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% MATH FOR A of A, by G and K, Third edition.

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{
\parskip 0pt
\baselineskip 10.5pt
\lineskip .5pt

\def \b{\par\noindent}
\def \i{\par}
\def \c{\par\vskip 6pt}

\b Authors:
\c
\b Daniel H. Greene
\b Computer Science Laboratory
\b Xerox Palo Alto Research Center
\b Stanford, California 94304, USA
\c
\b Donald E. Knuth
\b Department of Computer Science
\b Stanford University
\b Stanford, California 94305, USA

\par\vfil

\c {\sl First Edition, 1981.}
\i {\sl Second Edition, 1982.}
\i {\sl \hskip-1ptThird Edition, 1990.}

\par\vfil

{\frenchspacing
\b Library of Congress Cataloging in Publication Data
\c
\b Greene, Daniel H., 1955--
\i Mathematics for the analysis of algorithms.
\i (Progress in Computer Science ; v. 1)
\i Bibliography: p. 81
\i Includes index.
\i 1. Electronic digital computers---Programming.
\b 2. Algorithms. \ I. Knuth, Donald E. \ II. Title.
\b III. Series
\b QA76.6.G7423 $\qquad$ 1990 $\qquad$ 519.4 $\qquad$ 82-17718
\b ISBN 3--7643--3515--7

\par\vskip .2in

\b CIP---Kurztitelaufnahme der Deutschen Bibliothek
\c
\b Greene, Daniel H.:
\b Mathematics for the analysis of algorithms/
\b Daniel H. Greene ; Donald E. Knuth.--- 3. ed.
\b Boston ; Basel ; Stuttgart : Birkh\"auser, 1990.
\i (Progress in Computer Science ; Vol. 1)
\i ISBN 3--7643--3515--7
\b NE: Knuth, Donald E.:; GT

\par\vskip .2in
}
stored in a retrieval system, or transmitted, in any form or by any means,
electronic, mechanical, photocopying, recording or otherwise, without
prior permission of the copyright owner.
\c
\vskip1pt
\b ISBN 3--7643--3515--7
\b Printed in USA
}
\par\eject
\noheadtrue % This is the contents page
\null\vfill

\def\tc #1#2{\line{#1 \leaders\hbox to 10pt{\hfill .\hfill}\hfill\ #2}}

\def\oi{\hskip 10pt}

\def\ti{\hskip 20pt}

\def\ri{\hskip 30pt}

\yyskip
\tc{1. \ Binomial Identities}5
\tc{\oi 1.1 \ Summary of Useful Identities}5
\tc{\oi 1.2 \ Deriving the Identities}7
\tc{\oi 1.3 \ Inverse Relations}9
\tc{\oi 1.4 \ Operator Calculus}{12}
\tc{\oi 1.5 \ Hypergeometric Series}{13}
\tc{\oi 1.6 \ Identities with the Harmonic Numbers}{14}

\yyskip
\tc{2. \ Recurrence Relations}{15}
\tc{\oi 2.1 \ Linear Recurrence Relations}{15}
\tc{\ti 2.1.1 \ Finite History}{16}
\tc{\ri 2.1.1.1 \ Constant Coefficients}{16}
\tc{\ri 2.1.1.2 \ Variable Coefficients}{18}
\tc{\ti 2.1.2 \ Full History}{21}
\tc{\ri 2.1.2.1 \ Differencing}{21}
\tc{\ri 2.1.2.2 \ By Repertoire}{21}
\tc{\oi 2.2 \ Nonlinear Recurrence Relations}{25}
\tc{\ti 2.2.1 \ Relations with Maximum or Minimum Functions}{25}
\tc{\ti 2.2.2 \ Continued Fractions and Hidden Linear Recurrences}{29}
\tc{\ti 2.2.3 \ Doubly Exponential Sequences}{31}

\yyskip
\tc{3. \ Operator Methods}{35}
\tc{\oi 3.1 \ The Cookie Monster}{35}
\tc{\oi 3.2 \ Coalesced Hashing}{38}
\tc{\oi 3.3 \ Open Addressing: Uniform Hashing}{42}
\tc{\oi 3.4 \ Open Addressing: Secondary Clustering}{43}

\tc{4. \ Asymptotic Analysis}{46}
\tc{\oi 4.1 \ Basic Concepts}{46}
\tc{\ti 4.1.1 \ Notation}{47}
\tc{\ti 4.1.2 \ Bootstrapping}{47}
\tc{\ti 4.1.3 \ Dissecting}{48}
\tc{\ti 4.1.4 \ Limits of Limits}{49}
\tc{\ti 4.1.5 \ Summary of Useful Asymptotic Expansions}{51}
\tc{\ti 4.1.6 \ An Example from Factorization Theory}{52}
\tc{\oi 4.2 \ Stieltjes Integration and Asymptotics}{59}
\tc{\ti 4.2.1 \ $O$-notation and Integrals}{61}
\tc{\ti 4.2.2 \ Euler's Summation Formula}{62}
\tc{\ti 4.2.3 \ An Example from Number Theory}{64}
\tc{\oi 4.3 \ Asymptotics from Generating Functions}{69}
\tc{\ti 4.3.1 \ Darboux's Method}{69}
\tc{\ti 4.3.2 \ Residue Calculus}{72}
\tc{\ti 4.3.3 \ The Saddle Point Method}{74}

\yyskip
\tc{5. \ Bibliography}{81}

\yyskip
\tc{6. \ Appendices}{85}
\tc{\oi A. \ Schedule of Lectures, 1980}{85}
\tc{\oi B. \ Homework Assignments}{87}
\tc{\oi C. \ Midterm Exam I and Solutions}{88}
\tc{\oi D. \ Final Exam I and Solutions}{99}
\tc{\oi E. \ Midterm Exam II and Solutions}{105}
\tc{\oi F. \ Final Exam II and Solutions}{111}
\tc{\oi G. \ Midterm Exam III and Solutions}{115}
\tc{\oi H. \ Final Exam III and Solutions}{120}
\tc{\oi I. \ A Qualifying Exam Problem and Solution}{128}

\yyskip
\tc{7. Index}{133}
\eject
\global\noheadtrue % This is the preface page
\null\vfill
\centerline{\tit Preface}

This monograph is derived from an advanced course in computer science
at Stanford University on the analysis of algorithms.
The course presents examples
of the major paradigms used in the precise analysis of algorithms, emphasizing
some of the more difficult techniques. Much of~the material is drawn from the
starred sections of {\sl The Art of Computer Programming}, Volume~3 [Knuth III].

Analysis of algorithms, as a discipline, relies heavily on both computer science
and mathematics. This report is a mathematical look at the syn\-thesis---emphasizing
the mathematical perspective, but using motivation and examples from computer
science. It covers binomial identities, recurrence relations, operator methods
and asymptotic
analysis, hopefully in a format that is terse enough for easy reference and
yet detailed enough to be of use to those who have not attended the lectures.
However, it is assumed that the reader is familiar with the fundamentals of
complex variable theory and combinatorial analysis.

Winter 1980 was the fourth offering of Analysis of Algorithms, and credit is due to the
previous teachers and staff---Leo Guibas, Scott Drysdale, Sam Bent, Andy Yao,
and Phyllis Winkler---for their detailed
contributions to the documentation of the course.
Portions of earlier handouts are
incorporated in this monograph.
Harry Mairson, Andrei Broder, Ken Clarkson, and Jeff Vitter
and the preparation of these notes was also aided by the
facilities of Xerox corporation and the support
of NSF and Hertz graduate fellowships.
The material itself was typeset with the \TeX\ composition system,
using the Computer Modern family
of fonts recently developed with the {\manfnt METAFONT} system.

In this third edition we have made a few improvements to the exposition and
fixed a variety of minor errors. We have also added several
new appendices containing
exam problems from 1982 and 1988.

\vskip 8pt
\rightline{---D.H.G. and D.E.K.}
\eject
\advancepageno % page 4 is blank
\chap{Binomial Identities}
\major{Summary of Useful Identities}

So that the identities themselves do not become buried on an obscure page,
we summarize them immediately:

$$(x + y)↑n = \sum _k {n \choose k} x↑k y↑{n-k}, \qquad \cond{integer n\cr or n real and  \left| x/y \right| < 1\cr}\numeq$$

\noindent
$${r \choose k} = {r - 1 \choose k} + {r-1 \choose k-1}, \qquad \cond{real r\cr integer k\cr}\numeq$$

\noindent
$${n \choose k} = {n \choose n-k}, \qquad \cond{integer n \ge 0\cr integer k\cr}\numeq$$

\noindent
$${r \choose k} = {r \over k} {r-1 \choose k-1}, \qquad \cond{real r\cr integer k \ne 0\cr}\numeq$$

\noindent
$$\sum _{k = 0}↑{n} {r+k \choose k}={r + n + 1 \choose n}, \qquad \cond{real r\cr integer n \ge 0\cr}\numeq$$

\noindent
$$\sum _{k = 0}↑{n} {k \choose m} = {n+1 \choose m+1}, \qquad \cond{integer m, n \ge 0\cr}\numeq$$

\noindent
$${-r \choose k} = (-1)↑k {r + k - 1 \choose k}, \qquad \cond{real r\cr integer k\cr}\numeq$$

\noindent
$${r \choose m} {m \choose k} = {r \choose k} {r-k \choose m-k}, \qquad \cond{real r\cr integer m, k\cr}\numeq$$

\noindent
$$\sum _k {r \choose k} {s \choose n-k} = {r + s \choose n}, \qquad \cond{real r,s\cr integer n\cr}\numeq$$

\noindent
$$\sum _k {r \choose k} {s \choose n+k} = {r+s \choose r+n}, \qquad \cond{integer n, real s\cr integer r \ge 0\cr}\numeq$$

\noindent
$$\sum _k {r \choose k} {s+k \choose n} (-1)↑k = (-1)↑r {s \choose n-r}, \qquad \cond {integer n, real s\cr integer r\ge 0 \cr}\numeq$$

\noindent
$$\sum _{k=0} ↑r {r-k \choose m} {s+k \choose n} = {r+s+1 \choose m+n+1}, \qquad \cond{integer m,n,r,s \ge 0\cr n\ge s\cr}\numeq$$
Parameters called real here may also be complex.

One particularly confusing aspect of binomial coefficients is the ease
with which a familiar formula can be rendered unrecognizable by a few
transformations.  Because of this chameleon character there is no substitute
for practice of manipulations with binomial coefficients. The reader is referred
to Sections 5.1 and 5.2 of [GKP] for an explanation of the formulas above
and for examples of typical transformation strategy.

\vfill\eject
\major{Deriving the Identities}

Here is an easy way to understand many of the identities that do not include an
alternating $-1$.  The number of monotonic
paths through a rectangular lattice with sides
$m$ and $n$ is $m+n \choose m$. By cutting the lattice along different axes, and
counting the paths according to where they cross the cut, the identities are
derived. The pictures below show different ways of partitioning the paths and
the parameter $k$ used in the sum.

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\put(5,0){\line(0,1)3}
\put(0,0){\line(1,0)5}
\put(0,3){\line(1,0)5}
#\endpicture&\raise2pt\vbox{#}\cr
\noalign{\vskip15pt\vfill}
\put(0,3.8){\line(0,1){0.4}}
\put(2,3.8){\line(0,1){0.4}}
\put(0,4){\vector(1,0)2}
\put(0,4){\vector(-1,0)0}
\put(1,4.5){\makebox(0,0){$k$}}
\path{\up\rt\up\rt\up\rt\rt\rt}
\put(1.97,1.98){\line(0,1){1.04}}
\put(2.03,1.98){\line(0,1){1.04}}
&\halign{#\hfil\cr
A sum based on when the\cr
path hits the top edge\cr
derives identity \eq(1.5)\cr}\cr
\noalign{\vfill}
\put(-1.2,0){\line(1,0){0.4}}
\put(-1.2,2){\line(1,0){0.4}}
\put(-1,0){\vector(0,1)2}
\put(-1,0){\vector(0,-1)0}
\put(-1.4,1){\makebox(0,0){$k$}}
\path{\rt\up\rt\up\rt\rt\up\rt}
\put(2.47,-.4){\line(0,1){3.8}}
\put(2.53,-.4){\line(0,1){3.8}}
&\halign{#\hfil\cr
Counting paths according\cr
to when they cross a vertical\cr
line derives identity \eq(1.12)\cr}\cr
\noalign{\vfill}
\put(-1.2,1){\line(1,0){0.4}}
\put(-1.2,3){\line(1,0){0.4}}
\put(-1,1){\vector(0,1)2}
\put(-1,1){\vector(0,-1)0}
\put(-1.4,2){\makebox(0,0){$k$}}
\path{\rt\up\rt\rt\up\rt\up\rt}
\put(3.36,-.4){\line(-1,1){3.8}}
\put(3.44,-.4){\line(-1,1){3.8}}
&\halign{#\hfil\cr
Similarly, a sum based on\cr
a slanted line derives\cr
identity \eq(1.9)\cr}\cr
\noalign{\vfill}
}

More complicated identities can be derived by successive applications of the
identities given on pages 5 and 6. One example appears in
A trivial algorithm whose analysis isn't,'' by A. Jonassen and D. E. Knuth
[Jonassen~78], where the sum
$$S=\sum_k {m\choose k}\left(-{1\over 2}\right)↑k {2k\choose k} \numeq$$
is evaluated by a lengthy series of elementary transformations. Instead of
repeating that derivation, let us consider instead a derivation suggested by
I. Gessel. He attributes this elegant technique, the method of coefficients,''
to G. P. Egorychev.

\eject

First replace $k$ by $m-k$, giving
$$S=\sum_k {m\choose k} \left(-{1\over 2}\right)↑{m-k} {2m-2k\choose m-k}.\numeq$$
Using the notation
$[x↑n]\, f(x)$ for the coefficient of $x↑n$ in $f(x)$, we can express
portions of the sum with generating functions:
$${m\choose k} \left(-{1\over 2}\right)↑{-k} = [x↑k]\, (1-2x)↑m\numeq$$
$${2m-2k\choose m-k} = [y↑{m-k}]\, (1+y)↑{2m-2k}.\numeq$$
The whole sum is
$$S=\left(-{1\over 2}\right)↑m \sum_k [x↑k]\, (1-2x)↑m [y↑{m-k}]\, (1+y)↑{2m-2k}.\numeq$$
We can remove $[y↑{m-k}]\,$ from the sum by noting that
$[y↑{m-k}]\, =[y↑m]\, y↑k$:
$$S=\left(-{1\over 2}\right)↑m [y↑m]\, (1+y)↑{2m} \sum_k [x↑k]\, (1-2x)↑m \left({y\over (1+y)↑2}\right)↑k.\numeq$$
Finally, this seemingly aimless wandering comes to a glorious finish. The
sum in the last formula is a simple substitution for $x$,
since
$$\sum_k[x↑k]\, f(x) g(y)↑k=f\bigl( g(y)\bigr)\numeq$$
when $f$ is analytic. The solution follows immediately:
$$S= (-2)↑{-m} [y↑m]\, (1+y)↑{2m} \left( 1-{2y\over (1+y)↑2} \right)↑m = (-2)↑{-m} [y↑m]\, (1+y↑2)↑m;\shiftnumeq-1pt$$
\medskip
S= \left\{\,\vcenter{\halign{#,\hfil\qquad&#\hfil\cr 2↑{-m}{m\choose m/2}&m even;\cr \noalign{\vskip 3pt}\! 0&m odd.\cr}}\right.\numeq

A simpler approach to this problem has been pointed out by C.~C. Rousseau,
who observes that $2k\choose k$ is the coefficient of $x↑0$ in $(x+x↑{-1})↑{2k}$,
hence $S$ is the coefficient of $x↑0$ in $\bigl(1-(x+x↑{-1})↑2/2\bigr)↑m$.

\goodbreak
\vskip 8pt

From a theoretical standpoint, it would be nice to unify such identities
in one coherent scheme, much as the physicist seeks a unified field theory.
No single scheme covers everything, but there are several
meta'' concepts that explain the existence of large classes of binomial
identities. We will briefly describe three of these: inverse relations,
operator calculus, and hypergeometric series.

\vskip 20pt
\major{Inverse Relations}

One of the simplest set of inverse relations is the pair
$$a_n = \sum_k (-1)↑k {n \choose k} b_k, \qquad b_n = \sum_k (-1)↑k {n \choose k} a_k,\numeq$$
which follows from the orthogonal relation
$$[n=k]\, =\, \sum_{j=0}↑n (-1)↑{j+k} {n \choose j} {j \choose k}.\numeq$$
(A logical relation in brackets evaluates to 1 if true, 0 if false. We assume
that $n$ is a nonnegative integer.) This formula
is just a specialization of equation \eq(1.11) with $s$ equal to zero.  In general an
inverse relation will pair two series so that individual terms of one can be
computed from the terms of the other.  There will always be an associated
orthogonal relation.

In his book {\sl Combinatorial Identities}, John Riordan devotes several chapters to
inverse relations. Since
inverse relations are even more likely to change appearance than the binomial
identities we have seen already, care must be taken to recognize
relations that are basically the same. For this purpose
Riordan describes several transformations
and then groups equivalent inverse pairs into broad categories. His transformations
and classifications are summarized below.

Since we are working with a pair of equations,
we can move terms from one equation to another by replacements
like $b↑\prime_k = (-1)↑k b_k$, obtaining a new pair
$$a_n = \sum_k {n \choose k} b_k↑\prime, \qquad b_n↑\prime = \sum_k (-1)↑{k+n} {n \choose k} a_k.\numeq$$
An inverse relation corresponds to a pair of lower triangular matrices whose product
is the identity. By reflecting across the diagonal we can derive yet another pair
$$a_n = \sum_{k\ge n} {k \choose n} b_k, \qquad b_n = \sum_{k\ge n} (-1)↑{k+n}{k \choose n} a_k.\numeq$$
Finally, note that we can multiply both sides of the orthogonal relation \eq(1.23) by
almost any function that is unity when $n=k$, without affecting the orthogonal
character of the equation.

The last equation, \eq(1.25), has an extremely useful combinatorial sig\-nificance.
Suppose we  have a large collection of random events. Let $b_n$ be the
probability that {\sl exactly} $n$ events occur, and let $a_n$ be the
sum of the probability of $n$ simultaneous events taken over all selections
of $n$ events. Roughly speaking $a_n$ can be viewed
as a sloppy way of computing the probability that
exactly $n$ events occur since it makes no allowance for the possibility
of more than $n$ events. The left side of \eq(1.25) shows how $a_n$ is inflated.
However, $a_n$ is often easier to compute and the right hand side
of equation \eq(1.25), the principle of inclusion and exclusion,'' provides
a practical way of obtaining $b_n$.

Equations \eq(1.22), \eq(1.24) and \eq(1.25) belong to the simplest class of inverse
relations. [Riordan 68] lists several other classes
like the Chebyshev type:
$$a_n = \sum_{k=0}↑{\lfloor n/2 \rfloor} {n \choose k} b_{n-2k}, \qquad b_n = \sum_{k=0}↑{\lfloor n/2 \rfloor} (-1)↑k {n \over n-k}{n-k \choose k}a_{n-2k}.\numeq$$
Not surprisingly, these inverse relations are often associated with their name\-sakes
among the orthogonal polynomials used in interpolation.

The Gould class of inverse relations,
$$f_n = \sum_k (-1)↑k{n\choose k}{a+bk\choose n}g_k,\numeq$$
$$g_n{a+bn\choose n} =\sum_k(-1)↑k{a+bk-k\over a+bn-k}{a+bn-k\choose n-k}f_k,\numeq$$
has a very curious property. A Chinese mathematician L. Hsu recently
discovered that the binomial
coefficients containing $a$ and $b$ are inessential to the functioning of the
inversion. In fact if we choose $\{a_i\}$ and $\{b_i\}$ to be any two
sequences of numbers such that
$$\psi(x,n) = \prod_{i=1}↑n(a_i +b_ix) \ne 0, \qquad \cond{integer x, n\ge 0,\cr} \numeq$$
we obtain a general inversion:
$$f_n =\sum_k (-1)↑k{n\choose k} \psi(k,n) \, g_k,\numeq$$
$$g_n =\sum_k (-1)↑k {n\choose k}(a_{k+1}+ k \, b_{k+1})\psi(n,k+1)↑{-1}f_k .\numeq$$

Another well known pair of inverse relations uses Stirling numbers:
\eqalignno{\hskip-1.5em a_n&= \sum_{k=0}↑n (-1)↑{n-k} {n \brack k} b_k, \hskip .5em \cond { {n \brack k} \equiv  Stirling numbers of the first kind;\cr }&\anumeq\cr \hskip-1.5em b_n&= \sum_{k=0}↑n {n \brace k} a_k, \quad \cond { {n \brace k} \equiv  Stirling numbers of the second kind.\cr }&\anumeq\cr}
Here $a_n$ is usually $x↑{\underline n}$ and $b_n$ is $x↑n$, so that these
formulas convert between factorial powers and ordinary powers of $x$.

We cannot explore all the inverse relations here, but it is worth noting that many
generating functions can be converted to inverse relations.
A pair of power series $z(x)$ and $z↑\ast(x)$ such that $z(x) \, z↑\ast(x) = 1$ provides
a pair of relations:
$$a(x) = z(x) \, b(x), \quad \hbox{and} \quad b(x) = z↑\ast(x) \, a(x).\numeq$$
For example, we can let $z(x) = (1-x)↑{-p}$ and
$z↑\ast(x)=(1-x)↑p$; clearly $z(x) \, z↑\ast(x) = 1$, so we can proceed to compute
formulas for the coefficients in $a(x)$ and $b(x)$:
$$a_n = \sum_k (-1)↑k{-p \choose k} b_{n-k}, \qquad b_n = \sum_k (-1)↑k{p \choose k} a_{n-k}. \qquad \numeq$$
This pair is a member of the Gould class of inverse relations.

Inverse relations are partially responsible for the proliferation of
binomial identities. If one member of an inverse pair  can be embedded
in a binomial identity, then the other member of the pair will often provide
a new identity.
Inverse relations can also enter directly into the analysis of an algorithm.
The study of radix exchange sort, for example, uses the simple set of relations
\eq(1.22) introduced at the beginning of this section. For details see
[Knuth III; exercises 5.2.2--36 and 5.2.2--38].

\vfill\eject
\major{Operator Calculus}

There is a striking similarity between the integral
$$\int_a↑b x↑ndx =\left. x↑{n+1}\over n+1 \right|_a↑b \numeq$$
and the sum
$$\sum_{a\le x<b} x↑{\underline n} = \left. x↑{\underline {n+1}}\over n+1 \right|_a↑b,\numeq$$
where the underlined superscript,
$x↑{\underline n} = x(x-1)(x-2) \ldots (x-n+1)$,
denotes a falling factorial.
The latter sum is merely a variation of equation \eq(1.6) in a form that is easy
to remember. It is certainly easier to remember than the formula for sums of powers
found on page 51.

The similarity of equations \eq(1.36) and \eq(1.37) is a consequence of the facts that
$Dx↑n=nx↑{n-1}$ and $\Delta x↑{\underline n}=nx↑{\underline{n-1}}$, where $D$
and $\Delta$ are the operators of differentiation and difference that are
inverse to $\int$ and $\sum$:\xskip $D\,p(x)=p↑\prime(x)$ and
$\Delta \, p(x) = p(x+1)-p(x)$.
We can extend such analogies much further;
Rota, for example, gives the following generalization of Taylor's theorem:

\noindent{\bf Definitions.} Let $E↑a$ be the shift operator, $E↑a p(x) = p(x+a)$.
An operator $Q$ is a delta operator if it
is shift invariant ($Q \, E↑a = E↑a \, Q$) and if $Q @ x$ is a nonzero constant.
Such an operator has a sequence of basic polynomials defined as follows:
\par\vskip 6pt
\line{\hfill \vbox {\halign{\hfil# &#\hfil\cr
i)&$p_0(x) = 1$\cr
\noalign{\vskip 2pt}
ii)&$p_n(0) = 0,\quad n>0$\cr
\noalign{\vskip 2pt}
iii)&$Q \, p_n(x) = n \, p_{n-1}(x)$.\cr}
}\hfill}

\noindent
The third property means that whenever $Q$ is applied to its basic polynomials
the result is similar to $D$ applied to $1, x, x↑2, \ldots\,$. For example,
$\Delta$ is a delta operator with basic polynomials $x↑{\underline n} = x(x-1)\ldots (x-n+1)$.

\noindent{\bf Taylor's Theorem.}
$$T = \sum_k {a_k \over k!} Q↑k \numeq$$
where
\baselineskip 14pt \vbox{\halign{ #\hfil \cr T is any shift invariant operator;\cr Q is any delta operator with basic polynomials p_k(x);\cr a_k = T \, p_k(x) \bigr|_{x=0}.\cr}}
When $T = E↑a$ and $Q=D$, this reduces to the well known Taylor formula.
By changing $Q$ to $\Delta$, the difference operator, we obtain Newton's
expansion of $T=E↑a$,
$$p(x+a) = \sum_k {a↑{\underline k} \over k!} \Delta↑k p(x).\numeq$$
Newton's expansion is a useful tool for proving binomial identities.
Equa\-tion \eq(1.9), for example, is an expansion of $p(s+r)=(s+r)↑{\underline m}$.

A full exposition of operator calculus and its relation to binomial
identities can be found in [Rota 75].
The reader will also notice the close relationship between discrete and
continuous analysis in Chapter 2, where difference equations resemble
differential equations, and in Section 4.2 on Stieltjes integration,
where floor and ceiling functions are integrated'' to produce sums.

\vskip 10pt
\major{Hypergeometric Series}

The geometric series $1+z+z↑2+\cdots\,=1/(1-z)$ can be generalized to a
hypergeometric series
$$F(a,b;c;z) = 1 + {ab\over c}{z\over 1!} + {a(a+1)\,b(b+1)\over c(c+1)}{z↑2\over 2!} + {a↑{\overline n}b↑{\overline n}\over c↑{\overline n}}{z↑n\over n!} + \cdots,\numeq$$
where the overlined superscript
$a↑{\overline n} = a (a+1)(a+2)\ldots(a+n-1)$
signifies a rising factorial power. The semicolons in the parameter list
of $F$ indicate that there are two numerator
parameters ($a,b$) and one denominator parameter ($c$).
The hypergeometric series in this example
can be further generalized to an arbitrary number of numerator and denominator
parameters.

The standardization afforded by hypergeometric series has shed much light
on the theory of binomial identities.  For example, identities \eq(1.5), \eq(1.10)
and \eq(1.11) are all consequences of Vandermonde's theorem:
$$F(a,-n;c;1) = {(c-a)↑{\overline n}\over c↑{\overline n}}\qquad \cond{integer n > 0\cr}.\numeq$$
The negative integer argument $-n$ terminates the otherwise infinite series, allowing us to
express \eq(1.10) as a variation of this formula:
$${s↑{\underline n}\over n!}\hbox{}F(-r,-s+n;n+1;1)= {s↑{\underline n}\over n!}{(s+1)↑{\overline r}\over (n+1)↑{\overline r}}= {r+s\choose r+n}. \numeq$$
[Henrici I], and [GKP].

\vfill\eject

\major{Identities with the Harmonic Numbers}

Harmonic numbers occur frequently in the analysis of algorithms and there
are some curious identities that involve both binomial coefficients and
harmonic numbers. The commonly used identities are summarized here.

$$H_n = \sum_{k=1}↑n {1 \over k} \numeq$$

$$\sum_{k=1}↑n H_k = (n+1)H_n - n \numeq$$

$$\sum_{k=1}↑n {k \choose m} H_k = {n+1 \choose m+1} \left( H_{n+1} - { 1 \over m+1} \right) \numeq$$

$$\sum_{k=1}↑n {n \choose k} x↑k H_k = (x+1)↑n \left( H_n - \ln \left( 1 + {1\over x} \right) \right) + \epsilon, \qquad \cond {x>0 \cr 0 < \epsilon < {1\over x(n+1)}\cr} \numeq$$

$${ 1 \over (1-z)↑{m+1} } \ln \left( {1\over1-z} \right) = \sum_{n\ge 0} (H_{n+m} - H_m) {n+m \choose n} z↑n\numeq$$

$$\twolinenumeq{{1 \over (1-z)↑{m+1} } \ln \left( {1\over1-z} \right)↑2 =\sum _{n\ge 0} \left( (H_{n+m}-H_m)↑2\right.}{2pt}{ \left. - (H_{n+m}↑{(2)} - H_m↑{(2)}) \right) {n+m \choose n} \, z↑n}$$

The last two identities, along with a generalization to higher powers,
appear in [Zave 76].
We can regard them as identities valid for complex values of $m$, with
$H_{n+m}-H_m={1\over m+1}+{1\over m+2}+\cdots+{1\over m+n}$;
see the solution of problem 2(g), midterm exam II,
on pages 109--110 below.

\vskip 20pt
\chap{Recurrence Relations}
\major{Linear Recurrence Relations}

Recurrence relations are traditionally divided into two classes:
A recurrence with finite history'' depends on a fixed
number of earlier values,
$$x_n = f(x_{n-1}, x_{n-2}, \ldots, x_{n-m}), \qquad n\ge m. \numeq$$
An equation that depends on all preceding values has a full history.''

The simplest recurrences have a finite history, and $f$ is
a linear function with constant
coef\-ficients. Here the terminology parallels differential equation
theory; we distinguish between the homogeneous'' and the nonhomogeneous''
situations depending on the presence of an extra term $g(n)$:
$$c_0x_n + c_1x_{n-1} + \cdots + c_mx_{n-m} = g(n). \numeq$$

There are two classic treatises on the calculus of finite differences,
one by Jordan [Jordan~60] and the other by Milne-Thomson [Mil-Thom~33]. Although the
emphasis of these early works was on approximation and solution
of differential equations---problems in the mainstream
of numerical analysis rather
than analysis of algorithms---much can be learned from
this theory. We recommend a recent summary by Spiegel [Spiegel 71] and
{\sl An Introduction to Computational Combinatorics} by Page and Wilson
[Page 79].

