r-2} = a↓{r-3} + a↓{r-5}$ would imply that $a↓{r-5}
= 2↑{m-2}$, hence we must have $a↓{r-2} = a↓{r-3} + a↓{r-4}$.
As in a similar case treated in the text, the only possibility
is now seen to be $a↓{r-4} = 2↑{m-2} + 2↑{m-3}$, $a↓{r-3} = 2↑{m-2}
+ 2↑{m-3} + 2↑{d+1} + 2↑d$, $a↓{r-1} = 2↑m + 2↑{m-1} + 2↑{d+2}
+ 2↑{d+1}$, and even this possibility is impossible.

\ansno 16. $l↑B(n) = λ(n) + \nu (n) - 1$; so if $n = 2↑k$,
$l↑B(n)/λ(n) = 1$, but if $n = 2↑{k+1} - 1$, $l↑B(n)/λ(n) =
2$.

\ansno 17. Let $i↓1 < \cdots < i↓t$. Delete any intervals $I↓k$
that can be removed without affecting the union $I↓1 ∪ \cdots
∪ I↓t$.\xskip (The interval $(j↓k, i↓k]$ may be dropped out if either
$j↓{k+1} ≤ j↓k$ or $j↓1 < j↓2 < \cdots$ and $j↓{k+1}
≤ i↓{k-1}$.)\xskip Now combine overlapping intervals $(j↓1, i↓1]$,
$\ldotss$, $(j↓d, i↓d]$ into an interval $(j↑\prime , i↑\prime
] = (j↓1, i↓d]$ and note that
$$a↓{i↑\prime}<a↓{j↑\prime}(1 + \delta )↑{i↓1-j↓1+\cdots+i↓d-j↓d}
≤ a↓{j↑\prime}(1 + \delta )↑{2(i↑\prime-j↑\prime)},$$
since each point of $(j↑\prime , i↑\prime ]$ is
covered at most twice in $(j↓i, i↓1] ∪ \cdots ∪ (j↓d, i↓d]$.

\ansno 18. Call $f(m)$ a ``nice'' function if
$\biglp\log f(m)\bigrp /m → 0$ as $m→∞$. A
polynomial in↔$m$ is nice. The product of nice functions is
nice. If $g(m) → 0$ and $c$ is a positive constant, then $c↑{mg(m)}$
is nice; also ${2m\choose mg(m)}$ is nice, for by Stirling's
approximation this is equivalent to saying that $g(m)\log\biglp
1/g(m)\bigrp → 0$.

Now replace each term of the summation
by the maximum term that is attained for any $s$, $t$, $v$. The
total number of terms is nice, and so are ${m+s\choose t+v}$,
${t+v\choose v} ≤ 2↑{t+v}$, and $β↑{2v}$, because $(t +
v)/m → 0$. Finally, ${(m+s)↑2\choose t} ≤ (2m)↑{2t}\!/t!
< (4m↑2\!/t)↑te↑t$, where $(4e)↑t$ is nice; setting $t$ to its
maximum value $(1 - {1\over 2}ε)m/λ(m)$, we have the upper bound $(m↑2\!/t)↑t
= \biglp mλ(m)/(1 - {1\over 2}ε)\bigrp ↑t = 2↑{m(1-ε/2)}
\cdot f(m)$, where $f(m)$ is nice. Hence the entire sum is less
than $α↑m$ for large $m$, if $α = 2↑{1-\eta}$, $0 < \eta < {1\over
2}ε$.

\ansno 19. (a)\9 $M ∩ N$, $M ∪ N$, $M \uplus N$, respectively;
see Eqs.\ 4.5.2--6, 4.5.2--7.

(b)\9 $f(z)g(z)$,\xskip$\lcm\biglp f(z), g(z)\bigrp $,\xskip
$\gcd\biglp f(z), g(z)\bigrp $.\xskip (For the same reasons as (a),
because the monic irreducible polynomials over the complex numbers
are precisely the polynomials $z - \zeta$.)

(c)\9 \α{Commutative law}s $A \uplus B = B \uplus A$, $A ∪ B =
B ∪ A$, $A ∩ B = B ∩ A$. \α{Associative law}s $A \uplus (B \uplus C) = (A
\uplus B) \uplus C$, $A ∪ (B ∪ C) = (A ∪ B) ∪ C$, $A ∩ (B ∩ C) = (A ∩ B)
∩ C$. \α{Distributive laws} $A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)$, $A
∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)$, $A \uplus (B ∪ C) = (A \uplus B) ∪ (A
\uplus C)$, $A \uplus (B ∩ C) = (A \uplus B) ∩ (A \uplus C)$. \α{Idempotent} laws
$A ∪ A = A$, $A∩A = A$. \α{Absorption laws} $A ∪ (A ∩ B) = A$, $A ∩ (A
∪ B) = A$, $A ∩ (A \uplus B) = A$, $A ∪ (A \uplus B) = A \uplus B$. Identity
and zero laws $\emptyset \uplus A = A$, $\emptyset ∪A=A$, $\emptyset∩A=\emptyset$, 
where $\emptyset$ is the empty
multiset. \α{Counting law} $A \uplus B = (A ∪ B) \uplus (A ∩ B)$. Further
properties analogous to those of sets come from the \α{partial
ordering} defined by the rule $A \subset B$ iff $A ∩ B = A$ (iff
$A ∪ B = B$).

{\sl Notes:} Other common applications of multisets
are zeros and poles of meromorphic functions, invariants of
matrices in canonical form, invariants of finite Abelian groups,
etc.; multisets can be useful in combinatorial counting arguments
and in the development of measure theory. The terminal strings
of a noncircular \α{context-free grammar} form a multiset that
is a set if and only if the grammar is unambiguous. Although
\α{multiset}s appear frequently in mathematics, they often must
be treated rather clumsily because there is currently no standard
way to treat sets with repeated elements. Several mathematicians
have voiced their belief that the lack of adequate terminology
and notation for this common concept has been a definite handicap
to the development of mathematics.\xskip(A multiset is, of course,
formally equivalent to a mapping from a set into the nonnegative
integers, but this formal equivalence is of little or no practical
value for creative mathematical reasoning.)\xskip The author has discussed
this matter with many people in an attempt to find a good remedy.
Some of the names suggested for the concept were list, bunch, \α{bag},
heap, sample, weighted set, collection; but these words either
conflict with present terminology, have an improper connotation,
or are too much of a mouthful to say and to write conveniently.
It does not seem out of place to coin a new word for such an
important concept, and ``multiset'' has been suggested by N.
G. \α{de Bruijn}. The notation ``$A \uplus B$'' has been selected by
the author to avoid conflict with existing notations and to
stress the analogy with set union. It would not be as desirable
to use ``$A + B$'' for this purpose, since algebraists have
found that $A + B$ is a good notation for $\leftset α + β\relv α \in A\hbox{ and }
β \in B\rightset$. If $A$ is a multiset of nonnegative integers,
let $G(z) = \sum ↓{n\in A} z↑n$ be a \α{generating function} corresponding
to↔$A$.\xskip (Generating functions with nonnegative integer coefficients
obviously correspond one-to-one with multisets of nonnegative
integers.) If $G(z)$ corresponds to $A$ and $H(z)$ to $B$, then
$G(z) + H(z)$ corresponds to $A \uplus B$ and $G(z)H(z)$ corresponds
to $A + B$. If we form ``\α{Dirichlet}'' generating functions $g(z)
= \sum ↓{n\in A} 1/n↑z$, $h(z) = \sum ↓{n\in B} 1/n↑z$, then the product
$g(z)h(z)$ corresponds to the multiset product $AB$.

\ansno 20. Type 3: $(S↓0, \ldotss , S↓r) = (M↓{00}, \ldotss ,
M↓{r0}) = (\{0\}$, $\ldotss$, $\{A\}$, $\{A - 1, A\}$, $\{A - 1, A,
A\}$, $\{A - 1, A - 1, A, A, A\}$, $\ldotss$, $\{A + C - 3, A + C-3,A+C
- 2, A + C - 2, A + C - 2\})$.\xskip Type 5: $(M↓{00}, \ldotss , M↓{r0})
= (\{0\}$, $\ldotss$, $\{A\}$, $\{A - 1, A\}$, $\ldotss$, $\{A + C - 1,
A + C\}$, $\{A + C - 1, A + C - 1, A + C\}$, $\ldotss$, $\{A + C +
D - 1, A + C + D - 1, A + C + D\})$; $(M↓{01}, \ldotss , M↓{r1})
=(\emptyset$, $\ldotss$,$\emptyset$, $\emptyset$,
$\ldotss$, $\emptyset$, $\{A+C-2\}$, $\ldotss$,
$\{A + C + D - 2\})$, $S↓i = M↓{i0} \uplus M↓{i1}$.

\ansno 21. For example, let $u = 2↑{8q+5}$, $x = (2↑{(q+1)u} -
1)/(2↑u - 1) = 2↑{qu} + \cdots + 2↑u + 1$, $y = 2↑{(q+1)u} + 1$.
Then $xy = (2↑{2(q+1)u} - 1)/(2↑u - 1)$. If $n = 2↑{4(q+1)u}
+ xy$, we have $l(n) ≤ 4(q + 1)u + q + 2$ by Theorem↔F\null, but
$l↑\ast(n) = 4(q + 1)u + 2q + 2$ by Theorem↔H.

\ansno 22. Underline everything except the $u - 1$ insertions
used in the calculation of $x$.

\ansno 23. Theorem G (everything underlined).

\ansno 24. Use the numbers $(B↑{a↓i} - 1)/(B - 1)$, $0 ≤
i ≤ r$, underlined when $a↓i$ is underlined; and $c↓kB↑{i-1}(B↑{b↓j}
- 1)/(B - 1)$ for $0 ≤ j < t$, $0 < i ≤ b↓{j+1}-b↓j$, $1≤k ≤ l↑0(B)$, underlined
when $c↓k$ is underlined, where $c↓0$, $c↓1$, $\ldots$ is a minimum
length $l↑0$-chain for↔$B$. To prove the second inequality,
let $B = 2↑m$ and use (3).\xskip (The second inequality is rarely,
if ever, an improvement on Theorem↔G.)
% Vicinity of page 637
% folio 824 galley 6a 	*
\ansno 25. We may assume that $d↓k = 1$. Use the rule  R $A↓{k-1}
\ldotsm A↓1$, where $A↓j = \null$``XR'' if ${d↓j = 1}$, $A↓j = \null$``R''
otherwise, and where ``R'' means take the square root, ``X''
means multiply by $x$. For example, if $y = (.1101101)↓2$, the
rule is R R XR XR R XR XR.\xskip (There exist binary square-root extraction
algorithms suitable for computer \α{hardware}, requiring an execution
time comparable to that of division; computers with such hardware
could therefore calculate more general fractional powers using the
technique in this exercise.)

\ansno 26. If we know the pair $(F↓k, F↓{k-1})$, then we have $(F↓{k+1},
F↓k) = (F↓k + F↓{k-1}, F↓k)$ and $(F↓{2k}, F↓{2k-1}) = (F↑{2}↓{k}
+ 2F↓kF↓{k-1}, F↑{2}↓{k} + F↑{2}↓{k-1})$; so a binary method
can be used to calculate $(F↓n, F↓{n-1})$, using $O(\log n)$
arithmetic operations. Perhaps better is to use the pair of
values $(F↓k, L↓k)$, where $L↓k = F↓{k-1} + F↓{k+1}$ (cf.\ Section
4.5.4); then we have $(F↓{k+1}, L↓{k+1}) = \biglp {1\over 2}(F↓k + L↓k),
{1\over 2}(5F↓k + L↓k)\bigrp$, $(F↓{2k}, L↓{2k}) = \biglp F↓kL↓k, L↑{2}↓{k}
- 2(-1)↑k\bigrp $.

For the general \α{linear recurrence} $x↓n = a↓1x↓{n-1}
+ \cdots + a↓dx↓{n-d}$, we can compute $x↓n$ in $O(d↑3\log
n)$ arithmetic operations by computing the $n$th power of an
appropriate $d\times d$ matrix.\xskip [This observation is due to J. C.
P. \α{Miller} and D. J. S. \α{Brown}, {\sl Comp.\ J.} {\bf 9} (1966),
188--190.]

\ansno 27. First form the $2↑m - m - 1$ products $x↑{e↓1}↓{1}
\ldotss x↑{e↓m}↓{m}$, for all sequences of exponents such that
$0 ≤ e↓j ≤ 1$ and $e↓1 + \cdots +
e↓m ≥ 2$. Let $n↓j = (d↓{jλ} \ldotsm d↓{j1}d↓{j0})↓2$; to complete the
calculation, take $x↑{d↓{1λ}}↓{1} \ldotsm x↑{d↓{mλ}}↓{m}$,
then square and multiply by $x↑{d↓{1i}}↓{1} \ldotss x↑{d↓{mi}}↓{m}$,
for $i = λ - 1$, $\ldotss$, 1, 0.\xskip[Straus showed in
{\sl AMM \bf 71} (1964), 807--808, that $2λ(n)$ may be replaced by
$(1 + ε)λ(n)$ for any $ε > 0$, by generalizing this binary method
to $2↑k$-ary as in Theorem↔D\null. See exercise↔39 for later developments.]

\def\\{\mathbin{\char'562}}
\ansno 28. (a)\9 $x \\ y = x ∨ y ∨ (x + y)$, where
``$∨$'' is \α{logical ``or''}, cf.\ exercise 4.6.2--26; clearly $\nu
(x \\ y) ≤ \nu (x ∨ y) + \nu (x ∧ y) = \nu (x) + \nu (y)$.\xskip (b)
Note first that $A↓{i-1}/2↑{d↓{i-1}}\subset
A↓i/2↑{d↓i}$ for $1 ≤ i ≤ r$. Secondly, note that $d↓j = d↓{i-1}$ in
a nondoubling; for otherwise $a↓{i-1} ≥ 2a↓j ≥ a↓j + a↓k = a↓i$.
Hence $A↓j \subset A↓{i-1}$ and $A↓k \subset A↓{i-1}/2↑{d↓j-d↓k}
$.\xskip (c) An easy induction on $i$, except that
close steps need closer attention. Let us say that $m$ has property
$P(α)$ if the 1's in its binary representation all appear in
consecutive blocks of $≥α$ in a row. If $m$ and $m↑\prime$ have
$P(α)$, so does $m \\ m↑\prime $; if $m$ has $P(α)$ then $\rho
(m)$ has $P(α + \delta )$. Hence $B↓i$ has $P(1 + \delta c↓i)$.
Finally if $m$ has $P(α)$ then $\nu \biglp \rho (m)\bigrp ≤ (α +
\delta )\nu (m)/α$; for $\nu (m) = \nu ↓1 + \cdots + \nu ↓q$,
where each block size $\nu ↓j$ is $≥α$, hence $\nu \biglp \rho
(m)\bigrp ≤ (\nu ↓1 + \delta ) + \cdots + (\nu ↓q + \delta) ≤ (1
+ \delta /α)\nu ↓1 + \cdots + (1 + \delta /α)\nu ↓q$.\xskip (d) Let
$f = b↓r + c↓r$ be the number of nondoublings and $s$ the number
of small steps. If $f ≥ 3.271 \lg \nu (n)$ we have $s ≥ \lg
\nu (n)$ as desired, by (16). Otherwise we have $a↓i ≤ (1 +
2↑{-\delta})↑{b↓i}2↑{c↓i+d↓i}$ for $0 ≤ i ≤
r$, hence $n ≤ \biglp (1 + 2↑{-\delta})/2\bigrp ↑{b↓r}2↑r$,
and $r ≥ \lg n + b↓r - b↓r \lg(1 + 2↑{-\delta}) ≥ \lg n + \lg
\nu (n) - \lg(1 + \delta c↓r) - b↓r \lg(1 + 2↑{-\delta})$. Let
$\delta = \lceil \lg(f + 1)\rceil $; then $\ln(1 + 2↑{-\delta})
≤ \ln\biglp 1 + 1/(f + 1)\bigrp ≤ 1/(f + 1) ≤ \delta /(1 +
\delta f)$, and it follows that $\lg(1 + \delta x) + (f - x)\lg(1
+ 2↑{-\delta}) ≤ \lg(1 + \delta f)$ for $0 ≤ x ≤ f$. Hence finally
$l(n) ≥ \lg n + \lg \nu(n) - \lg\biglp 1 + (3.271\lg \nu
(n)) \lceil \lg(1 + 3.271 \lg \nu (n))\rceil\bigrp$.\xskip
[{\sl Theoretical Comp.\ Sci.\ \bf 1} (1975),
1--12.]

\ansno 29. In the paper just cited, \β{Sch\"onhage}
refined the method of exercise↔28 to prove that $l(n)
≥ \lg n + \lg \nu (n) - 2.13$ for all $n$. Can the remaining
gap be closed?