Within this section references are
given to additional examples of the solution of recurrence relations
from [Knuth I] and [Knuth III].
The last part of the section, on the repertoire approach to full
history equations, was introduced in a paper by D. Knuth and
A. Sch\"onhage [Knuth 78].
\minor{Finite History}

\sub{Constant Coefficients}

The constant coefficient problem is a beautiful example of the
use of generating functions to solve recurrence relations.
Rather than attempting to find $x_n$ directly, we construct a
function $G(z)$ with coefficients $x_n$ in its power series expansion:
$$G(z) = \sum_k x_k\,z↑k.\numeq$$
The recurrence relation is converted to an equation in $G(z)$ and solved
by whatever means are applicable.
This is best explained with an example,
$$x_{n+2}-3x_{n+1}+2x_n = n, \qquad x_0=x_1=1. \numeq$$
First we multiply by $z↑{n+2}$ and sum over all $n$, obtaining
$$\sum_{n\ge 0}x_{n+2}z↑{n+2}-3z\sum_{n\ge 0}x_{n+1}z↑{n+1} +2z↑2\sum_{n\ge 0}x_nz↑n =\sum_{n\ge 0}nz↑{n+2}. \numeq$$
The first sum is $G(z)$ missing its first two terms. The next two sums
are similarly close to $G(z)$, and the right side of the equation can
be expressed in closed form as $z↑3/(1-z)↑2$.
(This follows from the binomial theorem, equation \eq(1.1), when
$(x+y)↑n=(1-z)↑{-2}$. A list of standard closed forms for generating
functions appears in [GKP; Chapter 7].)

Putting everything together in one formula for $G(z)$ gives
$$G(z)-z-1-3z \left( G(z)-1 \right) +2z↑2G(z)= {z↑3 \over (1-z)↑2}. \numeq$$
And this is easy to solve for $G(z)$:
$$G(z)= {z↑3 \over (1-z)↑2(1-3z+2z↑2)} + {-2z+1\over (1-3z+2z↑2)}. \numeq$$
We would like to recover the coefficient of $z↑n$ in $G(z)$.
If the denominators of the fractions in $G(z)$ were linear, the
recovery problem would be simple: each term would be a geometric series.
This is not the case in the example we have, but by expressing our
solution for $G(z)$ in partial fractions we obtain a manageable form:
$$G(z)= {1\over 1-2z} +{1\over (1-z)↑2} - {1\over (1-z)↑3}.\numeq$$
Note that the only nonlinear denominators are higher powers
of a linear factor. These terms can be expanded with the binomial theorem, and
$x_n$ is easily computed:
$$x_n= 2↑n - {n↑2+n \over 2}. \numeq$$

Partial fractions are
powerful enough to deal with all linear recurrences with constant
coefficients. For simplicity, however, we will discuss
a different approach found in [Spiegel 71] and many of the older
references.
The approach is based on trial solutions
and is similar to the solution of differential equations.
In certain instances this second
approach will provide fast answers, but the rules often seem like black
fraction theory to understand why these rules of thumb'' work.

\bigskip
\noindent {\bf A) Homogeneous Equations.}

\noindent
$$c_0x_n + c_1x_{n-1} + \cdots +c_mx_{n-m} = 0, \qquad n\ge m.\numeq$$
We try $x_n = r↑n$, and obtain an $m\,$th degree polynomial in $r$.
Let $r_1, \ldots, r_m$ be the roots of this polynomial. The
general solution'' is
$$x_n = k_1r_1↑n +k_2r_2↑n + \cdots + k_mr_m↑n,\numeq$$
where the $k_i$ are constants determined by the initial values.

Multiple roots are accommodated by prefacing the terms of the
general solution with powers of $n$. Suppose that $r_1=r_2=r_3$;
then the adjusted solution would be
$$x_n = k_1r_1↑n + k_2n\,r_1↑n + k_3n↑2r_1↑n.\numeq$$

\vskip 10pt
\noindent{\bf B) Nonhomogeneous Equations.}

$$c_0x_n + c_1x_{n-1} + \cdots +c_mx_{n-m} = g(n). \numeq$$
First remove $g(n)$ and obtain the general solution to the
homogeneous equation.  Add to this homogeneous solution any particular''
solution to the nonhomogeneous equation.

A particular solution can be found by the method of undetermined
coefficients.'' The idea is to use a trial solution with unspecified
coefficients and then solve for these coefficients. The nature of the
trial solution depends on the form of $g(n)$:
\vbox{\baselineskip 0pt \lineskip 0pt \def\|{\vrule height 16pt depth 12pt} \halign{\quad\hfil#\hfil\quad&#&\quad\hfil#\hfil\quad\cr Form of g(n):&\|&Trial Solution:\cr \noalign{\hrule} \alpha↑n&\|&k\,\alpha↑n (multiply by n if \alpha is a root)\cr p(n)&\vrule height 8pt depth 12pt&polynomial of the same degree\cr} }
\sub{Variable Coefficients}

There is no guaranteed solution to the variable coefficient problem,
but there are several methods worth trying:

\noindent{\bf A) Summation Factors.}

If the equation is first order,''
$$a(n)x_n = b(n)\,x_{n-1} + c(n),\qquad n\ge 1,\numeq$$
then it can be reduced to a summation. First multiply both sides
by the summation factor
$$F(n)={\prod_{i=1_{\mathstrut}}↑{n-1} a(i) \over \prod_{j=1}↑{n↑{\mathstrut}} b(j)}.\numeq$$
Then the recurrence becomes
$$y_n = y_{n-1} + F(n)\,c(n), \numeq$$
where $y_n=b(n+1)\,F(n+1)\,x_n$.
The last recurrence allows us to express $x_n$ as a sum:
$$x_n = {x_0+\sum_{i=1}↑n F(i)\,c(i) \over b(n+1)\,F(n+1)}.\numeq$$
See [Knuth III; page 121] and [Lueker 80]
for illustrations of this technique.

\vskip 10pt
\noindent{\bf B) Generating Functions.}

Variable coefficients are amenable to a generating function attack.
If the coefficients are polynomials, the generating function is differentiated
to obtain the desired variability.
Let us look at a relatively simple problem
to get a feeling for what is involved:
$$(n+1)x_{n+1} -(n+r)x_n = 0.\numeq$$
Multiplying by $z↑n$ and summing over all $n$ will bring us closer to a formula
in $G(z)$:
$$\sum_n (n+1) x_{n+1}\,z↑n - \sum_n(n+r)x_n\,z↑n =0.\numeq$$
Using the derivative of $G(z)$ and multiplication by $z$ for shifting, we obtain
a differential equation,
$$(1-z)\,G↑\prime(z) - r\, G(z)=0.\numeq$$
In general any recurrence with coefficients that are polynomial in $n$
can be converted to a differential equation like \eq(2.20).
In this case, the coefficients of the solution,
$G(z)=(1-z)↑{-r}$,
can be recovered by the binomial theorem:
$$x_n = (-1)↑n{-r\choose n} = {r-1+n\choose n}.\numeq$$

More difficult problems will present the analyst with a wide variety
of equations in $G(z)$. While these are not always differential equations,
the reader is referred to [Boyce 69] for those differential equations that
do occur.

\vskip 10pt
\noindent{\bf C) Reduction of Order.}

If we are fortunate enough to factor the difference equation,
then we can attempt to solve each factor separately.
For example, the difference equation
$$y_{k+2} - (k+2)\,y_{k+1} + k\,y_k = k\numeq$$
can be written in operator notation:
$$\bigl( E↑2-(k+2)E +k \bigr) y_k =k. \numeq$$
And the operator can be factored so that
$$(E-1)(E-k)y_k=k.\numeq$$
If we first solve the equation
$$(E-1)z_k = k, \numeq$$
which has the simple answer $z_k={k\choose 2}$,
then we have reduced the order, leaving a first order equation:
$$(E-k)\, y_k = {k\choose 2}. \numeq$$
Using $F(n) = 1/n!$ as a summing factor, the last equation
can be solved:
$$y_n= {(n-1)! \over 2} \sum_{k=1}↑{n-3} {1\over k!}. \numeq$$
For simplicity we will omit the discussion of initial conditions;
see [Spiegel 71; page 176] for a solution of this example
with initial conditions $y_1=0$ and $y_2=1$.

\vskip 10pt
All three approaches to the variable coefficient problem have
serious short\-comings.  The summation factor may yield an inscrutable sum,
and the generating function can produce an equally intractable differential
equation. And alas, there is no certain way to factor an operator
equation to apply the reduction of order technique.  The variable coefficient
equation is a formidable problem; we will have to return to it later
in the exploration of asymptotic approximations.

\vfill\eject
\minor{Full History}

\sub{Differencing}

The differencing strategy eliminates full history by subtracting suitable
combinations of adjacent formulas. For example, [Knuth III; page 120]
solves the equation
$$x_n = f_n + {2\over n}\sum_{k=0}↑{n-1} x_k \numeq$$
by subtracting
$$n\,x_n = n\,f_n + 2\sum_{k=0}↑{n-1} x_k \numeq$$
from
$$(n+1)x_{n+1} = (n+1)f_{n+1} + 2\sum_{k=0}↑n x_k, \numeq$$
yielding a first order variable coefficient problem.
Note how the two formulas have been carefully rearranged to eliminate
the sum. In complex situations, several differences may be necessary
to remove the history. See, for example, [Knuth III; exercise
6.2.2--7].

\sub{By Repertoire}

In the next approach we take advantage of the linearity of the recurrence
and construct the desired solution from a repertoire of simple solutions.
Several recurrences in the analysis of algorithms have the form
$$x_n=a_n+\sum_{0\le k\le n}p_{nk}(x_k+x_{n-k}),\qquad \sum_kp_{nk}=1. \numeq$$
If we also know that
$$y_n=b_n+\sum_{0\le k\le n}p_{nk}(y_k+y_{n-k}), \numeq$$
then by linearity an equation with additive term $\alpha\,a_n + \beta\,b_n$
will have the solution $\alpha\,x_n + \beta\,y_n$.

The crucial idea is this: We choose $x_n$ first so as to make the
sum tractable, {\sl then} we see what additive term $a_n$ results from
the $x_n$. This is exactly backwards from the original problem,
where $a_n$ is given and $x_n$ is sought. However, once we have built up a
repertoire of enough additive terms, the original $a_n$ can
be constructed by linear combination.

For example, consider the recurrence associated with median-of-three
quicksort:
$$x_n=n+1+\sum_{1\le k\le n}{ {k-1\choose 1}{n-k\choose 1}\over {n\choose 3} } (x_{k-1} + x_{n-k}). \numeq$$
The ordinary quicksort algorithm is modified so that three elements are
chosen and the median of these three is used for the partitioning
phase of the algorithm. In ordinary quicksort each partition size is equally
likely. This modification makes it slightly more likely that the partition
will split into similar sized parts, because $p_{nk}={k-1\choose1}{n-k\choose1} /{n\choose3}$ is largest when $k$ is near $n/2$.

At first we observe that the sum is symmetric and we replace the additive
term, $n+1$, by $a_n$ in preparation for the repertoire approach:
$$x_n=a_n+{2\over{n\choose 3}}\sum_{1<k<n} (k-1)(n-k)x_{k-1}.\numeq$$
Choosing $x_n$ equal to the falling factorial $(n-1)↑{\underline s}$ makes
the sum in equation \eq(2.34) easy to compute:
\eqalignno{(n-1)↑{\underline s} &=a_n+{12\over n↑{\underline 3}} \sum_{1<k<n} (n-k)(k-1)↑{\underline{s+1}}\cr &=a_n+{12(s+1)!\over n↑{\underline 3}}\sum_{1<k<n}{n-k\choose 1}{k-1\choose s+1}\cr &=a_n+{12(s+1)!\over n↑{\underline 3}}{n\choose s+3};&\anumeq\cr a_n&=(n-1)↑{\underline s}- {12\over (s+2)(s+3)} (n-3)↑{\underline s}.&\anumeq\cr}
Now we have a family of solutions parameterized by $s$.
\vbox{\baselineskip 0pt \lineskip 0pt \def\|{\vrule height 16pt depth 12pt} \halign{\quad\hfil\displaystyle{#}\hfil\quad&#&% \quad\hfil\displaystyle{#}\hfil\quad&\quad\hfil\displaystyle{#}\hfil\quad\cr &\|&x_n&a_n\cr\noalign{\hrule} s=0&\|&1&-1\cr s=1&\|&(n-1)&2\cr s=2&\|&(n-1)(n-2)&{2n↑2+6n-26\over 5}\cr }}

However, the family is inadequate; it lacks a member with linear $a_n$. The
possibilities for $a_n$ jump from constant to $\Theta(n↑2)$ and unfortunately
the $a_n$ that we wish to reconstruct is $\Theta(n)$. On reflection, this is not
surprising. We expect the solution of this divide and conquer style of
iteration to be $O(n \log n)$ and yet we have limited the possibilities for
$x_n$ to polynomials in~$n$. So to expand our family of solutions we introduce
the harmonic numbers, $H_n$, which are also easy to sum and will contribute
$O(\log n)$ factors to the solutions. The new family is computed using
$x_n=(n-1)↑{\underline t}H_n$ in equation \eq(2.34) and solving for $a_n$.
\eqalignno{(n-1)↑{\underline t}H_n&=a_n+{12\over n↑{\underline 3}} \sum_{1<k<n}(n-k)(k-1)↑{\underline{t+1}}H_{k-1}\cr &=a_n+{12\over n↑{\underline 3}}\,\left(\sum_{1<k<n}n(k-1)↑{\underline{t+1}}H_{k-1} \right.\cr \noalign{\vskip-3pt}&\qquad\qquad\left.\null -\!\!\sum_{1<k<n}k↑{\underline{t+2}}H_k +\!\!\sum_{1<k<n}(k-1)↑{\underline{t+1}}\right)\cr\noalign{\vskip3pt} &=a_n+{12\over n↑{\underline 3}}\,\left({n↑{\underline{t+3}}\over t+2}\left( H_{n-1} -{1\over t+2}\right) \right.\cr &\qquad\qquad\left.\null -{n↑{\underline{t+3}}\over t+3}\left(H_n -{1\over t+3}\right)+{(n-1)↑{\underline{t+2}}\over t+2}\right).&\anumeq\cr}
Here we have used identity \eq(1.45) to evaluate the sums containing $H_n$.
The result can be simplified to
$$a_n=H_n\left( (n-1)↑{\underline t} -{12\over (t+2)(t+3)}(n-3)↑{\underline t} \right)+{12(2t+5)\over(t+2)↑2(t+3)↑2}(n-3)↑{\underline t}.\shiftnumeq2pt$$
This time, when we examine the small members of the family of solutions
we discover a fortunate alignment:
\vbox{\baselineskip 0pt \lineskip 0pt \def\|{\vrule height 13pt depth 9pt} \halign{\quad\hfil\displaystyle{#}\hfil\quad&#&% \quad\hfil#\hfil\quad&\quad\hfil#\hfil\quad\cr &\|&x_n&a_n\cr\noalign{\hrule} &\vrule height 3pt\cr t=0&\|&H_n&-H_n+{5\over3}\cr t=1&\|&(n-1)H_n&2H_n +{7\over 12}(n-3)\cr }}

The smallest two solutions for $a_n$ both have leading term $H_n$. By an appropriate
linear combination we can eliminate $H_n$ and obtain an $a_n$ that grows as order $n$:
$$x_n=(n+1)H_n \quad \swap \quad a_n={7n+19\over 12}.\numeq$$
The $s=0$ solution from the first family is used to adjust the constant term,
enabling us to reconstruct the $a_n$ given in the original problem:
$$\textstyle x_n={12\over 7}\bigl( (n+1)H_n + 1 \bigr) \quad \swap \quad a_n=n+1.\numeq$$
This solution for $x_n$ may not agree with the initial values $x_1$ and $x_2$.
To accommodate arbitrary initial values we need to discover two extra degrees
of freedom in the solution. One degree of freedom can be observed in the first
family of solutions. Combining $s=0$ with $s=1$ yields
$$x_n = n+1 \quad \swap \quad a_n=0.\numeq$$
So any multiple of $n+1$ can be added to the solution in equation \eq(2.40).

The second degree of freedom is not quite so obvious. Since ${a_n=0}$ we have
a simplified recurrence for $x_n$,
$$n↑{\underline 3} x_n = 12\sum_{1<k<n} (n-k)(k-1)x_{k-1}.\numeq$$
Using a generating function, $G(z)$, for the sequence $x_n$, the convolution
on the right of \eq(2.42) is represented by the product of $1/(1-z)↑2$ corresponding
to $(n-k)$ and $G↑\prime(z)$ corresponding to $(k-1)x_{k-1}$.
We obtain the differential equation
$$G↑{\prime\prime\prime}(z) ={12\over (1-z)↑2} G↑\prime(z).\numeq$$
The nature of the equation suggests a solution of the form $G(z)=(1-z)↑\alpha$,
and testing this solution yields $\alpha =-2$ or $5$. The case $\alpha =-2$
corresponds to our previous observation that multiples of $n+1$ do not affect
the solution. But for $\alpha =5$ we obtain an unusual solution that is zero after
its first five values:
$$x_1=-5,\quad x_2=10,\quad x_3=-10,\quad x_4=5,\quad x_5=-1.\numeq$$
This provides a second degree of freedom and gives the final solution
$$x_n={12\over 7}\bigl( (n+1)H_n +1\bigr) +c_1(n+1) +c_2(-1)↑n{5\choose n},\numeq$$
where $c_1$ and $c_2$ are determined by the initial conditions.

\vfill\eject
\major{Nonlinear Recurrence Relations}

Nonlinear recurrence relations are understandably more difficult than their
linear counterparts, and the techniques used to solve them are often less
systematic, requiring conjectures and insight rather than routine tools.  This
section explores two types of nonlinear recurrence relations, those with
maximum and minimum functions, and those with hidden or approximate linear
recurrences.

\minor{Relations with Maximum or Minimum Functions}

To solve a recurrence relation with $\max$ or $\min$ it is essential to know
where the $\max$ or $\min$ occurs.  This is not always obvious, since the
$\max$ (or $\min$) function may depend on earlier members of the sequence
whose character is initially unknown.  A typical solution strategy involves
computing small values with the recurrence relation until it is possible to
make a conjecture about the location of the $\max$ (or $\min$) at each
iteration. The conjecture is used to solve the recurrence
and then the solution is used to prove inductively that the conjecture is
correct.

This strategy is illustrated with the following example from the analysis of
an {\sl in situ\/} permutation algorithm [Knuth 71]. Briefly described, the problem arises
in a variation of the algorithm that searches both directions simultaneously
to verify cycle leaders.  To check a particular $j$, the algorithm first looks
at $p(j)$ and $p↑{-1}(j)$, then at $p↑2(j)$ and $p↑{-2}(j)$, etc., until either
encountering an element smaller than $j$, in which case $j$ is not a cycle
leader, or until encountering $j$ itself, in which case $j$ is a cycle leader
since the whole cycle has been scanned.

We wish to compute the worst case cost, $f(n)$, of ruling out all the
non-leaders in a cycle of size $n$.  A recurrence arises from the observation
that the second smallest element in the cycle partitions the problem.  For
convenience we place the cycle leader (the smallest element) at the origin and
assume that the second smallest element is in the $k$th location.
$$\hbox{(leader)}\ c_1\ c_2\ c_3 \ldots c_{k-1}\ \hbox{(second smallest)} \, c_{k+1}\ c_{k+2} \ldots c_{n-1} .\numeq$$
Any searching among $c_1\ldots c_{k-1}$ will not exceed the leader or the
second smallest element, so the worst case for this segment is identical to
the worst case for a cycle of size $k$.  Similarly the worst for
$c_{k+1}\ldots c_{n-1}$ is $f(n-k)$ and the cost of rejecting the second
smallest is $\min(k,n-k)$.  This gives:
$$f(n)=\max_k\bigl( f(k)+f(n-k)+\min(k,n-k)\bigr).\numeq$$

According to the strategy outlined above, our first step is to build up a
table that shows the values of $f(n)$ for small $n$, together with a list
of the values of $k$ where the maximum is achieved:
\vbox{\baselineskip 0pt \lineskip 0pt \def\|{\vrule height 12pt depth 8pt} \halign{\quad#\quad&\hfil#\hfil\quad&#&\quad\hfil#\hfil\quad\cr n&f(n)&\|&\hbox{location of the max }(k)\cr \noalign{\hrule} &&\vrule height 3pt\cr 1&0&\|&-\cr 2&1&\|&1 \cr 3&2&\|&1,2 \cr 4&4&\|&2 \cr 5&5&\|&1,2,3,4 \cr 6&7&\|&2,3,4 \cr 7&9&\|&3,4 \cr 8&12&\|&4 \cr}}
In some iterations the location of the $\max$ has many possibilities, but it
seems that $\lfloor n/2\rfloor$ is always among the candidates.  With the
conjecture that there is a maximum at $\lfloor n/2\rfloor$ the recurrence reduces to:
\eqalign{f(2m)&=2f(m)+m\cr f(2m+1)&=f(m)+f(m+1)+m.\cr} \numeq
The odd and even formulas are close enough to suggest differencing,
\eqalign{\Delta f(2n)&=f(2n+1)-f(2n)=f(n+1)-f(n)=\Delta f(n)\cr \Delta f(2n+1)&=f(2n+2)-f(2n+1)\cr &=f(n+1)-f(n)+1=\Delta f(n)+1.\cr}\numeq
In the differenced form the nature of $\Delta f(n)$ and $f(n)$ become clear:
$\Delta f(n)$ simply counts the number of ones in the binary representation of
$n$.  If we let $\nu(n)$ be the number of such $1$-bits then
$$f(n)=\sum_{0\le k<n}\nu(k)={1\over 2} n \log n + O(n).\numeq$$
(Digital sums like this play an important role in recurrence relations that
nearly split their sequences. The asymptotic study of $f(n)$ has a
confused history of independent discoveries [Stolarsky 77].  See
[DeLange 75] for detailed asymptotics, and see
[Knuth III; exercise 5.2.2--15] for a similar problem that depends
on the binary representation of its argument.)

To complete the solution of equation \eq(2.47) we must
prove our conjecture about the location of the $\max$,
or equivalently we must show that the two-parameter function
$$g(m,n)=f(m+n)-m-f(m)-f(n)\,, \qquad n\ge m\numeq$$
is always greater than or equal to zero.  Breaking this into odd and even
cases and using equation \eq(2.48) yields
\eqalign{g(2m,2n)&=2g(m,n) \cr g(2m+1,2n)&=g(m,n)+g(m+1,n) \cr g(2m,2n+1)&=g(m,n)+g(m,n+1) \cr g(2m+1,2n+1)&=1+g(m+1,n)+g(m,n+1) .\cr}\numeq
Now we can use boundary conditions that are derived from the definition of~$f$,
$$g(n,n)=0$$
$$g(n-1,n)=0,\numeq$$
to prove inductively that $g(m,n)\ge 0$.

In the example above, the conjecture about the location of the maximum is
straightforward and intuitive:  the worst case arises when the second element
is furthest from the leader so that it nearly bisects the cycle.  In other
examples the conjecture is more complicated.  Consider the recurrence
$$f(n)=1+\min_k\left({k-1 \over n}f(k-1)+{n-k \over n}f(n-k)\right)\,, \qquad f(1)=0\,,\numeq$$
which arises from a guessing game where one player tries to determine an
integer between $1$ and $n$.  After each guess the player is told whether
the guess is high, low, or right on.  The recurrence for $f(n)$ represents
the expected number of guesses necessary by the best possible strategy.

Once again intuition tells us that it is best to choose $k$ in the middle
of the interval, but strangely enough this is not always true.  The proper
conjecture for locating the minimum favors dividing the interval into odd
subproblems.  At $n=5$, for example, we should guess $4$ rather than $3$.

There are several general results that can help to locate the minimum.
Included below are the first few theorems from a paper by M. Fredman and
D. Knuth on recurrence relations with minimization [Fredman 74] that apply
to recurrences like
$$f(n+1)=g(n+1)+\min_k\bigl(\alpha f(k)+\beta f(n-k)\bigr)\numeq$$
with $\alpha$ and $\beta$ positive.
Equation \eq(2.54) above, when multiplied by $n$, is a member of this broad class.

\thbegin Definition.  A real valued function $g(n)$ is convex if
$\Delta↑2g(n)\ge 0$ for all $n$.  This means that
$$g(n+2)-g(n+1)\ge g(n+1)-g(n)\,, \qquad n \ge 0.\numeq$$

\thbegin Lemma.  Let $a(n)$ and $b(n)$ be convex functions.  Then the
minvolution'' defined by
$$c(n)=\min_{0\le k\le n}\bigl( a(k)+b(n-k)\bigr) \numeq$$
is also convex.  Moreover if $c(n)=a(k)+b(n-k)$ then
$$c(n+1)=\min\bigl( a(k)+b(n+1-k),\;a(k+1)+b(n-k)\bigr).$${\rm (In other words, the location
of the minimum does not shift drastically as $n$~increases.
The expression minvolution,'' coined by M. F. Plass, conveys the
similarity of formula  \eq(2.57) to the convolution of two sequences.)}

\yskip
This strong lemma has a very simple proof.  The process of constructing $c(n)$
can be viewed as a merging of the two sequences
$$\Delta a(0)\,,\ \Delta a(1)\,,\ \Delta a(2)\,,\ \ldots\numeq$$
and
$$\Delta b(0)\,,\ \Delta b(1)\,,\ \Delta b(2)\,,\ \ldots\,.\numeq$$
By hypothesis these two sequences are nondecreasing, so the merged sequence
$$\Delta c(0)\,,\ \Delta c(1)\,,\ \Delta c(2)\,,\ \ldots \numeq$$
is also nondecreasing, making $c(n)$ convex.

For any given $n$, the value of $c(n)$ is the sum of the $n$ smallest items in the two
sequences. The next value, $c(n+1)$,
will require one more item from either the $\Delta a$
sequence or the $\Delta b$ sequence, the smaller item determining whether or not
the location of the minimum shifts from $k$ to $k+1$.

\thbegin Theorem.  The function in equation $\eq(2.55)$ is convex provided that $g(n)$ is
convex and the first iteration of the recurrence is convex:
$$f(2)-f(1)\ge f(1)-f(0).\numeq$$

This theorem follows inductively; we assume that $f(1)\ldots f(n)$ are
convex, and apply the lemma to show that $f(n+1)$ will continue the convexity.

\yyskip
\minor{Continued Fractions and Hidden Linear Recurrences}
\mark{HIDDEN LINEAR RECURRENCES}

When the recurrence resembles a continued fraction, then a simple transformation
will reduce the problem to a linear recurrence relation.

We consider, as an example, the problem of counting the number of trees with
$n$ nodes and height less than or equal to $h$, denoted by $A_{nh}$.  For a
given height $h$ we can use the generating function
$$A_h(z)=\sum A_{nh}z↑n\numeq$$
to establish a recurrence.  A tree of height less than or equal to $h+1$
has a root and any number of trees of height $h$ or less.
\eqalign{A_{h+1}(z)&=z(1+A_h(z)+A_h(z)↑2+A_h(z)↑3+\cdots\,)\cr &=z/(1-A_h(z)).\cr}\numeq
The continued fraction flavor of this recurrence,
$$A_{h+1}={z \over 1-\displaystyle{z\mathstrut\over 1-A_{h-1}(z)}}\,,\numeq$$
suggests the transformation
$$A_h(z)={z\,P_h(z) \over P_{h+1}(z)},\numeq$$
which yields a linear recurrence relation:
$$P_{h+1}(z)=P_h(z)-z\,P_{h-1}(z),\qquad P_0(z)=0,\quad P_1(z)=1.\numeq$$
By standard techniques for quadratic linear relations we obtain
$$P_h(z)={1\over\sqrt{@1-4z}↑{\mathstrut}} \left(\left({1+\sqrt{@1-4z} \over 2}\right)↑{\!h} -\left({1-\sqrt{@1-4z} \over 2}\right)↑{\!h}\right).\numeq$$
The remainder of the analysis of ordered trees, in which the coefficients
of $P_h(z)$ are investigated further,
does not bear directly on
nonlinear recurrences, so we refer the reader to [deBruijn 72] for complete
details.

It is worth noting that in seeking a transformation we were lead to a ratio
of polynomials, equation \eq(2.65), by the continued fraction nature of the recurrence.
In the example above, the regularity of recurrence allowed us to use only one
family of polynomials, $P_h(z)$.  The underlying continued fraction theory
that suggests this solution and accommodates less regular continued fractions
uses two families. In general, the $n$th convergent,''
$$f_n=a_0+{b_1\over a_1+\displaystyle{b_2↑{\mathstrut}\over a_2↑{\mathstrut}+\, \lower6pt\hbox{\displaystyle\ddots+{b_n\over a_n}}}},\numeq$$
is equal to
$$f_n=p_n/q_n \numeq$$
where $p_n$ and $q_n$ have linear recurrence relations:
\vcenter{\halign{#\qquad&#\qquad&#\cr p_n=a_np_{n-1}+b_np_{n-2}&p_0=a_0&p_1=a_1a_0+b_1\cr q_n=a_nq_{n-1}+b_nq_{n-2}&q_0=1&q_1=a_1.\hfill\cr}}\numeq
This theory, found for example in Chapter 10 of Hardy and Wright [Hardy 79],
assures us that we could reduce a less regular recurrence like
$$f_h(z)={z\over 1-\displaystyle{z↑{2↑{\mathstrut}}\over 1-f_{h-1}(z)}}\numeq$$
to a problem with two linear recurrence relations.