\ansno 30. $n = 31$ is the smallest example;
$l(31) = 7$, but 1, 2, 4, 8, 16, 32, 31 is an addition-subtraction
chain of length 6.\xskip
[After proving Theorem E\null, \α{Erd\H os} stated that the
same result holds
also for addition-subtraction chains. Sch\"onhage has
extended the lower bound of exercise↔28 to addition-subtraction
chains, with $\nu (n)$ replaced by $\=\nu (n) =\null$minimum
number of nonzero digits to represent $n = (n↓q \ldotsm n↓0)↓2$
where each $n↓j$ is $-1$, 0, or $+1$. This quantity $\=\nu(n)$ is the number
\inx{negative digits}
of 1's, in the ordinary binary representation of $n$, that
are immediately preceded by 0 or by the string $00(10)↑k1$ for
some $k ≥ 0$.]

\ansno 32. First compute $2↑i$ for $1≤i≤λ(n↓m)$, then compute each $n=n↓j$
by the following variant of the $2↑k$-ary method: For all odd $i<2↑k$, compute
$f↓i=\sum\leftset 2↑{kt+e}\relv d↓t=2↑ei\rightset$ where $n=(\ldotsm d↓1d↓0)↓{2↑k}$,
in at most $\lfloor{1\over k}\lg n\rfloor$ steps; then compute $n=\sum if↓i$ in
at most $\sum l(i)+2↑{k-1}$ further steps. The number of steps per $n↓j$ is
$≤\lfloor{1\over k}\lg n\rfloor+O(k2↑k)$, and this is $λ(n)/λλ(n)+O\biglp
λ(n)λλλ(n)/λλ(n)↑2\bigrp$ when $k=\lfloor\lg\lg n-3\lg\lg\lg n\rfloor$.

[A generalization of Theorem↔E gives the corresponding lower bound.\xskip
Reference: {\sl SIAM J. Computing \bf5} (1976), 100--103. See also exercise↔39.]

\ansno 33. The following construction due to D. J. \α{Newman} provides the best
upper bound currently known: Let $k=p↓1\ldotsm p↓r$ be the product of the first
$r$ primes. Compute $k$ and all quadratic residues mod $k$ by the method of
exercise↔32, in $O(2↑{-r}k\log k)$ steps (because there are approximately $2↑{-r}k$
\α{quadratic residues}). Also compute all multiples of $k$ that are $≤m↑2$, in about
$m↑2\!/k$ further steps. Now $m$ additions suffice to compute $1↑2$, $2↑2$,
$\ldotss$, $m↑2$. We have 
$k=\exp\biglp p↓r+O\biglp p↓r/(\log p↓r)↑{1000}\bigrp\bigrp$ where $p↓r$ is
given by the formula in exercise 4.5.4--35; so by choosing 
$$\textstyle r=\hbox{\:u\char'142}\biglp1+(1+
{1\over2}\ln2)/\!\lg\lg m\bigrp\ln m/\!\ln\ln m\hbox{\:u\char'143}$$ it follows
that $l(1↑2,\ldotss,m↑2)=m+O\biglp m\cdot\exp(-{1\over2}\ln2\ln m/\!\ln\ln m)
\bigrp$.

On the other hand, D. \α{Dobkin} and R. \α{Lipton} have shown that, for any $ε>0$,
$l(1↑2,\ldotss,m↑2)>m+m↑{2/3-ε}$ when $m$ is sufficiently large [{\sl SIAM J.
Computing \bf9} (1980), 121--125].

\ansno 35. See {\sl Discrete Math.\ \bf23} (1978), 115--119.

\ansno 36. Eight; there are four ways to compute $39=12+12+12+3$ and two ways to
compute $79=39+39+1$.

\ansno 37. The statement is true. The labels in the reduced graph of the binary
chain are $\lfloor n/2↑k\rfloor$ for $k=e↓0$, $\ldotss$,↔0; they are 1, 2, $\ldotss$,
$2↑{e↓0}$, $n$ in the dual graph.\xskip[Similarly, the right-to-left $m$-ary method
of exercise↔9 is the dual of the left-to-right method.]

\ansno 38. $2↑t$ are equivalent to the binary chain; it would be $2↑{t-1}$ if
$e↓0=e↓1+1$. The number of chains equivalent to the scheme of Algorithm↔A is the
number of ways to compute the sum of $t+2$ numbers of which two are identical.
This is ${1\over2}f↓{t+1}+{1\over2}f↓t$, where $f↓m$ is the number of ways to
compute the sum of $m+1$ distinct numbers. When we take \α{commutativity} into
account, we see that $f↓m$ is $2↑{-m}$ times $(m+1)!$ times the number of
\α{binary trees} on $m$ nodes,
\inx{Catalan numbers}
so $f↓m=(2m-1)(2m-3)\ldotsm1$.

\ansno 39. The quantity $l\biglp[n↓1,n↓2,\ldotss,n↓m]\bigrp$ is the minimum of
$\hbox{arcs}-\hbox{vertices}+m$ taken over all directed graphs having $m$
vertices $s↓j$ whose in-degree is zero and one vertex↔$t$ whose out-degree is
zero, where there are exactly $n↓j$ oriented paths from $s↓j$ to $t$ for ${1≤j≤m}$.
The quantity $l(n↓1,n↓2,\ldotss,n↓m)$ is the minimum of $\hbox{arcs}-\hbox{vertices}
+1$ taken over all directed graphs having one vertex $s$ whose in-degree is zero
and $m$ vertices $t↓j$ whose out-degree is zero, where there are exactly $n↓j$
oriented paths from $s$ to $t↓j$ for ${1≤j≤m}$. These problems are \α{dual} to each
other, if we change the direction of all the arcs.

{\sl Note:} C. H. \α{Papadimitriou} points out that this is a special case of a
much more general theorem. Let $N=(n↓{ij})$ be an $m\times p$ matrix of nonnegative
integers having no row or column entirely zero. We can define $l(N)$ to be the
minimum number of multiplications needed to compute the set of \α{monomial}s
$\leftset x↓1↑{n↓{1j}}\ldotss x↓m↑{n↓{mj}}\relv 1≤j≤p\rightset$. Now $l(N)$ is
also the minimum of $\hbox{arcs}-\hbox{vertices}+m$ taken over all directed
graphs having $m$ vertices $s↓i$ whose in-degree is zero and $p$ vertices $t↓j$
whose out-degree is zero, where there are exactly $n↓{ij}$ oriented paths from
$s↓i$ to $t↓j$ for each $i$ and $j$. By duality we have $l(N)=l(N↑T)+m-p$.

N. \α{Pippenger} has proved a comprehensive generalization of the results of
exercises 27 and↔32. Let $L(m,p,n)$ be the maximum of $l(N)$ taken over all
$m\times p$↔matrices↔$N$ of nonnegative integers $n↓{ij}≤n$. Then $L(m,p,n)=
\min(m,p)\lg n+H/\!\lg H+{O\biglp m+p+H(\log\log H)↑{1/2}(\log H)↑{-3/2}\bigrp}$,
where $H=mp\lg(n+1)$.\xskip[{\sl SIAM J. Com\-puting \bf9} (1980), 230--250.]

\ansno 40. By exercise 39, it suffices to show that $l(m↓1n↓1+\cdots+m↓tn↓t)≤
l(m↓1,\ldotss,m↓t)+l\biglp[n↓1,\ldotss,n↓t]\bigrp$. But this is clear, since
we can first form $\{x↑{m↓1},\ldotss,x↑{m↓t}\}$ and then compute the monomial
$(x↑{m↓1})↑{n↓1}\ldotss(x↑{m↓t})↑{n↓t}$.

{\sl Notes:} Theorem F is a special case of this result, since we clearly
have $l(2↑m,x)
≤m+\nu(x)-1$ when $λ(x)≤m$. One strong way to state Olivos's theorem is that if
$a↓0$, $\ldotss$, $a↓r$ and $b↓0$, $\ldotss$, $b↓s$ are any addition chains,
then $l\biglp\sum c↓{ij}a↓ib↓j\bigrp≤{r+s+\sum c↓{ij}-1}$ for any $(r+1)\times(s+1)$
matrix of nonnegative integers↔$c↓{ij}$.

\ansno 41. [To appear.]\xskip
The stated formula can be proved whenever $A≥9m↑2$. Since this is a
polynomial in $m$, and since the problem of finding a minimum vertex cover is
NP↔hard (cf.\ Section↔7.9), 
the problem of computing $l(n↓1,\ldotss,n↓m)$ is \α{NP complete}.\xskip
[It is unknown whether or not the problem of computing
$l(n)$ is NP↔complete.]
% Vicinity of page 639
% folio 826 galley 6b 	*
\ansbegin{4.6.4}

\ansno 1. Set $y ← x↑2$, then compute
$\biglp (\ldotsm(u↓{2n+1}y + u↓{2n-1})y + \cdotss)y+u↓1\bigrp x$.

\ansno 2. Replacing $x$ in (2) by the polynomial $x + x↓0$ leads
to the following procedure:

\yyskip\hangindent 40pt\noindent\hbox to 40pt{\hfill\bf G1. }
Do step G2 for $k = n$, $n - 1$, $\ldotss$,
0 (in this order), and stop.

\yskip\hangindent 40pt\noindent\hbox to 40pt{\hfill\bf G2. }
Set $v↓k ← u↓k$, and then set
$v↓j ← v↓j + x↓0v↓{j+1}$ for $j = k$, $k + 1$, $\ldotss$, $n - 1$.\xskip
(When $k = n$, this step simply sets $v↓n ← u↓n$.)\quad\blackslug

\yyskip\noindent The computations turn out to be
identical to those in H1 and H2, but performed in a different order.
$\biglp$This
application was, in fact, \α{Newton}'s original motivation for using
scheme↔(2).$\bigrp$

\ansno 3. The coefficient of $x↑k$ is a polynomial in $y$ that
may be evaluated by Horner's rule: $\biglp \ldotsm(u↓{n,0}x
+ (u↓{n-1,1}y + u↓{n-1,0}))x + \cdotss\bigrp x + \biglp (\ldotsm
(u↓{0,n}y + u↓{0,n-1})y + \cdotss)y + u↓{0,0}\bigrp $.\xskip [For
\inx{homogeneous polynomial}
a ``homogeneous'' polynomial, such as $u↓nx↑n + u↓{n-1}x↑{n-1}y
+ \cdots + u↓1xy↑{n-1} + u↓0y↑n$, another scheme is more efficient:
if $0<|x|≤|y|$, first divide $x$ by $y$, evaluate a polynomial in $x/y$, then
multiply by $y↑n$.]

\ansno 4. Rule (2) involves $4n$ or $3n$ real multiplications
and $4n$ or $7n$ real additions; (3) is worse, it takes 4$n
+ 2$ or $4n + 1$ mults, $4n + 2$ or $4n + 5$ adds.

\ansno 5. One multiplication to compute $x↑2$; $\lfloor
n/2\rfloor$ multiplications and $\lfloor n/2\rfloor$ additions
to evaluate the first line; $\lceil n/2\rceil$ multiplications
and $\lceil n/2\rceil - 1$ additions to evaluate the second
line; and one addition to add the two lines together. Total:
$n + 1$ multiplications and $n$ additions.

\ansno 6. \hbox to 21pt{\hfill\bf J1. }Compute and store the values $x↑{2}↓{0}$,
$\ldotss$, $x↑{\lceil n/2\rceil}↓{0}$.

\yskip\noindent\hbox to 40pt{\hfill\bf J2. }Set $v↓j ← u↓jx↑{j-\lfloor n/2\rfloor
}↓{0}$ for $0 ≤ j ≤ n$.

\yskip\noindent\hbox to 40pt{\hfill\bf J3. }For $k = 0$, 1, $\ldotss$, $n - 1$,
set $v↓j ← v↓j + v↓{j+1}$ for $j = n - 1$, $\ldotss$, $k + 1$, $k$.

\yskip\noindent\hbox to 40pt{\hfill\bf J4. }Set $v↓j ← v↓jx↑{\lfloor n/2\rfloor
-j}↓{0}$ for $0 ≤ j ≤ n$.\quad\blackslug

\yyskip\noindent There are $(n↑2 + n)/2$ additions, $n + \lceil
n/2\rceil - 1$ multiplications, $n$ divisions. Another multiplication
and division can be saved by treating $v↓n$ and $v↓0$ as special
cases.\xskip {\sl Reference:  SIGACT News \bf 7}, 3 (Summer 1975),
32--34.

\ansno 7. Let $x↓j = x↓0 + jh$, and consider (42), (44). Set
$y↓j ← u(x↓j)$ for $0 ≤ j ≤ n$. For $k = 1$, 2, $\ldotss$, $n$ (in
this order), set $y↓j ← y↓j - y↓{j-1}$ for $j = k$, $k + 1$, $\ldotss
$, $n$ (in this order). Now $β↓j = y↓j$ for all $j$.

\ansno 8. See (43).

\ansno 9. [{\sl Combinatorial Mathematics} (Buffalo: Math.\ Assoc.\ of
America, 1963), 26--28.]\xskip This formula can be regarded as
an application of the principle of inclusion and exclusion (Section
1.3.3), since the sum of the terms for $n - ε↓1 - \cdots - ε↓n
= k$ is the sum of all $x↓{1j↓1}x↓{2j↓2}\ldotsm
x↓{nj↓n}$ for which $k$ values of the $j↓i$ do not appear.
A direct proof can be given by observing that the coefficient
of $x↓{1j↓1}\ldotsm x↓{nj↓n}$ is
$$\sum (-1)↑{n-ε↓1-\cdots-ε↓n\,}ε↓{j↓1}\ldotsm ε↓{j↓n};$$
if the $j$'s are distinct, this equals unity, but
if $j↓1$, $\ldotss$, $j↓n ≠ k$ then it is zero, since the terms
for $ε↓k = 0$ cancel the terms for $ε↓k = 1$.

To evaluate the sum efficiently, we can start
with $ε↓1 = 1$, $ε↓2 = \cdots = ε↓n = 0$, and we can then proceed
through all combinations of the $ε$'s in such a way that only
one $ε$ changes from one term to the next.\xskip(See ``\α{Gray code}''
in Chapter↔7.)\xskip The work to compute the first term is $n - 1$
multiplications; the subsequent $2↑n - 2$ terms each involve
$n$ additions, then $n - 1$ multiplications, then one more addition.
Total: $(2↑n - 1)(n - 1)$ multiplications, and $(2↑n - 2)(n
+ 1)$ additions. Only $n + 1$ temp storage locations are needed,
one for the main partial sum and one for each factor of the
current product.
% Vicinity of page 640
% folio 828 galley 7 	*
\ansno 10. $\sum ↓{1≤k<n} (k + 1){n\choose k+1} = n(2↑{n-1}
- 1)$ multiplications and $\sum ↓{1≤k<n} k{n\choose k+1} =
{n2↑{n-1} - 2↑n} + 1$ additions. This is approximately half as
many arithmetic operations as the method of exercise↔9, although
it requires a more complicated program to control the sequence.
Approximately ${n\choose \lceil n/2\rceil } + {n\choose \lceil
n/2\rceil -1}$ temporary storage locations must be used, and
this grows exponentially large (on the order of $2↑n/\sqrt{n}\,$).

The method in this exercise is equivalent to the
unusual matrix factorization of the permanent function given
by \α{Jurkat} and \α{Ryser} in {\sl J. Algebra \bf 5} (1967), 342--357.
It may also be regarded as an application of (39) and (40), in
an appropriate sense.

\ansno 12. Here is a brief summary of progress on this famous research
\inxf{matrix multiplication}
problem: J. \α{Hopcroft} and L. R. \α{Kerr} proved, among other
things, that seven multiplications are necessary in $2 \times 2$ matrix
multiplication [{\sl SIAM J. Appl.\ Math.\ \bf 20} (1971), 30--36].
R. L. \α{Probert} showed that all 7-multiplication schemes, in
which each multiplication takes a linear combination of elements
from one matrix and multiplies by a linear combination of elements
from the other, must have at least 15 additions [{\sl SIAM J.
Computing \bf5} (1976), 187--203]. For $n=3$, the best method known is due to
J. D. \α{Laderman} [{\sl Bull.\ Amer.\ Math.\ Soc.\ \bf 82}
(1976), 126--128], who showed that 23 noncommutative multiplications
suffice. His construction has been generalized by Ondrej \α{S\'ykora,} who exhibited a
method requiring $n↑3-(n-1)↑2$ noncommutative multiplications and
$n↑3-n↑2+11(n-1)↑2$\linebreak
 additions, a result that also reduces to (36) when $n=2$
[{\sl Lecture Notes in Comp.\ Sci.\ \bf53} (1977), 504--512].
The best lower bound known to hold for all $n$ is the fact that
$2n↑2-1$ nonscalar multiplications are necessary [Jean-Claude \α{Lafon} and S. \α{Winograd},
{\sl Theoretical Comp.\ Sci.}, to appear]. 
\α{Pan} has generalized this to a lower bound of $mn+ns+m-n-1$ in the $m\times
n\times s$ case [see {\sl SIAM J. Computing \bf9} (1980), 341].
The best upper bounds known for large
$n$ were changing rapidly as this book was being prepared for publication;
see exercises 60--64.

\ansno 13. By summing geometric series, we find that $F(t↓1, \ldotss , t↓n)$ equals
$$\textstyle\sum ↓{0≤s↓1<m↓1,\ldots,0≤s↓n<m↓n}
\exp\biglp -2πi(s↓1t↓1/m↓1 + \cdots + s↓nt↓n/m↓n)f(s↓1, \ldotss
, s↓n)\bigrp /m↓1 \ldotsm m↓n.$$
The inverse transform times $m↓1 \ldotsm m↓n$ can be found by
doing a regular transform and interchanging $t↓j$ with $m↓j
- t↓j$ when $t↓j ≠ 0$; cf.\ exercise 4.3.3--9.

[If we regard $F(t↓1, \ldotss , t↓n)$
as the coefficient of $x↑{t↓1}↓{1} \ldotss x↑{t↓n}↓{n}$ in a
multivariate polynomial, the finite Fourier transform amounts
to evaluation of this polynomial at roots of unity, and the
inverse transform amounts to finding the \α{interpolating} polynomial.]

\ansno 14. Let $m↓1=\cdots=m↓n=2$, $F(t↓1, t↓2, \ldotss , t↓n) = F(2↑{n-1}t↓n
+ \cdots + 2t↓2 + t↓1)$, and $f(s↓1, s↓2, \ldotss , s↓n) = f(2↑{n-1}s↓1
+ 2↑{n-2}s↓2 + \cdots + s↓n)$; note the reversed treatment between
$t$'s and $s$'s. Also let $g↓k(s↓k, \ldotss , s↓n, t↓k)$ be $\omega$
raised to the $2↑{k-1}t↓k(s↓n + 2s↓{n-1} + \cdots + 2↑{n-k}s↓k)$
power.

At each iteration we essentially take $2↑{n-1}$ pairs of complex numbers $(α,β)$ and
replace them by $(α+\zeta β,α-\zeta β)$, where $\zeta$ is a suitable power of
$\omega$, hence $\zeta=\cos\theta+i\sin\theta$ for some $\theta$.
If we take advantage of simplifications when $\zeta=\pm1$ or $\pm i$, the total
work comes to $\biglp(n-3)\cdot 2↑{n-1}+2\bigrp$ complex multiplications and $n\cdot
2↑n$ complex additions; the techniques of exercise↔41 can be used to reduce the
real multiplications and additions used to implement these complex operations.

The number of complex multiplications can be reduced about 25 per cent without
changing the number of additions by combining passes $k$ and $k+1$ for $k=1$, 3,
$\ldotss$; this means that $2↑{n-2}$ quadruples $(α,β,\gamma,\delta)$ are being
replaced by $(α+\zetaβ+\zeta↑2\gamma+\zeta↑3\delta$, $α+i\zetaβ-\zeta↑2\gamma-
i\zeta↑3\delta$, $α-\zetaβ+\zeta↑2\gamma-\zeta↑3\delta$, $α-i\zetaβ-\zeta↑2\gamma
+i\zeta↑3\delta)$. The number of complex multiplications when $n$ is even is
thereby reduced to $(3n-2)2↑{n-3}-5\lfloor2↑{n-1}/3\rfloor$.

These calculations assume that the given numbers $F(t)$ are complex. If the $F(t)$
are real, then $f(s)$ is the complex conjugate of $f(2↑n-s)$, so we can avoid the
redundancy by computing only the $2↑n$ independent real numbers $f(0)$,
$\real f(1)$, $\ldotss$,↔$\real f({2↑{n-1}-1})$, $f(2↑{n-1})$, $\imag 
f(1)$, $\ldotss$, $\imag f(2↑{n-1}-1)$. The entire calculation in this case can
be done by working with $2↑n$ real values, using the fact that $f↑{[k]}(s↓{n-k+1},
\ldotss,s↓n,t↓1,\ldotss,t↓{n-k})$ will be the complex conjugate of $f↑{[k]}(
s↓{n-k+1}↑\prime,\ldotss,s↓n↑\prime,t↓1,\ldotss,t↓{n-k})$ when $(s↓1\ldotsm s↓n)↓2
+(s↓1↑\prime\ldotsm s↓n↑\prime)↓2≡0\modulo{2↑n}$. About half as many multiplications
and additions are needed as in the complex case.

[The fast Fourier
transform algorithm is essentially due to C. \α{Runge} and H. \α{K\"onig}
in 1924, and it was generalized by J. W. \α{Cooley} and J.
W. \α{Tukey,} {\sl Math.\ Comp.\ \bf 19} (1965), 297--301. Its interesting
history has been traced by J. W. Cooley, P. A. W. \α{Lewis,} and P.
D. \α{Welch,} {\sl Proc.\ IEEE} {\bf 55} (1967), 1675--1677. Details
concerning its use have been discussed by R. C. \α{Singleton,} {\sl CACM
\bf 10} (1967), 647--654; M. C. \α{Pease,} {\sl JACM \bf 15} (1968),
252--264; G. D. \α{Berglund,} {\sl Math.\ Comp.\ \bf 22} (1968), 275--279,
{\sl CACM \bf 11} (1968), 703--710; A. M. \α{Macnaghten} and C. A. R. \α{Hoare,} {\sl Comp.\
J. \bf20} (1977), 78--83. See also exercises 53, 57, and↔59.]

\ansno 15. (a)\9 The hint follows by integration and induction.
Let $f↑{(n)}(\theta )$ take on all values between $A$ and $B$
inclusive, as $\theta$ varies from $\min(x↓0, \ldotss , x↓n)$
to $\max(x↓0, \ldotss , x↓n)$. Replacing $f↑{(n)}$ by each of these bounds,
in the stated integral, yields $A/n! ≤ f(x↓0, \ldotss , x↓n)
≤ B/n!$.\xskip (b)↔It suffices to prove this for $j = n$. Let $f$
be Newton's interpolation polynomial, then $f↑{(n)}$ is the
constant $n! α↓n$.

\ansno 16. Carry out the multiplications and additions of (43)
as operations on polynomials.\xskip (The special case $x↓0 = x↓1 =
\cdots = x↓n$ is considered in exercise↔2. We have used this
method in step C8 of Algorithm↔4.3.3C.)

\ansno 17. T. M. \α{Vari} has shown that $n - 1$ multiplications
are necessary, by proving that $n$ multiplications are necessary to compute
$x↑{2}↓{1} + \cdots + x↑{2}↓{n}$ [Cornell Computer Science Report
120 (Jan. 1972)].

\ansno 18. $α↓0 = {1\over 2}(u↓3/u↓4 + 1)$, $β = u↓2/u↓4 - α↓0(α↓0
- 1)$, $α↓1 = α↓0β - u↓1/u↓4$, $α↓2 = β - 2α↓1$, $α↓3 = u↓0/u↓4 -
α↓1(α↓1 + α↓2)$, $α↓4 = u↓4$.

\ansno 19. Since $α↓5$ is the leading coefficient, we may assume
without loss of generality that $u(x)$ is monic (i.e., $u↓5
= 1$). Then $α↓0$ is a root of the cubic equation ${40z↑3 - 24u↓4z↑2}
+ (4u↑{2}↓{4} + 2u↓3)z + (u↓2 - u↓3u↓4) = 0$; this equation
always has at least one real root, and it may have three. Once
$α↓0$ is determined, we have $α↓3 = u↓4 - 4α↓0$, $α↓1 = u↓3 -
4α↓0α↓3 - 6α↑{2}↓{0}$, $α↓2 = u↓1 - α↓0(α↓0α↓1 + 4α↑{2}↓{0}α↓3
+ 2α↓1α↓3 + α↑{3}↓{0})$, $α↓4 = u↓0 - α↓3(α↑{4}↓{0} + α↓1α↑{2}↓{0}
+ α↓2)$.

For the given polynomial we are to solve the cubic
equation $40z↑3 - 120z↑2 + 80z = 0$; this leads to three solutions
$(α↓0, α↓1, α↓2, α↓3, α↓4, α↓5) = (0, -10, 13, 5, -5, 1)$, $(1,
-20, 68, 1, 11, 1)$, $(2, -10, 13, -3, 27, 1)$.

\ansno 20. $\vtop{\halign{\tt#\hfill\quad⊗\tt#\hfill\qquad⊗#\hfill\cr
LDA⊗X\cr
FADD⊗=$α↓3$=\cr
STA⊗TEMP1\cr
FADD⊗=$α↓0$-$α↓3$=\cr
STA⊗TEMP2\cr
FMUL⊗TEMP2\cr
\lower6pt\null STA⊗TEMP2\cr}}\hskip50pt
\vtop{\halign{\tt#\hfill\quad⊗\tt#\hfill\qquad⊗#\hfill\cr
FADD⊗=$α↓1$=\cr
FMUL⊗TEMP2\cr
FADD⊗=$α↓2$=\cr
FMUL⊗TEMP1\cr
FADD⊗=$α↓4$=\cr
FMUL⊗=$α↓5$=⊗\quad\blackslug\cr}}$

\ansno 21. $z = (x + 1)x - 2$, $w = (x + 5)z + 9$, $u(x) =
(w + z - 8)w - 8$; or $z = (x + 9)x + 26$, $w = (x - 3)z + 73$,
$u(x) = (w + z - 24)w - 12$.

\ansno 22. $α↓6 = 1$, $α↓0 = -1$, $α↓1 = 1$, $β↓1 = -2$, $β↓2 = -2$,
$β↓3=-2$,
$β↓4 = 1$, $α↓3 = -4$, $α↓2 = 0$, $α↓4 = 4$, $α↓5 = -2$. We form $z =
(x - 1)x + 1$, $w = z + x$, and $u(x) = \biglp (z - x - 4)w +
4\bigrp z - 2$. In this
special case we see that one of the seven additions can be saved if we compute
$w=x↑2+1$, $z=w-x$.

\ansno 23. (a)\9 We may use induction on $n$; the result is trivial
if $n < 2$. If $f(0) = 0$, then the result is true for the polynomial
$f(z)/z$, so it holds for $f(z)$. If $f(iy) = 0$ for some real
$y ≠ 0$, then $g(\pm iy) = h(\pm iy) = 0$; since the result
is true for $f(z)/(z↑2 + y↑2)$, it holds also for $f(z)$. Therefore
we may assume that $f(z)$ has no roots whose real part is zero.
Now the net number of times the given path circles the origin
is the number of roots of $f(z)$ inside the region, which is
at most 1. When $R$ is large, the path $f(Re↑{it})$ for $π/2
≤ t ≤ 3π/2$ will circle the origin clockwise approximately $n/2$
times; so the path $f(it)$ for $-R ≤ t ≤ R$ must go counterclockwise
around the origin at least $n/2 - 1$ times. For $n$ even, this
implies that $f(it)$ crosses the imaginary axis at least $n
- 2$ times, and the real axis at least $n - 3$ times; for $n$
odd, $f(it)$ crosses the real axis at least $n - 2$ times and
the imaginary axis at least $n - 3$ times. These are roots respectively
of $g(it) = 0$, $h(it) = 0$.

(b)\9 If not, $g$ or $h$ would have a root of the
form $a + bi$ with $a ≠ 0$ and $b ≠ 0$. But this would imply
the existence of at least three other such roots, namely $a
- bi$ and $-a \pm bi$, while $g(z)$ and $h(z)$ have at most $n$ roots.
% Vicinity of page 643
% folio 832 galley 8 	*
\ansno 24. The roots of $u$ are $-7$, $-3 \pm i$, $-2 \pm i$, and $-1$;
permissible values of $c$ are 2 and↔4 (but {\sl not\/} 3, since
$c = 3$ makes the sum of the roots equal to zero).\xskip Case 1, $c
= 2$: $p(x) = (x + 5)(x↑2 + 2x + 2)(x↑2 + 1)(x - 1) = x↑6 + 6x↑5
+ 6x↑4 + 4x↑3 - 5x↑2 - 2x - 10$; $q(x) = 6x↑2 + 4x - 2 = 6(x +
1)(x - {1\over 3})$. Let $α↓2 = -1$, $α↓1 = {1\over 3}$; $p↓1(x)
= x↑4 + 6x↑3 + 5x↑2 - 2x - 10 = (x↑2 + 6x + {16\over 3})(x↑2
- {1\over 3}) - {74\over 9}$; $α↓0 = 6$, $β↓0 = {16\over 3}$, $β↓1
= -{74\over 9}$.\xskip Case 2, $c = 4$: A similar analysis gives $α↓2
= 9$, $α↓1 = -3$, $α↓0 = -6$, $β↓0 = 12$, $β↓1 = -26$.

\ansno 25. $β↓1 = α↓2$, $β↓2 = 2α↓1$, $β↓3 = α↓7$, $β↓4 = α↓6$, $β↓5
= β↓6 = 0$, $β↓7 = α↓1$, $β↓8 = 0$, $β↓9 = 2α↓1 - α↓8$.

\ansno 26. (a)\9 $λ↓1 = α↓1 \times λ↓0$, $λ↓2 = α↓2 + λ↓1$, $λ↓3
= λ↓2 \times λ↓0$, $λ↓4 = α↓3 + λ↓3$, $λ↓5 = λ↓4 \times λ↓0$, $λ↓6
= α↓4 + λ↓5$.\xskip (b) $\kappa ↓1 = 1 + β↓1x$, $\kappa ↓2 = 1 + β↓2\kappa
↓1x$, $\kappa ↓3 = 1 + β↓3\kappa ↓2x$, $u(x) = β↓4\kappa ↓3 = β↓1β↓2β↓3β↓4x↑3
+ β↓2β↓3β↓4x↑2 + β↓3β↓4x + β↓4$.\xskip (c) If any coefficient is zero,
the coefficient of $x↑3$ must also be zero in (b), while (a)
yields an arbitrary polynomial $α↓1x↑3 + α↓2x↑2 + α↓3x + α↓4$
of degree $≤3$.

\ansno 27. Otherwise there would be a nonzero polynomial $f(q↓n,
\ldotss , q↓1, q↓0)$ with integer coeffi\-cients such that $q↓n
\cdot f(q↓n, \ldotss , q↓1, q↓0) = 0$ for all sets $(q↓n, \ldotss
, q↓0)$ of real numbers. This cannot happen, since it is easy
to prove by induction on $n$ that a nonzero polynomial always
takes on some nonzero value.\xskip (Cf.\ exercise 4.6.1--16.
However, this result is false
for {\sl finite} fields in place of the real numbers.)

\ansno 28. The indeterminate quantities $α↓1$, $\ldotss$, $α↓s$
form an algebraic basis for the polynomial domain $Q[α↓1, \ldotss , α↓s]$, where $Q$
is the field of rational numbers. Since $s + 1$ is greater than
the number of elements in a basis, the polynomials $f↓j(α↓1,
\ldotss , α↓s)$ are algebraically dependent; this means
that there is a nonzero polynomial $g$ with rational coefficients
such that $g\biglp f↓0(α↓1, \ldotss , α↓s), \ldotss , f↓s(α↓1,
\ldotss , α↓s)\bigrp$ is identically zero.

\ansno 29. Given $j↓0$, $\ldotss$, $j↓t \in \{0, 1, \ldotss , n\}$,
there are nonzero polynomials with integer coefficients such
that $g↓j(q↓{j↓0}, \ldotss , q↓{j↓t}) = 0$ for all
$(q↓n, \ldotss , q↓0)$ in $R↓j$, $1 ≤ j ≤ m$. The product $g↓1g↓2
\ldotsm g↓m$ is therefore zero for all $(q↓n, \ldotss , q↓0)$
in $R↓1 ∪ \cdots ∪ R↓m$.

\ansno 30. Starting with the construction in Theorem↔M\null, we will
prove that $m↓p + (1 - \delta ↓{0m↓c})$ of the $β$'s
may effectively be eliminated: If $\mu↓i$ corresponds to
a parameter multiplication, we have $\mu ↓i = β↓{2i-1} \times
(T↓{2i} + β↓{2i})$; add $cβ↓{2i-1}β↓{2i}$ to each $β↓j$ for
which $c\mu ↓i$ occurs in $T↓j$, and replace $β↓{2i}$ by zero.
This removes one parameter for each parameter multiplication. If $\mu ↓i$ is the
first chain multiplication, then $\mu ↓i$ is the first chain
multiplication, then $\mu ↓i = (\gamma ↓1x + \theta ↓1 + β↓{2i-1})
\times (\gamma ↓{2x} + \theta ↓2 + β↓{2i})$, where $\gamma ↓1$,
$\gamma ↓2$, $\theta ↓1$, $\theta ↓2$ are polynomials in $β↓1$, $\ldotss
$, $β↓{2i-2}$ with integer coefficients. Here $\theta ↓1$ and
$\theta ↓2$ can be ``absorbed'' into $β↓{2i-1}$ and $β↓{2i}$,
respectively, so we may assume that $\theta ↓1 = \theta ↓2 =
0$. Now add $cβ↓{2i-1}β↓{2i}$ to each $β↓j$ for which $c\mu
↓i$ occurs in $T↓j$; add $β↓{2i-1}\gamma ↓2/\gamma ↓1$ to $β↓{2i}$;
and set $β↓{2i-1}$ to zero. The result set is unchanged by this
elimination of $β↓{2i-1}$, except for the values of $α↓1$, $\ldotss
$, $α↓s$ such that $\gamma ↓1$ is zero.\xskip [This proof is essentially
due to V. \t Ia.\ \α{Pan,} {\sl Russian Mathematical Surveys} {\bf 21},1
(1966), 105--136.]\xskip The latter case can be handled as in
the proof of Theorem↔A\null, since the polynomials with $\gamma ↓1
= 0$ can be evaluated by eliminating $β↓{2i}$ (as in the first
construction, where $\mu ↓i$ corresponds to a parameter multiplication).