Besides continued fractions, there are many other types of nonlinear
recurrence relations that
are only thinly disguised linear recurrences.
A few examples are summarized here:
\vbox{\baselineskip 0pt \lineskip 0pt \def\|{\vrule height 16pt depth 12pt} \halign{\hfil#\hfil\quad&#&\quad\hfil\displaystyle{#}\hfil\cr \hbox{Original recurrence}&\|&\hbox{Linear variation}\cr \noalign{\hrule} &\vrule height 4pt\cr f_n=f_{n-1}-f_nf_{n-1}&\|&{1\over f_{n-1}} = {1 \over f_n}-1 \cr f_n=f_{n-1}↑3/f_{n-2}&\|&\ln f_n=3\ln f_{n-1}-\ln f_{n-2} \cr f_n-f_{n-1}f_n-zf_n=z-zf_{n-1}&% {\vrule height 16pt depth 32pt}% &f_n={z \over 1-\displaystyle{z\mathstrut\over 1-f_{n-1}}}\cr f_n=7f_{n/2}+n↑2&{\vrule height 18pt depth 12pt}% &g_k=g_{k-1}+\left(4\over7\right)↑{\!k},\quad g_k={f_{2↑k}\over 7↑k}\cr }}
The last flavor of recurrence occurs frequently in the analysis of divide and
conquer algorithms.

\yyskip
\minor{Doubly Exponential Sequences}

In the preceding section we explored nonlinear recurrences that contained
hidden linear relations. We turn now to a slightly different situation,
where the nonlinear recurrence contains a very close approximation
to a linear recurrence relation.

A surprisingly large number of nonlinear recurrences fit the pattern
$$x_{n+1}=x_n↑2+g_n,\numeq$$
where $g_n$ is a slowly growing function of $n$, possibly depending on the
earlier members of the
sequence.  As we follow the solution to \eq(2.72) found in an article by Aho and
Sloane [Aho 73], the exact requirements on $g_n$ will become clear.

We begin by taking the logarithm of \eq(2.72) and discovering a nearly linear formula,
$$y_{n+1}=2y_n+\alpha_n,\numeq$$
where $x_n$ and $g_n$ are replaced by
\eqalignno{y_n&=\ln x_n;&\anumeq\cr\noalign{\nobreak} \alpha_n&=\ln\left(1+{g_n\over x_n↑2}\right).&\anumeq\cr}
By using logarithms we have made our first assumption, namely that the $x_n$
are greater than zero.

If we unroll the recurrence for $y_n$ we obtain
$$y_n=2↑n\left(y_0+{\alpha_0\over 2}+{\alpha_1\over 2↑2}+\cdots +{\alpha_{n-1}\over 2↑n}\right).\numeq$$
It is now convenient to extend the series in $\alpha_k$ to infinity:
$$Y_n=2↑ny_0+\sum_{k=0}↑\infty 2↑{n-1-k}\alpha_k\numeq$$
$$r_n=Y_n-y_n=\sum_{k=n}↑\infty 2↑{n-1-k}\alpha_k .\numeq$$
This extension is helpful only when the series converges rapidly, so we make a
second assumption:  The $g_n$ are such that
$$\left|\alpha_n\right| \ge \left|\alpha_{n+1}\right| \qquad\hbox{for }n\ge n_0.\numeq$$
With this second assumption $Y_n$ is well defined and the error $\left| r_n\right|$
is bounded by the first term $\left|\alpha_n\right|$;
we can exponentiate and recover the original solution:
$$x_n=e↑{Y_n-r_n}=K↑{2↑n}\cdot e↑{-r_n}\numeq$$
where
$$K=x_0 \exp\biggl(\sum_{k=0}↑\infty 2↑{-k-1}\alpha_k\biggr).\numeq$$
Since the $\alpha_k$ usually depend on the $x_k$, equation \eq(2.80) is not a legitimate
closed form solution.  Nevertheless, the solution does show that there exists
a constant $K$, perhaps hard to compute, that characterizes the sequence $x_n$.
In some cases it is possible
to determine the exact value of $K$.

A curious aspect of equation \eq(2.80) is the closeness of $K↑{2↑n}$ to the true
solution; as we will see shortly, $e↑{-r_n}$ usually makes a negligible contribution.
To demonstrate this, we will introduce a third assumption:
$$\textstyle \left| g_n\right| <{1\over 4}x_n \xskip\hbox{and }x_n\ge 1 \xskip\hbox{for }% n\ge n_0.\numeq$$

We wish to explore the closeness of $X_n=K↑{2↑n}$ to the exact solution~$x_n$.
Since $\left| r_n\right| \le \left| \alpha_n\right|$, we have
$$x_ne↑{-\left|\alpha_n\right|}\le X_n\le x_ne↑{\left|\alpha_n\right|}.\numeq$$
Expanding the right side of this equation (by taking care of the case where
$\alpha_n<0$ with the identity $(1-u)↑{-1}\le 1+2u$ for $0\le u\le 1/2$, using the third
assumption) yields a new bound:
$$X_n\le x_n+{2\left| g_n\right|\over x_n}.\numeq$$
Similarly,
$$X_n\ge x_ne↑{-\left|\alpha_n\right|} \ge x_n\left(1-{\left| g_n\right|\over x_n↑2}\right)=x_n-{\left| g_n\right|\over x_n}.\numeq$$
Finally, the assumption $\left| g_n\right| <{1\over 4} x_n$ permits us to claim that
$$\left| x_n-X_n\right|<{1\over 2}.\numeq$$
So in cases where we know that $x_n$ is an integer the solution is
$$x_n=\hbox{ nearest integer to }K↑{2↑n},\quad\hbox{for }n\ge n_0.\numeq$$

\yyskip
Here are several recurrence relations that fit the general pattern given by
equation \eq(2.72):

\yskip
\noindent 1)  Golomb's Nonlinear Recurrences.

\noindent
$$y_{n+1}=y_0\,y_1\,\ldots\,y_n+r,\qquad y_0=1.\numeq$$
This definition is equivalent to the finite-history recurrence
$$y_{n+1}=(y_n-r)y_n+r,\qquad y_0=1,\quad y_1=r+1.\numeq$$
And when the square is completed with the following substitution
$$x_n=y_n-{r\over 2}\numeq$$
$$x_{n+1}=x_n↑2+{r\over 2}-{r↑2\over 4}\numeq$$
the recurrence becomes an obvious member of the family just solved.  Since the
$g_n$ term is constant, it is easy to verify that all the assumptions are
satisfied.

In the special cases $r=2$ and $r=4$, the constant $k$ is known to
be~equal to
$\sqrt{2}$ and the golden ratio respectively.  In other cases the constant can
be estimated by iterating the recurrence and solving for $k$.  The doubly
exponential growth of the sequence makes such estimates converge rapidly;
but it also makes the estimates inexact for further terms in the
sequence.

\yskip
\noindent 2)  Balanced Trees.

The following recurrence, given in [Knuth III; Section 6.2.3], counts the
number of balanced binary trees of height $n$.
$$y_{n+1}=y_n↑2+2y_ny_{n-1}.\numeq$$
When we make the transformation
$x_n=y_n+y_{n-1}$
the recurrence appears in a more complex yet more tractable form,
$$x_{n+1}=x_n↑2+2y_{n-1}y_{n-2}.\numeq$$
Here the $g_n$ term is not constant, but grows slowly enough
$(2y_{n-1}y_{n-2} \ll y_n<x_n)$ to meet the requirements on $g_n$.  We can
assert that there exists a $k$ such that
$$x_n=\lfloor k↑{2↑n}\rfloor\numeq$$
and
$$y_n=\lfloor k↑{2↑n}\rfloor-\lfloor k↑{2↑n-1}\rfloor + \cdots \pm\lfloor k\rfloor\pm 1.\numeq$$
(The use of the floor function in place of the nearest integer is a
consequence of $g_n$ being positive, making $k↑{2↑n}$ always slightly
larger than the correct value.)

We conclude with two recurrences mentioned in Aho and Sloane:
$$y_{n+1}=y_n↑3-3y_n\numeq$$
$$y_{n+1}=y_ny_{n-1}+1.\numeq$$
Strictly speaking, these relations do not fit the pattern solved at the
beginning of this section.  However, the techniques developed earlier are
equally applicable.  After taking logarithms both
recurrences become nearly linear.  Equation \eq(2.97), for example, has a
Fibonacci-like solution:
$$y_n=\lfloor k_0↑{F_{n-1}}k_1↑{F_n}\rfloor,\numeq$$
where $F_n=F_{n-1}+F_{n-2}$.

\yyskip
\chap{Operator Methods}

\def \ba #1{\par\vskip 5pt
\vbox{\line{\hfill\vbox{
\hfill}}\vskip 5pt}

\def \famly #1{\par\vskip 5pt \vbox{\line{\hfill\vbox{
\halign{\hfil$##$$\;&##\hfil\qquad&\hfil##$$\;$&$##$\hfil\cr #1}
}\hfill}}\vskip 5pt}

The following analysis of hashing, based on unpublished notes by
Michael Paterson,
relies on two concepts: eigenoperators and what he calls
induction from the other end.''
The cookie monster example below illustrates the value of finding an eigenoperator.
Induction at the other end'' will appear later when we apply the
techniques to various hashing schemes.

Consider a monster whose ability to catch cookies is proportional to
its current size. When we throw a cookie and the monster has $k$ cookies in its
belly, with probability $pk$ the monster grows to size $k+1$. (We assume that
$pk\le 1$.)

Let $g_{nk}$ be the probability that the monster is size $k$ after $n$ cookies
have been thrown. We construct a generating function,
$$g_n(x)=\sum_k g_{nk\,}x↑k,\numeq$$
that represents the distribution after $n$ cookies. Initially the monster has
somehow eaten one
cookie, that is, $g_0(x)=x$.

Given $g_n(x)$, an operator'' will provide a means of obtaining $g_{n+1}(x)$.
First let's look at the impact of a cookie on an individual term.
\vbox{\baselineskip 0pt \lineskip 0pt \def\|{\vrule height 12pt depth 8pt} \halign{\quad\hfil#\hfil\quad&#&\quad\hfil#\hfil\quad\cr Before:&\|&After:\cr\noalign{\hrule} x↑k&\|&pk\,x↑{k+1}+(1-pk)\,x↑k\cr &\|&or \ x↑k+p(x-1)k\,x↑k\cr}}
This change is captured by the operator $\Phi = 1+p\,(x-1)xD$, where $D$
is the derivative. Applying $\Phi$ repeatedly gives $g_n(x)=\Phi↑n g_0(x)$.

Before proceeding further it is helpful to review some facts about oper\-ators.
We will be using the following:
\ba{$D$&derivative\cr
$U$&evaluate at $x=1$\cr
$Z$&evaluate at $x=0$\cr
$U\!D$&obtain the mean $f↑\prime(1)$\cr
$U_n$&shorthand for $U\!D↑n$\cr
$x↑n$&multiply by $x↑n$\cr}
It will be important to understand how operators commute with one another. For
example,
$$D\,x↑n\,f(x)=n\,x↑{n-1}f(x)+x↑n\,D\,f(x),\numeq$$
so we can move $D$ past $x↑n$ by the formula:
$$Dx↑n = x↑nD+n\,x↑{n-1}.\numeq$$
This generalizes to arbitrary polynomials $r(x)$:
$$D\,r(x)=r(x)\,D+r↑\prime(x).\numeq$$

Another useful fact about operators is the relation
$$U_nx=U_n+n\,U_{n-1}\numeq$$
or
$$U_n(x-1)=n\,U_{n-1}.\numeq$$
This can be shown by commuting $x$ with each of the $D$ operators in $U_nx$.

Returning to the cookie monster, we would like to obtain the mean size
of the monster after $n$ cookies:
$$U_1\,g_n(x)=U_1\,\Phi↑n\,g_0(x).\numeq$$
Here is where commuting is important, since it would be nice to be able to
move $U_1$ past $\Phi$. Applying $U_1$ to $\Phi$ gives
\eqalignno{ U\!D\,\Phi&=U\!D(1+p(x-1)xD)\cr &=U\left( D+p(x-1)xD↑2+p(2x-1)D \right)\cr &=(1+p)\,U\!D.&\anumeq\cr}
So $U\!D$ is an eigenoperator of $\Phi$, and by this self-replication we can
compute the mean:
$$U_1\Phi↑ng_0(x)=(1+p)↑nU_1g_0(x) =(1+p)↑n.\numeq$$

\def\var{\hbox{\rm Var\/}}
The variance is obtained with $U_2$, since $\var(g) = g↑{\prime\prime}(1) + g↑\prime(1) - \left( g↑\prime(1) \right)↑2$ and $U_2 g_n(x)= g↑{\prime\prime}_n(1)$.
Unfortunately, $U_2$ does not have the nice eigen\-operator property
that $U_1$ possesses; we have
\baselineskip 15pt\eqalignno{ U_2\Phi&=U(D↑2+p(x-1)xD↑3+2p(2x-1)D↑2+2pD)\cr &=U\!D↑2+2pU\!D↑2+2pU\!D\cr &=(1+2p)U_2+2pU_1.&\anumeq\cr}
However, by a suitable linear combination with $U_1$, we do obtain an eigen\-operator:
$$(U_2+2U_1)\Phi = (1+2p)(U_2+2U_1).\numeq$$
In fact there is a whole family of eigenoperators given by the scheme
\famly{V_1\Phi&=(1+p)V_1&V_1&=U_1\cr
V_2\Phi&=(1+2p)V_2&V_2&=U_2+2U_1\cr
V_3\Phi&=(1+3p)V_3&V_3&=U_3+6U_2+6U_1\cr
V_n\Phi&=(1+np)V_n&V_n&=U_nx↑{n-1}.\cr}
\noindent
This can be shown with equations \eq(3.6) and \eq(3.3):
\baselineskip 15pt\eqalignno{V_n\Phi&=U_nx↑{n-1}\left( 1+p(x-1)x\,D \right)\cr &=V_n + U_np(x-1)x↑nD\cr &=V_n + pnU_{n-1}(Dx↑n-nx↑{n-1})\cr &=V_n + pn(U_nx-n\,U_{n-1})x↑{n-1}\cr &=V_n + pnV_n.&\anumeq\cr}
In principle we can therefore recover all the higher moments of the distribution
using the $V_i$. The variance, for example, is computed with $V_2$:
\eqalignno{V_2\Phi↑ng_0(x)&=(1+2p)↑nV_2x=2(1+2p)↑n\cr \noalign{\vskip 5pt} U_2\Phi↑ng_0(x)&=(V_2-2V_1)\Phi↑ng_0(x)\cr &=2(1+2p)↑n-2(1+p)↑n\cr \noalign{\vskip 5pt} \var(g_n)&=g_n↑{\prime\prime}(1)+g↑\prime_n(1)-\left( g↑\prime_n(1) \right)↑2\cr &=2(1+2p)↑n-(1+p)↑n-(1+p)↑{2n}.&\anumeq\cr}
\major{Coalesced Hashing}

A moment's reflection indicates that the behavior of
the cookie monster is very closely related to certain kinds of hashing.
When keys collide, a long chain develops, and the likelihood of
hitting the chain increases. Suppose we resolve collisions by finding
the first free spot at the left end of the table and by linking
this spot on the end of the chain. As the algorithm proceeds we
will have a distribution of monsters of various sizes. Let
$$g_n(x)=\sum_k(\hbox{expected number of chains of length k})\,x↑k. \numeq$$
Once again we would like to find an operator that describes the addition

Because we are dealing with expected values, the general term will behave
like a single cookie monster, even though there may be several monsters
involved. Here $p=1/m$, where $m$ is the number of slots in the hash table.
So the probability  of a chain of length $k$ growing to $k+1$ is $pk$,
the probability of hitting the chain. However, the computation of the
constant term in the generating function presents new difficulties.
The expected number of empty chains is just $m-n$, so the operator
must be:
$$\Psi = \Phi + \hbox{(fudge the constant term to m-n)}.\numeq$$
Without fudging, $\Phi$ applied to the constant term of $g_n(x)$ is
$\Phi(m-n)=m-n$. The correct change should be:
\vbox{\baselineskip 0pt \lineskip 0pt \def\|{\vrule height 12pt depth 8pt} \halign{\quad\hfil#\hfil\quad&#&\quad\hfil#\hfil\quad\cr Before:&\|&After:\cr\noalign{\hrule} m-n&\|&(m-n-1) + (1-np)\,x\cr }}
We can patch $\Phi$ using the evaluate-at-zero operator, $Z$:
$$\Psi = \Phi + p(x-1)Z - p\,U_1.\numeq$$
Note that $Z$ applied to $g_n(x)$ gives $m-n$ and $U_1$ gives $n$, so $\Psi$
performs properly on the constant term:
\baselineskip 15pt\eqalign{\Psi (m-n) &= m-n +p(x-1)(m-n)-pn\cr &=m-n-mp+(1-n\,p)\,x.\cr}\numeq
(Recall that $mp=1$.)
Using $Z$ and $U_1$ for this fudge might seem at first like a difficult way of
accomplishing a simple fix, but it is important that the change be done
entirely with linear operators.

Now $g_n(x)$ is given by
$$g_n(x)=\Psi↑n g_0(x),\qquad g_0(x)=m. \numeq$$
As before, we seek an eigenoperator of $\Psi$; the application
of $U_1$ to  $\Psi$ gives
$$U_1\Psi = (1+p)U_1 + pZ. \numeq$$
There is no systematic way to find eigenoperators, but the presence of $Z$
suggests trying
$$Z\,\Psi = (1-p)\,Z- p\, U_1. \numeq$$
We see now that the following linear combination is an eigenoperator:
$$(U_1 +Z)\Psi = (U_1 + Z). \numeq$$

The mean'' in this problem is not particularly interesting, since $U_1$ applied
to $g_n(x)$ is just $n$, and the eigenoperator confirms this fact:
$$(U_1+Z) g_n(x)=1↑nm;\numeq$$
$$Zg_n(x) = m-n.\numeq$$
The power of the eigenoperator lies instead in the computation of the expected
number of collisions on the $(n+1)$st insertion. Let
$$h_n(x)=\smash{\sum_k\bigl(\hbox{probability of k collisions on the (n+1)\hskip 1pt st insertion}\bigr)\, x↑k.}\numeq$$
\vskip2pt\noindent
The $x↑k$ term in $g_n(x)$ will contribute $(x↑k+x↑{k-1}+\cdots+x)p$ to $h_n(x)$,
because each item in a $k$-chain is equally likely to be hit, yet they are
at different distances from the end of the chain.

We want to compute $U_1h_n(x)$ based on $g_n(x)$. Applying $U_{r+1}$ to a
polynomial, and taking liberties with the constant term, gives
\baselineskip 15pt\eqalignno{U_{r+1}(x↑{k+1})&=U_{r+1}(x↑{k+1}-1)\cr &=U_{r+1}(x-1)(1+x+\cdots+x↑k)\cr &=(r+1)U_r(1+x+\cdots+x↑k)\cr &=(r+1)U_r(x+x↑2+\cdots+x↑k).&\anumeq\cr}
(These liberties are justified because we are applying $U$ to the poly\-no\-mial argument\vadjust{\goodbreak}
$x↑{k+1}$; we are not commuting $U$ with the operator $x↑{k+1}$ as in equation~\eq(3.3).)
Using $r=1$ relates $g$ and $h$,
$$U_1h_n(x)={p\over 2}U_2xg_n(x).\numeq$$
Since $U_2x=U_2+2 U_1$, and since
$U_1$ is easy to compute, we now seek an eigen\-operator
of $\Psi$ that contains $U_2$.
Here is an appropriate family of eigenoperators:
\famly{C_2\Psi&=(1+2p)C_2&C_2&=V_2-{1\over 2}(U_1-Z)\cr
\noalign{\vskip 2pt}
C_3\Psi&=(1+3p)C_3&C_3&=V_3-{2\over 3}(U_1-2Z)\cr
\noalign{\vskip 2pt}
C_n\Psi&=(1+np)C_n&C_n&=V_n-{(n-1)!\over n}(U_1-(n-1)Z).\cr}
This enables us to find all the higher moments of the distribution
of collisions necessary to insert the $(n+1)$st element.
For instance, the mean number of collisions is obtained with the $C_2$
operator:
\baselineskip 15pt \eqalignno{U_1h_n(x)&={p\over 2}(U_2 + 2U_1)\,g_n(x)\cr &={p\over 2}\left( C_2 +{U_1\over 2}-{Z\over 2} \right)\,g_n(x) \cr &={1\over 2m}\left(\left(1+{2\over m}\right)↑n{m\over 2} -{m\over 2}+n\right)\,.&\anumeq\cr}

\vskip 10pt
The reader might have noticed that the last analysis takes no account of the
time necessary to find the first free cell on the left end of the array.
Suppose that the hashing algorithm uses a pointer to keep track of the previous
free cell. After each collision the pointer is moved rightward until a new free
cell is discovered. The algorithm is modeled by the following game. We start
with an empty array and a pointer at zero. The game requires $n$ R-steps,''
after which we compute the distance from the pointer to the next free cell.
When there are $j$ unoccupied cells, an R-step'' occupies an empty cell
with probability $pj$ or occupies the leftmost free cell with
probability $(1-pj)$. The second case corresponds to a collision, and the
pointer is set to the recently occupied cell. The final score of the game,
the distance between the pointer and the next free cell, gives the cost
of finding an empty cell for a future collision.

Once again we use a generating function. Let $G_{mn}(z)$ be
$$\smash{\sum_k \hbox{(probability that the score is k in an m array after n R-steps)}\,z↑k.}\numeq$$
\smallskip\noindent
We seek an operator to construct $G_{mn}$ from smaller problems,
this time with a different style of induction. Suppose we have a
sequence of R-steps:
$$3\quad 1\quad 4\quad C \quad 7$$
The numbers indicate cells occupied, and $C$  represents a collision where
the leftmost free cell is occupied and the pointer adjusted.
Every such sequence of steps has a certain probability of occurring, and
leads to a certain score, as defined above.

Rather than
add a new element to the end of the sequence we place it at the beginning,
hence the expression induction at the other end.''
Specifically, we will add a new key and a new cell to the array. The key
can fit anywhere in the old array, so we can describe it as the addition
of $k \in \left\{ {1\over 2}, {3\over 2},\ldots,{2m+1\over 2}, C \right\}$
at the beginning of the sequence, and a renumbering to make the sequence
integer again.

For example, consider the R-steps given above, and assume that
the array size is $m=7$. When the $C$ arrives cell $1$ is occupied, so it
lands in cell $2$. At the end of the game the next free cell is $5$, so
the score is $3$. Here are the possible changes, depending which new
R-step we place at the beginning of the sequence:
\par\vskip 10pt\vbox{\line{\hfill
\def \qu #1#2#3#4#5#6#7#8{$#1$&$#2$&$#3\;#4\;#5\;#6\;#7$&$#8$\cr}
\vbox{\halign{
Probability:&New First Element:&Remaining Sequence:&Score:\cr
\noalign{\vskip 2pt}
\qu p1425C83
\qu p2415C83
\qu p3415C84
\qu p4315C84
\qu p5314C84
\qu p6314C83
\qu p7314C83
\qu p8314C73
\qu {1-8p}C425C83
}}\hfill}}\vskip 10pt

How will this affect the final score? The score is the length
of the region between the pointer and the next free cell. If the new key
lands in this region the score is increased by one, otherwise the score
remains unchanged. Since the probability of hitting this region is proportional
to the region size, the cookie monster rears his ugly head and with the
familiar $\Phi$ operator he devours the rest of this analysis:
$$G_{m\,n}(x)=\Phi\,G_{m-1,n-1}(x)=\Phi↑n x.\numeq$$
([Knuth III; exercise 6.4--41] has a less elegant solution to this problem,
and says, Such a simple formula deserves a simpler proof!'')

\vfill\eject

Let us consider a slight variation on the previous game. Instead of
an R-step we use a T-step that fills an empty cell at random and
leaves the pointer at the left end of the array. The final score
is the distance from the left end of the array to the first free cell.

Motivation for this new game comes from the slightly unrealistic
assumption that each key has a random permutation for a probe sequence.
The key pursues its probe sequence until it finds an empty cell. This
assumption, usually called uniform hashing, will be refined later
when we discuss secondary clustering.

We would like to determine the expected
number of entries that the $(n+1)$st element must examine
in its probe sequence. We are free to assume that this element has
$1, 2, 3,\ldots$ for a probe sequence by rearranging the array if necessary
so that this is true. Then the $(n+1)$st insertion requires finding the leftmost
free cell, and this is equal to the score of the T-step game described
above.

Using induction at the other end, we run into the cookie monster once again.
This time he has occupied the cells at the beginning of the array.
However, we must be careful about the probability $p$. The probability
of landing in a given cell is $1/m$, so the operator is
$$\Phi_m=1+{1\over m}(x-1)xD. \numeq$$
Remember that induction at the other end adds both a new key and a new array
slot, so that the probability changes and we must parameterize the
operator $\Phi$ with $m$.

With this parameterized operator, the generating function for monster size
is given by
$$G_{mn}(x)=\Phi_m\Phi_{m-1}\ldots\Phi_{m-n+1}x.\numeq$$
$V_1$ and $V_2$ are still eigenoperators; they give products that
telescope nicely. For example, the average number of probes used
to insert the $(n+1)$st element is
\lineskip 3pt\eqalignno{ V_1G_{mn}(x)&=\left(1+{1\over m}\right)\left(1+{1\over m-1}\right)\ldots \left(1+{1\over m-n+1}\right)\cr \noalign{\vskip 3pt} &={m+1 \over m-n+1}.&\anumeq\cr}
And since all the $V_i$ telescope
there is a systematic way of computing the mean and variance of the
probes necessary to insert the $(n+1)$st element.

\vskip 20pt

In the secondary clustering model each key is mapped to a single hash value,
then the hash value provides  a random permutation for the probe sequence.
Rather than each key having its own random probe sequence, the keys share
probe sequences with those keys mapping to the same hash value. The hash
values and the probe sequences are still random, but the additional
sharing makes collisions more likely.

This time the game we play has an S-rule: If the leftmost cell is
unoccupied use rule S0 otherwise use S1. Rule S0 occupies an empty
cell at random. Rule S1 has a choice: With probability $p$ it occupies
the leftmost empty cell, and with probability $q=1-p$ it occupies
any empty cell at random.

The S-rule captures the somewhat subtle behavior of
secondary clustering. We assume without loss of
generality that each key hashing to the leftmost cell has probe
sequence 1, 2, 3, \dots. Then in rule S1 with probability $p$ we
hash to the leftmost cell, reuse the same hash sequence 1, 2, 3, \dots,
and occupy the leftmost empty cell.

The score is the distance to the first free cell, and we have two score-%
generating functions for the two rules: $H_{mn}(x)$ for S0 and
$G_{mn}(x)$ for S1. Let's look first at $G_{mn}(x)$:
$$G_{mn}(x)=(px+q\Phi_m)\,G_{m-1,n-1}(x). \numeq$$
The operator for $G$ is derived as before by using induction from the other end.''
With probability $p$ the key lands at location one and increases the
monster by one. With probability $q$ we play the old cookie monster
game by adding a key at random.

There is a fine distinction among probabilities in this operator: The probability
$p$ is fixed at $1/m$ before the induction step and remains fixed as
$m$ decreases. The operator $\Phi_m$, however, is parameterized with $m$, so the probability
in this operator increases with smaller $m$. The distinction is precisely
what we want, since the probability of a new key sharing the same
probe sequence with a particular old key is fixed at $1/m$ throughout
the process.

The quantity  $\Omega_m=px+q\Phi_m$ in \eq(3.33) does not have an eigenoperator,
but it does have a sliding'' operator:
$$(U_1-(m+1)\,U_0)\,\Omega_m=(1+{q\over m}) (U_1-m\,U_0). \numeq$$
The sliding operator $A_m = U_1-m\,U_0$ changes its parameter by one
when it commutes with $\Omega_m$, and this behavior is just as valuable
as an eigenoperator when we want to compute $U_1$:
$$\vcenter{\twoline{A_{m+1}G_{mn}(x)=\left(1+{q\over m}\right)\left(1+{q\over m-1}\right)}{ 5pt}{\ldots \left(1+{q\over m-n+1}\right)A_{m-n+1}x} }\numeq$$
$$\vcenter{\twoline{U_1G_{mn}(x)=\left(1+{q\over m}\right)\left(1+{q\over m-1}\right)}{ 5pt}{\ldots \left(1+{q\over m-n+1}\right)(n-m) + (m+1)}}\numeq$$

Now we can turn our attention to $H_{mn}(x)$ and rule S0. Until the
first cell is occupied this also behaves like a cookie monster.
Once the first cell is hit, we switch to $G_{mn}(x)$.
Using induction at the other end, this gives the recurrence:
$$H_{mn}(x) = \Phi_mH_{m-1,n-1}(x)-{x\over m}H_{m-1,n-1}(x) +{x\over m}G_{m-1,n-1}(x). \numeq$$
The middle term corresponds to a mistaken use of $H$ by the $\Phi_m$
operator in the case of an occupied first cell.

Since the game begins in S0, $H_{mn}$ is the desired generating function for the
whole game, and we would like to find its mean, $U_1H_{mn}(x)$:
\eqalignno{U_1H_{mn}(x)&=\left(1+{1\over m}\right)U_1H_{m-1,n-1} -{U_1x \over m}H_{m-1,n-1}+{U_1x \over m}G_{m-1,n-1}\hskip-2em\cr &=U_1H_{m-1,n-1} + {U_1\over m}G_{m-1,n-1}.&\anumeq\cr}
A similar recurrence for $G_{mn}$ can be deduced from equation \eq(3.34):
$$U_1G_{mn}(x)=\left(1+{q\over m}\right) U_1 G_{m-1,n-1} + p.\numeq$$
The situation calls for a new operator trick. Note that a linear combination
of $H$ and $G$ replicates itself:
$$U_1\left(H_{mn}-{1\over q}G_{mn}\right) =U_1\left(H_{m-1,n-1}-{1\over q}G_{m-1,n-1}\right)-{p\over q}. \numeq$$
Furthermore, the term ${p_{\mathstrut}\over q}$ is independent  of $m$, so we have
$$U_1\left(H_{mn}-{1\over q}G_{mn}\right) =U_1\left(H_{m-n,0}-{1\over q}G_{m-n,0}\right)-{n\,p\over q}. \numeq$$
Given the boundary conditions $H_{m,0}=G_{m,0}=x$ and the previously
computed $U_1G_{mn}$, we can determine $U_1H_{mn}$:
\baselineskip 15pt \eqalignno{U_1H_{mn}&=U_1\left(H_{m-n,0}-{1\over q}G_{m-n,0} +{1\over q}G_{mn}\right) - {np\over q}\cr &=1+{1\over q}\left(m-np+(n-m)\prod_{k=m-n+1}↑{m}\left(1+{q\over k}\right) \right).&\anumeq\cr}
(It is interesting to compare the solution above with the brute-force
approach to hashing found in [Knuth III; exercise 6.4--44].)

The last operator trick bears a strong resemblance to the earlier use
of eigen\-operators and sliding operators. In all of these cases we
moved through the recurrence by a self-replicating process. The power
of operator methods lies in their ability to hide unimportant
details so that this kind of self-replication becomes apparent;
therefore quantities like means and variances become relatively easy to compute.

\vfill
\chap{Asymptotic Analysis}
\major{Basic Concepts}

There is no guarantee that the study of algorithms will produce sums and
recurrences with straightforward closed form solutions.  In fact much of
the adventure of analysis of algorithms lies in the variety of mathematics
to which researchers are drawn (at times kicking and screaming) in their
attempts to understand algorithms.  Frequently the researchers will turn to
asymptotic analysis.

Asymptotic analysis attempts to find a solution that closely approxi\-mates
the exact solution.  Often the relative error of this approximation becomes
small for large values of the parameters involved.  We will attempt to
discover as thorough an asymptotic approximation as possible.
For example, instead of knowing that an algorithm runs in $O(n↑2)$ time
it will be far more satisfying to know that the running time is $3n↑2+7n+O(1)$.

Giving attention to asymptotic detail has several rewards.  Frequently the
approximate solution converges so rapidly that the researcher can test a
few small cases and have immediate confirmation of the correctness of a
solution.
It is important in practice to know more than the leading term, since
$1.8\ln n+20$ will be smaller than $2\ln n+10$ only when $n>e↑{50}$.
Moreover, the pursuit of additional asymptotic terms usually
leads to more general and powerful mathematical techniques.

The purpose of this chapter is to introduce the basic tools of asymp\-totics:
$O$-notation, bootstrapping, and dissecting.  The first few sections will
describe these ideas briefly, and the last section includes the derivation
of an asymptotic result that is difficult as a whole, but basic at each step.

\minor{Notation}

\noindent Definition of $O$ or $\preceq$.

We say that $f(n)=O(g(n))$ $\bigl( \hbox{or } f(n)\preceq g(n)\bigr)$
as $n\to\infty$ if there exist integers
$N$ and $K$ such that
$\bigl| f(n)\bigr|\le K\,\bigl|g(n)\bigr|$
for all $n\ge N$.

\yskip
\noindent Definition of $\Omega$ or $\succeq$.

In a similar vein, $f(n)=\Omega(g(n))$ $\bigl( \hbox{or } f(n)\succeq g(n)\bigr)$
as $n\to\infty$
if there exist integers $N$ and $K$ such that $\bigl|f(n)\bigr|\ge K\,\bigl|g(n)\bigr|$ for all $n\ge N$.

When both of these definitions apply, the situation is denoted by $f(n)=\Theta(g(n))$
or $f(n)\asymp g(n)$ [Knuth 76b].

There are similar definitions for little $o$ notation.  For example,
$f(n)=o(g(n))$ or $f(n)\prec g(n)$ whenever $\lim_{n\to \infty }f(n)/g(n)=0$.
There is also a notation for equivalence, $f(n) \sim g(n)$ if
$\lim_{n\to \infty }f(n)/g(n)=1$.  However, in general we will avoid these notations
because they do not capture information about the rate of convergence of the
limits involved. We prefer to use a strong assertion like $O(n↑{-1/2})$
instead of a weak one like $o(1)$.

\yskip
\minor{Bootstrapping}

Bootstrapping is helpful in situations where there is an implicit equation for
a given function of interest.  By repeatedly feeding asymptotic information about
the function back into the equation the approximation is steadily improved.
Here is an example from [deBruijn 70]:
$$f(t)e↑{f(t)}=t,\qquad t\to\infty.\numeq$$
The formula can be rewritten as
$$f(t)=\ln t-\ln f(t).\numeq$$
We prime the pump'' by observing that for $t>e$ we have $f(t)>1$.  Using this in
equation \eq(4.2) gives
$$f(t)=O(\ln t).\numeq$$
Inserting the approximation again into \eq(4.2) yields a better result:
$$f(t)=\ln t+O(\ln \ln t).\numeq$$
Once again we feed this result back into equation \eq(4.2) to improve the
result further,
\eqalignno{f(t)&=\ln t - \ln \ln t - \ln\left(1+O\left({\ln \ln t\over \ln t}\right)\right)\cr \noalign{\vskip5pt} &= \ln t - \ln \ln t+O\left({\ln \ln t\over \ln t}\right).&\anumeq\cr}
In this manner the approximation can be bootstrapped to any degree of
accuracy.

\yyskip

\minor{Dissecting}

Dissecting is applied chiefly to sums and integrals.  In a typical situation
a sum is given over a large range, and the summand has several components.
No single component of the summand is small throughout the range, but if the
range is dissected into pieces  then each piece becomes small (for a variety
of different reasons) and in this fashion the whole sum is shown to be small.

The dissection technique can be illustrated by the sum
$$f(n)=\sum_{3\le d\le n/2}{1\over d\,(n/d)↑d}.\numeq$$
We break the sum into three intervals.  When $3\le d\le 8$ the sum is less than
$$\sum_{3\le d\le 8}{1\over 3(n/8)↑3}=O(n↑{-3}).\numeq$$
Note that the $d$'s in the original formula are replaced by $3$ or $8$ in
equation~\eq(4.7), depending on their worst possible effect on the sum.  Then the
constant number of terms in the sum allows us to claim a $O(n↑{-3})$ bound.

On the second interval, $8\le d\le \sqrt{n}$, we do a similar replacement of~$d$ by
its extreme values so that the sum is less than
$$\sum_{8\le d\le \sqrt{n}}\,{1\over 8(n/\sqrt{n}\,)↑8}=O(n↑{-4}\sqrt{n}\,).\numeq$$
Here the $O(n↑{-4}\sqrt{n}\,)$ bound is caused by $O(\sqrt{n}\,)$ terms of size
at most $O(n↑{-4})$.

The sum over the
remaining interval, $\sqrt{n}\le d\le n/2$, is extremely small, since it is
less than
$$\sum_{\sqrt{n}\le d\le n/2}{1\over \sqrt{n}\,2↑{\sqrt{n}}} =O\left({\sqrt{n}\over 2↑{\sqrt{n}}}\right).\numeq$$
Combining the three intervals, we conclude that the whole sum is $O(n↑{-3})$.

It is clear from the example above that the difficulty of dissecting lies in the
choice of intervals.  The division points $8$ and $\sqrt{n}$ are not sacred:
$10$ and $↑3\hskip -5pt \sqrt{n}$,
for example, work equally well. Nevertheless the choice
of $8$ and $\sqrt{n}$ is somewhat of an art requiring insight into the
behavior of the summand throughout the entire interval.

\yskip
\minor{Limits of Limits}

Occasionally an asymptotic argument will involve two or more limiting processes.
The ordering of the limits is often critical, and it is useful to know when
the exchange of limits is permissible.  In simple situations like
$$\sum_{n=0}↑\infty \sum_{m=0}↑\infty a_{mn}\numeq$$
the absolute convergence of the $a_{mn}$ allows the series to be rearranged
at will.  We could, for example, sum on $n$ before $m$.

Later in this chapter we need to change limits in more delicate
cir\-cum\-stances.  In particular, we want to invert the following theorem:

\thbegin Abelian Theorem.  If
$$\lim_{n\to \infty } \sum_{k=0}↑n a_k=A$$
then
$$\lim_{z\to 1-} \,\lim_{n\to \infty } \sum_{k=0}↑n a_kz↑k=A.$$
{\rm(In this limit and hereafter we assume that $z$ approaches unity from below.)}

The converse statement is not always true:

\thbegin False Conjecture.  If
$$\lim_{z\to 1-} \,\lim_{n\to \infty } \sum_{k=0}↑n a_kz↑k=A$$
then
$$\lim_{n\to \infty } \sum_{k=0}↑n a_k=A.$$

N. G. de Bruijn gives the following counterexample.  Let
$$f(z)= {1-z\over 1+z}= 1-2z+2z↑2-2z↑3+\cdots \numeq$$
and let $a_k$ be the coefficients of the power series expansion for $f(z)$.
The series converges absolutely within a circle of radius one around the
origin, and its limit at one is zero:
$$\lim_{z\to 1-}f(z)=0.\numeq$$
But the partial sums of $a_k$ will never converge to zero:
$$a_0+a_1+\cdots +a_n= (-1)↑n.\numeq$$

Tauber supplied an additional requirement to invert Abel's theorem.  He
stipulated that $a_k$ must be $o(k↑{-1})$.  Hardy and
Littlewood subsequently weakened this condition to $a_k>-C\,k↑{-1}$ for some $C>0$,
although the theorem is still labeled Tauberian because of the general
flavor of the result.  Tauberian theorems supply the conditions necessary
to invert Abelian theo\-rems.

\thbegin Tauberian Theorem.  If
$$\lim_{z\to 1-} \,\lim_{n\to \infty } \sum_{k=0}↑n a_kz↑k=A$$
and if $a_k>-Ck↑{-1}$ for some $C>0$, then
$$\lim_{n\to \infty } \sum_{k=0}↑n a_k=A.$$

For a collection of deeper Tauberian theorems see [Hardy 49; page 154].
\minor{Summary of Useful Asymptotic Expansions}

In the formulas below, $n$ tends to infinity and $\epsilon$ tends to zero.
$$H_n =\ln n + \gamma +{1\over 2n}-{1\over 12n↑2}+O(n↑{-4})\numeq$$
$$n!=\sqrt{2\pi n}\,\left( n\over e\right)↑n\left(1+{1\over 12n} +{1\over 288n↑2}+O(n↑{-3})\right)\numeq$$
$$\ln(1+\epsilon)=\epsilon-{\epsilon↑2\over2}+{\epsilon↑3\over 3} -{\epsilon↑4\over4}+\cdots +(-1)↑{m-1}{\epsilon↑m\over m}+O(\epsilon↑{m+1})\numeq$$
\eqalign{\sum_{k=1}↑n k↑m&= {B_{m+1}(n)-B_{m+1} \over m+1}\qquad\cond{integer m,n>0\cr}\cr &={n↑{m+1}\over m+1} + {n↑m \over 2} + {mn↑{m-1}\over 12} +O(n↑{m-2}),\qquad\cond{m>1\cr}\cr}\numeq
($B_i(x)$ and $B_i$ are the Bernoulli polynomials and numbers, see page 63.)
$$\sum_{k=n_0}↑n {1\over k\,\ln k\,\ln\ln k\,\ldots \bigl(\ln↑{(i)}k\bigr)↑{1+\epsilon}}=O(1),\qquad\cond{\epsilon>0\cr}\numeq$$
The last equation represents the turning point for sums. When $\epsilon=0$
the sums will diverge. For example, the sums
$$\sum{1\over k},\qquad\sum{1\over k\ln k},\quad\hbox{and}\quad \sum{1\over k\ln k\,\ln\ln k}$$
are all unbounded.

There are several ways to obtain crude estimates.
One involves the replacement of sums by their integral counterparts.
In Section 4.2.2 on Euler's summation formula
we will see when this substitution is valid, and how to refine the
results of the approximation.
Another estimate applies to random variables with mean $\mu$ and variance
$\sigma↑2$. Chebyshev's inequality tells us that
$$\hbox{Prob}\left(\left| X-\mu \right| \ge t\right)\le {\sigma↑2\over t↑2}.\numeq$$
In Section 4.3.3 we will develop detailed formulas for the case where
$X$ is a sum of independent random variables.

\vfill\eject
\minor{An Example from Factorization Theory}

We turn now to the problem of computing the probability that a polynomial of
degree $n$ has irreducible factors of distinct degrees modulo a large prime
$p$, a situation that is advantageous for certain factoring algorithms
[Knuth II; pages 429--431].
The probability that an $n$th degree polynomial is itself irreducible mod~$p$ is
$${1\over n}+O(p↑{-n/2}).\numeq$$
(This result is proved, for example, in
[Knuth II; exercise 4.6.2--4].) The modulus, $p$, is unimportant, so we let $p$ go
to infinity and use probability $1/n$ as a foundation for the more
difficult problem of factoring into distinct-degree polynomials.

The solution relies on a partition-style generating function.  The coeffi\-cient
of $z↑n$ in
$$h(z)=\prod_{k\ge 1}\left(1+{z↑k\over k}\right)\numeq$$
is the desired solution, that is, the probability of a distinct-degree factorization.
To see this, note that if $h_n$ is the coefficient of $z↑n$, $h_n$ will be a
sum of terms like
$${z↑{k_1}\over k_1} {z↑{k_2}\over k_2} \cdots {z↑{k_m}\over k_m}\numeq$$
where each of the $k$'s is distinct.  Each term like \eq(4.22) corresponds to a
partition of $n$ into distinct integers $k_1, k_2, \ldots,k_m$.
Suppose we are to construct a
polynomial of size $n$ by multiplying polynomials of sizes $k_1,k_2,\ldots,k_m$.
(We assume that these small polynomials and the large polynomial are all monic.
Other leading coefficients do not affect the results that follow.)
There are
$p↑{k_1}$ polynomials of degree $k_1$.  Of these $p↑{k_1}\!/k_1$ are
irreducible, by our assumption.
Treating each polynomial this way gives a total of
$${p↑{k_1}\over k_1} {p↑{k_2}\over k_2} \cdots {p↑{k_m}\over k_m}= {p↑n\over k_1\,k_2\ldots k_m}\numeq$$
polynomials whose irreducible factors have the appropriate sizes.  Since there
are a total of $p↑n$ monic polynomial of size $n$, this means that the
coefficient
$${1\over k_1\,k_2\ldots k_m}\numeq$$
in equation \eq(4.22) is the probability of obtaining a factorization into
irreducible parts of distinct sizes $k_1, k_2, \ldots,k_m$.

The whole of $h_n$ consists of all possible partitions, each contributing a
term of the form \eq(4.22), and since all of the events are disjoint these
probabilities are summed.  Thus the generating function properly determines~$h_n$,
the limiting probability that a polynomial of degree $n$ factors into irreducible parts
of distinct sizes modulo a large prime.

Equation \eq(4.21) does not give us a closed form for $h_n$, and there does not
seem to be one, so instead we seek
an asymptotic formula as $n\to \infty$.  Taking logarithms and expanding each
logarithm yields
$$h(z)=\exp\,\biggl(\sum_{k\ge 1}\left({z↑k\over k}-{z↑{2k}\over 2k↑2} +{z↑{3k}\over 3k↑3}- \cdots\right)\biggr).\numeq$$
For $z<1$, the series converges absolutely, permitting us to rearrange it as
necessary.  Our strategy will be to split the larger terms off from the
beginning of the series, and sum them separately.  First we have
\eqalign{h(z)&=\exp\,\biggl(\sum_{k\ge 1}{z↑k\over k}+\sum_{k\ge 1}\! \left(-{z↑{2k}\over 2k↑2}+{z↑{3k}\over 3k↑3}-\cdots\right)\biggr)\cr \noalign{\vskip 5pt} &={1\over 1-z}g(z),\cr}\numeq
where
$$g(z)=\exp\,\biggl( \sum_{k\ge 1}\left(-{z↑{2k}\over 2k↑2}+{z↑{3k}\over 3k↑3}-\cdots\right)\biggr).\numeq$$
In this form, $h_n$ is the partial sum of the $g_j$ coefficients in $g(z)$:
$$h_n=\sum_{0\le j\le n}g_j .\numeq$$
We will see later that the Tauberian limit theorem applies, hence
\eqalignno{\lim_{n\to \infty }h_n&=\lim_{z\to 1-}g(z)\cr \noalign{\vskip 5pt} &=\exp\,\biggl(\sum_{k\ge 1}\left(-{1\over 2k↑2}+{1\over 3k↑3}-\cdots\right)\biggr)\cr \noalign{\vskip 5pt} &=\exp\,\biggl(\sum_{k\ge 1}\left(\ln\left(1+{1\over k}\right)-{1\over k}\right)\biggr)\cr \noalign{\vskip 5pt} &=\exp\left(\lim_{n\to \infty }\bigl(@\ln (n+1)-H_n\bigr)\right)\cr \noalign{\vskip 5pt} &=e↑{-\gamma} .&\anumeq\cr}
Euler's constant, $\gamma$, appears mysteriously from the asymptotics for
$H_n$, the harmonic numbers:
$$H_n=\ln n+\gamma +{1\over 2n}+O(n↑{-2}).\numeq$$

Unfortunately part of the mystery lies in how fast $h_n$ converges to this strange
constant $e↑{-\gamma}$.  For error bounds, the Tauberian limit theorem is not
particularly helpful.  We must split another term off of the series in equation
\eq(4.25), and continue with a more detailed analysis:
$$g(z)=p(z↑2)\,q(z)\numeq$$
where
\eqalign{p(z)&=\exp\,\biggl(-{1\over 2}\sum_{k\ge 1} {z↑k\over k↑2}\biggr)\cr q(z)&=\exp\,\biggl(\sum_{k\ge 1}\left({z↑{3k}\over 3k↑3}-{z↑{4k}\over 4k↑4}+\cdots \right)\biggr).\cr}\numeq

First we attack $p(z)$ by deriving a recurrence relation for its coefficients:
$$p↑{\prime}(z)=p(z)\,\biggl(-{1\over 2}\sum_{k\ge 1}{z↑{k-1}\over k}\biggr)\numeq$$
$$-2n\,p_n=\sum_{0\le k<n}{p_k\over n-k}.\numeq$$
With this implicit formula we can use bootstrapping to derive a good estimate
for $p_n$.  To prime the pump,'' it is easy to verify inductively that
$p_n=O(1)$.  Using this crude estimate in equation \eq(4.34),
$$-2n\,p_n=\sum_{0\le k<n}{O(1)\over n-k} ,\numeq$$
and replacing the right side with the asymptotics for the harmonic numbers,
$O(\log n)$, gives an improved estimate of $p_n$:
$$p_n=O\left({\log n\over n}\right).\numeq$$
A further iteration of bootstrapping yields
$$\postdisplaypenalty=-500 p_n=O\left({\log n\over n}\right)↑2.\numeq$$
At this point our estimate of $p_n$ is good enough to begin dissecting the
sum in equation \eq(4.34).  We wish to introduce more than a $O$-term in the
asymptotics for $p_n$, so we remove the dominant part of the series in a form
that is easy to sum:
\eqalignno{-2np_n&=\sum_{0\le k<n}{p_k\over n}+\sum_{0\le k<n}p_k \left({1\over n-k}-{1\over n}\right)\cr \noalign{\vskip5pt} &={1\over n}\sum_{k\ge 0}p_k-{1\over n}\sum_{k\ge n}p_k+{1\over n}\sum_{0\le k<n}p_k \left({k\over n-k}\right)\cr \noalign{\vskip5pt} &={1\over n}p(1)-{1\over n}\sum_{k\ge n}O\left({\log k\over k}\right)↑2 +{1\over n}\sum_{0\le k<n}O\left({(\log k)↑2\over k(n-k)}\right)\cr \noalign{\vskip5pt} &={1\over n}e↑{-\pi↑2/12}+O\left({(\log n)↑3\over n↑2}\right).&\anumeq\cr}
In the last step we computed $p(1)$ by summing the infinite series
$$\sum_{k\ge 1}{1\over k↑2}=\zeta(2)={\pi↑2\over 6}.\numeq$$
We estimated the sum $\sum_{k\ge n}O\bigl({\log k\over k}\bigr)↑2$ by
considering its integral counterpart
$$\int_n↑\infty \left({\log x\over x}\right)↑2dx=O\left({(\log n)↑2\over n}\right). \numeq$$
\smallskip\noindent
And we estimated the remaining sum by computing with partial fractions:
\eqalignno{\sum_{0\le k<n}O\left({(\log k)↑2\over k(n-k)}\right) &=O\left((\log n)↑2 \sum{1\over k(n-k)}\right)\cr \noalign{\vskip5pt} &=O\left({(\log n)↑2\over n}\sum\left({1\over k}+{1\over n-k}\right)\right)\cr \noalign{\vskip5pt} &=O\left({(\log n)↑3\over n}\right).&\anumeq\cr}

Returning to equation \eq(4.38), we now have a refined estimate of $p_n$,
$$p_n={-e↑{-\pi↑2/12}\over 2n↑2}+O\left({\log n\over n}\right)↑3.\numeq$$
This expression can be bootstrapped
through another iteration to obtain the slightly better approximation
$$p_n={-e↑{-\pi↑2/12}\over 2n↑2}+O\left({\log n\over n↑3}\right).\numeq$$

Now that $p(z)$ is well understood, we turn our attention to the $q(z)$ portion
remaining in equation \eq(4.31).  This time we split away the terms with $k=1$, so
that
$$q(z)=s(z)\,r(z)\numeq$$
where
\eqalign{s(z)&=\exp\left({z↑3\over 3}-{z↑4\over 4}+{z↑5\over 5}\cdots\right)\cr r(z)&=\exp\,\biggl(\sum_{k\ge 2}\left({z↑{3k}\over 3k↑3}-{z↑{4k}\over 4k↑4}+\cdots \right)\biggr) .\cr}\numeq
The expression for $s(z)$ can be reworked,
\eqalignno{s(z)&=\exp\left(\ln(1+z)-z+{z↑2\over 2}\right)\cr &=(1+z)e↑{-z+z↑2\!/2}.&\anumeq\cr}
From this we conclude that the coefficients, $s_n$, are exponentially small.

In $r(z)$, we collect terms with similar powers:
$$r(z)=\exp\,\biggl(\sum_{k\ge 3}z↑k\sum_{\scriptstyle 3\le d\le k/2\atop \scriptstyle d\,\hbox{\sevenrm divides} \, k}{\pm1\over d\,(k/d)↑d}\biggr) .\numeq$$
The inside sum is $O(k↑{-3})$. (This follows from the example used in Section
4.1.3 to illustrate dissecting sums.)  Differentiating the formula for $r(z)$
and equating coefficients gives a recurrence relation for $r_n$:
$$r↑{\prime}(z)=r(z)\sum_{k\ge 3}k\,z↑{k-1}O(k↑{-3})\numeq$$
$$nr_n=\sum_{0\le k<n}r_kO\left({1\over n-k}\right)↑2 .\numeq$$
This recurrence can be bootstrapped repeatedly to give the successive bounds
$r_n=O(1)$, $r_n=O(n↑{-1})$, $r_n=O(n↑{-2})$, and $r_n=O(n↑{-3})$.

We have shattered our original problem into numerous fragments, but we
have been able to deal effectively with each piece. Now
we can begin to assemble the final result.

The pieces $r(z)$ and $s(z)$ combine to form $q(z)$ with coefficients
$$q_n=\sum_{0\le k\le n}r_ks_{n-k} .\numeq$$
This is a convolution of two series that are $O(n↑{-3})$, so the result is also
$O(n↑{-3})$.  (To see this, divide the range into two parts,
$0\le k\le n/2$ and $0\le n-k\le n/2$. What
requirements on $f(n)$ suffice to make
the convolution of two series that are $O(f(n))$ also
$O(f(n))$?)

Next $q(z)$ and $p(z)$ combine to form $g(z)$:
$$g(z)=p(z↑2)\,q(z)\numeq$$
$$g_n=\sum_{2k+l=n}p_kq_{l}.\numeq$$
Then $g_n$ is summed to obtain $h_n$:
$$h_n=\sum_{j\le n}g_j=\sum_{2k+l\le n}p_kq_{l} .\numeq$$
We already know that the series on the right side of equation \eq(4.53), when
extended to infinity, converges to $e↑{-\gamma}$, so we focus our attention
on the tail:
\eqalignno{h_n&=e↑{-\gamma}-\sum_{2k+l>n}p_kq_{l}\cr \noalign{\vskip5pt} &=e↑{-\gamma}-\biggl(\,\sum_{l\ge 0}q_{l}\,\biggl(\sum_{2k>n}p_k+\!\! \sum_{n-l<2k\le n}p_k\biggr)\biggr).&\anumeq\cr}
Using our earlier result for $p_k$, we can estimate the two internal sums. First
\eqalignno{\sum_{2k>n}p_k&=\sum_{2k>n}{-p(1)\over 2k↑2}+\sum_{2k>n}\! O\left({\log k\over k↑3}\right)\cr \noalign{\vskip5pt} &={-p(1)\over n}+O\left({\log n\over n↑2}\right).&\anumeq\cr}
Here we have used $p(1)$ instead of $e↑{-\pi↑2/12}$.  This will prove useful
when $p(1)$ and $q(1)$ combine to give $e↑{-\gamma}$.  In the last step we
applied Euler's summation formula to both sums.

The other sum in equation \eq(4.54) can be bounded by splitting it into two ranges,
\eqalignno{\sum_{l\ge 0}q_{l}\sum_{n-l<2k\le n}p_k &=\sum_{0\le l<n/2}q_{l}\sum_{n-l<2k\le n}p_k +\sum_{l\ge n/2}q_{l}\sum_{n-l\le 2k\le n}p_k\cr \noalign{\vskip5pt} &=O\biggl(\sum_{0\le l<n/2}q_{l}\cdot l\cdot\left| p_{n/4}\right| +\sum_{l\ge n/2}q_{l}\cdot\left| p(1)\right|\biggr)\cr \noalign{\vskip5pt} &=O(n↑{-2}).&\anumeq\cr}
Now that we have bounded all parts of equation \eq(4.54), we can finally
compute $h_n$:
\eqalignno{h_n&=e↑{-\gamma}+{p(1)q(1)\over n}+O\left({\log n\over n↑2}\right)\cr &=e↑{-\gamma}+{e↑{-\gamma}\over n}+O\left({\log n\over n↑2}\right).&\anumeq\cr}

Similar but simpler methods show that $g_n=O(n↑{-1})$, so that our earlier use of
the Tauberian theorem was indeed justified.

\vfill\eject
\major{Stieltjes Integration and Asymptotics}

Integrals are useful tools in asymptotics since they can be used to approximate
discrete sums, and it is helpful to understand how an integral interacts
with $O$-notation. For this reason we shall study the Stieltjes integral.
The following definition and its immediate consequences are developed
in [Apostol 57]:

\noindent{\bf Definition.}

1) Let $f$ and $g$ be real-valued functions on $[a,b]$.

2) Let $P$ be a partition of $[a,b]$ into $a=x_0<x_1<\ldots<x_n=b$.

3) Define a sum,
$$S(P)=\sum_{0\le k<n} f(t_k) \left( g(x_{k+1})-g(x_k) \right),\qquad t_k \in [x_k, x_{k+1}] \numeq$$

4) Then $A$ is the value of the Stieltjes integral $\int_a↑b f(t) \,dg(t)$
if and only if for all $\epsilon >0$ there exists a $P_\epsilon$ such that
all refinements $P$ of $P_\epsilon$ lead to sums near~$A$, that is,
$\left| S(P)-A \right| < \epsilon$.

\noindent{\bf Consequences.}

1) The Stieltjes integral has at most one value.

2) The Stieltjes integral is linear in $f$ and $g$.

3) Adjacent intervals can be combined, $\int_a↑b+\int_b↑c=\int_a↑c$.

4) (Integration by parts.) If $\int_a↑b f(t)\,dg(t)$ exists then $\int_a↑b g(t) \,df(t)$ exists and the sum of these two integrals
is $\left.f(t)g(t)\right|_a↑b$.

5) (Change of variables by a continuous nondecreasing function $h$.)
$$\int_a↑b f(h(t))\,dg(h(t)) =\int_{h(a)}↑{h(b)} f(t)\,dg(t).\numeq$$

6) If $\int_a↑b f(t)\,dg(t)$ exists and $g↑\prime(t)$ is continuous on $[a,b]$,
then
$$\int_a↑b f(t)\, dg(t)=\int_a↑b f(t) g↑\prime(t)\,dt. \numeq$$

7) If $a$ and $b$ are integers and
$f$ is continuous from the right at integer points then
$$\int_a↑b f(t) \,dg(\lceil t \rceil) =\sum_{a\le k<b} f(k)\,\Delta g(k), \qquad \Delta g(k)=g(k+1)-g(k).\numeq$$

7$↑\prime$) If $a$ and $b$ are integers and
$f$ is continuous from the left at integer points, then
$$\int_a↑b f(t) \,dg(\lfloor t \rfloor) =\sum_{a<k\le b} f(k)\,\nabla g(k), \qquad \nabla g(k)=g(k)-g(k-1).\numeq$$

8) If $a$ and $b$ are integers and $g$ is continuous from the left at integer points,
then
$$\int_a↑b f(\lfloor t \rfloor) \,dg(t) = \sum_{a\le k<b} f(k)\,\Delta g(k). \numeq$$

9) (Derivative of the integral.)
$$\int_a↑b f(t) \,d\!\int_a↑t\! g(u)\,dh(u) = \int_a↑b f(t)\,g(t)\,dh(t). \numeq$$

10) If $\int_a↑b f(t)\,dg(t)$ exists
then $\int f(t)\,dg(t)$ exists for all subintervals of $[a,b]$.