\ansno 31. Otherwise we could add one parameter multiplication
as a final step, and violate Theorem↔C\null.\xskip (The exercise is an
improvement over Theorem↔A\null, in this special case, since there
are only $n$ degrees of freedom in the coefficients of a monic
polynomial of degree $n$.)

\ansno 32. $λ↓1 = λ↓0 \times λ↓0$, $λ↓2 = α↓1 \times λ↓1$, $λ↓3
= α↓2 + λ↓2$, $λ↓4 = λ↓3 \times λ↓1$, $λ↓5 = α↓3 + λ↓4$. We need
at least three multiplications to compute $u↓4x↑4$ (see Section
4.6.3), and at least two additions by Theorem↔A.

\ansno 33. We must have $n + 1 ≤ 2m↓c + m↓p + \delta ↓{0m↓c}
$, and $m↓c + m↓p = (n + 1)/2$; so there are no parameter multiplications.
Now the first $λ↓i$ whose leading coefficient (as a polynomial
in $x$) is not an integer must be obtained by a chain addition;
and there must be at least $n + 1$ parameters, so there are
at least $n + 1$ parameter additions.

\ansno 34. Transform the given chain step by step, and also
define the ``content'' $c↓i$ of $λ↓i$, as follows:\xskip (Intuitively,
$c↓i$ is the leading coefficient of $λ↓i$.)\xskip Define $c↓0 = 1$.
\xskip (a) If the step has the form $λ↓i = α↓j + λ↓k$, replace it by
$λ↓i=β↓j+λ↓k$, where $β↓j=α↓j/c↓k$; and define $c↓i = c↓k$.\xskip (b) If the step
has the form $λ↓i = α↓j - λ↓k$, replace it by $λ↓i = β↓j + λ↓k$,
where $β↓j = -α↓j/c↓k$; and define $c↓i = -c↓k$.\xskip (c) If the
step has the form $λ↓i = α↓j \times λ↓k$, replace it by $λ↓i
= λ↓k$ (the step will be deleted later); and define $c↓i = α↓jc↓k$.\xskip
(d) If the step has the form $λ↓i = λ↓j \times λ↓k$, leave it
unchanged; and define $c↓i = c↓jc↓k$.

After this process is finished, delete all steps
of the form $λ↓i = λ↓k$, replacing $λ↓i$ by $λ↓k$ in each future
step that uses $λ↓j$. Then add a final step $λ↓{r+1} = β \times
λ↓r$, where $β = c↓r$. This is the desired scheme, since it
is easy to verify that the new $λ↓i$ are just the old ones divided
by the factor $c↓i$. The $β$'s are given functions of the $α$'s;
division by zero is no problem, because if any $c↓k = 0$ we
must have $c↓r = 0$ (hence the coefficient of $x↑n$ is zero),
or else $λ↓k$ never contributes to the final result.

\ansno 35. Since there are at least five parameter steps, the
result is trivial unless there is at least one parameter multiplication;
considering the ways in which three multiplications can form
$u↓4x↑4$, we see that there must be one parameter multiplication
and two chain multiplications. Therefore the four addition-subtractions
must each be parameter steps, and exercise↔34 applies. We can
now assume that only additions are used, and that we have a
chain to compute a general {\sl monic} fourth-degree polynomial
with {\sl two} chain multiplications and four parameter additions.
The only possible scheme of this type that calculates a fourth-degree
polynomial has the form
$$\baselineskip12pt\teqalign{λ↓1⊗= α↓1 + λ↓0\cr
λ↓2⊗= α↓2 + λ↓0\cr
λ↓3⊗=λ↓1\timesλ↓2\cr
λ↓4⊗=α↓3+λ↓3\cr
λ↓5⊗=α↓4+λ↓3\cr
λ↓6⊗=λ↓4\timesλ↓5\cr
λ↓7⊗=α↓5+λ↓6\cr}$$
Actually this chain has one addition too many,
but any correct scheme can be put into this form if we restrict
some of the $α$'s to be functions of the others. Now $λ↓7$
has the form $(x↑2 + Ax + B)(x↑2 + Ax + C) + D = x↑4 + 2Ax↑3
+ (E + A↑2)x↑2 + EAx + F$, where $A = α↓1 + α↓2$, 
$B = α↓1α↓2 + α↓3$, $C = α↓1α↓2 + α↓4$, $D = α↓6$, $E = B + C$,
$F = BC + D$; and since this involves only three independent
parameters it cannot represent a general monic fourth-degree
polynomial.

\ansno 36. As in the solution to exercise↔35, we may assume that the chain computes
a general monic polynomial of degree six, using only three chain multiplications and
six parameter additions. The computation must take one of two general forms
$$\baselineskip12pt
\hbox to size{$\hfill\teqalign{λ↓1⊗=α↓1+λ↓0\cr
λ↓2⊗=α↓2+λ↓0\cr
λ↓3⊗=λ↓1\timesλ↓2\cr
λ↓4⊗=α↓3+λ↓0\cr
λ↓5⊗=α↓4+λ↓3\cr
λ↓6⊗=λ↓4\timesλ↓5\cr
λ↓7⊗=α↓5+λ↓6\cr
λ↓8⊗=α↓6+λ↓6\cr
λ↓9⊗=λ↓7\times λ↓8\cr
λ↓{10}⊗=α↓7+λ↓9\cr}\hfill
\hfill\teqalign{λ↓1⊗=α↓1+λ↓0\cr
λ↓2⊗=α↓2+λ↓0\cr
λ↓3⊗=λ↓1\timesλ↓2\cr
λ↓4⊗=α↓3+λ↓3\cr
λ↓5⊗=α↓4+λ↓3\cr
λ↓6⊗=λ↓4\timesλ↓5\cr
λ↓7⊗=α↓5+λ↓3\cr
λ↓8⊗=α↓6+λ↓6\cr
λ↓9⊗=λ↓7\times λ↓8\cr
λ↓{10}⊗=α↓7+λ↓9\cr}\hfill$}$$
where, as in exercise↔35, an extra addition has been
inserted to cover a more general case. Neither of these schemes
can calculate a general sixth-degree monic polynomial, since
the first case is a polynomial of the form
$$(x↑3 + Ax↑2 + Bx + C)(x↑3 + Ax↑2 + Bx + D) + E,$$
and the second case (cf.\ exercise↔35) is a polynomial of the form
$$\biglp x↑4 + 2Ax↑3 + (E + A↑2)x↑2 + EAx + F\bigrp (x↑2 +
Ax + G) + H;$$
both of these involve only five independent parameters.

\ansno 37. Let $u↑{[0]}(x) = u↓nx↑n + u↓{n-1}x↑{n-1} + \cdots
+ u↓0$, $v↑{[0]}(x) = x↑n + v↓{n-1}x↑{n-1} + \cdots + v↓0$. For
$1 ≤ j ≤ n$, divide $u↑{[j-1]}(x)$ by the monic polynomial $v↑{[j-1]}(x)$,
obtaining $u↑{[j-1]}(x) = α↓jv↑{[j-1]}(x) + β↓jv↑{[j]}(x)$.
Assume that a monic polynomial $v↑{[j]}(x)$ of degree $n - j$
exists satisfying this relation; this will be true for almost
all rational functions.
Let $u↑{[j]}(x) = v↑{[j-1]}(x) - xv↑{[j]}(x)$.
These definitions imply that deg$(u↑{[n]}) < 1$, so we may let
$α↓{n+1} = u↑{[n]}(x)$.

For the given rational function we have
$$\vbox{\halign{$\ctr{#}$⊗\qquad$\ctr{#}$⊗\qquad$\ctr{#}$⊗\qquad$\ctr{#}$\cr
α↓j⊗β↓j⊗v↑{[j]}(x)⊗u↑{[j]}(x)\cr
\noalign{\vskip3pt}
1⊗2⊗x + 5⊗3x + 19\cr
3⊗4⊗1⊗5\cr}}$$
so $u↑{[0]}(x)/v↑{[0]}(x) = 1 + 2/\biglp x + 3 + 4/(x + 5)\bigrp $.

{\sl Notes:} A general rational function of the stated
form has $2n + 1$ ``degrees of freedom,'' in the sense that
it can be shown to have $2n + 1$ essentially independent parameters.
If we generalize polynomial chains to ``\α{arithmetic chains},''
which allow division operations as well as addition, subtraction,
and multiplication, we can obtain the following results with
slight modifications to the proofs of Theorems A and↔M\null: {\sl
An arithmetic chain with $q$ addition-subtraction steps has at
most $q + 1$ degrees of freedom. An arithmetic chain with $m$
multiplication-division
steps has at most $2m + 1$ degrees of freedom.} Therefore an arithmetic
chain that computes almost all rational functions of the stated
form must have at least $2n$ addition-subtractions, and $n$
multiplication-divisions; the method in this exercise is ``optimal.''
% Vicinity of page 645
% folio 835 galley 9a 	*
\ansno 38. The theorem is certainly true if $n = 0$. Assume
that $n$ is positive, and that a polynomial chain computing
$P(x; u↓0, \ldotss , u↓n)$ is given, where each of the parameters
$α↓j$ has been replaced by a real number. Let $λ↓i = λ↓j \times
λ↓k$ be the first chain multiplication step that involves one
of $u↓0$, $\ldotss$, $u↓n$; such a step must exist because of the
rank of $A$. Without loss of generality, we may assume that
$λ↓j$ involves $u↓n$; thus, $λ↓j$ has the form $h↓0u↓0 + \cdots
+ h↓nu↓n + f(x)$, where $h↓0$, $\ldotss$, $h↓n$ are real, $h↓n ≠
0$, and $f(x)$ is a polynomial with real coefficients.\xskip$\biglp
$The $h$'s and the coefficients of $f(x)$ are derived from the
values assigned to the $α$'s.$\bigrp$

Now change step $i$ to $λ↓i = α \times λ↓k$, where
$α$ is an arbitrary real number.\xskip (We could take $α = 0$; general
$α$ is used here merely to show that there is a certain amount
of flexibility available in the proof.)\xskip Add further steps to
calculate
$$λ = \biglp α - f(x) - h↓0u↓0 - \cdots - h↓{n-1}u↓{n-1}\bigrp/h↓n;$$
these new steps involve only additions and parameter
multiplications (by suitable new parameters). Finally, replace
$λ↓{-n-1} = u↓n$ everywhere in the chain by this new element↔$λ$.
The result is a chain that calculates
$$Q(x; u↓0, \ldotss , u↓{n-1}) = P\biglp x; u↓0, \ldotss , u↓{n-1},
(α - f(x) - h↓0u↓0 - \cdots - h↓{n-1}u↓{n-1})/h↓n\bigrp ;$$
and this chain has one less chain multiplication.
The proof will be complete if we can show that $Q$ satisfies
the hypotheses. The quantity $\biglp α - f(x)\bigrp /h↓n$ leads
to a possibly increased value of $m$, and a new vector $B↑\prime
$. If the columns of $A$ are $A↓0$, $A↓1$, $\ldotss$, $A↓n$ (these
vectors being linearly independent over the reals), the new
matrix $A↑\prime$ corresponding to↔$Q$ has the column vectors
$$A↓0 - (h↓0/h↓n)A↓n,\qquad \ldotss ,\qquad A↓{n-1} - (h↓{n-1}/h↓n)A↓n,$$
plus perhaps a few rows of zeros to account for
an increased value of $m$, and these columns are clearly also
linearly independent. By induction, the chain that computes↔$Q$
has at least $n - 1$ chain multiplications, so the original
chain has at least $n$.

[Pan showed also that the use of division would give no improvement;
cf.\ {\sl Problemy Kibernetiki \bf7} (1962), 21--30.
Generalizations to the computation of several
polynomials in several variables, with and without various kinds
of preconditioning, have been given by S. \β{Winograd,} {\sl Comm.\
Pure and Applied Math.\ \bf 23} (1970), 165--179.]

\ansno 39. By induction on $m$. Let $w↓m(x) = x↑{2m} + u↓{2m-1}x↑{2m-1}
+ \cdots + u↓0$, $w↓{m-1}(x) = x↑{2m-2} + v↓{2m-3}x↑{2m-3} + \cdots
+ v↓0$, $a = α↓1 + \gamma ↓m$, $b = α↓m$, and let $$\textstyle f(r) = \sum ↓{i,j≥0}
(-1)↑{i+j}{i+j\choose j}u↓{r+i+2j\,}a↑ib↑j.$$It follows that $v↓r
= f(r + 2)$ for $r ≥ 0$, and $\delta ↓m = f(1)$. If $\delta
↓m = 0$ and $a$ is given, we have a polynomial of degree $m
- 1$ in $b$, with leading coefficient $\pm (u↓{2m-1} - ma) =
\pm (\gamma ↓2 + \cdots + \gamma ↓m - m\gamma ↓m)$.

In Motzkin's unpublished notes he arranged to
make $\delta ↓k = 0$ almost always, by choosing $\gamma$'s
so that this leading coefficient is $≠0$ when $m$ is
even and $=0$ when $m$ is odd; then we almost always can let $b$
be a (real) root of an odd-degree polynomial.

\ansno 40. No; S. Winograd found a way to compute all polynomials
of degree 13 with only 7 (possibly complex) multiplications
[{\sl Comm.\ Pure and Applied Math.\ \bf 25} (1972), 455--457].
L. \α{Revah} found schemes that evaluate almost all polynomials
of degree $n ≥ 9$ with $\lfloor n/2\rfloor + 1$ (possibly complex)
multiplications [{\sl SIAM J. Computing} {\bf 4} (1975), 381--392];
she also showed that when $n=9$ it is possible to achieve $\lfloor n/2\rfloor+1$
multiplications only with at least $n+3$ additions. By appending sufficiently
many additions (cf.\ exercise↔39), the ``almost all'' and ``possibly complex''
provisos disappear.\xskip V. \t Ia.\ \α{Pan} [{\sl Proc.\ ACM Symp.\
Theory Comp.\ \bf10} (1978), 162--172; IBM Research Report RC7754 (1979)]
found schemes with $\lfloor n/2\rfloor+1$ (complex)
multiplications and the minimum number $n+2+\delta↓{n9}$ of (complex) additions, for
all odd $n≥9$; his method for $n=9$ is
$$\baselineskip14pt\cpile{v(x)=\biglp(x+α)↑2+β\bigrp(x+\gamma),\qquad
w(x)=v(x)+x,\cr t(x)=\biglp v(x)+\delta\bigrp\biglp w(x)+ε\bigrp-\biglp
v(x)+\delta↑\prime\bigrp\biglp w(x)+ε↑\prime\bigrp,\cr
u(x)=\biglp v(x)+\zeta\bigrp\biglp t(x)+\eta\bigrp+\kappa.\cr}$$
The minimum number of {\sl real\/} additions necessary, when the minimum number of
(real) multiplications is achieved, remains unknown for $n≥9$.

\ansno 41. $a(c + d) - (a + b)d + i\biglp a(c + d) + (b - a)c\bigrp$.\xskip
[Beware numerical instability. Three multiplications are necessary, since complex
multiplication is a special case of (69) with $p(u)=u↑2+1$.
Without the restriction on additions there are
other possibilities. For example, the symmetric formula $ac-bd+i\biglp(a+b)(c+d)-ac-bd
\bigrp$ was suggested by Peter \α{Ungar} in 1963; cf.\ Eq.\ 4.3.3--2 with $2↑n$
replaced by $i$. See I. \α{Munro}, {\sl Proc.\ ACM Symp.\ Theory Comp.\ \bf3}
(1960), 40--44; S. Winograd, {\sl Linear Alg.\ Appl.\ \bf4} (1971), 381--388.]

Alternatively, if $a↑2+b↑2=1$ and $t=(1-a)/b=b/(1+a)$, the algorithm ``$w=c-td$,
$v=d+bw$, $u=w-tv$'' for calculating the product $(a+bi)(c+di)=u+iv$ has been
suggested by Oscar \α{Buneman} [{\sl J. Comp. Phys.\ \bf12} (1973), 127--128]. In
this method if $a=\cos\theta$ and $b=\sin\theta$, we have $t=\tan(\theta/2)$.