11) $\int_a↑b f(t)\,dg(t)$ exists if $f$ is continuous and $g$ is of bounded
variation.

By bounded variation we mean that $\int_a↑b \left| dg(t) \right|$
exists. Intuitively this implies that the variation,
$\sum\left| g(x_{k+1})-g(x_k)\right|$,
gets small as the partition~$P$ gets small. Continuity is not enough, since
$f(t)=g(t)=\sqrt{t} \cos(1/t)$ has no Stieltjes integral in intervals that
include~0.

12) (Summation by parts.) Combining consequences 4, 7, and 7$↑\prime$,
we obtain a very useful formula when $a$ and~$b$ are integers:
$$\sum_{a\le k<b}f(k)\,\Delta g(k) = \left. f(k)g(k) \right| _a↑b - \sum_{a<k\le b}g(k)\,\nabla f(k).\numeq$$

\vfill\eject
\minor{\bi O-notation and Integrals}
\mark{$O$-NOTATION AND INTEGRALS}

The basic properties of Stieltjes integration allow us to derive
two theo\-rems stipulating when $O$ can be removed from an integral.

\thbegin Theorem 1.
$$\int_a↑b O\bigl( f(t) \bigr)\,dg(t)=O\biggl(\int_a↑b f(t)\,dg(t) \biggr)\numeq$$
if $g$ is monotone increasing, $f$ is positive, and both integrals exist.

\thbegin Proof. {\rm
Recall that $a(t)=O\left( f(t) \right)$ means that there is
a constant $M$ such that $\left| a(t) \right| <M f(t)$.
Since $f(t)$ and $dg(t)$ are nonnegative by hypothesis, we can
bound the integral by $\int_a↑b Mf(t)\,dg(t)$ and move $M$ outside to derive the
theorem.}

\thbegin Theorem 2.
$$\int_a↑b f(t)\,dO\bigl( g(t) \bigr) =O\bigl( f(a)g(a) \bigr)+ O\bigl( f(b)g(b) \bigr) + O\biggl( \int_a↑b f(t)\,dg(t) \biggr) \numeq$$
when $f$ and $g$ are monotone increasing positive functions and the integrals
exist.

\thbegin Proof. {\rm
Let $b(t)$ be the function that is $O\bigl( g(t)\bigr)$. We can integrate by parts
and obtain
$$\int_a↑b f(t)\,db(t) = \left.f(t)b(t)\right|_a↑b - \int_a↑b b(t)\,d_{}f(t).$$
Theorem 1 applies to the last integral, hence we have
$$\int_a↑b f(t)\,dO\bigl( g(t) \bigr) =O\bigl( f(a)g(a) \bigr)+ O\bigl( f(b)g(b)\bigr) + O\biggl( \int_a↑b g(t)\,df(t) \biggr). \numeq$$
Integration by parts is used again to exchange $f$ and $g$,
completing the proof of Theorem 2.}

\vfill\eject
\minor{Euler's Summation Formula}

Stieltjes integration provides a theoretical framework for the approx\-i\-ma\-tion
of sums by integrals. Suppose we wish to approximate the sum of~$f(k)$.
We can begin with consequence number 7,
$$\sum_{a\le k<b} f(k) = \int_a↑b f(t) d \lceil t \rceil.\numeq$$
Using the linearity property, the right-hand side can be expanded to
$$\int_a↑b f(t)\, dt -\int_a↑b f(t)\,d \left(\textstyle t - \lceil t \rceil +{1 \over 2} \right) +\int_a↑b f(t) \, d\left(\textstyle+{1\over 2} \right).\numeq$$
The first integral is a rough approximation to the sum; the second
integral will allow us to refine the approximation; and the third integral
is zero. A new term of Euler's summation formula appears when we integrate
the second term by parts:
\eqalignno{\sum_a↑b f(k)&= \int_a↑b f(t)\,dt - \left. f(t) \left( t-\lceil t \rceil +{1\over2} \right) \right|_a↑b +\int_a↑b \left( t-\lceil t \rceil + {1\over 2} \right) \, df(t) \cr &= \int_a↑b f(t)\,dt - \left. {1\over 2} f(t) \right|_a↑b +\int_a↑b \left( t-\lceil t \rceil + {1\over 2} \right) \, df(t).&\anumeq\cr}
On the interval $[n,n+1]$ the last integral can be rewritten to read
$$\int_n↑{n+1} \left(t-n-{1\over2}\right)df(t) = \int_n↑{n+1} f↑\prime(t)\,d\left( {(t{-}n)↑2-(t{-}n)+1/6\over 2}\right).\numeq$$
And we can iterate this process, integrating by parts, and exchanging the
r\↑oles of $f$ and $g$ in the new integral $\int g\,df$.

There are several requirements necessary for such an iteration to work
prop\-erly, and if we explore these requirements the mystery of the constants
$1/2$ and $1/6$ will be revealed. First of all, we assume that $f↑\prime(t)$
exists. In fact each iteration will require a higher derivative of $f(t)$.
The second requirement enters when we integrate'' the factor $(t-n-1/2)$ and
obtain
$\bigl( (t-n)↑2 - (t-n) + 1/6 \bigr)/2$. This change is made
on each interval ${[n, n+1]}$, and from these segments the whole range is assembled.
It is fortunate that
$\left( (t-n)↑2 - (t-n) + 1/6 \right)/2$ has the same value at $n$
and ${n+1}$, so that the assembled integral $\int_a↑b f↑\prime(t) \,dg(t)$ has a
continuous function in the position of $g(t)$. Any discontinuities in $g$ would
make significant and unwanted contributions to the Stieltjes integral.
The constant $1/2$' in $(t-n-1/2)$ is responsible for the continuity, and
the constant $1/6$' will guarantee a similar continuity when we integrate
the polynomial again in the next iteration. We have a family of polynomials,
$B_n(t-\lfloor t \rfloor)$, with the continuity condition $B_n(0)=B_n(1)$
holding at the endpoints for $n>1$, satisfying the
derivative relation $B_n↑\prime(x)=nB_{n-1}(x)$. These two requirements are
sufficient to determine the Bernoulli polynomials:
\eqalign{B_1(x)&=x-1/2\cr B_2(x)&=x↑2-x+1/6\cr B_3(x)&=x↑3-(3/2)x↑2+(1/2)x\cr \noalign{\vskip 5pt} B_n(x)&=\sum_k {n\choose k}B_k x↑{n-k}\cr}\numeq
The constants $B_k$ in the sum are the Bernoulli numbers:
$$B_0=1,\quad B_1=-1/2, \quad B_2=1/6, \quad B_3 = 0,\quad B_4=-1/30 \numeq$$
And these coefficients appear in the final summation formula:
$$\twolinenumeq{\hskip-9pt\sum_{a\le k<b} \!f(k) = \int_a↑b \!f(t)\,dt +\left.\vbox to 10pt{}B_1f(t)\right|_a↑b + \left.{B_2\over 2!}f↑\prime(t) \right|_a↑b + \cdots+\left.{B_{2m}\over (2m)!}f↑{(2m-1)}(t)\right|_a↑b}{6pt}{ \null+\int_a↑b{B_{2m+1}\bigl(t-\lfloor t\rfloor\bigr)\over(2m+1)!}f↑{(2m+1)}(t)\,dt.}$$
(Strictly speaking, the sum implicitly represented by dots here has
alternating signs,
$$\left.{B_2\over 2!}f↑\prime(t) \right|_a↑b -\left.{B_3\over 3!}f↑{\prime\prime}(t) \right|_a↑b +\left.{B_4\over 4!}f↑{\prime\prime\prime}(t) \right|_a↑b - \cdots+\left.{B_{2m}\over (2m)!}f↑{(2m-1)}(t)\right|_a↑b;$$
but these signs are immaterial because the odd-numbered coefficients
$B_3$, $B_5$, $B_7$, \dots\ are all zero. See [GKP; Section 6.5] for
further discussion of Bernoulli numbers.)

\vfill\eject
\minor{An Example from Number Theory}

Suppose we have an integer $n$ chosen at random from the interval $[1,x]$.
The average number of distinct prime factors of $n$ is given
by the formula
$${1\over x}\sum_{n\le x}\sum_{p \divides n}1={1\over x}\sum_{p\le x}\left\lfloor x\over p \right\rfloor.\numeq$$
(Hereafter $p$ will denote a prime. The notation $p\divides n$ means $p$
divides $n$.'')
Ignoring the slight aberration caused by the floor function, the quantity
of interest in the formula above is the sum of the reciprocals of primes~$\le x$.
We turn now to this restricted problem, where we
will make several uses of Stieltjes
integration. Initially, we can express the sum as an integral:
$$\sum_{p \le x} {1\over p} = \int_{1.5}↑x {1\over t} \,d \pi (t), \qquad \pi (t)=\sum_{p \le t} 1.\numeq$$
Here $\pi (t)$ is a step function that changes only at the primes.
The function $L(x)$ given by
$$L(x)=\int_{1.5}↑x {dt \over \ln t} \numeq$$
is known to give a close approximation to $\pi (x)$:
$$\pi (x)=L(x) + O \left( x \, e↑{-c \sqrt{@\log x}} \right).\numeq$$
(This strong form of the prime number theorem is due to de la Vall\'ee
Poussin in the 19th century; cf.\ [Knuth 76a].)
By using $L(x)$ for $\pi (x)$ and applying Theorem 2 to remove $O$
from the integral, we obtain an asymptotic estimate:
\eqalignno{\sum_{p\le x} {1\over p} &= \int_{1.5}↑x {dt\over t \ln t} +\int_{1.5}↑x {1\over t}\, d@O\!\left( t \, e↑{-c \sqrt{@\log t}}\right)\cr &= \ln \ln x + O(1).&\anumeq\cr}

\vskip 5pt

Although we have no analog of Euler's summation formula for sums over primes,
there is a roundabout way of improving this estimate. Using reasoning similar
to that used above
we can compute further sums:
$$C_m(x) = \sum_{p\le x} {(\ln p)↑m\over p} = {(\ln x)↑m\over m} + O(1), \qquad m\ge 1.\numeq$$
Then by Consequence 9, our original sum can be expressed as
$$C_0(x)=\int_{1.5}↑x {d \pi (t)\over t} =\int_{1.5}↑x{1\over (\ln t)↑m} \, d\int_{1.5}↑t {(\ln u)↑m\,d\pi (u) \over u} =\int_{1.5}↑x {dC_m(t)\over (\ln t)↑m} .\numeq$$
And the last integral submits to integration by parts,
\eqalignno{C_0(x)&=\left.{C_m(t)\over(\ln t)↑m} \right|_{1.5}↑x + m\int_{1.5}↑x {C_m(t)\,dt \over t (\ln t)↑{m+1}} \cr &={1\over m}+O\left( (\ln x)↑{-m} \right) + \int_{1.5}↑x {dt\over t\ln t} + m\int_{1.5}↑x {O(1)\,dt\over t (\ln t)↑{m+1} }\cr &= \ln \ln x + M + O\left( (\log x)↑{-m} \right), \hbox{ for some constant M}.&\anumeq\cr}
This analysis applies to all $m>0$, so we have proved a rather strong result
about the asymptotics of the sum of reciprocal primes.
However, the strength of the result makes the exact value of $M$ a tantalizing
question.

We can evaluate $M$ by making use of the Riemann zeta function and M\"obius
inversion. The zeta function is related to prime numbers by
$$\zeta(s)=\sum_{n\ge 1} {1\over n↑s} =\prod_p\left({1\over 1-p↑{-s}}\right) =\prod_p\left( 1+p↑{-s}+p↑{-2s}+ \cdots\, \right), \quad\! s>1. \shiftnumeq4pt$$
Following Euler, we will find it useful to work with the logarithm of this equation,
$$\ln\zeta(s)=\sum_p\left({1\over p↑s}+{1\over 2p↑{2s}}+\cdots\,\right) =\Sigma(s)+{1\over 2}\Sigma(2s)+{1\over 3}\Sigma(3s)+\cdots\,,\numeq$$
where $\Sigma(s)=\sum_p p↑{-s}$. We are interested in the partial sums of the
divergent series $\Sigma(1)$, and we can get information about them by considering
the convergent series $\Sigma(s)$ for $s > 1$.

The M\"obius function, defined by
\mu(n)=\left\{\,\vcenter{\halign{#,\hfil\qquad&#\hfil\cr 1&if n=1;\cr 0&if n has a squared factor;\cr (-1)↑k&if n has k distinct prime factors;\cr}}\right.\numeq
will invert formulas such as \eq(4.85) above. The common form of M\"obius inversion is
$$g(n)=\sum_{d\divides n} f(d) \qquad \swap \qquad f(n)=\sum_{d\divides n} \mu(d) g({n\over d}).\numeq$$
But for our purposes we need another formulation,
$$g(x)=\sum_{m=1}↑\infty f(mx) \qquad \swap \qquad f(x)=\sum_{n=1}↑\infty \mu(n)g(nx). \numeq$$
This allows us to express $\Sigma(s)$ in terms of $\zeta(s)$,
$$\Sigma(s) = \sum_n \mu(n) {\ln \zeta(ns)\over n}.\numeq$$
Since $\zeta(s)=1+O\left( 2↑{-s} \right)$ this last sum converges quickly
to $\Sigma(s)$; we have a rapid way to evaluate $\Sigma(s)$ that
will prove useful later when we express $M$ in terms of $\Sigma(s)$.
(These properties of the zeta and M\"obius functions can be found, for example,
in [Hardy 79; pp.\ 233--259].)

Let us pause a moment to plot strategy. We are interested in $\Sigma(1)$,
but the formula above is valid only for $s>1$. We could look at $\Sigma(1+\epsilon)$
and let $\epsilon \to 0$,
$$\Sigma(1+\epsilon)=\ln\zeta(1+\epsilon)-{1\over 2}\Sigma(2+2\epsilon) -{1\over 3}\Sigma(3+3\epsilon)+\cdots\,. \numeq$$
Standard references like
[Hardy 79] give $\epsilon↑{-1}+O(1)$ for the asymptotics of $\zeta(1+\epsilon)$
near $1$, so this simplifies to
$$\Sigma(1+\epsilon)=-\ln\epsilon -\sum_{n=2}↑\infty {\Sigma(n+n\epsilon)\over n} +O(\epsilon).\numeq$$
Unfortunately this formula blows up in a different sense than our original
expression,
$$C_0(x)=\sum_{p\le x} {1\over p} = \ln\ln x +M+ O\left( (\log x)↑{-m} \right),\numeq$$
does. So we cannot simply cancel the leading terms of
the two formulas to obtain information about $M$.
Instead we must rework the $C_0$ formula to depend on $\epsilon$.

To rework $C_0$, we introduce $\epsilon$ so that $x$ can be sent to
infinity,
$$\sum_p{1\over p↑{1+\epsilon}}= \int_{1.5}↑\infty {d\pi (t)\over t↑{1+\epsilon}} = \lim_{x\to \infty }\int_{1.5}↑x {dC_0(t)\over t↑\epsilon}; \numeq$$
here again we have used Consequence 9 to replace $d\pi (t)$ with $t\,dC_0(t)$.
Integrating by parts gives
\eqalign{\Sigma(1+\epsilon)&=\lim_{x\to \infty }\left({C_0(x)\over x↑\epsilon} - \int_{1.5}↑x C_0(t) d\left(t↑{-\epsilon}\right)\right).\cr }\numeq
Now the old asymptotics for $C_0$ will replace $C_0$ in the integral.
By these same asymptotics $C_0(x)/x↑\epsilon$ vanishes. This leaves
$$\Sigma(1+\epsilon)=\epsilon\,\left(\int_{1.5}↑\infty \left(@\ln\ln t +M+ O\left( (\log t)↑{-1} \right) \right) {dt\over t↑{1+\epsilon}}\right).\numeq$$
Next we substitute $e↑{u/\epsilon}$ for $t$, obtaining
$$\Sigma(1+\epsilon)=\int_{\epsilon\ln 1.5}↑\infty e↑{-u}\left( \ln u -\ln \epsilon +M+O\left({\epsilon\over u}\right) \right) \, du. \numeq$$
Most terms of this integral are easy to deal with, except $e↑{-u}\ln u$
which can be expressed in terms of the exponential integral:
$$\int_a↑\infty e↑{-u}\ln u \,du = e↑{-a}\ln a +\int_a↑\infty {e↑{-u}\over u} \, du. \numeq$$
For small $a$ the exponential integral has well understood asymptotics,
$$E_1(a)=\int_a↑\infty {e↑{-u}\over u} du = -\ln a -\gamma +O(a). \numeq$$
Applying our knowledge of $E_1(a)$ to equation \eq(4.96) gives
\eqalignno{ \Sigma(1+\epsilon)&= (1.5)↑{-\epsilon}(\ln \epsilon +\ln\ln 1.5)+E_1(\epsilon \ln 1.5)\cr & \qquad -(1.5)↑{-\epsilon} \ln \epsilon +(1.5)↑{-\epsilon}M+O\left(\epsilon E_1(\epsilon\ln 1.5)\right)\cr \noalign{\vskip 5pt} &= -\ln \epsilon -\gamma + M + O\left(\epsilon\ln{1\over \epsilon}\right).&\anumeq\cr}
Now we can compare this reworked formula with our previous expression \eq(4.91) for
$\Sigma(1+\epsilon)$, to derive the desired formula for $M$:
$$M= \gamma - {1\over 2}\Sigma(2)-{1\over 3}\Sigma(3)-\cdots\,.\numeq$$
Since $\Sigma(s) = O\left( 2↑{-s} \right)$ this series converges rapidly;
the precise value of $M$ is 0.26149 72128 47643 ([Mertens 1874], [Knuth 76a]).

Returning to the question raised at the beginning of the section,
we find that the average number of distinct prime factors of $n$ can
be computed from the results above:
\baselineskip 15pt \eqalignno{ {1\over x}\sum_{p\le x} \left\lfloor {x\over p} \right\rfloor &= {1\over x}\sum_{p\le x}\left({x\over p} +O(1)\right)\cr &=\sum_{p\le x}{1\over p} + O\left(1\over \log x\right)\cr &=\ln \ln x +M+O\left(1\over \log x\right).&\anumeq\cr}

\vfill\eject
\major{Asymptotics from Generating Functions}

Frequently a combinatorial argument will produce a generating function,
$G(z)$, with interesting coefficients that have no simple closed form.
This section will address two popular techniques for obtaining the asymp\-totics
of these $g_n$ for large~$n$,
given $G(z)=\sum g_nz↑n$. The choice of technique
depends on the nature of $G(z)$: If $G(z)$ has singularities, then
a Darboux approach can use these singularities to obtain the asymptotics of $g_n$.
On the other hand, if $G(z)$ converges everywhere we employ the saddle point method
to find and evaluate a contour integral.

\minor{Darboux's Method}

When $G(z)$ converges in a circle of radius $R$,
the sum $\sum \left| g_n r↑n \right|$ converges
absolutely for $r<R$ and this is possible only if $g_n=O\left( r↑{-n} \right)$. This basic fact
about series suggests the following (somewhat idealized)
approach to the asymptotics of \penalty 999  $g_n$.
When $G(z)$ has a singularity at radius
$R$ we find a function $H(z)$ with well known coefficients that has the same
singularity. Then $G(z)-H(z)$ will often have a greater radius of convergence, $S$, and
$g_n$ will be well approximated by $h_n$:
$$g_n = h_n + O\left( s↑{-n} \right), \qquad s<S.\numeq$$
The process is repeated until $S$ is extended far enough to provide a small
error bound.

The method depends critically on finding a comparison function $H(z)$ with
well known coefficients. If we are attempting to cancel an ordinary pole
at $z=a$ in $G(z)$, then $H(z)$ is easy to construct since $G(z)$ will have the
Laurent form
$$G(z)={C_{-m}\over (z-a)↑m}+\cdots+{C_{-1}\over(z-a)}+ C_0+C_1(z-a)+\cdots\,.\numeq$$
For $H(z)$ we use the terms with negative powers of $(z-a)$ in the expansion:
$$H(z)={C_{-m}\over (z-a)↑m}+{C_{-m+1}\over (z-a)↑{m-1}}+\cdots+{C_{-1} \over (z-a)}.\numeq$$
The coefficients of $H(z)$ can be obtained with the binomial expansion of
$$(z-a)↑{-j}=(-a)↑{-j}\sum_k {-j\choose k} \left( -z\over a \right)↑k.\numeq$$
(See [Knuth III; pp.\ 41--42] for an illustration of Darboux's
technique applied to a function where the singularities are poles.)

Algebraic singularities are considerably harder to remove;
in fact we will only be able to improve'' the singularity
in a vague sense that will become clear shortly. By algebraic
we mean that $G(z)$ can be expressed as a finite sum of terms of the form
$$(z-a)↑{-w} g(z), \quad w \hbox{ complex},\quad g(z) \hbox{ analytic at a}. \numeq$$
For example,
$$\sqrt{@1-z} = \sum_n {1/2 \choose n} (-z)↑n \numeq$$
has an algebraic singularity at $z=1$, although in this case the function
also has a binomial expansion so that Darboux's method is unnecessary.
Darboux's technique will be illustrated with the function
$$G(z)=\sqrt{(1-z)(1-\alpha z)}, \qquad \alpha < 1. \numeq$$
(See [Knuth I; exercise 2.2.1--12].)
We need a comparison function that will attack the singularity at
$z=1$, so we first expand
$$\sqrt{@1- \alpha z} = \sqrt{@1-\alpha} + C_1 (1-z) + C_2 (1-z)↑2 + \cdots\,.\numeq$$
The first term of the expansion suggests choosing the comparison function
$$H(z)= \sqrt{@1-z}\,\sqrt{@1- \alpha}\,; \numeq$$
further terms of the expansion can be used to improve the estimate.
Let us see how well $H(z)$ performs by itself:
\eqalignno{G(z)-H(z)&=\sqrt{@1-z}\,\left(\sqrt{@1-\alpha z} -\sqrt{@1-\alpha}\,\right)\cr &=\alpha (1-z)↑{3/2}\,\left( 1\over \sqrt{@1-\alpha z}+\sqrt{@1-\alpha}↑{\mathstrut} \right)\cr &=A(z) \, B(z)&\anumeq\cr}
where
\eqalign{A(z)&=\alpha (1-z)↑{3/2}\cr B(z)&=1/\bigl(\sqrt{@1-\alpha z} + \sqrt{@1-\alpha}\,\bigr).\cr}\numeq
Note that we have not removed the singularity at $z=1$, but instead we
have improved'' the singularity from $(1-z)↑{1/2}$ to $(1-z)↑{3/2}$.
This im\-provement is strong enough to make $H(z)$ a good approximation
to~$G(z)$. The error is the coefficient of $z↑n$ in  $A(z) B(z)$.
The power series
$B(z)$ has a radius of convergence greater than 1, and so $b_n=O(r↑{-n})$
for some $r>1$. Furthermore $A(z)$ can be expanded,
$$A(z)=\alpha \sum_{n\ge 0} {3/2 \choose n}(-z)↑n =\alpha \sum_{n\ge 0}{n-5/2 \choose n} z↑n, \numeq$$
and this gives $a_n=\alpha {n-5/2 \choose n}=O\left(n↑{-5/2}\right)$.
To derive the error bound we proceed as in Section 4.1 to
split the convolution of $A(z)$ and $B(z)$ into two sums:
\eqalign{\sum_{0\le k\le n/2} a_k b_{n-k} &= O\left( r↑{-n/2} \right)\cr \sum_{n/2<k\le n} a_k b_{n-k}&= O\left(n↑{-5/2}\right).\cr}\numeq
Thus we may assert that
$$g_n= \sqrt{@1-\alpha}\, (-1)↑n{1/2@\choose n} + O\left( n↑{-5/2} \right).\numeq$$

In retrospect, our derivation of $g_n$ is simply an expansion of $G(z)$ about
$z=1$. The error term is tricky, but depends on increasing the exponent of
$(1-z)$ from $1/2$ to $3/2$. In fact a similar exponent dependency appears in the
statement of Darboux's theorem below. The notion of weight is introduced
and we improve'' the singularities by decreasing their weight:

\thbegin Theorem.
Suppose $G(z)=\sum_{n\ge 0} g_nz↑n$ is analytic near 0 and has only algebraic
singularities on its circle of convergence. The singularities, resembling
$$(1-z/\alpha)↑{-w} h(z), \numeq$$
are given weights equal to the real parts of their $w$'s. Let $W$ be the maximum
of all weights at these singularities. Denote by $\alpha_k$, $w_k$, and $h_k(z)$
the values of $\alpha$, $w$, and $h(z)$ for those terms of the form $\eq(4.116)$ of
weight $W$. Then
$$g_n={1\over n}\sum_k{h_k(\alpha_k)n↑{w_k}\over \Gamma(w_k)\alpha_k↑n}+ o\left(s↑{-n}n↑{W-1}\right), \numeq$$
where $s=\left| \alpha_k \right|$, the radius of convergence of $G(z)$, and
$\Gamma(z)$ is the Gamma function.

This version of Darboux's theorem, found in [Bender 74], gives the first term
of the asymptotics by diminishing all the heavy weight singularities. The process
can be repeated, resulting in the slightly more complicated statement of the
theorem found in [Comtet 74].
An ordinary pole corresponds to integer $w$ in the theorem,
in which case repeated application will eventually reduce $w$ to $0$,
eliminating the singularity completely, since the
values $w=0,-1,-2,\ldots$ are not singularities. The elementary method we have
illustrated in our analysis of \eq(4.108) is powerful enough to prove
Darboux's general theorem [Knuth~89].

\vskip 20pt
\minor{Residue Calculus}

The residue theorem states that the integral
around a closed curve in the complex plane
can be computed from the residues at the enclosed
poles:
$${1\over 2\pi i}\oint_C f(z)\,dz = \sum \hbox{(residues at enclosed poles)}. \numeq$$
Here the residue of $f(z)$ at $a$ is defined to be the coefficient $C_{-1}$
in the Laurent expansion of $f(z)$:
$$f(z)={C_{-m}\over(z-a)↑m}+\cdots+ {C_{-1}\over (z-a)} + C_0 +C_1(z-a)+\cdots\,. \numeq$$
Residues are relatively easy to compute. If $m=1$ then the pole is first order
and the residue is given by
$$C_{-1}=\lim_{z\to a} (z-a)f(z).\numeq$$
The limit usually succumbs to repeated application of l'Hospital's rule:
$$\lim_{z\to a}{g(z)\over h(z)} = \lim_{z\to a}{g↑\prime(z)\over h↑\prime(z) }. \numeq$$
A pole is considered to be
of order $m$ if $\lim_{z\to a} (z-a)↑m f(z)$ is a nonzero constant, but this
limit does not tell us anything about the residue when $m>1$.
Differentiation can be used to isolate the correct coefficient:
$$C_{-1}={1\over (m-1)!} \lim_{z\to a} {d↑{@m-1}\over dz↑{m-1} } \left( (z-a)↑mf(z) \right).\numeq$$
In practice, however, it is often faster to deduce the behavior
of $f(z)$ near $z=a$ by substituting $z=a+w$ and expanding in
powers of~$w$, then to obtain the coefficient of $w↑{-1}$
by inspection.

Traditionally the residue theorem is given as an easy way to compute the integral
on a closed curve. In asymptotics
we often use the formula backwards, placing the combinatorial quantity
of interest in the residue and then evaluating the integral.

For example, suppose we have a double generating function,
$$F(w,z)=\sum_{m,n} a_{mn}w↑mz↑n,\numeq$$
and we wish to compute a generating function for the diagonal elements,
$$G(z)=\sum_{n} a_{nn} z↑n. \numeq$$
Terms with $n=m$ are moved to the coefficient of $t↑{-1}$, where
they become the residue:
\eqalignno{ {1\over 2\pi i}\oint F(t,z/t)\,{dt\over t} &= {1\over 2\pi i}\oint \left( \sum_{m,n} a_{mn} t↑m \left(z\over t\right)↑n\right) \,{dt\over t}\cr &={1\over 2\pi i} \sum_{m,n} \oint a_{mn} t↑{m-n} z↑{n} {dt\over t}\cr &=\sum_n a_{nn}z↑n = G(z).&\anumeq\cr}
This interchange of summation and integration is legitimate if the series converges uniformly,
so the path of integration
must be chosen to make both $\left| t \right|$ and $\left| z/t \right|$ sufficiently small.

A classic illustration of the diagonalization of power series begins with
$$F(w,z)=\sum_{m,n\ge 0}{m+n\choose n}\,w↑mz↑n ={1\over 1-w-z}.\numeq$$
We seek an expression for the generating function
$$G(z)=\sum_n {2n\choose n} z↑n.\numeq$$
Using the formula derived above,
$$G(z)={1\over 2\pi i}\oint_C {dt\over (1-t-z/t)t}. \numeq$$
If $C$ is chosen to be a small curve around the origin, it encloses
the first order pole at
$$t={1-\sqrt{@1-4z} \over 2}. \numeq$$
Here the residue is $(1-4z)↑{-1/2}$ so the value of the integral is
$$G(z)= {1\over \sqrt{@1-4z}↑{\mathstrut}}.\numeq$$

For a second illustration of diagonalization
consider the problem of obtaining the termwise product of two power
series,
$$A(z)=\sum a_nz↑n, \quad \hbox{and} \quad B(z)=\sum b_nz↑n. \numeq$$
Using the result derived in \eq(4.125), we obtain the Hadamard product:
$$G(z)=\sum a_nb_nz↑n ={1\over 2\pi i}\oint A(t)B\left( z\over t \right) {dt\over t}. \numeq$$

\vskip 20pt

Our next example
makes use of several standard techniques that de\-serve attention
before we begin the actual problem. Initially, we will use the
residue theorem backwards:
$$g_n={1\over 2\pi i}\oint {G(z) dz\over z↑{n+1}}. \numeq$$
A generating function $G(z)$ is given,
and we assume that it is free of singularities
(otherwise a Darboux attack would provide the asymptotics) so that
the only constraint on the path of integration is that it encloses
the origin. A wise choice for this path allows the integral to
be easily estimated, and a good heuristic for choosing paths
is the saddle point method. The idea is to run the path of
integration through a saddle point, which is defined to be a place
where the derivative of the integrand
is zero. Like a lazy hiker, the path then crosses the
ridge at a low point; but unlike the hiker, the best path takes
the steepest ascent to the ridge. In fact, for our purposes, this property
is far more important than crossing the ridge at the lowest point.

Once we have chosen a path of integration another technique,
Laplace's method for integrals, is frequently helpful.
The integral will be concentrated in a small interval, but will include
negligible tails extending over the whole region. Laplace's method removes
these tails and replaces them by a different small function that is
convenient for the evaluation of the integral. Both the old tails
and the new tails must be shown to be insignificant to the result of the
evaluation.

As an example problem, we will derive a strong version of the central
limit theorem, which states that the mean of a large number of drawings
from an arbitrary distribution is normally distributed:
$$\twolinenumeq{ \hbox{Prob} \left(\mu-{\alpha \sigma\over \sqrt{n} } < {X_1+X_2+\cdots+X_n\over n} < \mu+{\beta \sigma \over \sqrt{n} } \right) =}{4pt}{{1\over \sqrt{2\pi }}\int_{-\alpha}↑\beta e↑{-z↑2/2}dz \left( 1+O(n↑{-1}) \right).}$$
Here the $X_i$ are arbitrary but identically distributed
random variables with  mean $\mu$ and
standard deviation $\sigma$. With several minor restrictions we can
prove an even stronger result, clarifying exactly how fast an
arbitrary integer-valued random variable converges to the normal
distribution.

\noindent
{\bf Assumption 1.} The $X_i$ are drawn from an integer-valued distribution
with gener\-ating function $g(z)$:
$$g(z)=\sum_{k\ge 0} p_kz↑k, \qquad p_k=\hbox{Prob}(X_i=k).\numeq$$
We assume that $g(z)$ is analytic for $\left| z \right| < 1+\delta$, and
since $g(1)=1$ for a probability distribution, we may conclude that
$\ln g(e↑t)$ is analytic at $t=0$. This allows us to characterize $g(z)$
by its Thiele expansion,
$$g(e↑t)=\exp\left(\mu t +{\sigma↑2 t↑2\over 2!}+{\kappa_3t↑3\over 3!}+ {\kappa_4t↑4\over 4!}+\cdots\, \right). \numeq$$
where $\kappa_j$ is the $j\,$th semi-invariant of $g(z)$.

\noindent {\bf Assumption 2.} $g(0)$ must be nonzero,
that is $p_0 \not= 0$. This is not a restriction, since we
can translate the generating function to $z↑{-m}g(z)$, where $p_m$ is the
first nonzero coefficient.

\noindent {\bf Assumption 3.} The greatest common
divisor of all $k$ with $p_k \not = 0$ must be 1.
This is also not a restriction since we may analyze $g(z↑{1/m})$, where
$m$ is the greatest common divisor of the $k$ such that $p_k \not= 0$.

\goodbreak
The sum of $n$ drawings from $g(z)$ has distribution $g(z)↑n$. We wish
to understand the behavior of this sum near its mean, $\mu n$, so we
define
$$A_{n,r} = \hbox{coefficient of } z↑{\mu n +r} \hbox{ in } g(z)↑n,\numeq$$
where $r$ is chosen to make $\mu n + r$ an integer. By the residue theorem,
$$A_{n,r}={1\over 2\pi i}\oint {g(z)↑n dz\over z↑{\mu n+r+1} }. \numeq$$
The saddle point is near $z=1$, so we choose a path of integration
with radius $1$ enclosing the origin, and substitute $z=e↑{it}$:
$$A_{n,r}={1\over 2\pi }\int_{-\pi }↑\pi {g(e↑{it})↑n dt\over e↑{it(\mu n +r)}}.\numeq$$
Assumption 3 implies that the terms of $g(e↑{it})$ will all be in phase
only when $t=0$, so $\left| g(e↑{it}) \right| < 1$, except when $t=0$, in which case
$g(1)=1$. Raising $g(e↑{it})$ to the $n$th power makes the tails
of the integral exponentially small, and leaves the primary contribution
at $t=0$. In particular, whenever we choose a $\delta > 0$ there exists
an $\alpha \in [0,1)$ such that $\left| g(e↑{it}) \right| < \alpha$
for ${\delta \le \left| t \right| \le \pi }$, and this means that
$$A_{n,r} = {1\over 2\pi }\int_{-\delta}↑\delta {g(e↑{it})↑n dt\over e↑{it(\mu n +r)}} + O(\alpha↑n). \numeq$$

Laplace's technique suggests that we chop off the tails and replace them
with more agreeable functions. We will make three passes at the
present tails, refining the interval each time, before adding new tails.
First we set $\delta_1$ small enough so that the Thiele expansion for $g(e↑t)$
is valid, and then expand
$${g(e↑{it})↑n\over e↑{it (\mu n+r)}}=\exp\left( -irt-{\sigma↑2t↑2n\over 2!} -{i\kappa_3t↑3n\over 3!}+{\kappa_4t↑4n\over 4!}+\cdots\, \right).\numeq$$
Next we set $\delta_2$ smaller than $\delta_1$ so that the first two
terms in the expansion dominate the remaining terms:
$$\left| -{i\kappa_3t↑3n\over 3!}+{\kappa_4t↑4n\over 4!}+\cdots\, \right| < {\sigma↑2 \left| t↑2 \right| n\over 6},\qquad \hbox{for |t| < \delta_2.}\numeq$$
These first two refinements permit a third refinement from
$[-\delta_2,\delta_2]$ to $[-n↑{-1/2+\epsilon}, n↑{-1/2+\epsilon}]$.
(The role of the mysterious epsilon will become apparent shortly.)
The error introduced by this refinement is the sum of two terms like
$$\int_{n↑{-1/2+\epsilon}}↑{\delta_2} \left|\exp\left(-irt-{\sigma↑2t↑2n\over 2!} \right) \right| \cdot \left|\exp\left(-{i\kappa_3t↑3n\over 3!}+{\kappa_4t↑4n\over 4!}+\cdots\,\right) \right|dt.\shiftnumeq-6pt$$
\smallskip\noindent
The $-irt$ in the first term contributes an irrelevant phase, and the second
term is bounded by equation \eq(4.142), so the error is exponentially small:
$$\int_{n↑{-1/2+\epsilon}}↑{\delta_2} \exp\left( -{\sigma↑2t↑2n\over 2 }+{\sigma↑2t↑2n\over 6}\right)\, dt \le \delta_2 e↑{-\sigma↑2n↑{2\epsilon}/3}.\numeq$$
The reason for choosing $n↑{-1/2+\epsilon}$ should now be clear. In the last
step we bounded the integral by its largest value, substituting $n↑{-1/2+\epsilon}$
for $t$ in the integrand. The $n↑{-1/2}$ exactly cancels the $n$ associated with
$t↑2$ in the integrand, so $\epsilon$ becomes the straw that breaks the camel's
back'' and drives the integral to zero.

We can summarize the progress so far by claiming that there exists an
$\alpha\in(0,1)$ such that, for all $\epsilon > 0$,
$$A_{n,r}={1\over 2\pi }\int_{-n↑{-1/2+\epsilon}}↑{n↑{-1/2+\epsilon}} {g(e↑{it})↑ndt\over e↑{it(n\mu + r)} } +O\left(\alpha↑{n↑{2\epsilon}}\right). \numeq$$
Within such a small interval, the first two terms of the Thiele expansion
are of principal importance:
$$A_{n,r}={1\over 2\pi }\int_{-n↑{-1/2+\epsilon}}↑{n↑{-1/2+\epsilon}} \exp \left(-irt-{\sigma↑2 t↑2 n\over 2}+O\left(n↑{3\epsilon-1/2}\right)\right)dt +O\left(\alpha↑{n↑{2\epsilon}}\right). \shiftnumeq-1pt$$
\smallskip\noindent
At this point we are ready to add new tails to the integral, using the
first two terms of the Thiele expansion as a convenient function.
The new tails are exponentially small,
$$\twoline{\left| \int_{n↑{-1/2+\epsilon}}↑\infty \exp\left(-irt-{\sigma↑2 t↑2 n\over 2}\right) dt\, \right|}{5pt}{ \le \int_{n↑{-1/2+\epsilon}}↑\infty \exp\left(-\sigma↑2 t↑2 n\over 2\right)dt =O\left( \exp \left(-\sigma↑2n↑{2\epsilon} \over 2 \right)\right).}$$
It is then an easy matter to evaluate the integral on $[-\infty ,\infty ]$ by completing the
square in the exponent:
\eqalignno{A_{n,r}&={1\over 2\pi }\int_{-\infty }↑\infty \exp \left(-irt-{\sigma↑2 t↑2 n\over 2}\right) \left(1+O\left(n↑{3\epsilon-1/2}\right)\right) \,dt +O\left(\alpha↑{n↑{2\epsilon}}\right) \cr &={1\over \sigma \sqrt{2\pi n}}\exp \left( -r↑2\over 2 \sigma↑2 n \right) +O\left(n↑{3\epsilon -1}\right).&\anumeq\cr}
Note that the constant implied by this $O$ depends on $g(t)$ and $\epsilon$
but not on $n$ or $r$.

We can improve this result by using more of the Thiele expansion
in equation \eq(4.146). This will require integrating terms like
$$\int_{-\infty }↑\infty e↑{iat-bt↑2}t↑k dt. \numeq$$
Completing the square in the exponent and expanding $t↑k$ with the
binomial theorem will lead to still more terms of the form
$$\int_{-\infty }↑\infty e↑{-u↑2} u↑j du. \numeq$$
For odd $j$ the integral vanishes, and for even $j$ it can be transformed
to the Gamma function by the substitution $v=u↑2$.
With this machinery we can extend our estimate, obtaining for example
$$A_{n,r}={1\over \sigma \sqrt{2\pi n}}\exp \left( -r↑2\over 2 \sigma↑2 n \right) \biggl(1-{\kappa_3\over 2\sigma↑4}\left(r\over n\right) +{\kappa_3\over6\sigma↑6}\left(r↑3\over n↑2\right)\biggr)+O(n↑{-3/2}).\shiftnumeq4pt$$
The coefficient of the general term in the expansion, $r↑R\!/n↑N$, is given by
$$\smash{ \sum_{S\ge 0}{(-1)↑S (R+2S)↑{\underline{2S}} \over \sigma↑{2(R+S)}2↑S S!}\!\!\! \sum_{\scriptstyle p_3+p_4+p_5+\cdots\,=R-N+S\atop \scriptstyle 3p_3+4p_4+5p_5+\cdots\,=R+2S} {1\over p_3!}\left(\kappa_3\over 3!\right)↑{p_3} {1\over p_4!}\left(\kappa_4\over 4!\right)↑{p_4}\!\!\!\ldots;}\numeq$$
\bigskip\noindent
such terms are present for $0\le R\le {3\over2}N$.

This is a very strong result;
the central limit theorem follows immediately by summing
on $r$, and if necessary, we have a detailed understanding of
the asymptotic behavior of individual terms of the distribution.
Unfortunately the formula above suffers from a weakness that is common to
central limit theorems: Its range is limited.
Note that since the error term is polynomial in $n$, the estimate
of $A_{n,r}$ is useful only when $r=O\left(\sqrt{n}\,\right)$.
This is not surprising, since we were sloppy in our choice of the
path of integration; it goes through the lowest portion of the saddle point
only when $r=0$, and becomes progressively worse for larger $r$.
The obvious remedy would be to change the path of integration when $r$
exceeds $\sqrt{n}$. However, we will see in a moment that
the distribution itself can be shifted.
This often proves to be easier than repeating the derivation with a
different path of integration, although both techniques are essentially the same.

The coefficient of $z↑m$ in $g(z)↑n$ can be obtained with the formula
$$[z↑m]\, \left( g(z)↑n \right) = {g(\alpha)↑n \over \alpha↑m} \,[z↑m]\, \left( g(\alpha z)\over g(\alpha) \right)↑n. \numeq$$
The right side of the equation seems like an unnecessary complication
of the left side, since both require extracting the coefficient of $z↑m$;
but in fact the right side has an extra degree of freedom represented by
$\alpha$, which allows us to shift the mean of the distribution to a value
close to $m/n$. We do this by choosing $\alpha$ so that
$${\alpha g↑\prime(\alpha) \over g(\alpha)} = {m\over n}. \numeq$$

Take, for a simple example, the problem of finding the coefficient of $z↑{n/3}$
in the binomial distribution with parameter 1/2,
$$[z↑{n/3}]\, \left(1+z\over 2\right)↑n.\numeq$$
The coefficient of interest is at a distance $\gg \sqrt n$ from the mean, $n/2$,
so equation \eq(4.150) is useless until we shift the mean to $n/3$ by appropriate
choice of $\alpha$:
$${\alpha g↑\prime(\alpha) \over g(\alpha)} = {\alpha \over 1+\alpha} ={1\over 3},\numeq$$
$$\alpha = {1\over 2}. \numeq$$
The new distribution,
$$\textstyle \left( {2\over 3} + {1\over 3} z \right), \numeq$$
has mean $\mu = 1/3$, standard deviation $\sigma = \sqrt 2 /3$,
$\kappa_3=2/27$, and $\kappa_4=-2/27$.
We apply \eq(4.150) and \eq(4.151) with $r=0$, obtaining
\eqalignno{[z↑{n/3}]\, \textstyle \left( {2\over 3}+{1\over 3} z\right)↑n &=A_{n,0}\cr\noalign{\vskip 3pt} &= {3\over 2 \sqrt{\pi n} }\left(1-{7\over 24n}\right)+O(n↑{-5/2}).&\anumeq\cr}
Multiplying by the $g(\alpha)↑n/\alpha↑m$ factor found in equation \eq(4.152)
gives a solution to the original problem of estimating the probability
of exactly $n/3$ heads appearing in $n$ tosses of a fair coin:
$$2↑{n/3}\,\left(3\over 4\right)↑n {3\over 2 \sqrt{\pi n}} \left(1-{7\over 24n}+O(n↑{-2})\right).\numeq$$

Lest the reader be left with the impression that shifting the mean
is a panacea for all range problems, several difficulties should be
mentioned. In equation \eq(4.155) we were fortunate to find $\alpha$ constant.
In general $\alpha$ will have some dependency on $n$, which in turn will
make the mean, standard deviation, and other semi-invariants dependent on $n$.
Our derivation of a strong version of the central limit theorem made no allowance
for this dependency, and must be reworked to accommodate the specific problem.
In particular, the application of Laplace's method (shaving the tails of
the integral and adding new tails) is likely to be affected by the
new variations. Nevertheless, shifting the mean is still useful as a clear guide
for the asymptotic derivation, and the reader will find it interesting to
derive asymptotic formulas for Stirling numbers in this way.

\vskip 20pt

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\def \bibex #1 #2 #3 #4 #5{\par\vskip 5pt plus2pt minus1pt\vbox{\halign{\hbox to 70pt{##\hfill}&##\hfil\cr
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{The Expected Linearity of a Simple Equivalence Algorithm}
{{\sl Theoretical Computer Science\/} 6(3):281--315, 1978}

\bib{[Knuth 89]}
{Knuth, D. and Wilf, H.}
{A Short Proof of Darboux's Lemma}
{{\sl Applied Mathematics Letters\/} 2(2):139--140, 1989}

\bib{[Lueker 80]}
{Lueker, G. S.}
{Some Techniques for Solving Recurrences}
{{\sl Computing Surveys\/} 12(4):419--436, 1980}

\bibex{[Mertens 1874]}
{Mertens, F.}
{Ein Beitrag zur analytischen Zahlentheorie}
{{\sl Journal f\"ur die reine und angewandte Mathematik\/}}
{78:46--62, 1874}

\bib{[Mil-Thom 33]}
{Milne-Thomson, L. M.}
{\sl The Calculus of Finite Differences}
{Macmillan, 1933}

\bib{[Odlyzko 88]}
{Odlyzko, A. and Wilf, H.}
{The Editor's Corner: $n$ Coins in a Fountain}
{{\sl The American Mathematical Monthly\/} 95(9):840--843, 1988}

\bib{[Olver 74]}
{Olver, F. W. J.}
{\sl Asymptotics and Special Functions}

\bib{[Page 79]}
{Page, E. and Wilson, L.}
{\sl An Introduction to Computational Combinatorics}
{Cambridge University Press, 1979}

\bib{[Riordan 68]}
{Riordan, J.}
{\sl Combinatorial Identities}
{John Wiley and Sons, 1968}

\bibex{[Rota 75]}
{Rota, G.}
{with Doubilet, Greene, Kahaner, Odlyzko, and Stanley}
{\sl Finite Operator Calculus}

\bibex{[Sedgewick 75]}
{Sedgewick, R.}
{\sl Quicksort}
{Ph.D. Dissertation, Stanford, 1975}
{Garland Publishing, 1980}

\bib{[Spiegel 71]}
{Spiegel, M.}
{\sl Calculus of Finite Differences and Difference Equations}
{Schaum's Outline Series, McGraw-Hill, 1971}

\bibex{[Stolarsky 77]}
{Stolarsky, K. B.}
{Power and Exponential Sums of Digital Sums}
{Related to Binomial Coefficient Parity}
{{\sl SIAM Journal on Applied Mathematics\/} 32(4):717--730, 1977}

\bib{[Whittaker 40]}
{Whittaker, E. T. and Watson, G. N.}
{\sl A Course of Modern Analysis}
{Cambridge, 1940}

\bib{[Zave 76]}
{Zave, D. A.}
{A Series Expansion Involving the Harmonic Numbers}
{{\sl Information Processing Letters\/} 5(1):75--77, 1976}

\vfill\eject
\pschap{Appendices}
\psmajor{Appendix A: Schedule of Lectures, 1980}
\mark{lectures}

\parskip 1pt

\def\inafter #1 #2{\par\vskip 14pt\noindent \hbox to 40pt{#1\hfill}#2}
\def\ins #1{\par\noindent \hskip 40pt #1}
\def\insi #1{\par\vskip 3pt\noindent \hskip 40pt {\sl #1}}

\inafter{1 \&\ 2}
{Analysis of an {\sl in situ\/} permutation algorithm.}
\insi{Ref: [Knuth 71]}

\inafter{3}
{Permutations with $k$ inversions.}
\ins{Generating skewed distributions.}
\ins{(D. Greene, lecturer)}
\insi{Ref: [Knuth III; 5.1.1--14 and 5.1.1--18]}

\inafter{4 \&\ 5}
{Analysis of insertion sort and Shell's sort.}
\insi{Ref: [Knuth III; 5.2.1]}

\inafter{6}
{The principle of postponed information (late binding).}
\ins{Dijkstra's algorithm for shortest paths.}
\ins{Quicksort.}
\insi{Ref: [Knuth 76c], [Knuth III; 5.2.2]}

\inafter{7 \&\ 8}
{Quicksort.}
\insi{Ref: [Knuth III; 5.2.2], [Sedgewick 75]}

\inafter{9 \&\ 10}
{Paterson's technique for hashing analysis.}
\insi{Ref: Chapter 3 \ Operator Methods}

\inafter{11}
{Ordered hash tables.}
\insi{Ref: [Amble 74]}

\inafter{12}
{Recurrence relations with minimization.}
\insi{Ref: Section 2.2.1 \ Relations with Max or Min Functions}

\inafter{13}
{Introduction to asymptotics.}
\insi{Ref: Section 4.1 \ Basic Concepts}

\inafter{14}
{The use of Stieltjes integration in asymptotics.}
\insi{Ref: Section 4.2 \ Stieltjes Integration}

\inafter{15}
{Mellin transforms and the Gamma function technique.}
\ins{(L. Ramshaw, lecturer)}
\insi{Ref: [Knuth III, 129--134] and}
\par\noindent\hskip 40pt {\sl work in progress by L. Guibas, L. Ramshaw, and R. Sedgewick} % that work was never published ---L Ramshaw, Jan 90

\inafter{16}
{Stieltjes integration applied to a sum of reciprocal primes.}
\insi{Ref: Section 4.2.3 \ An Example from Number Theory}

\inafter{17}
{Introduction to residue calculus.}
\ins{Darboux's approach to generating functions with singularities.}
\insi{Ref: Section 4.3 \ Asymptotics from Generating Functions}

\inafter{18}
{Saddle points and Laplace's method for obtaining asymptotics.}
\ins{(D. Greene, lecturer)}
\insi{Ref: Section 4.3.3 \ The Saddle Point Method}

\inafter{19 \&\ 20}
{The Hungarian method.}
\insi{Ref: [Erd\H os 59], [Erd\H os 60], [Knuth 76c]}

\vfill\eject
\psmajor{Appendix B: Homework Assignments}
\mark{homework assignments}

The homework problems and their solutions appear in [Knuth III].

\inafter{1}
{5.1.1--8 [M 24]}
\ins{5.1.1--15 [M 23]}
\ins{5.1.1--16 [M 25]}

\noindent
Show that permutations obtainable with a stack'' (namely $a_1\,a_2 \ldots a_n$
of $\{1, 2, \ldots, n\}$ where $i<j<k \Rightarrow \lnot\, (a_j<a_k<a_i)$, see exercise 2.2.1--5)
can be characterized in terms of inversion tables. Find and prove a simple
property of the inversion table $C_1\,C_2\ldots C_n$ that holds if and only if the permutation
is obtainable with a stack. (Note: This was intended to be exercise 5.1.1--21,
and in fact [Knuth III] contains the answer but not the exercise!)

\inafter{2}
{5.2.1--5 [M 27]}
\ins{5.2.1--14 [M 24]}
\ins{5.2.1--37 [M 25]}

\inafter{3}
{5.2.2--7 [M 28]}
\ins{5.2.2--14 [M 21]}
\ins{5.2.2--20 [M 20]}
\ins{5.2.2--22 [M 25]}

\inafter{4}
{6.2.1--25 [M 25]}
\ins{6.2.2--6 [M 26]}
\ins{6.2.2--7 [M 30]}

\inafter{5}
{6.4--27 [M 27]}
\ins{6.4--34 [M 22]}
\ins{6.4--49 [HM 24]}

\inafter{6}
{5.1.4--31 [HM 30]}
\ins{5.2.2--57 [HM 24]}

\inafter{7}
{5.1.3--10 [HM 30]}
\ins{5.2.2--54 [HM 24]}

\inafter{8}
{6.3--34 [HM 40]}

\vfill\eject
\psmajor{Appendix C: Midterm Exam I and Solutions}
\psminor{Midterm Exam I}
\noindent Problem 1.  (a) [10 points]  How many permutations on
$\{1,2,\ldots,n\}$ are sorted by at most two bubble sort'' passes?
Example:
\vbox{\halign{\hfil# &\xskip \hfil#\hfil &\xskip \hfil#\hfil &\xskip \hfil#\hfil &\xskip \hfil#\hfil &\xskip \hfil#\hfil &\xskip \hfil#\hfil &\xskip \hfil#\hfil &\xskip \hfil#\hfil &\xskip \hfil#\hfil\cr given&3&1&2&9&6&4&5&8&7\cr first pass&1&2&3&6&4&5&8&7&9\cr second pass&1&2&3&4&5&6&7&8&9\cr}}
(A bubble-sort pass interchanges $K_j \swap K_{j+1}$ iff $K_j>K_{j+1}$
for $j$ running from $1$ up to $n-1$.)

\yskip
\noindent (b) [40 points]  How many permutations on $\{1,2,\ldots,n\}$ are
sorted by one double-pass of the cocktail-shaker sort''?  Example:
\vbox{\halign{\hfil# &\xskip \hfil#\hfil &\xskip \hfil#\hfil &\xskip \hfil#\hfil &\xskip \hfil#\hfil &\xskip \hfil#\hfil &\xskip \hfil#\hfil &\xskip \hfil#\hfil &\xskip \hfil#\hfil &\xskip \hfil#\hfil\cr given&2&7&3&1&4&6&9&8&5\cr left-to-right pass&2&3&1&4&6&7&8&5&9\cr right-to-left-pass&1&2&3&4&5&6&7&8&9\cr}}
(The cocktail shaker sort alternates between bubble-sort passes and similar
passes in which $j$ goes {\sl down} from $n-1$ to $1$.)

\yyskip
\noindent Problem 2.  Dave Ferguson's scheme for representing binary trees
[exercise 2.3.1--37] would store the binary search tree
$$\unitlength=24pt \beginpicture(6,5)(0,0) \put(0,2){\circle1} \put(0,2){\makebox(0,0){a}} \put(2,4){\circle1} \put(2,4){\makebox(0,0){b}} \put(4,2){\circle1} \put(4,2){\makebox(0,0){c}} \put(6,0){\circle1} \put(6,0){\makebox(0,0){d}} \put(1.646,3.646){\vector(-1,-1){1.292}} \put(2.354,3.646){\vector(1,-1){1.292}} \put(4.354,1.646){\vector(1,-1){1.292}} \endpicture$$
\vfill\eject

\noindent in five locations, e.g., as follows:

\vskip 20pt
\noalign{\vskip 5pt}
1&b&2\cr
2&a&$\Lambda$\cr
3&c&4\cr
4\cr
5&d&$\Lambda$\cr}}\hfill}}
\vskip 20pt

\noindent The standard tree search and insertion procedure [Algorithm 6.2.2T] can
obviously be adapted to use this representation.

Let $p_{nk}$ be the probability that a binary search tree, constructed by
starting with the empty tree and inserting $n$ randomly ordered keys, will
occupy $2n+1-2k$ locations under Ferguson's scheme; and let
$P_n(z)=\sum_{k\ge 0}p_{nk}z↑k$ be the corresponding generating function.
For example, we have $P_1(z)=P_2(z)=z$ and $P_3(z)={2\over 3}z+{1\over 3}z↑2$.

\yskip
\noindent (a) [10 points]  Find a differential operator $\Phi_n$ such that
$P_{n+1}(z)=\Phi_nP_n(z)$ for all $n\ge 1$.

\yskip
\noindent (b) [15 points]  Let $D$ be the operator $d/dz$, and let $U$ be
the operator that sets $z=1$, so that $U\!DP_n(z)=P_n↑{\prime}(1)$ is the
mean value of $k$.  Show that this mean value can be expressed as a simple
function of $n$.

\yskip
\noindent (c) [25 points]  Extending the result of (b), find the variance
of $k$ as a function of $n$.

\yyskip
\noindent Problem 3.  [100 points]

\def\ty #1{\hbox{\tt #1}\xskip}
\def\br #1{\hbox{$\langle$#1$\rangle$}\xskip}

Consider an electric typewriter that has exactly 40 keys
and an infinite carriage. The keys are:
$$\;\ty{a}\ty{b}\ty{c}\ldots\xskip\ty{0}\ty{1}\ty{2}\ldots\xskip\ty{9} \br{period}\br{space}\br{backspace}\br{carriage-return}$$
A monkey types at random, starting at the beginning of a line, until hitting
$\langle$carriage-return$\rangle$ for the first time; this scares him, so he
runs off to eat a banana.

\yskip
\noindent (a) Determine the generating function $G(z)=\sum_{n\ge 0}g_nz↑n$,
where $g_n$ is the number of keystroke sequences of length $n$ that leave
the word {\tt ape}'' at the beginning of the line (and no other nonblank symbols).

For example, here is one such sequence of length 12 ($\langle$bs$\rangle$ stands for
backspace):
$$\br{space}\ty{p}\br{bs}\br{bs}\br{bs}\ty{a}\br{space}\ty{e}\br{bs}\br{bs} \ty{p}\br{carriage-return}$$
(Note that $\langle$backspace$\rangle$ at the beginning of a line has no
effect, and characters may be overstruck.)

\noindent $\Rightarrow$ You need not display $G(z)$ explicitly; it suffices
to specify equations that determine this function uniquely.

\yskip
\noindent (b) What is the probability that the monkey types {\tt ape}'' in this
way?  (If possible, give your answer as an explicit real number, and explain
how you got it.)

\yskip
\noindent[In case you dislike working on this problem, you might contemplate the
probability that the monkey types certain FOUR-letter words.]

\yyskip
\noindent{\bf Solutions to Midterm Exam I}
\par\vskip .2\vu

\def\<{$\langle$} \def\>{$\rangle$}
\def\.#1{\hbox{\tt #1}}

\rm

\pssub{Solution to Problem 1.}
(a) According to [Knuth III; page 108], we want to count how many inversion tables
$b_1\ldots b_n$ have all $b_j\le 2$; this is clearly $3↑{n-2}{\cdot}2$,
for $n\ge 2$.

(b) Call the inversion table $b_1\ldots b_n$ {\sl easy} if a cocktail-style
double-pass will sort the corresponding permutation. It turns out that there is
a fairly nice way to characterize such inversion tables: $b_1\ldots b_n$ is
easy if and only if it is a valid inversion table such that either
\baselineskip 14pt\vbox{\halign{\hfill# &#\hfill\cr \noalign{\smallskip} &b_1=0 and b_2\ldots b_n is easy,\cr or&b_1=1 and b_2\ldots b_n is easy,\cr or&b_1=2 and b_2\le 1 and b_3\ldots b_n is easy,\cr or&b_1=3 and b_2\le 1 and b_3\le 1 and b_4\ldots b_n is easy,\cr or&\ldots\cr or&b_1=n-1 and b_2\le 1 and \ldots and b_{n-1}\le 1 and b_n is easy.\cr}}
[{\sl Outline of proof:\/} Suppose $b_1=k>0$. After one left-to-right pass,
there are $k-1$ inversions of element 1, and at this stage the permutation
must begin with
$2\ldots k\,1$ if it is to be sorted in one right-to-left pass.]