[Helmut \α{Alt} and Jan \α{van Leeuwen} have shown that four real multiplications
or divisions are necessary for computing $1/(a+bi)$, and four are sufficient
\inx{division of complex numbers}
for computing $a/(b+ci)$. It is unknown whether $(a+bi)/(c+di)$ can be computed
with only five multiplications or divisions.] % to appear

\ansno 42. (a)\9 Let $π↓1$, $\ldotss$, $π↓m$ be the $λ↓i$'s that correspond
to chain multiplications; then $π↓i = P↓{2i-1} \times P↓{2i}$
and $u(x) = P↓{2m+1}$, where each $P↓j$ has the form $β↓j +
β↓{j0}x + β↓{j1}π↓1 + \cdots + β↓{jr(j)}π↓{r(j)}$, where $r(j)
≤ \lceil j/2\rceil - 1$ and each of the $β↓j$ and $β↓{jk}$ is
a polynomial in the $α$'s with integer coefficients. We can
systematically modify the chain (cf.\ exercise↔30) so that $β↓j
= 0$ and $β↓{jr(j)} = 1$, for $1 ≤ j ≤ 2m$; furthermore we can
assume that $β↓{30} = 0$. The result set now has at most $m
+ 1 + \sum ↓{1≤j≤2m} (\lceil j/2\rceil - 1) = m↑2 + 1$ degrees
of freedom.

(b)\9 Any such polynomial chain with at most $m$
chain multiplications can be simulated by one with the form
considered in (a), except that now we let $r(j) = \lceil j/2\rceil
- 1$ for $1 ≤ j ≤ 2m + 1$, and we do not assume that $β↓{30}
= 0$ or that $β↓{jr(j)} = 1$ for $j ≥ 3$. This single canonical
form involves $m↑2 + 2m$ parameters. As the $α$'s run through all integers
and as we run through all chains, the $β$'s run through at most
$2↑{m↑2+2m}$ sets of values mod 2, hence the result
set does also. In order to obtain all $2↑n$ polynomials of degree
$n$ with 0--1 coefficients, we need $m↑2 + 2m ≥ n$.

(c)\9 Set $m ← \lfloor \sqrt{n}\rfloor$ and compute
$x↑2$, $x↑3$, $\ldotss$, $x↑m$. Let $u(x) = u↓{m+1}(x)x↑{(m+1)m} +
\cdots + u↓1(x)x↑m + u↓0(x)$, where each $u↓j(x)$ is a polynomial
of degree $≤m$ with integer coefficients (hence it can be evaluated
without any more multiplications). Now evaluate $u(x)$ by rule
(2) as a polynomial in $x↑m$ with known coefficients.\xskip (The number
of additions used is approximately the sum of the absolute values
of the coefficients, so this algorithm is efficient on 0--1
polynomials. Paterson and Stockmeyer also gave another algorithm
that uses about $\sqrt{2n}$ multiplications.)

Reference: {\sl SIAM J. Computing \bf2} (1973), 60--66; see also
J. E. \α{Savage,} {\sl SIAM J. Computing \bf3} (1974), 150--158.
For analogous results about
additions, see \α{Borodin} and \α{Cook},
{\sl SIAM J. Computing \bf5} (1976), 146--157; \α{Rivest} and \α{Van de
Wiele}, {\sl Inf.\ Proc.\ Letters \bf8} (1979), 178--180.]

\ansno 43. When $a↓i = a↓j + a↓k$ is a step in some optimal
addition chain for $n + 1$, compute $x↑i = x↑jx↑k$ and $p↓i=
p↓kx↑j + p↓j$, where $p↓i = x↑{i-1} + \cdots + x + 1$; omit
the final calculation of $x↑{n+1}$. We save one multiplication
whenever $a↓k = 1$, in particular when $i = 1$.\xskip $\biglp$Cf.\
exercise 4.6.3--31 with $ε = {1\over 2}$.$\bigrp$

\ansno 44. It suffices to show that $(T↓{ijk})$'s rank is {\sl at most} that of
$(t↓{ijk})$, since we can obtain $(t↓{ijk})$ back from $(T↓{ijk})$ by transforming
it in the same way with $F↑{-1}$, $G↑{-1}$, $H↑{-1}$. If $t↓{ijk}=\sum↓{1≤l≤r}
a↓{il\,}b↓{jl\,}c↓{kl}$ then it follows immediately that
$$\textstyle T↓{ijk}=\sum↓{1≤l≤r}\biglp\sum↓{1≤p≤m}F↓{ip\,}a↓{pl}\bigrp
\biglp\sum↓{1≤q≤n}G↓{jq\,}b↓{ql}\bigrp
\biglp\sum↓{1≤r≤s}H↓{kr\,}c↓{rl}\bigrp.$$

[H. F. \α{de Groote} has proved that all normal schemes that yield $2\times2$ matrix
products with seven chain multiplications are equivalent, in the sense that they can
be obtained from each other by nonsingular matrix multiplication as in this
exercise. In this sense \α{Strassen}'s algorithm is unique.]

\ansno 45. By exercise↔44 we can add any multiple of a row, column, or plane to
another one without changing the rank; we can also multiply a row, column,
or plane by a nonzero constant, or transpose the tensor. A sequence of such
operations can always be found to reduce a give $2\times2\times2$ tensor to one
of the forms ${0\,0\choose0\,0}{0\,0\choose0\,0}$, ${1\,0\choose0\,0}{0\,0\choose
0\,0}$, ${1\,0\choose0\,1}{0\,0\choose0\,0}$, ${1\,0\choose0\,0}{0\,0\choose0\,1}$,
${1\,0\choose0\,1}{0\,1\choose q\,r}$. The last tensor has rank 3 or 2 according as
the polynomial $u↑2-ru-q$ has one or two irreducible factors in the field of
interest, by Theorem↔W $\biglp$cf.↔(72)$\bigrp$.

\ansno 46. A general $m\times n\times s$ tensor has $mns$ degrees of freedom. By
exercise↔28 it is impossible to express all $m\times n\times s$ tensors in terms
of the $(m+n+s)r$ elements of a realization $A$, $B$, $C$ unless $(m+n+s)r≥mns$.
On the other hand, assume that $m≥n≥s$. The rank of an $m\times n$ matrix is at most
$n$, so we can realize any tensor in $ns$ chain multiplications by realizing each
matrix plane separately.\xskip[Exercise↔45 shows that this lower bound on the maximum
tensor rank is not best possible, nor is the upper bound. Thomas D. \α{Howell}
(Ph.\ D. thesis, Cornell Univ., 1976) has shown that there are tensors of rank
$≥\lceil mns/(m+n+s-2)\rceil$ over the complex numbers.]

\ansno 48. If $A$, $B$, $C$ and $A↑\prime$, $B↑\prime$, $C↑\prime$ are realizations
of $(t↓{ijk})$ and $(t↓{ijk}↑\prime)$ of respective lengths $r$ and $r↑\prime$, then
$A↑{\prime\prime}=A\oplus A↑\prime$,
$B↑{\prime\prime}=B\oplus B↑\prime$,
$C↑{\prime\prime}=C\oplus C↑\prime$,
and
$A↑{\prime\prime\prime}=A\otimes A↑\prime$,
$B↑{\prime\prime\prime}=B\otimes B↑\prime$,
$C↑{\prime\prime\prime}=C\otimes C↑\prime$,
are realizations of $(t↓{ijk}↑{\prime\prime})$ and $(t↓{ijk}↑{\prime\prime\prime})$
of respective lengths $r+r↑\prime$ and $r\cdot r↑\prime$.

{\sl Note:} Many people have made the natural conjecture that \def\\{\hbox{rank}}
$\\\biglp(t↓{ijk})\oplus(t↓{ijk}↑\prime)\bigrp=\\(t↓{ijk})+\\(t↓{ijk}↑\prime)$,
but the construction in exercise 60(b) makes this seem much less plausible than
it once was.

\ansno 49. By Lemma↔T\null, rank$(t↓{ijk})≥\hbox{rank}(t↓{i(jk)})$. Conversely if
$M$ is a matrix of rank $r$ we can transform it by row and column operations,
finding nonsingular matrices $F$ and $G$ such that $FMG$ has all entries 0 except
for $r$ diagonal elements that are↔1; cf.\ Algorithm↔4.6.2N\null. The tensor rank
of $FMG$ is therefore $≤r$; and it is the same as the tensor rank of $M$, by
exercise↔44.

\ansno 50. Let $i=\langle i↑\prime,i↑{\prime\prime}\rangle$ where $1≤i↑\prime≤m$
and $1≤i↑{\prime\prime}≤n$; then $t↓{\langle i↑\prime,i↑{\prime\prime}\rangle jk}
=\delta↓{i↑{\prime\prime}j}\delta↓{i↑\prime k}$, and it is clear that
rank$(t↓{i(jk)})=mn$ since $(t↓{i(jk)})$ is a permutation matrix. By Lemma↔L\null,
rank$(t↓{ijk})≥mn$. Conversely, since $(t↓{ijk})$ has only $mn$ nonzero entries,
its rank is clearly $≤mn$.\xskip(There is consequently no normal scheme requiring
fewer than the $mn$ obvious multiplications. There is no such abnormal scheme
either [{\sl Comm.\ Pure and Appl.\ Math.\ \bf3} (1970), 165--179]. But some
savings can be achieved if the same matrix is used with $s>1$ different column
vectors, since this is equivalent to $(m\times n)$ times $(n\times s)$ matrix
multiplication.)

\ansno 51. (a)\9 $s↓1=y↓0+y↓1$, $s↓2=y↓0-y↓1$; $m↓1={1\over2}(x↓0+x↓1)s↓1$,
$m↓2={1\over2}(x↓0-x↓1)s↓2$; $w↓0=m↓1+m↓2$, $w↓1=m↓1-m↓2$.\xskip(b) Here are some
intermediate steps, using the methodology in the text: $\biglp(x↓0-x↓2)+(x↓1-x↓2)u
\bigrp\biglp(y↓0-y↓2)+(y↓1-y↓2)u\bigrp\mod(u↑2+u+1)=\biglp(x↓0-x↓2)(y↓0-y↓2)-
(x↓1-x↓2)(y↓1-y↓2)\bigrp+\biglp(x↓0-x↓2)(y↓0-y↓2)-(x↓1-x↓0)(y↓1-y↓0)\bigrp u$.
The first realization is
$$\def\\#1.#2.#3.{\biggglp\,\vcenter{\vskip-4pt
\halign{##\cr\vbox to 9pt{}#1\cr#2\cr#3\cr}}\,\bigggrp}
\\1 1 \=1 0.1 0 1 1.1 \=1 0 \=1.,\qquad
\\1 1 \=1 0.1 0 1 1.1 \=1 0 \=1.,\qquad
\\1 1 1 \=2.1 1 \=2 1.1 \=2 1 1.\times{1\over3}.$$
The second realization is
$$\def\\#1.#2.#3.{\biggglp\,\vcenter{\vskip-4pt
\halign{##\cr\vbox to 9pt{}#1\cr#2\cr#3\cr}}\,\bigggrp}
\\1 1 1 \=2.1 1 \=2 1.1 \=2 1 1.\times{1\over3},\qquad
\\1 1 \=1 0.1 \=1 0 \=1.1 0 1 1.,\qquad
\\1 1 \=1 0.1 0 1 1.1 \=1 0 \=1..$$
The resulting algorithm computes $s↓1=y↓0+y↓1$, $s↓2=y↓0-y↓1$, $s↓3=y↓2-y↓0$,
$s↓4=y↓2-y↓1$, $s↓5=s↓1+y↓2$; $m↓1={1\over3}(x↓0+x↓1+x↓2)s↓5$,
$m↓2={1\over3}(x↓0+x↓1-2x↓2)s↓2$, $m↓3={1\over3}(x↓0-2x↓1+x↓2)s↓3$,
$m↓4={1\over3}(-2x↓0+x↓1+x↓2)s↓4$; $t↓1=m↓1+m↓2$, $t↓2=m↓1-m↓2$, $t↓3=m↓1+m↓3$,
$w↓0=t↓1-m↓3$, $w↓1=t↓3+m↓4$, $w↓2=t↓2-m↓4$.

\def\dprime{{\prime\prime}}
\ansno 52. Let $i=\langle i↑\prime,i↑\dprime\rangle$ when $i\mod n↑\prime=i↑\prime$
and $i\mod n↑\dprime=i↑\dprime$. Then we wish to compute $w↓{\langle k↑\prime,
k↑\dprime\rangle}=\sum x↓{\langle i↑\prime,i↑\dprime\rangle}y↓{\langle j↑\prime,
j↑\dprime\rangle}$ summed for $i↑\prime+j↑\prime≡k↑\prime\modulo{n↑\prime}$ and
$i↑\dprime+j↑\dprime≡k↑\dprime\modulo{n↑\dprime}$. This can be done by applying
the $n↑\prime$ algorithm to the $2n↑\prime$ vectors $X↓{i↑\prime}$ and $Y↓{j↑\prime
}$ of length $n↑\dprime$, obtaining the $n↑\prime$ vectors $W↓{k↑\prime}$. Each
vector addition becomes $n↑\dprime$ additions, each parameter multiplication
becomes $n↑\dprime$ parameter multiplications, and each chain multiplication of
vectors is replaced by a cyclic convolution of degree $n↑\dprime$.\xskip [If
the subalgorithms use the minimum number of chain multiplications, this algorithm
uses $2\biglp n↑\prime-d(n↑\prime)\bigrp\biglp n↑\dprime-d(n↑\dprime)\bigrp$ more
than the minimum, where $d(n)$ is the number of divisors of $n$.]

\ansno 53. (a)\9 Let $n(k)=(p-1)p↑{e-k-1}=\varphi(p↑{e-k})$ for $0≤k<e$, and $n(k)=
1$ for $k≥e$. Represent the numbers $\{1,\ldotss,m\}$ in the form $a↑ip↑k\modulo m$,
where $0≤k≤e$ and $0≤i<n(k)$,
and $a$ is a fixed primitive element modulo $p↑e$. For example,
when ${m=9}$ we can let $a=2$; the values are $\{2↑03↑0,2↑13↑0,2↑03↑1,2↑23↑0,2↑53↑0,
2↑13↑1,2↑43↑0,2↑33↑0,2↑03↑2\}$. Then $f(a↑ip↑k)=\sum↓{0≤l≤e}\sum↓{0≤j<n(l)}
\omega↑{g(i,j,k,l)}F(a↑jp↑l)$ where $g(i,j,k,l)=a↑{i+j}p↑{k+l}$.

We shall compute $f↓{ikl}=\sum↓{0≤j<n(l)}\omega↑{g(i,j,k,l)}F(a↑jp↑l)$ for $0≤i<
n(k)$ and for each $k$, $l$. This is a cyclic convolution of degree $n(k+l)$ on
the values $x↓i=\omega↑{a↑ip↑{k+l}}$ and $y↓s=\sum↓{0≤j<n(l),\,s+j≡0\,\modulo{n(k+l)}}
F(a↑jp↑l)$, since $f↓{ikl}=\sum x↓ry↓s$ summed over $r+s≡i$ $\biglp$modulo
$n(k+l)\bigrp$. The Fourier transform is obtained by summing appropriate
$f↓{ikl}$'s.\xskip[{\sl Note:} When linear combinations of the $x↓i$ are
formed, e.g., as in (67), the result will be purely real or purely imaginary,
when the cyclic convolution algorithm has been constructed by using rule (57) with
$u↑{n(k)}-1=(u↑{n(k)/2}-1)(u↑{n(k)/2}+1)$. The reason is that reduction mod
$(u↑{n(k)/2}-1)$ produces a polynomial with real coefficients $\omega↑j+\omega↑{-j}$
while reduction mod $(u↑{n(k)/2}+1)$ produces a polynomial with imaginary
coefficients $\omega↑j-\omega↑{-j}$.]

When $p=2$ an analogous construction applies, using the representation $(-1)↑ia↑j
2↑k\modulo m$, where $0≤k≤e$ and $0≤i≤\min(e-k,1)$ and $0≤j<2↑{e-k-2}$. In this
case we use the construction of exercise↔52 with $n↑\prime=2$ and $n↑\dprime=
2↑{e-k-2}$; although these numbers are not relatively prime, the construction does
yield the desired direct product of cyclic convolutions.

(b)\9 Let $a↑\prime m↑\prime+a↑\dprime m↑\dprime=1$; and let $\omega↑\prime=
\omega↑{a↑\dprime m↑\dprime}$, $\omega↑\dprime=\omega↑{a↑\prime m↑\prime}$. Define
$s↑\prime=s\mod m↑\prime$, $s↑\dprime=s\mod m↑\dprime$, $t↑\prime=t\mod m↑\prime$,
$t↑\dprime=t\mod m↑\dprime$, so that $\omega↑{st}=(\omega↑\prime)↑{s↑\prime
t↑\prime}(\omega↑\dprime)↑{s↑\dprime t↑\dprime}$. It follows that $f(s↑\prime,s↑
\dprime)=\sum↓{0≤t↑\prime<m↑\prime,\,0≤t↑\dprime<m↑\dprime}(\omega↑\prime)↑{s↑
\prime t↑\prime}(\omega↑\dprime)↑{s↑\dprime t↑\dprime}F(t↑\prime,t↑\dprime)$; in
other words, the one-dimensional Fourier transform on $m$ elements is actually a
two-dimensional Fourier transform on $m↑\prime\times m↑\dprime$ elements, in
slight disguise.