We now find that the number of easy permutations for $n\ge 2$ satisfies
$$x_n=x_{n-1}+x_{n-1}+2x_{n-2}+4x_{n-3}+\cdots$$
where we set $x_1=1$ and $x_j=0$ for $j\le 0$. It follows that ${1\over2}x_{n+1}-x_n =x_n-x_{n-1}$, i.e., $x_{n+1}=4x_n-2x_{n-1}$. The solution to this linear
recurrence is $x_n={1\over2}\bigl((2+\sqrt2\,)↑{n-1}+(2-\sqrt2\,)↑{n-1}\bigr)$.

Another solution appears in [Knuth III; exercises 5.4.8--8, 9].
\pssub{Solution to Problem 2.}
If there are $k$ childless nodes in the tree,
Ferguson's scheme requires $2n+1-2k$ locations: one for the root and
$2(n-k)$ for the children of nodes. Going from $P_n$ to $P_{n+1}$, the value
of $k$ is unchanged if the new node replaces one of the $2k$ children of the
$k$ childless nodes, otherwise it increases by 1; hence
$$P_{n+1}(z)=\sum_k p_{nk}\,\biggl({2k\over n+1}z↑k+\biggl((1-{2k\over n+1}\biggr)\, z↑{k+1}\biggr)\,.$$

{\baselineskip 14pt\noindent
The corresponding differential operator is $\Phi_n=z+{2\over n+1}z(1-z)D$; we have
$$P_n(z)=\Phi_{n-1}\ldots \Phi_0P_0(z)\qquad\hbox{where P_0(z)=1.}$$
To get the mean $x_n$, we note that
\eqalign{D\Phi_n&=\textstyle1+zD+{2\over n+1}\bigl((1-2z)D+z(1-z)D↑2\bigr),\cr U\!D\Phi_n&=\textstyle U+U\!D+{2\over n+1}(-U\!D)=U+{n-1\over n+1}U\!D.\cr}
Hence $x_{n+1}=U\!DP_{n+1}(z)=(U+{n-1\over n+1}@U\!D)P_n(z)=1+{n-1\over n+1}x_n$, and
this recurrence has the solution $x_n=(n+1)/3$ for $n\ge 2$.

Similarly, to get the variance we find
$$U\!D↑2\Phi_n=\textstyle 2{n-1\over n+1}@\!D+{n-3\over n+1}U\!D↑2.$$
Let $y_n=P_{\!n}↑{\prime\prime}(1)$, so that
$$y_{n+1}=\textstyle2{n-1\over n+1}x_n+{n-3\over n+1}y_n={2\over3}(n-1)+ {n-3\over n+1}y_n.$$
Applying a summation factor as on page 18,
we get $z_{n+1}=(n+1)↑{\underline4}y_{n+1}= {2\over3}(n+1)↑{\underline4}(n-1)+n↑{\underline4}y_n={2\over3}(n+1)↑{\underline5} +{4\over3}(n+1)↑{\underline4}+z_n$ and $z_3=0$. Therefore $z_n={2\over18}(n+1)↑ {\underline6}+{4\over15}(n+1)↑{\underline5}$ and $y_n={2\over18}(n+1)(n-4)+ {4\over15}(n+1)$ for $n\ge 4$. The variance is $y_n+x_n-x_n↑2=(n+1)\bigl({2\over18} (n-44)+{4\over15}+{1\over3}-{1\over9}(n+1)\bigr)={2\over45}(n+1)$ for $n\ge 4$.

}\vskip 6pt
[{\sl Note:\/} A completely different approach could also be used to get the mean
and variance, using what might be called induction at the other end.' By
considering the various choices of root nodes, we have the recurrence
$$P_n(z)={1\over n}\bigl( P_0(z)P_{n-1}(z)+P_1(z)P_{n-2}(z)+\cdots+ P_{n-1}(z)P_0(z)\bigr)$$
for $n\ge 2$. Let $\bi P(w)=\sum_{n\ge 0}w↑nP_n(z)$; this recurrence leads to the
differential equation $\bi P↑{@\prime}=\bi P↑2+P_1(z)-P_0(z)↑2=\bi P↑2+z-1$, and
the solution is
$$\bi P(w)=\sqrt{z-1}\,\tan\,\biggl( w\sqrt{z-1}+\mathop{\hbox{arctan}} {1\over\sqrt{z-1}}\biggr) ={1+(z-1)\bi T(w)\over 1-\bi T(w)},$$
where $\bi T(w)=\bigl(\tan w\sqrt{z-1}\,\bigr)/\sqrt{z-1}$. By rewriting the
solution so as to avoid the square roots, we obtain
$$\bi P(w)={1+\sum_{k\ge 0}t_{2k+1}(z-1)↑{k+1}w↑{2k+1}\over 1-\sum_{k\ge 0}t_{2k+1}(z-1)↑kw↑{2k+1}}$$
$$\hbox{where \tan x=\sum_{k\ge 0}t_{2k+1} x↑{2k+1}.}$$
This can be expanded in powers of $z-1$, using the values $t_1=1$, $t_3={1\over3}$,
$t_5={2\over15}$, to get
$$\twoline{\bi P(w)={1\over1-w}+\biggl({1\over3(1-w)↑2}+{w-1\over3}\biggr)(z-1) }{4pt}{\null+\biggl({1\over9(1-w)↑3}-{1\over5(1-w)↑2}+{1\over9}-{(1-w)↑3\over45} \biggr)(z-1)↑2+\cdots\,.}$$
So $\sum\!P_n(1)w↑n=1/(1-w)$, $\sum\!P_{\!n}↑\prime(1)w↑n\!=\!{1\over3}(1-w)↑{-2}+ {1\over3}(w-1)$, $\sum\!{1\over2}P_{\!n}↑{\prime\prime}(1)w↑n$ is the coefficient
of $(z-1)↑2$, and we find the variance in a few more steps. But this method of
solution does not follow the operator approach that was specified in the
problem statement.]
\pssub{Solution to Problem 3.}
It is convenient to consider the related function $G(x_1,x_2,x_3)$ that allows
exactly $x_j$ characters other than \<backspace\>\ and \<carriage return\>\
to be typed in column $j$. Then by inclusion and exclusion,
$$\twoline{G=G(2,2,2)-G(2,2,1)-G(2,1,2)-G(1,2,2)}{2pt}{\null +G(2,1,1)+G(1,2,1)+G(1,1,2)-G(1,1,1)}$$
enumerates sequences that include all three of the letters \.a, \.p, \.e.

In order to avoid infinitely many equations, we consider first the set of all
sequences of spaces and backspaces that begin in some column $j>3$ and end
in column $j-1$ without going left of column $j$ until the very last step.
The context-free grammar
$$L\;\gets \;\hbox{\<backspace\>}\;\mid\;\hbox{\<space\>}\,L\,L$$
unambiguously describes such sequences, hence
$$L(z)=z+zL(z)↑2$$
is the generating function $\{\, z↑{|\sigma|}\mid \sigma$ in $L\,\}$,
and we have
$$L(z)=\bigl( 1-\sqrt{@1-4z↑2}\,\bigr)/2z=z+z↑3+2z↑5+5z↑7+\cdots\,.$$
Similarly let $Q(z)$ enumerate sequences of spaces and backspaces that begin in
some column $j>3$ and never go left of column $j$; the unambiguous grammar
$$Q\;\gets \;\hbox{\<empty\>}\;\mid\;\hbox{\<space\>}\,Q\;\mid\; \hbox{\<space\>}\,L\,Q$$
proves that
\eqalign{Q(z)&=1+zQ(z)+zL(z)Q(z),\cr Q(z)&=1/\bigl(1-z-zL(z)\bigr)=1+z+2z↑2+3z↑3+6z↑4+\cdots\,.\cr}
[Incidentally, simple algebraic manipulations yield the identity
$$Q(z)=\penalty -50 \bigl(1-L(z)\bigr)/ (1-2z),$$a formula equivalent to $Q(z)+L(z)=1+2zQ(z)$. A direct proof of the
latter equation follows from the observation that every $Q$ or $L$ is either
empty or has the form $Q$\<space\>\ or $Q$\<backspace\>.]

Now let $G_j(z)$ be the generating function we seek when the typewriter
starts $j$ positions from the left, so that $G(z)=G_0(z)$. We have
\global\setbox1=\hbox{\eqalign{G_0(z)&=z+zG_0(z)+x_1zG_1(z),\cr G_1(z)&=z+zG_0(z)+x_2zG_2(z),\cr G_2(z)&=z+zG_1(z)+x_3zG_3(z),\cr}} \copy1
by considering sequences that begin with \<carriage return\>, \<backspace\>,
or something else, respectively. Furthermore
$$\hbox to\wd1{G_3(z)=L(z)G_2(z)+Q(z)z,\hfill}$$
since each sequence starting in column 4 either returns to column 3 or doesn't.
The solution to this tridiagonal system of linear equations is the desired
generating function $G(x_1,x_2,x_3)$.

The probability of any given sequence of keystrokes of length $n$ is $1/40↑n$,
if we stop the sequence at the first \<carriage return\>, and such sequences
are mutually exclusive. So the probability of typing \.{ape} is $G(1/40)$.

We have now derived all that was needed to satisfy the stated problem requirements, but it
is interesting to go further and coax {\rh macsyma} to obtain reasonably simple
formulas for the answer. See the attached transcript; it turns out that
$$G(z)={x_1x_2x_3z↑4Q(z)-(x_1z+1)x_3z↑2L(z)+(x_1-1)x_2z↑3+x_1z↑2+z\over (x_1z↑3+z↑2-z)x_3L(z)+x_2z↑3-(x_1+x_2)z↑2-z+1}.$$
And after inclusion and exclusion have removed the $x_i$, the
generating function for \.{ape} sequences begins as follows:
$$z↑4+3z↑5+15z↑6+44z↑7+163z↑8+472z↑9+1550z↑{10}+\cdots\,.$$
The exact probability turns out to be
{\vbox{\halign{#\cr 2999609859061393872672851275099904646499040221\sqrt{399}\hfill\quad\cr \quad\hfill\null - 59877588713530290411629940237569556287667865416\cr}}\over 93355082549464520980187663115368403895040545354710},
which is approximately .0000004238793706620676.

Jorge Stolfi pointed out that we could allow \.o'' to be typed in the second
column on many typewriters, since the ink in \.o'' might be a subset of
the ink in \.p''. In this case the answer would be
\eqalign{G&=G(2,3,2)-G(2,3,1)-G(2,2,2)-G(1,3,2)\cr &\hskip6em\null+G(2,2,1)+G(1,3,1)+G(1,2,2)-G(1,2,1)\cr &=z↑4+3z↑5+17z↑6+52z↑7+215z↑8+664z↑9+2406z↑{10}+\cdots\cr}
and the \.{ape} probability would rise to about .0000004244.
\par\vfill\eject
{\vermode
\null

:macsyma
\good
This is MACSYMA 292
\good
FIX292 14 DSK MACSYM being loaded
\good
(C1) solve(L=z+z*L**2,L);
\good
SOLVE FASL DSK MACSYM being loaded
Solution:
\good
2
SQRT(1 - 4 Z ) - 1
(E1)                       L = - ------------------
2 Z
\good
2
SQRT(1 - 4 Z ) + 1
(E2)                        L = ------------------
2 Z
(D2)                               [E1, E2]
\good
(C3) solve(Q=1+z*Q+z*L*Q,Q);
Solution:
\good
1
(E3)                         Q = - -------------
(L + 1) Z - 1
(D3)                                 [E3]
\good
(C4) algebraic:true;
(D4)                                 TRUE
\good
(C5) g0=z+z*g0+x1*z*g1;
(D5)                       G0 = G1 X1 Z + G0 Z + Z
\good
(C6) g1=z+z*g0+x2*z*g2;
(D6)                       G1 = G2 X2 Z + G0 Z + Z
\good
(C7) g2=z+z*g1+x3*z*g3;
(D7)                       G2 = G3 X3 Z + G1 Z + Z
\good
(C8) g3=L*g2+z*Q;
(D8)                           G3 = Q Z + G2 L
\good
(C9) solve([d5,d6,d7,d8],[g0,g1,g2,g3]);
\vfill\eject
Solution:
\good
4                          3      2
Q X2 Z  + ((- Q - L) X1 - Q X2) Z  - Q Z  + (Q + L) Z
(E9)  G3 = ------------------------------------------------------------
3                     2
(L X1 X3 + X2) Z  + (L X3 - X2 - X1) Z  + (- L X3 - 1) Z + 1
\good
4                  3                2
Q X2 X3 Z  + X2 (1 - Q X3) Z  + (L X3 - X2) Z  - Z
(E10) G1 = - ------------------------------------------------------------
3                     2
(L X1 X3 + X2) Z  + (L X3 - X2 - X1) Z  + (- L X3 - 1) Z + 1
\good
4                3         2
Q X1 X3 Z  + (Q X3 + X1) Z  - Q X3 Z  - Z
(E11) G2 = - ------------------------------------------------------------
3                     2
(L X1 X3 + X2) Z  + (L X3 - X2 - X1) Z  + (- L X3 - 1) Z + 1
\good
4                          3                2
Q X1 X2 X3 Z  + (X1 (X2 - L X3) - X2) Z  + (X1 - L X3) Z  + Z
(E12) G0 = -------------------------------------------------------------
3                     2
(L X1 X3 + X2) Z  + (L X3 - X2 - X1) Z  + (- L X3 - 1) Z + 1
(D12)                       [[E9, E10, E11, E12]]
\good
(C13) g(x1,x2,x3):=([t],t:ratsimp(ev(g0,e12,e3,eval)),ratsimp(ev(t,e1)));
\good
(D13) G(X1, X2, X3) := ([T], T : RATSIMP(EV(G0, E12, E3, EVAL)),
\good
RATSIMP(EV(T, E1)))
\good
(C14) g(1,1,1);
\good
Z
(D14)                             - -------
2 Z - 1
\good
+g(2,1,1)+g(1,2,1)+g(1,1,2)-g(1,1,1);
\good
(D15)
\good
7               2      6      4        6      5      4      3    2
8 Z  + SQRT(1 - 4 Z ) (8 Z  - 4 Z ) - 12 Z  - 4 Z  - 4 Z  + 6 Z  + Z  - Z
- -------------------------------------------------------------------------
7      6       5       4       3      2
40 Z  - 4 Z  - 32 Z  - 12 Z  + 26 Z  - 3 Z  - 4 Z + 1
\good
7               2      6      5      4       6       5      4      3
+ (18 Z  + SQRT(1 - 4 Z ) (2 Z  + 2 Z  - 2 Z ) + 5 Z  - 19 Z  - 6 Z  + 8 Z
\good
2           7       6       5      4       3      2
+ Z  - Z)/(34 Z  + 11 Z  - 44 Z  - 9 Z  + 26 Z  - 3 Z  - 4 Z + 1)
\good
7               2      6      4       6      5      4      3      2
8 Z  + SQRT(1 - 4 Z ) (4 Z  - 2 Z ) - 8 Z  + 4 Z  - 9 Z  + 4 Z  + 2 Z  - Z
+ --------------------------------------------------------------------------
7      6      5       4       3    2
16 Z  - 8 Z  - 2 Z  - 19 Z  + 20 Z  - Z  - 4 Z + 1
\good
7               2    6    5    4       6      5       4      3      2
- (6 Z  + SQRT(1 - 4 Z ) (Z  + Z  - Z ) + 5 Z  - 4 Z  - 10 Z  + 5 Z  + 2 Z
\good
7      6       5       4       3    2
- Z)/(10 Z  + 7 Z  - 14 Z  - 16 Z  + 20 Z  - Z  - 4 Z + 1)
\good
5               2    4    3       4      3      2
4 Z  + SQRT(1 - 4 Z ) (Z  - Z ) - 8 Z  + 3 Z  + 2 Z  - Z
- --------------------------------------------------------
5       4       3    2
10 Z  - 21 Z  + 14 Z  + Z  - 4 Z + 1
\good
3             2       4      3    2        2             2       3
2 Z  SQRT(1 - 4 Z ) + 4 Z  - 4 Z  - Z  + Z   Z  SQRT(1 - 4 Z ) - 4 Z  + Z
- ------------------------------------------ + ----------------------------
5       4       3    2                       3      2
16 Z  - 24 Z  + 14 Z  + Z  - 4 Z + 1         10 Z  - 5 Z  - 2 Z + 1
\good
Z
+ -------
2 Z - 1
\good
\good

HAYAT FASL DSK MACSYM being loaded
4      5       6       7        8        9         10
(D16)/T/ Z  + 3 Z  + 15 Z  + 44 Z  + 163 Z  + 472 Z  + 1550 Z   + . . .
\good
\good
(D17) (2999609859061393872672851275099904646499040221 SQRT(399)
\good
- 59877588713530290411629940237569556287667865416)
\good
/93355082549464520980187663115368403895040545354710
\good
(C18) factor(denom(\%));
\good
(D18) 2 3 5 11 17 19 23 29 53 59 79 167 211 457 6673 7019 9199 20773 28559
\good
1291357141
\good
(C19) bfloat(d17);
\good

FLOAT FASL DSK MACSYM being loaded
(D19)                        4.238793706620676B-7
\good
(C20) time(d14,d15,d17);
\good
TIME or [TOTALTIME, GCTIME] in msecs.:
(D20)             [[1813, 914], [13204, 5191], [1595, 537]]

}
\bigskip
\noindent{\sl Acknowledgment:\/}
The {\rh macsyma} system, developed by the Mathlab group at M.I.T., had the
support of U.S. Energy Research and Development contract number E(11--1)--3070
and National Aeronautics and Space Administration grant number NSG 1323.

\vfill\eject
\psmajor{Appendix D: Final Exam I and Solutions}
\psminor{Final Exam I}
\noindent Problem 1.  [50 points]  Find the asymptotic value of
$\prod_{0\le k\le n}{n\choose k}$ to with\-in a relative error of
$O(1/n)$ as $n\to \infty$.  [In other words, your answer should have the form
$f(n)(1+O(1/n))$ for some explicit'' function $f$.]

\yyskip
\noindent Problem 2.  [100 points]  Let us say that the positive integer $n$
is {\sl unusual\/} if its largest prime factor is at least $\sqrt{n}$.  Thus,
a prime number is unusual, as is the product of two primes.  (The number $1$
is also highly unusual, since it is a positive integer for which the
definition makes no sense.)

Determine the asymptotic number of unusual integers $n$ in the range
$1<n\le N$, as $N\to \infty$, with an absolute error of $O(N/(\log N)↑2)$.

{\sl Hint}:  Count separately the unusual integers in the stated range
whose largest prime factor is $\le \sqrt{N}$ [this part of the problem is
worth 35 points] and those having a prime factor $>\sqrt{N}$ [this part is
worth 65 points].

Additional credit will be given for answers that are correct to within
$O(N/(\log N)↑3)$.  But you are advised to do problem 3 first before trying
to get extra  credit on problem 2.

\yyskip
\noindent Problem 3.  [150 points]  The following algorithm for traversing
a binary tree in preorder is analogous to Algorithm 2.3.1T of [Knuth~I]
for inorder
traversal, except that fewer items are put onto the stack:

\yskip

\def\.#1{\hbox{\tt#1}}
\algstep P1. [Initialize.] Set stack $\.A$ empty, and set the link variable
$\.P\gets \.T$.

\algstep P2. [$\.P=\Lambda$?]  If $\.P=\Lambda$, go to step P5.

\algstep P3. [Visit $\.P$.]  (Now $\.P$ points to a nonempty binary tree that
is to be traversed.) Visit'' \.{NODE(P)}.

\algstep P4. [Stack $\Leftarrow$ \.{RLINK(P)}.]  If $\.{RLINK(P)}\ne \Lambda$, set
$\.A\Leftarrow \.{RLINK(P)}$, i.e., push the value of \.{RLINK(P)} onto stack $\.A$.
Then set $\.P\gets \.{LLINK(P)}$ and return to step P2.

\algstep P5.  [$\.P\Leftarrow \null$Stack.]  If stack $\.A$ is empty, the algorithm terminates;
otherwise set $\.P\Leftarrow \.A$ and return to step P3.\quad \blackslug.

\yskip
Your problem is to solve the analog of exercise 2.3.1--11 for this algorithm:
What is the average value of the largest stack size occurring during the
execution of Algorithm P, as a function of $n$, when all $n$-node binary
$O(n↑{-1/2}\log n)$.

\psminor{Solutions to Final Exam I}

\lineskiplimit 2pt
\lineskip 2pt

\pssub{Solution to Problem 1.}
Well, we have $\ln\prod_{0\le k\le n}{n\choose k}= \sum_{0\le k\le n}\bigl(@\ln n!-\ln k!-\ln (n-k)!\bigr)= 2\sum_{1\le k\le n}k\ln k-{(n+1)}\sum_{1\le k\le n}\ln k$. By Euler's summation formula
(cf.\ [Knuth~I; exercise 1.2.11.2--7 and Eq.\ 1.2.11.2--18]) this is
\lineskip 3pt\vbox{\halign{\line{ #}\cr \quad 2\left(\textstyle{1\over2}n↑2\ln n+{1\over2}n\ln n+{1\over12}\ln n -{1\over4}n↑2+\ln A+\displaystyle O\left(1\over n↑2\right)\right)\hfill\cr \hfill\null-(n+1)\,\left(\textstyle n\ln n+{1\over2}\ln n-n+\ln\sqrt{2\pi } +\displaystyle{1\over12n}+O\left(1\over n↑3\right)\right)\qquad\cr \hfill=\textstyle{1\over2}n↑2-{1\over2}n\ln n+\bigl(1-\ln\sqrt{2\pi }\,\bigr) \,n -{1\over3}\ln n-{1\over12}+2\ln A-\ln\sqrt{2\pi }+O(n↑{-1}),\quad\cr}}
where $A$ is Glaisher's constant. Since $\ln A={1\over12}\bigl(\gamma- \zeta↑\prime(2)/\zeta(2)+\ln2\pi \bigr)$, a formula that can be found either
in [deBruijn~70; \S3.7] or by using the
Abel-Plana'' formula as described in
[Olver 74; \S8.3.3], the answer is
$$\twoline{\textstyle\exp\bigl({1\over2}n↑2-{1\over2}n\ln n+(1-{1\over2}\ln2\pi )n- {1\over3}\ln n-{1\over12}}{ 2pt}{\textstyle\null+{1\over6}\gamma-\zeta↑\prime(2)/\pi ↑2-{1\over3}\ln 2\pi \bigr) \bigl(1+O(1/n)\bigr).}$$

\pssub{Solution to Problem 2.}
(a) The unusual numbers $n\le N$ whose largest prime factor is a
given prime $p\le \sqrt N$ are the $p$ numbers $p$, $2p$, \dots, $p↑2$.
So there are $\sum_{p\le \sqrt N}\,p$ unusual numbers of type (a). This is
$$\int_{\sqrt2}↑{\sqrt N}t\,d\pi (t)=\int_{\sqrt2}↑{\sqrt N}t\,dL(t)+ \int_{\sqrt2}↑{\sqrt N}t\,dO\bigl( t/(\log t)↑{1000}\bigr)$$ where $L(t)= \int_2↑t d_{\null}u/\!\ln u$. The second integral is
$$\twoline{O\bigl(\sqrt N\cdot\sqrt N/(\log\sqrt N)↑{1000}\bigr)+ O\bigl(\int_{\sqrt{2}}↑{N↑{1/3}}(t/(\log t)↑{1000})\,dt\bigr)+}{ 5pt}{O\bigl(\int_{N↑{1/3}}↑{N↑{1/2}}(t/(\log t)↑{1000})\,dt\bigr)}$$
so it is $O\bigl( N\!/(\log N)↑{1000}\bigr)$. The first integral is
$$\int_{\sqrt2}↑{\sqrt N}t\,dt/\!\ln t= \int_2↑N d_{\null}u/\!\ln u=L(N)$$
[so, curiously, $\sum_{p\le \sqrt N}\,p\approx\sum_{p\le N}1$], which integrates by parts
into the well known asymptotic form
$$\twoline{N\!/\!\ln N+N\!/(\ln N)↑2+2!\,N\!/(\ln N)↑3+}{ 5pt}{\cdots+998!\,N\!/(\ln N)↑{999}+O\bigl( N\!/(\log N)↑{1000}\bigr).}$$

\vskip 3pt plus 6pt
(b) The unusual numbers $n\le N$ whose largest prime factor is a given prime $p>\sqrt N$
are the $\lfloor N\!/p\rfloor$ numbers $p$, $2p$, \dots, $\lfloor N\!/p\rfloor p$. So there are $\sum_{p>\sqrt N}\lfloor N\!/p\rfloor$ unusual numbers of
type (b). This equals $\int_{\sqrt N}↑\infty \lfloor N\!/t\rfloor\,d\pi (t)= \int_{\sqrt N}↑\infty \lfloor N\!/t\rfloor\,dL(t)+ \int_{\sqrt N}↑\infty \lfloor N\!/t\rfloor\,dO\bigl( t/(\log t)↑{1000}\bigr),$
and the second integral is
$$O\left( N\!/\bigl(\log\sqrt N\,\bigr)↑{1000}\right)+O\biggl( N \int_{\sqrt N}↑\infty {dt\over t(\log t)↑{1000}}\biggr)$$
so it is $O\bigl( N\!/(\log N)↑{999}\bigr)$. The first integral is
\lineskip3pt\eqalign{\int_{\sqrt N}↑N\left\lfloor N\over t\right\rfloor{dt\over\ln t}&= \int_{\sqrt N}↑N\left({N\over t}-\left\{N\over t\right\}\right){dt\over\ln t}\cr &=\left.\vbox to 12pt{}N\ln\ln t\,\right|↑N_{\sqrt N}-N\int_1↑{\sqrt N}{\{u\}\,d_{\null}u \over u↑2\ln(N\!/u)},\cr}
where $\{x\}=x-\lfloor x\rfloor$ and $u=N\!/t$. Since $\ln\ln N-\ln\ln\sqrt N= \ln 2$, we get
$$N\ln2-{N\over\ln N}\int_1↑{\sqrt N}{\{u\}\,d_{\null}u\over u↑2}\left(1+ {\ln u\over\ln N}+O\left(\ln u\over\ln N\right)↑2\right).$$
Now $\int_1↑\infty \bigl({\ln u\over u}\bigr)↑2d_{\null}u$ exists, so the $O$ term can be dropped.
We have
\baselineskip15pt\lineskip3pt \eqalign{\int_1↑{\sqrt N}{\{u\}\,d_{\null}u\over u↑2}&=\sum_{1\le k<\sqrt N}\,\int_k↑{k+1} {(u-k)\,d_{\null}u\over u↑2}+O\left(1\over N\right)\cr &=\sum_{1\le k<\sqrt N}\left(\ln(k+1)-\ln k-{1\over k+1}\right)+O\left(1\over N\right)\cr &=\ln\sqrt N-H_{\sqrt N}+1+O\bigl(1/\sqrt N\,\bigr)\cr &=1-\gamma+O(N↑{-1/2}).\cr}
Similarly (here comes the extra credit' part that nobody got)
$$\twoline{\int_1↑{\sqrt N}{\{u\}\ln u\,d_{\null}u\over u↑2}}{ 5pt}{={1\over2}\ln↑2\sqrt N+\ln\sqrt N -H_{\sqrt N}+1-\!\!\!\!\sum_{1\le k<\sqrt N}{\ln(k+1)\over k+1}+O\left(\log N\over\sqrt N \right).}$$
Let $$\zeta(1+\epsilon )={1\over\epsilon }+\gamma_0-\gamma_1\epsilon +{\gamma_2\epsilon ↑2\over2!}-{\gamma_3\epsilon↑3\over3!}+\cdots\,.$$
According to the formula in the answer to [Knuth III; exercise 6.1--8],
$$\sum_{1\le k\le m}{1\over k↑{1+\epsilon }}=\zeta(1+\epsilon )+{m↑{-\epsilon }\over-\epsilon }+O(m↑{-1-\epsilon }).$$
Expanding both sides in powers of $\epsilon$ and equating like coefficients yields
$$\sum_{1\le k\le m}{(\ln k)↑r\over k}={(\ln m)↑{r+1}\over r+1}+\gamma_r +O\biggl({(\ln m)↑r\over m}\biggr).$$
Thus, $\int_1↑{\sqrt N}\{u\}\ln u\,d_{\null}u/u↑2=1-\gamma_1-\gamma_0+O(N↑{-1/2}\log N)$,
and there are $N\ln 2+{(\gamma-1)}N\!/\!\ln N+(\gamma+\gamma_1-1)N\!/(\ln N)↑2+ O\bigl( N\!/(\log N)↑3\bigr)$ unusual integers of type (b); they aren't so unusual
after all.

\pssub{Solution to Problem 3.}
Let the maximum stack size required by Algorithm P to traverse a
binary tree be called its hite,'' and let the analogous quantity for Algo\-rithm~T
be the height.'' Let ${\bar a}_{nk}$ be the number of binary trees with $n$ nodes
whose hite is at most $k$. If ${\bar g}_k(z)=\sum_n{\bar a}_{nk}z↑n$, we find
${\bar g}_0(z)=1/(1-z)$ and ${\bar g}_k(z)=1+z{\bar g}_{k-1}(z){\bar g}_k(z)+z{\bar g}_k(z)-z{\bar g}_{k-1}(z)$. \ (The first term is for an empty binary tree, the next for a binary tree with
left subtree hite $<k$ and right subtree hite $\le k$, and the last two are for a
binary tree with left subtree hite $=k$ and an empty right subtree.) \
Thus$${\bar g}_k(z)={1-z{\bar g}_{k-1}(z)\over1-z{\bar g}_{k-1}(z)-z}={1\over1-\displaystyle {z\mathstrut\over1-z{\bar g}_{k-1}(z)}},$$
and it follows immediately that ${\bar g}_k(z)=g_{2k+1}(z)$. From this surprising
relation we conclude that the number of binary trees of hite $k$ is the same as
the number of binary trees of height $2k$ or $2k+1$. \ [It is interesting to
find a one-to-one correspondence between these two sets; see the note on the
last page. We don't want to spoil things for you by giving the correspondence
before you've had a chance to find it for yourself, since this makes a very
nice little problem.] \ A binary tree of height $h$ corresponds to a binary tree of
hite ${1\over2}h$ or ${1\over2}h-{1\over2}$, so we expect the average hite to
be approximately half of the average height, minus $1\over4$. This in fact is
what happens, but the point of the problem is to prove it rigorously with
analytic techniques.