We shall deal with ``normal'' algorithms consisting of (i) a number of sums $s↓i$
of the $F$'s and $s$'s; followed by (ii) a number of products $m↓j$, each of which
is obtained by multiplying one of the $F$'s or $S$'s by a real or imaginary number
$α↓j$; followed by (iii) a number of further sums $t↓k$, each of which is formed
from $m$'s or $t$'s (not $F$'s or $s$'s). The final values must be $m$'s or $t$'s.
For example, the ``normal'' Fourier transform scheme for $m=5$ constructed from
(67) and the method of part (a) is as follows: $s↓1=F(1)+F(4)$, $s↓2=F(3)+F(2)$,
$s↓3=s↓1+s↓2$, $s↓4=s↓1-s↓2$, $s↓5=F(1)-F(4)$, $s↓6=F(2)-F(3)$, $s↓7=s↓5-s↓6$;
$m↓1={1\over4}(\omega+\omega↑2+\omega↑4+\omega↑3)s↓3$, $m↓2={1\over4}(\omega-
\omega↑2+\omega↑4-\omega↑3)s↓4$, $m↓3={1\over2}(\omega+\omega↑2-\omega↑4-\omega↑3)
s↓5$, $m↓4={1\over2}(-\omega+\omega↑2+\omega↑4-\omega↑3)s↓6$, $m↓5={1\over2}(
\omega↑3-\omega↑2)s↓7$, $m↓6=1\cdot F(5)$, $m↓7=1\cdot s↓3$; $t↓0=m↓1+m↓6$, $t↓1=
t↓0+m↓2$, $t↓2=m↓3+m↓5$, $t↓3=t↓0-m↓2$, $t↓4=m↓4-m↓5$, $t↓5=t↓1+t↓2$, $t↓6=t↓3+t↓4$,
$t↓7=t↓1-t↓2$, $t↓8=t↓3-t↓4$, $t↓9=m↓6+m↓7$. Note the multiplication by 1 shown in
$m↓6$ and $m↓7$; this is required by our conventions, and it is important to
include such cases for use in recursive constructions (although the multiplications
need not really be done). Here $m↓6=f↓{001}$, $m↓7=f↓{010}$, $t↓5=f↓{000}+f↓{001}
=f(2↑0)$, $t↓6=f↓{100}+f↓{101}=f(2↑1)$, etc. We can improve the scheme by
introducing $s↓8=s↓3+F(5)$, replacing $m↓1$ by $\biglp{1\over4}(\omega+\omega↑2
+\omega↑4+\omega↑3)-1\bigrp s↓3$ [this is $-{5\over4}s↓3$], replacing $m↓6$ by
$1\cdot s↓8$, and deleting $m↓7$ and $t↓9$; this saves one of the trivial
multiplications by 1, and it will be advantageous when the scheme is used to
build larger ones. In the improved scheme, $f(5)=m↓6$, $f(1)=t↓5$, $f(2)=t↓6$,
$f(3)=t↓8$, $f(4)=t↓7$.

Now suppose we have normal one-dimensional schemes for $m↑\prime$ and $m↑\dprime$,
using respectively $(a↑\prime,a↑\dprime)$ complex additions, $(t↑\prime,t↑\dprime)$
trivial multiplications by $\pm1$ or $\pm i$, and a total of $(c↑\prime,c↑\dprime)$
complex multiplications including the trivial ones.\xskip(The nontrivial complex
multiplications are all ``simple'' since they involve only two real multiplications
and no real additions.)\xskip We can construct a normal scheme for the
two-dimensional $m↑\prime\times m↑\dprime$ case by applying the $m↑\prime$ scheme
to vectors $F(t↑\prime,\ast)$ of length $m↑\dprime$. Each $s↓i$ step becomes
$m↑\dprime$ additions; each $m↓j$ becomes a Fourier transform on $m↑\dprime$
elements, but with all of the $α$'s in this algorithm multiplied by $α↓j$; and
each $t↓k$ becomes $m↑\dprime$ additions. Thus the new algorithm has $(a↑\prime
m↑\dprime+c↑\prime a↑\dprime)$ complex additions, $t↑\prime t↑\dprime$ trivial
multiplications, and a total of $c↑\prime c↑\dprime$ complex multiplications.

Using these techniques, Winograd has found normal one-dimensional schemes
for the following small values of $m$ with the following costs $(a,t,c)$:
$$\baselineskip13pt
\vbox{\halign{$m=#\hfill$⊗\quad$(\hfill#$,⊗$\,#,$⊗$\,\hfill#)$⊗\hskip 100pt
$m=#\hfill$⊗\quad$(\hfill#$,⊗$\,#,$⊗$\,\hfill#)$\cr
2⊗2⊗2⊗2⊗7⊗36⊗1⊗9\cr
3⊗6⊗1⊗3⊗8⊗26⊗6⊗8\cr
4⊗8⊗4⊗4⊗9⊗46⊗1⊗12\cr
5⊗17⊗1⊗6⊗16⊗74⊗8⊗18\cr}}$$
By combining these schemes as described above, we obtain methods that use fewer
arithmetic operations than the ``\α{fast Fourier transform}'' (FFT) discussed in
exercise↔14. For example, when $m=1008=7\cdot9\cdot16$, the costs come to
$(17946,8,1944)$, so we can do a Fourier transform on 1008 complex numbers with
3872 real multiplications and 35892 real additions.
It is possible to improve on Winograd's method for combining relatively prime
moduli by using multidimensional convolutions, as shown by \α{Nussbaumer} and
\inx{convolutions, multidimensional}
\α{Quandalle} in {\sl IBM J. Res.\ and Devel.\ \bf22} (1978), 134--144; their
ingenious approach reduces the amount of computation needed for 1008-point
complex Fourier transforms to 3084 real multiplications and 34668 real additions.
 By contrast, the FFT on 1024
complex numbers involves 14344 real multiplications and 27652 real additions.
If the two-passes-at-once improvement in the answer to exercise↔14 is used,
however, the FFT on 1024 complex numbers needs only 10936 real multiplications and
25948 additions, and it is not difficult to implement. Therefore the subtler methods
are faster only on machines that take significantly longer to multiply than to add.

[References: {\sl Proc.\ Nat.\ Acad.\ Sci.\ USA \bf73} (1976), 1005--1006;
{\sl Math.\ Comp.\ \bf32} (1978), 175--199; {\sl Advances in Math.\ \bf32} (1979),
83--117.]

\ansno 54. $\max\biglp 2e↓1\hbox{deg}(p↓1)-1,\ldotss,2e↓q\hbox{deg}(p↓q)-1,q+1)$.

\ansno 55. $2n↑\prime-q↑\prime$, where $n↑\prime$ is the degree of the minimum
polynomial of $P$ (i.e., the monic
polynomial $\mu$ of least degree such that $\mu(P)$ is the zero matrix) and
$r↑\prime$ is the number of distinct irreducible factors it has.\xskip(Reduce $P$
by similarity transformations.)

\ansno 56. Let $t↓{ijk}+t↓{jik}=\tau↓{ijk}+\tau↓{jik}$, for all $i$, $j$, $k$. If
$A$, $B$, $C$ is a realization of $(t↓{ijk})$ of rank $r$, then $\lower6pt\null
\sum↓{1≤l≤r}c↓{kl\,}\biglp\sum a↓{il\,}x↓i\bigrp\biglp\sum b↓{jl\,}x↓j\bigrp=
\sum↓{i,j}t↓{ijk}x↓ix↓j=\sum↓{i,j}\tau↓{ijk}x↓ix↓j$ for all $k$. Conversely, let the
$l$th chain multiplication of a polynomial chain, for $1≤l≤r$, be the product
$\biglp α↓l+\sum α↓{il\,}x↓i\bigrp\biglp β↓l+\sum β↓{jl}x↓j\bigrp$, where
$α↓l$ and $β↓l$ denote possible constant terms and/or nonlinear terms. All terms
of degree 2 appearing at any step of the chain can be expressed as a linear
combination $\sum↓{1≤l≤r}c↓{l\,}\biglp\sum a↓{il\,}x↓i\bigrp\biglp\sum
b↓{jl\,}x↓j\bigrp$; hence the chain defines a tensor $(t↓{ijk})$ of rank $≤r$
such that $t↓{ijk}+t↓{jik}=\tau↓{ijk}+\tau↓{jik}$. This establishes the hint. Now
\def\\{\hbox{rank}}$\\(\tau↓{ijk}+\tau↓{jik})=\\(t↓{ijk}+t↓{jik})≤\\(t↓{ijk})+\\(t↓
{jik})=2\,\\(t↓{ijk})$.

A bilinear form in $x↓1$, $\ldotss$, $x↓m$, $y↓1$, $\ldotss$, $y↓n$ is a
quadratic form in $m+n$ variables, where $\tau↓{ijk}=t↓{i,j-m,k}$ for $i≤m$ and $j>m$,
otherwise $\tau↓{ijk}=0$. Now $\\(\tau↓{ijk})+\\(\tau↓{jik})≥\\(t↓{ijk})$, since we obtain
a realization of $(t↓{ijk})$ by suppressing the last $n$ rows of $A$ and the first
$m$ rows of $B$ in a realization $A$, $B$, $C$ of $(\tau↓{ijk}+\tau↓{jik})$.

\ansno 57. Let $N$ be the smallest power of 2 that exceeds
$2n$, and let $u↓{n+1} = \cdots = u↓{N-1} = v↓{n+1} = \cdots
= v↓{N-1} = 0$. If $U↓s = \sum ↓{0≤t<N} \omega ↑{st}u↓t$, $V↓s
= \sum ↓{0≤t<N} \omega ↑{st}v↓t$, ${0 ≤ s < N}$, $\omega = e↑{2πi/N}$,
then $\sum ↓{0≤s<N} \omega ↑{-st}U↓sV↓s = N \sum u↓{t↓1}v↓{t↓2}$, where
the latter sum is taken over all $t↓1$ and $t↓2$
with $0 ≤ t↓1, t↓2 < N$, $t↓1 + t↓2 ≡ t\modulo N$. The terms
vanish unless $t↓1 ≤ n$ and $t↓2 ≤ n$, so $t↓1 + t↓2 < N$; thus the
sum is the coefficient of $z↑t$ in the product $u(z)v(z)$. If
we use the method of exercise↔14 to compute the Fourier transforms and
the inverse transforms, the number of complex operations is
$O(N\log N) + O(N \log N) + O(N) + O(N\log N)$; and $N
< 4n$.\xskip [Cf.\ Section 4.3.3 and the paper by J. M. \α{Pollard,} {\sl Math.\
Comp.\ \bf 25} (1971), 365--374.]

When multiplying integer polynomials, it is possible to use an {\sl integer} number
$\omega$ that is of order $2↑t$ modulo a prime $p$, and to
determine the results modulo sufficiently many primes. Useful
\inx{primes, useful}
primes in this regard, together with their least primitive roots↔$r$
(from which we take $\omega = r↑{(p-1)/2↑t}\mod
p$ when $p \mod 2↑t = 1)$, can be found as described in Section
4.5.4. For $t = 9$, the ten largest cases $<2↑{35}$ are $p = 2↑{35}
- 512a + 1$, where $(a, r) = (28, 7)$, $(31, 10)$, $(34, 13)$, $(56,
3)$, $(58, 10)$, $(76, 5)$, $(80, 3)$, $(85, 11)$, $(91, 5)$, $(101, 3)$;
the ten largest cases $<2↑{31}$ are $p = 2↑{31} - 512a + 1$, where
$(a, r) = (1, 10)$, $(11, 3)$, $(19, 11)$, $(20, 3)$, $(29, 3)$, $(35,
3)$, $(55, 19)$, $(65, 6)$, $(95, 3)$, $(121, 10)$. For larger $t$,
all primes $p$ of the form $2↑tq + 1$ where $q < 32$ is odd
and $2↑{24} < p < 2↑{36}$ are given by $(p - 1, r) = (11 \cdot
2↑{21}, 3)$, $(25 \cdot 2↑{20}, 3)$, $(27 \cdot 2↑{20}, 5)$, $(25
\cdot 2↑{22}, 3)$, $(27 \cdot 2↑{22}, 7)$, $(5 \cdot 2↑{25}, 3)$,
$(7 \cdot 2↑{26}, 3)$, $(27 \cdot 2↑{26},
13)$, $(15 \cdot 2↑{27}, 31)$, $(17 \cdot 2↑{27}, 3)$, $(3 \cdot 2↑{30},
5)$, $(13 \cdot 2↑{28}, 3)$, $(29 \cdot 2↑{27}, 3)$, $(23 \cdot
2↑{29}, 5)$. Some of the latter primes can be used with $\omega
= 2↑e$ for appropriate small $e$. For a discussion of such primes, see R. M.
\α{Robinson,} {\sl Proc.\ Amer.\ Math.\ Soc.\ \bf9} (1958), 673--681; S. W. \α{Golomb},
{\sl Math.\ Comp.\ \bf30} (1976), 657--663.

However, the method of exercise 59 will almost always be preferable in practice.

\ansno 58. (a)\9 In general if $A$, $B$, $C$ realizes $(t↓{ijk})$, then
$(x↓1,\ldotss,x↓m)A$, $B$, $C$ is a realization of the $1\times n\times s$ matrix
whose entry in row $j$, column $k$ is $\sum x↓it↓{ijk}$. So there must be at
least as many nonzero elements in $(x↓1,\ldotss,x↓m)A$ as the rank of this matrix.
In the case of the $m\times n\times(m+n-1)$ tensor corresponding to polynomial
multiplication of degree $m-1$ by degree $n-1$, the corresponding matrix has
rank $n$ whenever $(x↓1,\ldotss,x↓m)≠(0,\ldotss,0)$. A similar statement holds
with $A\xch B$ and $m\xch n$.\xskip[In particular, if we work over the field of
2 elements, this says that the rows of $A$ modulo 2 form a ``linear code'' of
$m$ vectors having distance at least $n$, whenever $A$, $B$, $C$ is a realization
consisting entirely of integers. This observation, due to R. W. \α{Brockett} and
D. \α{Dobkin} [{\sl Linear Algebra and its Applic.\ \bf19} (1978), 207--235,
Theorem↔14], can be used to obtain nontrivial lower bounds on the rank over the integers.
For example, M.↔R. \α{Brown} and D. Dobkin have used it to show that realizations 
of $n\times n$ polynomial multiplication over the integers must have $t≥3.52n$,
for all sufficiently large $n$; see {\sl IEEE Trans.\ \bf C--29} (1980),
337--340.]

\yskip
(b) $\left(\vcenter{\halign{#\cr
1 0 0 0 0 1 1 1\cr 0 1 0 0 1 1 0 1\cr 0 0 1 1 0 0 1 1\cr}}\right)$,
$\left(\vcenter{\halign{#\cr
1 0 0 0 0 1 1 1\cr 0 1 0 0 0 1 0 1\cr 0 0 1 0 0 0 1 1\cr 0 0 0 1 1 0 0 1\cr}}
\right)$,
$\left(\vcenter{\halign{#\cr 1 0 0 0 0 0 0 0\cr
\=1 \=1 0 0 0 1 0 0\cr \=1 1 \=1 0 0 0 1 0\cr 1 0 0 \=1 \=1 \=1 \=1 1\cr
0 0 1 0 1 0 0 0\cr 0 0 0 1 0 0 0 0\cr}}\right)$.

\ansno 59. [{\sl IEEE Trans.\ \bf ASSP--28} (1980), 205--215.]\xskip
Note that cyclic convolution is polynomial multiplication mod $u↑n-1$, and
negacyclic convolution is \α{polynomial multiplication} mod $u↑n+1$.
Let us now change notation, replacing $n$ by $2↑n$; we shall consider
\α{recursive} algorithms for cyclic and negacyclic convolution $(z↓0,\ldotss,z↓{2↑n-1})$
of $(x↓0,\ldotss,x↓{2↑n-1})$ with $(y↓0,\ldotss,y↓{2↑n-1})$. The algorithms
are presented in unoptimized form, for brevity and ease in exposition; the reader
who implements them will notice that many things can be streamlined.

\algstep C1. [Test for simple case.] If $n=1$, set $z↓0←x↓0y↓0+x↓1y↓1$,
$z↓1←(x↓0+x↓1)(y↓0+y↓1)-z↓0$, and terminate. Otherwise set $m←2↑{n-1}$.

\algstep C2. [Remainderize.] For $0≤k<m$, set $(x↓k,x↓{m+k})←(x↓k+x↓{m+k},
x↓k-x↓{m+k})$ and $(y↓k,y↓{m+k})←(y↓k+y↓{m+k},y↓k-y↓{m+k})$.\xskip
$\biglp$Now we have $x(u)\mod(u↑m-1)=x↓0+\cdots+x↓{m-1}u↑{m-1}$ and $x(u)\mod(u↑m+1)=
x↓m+\cdots+x↓{2m-1}$; we will compute $x(u)y(u)\mod(u↑m-1)$ and $x(u)y(u)\mod
(u↑m+1)$, then we will combine the results by (57).$\bigrp$

\jpar1000
\algstep C3. [Recurse.] Set $(z↓0,\ldotss,z↓{m-1})$ to the cyclic convolution
of $(x↓0,\ldotss,x↓{m-1})$ with $(y↓0,\ldotss,y↓{m-1})$. Also set
$(z↓m,\ldotss,z↓{2m-1})$ to the negacyclic convolution of $(x↓m,\ldotss,x↓{2m-1})$ with $(y↓m,\ldotss,
y↓{2m-1})$.

\jpar2
\algstep C4. [Unremainderize.] For $0≤k<m$, set $(z↓k,z↓{m+k})←
{1\over2}(z↓k+z↓{m+k},z↓k-z↓{m+k})$. Now $(z↓0,\ldotss,z↓{m-1})$ is the
desired answer.\quad\blackslug

\yskip
\algstep N1. [Test for simple case.] If $n=1$, set $t←x↓0(y↓0+y↓1)$, $z↓0←t-(x↓0+
x↓1)y↓1$, $z↓1←t+(x↓1-x↓0)y↓0$, and terminate. Otherwise set $m←2↑{\lfloor n/2
\rfloor}$ and $r←2↑{\lceil n/2\rceil}$.\xskip
$\biglp$The following steps use $2↑{n+1}$ auxiliary variables $X↓{ij}$
for $0≤i<2m$ and $0≤j<r$, to represent $2m$ polynomials $X↓i(w)=X↓{i0}+X↓{i1}w+
\cdots+X↓{i(r-1)}w↑{r-1}$; similarly, there are $2↑{n+1}$ auxiliary variables↔$Y↓{ij}$.$\bigrp$

\algstep N2. [Initialize auxiliary polynomials.] Set $X↓{ij}←X↓{(i+m)j}←x↓{mj+i}$
$Y↓{ij}←Y↓{(i+m)j}←y↓{mj+i}$, for $0≤i<m$ and $0≤j<r$.\xskip
$\biglp$At this point we have $x(u)=X↓0(u↑m)+uX↓1(u↑m)+\cdots+u↑{m-1}X↓{m-1}(u↑m)$,
and a similar formula holds for↔$y(u)$. Our strategy will be to multiply these
polynomials modulo $(u↑{mr}+1)=(u↑n+1)$, by operating modulo $(w↑r+1)$ on the
polynomials $X(w)$ and $Y(w)$, finding their cyclic correlation of length
$2m$ and thereby obtaining $x(u)y(u)=Z↓0(u↑m)+uZ↓1(u↑m)+\cdots+u↑{2m-1}Z↓{2m-1}(u↑m)$.$\bigrp$

\jpar100
\algstep N3. [Transform.] $\biglp$Now we will essentially do a \α{fast Fourier
transform} on the polynomials $(X↓0,\ldotss,X↓{m-1},0,\ldotss,0)$ and
$(Y↓0,\ldotss,Y↓{m-1},0,\ldotss,0)$, using $w↑{r/m}$ as a $(2m)$th root of unity.
This is efficient, because multiplication by a power of $w$ is not really a
multiplication at all.$\bigrp$ For $j=\lfloor n/2\rfloor-1$, $\ldotss$, 1,↔0
(in this order), do the following for all $m$ binary numbers $s+t=
(s↓{\lfloor n/2\rfloor}\ldotsm s↓{j+1}0\ldotsm0)↓2+(0\ldotsm0t↓{j-1}\ldotsm t↓0)↓2$:
Replace $\biglp X↓{s+t}(w),X↓{s+t+2↑j}(w)\bigrp$ by the pair of polynomials
$\biglp X↓{s+t}(w)+w↑{(r/m)(s/2)}X↓{s+t+2↑j}(w),
X↓{s+t}(w)-w↑{(r/m)(s/2)}X↓{s+t+2↑j}(w)\bigrp$.\xskip
$\biglp$See Section 4.3.3 and
Eq.\ 4.3.3--33. The operation $X↓i(w)←X↓i(w)+w↑kX↓l(w)$ means,
more precisely, that we set $X↓{ij}←X↓{ij}+X↓{l(j+k)}$ if $j+k<r$, otherwise
$X↓{ij}←X↓{ij}-X↓{l(j+k-r)}$, for $0≤j<r$.$\bigrp$ Do the same transformation
on the $Y$'s.

\jpar2
\algstep N4. [Recurse.] For $0≤i<2m$, set $(Z↓{i0},\ldotss,Z↓{i(r-1)})$ to the
negacyclic convolution of
$(X↓{i0},\ldotss,X↓{i(r-1)})$ and
$(Y↓{i0},\ldotss,Y↓{i(r-1)})$.

\algstep N5. [Untransform.] For $j\!=\!0$, 1, $\ldotss$, $\lfloor n/2\rfloor$
(in this order), set $\biglp Z↓{s+t}(w),Z↓{s+t+2↑j}(w)\bigrp\!←{1\over2}\biglp
Z↓{s+t}(w)+Z↓{s+t+2↑j}(w),
w↑{-(r/m)(s/2)}(Z↓{s+t}(w)-Z↓{s+t+2↑j}(w))\bigrp$, for all $m$ choices of
$s$ and $t$ as in step↔N3.

\algstep N6. [Repack.] (Now we have accomplished the goal stated at the end of
step↔N2, since it is easy to show that the transform of the $Z$'s is the
product of the transforms of the $X$'s and the $Y$'s.)\xskip
Set $z↓i←Z↓{i0}-Z↓{(m+i)(r-1)}$ and $z↓{mj+i}←Z↓{ij}+Z↓{(m+i)(j-1)}$ for
$0<j<r$, for $0≤i<m$.\quad\blackslug

\yyskip
It is easy to verify that at most $n$ extra bits of precision are needed for the
intermediate variables in this calculation; i.e., if $|x↓i|≤M$ for $0≤i<2↑n$
at the beginning of the algorithm, then all of the $x$ and $X$ variables will be
bounded by $2↑nM$ throughout.

Algorithm N performs $A↓n$ addition-subtractions, $D↓n$ halvings, and
$M↓n$ multiplications, where $A↓1=5$, $D↓1=0$, $M↓1=3$; for $n>1$ we
have $A↓n=\lfloor n/2\rfloor2↑{n+2}+2↑{\lfloor n/2\rfloor+1}
A↓{\lfloor n/2\rfloor}+\biglp\lfloor n/2\rfloor+1\bigrp2↑{n+1}+2↑n$,
$D↓n=2↑{\lfloor n/2\rfloor+1}D↓{\lfloor n/2\rfloor}+\biglp\lfloor n/2\rfloor+1\bigrp2↑{n+1}$,
and $M↓n=2↑{\lfloor n/2\rfloor+1}M↓{\lfloor n/2\rfloor}$. The solutions are
$A↓n=11\cdot2↑{n-1+\lceil\lg n\rceil}-3\cdot2↑n+6\cdot2↑nS↓n$,
$D↓n=4\cdot2↑{n-1+\lceil\lg n\rceil}-2\cdot2↑n+2\cdot2↑nS↓n$,
$M↓n=3\cdot2↑{n-1+\lceil\lg n\rceil}$; here $S↓n$ satisfies the recurrence
$S↓1=0$, $S↓n=2S↓{\lceil n/2\rceil}+\lfloor n/2\rfloor$, and it is not difficult
\inx{analysis of algorithms}
to prove the inequalities
${1\over2}n\lceil\lg n\rceil≤S↓n≤{1\over2}n\lg n+n$. Algorithm↔C
does approximately the same amount of work as Algorithm↔N.

It would be interesting to find a simpler way to carry out the additions and
subtractions in step↔N3 (and the reverse operations in↔N5), perhaps analogous
to \α{Yates}'s method. The operation $X↓i←X↓i+w↑kX↓l$ sketched
above can be done
with a procedure that generalizes the data-rotation algorithm of \α{Fletcher} and
\α{Silver} in {\sl CACM \bf9} (1966), 326, but there might be a better way.

\ansno 60. (a)\9 In $\Sigma↓1$, for example, we can group all terms having a
common value of $j$ and↔$k$ into a single trilinear term; this gives $\nu↑2$
trilinear terms when $(j,k)\in E{\times}E$, plus $\nu↑2$ when $(j,k)\in E{\times}O$
and $\nu↑2$ when $(j,k)\in O{\times}E$. When $\s\jit=k$ we can also include
$-x↓{jj\,}y↓{j\s\jit\,}z↓{\s\jit j\,}$ in $\Sigma↓1$, free of charge.\xskip
[In the case $n=10$, the method multiplies 10 by 10 matrices with 710 noncommutative
multiplications; this is fewer than required by any other known method, although
\α{Winograd}'s scheme (35) uses only 600 when commutativity is allowed.]

(b)\9 Here we simply let $S$ be all the indices $(i,j,k)$ of one problem,
$\s S$ the indices of the other.\xskip[When $m=n=s=10$, the result is quite
surprising: We can multiply two separate $10\times10$ matrices with 1300
noncommutative multiplications, while no scheme is known that would multiply
each of them with↔650.]

(c)\9 Corresponding to the left-hand side of the stated identity we get the
terms$$\!x↓{i+ε,j+\zeta\,}y↓{j+\zeta,k+\eta\,}z↓{k+\eta,i+ε}+x↓{j+\eta,k+ε\,}
y↓{k+ε,i+\zeta\,}z↓{i+\zeta,j+\eta}+x↓{k+\zeta,i+\eta\,}y↓{i+\eta,j+ε\,}
z↓{j+ε,k+\zeta}\!\hskip-.33pt$$
summed over $(i,j,k)\in S$ and $0≤ε,\zeta,\eta≤1$, so we get all
the trilinear terms of the form $x↓{ij\,}y↓{jk\,}z↓{ki}$ except when $\lceil i/2
\rceil=\lceil j/2\rceil=\lceil k/2\rceil$; however, these missing terms can
all be included in $\Sigma↓1$, $\Sigma↓2$, or $\Sigma↓3$. The sum $\Sigma↓1$
turns out to include terms of the form $x↓{i+ε,j+\zeta\,}y↓{i+\eta,j+ε}$ times
some sum of $z$'s, so it contributes $8\nu↑2$ terms to the trilinear realization;
and $\Sigma↓2$, $\Sigma↓3$ are similar. To verify that the $aB\Cscr$ terms
cancel out, note that they are $\sum(-1)↑{\zeta+\eta}x↓{i+ε,j+\zeta\,}
y↓{k+ε,i+\zeta\,}z↓{j+ε,k+\zeta}$, so $\eta=1$ cancels with $\eta=0$.\xskip
[This technique leads to asymptotic improvements over Strassen's method whenever
${1\over3}n↑3+6n↑2-{4\over3}n<n↑{\lg7}$, namely when $36≤n≤184$, and it was the first
construction known to break the $\lg7$ barrier. Reference: {\sl SIAM J.
Computing \bf9} (1980), 321--342.]

\def\\{\mathop{\hbox{rank}}}
\ansno 61. (a)\9 Replace $a↓{ij}(u)$ by $ua↓{il}(u)$.\xskip
(b) Let $a↓{il}(u)=a↓{il\mu}u↑\mu$, etc., in a polynomial realization of 
length $r=\\↓d(t↓{ijk})$. Then $t↓{ijk}=\sum↓{\mu+\nu+\sigma=d}\sum↓{1≤l≤r}a↓{il\mu}
b↓{jl\nu}c↓{kl\sigma}$.\xskip[This result can be improved to $\\(t↓{ijk})≤(2d+1)
\\↓d(t↓{ijk})$ in an infinite field, because the trilinear form
$\sum↓{\mu+\nu+\sigma}a↓\mu b↓\nu c↓\sigma$ corresponds to multiplication of
polynomials modulo $u↑{d+1}$, as pointed out by \α{Bini} and \α{Pan}.]\xskip
(c,d) This is clear from the realizations in exercise↔48.

\ansno 62. The rank is 3, by the method of proof in Theorem↔W with $P=\left(
{0\atop0}\,{1\atop0}\right)$. The border rank cannot be↔1, since we cannot have
$a↓1(u)b↓1(u)c↓1(u)≡a↓1(u)b↓2(u)c↓2(u)≡u↑d$ and $a↓1(u)b↓2(u)c↓1(u)≡a↓1(u)b↓1(u)
c↓2(u)≡0\modulo{u↑{d+1}}$. The border rank is↔2 because of the realization
$\left({1\atop u}\,{1\atop0}\right)$, $\left({u\atop1}\,{0\atop1}\right)$,
$\left({1\atop0}\,{-1\atop\hfill u}\right)$.

\ansno 63. (a)\9 Let the elements of $T(m,n,s)$ and $T(M,N,S)$ be denoted by
$t↓{\langle i,j↑\prime\rangle\langle j,k↑\prime\rangle\langle k,i↑\prime\rangle}$
and
$T↓{\langle I,J↑\prime\rangle\langle J,K↑\prime\rangle\langle K,I↑\prime\rangle}$,
respectively. Each element
$\Tscr↓{\langle \Iscr,\Jscr↑\prime\rangle\langle \Jscr,\Kscr↑\prime\rangle\langle
\Kscr,\Iscr↑\prime\rangle}$ of the direct product, where $\Iscr=\langle i,I\rangle$,
$\Jscr=\langle j,J\rangle$, and $\Kscr=\langle k,K\rangle$, is equal to
$t↓{\langle i,j↑\prime\rangle\langle j,k↑\prime\rangle\langle k,i↑\prime\rangle}
T↓{\langle I,J↑\prime\rangle\langle J,K↑\prime\rangle\langle K,I↑\prime\rangle}$
by definition, so it is↔1 iff $\Iscr↑\prime=\Iscr$ and $\Jscr↑\prime=\Jscr$ and
$\Kscr↑\prime=\Kscr$.

(b)\9 We have $M(mns)≤r↑3$, since $T(mns,mns,mns)=T(m,n,s)\otimes T(n,s,m)\otimes
T(s,m,n)$. If $M(P)≤R$ we have $M(P↑h)≤R↑h$ for all $h$, and it follows that
$M(N)≤M(P↑{\lceil
\log↓PN\rceil})≤R↑{\lceil\log↓PN\rceil}≤RN↑{\log R/\!\log P}$.\xskip [This
result appears in \α{Pan}'s 1972 paper.]

(c)\9We have $M↓d(mns)≤r↑3$ for some $d$, where $M↓d(n)=\\↓d\biglp T(n,n,n)\bigrp$.
If $M↓d(P)≤R$ we have $M↓{hd}(P↑h)≤R↑h$ for all $h$, and the stated formula follows
since $M(P↑h)≤{hd+2\choose2}R↑h$
by exercise↔61. In an infinite field we save a factor of $\log N$.\xskip [This
result is due to \α{Bini} and \α{Sch\"onhage}, 1979.]

(d)\9Let $P=mns$; we can perform $p↑{3h}$ independent $P↑h\times P↑h$ matrix
multiplications with at most ${hd+2\choose2}p↑{3h}r↑{3h}$ noncommutative scalar
multiplications. Reduce $M(P↑{2h})$ to $M(P↑h)$ matrix multiplications of size
$P↑h\times P↑h$; thus we have $M(P↑{2h})≤{hd+2\choose2}r↑{3h}M(P↑h)\*
\biglp1+O(p/r)↑{3h}\bigrp$.
Iterating this recurrence gives $$M(N)=O(N↑{β(m,n,s,r)})\cdot\exp\biglp
O(\log\log n)↑2\bigrp.$$

\ansno 64. (a)\9 Let $a=x↓{ij}$, $A=uX↓{ki}$, $b=y↓{jk}$, $B=Y↓{ij}$, $c=uz↓{ki}$,
$C=Z↓{jk}$, so that $\Sigma↓2=O(u↑2)$ can be eliminated.\xskip[When $m=s=7$ and
$n=1$, this gives $M(N)=O(N↑{2.66})$.]\xskip(b) Take $α↓1=s-1$, $α↓2=-α↓1↑{-2}$,
$α↓3=α↓4=-1$, $α↓5=1$, $α↓6=s-1$, and $d≥4$.\xskip[We assume that $α↑{-1}$ exists
in the field.]\xskip
(c) Taking the direct product of $T(m,n,2s)\oplus T(2n,s,m)\oplus T(s,2m,n)$ with
itself $3h$ times gives a tensor whose border rank is at most $\biglp2(m+1)n
(s+2)\bigrp↑{3h}$. This tensor is the direct sum of $3↑{3h}$ terms of the
form $T\biglp m↑i(2n)↑js↑k,n↑is↑j(2m)↑k,(2s)↑im↑jn↑k\bigrp$ where $i+j+k=3h$,
and $(3h)!/h!↑3$ of these have $i=j=k=h$. Thus if we let
$P=(2mns)↑h$ and $p=(3h)!/h!↑3$,
the border rank of $pT(P,P,P)$ is at most $pr$, where$$r=\biglp2(m+1)n
(s+2)\bigrp↑{3h}/p.$$Exercise 63(d) now implies that $M(N)=O(N↑{β(P,P,P,r)+ε})$
for all $ε>0$; here $P$ and↔$r$ are functions of $h$. We complete the
proof by letting $h$ be large in $β(P,P,P,r)=\log r/\!\log P=\biglp3h\log2(m+1)n
(s+2)-3h\log3+O(\log h)\bigrp/(3h\log2mns)$,
which equals $β\biglp m,n,2s,{2\over3}(m+1)n(s+2)\bigrp
+O\biglp(\log h)/h\bigrp$.\xskip[The best value is
obtained for $m=5$, $n=1$, $s=11$, $β=3\log↓{110}52<2.522$.]
% Vicinity of page 655
% folio 840 galley 9b 	*
\ansbegin{4.7}

\ansno 1. Find the first nonzero coefficient
$V↓m$, as in (4), and divide both $U(z)$ and $V(z)$
by↔$z↑m$ (shifting the coefficients $m$ places to the left). The
quotient will be a power series iff $U↓0 = \cdots = U↓{m-1}
= 0$.

\ansno 2. We have
$V↑{n+1}↓{\!0}W↓n = V↑{n}↓{\!0}U↓n - (V↑{1}↓{\!0}W↓0)(V↑{n-1}↓{\!0}V↓n)
- (V↑{2}↓{\!0}W↓1)(V↑{n-2}↓{\!0}V↓{n-1}) - \cdots - (V↑{n}↓{\!0}W↓{n-1})(V↑{0}↓{\!0}
V↓1)$.
Thus, we can start by replacing $(U↓j, V↓j)$ by $(V↑{j}↓{\!0}U↓j,
V↑{j-1}↓{\!0}V↓j)$ for ${j ≥ 1}$, then set $W↓n ← U↓n - \sum ↓{0≤k<n}
W↓kV↓{n-k}$ for $n ≥ 0$, finally replace $W↓j$ by $W↓j/V↑{j+1}↓{\!0}$
for $j ≥ 0$. Similar techniques are possible in connection with other algorithms
in this section.

\ansno 3. Yes. When $α = 0$, it is easy to prove by induction
that $W↓1 = W↓2 = \cdots = 0$. When $α = 1$, we find $W↓n =
V↓n$, by the ``cute'' identity
$$\sum ↓{1≤k≤n}\left({k - (n - k)\over n}\right)V↓kV↓{n-k} = V↓0V↓n.$$

\ansno 4. If $W(z) = e↑{V(z)}$, then $W↑\prime (z)
= V↑\prime (z)W(z)$; we find $W↓0 = 1$, and
$$W↓n = \sum ↓{1≤k≤n} {k\over n} V↓kW↓{n-k},\qquad\hbox{for }n ≥ 1.$$
If $W(z) = \ln\biglp 1 + V(z)\bigrp $, then $W↑\prime
(z) + W↑\prime (z)V(z) = V↑\prime (z)$; the rule is $W↓0 = 0$,
and $W↓1 + 2W↓2z + \cdots = V↑\prime (z)/\biglp 1 + V(z)\bigrp$.

[By exercise↔6, the logarithm can be obtained
to order $n$ in $O(n\log n)$ operations.\xskip R. P. \α{Brent} observes
that $\exp\biglp V(z)\bigrp$ can also be calculated with this
asymptotic speed by applying \α{Newton's method} to $f(x) = \ln
x - V(z)$; therefore general exponentiation $\biglp 1 + V(z)\bigrp ↑α = \exp\biglp
α \ln(1 + V(z))\bigrp$ is $O(n \log n)$ too. Reference: {\sl Analytic
Computational Complexity}, ed.\ by J. F. \α{Traub} (New York: Academic
Press, 1975), 172--176.]

\ansno 5. We get the original series back again. This can be
used to test a reversion algorithm.

\ansno 6. $\varphi (x) = x + x\biglp 1 - xV(z)\bigrp $, cf.\
Algorithm↔4.3.3R\null. Thus after $W↓0$, $\ldotss$, $W↓{N-1}$ are known,
the idea is to input $V↓N$, $\ldotss$, $V↓{2N-1}$, compute ${(W↓0 + \cdots +
W↓{N-1}z↑{N-1})}\*{(V↓0 + \cdots + V↓{2N-1}z↑{2N-1})} = 1 + R↓0z↑N
+ \cdots + R↓{N-1}z↑{2N-1} + O(z↑{2N})$, and
let $W↓N + \cdots + W↓{2N-1}z↑{N-1}
= -(W↓0 + \cdots + W↓{N-1}z↑{N-1})(R↓0 + \cdots + R↓{N-1}z↑{N-1})
+ O(z↑N)$.\xskip [{\sl Numer.\ Math.\ \bf 22} (1974), 341--348; this algorithm
was, in essence, first published by M. \α{Sieveking,} {\sl Computing
\bf 10} (1972), 153--156.]\xskip Note that the total time for $N$
coefficients is $O(N \log N)$ if we use ``fast'' polynomial
multiplication (exercise 4.6.4--57).

\ansno 7. $W↓n = {mk\choose k}/n$ when $n = (m - 1)k + 1$, otherwise
0.\xskip(Cf.\ exercise 2.3.4.4--11.)
% Vicinity of page 656
% folio 842 galley 10a 	*
\ansno 8. Input $G↓1$ in step L1, and $G↓n$ in step L2. In
step L4, the output should be $(U↓{n-1}G↓1 + 2U↓{n-2}G↓2 + \cdots
+ nU↓0G↓n)/n$.\xskip (The running time of the order $N↑3$ algorithm
is hereby increased by only order $N↑2$. The value $W↓1 = G↓1$
might be output in step↔L1.)

{\sl Note:} Algorithms T and N determine $V↑{-1}\biglp
U(z)\bigrp$; the algorithm in this exercise determines $G\biglp
V↑{-1}(z)\bigrp $, which is somewhat different. Of course, the
results can all be obtained by a sequence of operations of reversion
and \α{composition} (exercise↔11), but it is helpful to have more
direct algorithms for each case.

\ansno 9. $\vtop{\halign{$# $⊗$\ctr{#}$\quad⊗$\ctr{#}$\quad⊗$\ctr{#}$\quad
⊗$\ctr{#}$\quad⊗$\ctr{#}$\cr
⊗n = 1⊗n = 2⊗n = 3⊗n = 4⊗n = 5\cr
\noalign{\vskip3pt}
T↓{1n}⊗1⊗1⊗2⊗5⊗14\cr
T↓{2n}⊗⊗1⊗2⊗5⊗14\cr
T↓{3n}⊗⊗⊗1⊗3⊗\99\cr
T↓{4n}⊗⊗⊗⊗1⊗\94\cr
T↓{5n}⊗⊗⊗⊗⊗\91\lower3pt\null\cr}}$

\ansno 10. Form $y↑{1/α} = x(1 + a↓1x + a↓2x↑2 + \cdotss
)↑{1/α} = x(1 + c↓1x + c↓2x↑2 + \cdotss)$ by means of Eq.\ (9);
then revert the latter series.\xskip (See the remarks following Eq.\
1.2.11.3--11.)

\ansno 11. Set $W↓0 ← U↓0$, and set $(T↓k, W↓k) ← (V↓k, 0)$
for $1 ≤ k ≤ N$. Then for $n = 1$, 2, $\ldotss$, $N$, do the following:
Set $W↓j ← W↓j + U↓nT↓j$ for $n ≤ j ≤ N$; and then set $T↓j
← T↓{j-1}V↓1 + \cdots + T↓nV↓{j-n}$ for $j = N$, $N - 1$, $\ldotss
$, $n + 1$.

Here $T(z)$ represents $V(z)↑n$. An {\sl on-line}
power series algorithm for this problem, analogous to Algorithm↔T\null,
could be constructed, but it would require about $N↑2\!/2$
storage locations. There is also an \α{on-line algorithm} that
solves this exercise and needs only $O(N)$ storage locations:
We may assume that $V↓1 = 1$, if $U↓k$ is replaced by $U↓kV↑{k}↓{\!1}$
and $V↓k$ is replaced by $V↓k/V↓1$ for all $k$. Then we may
revert $V(z)$ by Algorithm↔L\null, using its output as input to the
algorithm of exercise↔8 with $G↓1 = U↓1$, $G↓2 = U↓2$, etc., thus
computing $U\biglp (V↑{-1})↑{-1}(z)\bigrp - U↓0$.

\α{Brent} and \α{Kung} have constructed several algorithms
that are asymptotically faster. For example, we can evaluate
$U(x)$ for $x = V(z)$ by a slight variant of exercise 4.6.4--42(c),
doing about $2\sqrt{n}$ chain multiplications of cost $M(n)$
and about $n$ parameter multiplications of cost $n$, where $M(n)$ is the
number of operations needed to multiply power series to order $n$; the total
time is therefore $O\biglp \sqrt{n}M(n) + n↑2\bigrp = O(n↑2)$.
A still faster method can be based on the identity $U\biglp
V↓0(z) + z↑mV↓1(z)\bigrp = U\biglp V↓0(z)\bigrp + z↑mU↑\prime
\biglp V↓0(z)\bigrp V↓1(z) + z↑{2m}U↑{\prime\prime}\biglp V↓0(z)\bigrp
V↓1(z)↑2\!/2! + \cdotss$, extending to about $n/m$ terms, where
we choose $m \approx \sqrt{n/\!\log n}$; the first term $U\biglp
V↓0(z)\bigrp$ is evaluated in $O\biglp mn(\log n)↑2\bigrp$
operations using a method somewhat like that in exercise 4.6.4--43.
Since we can go from $U↑{(k)}\biglp V↓0(z)\bigrp$
to $U↑{(k+1)}\biglp V↓0(z)\bigrp$ in $O(n \log n)$ operations
by differentiating and dividing by $V↑\prime↓{\!0}(z)$,
the entire procedure takes $O\biglp mn(\log n)↑2 + (n/m)\,n
\log n) = O(n \log n)↑{3/2}$ operations.\xskip[{\sl JACM \bf25} (1978), 581--595.]

\ansno 12. Polynomial division is trivial unless $m ≥ n ≥ 1$.
Assuming the latter, the equation $u(x) = q(x)v(x) + r(x)$ is
equivalent to $U(z) = Q(z)V(z) + z↑{m-n+1}R(z)$ where $U(x)
= x↑mu(x↑{-1})$, $V(x) = x↑nv(x↑{-1})$, $Q(x) = x↑{m-n}q(x↑{-1})$, and
\inx{reverse of a polynomial}
$R(x) = x↑{n-1}r(x↑{-1})$ are the ``reverse'' polynomials of
$u$, $v$, $q$, and $r$.

To find $q(x)$ and $r(x)$, compute the first $m
- n + 1$ coefficients of the power series $U(z)/V(z) = W(z)
+ O(z↑{m-n+1})$; then compute the power series $U(z) - V(z)W(z)$,
which has the form $z↑{m-n+1}T(z)$ where $T(z) = T↓0 + T↓1z
+ \cdotss$. Note that $T↓j = 0$ for all $j ≥ n$; hence $Q(z)
= W(z)$ and $R(z) = T(z)$ satisfy the requirements.

\ansno 13. Apply exercise 4.6.1--3 with $u(z)=z↑N$ and
$v(z)=W↓0+\cdots+W↓{N-1}z↑{N-1}$; the desired approximations are the values of
$v↓3(z)/v↓2(z)$ obtained during the course of the algorithm. Exercise 4.6.1--26
tells us that there are no further possibilities with relatively prime
numerator and denominator. If each $W↓i$ is an integer, an all-integer
\inx{polynomial remainder sequence}
extension of Algorithm↔4.6.1C will have the desired properties.

{\sl Notes:} The case $N=2n+1$ and deg$(w↓1)=\hbox{deg}(w↓2)=n$ is of
particular interest, since it is equivalent to a so-called \α{Toeplitz} system.
The method of this exercise can be generalized to arbitrary
rational \α{interpolation} of the
form $W(z)≡p(z)/q(z)$ ${\biglp \hbox{modulo }(z-z↓1)\ldotsm(z-z↓N)\bigrp}$,
where the $z↓i$'s need not be distinct; thus, we can specify
the value of $W(z)$ and some of its derivatives at several points.
See Fred G. \α{Gustavson} and David Y. Y. \α{Yun},
IBM Research Report RC7551 (1979). % to appear

\ansno 14. If $U(z)=z+U↓kz↑k+\cdots$ and $V(z)=z↑k+V↓{k+1}z↑{k+1}+\cdotss$, we
find that the difference
$V\biglp U(z)\bigrp-U↑\prime(z)V(z)$ is $\sum↓{j≥1}z↑{2k+j-1}j\biglp
U↓kV↓{k+j}-U↓{k+j}+($polynomial involving only $U↓k$, $\ldotss$, $U↓{k+j-1}$,
$V↓{k+1}$, $\ldotss$, $V↓{k+j-1})\bigrp$; hence $V(z)$ is unique if $U(z)$ is
given and $U(z)$ is unique if $V(z)$ and $U↓k$ are given.

The solution depends on two auxiliary algorithms, the first of which
solves the equation $V\biglp z+z↑kU(z)\bigrp=\biglp 1+z↑{k-1}W(z)\bigrp V(z)+
z↑{k-1}S(z)+O(z↑{k-1+n})$ for $V(z)=V↓0+V↓1z+\cdots+V↓{n-1}z↑{n-1}$, given
$U(z)$, $W(z)$, $S(z)$, and $n$. If $n=1$, let $V↓0=-S(0)/W(0)$; or let $V↓0=1$
when $S(0)=W(0)=0$. To go from $n$ to $2n$, let
$$\eqalign{V\biglp z+z↑kU(z)\bigrp⊗=\biglp
1+z↑{k-1}W(z)\bigrp V(z)+z↑{k-1}S(z)-z↑{k-1+n}R(z)+O(z↑{k-1+2n}),\cr
1+z↑{k-1}\A W(z)⊗=\biglp z/(z+z↑kU(z))\bigrp↑n\biglp 1+z↑{k-1}W(z)\bigrp,\cr
\A S(z)⊗=\biglp z/(z+z↑kU(z))\bigrp↑nR(z),\cr
\noalign{\vskip 11pt plus 3pt minus 8pt
\hbox{and let $\A V(z)=V↓n+V↓{n+1}z+\cdots+V↓{2n-1}z↑{n-1}$ satisfy}
\vskip 11pt plus 3pt minus 8pt}
\;\A V\biglp z+z↑kU(z)\bigrp⊗=\biglp 1+z↑{k-1}\A W(z)\bigrp
\A V(z)+z↑{k-1}\A S(z)+O(z↑{k-1+n}).\cr}$$

The second algorithm solves $W(z)U(z)+zU↑\prime(z)=V(z)+O(z↑n)$ for $U(z)=U↓0+
U↓1z+\cdots+U↓{n-1}z↑{n-1}$, given $V(z)$, $W(z)$, and $n$. If $n=1$, let
$U↓0=V(0)/W(0)$, or let $U↓0=1$ in case $V(0)=W(0)=0$. To go from $n$ to $2n$, let
$W(z)U(z)+zU↑\prime(z)=V(z)-z↑nR(z)+O(z↑{2n})$, and let $\A U(z)=U↓n+\cdots+
U↓{2n-1}z↑{n-1}$ be the solution to the equation
$\biglp n+W(z)\bigrp\A U(z)+z{\A U}↑\prime(z)=R(z)
+O(z↑n)$.

Resuming the notation of (27), the first algorithm can be used to solve
$\A V\biglp U(z)\bigrp=U↑\prime(z)\biglp z/U(z)\bigrp↑k\A V(z)$ to any desired
accuracy, and we set $V(z)=z↑k\A V(z)$. To find $P(z)$, suppose we have $V\biglp
P(z)\bigrp=P↑\prime(z)V(z)+O(z↑{2k-1+n})$, an equation that holds for $n=1$
when $P(z)=z+αz↑k$ and $α$ is arbitrary. 
We can go from $n$ to $2n$ by letting $V\biglp P(z)\bigrp
=P↑\prime(z)V(z)+z↑{2k-1+n}R(z)+O(z↑{2k-1+2n})$ and replacing $P(z)$ by $P(z)+
z↑{k+n}\A P(z)$, where the second algorithm is used to find the polynomial
$\A P(z)$ such that
\def\Phat{\save7\hbox{$P$}\vbox to 1ht7{\vss\hbox{$\A{\box7}$}}}
$\biglp k+n-zV↑\prime(z)/V(z)\bigrp\A P(z)+z\Phat↑\prime(z)=\biglp z↑k/V(z)\bigrp
R(z)+O(z↑n)$.
\vfill\end