Following [Knuth I; exercise 2.3.1--11] and [deBruijn 72], ${\bar a}_{nk}=a_{n(2k+1)}$ is the
coefficient of $u↑n$ in
$$(1-u)(1+u)↑{2n}{1-u↑{2k+2}\over 1-u↑{2k+3}},$$
and ${\bar b}_{nk}={\bar a}_{nn}-{\bar a}_{nk}$ is the coefficient of $u↑{n+1}$ in
$$(1-u)↑2(1+u)↑{2n}{u↑{2k+3}\over 1-u↑{2k+3}}.$$
Thus ${\bar s}_n=\sum_{k\ge 1}k({\bar a}_{nk}-{\bar a}_{n(k-1)})= \sum_{k\ge 0}{\bar b}_{nk}$ is the coefficient of $u↑{n+1}$ in
$$(1-u)↑2(1+u)↑{2n}\sum_{k\ge 0}{u↑{2k+3}\over 1-u↑{2k+3}}.$$
Let us add ${\bar b}_{n(-1)}=a_{nn}$ for convenience; $s_n+a_{nn}$ is the coefficient
of $u↑{n+1}$ in
$$(1-u)↑2(1+u)↑{2n}\sum_{k\hbox{\sevenrm\ odd}}{u↑k\over 1-u↑k},$$
which is
the sum in Eq.\ (23) of the cited paper but with $d(k)$ replaced by ${\bar d}(k)$,
the number of {\sl odd\/} divisors of $k$. We have
$$\sum_{k\ge 1}{\bar d}(k)/k↑z=\zeta(z)\bigl(\sum_{k\hbox{\sevenrm\ odd}}1/k↑z\bigr) =\zeta(z)\bigl( \zeta(z)-2↑{-z}\zeta(z)\bigr),$$
so ${\bar s}_n+a_{nn}$ is obtained by the method in the paper except that we have an
additional factor $(1-2↑{b-2z})$ in the integral (29). The residue at the
double pole now becomes $$n↑{(b+1)/2}\,\Gamma\bigl({1\over2}(b+1)\bigr)\bigl( {1\over8}\ln n+{1\over8}\psi\bigl({1\over2}(b+1)\bigr)+{1\over2}\gamma+{1\over4} \ln 2\bigr)$$ and at $z=-k$ it is $1-2↑{2k+b}$ times the value in (31). The
answer we seek comes to $-1+(n+1)\bigl((-2/n){\bar g}_0(n)+(4/n↑2){\bar g}_2(n)+O(n↑{-3/2} \log n)\bigr)={1\over2}\sqrt{\pi n}-1+O(n↑{-1/2}\log n)$.

\vfill\eject
\def\{\hbox{\bf#1\hskip.5pt}}
\def\\#1{\hbox{\sl#1\/\hskip1pt}}
The promised correspondence is given by the following recursive procedure:
\vbox{\halign{#\hfill\cr \&{ref\/}(\\{node}) \&{procedure} \\{transform}(\&{ref\/}(\\{node}) \&{value} p);\cr \&{begin} \&{ref\/}(\\{node}) q,r;\cr \&{if} p=\\{null} \&{then} \&{return}(\\{null});\cr r\gets \\{left}(p); \\{left}(p)\gets \\{transform}(\\{right}(p)); \\{right}(p)\gets \\{null};\cr \&{if} r=\\{null} \&{then} \&{return}(p);\cr q\gets r; \&{while} \&{true} \&{do}\cr \quad\&{begin} \\{left}(q)\gets \\{transform}(\\{left}(q));\cr \quad\&{if} \\{right}(q)=\\{null} \&{then} \&{done};\cr \quadq\gets \\{right}(q);\cr \quad\&{end};\cr \\{right}(q)\gets p; \&{return}(r);\cr \&{end}.\cr}}
It can be shown that the transformed tree has the following strong
property: Let $s$ be the height of the stack when Algorithm T puts a pointer to
a given node of $B$ onto its stack, and let $s↑\prime$ be the height of the stack
after step P4 just following the time Algorithm P visits this same node in the
transformed tree $B↑\prime$. Then $s↑\prime=\lfloor s/2\rfloor$. Thus the
stack size during the traversal of $B↑\prime$ in preorder is almost exactly
half the stack size during the inorder traversal of $B$, and we have
a relation between the average as well as the maximum stack
sizes.

\par\vfill\eject
\psmajor{Appendix E: Midterm Exam II and Solutions}
\psminor{Midterm Exam II}

\def\yskip{\vskip 3pt plus 1pt minus 1pt}
\def\prob #1 (#2). {\par\yskip\noindent{\bf Problem #1. }[#2]\ }
\def\part #1 (#2). {\par\yskip\noindent{(#1) }[#2]\ }
\def\ans #1. {\pssub{Solution to problem #1.}}
\def\\{\,.\,.\,}
\def\bigslash{\big/}

\prob 1 (50 points).

Continuing the analysis of secondary clustering in \S3.4,
find a sliding operator'' for $\Omega_m$ that allows
$U_2G_{mn}(x)$ to be computed. Also find an analog of \eq(3.40) that
allows $U_2H_{mn}(x)$ to be computed. Express $G↑{\prime\prime}_{mn}(1)$
and $H↑{\prime\prime}_{mn}(1)$ as simple'' functions of $m$, $n$, $P_1$,
and $P_2$, where
$$P_k=\left(1+{kq\over m}\right)\left(1+{kq\over m-1}\right)\,\ldots\, \left(1+{kq\over m-n+1}\right).$$

\prob 2 (150 points total, distributed non-uniformly as shown below).

A student named J. H. Quick woke up one morning with an idea for a new kind of
binary search tree. He had learned about the advantages of late binding''
in his studies of computer science, and he thought: {\sl Why should I use
the first key to decide how the rest of the tree will be partitioned? I could
do better by postponing that decision and letting further keys influence
what happens.}'' Running to his interactive workstation, he hastily prepared
a file containing a description of his new data structure, which he
chose to call Late Binding Trees (LBTs); and then he ate breakfast.

Unfortunately there is not room here to describe the subsequent events in
Quick's life. The story about his fateful encounters with the Chuvstvenni
sisters in Gstaad, who vowed to stop at nothing until they had learned his
secret, will probably never be told. Let us rather turn our attention to the
specifics of LBTs, suppressing the details of how this information was
learned.

There are two types of nodes: branch nodes and leaves. A {\sl branch node\/}
contains two keys $a$ and $b$, where $a<b$, and it also contains two links
$l$ and $r$ that point to subtrees. All keys in the $l$ subtree are $\le a$,
and all keys in the $r$ subtree are $\ge b$. Such a node can be represented by
the notation $(a\\b)$', having its subtrees drawn below. A {\sl leaf node\/}
contains a full record, including a key $a$; such a node can be represented by
$[a]$'.

LBTs are never empty; they start out with a single (leaf) node. One of the
nodes in the left
subtree of a branch node $(a\\b)$ is the leaf node $[a]$; similarly, the
right subtree of $(a\\b)$ always
contains $[b]$. If we want to insert a new record with
key $x$ into a given LBT, we proceed as follows, assuming that $x$ is
different from all keys already in the tree:

\yskip\item{(1)} If the LBT is $[a]$, and if $a<x$, change the LBT to
$(a\\x)$, with left subtree $[a]$ and right subtree $[x]$. A similar construction
with $a$ and $x$ interchanged is used if $x<a$.

\yskip\item{(2)} If the LBT has root $(a\\b)$ and if $x<a$, insert the new
record into the left subtree, using the same method recursively.

\yskip\item{(3)} If the LBT has root $(a\\b)$ and if $x>b$, insert the new
record into the right subtree, using the same method recursively.

\yskip\item{(4)} If the LBT has root $(a\\b)$ and if $a<x<b$, flip a truly
random coin. If it comes up heads, change the root to $(x\\b)$ and insert the
new record in the left subtree; otherwise change the root to $(a\\x)$ and
insert the new record in the right subtree.

\yskip\noindent The idea is therefore to keep track of a range of possible
splitting keys in the root of the tree, instead of deciding prematurely
on a particular one.

The purpose of this problem is to learn something about the analysis of
algorithms by analyzing the average total external path length of LBTs,
assuming that LBTs are created by inserting records in random order. The
total external path length is the sum, over all leaves, of the distance from
the root to the leaf. Let $n$ be the number of leaves.
Then if $n=1$, the total external path length is
always 0; if $n=2$, it is always 2; if $n=3$, it is always 5; and if
$n=4$, it is either 8 or 9.

\part a (15 points). Suppose that the root of an LBT is $(k\\k+1)$, after
inserting $n$ keys $x_1\ldots x_n$ that form a random permutation of
$\{1,\ldots,n\}$. (In other words, the LBT starts out containing only $[x_1]$,
then $x_2$ is inserted, and so on; there are $n$ leaves after $x_n$
has been inserted.) The left subtree of the root is the LBT formed by the
permutation $y_1\ldots y_k$ of $\{1,\ldots,k\}$ consisting of the $x_i$
that are $\le k$; the right subtree is the LBT formed by the permutation
$z_1\ldots z_{n-k}$ of $\{k+1,\ldots,n\}$ consisting of the remaining~$x_i$.

Prove that the permutations $y_1\ldots y_k$ are not uniformly distributed; if $y_1\ldots y_k$ has $t$ left-to-right maxima, it
occurs with probability $2↑{k-t}$ times the probability that the identity
permutation $1\ldots k$ occurs. Similarly, the permutations $z_1\ldots z_{n-k}$
are not uniformly random; their distribution depends on left-to-right
minima.

\part b (15 points). Let $p_{nk}$ be the probability that the root of an
LBT will be $(k\\k+1)$, after inserting $n$ keys that are in uniformly
random order. Find a formula for $p_{nk}$.

\part c  (20 points). Let us say that permutations on $\{1,\ldots,n\}$ are
$U$-distributed if all permutations are equally likely; they are $L@$-distributed
if they occur with probability proportional to $2↑{-t}$, where $t$ is the
number of left-to-right maxima; they are $R$-distributed if they occur with
probability proportional to $2↑{-s}$, where $s$ is the number of
left-to-right minima; and they are $X$-distributed if they occur with probability
proportional to $2↑{-s-t}$.

Part (a) showed that the left and right subtrees of LBTs constructed from
$U$-distributed permutations are respectively $L@$- and $R$-distributed. Prove
that if we start with $L@$-, $R$-, or $X$-distributed permutations, the
subtrees are constructed from $(L,X)$, $(X,R)$, or $(X,X)$-distributed
permutations, respectively.

\part d (5 points). Let $U_n$, $L_n$, $R_n$, and $X_n$ be the average
total external path length of the LBTs formed by distributions $U$, $L$,
$R$, $X$. Prove that, for all $n\ge 2$, we have
\eqalign{ U_n&=n+\textstyle\sum_{1\le k<n}p_{nk}(L_k+R_{n-k}),\cr L_n&=n+\textstyle\sum_{1\le k<n}q_{nk}(L_k+X_{n-k}),\cr R_n&=n+\textstyle\sum_{1\le k<n}q_{n(n-k)}(X_k+R_{n-k}),\cr X_n&=n+\textstyle\sum_{1\le k<n}r_{nk}(X_k+X_{n-k}),\cr}
where $q_{nk}$ and $r_{nk}$ are the respective probabilities that $L@$- and
$X$-distributed LBTs have $(k\\k+1)$ at the root.

\part e (20 points). Prove that $q_{nk}={k-1/2@\choose k-1} \bigslash{n-1/2@\choose n-2}$ and $r_{nk}=1/(n-1)$, for $1\le k<n$.

\part f (5 points). Prove that $X_n=2nH_n-2n$.

\part g (20 points).
Prove that $$\sum_{1\le k<n}q_{nk}X_{n-k}= \textstyle{4\over5}(n+{1\over2})(H_{n+1/2}-H_{5/2}).$$[{\sl Hint:\/} Show that
Eq.\ \eq(1.47) can be used for non-integer $m$.]

\part h (25 points). Solve the recurrence for $L_n$ that appears in part (d),
using the repertoire method to study recurrences of the form $x_n= a_n+\sum_{1\le k<n}q_{nk}x_k$.

\part i (25 points). Prove that $U_n=(2n+{1\over2})H_n-{13\over6}n -{5\over12}$.

\vskip 10pt
\psminor{Solutions to Midterm Exam II}
\ans 1.
We have $U_1x=U_1+U_0$ and $U_2x=U_2+2U_1$ by \eq(3.5), hence $U_0\Omega_m= U_0$, $U_1\Omega_m=(1+{q\over m})U_1+pU_0$, $U_2\Omega_m=(1+{2q\over m})U_2 +2(p+{q\over m})U_1$. Let $B_m$ be the operator $U_2-2mU_1+(m+1)mU_0$;
it turns out that $B_{m+1}\Omega_m=(1+{2q\over m})B_m$. Therefore
$B_{m+1}G_{mn}(x)=P_2B_{m-n+1}x=(m-n+1)(m-n)P_2$. Furthermore
$B_{m+1}G_{mn}(x)=U_2G_{mn}(x)-2(m+1)\bigl( m+1-(m-n)P_1)+(m+2)(m+1)$, by
\eq(3.36), and it follows that
$$U_2G_{mn}(x)=(m-n+1)(m-n)P_2-2(m+1)(m-n)P_1+m(m+1).$$
How about $H$? Well,$$\textstyle U_2H_{mn}=(1+{1\over m})U_2H_{m-1,n-1}+({1\over m}U_2+ {2\over m}U_1)G_{m-1,n-1},$$ and it turns out that
$$\twoline{\textstyle{1\over m+1}U_2(H_{mn}+{1\over p-q}G_{mn})- {2p\over p-q}U_1H_{mn}}{3pt}{\textstyle= {1\over m}U_2(H_{m-1,n-1}+{1\over p-q}G_{m-1,n-1})- {2p\over p-q}U_1H_{m-1,n-1}.}$$
Thus the quantity
${1\over m+1}U_2(H_{mn}+{1\over p-q}G_{mn})-{2p\over p-q}U_1H_{mn}$ turns
out to be equal to ${1\over m-n+1}U_2(H_{m-n,0}+{1\over p-q}G_{m-n,0})-{2p\over p-q}U_1H_{m-n,0}= -{2p\over p-q}$, and we can plug into our formula for $U_2G_{mn}$ to obtain
$$\textstyle U_2H_{mn}={m+1\over q}\bigl( m-np-2(m-n)P_1\bigr) +{(m+1)(m-n)\over p-q}\,\left({p\over q}-{m-n+1\over m+1}P_2\right)\,.$$
Note that when $p=q={1\over2}$ the latter term becomes 0/0, so we need
a separate formula for this case. By differentiating $P_2$ with respect to
$q$ we find that ${p\over q}-{m-n+1\over m+1}P_2=(1-2q)(H_{m+1}-H_{m-n+1}+2) +O(1-2q)↑2$ as $q\to {1\over2}$, hence the value of $U_2H_{mn}$ involves
harmonic numbers when $q={1\over2}$.
\ans 2. Let us use the shorter terms leftmax'' and leftmin'' for
left to right maximum'' and left to right minimum,'' respectively.

(a)
In order to obtain $y_1\ldots y_k$ and $z_1\ldots z_{n-k}$, we
need $x_1x_2=y_1z_1$ or $z_1y_1$, and the remaining $x$'s must contain
$y_2\ldots y_k$ and $z_1\ldots z_{n-k}$ merged together in some way.
When $x_i$ is being inserted, if it is a $y_j$ we put it in the left subtree
with probability ${1\over2}$ if that $y_j$ is a leftmax, and if it is a $z_j$
we put it in the right subtree with probability $1\over2$ if $z_j$ is a
leftmin. Otherwise the probability is 1; and if the coin flip goes the wrong
way, we don't get $(k\\k+1)$ at the root. Thus the probability of obtaining
$y_1\ldots y_k$ is proportional to $2↑{-t}$.

(b) For each pair of permutations $y_1\ldots y_k$ and $z_1\ldots z_{n-k}$
having respectively $t$ leftmaxes and $s$ leftmins, and for each of the
$2{n-2@\choose k-1}$ ways to merge these together as $x_1\ldots x_n$, the
probability of sending $y_1\ldots y_k$ to the left and $z_1\ldots z_{n-k}$
to the right is $2↑{2-t-s}$. Therefore $p_{nk}$ is $2{n-2@\choose k-1}$ times
$\sum_{y,z}2↑{2-t(y)-s(z)}$, divided by $n!$.

Now the generating function for leftmaxes is $$\textstyle \sum_y z↑{t(y)}=z(1+z)\ldots (k-1+z),$$by considering the inversion tables, hence $\sum_y 2↑{1-t(y)}= (k-{1\over2})↑{\underline{k-1}}$. It follows that$$\textstyle p_{nk}=2{n-2@\choose k-1} (k-{1\over2})↑{\underline{k-1}}(n-k-{1\over2})↑{\underline{n-k-1}}/n!= {k-1/2@\choose k-1}{n-k-1/2@\choose n-k-1}/{n\choose2}.$$
Ken Clarkson also found the curious formula $$\textstyle p_{nk}=8{k-1/2@\choose n}{n-2@\choose k-1}(-1)↑{n-k}.$$

(c) The leftmins in $x_1\ldots x_n$ all occur in $y_1\ldots y_k$,
and the leftmaxes all occur in $z_1\ldots z_{n-k}$, except perhaps for
the very first ones. Thus, the probability of obtaining a particular
permutation $y$ is equal to $2{n-2@\choose k-1}$ times $\sum_z 2↑{2-t(y)-s(z)}p(x)$,
where $p(x)$ is the probability that $x_1\ldots x_n$ is input. If we
assume (as we may) that $x_1<x_2$, then $p(x)$ is proportional to
$2↑{-t(z)}$, $2↑{-s(y)}$, $2↑{-t(z)-s(y)}$ in distributions L, R,~X. The
result is proportional to $2↑{-t(y)}$, $2↑{-s(y)-t(y)}$, $2↑{-s(y)-t(y)}$,
so the left subtrees have distributions L, X,~X. The right subtrees are similar.

(d)
The total path length is $n$ plus the total path length of the left
subtree plus the total path length of the right subtree. So, with
probability $p_{nk}$, we obtain a contribution of $n+L_k+R_{n-k}$ to the
average total path length. The duality between left and right shows that
$q_{k(n-k)}$ is the probability that an R-distributed LBT has $(k\\k+1)$ at the
root. It follows that $L_n=R_n$ (which was obvious).

(e)
By part (c), the probability $q_{nk}$ is proportional to the double sum $\sum_{y,z} 2↑{-t(y)-s(z)-t(z)}{n-2@\choose k-1}$, where the constants of proportionality
for fixed $n$ are independent of $k$. The generating function
$\sum_y z↑{s(y)+t(y)}$ is equal to $(z↑2)(2z)(1+2z)\ldots (k-2+2z)$, hence
$\sum_y 2↑{-s(y)-t(y)}={1\over 4}(k-1)!$. Thus $q_{nk}$ is proportional to
$k-1/2@\choose k-1$ and $r_{nk}$ is independent of $k$; it only remains to
find the constants of proportionality so that $\sum q_{nk}=\sum r_{nk}=1$.
See equation \eq(1) below.

(f) We have $X_n=C_{n-1}$ in standard quicksort [GKP; \eq(2.12), \eq(2.14)].

(g)
We have $(1-z)↑{-1-m}=\sum_{n\ge 0}{n+m\choose n}z↑n$, for all
com\-plex $m$, by the binomial theorem. Differentiating with respect to $m$
(this idea was suggested by John Hobby), we obtain \eq(1.47): $$(1-z)↑{-1-m}\ln(1-z)↑{-1} =\sum_{n\ge 0}{n+m\choose n}(H_{n+m}-H_m)z↑n.$$

Let us now tabulate a bunch of formulas that follow immediately from
this identity, since the formulas will prove useful in the sequel.
All sums are over the range $1\le k<n$. We use the facts that ${k-1/2@\choose k-1} (k-1)={3\over2}{k-1/2@\choose k-2}$, that $k-1/2@\choose k-1$ is the coefficient
of $z↑{k-1}$ in $(1-z)↑{-3/2}$, etc.
\baselineskip18pt\lineskip3pt\lineskiplimit3pt \vbox{\tabskip 0pt plus 1000pt minus 1000pt \halign to \hsize{\hfill{#}\tabskip 0pt &{\null#}\hfill\tabskip 0 pt plus 1000pt minus 1000pt &\hfill \eq#\tabskip 0pt\cr \sum{k-1/2@\choose k-1}&={n-1/2@\choose n-2}&(1)\cr \sum{k-1/2@\choose k-1}(k-1)&={3\over2}{n-1/2@\choose n-3}&(2)\cr \sum{k-1/2@\choose k-1}(H_{k-1/2}-H_{1/2})&={n-1/2@\choose n-2}(H_{n-1/2}-H_{3/2}) &(3)\cr \sum{k-1/2@\choose k-1}(n-k)(H_{n-k}-H_1)&={n+1/2@\choose n-2}(H_{n+1/2}-H_{5/2}) &(4)\cr \sum{k-1/2@\choose k-1}(k-1)(H_{k-1/2}-H_{3/2})&={3\over2}{n-1/2@\choose n-3} (H_{n-1/2}-H_{5/2})&(5)\cr \sum{k-1/2@\choose k-1}{n-k-1/2@\choose n-k-1}(k-1)&={3\over2}{n\choose n-3}&(6)\cr \sum{k-1/2@\choose k-1}{n-k-1/2@\choose n-k-1}(H_{k-1/2}-H_{1/2})&= {n\choose n-2}(H_n-H_2)&(7)\cr \hskip-2em\sum{k-1/2@\choose k-1}{n-k-1/2@\choose n-k-1}(k-1)(H_{k-1/2}-H_{3/2})&= {3\over2}{n\choose n-3}(H_n-H_3)&(8)\cr }}\lineskiplimit0pt
Each of these identities is obtained by looking at the coefficients of the
product of two generating functions.

The answer to part (g) comes from \eq(4), after multiplying by $2/{n-1/2@\choose n-2}$.

(h)
We need to solve $L_n=n+{4\over5}(n+{1\over2})(H_{n+1/2}-H_{5/2})+ \sum q_{nk}L_k$, for $n\ge 2$. Trying $x_n=n-1$ in $x_n=a_n+\sum q_{nk}x_k$
gives $a_n={2\over5}n+{1\over5}$ for $n\ge 2$, by \eq(2), since $n-1=a_n+{3\over2} {n-1/2@\choose n-3}/{n-1/2@\choose n-2}=a_n+{3\over5}(n-2)$. Similarly, trying
$x_n=H_{n-1/2}-H_{1/2}$ gives $a_n={2\over3}$, by \eq(3); and $x_n=(n-1)(H_ {n-1/2}-H_{3/2})$ gives $a_n={2\over5}(n+{1\over2})(H_{n-1/2}-H_{5/2})+ {2\over5}(n-1)$ by~\eq(5). Taking an appropriate linear combination of all this
yields the solu\-tion $L_n=(2n+{1\over4})(H_{n-1/2}-H_{1/2})-{5\over6}(n-1)$.

(i)
We have $U_n=n+2\sum p_{nk}L_k$. Write $L_k=2(k-1)\*(H_{k-1/2}-H_{3/2}) +{9\over4}(H_{k-1/2}-H_{1/2})+{1\over2}(k-1)$ and use \eq(8), \eq(7), \eq(6), to get
$$\textstyle U_n=n+2(n-2)(H_n-{11\over6})+{9\over2}(H_n-{3\over2}) +{1\over2}(n-2).$$

We may conclude that LBTs do not deserve to be implemented; they offer us
instructive insights into discrete mathematics and the analysis of
algorithms, but they will never become known as Quicksearch. It is somewhat
surprising that $U_n\le L_n\le X_n$, since a reluctance to insert extreme''
elements might be thought to make the inequalities go the other~way.

\vfill\eject
\psmajor{Appendix F: Final Exam II and Solutions}
\psminor{Final Exam II}

\prob 1 (75 points).

Find the asymptotic value of $S_n=\sum_{0<k<n}H_k/(2n-k)$, correct to
terms of $O(n↑{-3/2})$.

\yskip\prob 2 (100 points total).

Let $a_n$ be the number of paths from $(0,0)$ to $(n,n)$ on a grid, where we
are allowed to go at each step from $(i,j)$ to $(i,j+1)$ or $(i+1,j)$ or
$(i+1,j+1)$. Thus, $(a_0,a_1,a_2,a_3,\ldots\,)=(1,3,13,63,\ldots\,)$.

\part a (50 points). Let $A(z)=\sum_n a_nz↑n$. Use the method of \eq(4.125)
to prove that $A(z)=1/\sqrt{@1-6z+z↑2}$.

\part b (50 points). Find the asymptotic value of $a_n$ as $n\to \infty$, giving
explicit values of constants $c$, $p$, and $\theta$ such that
$a_n=cn↑p\theta↑n+O(n↑{p-1}\theta↑n)$.

\yskip\prob 3 (125 points total).

A certain professor gives final exams that contain an infinite number of
problems. In order to solve each problem, the student must have
solved all of the preceding problems perfectly,
since the professor stops grading an
exam as soon as he finds a mistake. Each student has probability $p$ of
getting any particular problem right, independently of the other students, and
independently of the problem number. For example, if $p=\smash{1\over2}$, there is
probability $2↑{-n-1}$ that a particular exam will have exactly
$n$ problems right.

The professor gives an A$↑+$ to the student who solves the most problems,
provided that only one student had the maximum score. Otherwise nobody
in the class gets A$↑+$.

\part a (25 points). Write down an expression for the probability that
an A$↑+$ is given when $n$ students take the exam. (Your expression can
be left in the form of a summation, since there appears to be no closed
form'' for the probability in question.)

\part b (100 points). Find the asymptotic behavior of the probability that
an A$↑+$ is given after $n$ students take the exam, for fixed $p$ as $n\to \infty$.
Assume that $0<p<1$.

\yskip\noindent
{\bf Important note:} You must solve problem 3a correctly if you want to
get any credit for problem 3b. Make sure that your formula gives the
value $2p(1+p)↑{-1}$ when $n=2$ and the value $3p(1+p↑2)(1+p)↑{-1} (1+p+p↑2)↑{-1}$ when $n=3$, before you tackle the asymptotics.

\vfill\eject
\psminor{Solutions to Final Exam II}
\ans 1.
Summing by parts yields $$2S_n=\sum_{0<k<2n}H_k/(2n-k)-H_n↑2,$$which equals
$H_{2n}↑2-H_{2n}↑{(2)}-H_n↑2$ by \eq(1.48).
Now$$\textstyle H_{2n}↑2-H_n↑2=\bigl(@\ln n+\ln2+\gamma+{1\over4n}+O(n↑{-2})\bigr)↑2- \bigl(@\ln n+\gamma+{1\over2n}+O(n↑{-2})\bigr)↑2,$$and $H_{2n}↑{(2)}={1\over6}\pi↑2 -{1\over2n}+O(n↑{-2})$ by [Knuth III; exercise 6.1--8]. Multiplying out and collecting
terms yields
$$\twoline{(\ln2)(\ln n)+\gamma\ln2+{1\over2}(\ln2)↑2-{1\over12}\pi↑2}{5pt }{-{1\over4}n↑{-1}\ln n+{1\over4}n↑{-1}(\ln2+1-\gamma)+O(n↑{-2}\log n).}$$
[This problem was too easy. It would have been better to ask
for the asymptotics of, say, $\sum_{1\le k\le n-\sqrt n}H_k/(2n-k)$. Then the
asymptotics could be worked out most easily by using the identity
$\sum_{1\le k\le m}H_k/(n-k)=\sum_{1\le k\le m}H_{n-k}/k-H_mH_{n-m-1}$.]

\vskip .25in
\ans 2.
Set $F(w,z)=\sum a_{mn}w↑mz↑n$, where $a_{mn}$ is the number of paths from
$(0,0)$ to $(m,n)$. Then $F=1+wF+zF+wzF$, so we have $F(w,z)=(1-w-z-wz)↑{-1}$.
The diagonal terms are
$$A(z)={1\over2\pi i}\oint F(t,z/t){dt\over t} = {1\over2\pi i}\oint {dt\over t-t↑2-z-zt}.$$
The denominator can be written in factored form, $-\bigl( t-r(z)\bigr)\bigl( t-s(z)\bigr)$, where
$r(z)={1\over2}\bigl(1-z+\sqrt{@1-6z+z↑2}\,\bigr)$ and
$s(z)={1\over2}\bigl(1-z-\sqrt{@1-6z+z↑2}\,\bigr)$.

Let $|z|$ be small, so that $r(z)$ is near 1 and $s(z)$ is near 0.
Integrate around a contour with small $|t|$ that encloses the point $s(z)$;
then make $|z|$ and $|s(z)|$ even smaller so that $|z/t|$ is small enough
to guarantee absolute convergence of $\sum a_{mn}t↑m(z/t)↑n$. (It is clear
that $a_{mn}\le 3↑{m+n}$, so such a contour exists.) The result is $A(z)= \hbox{residue at }s(z)=1/\bigl( r(z)-s(z)\bigr)$.

Now $A(z)=1/\sqrt{(1-\theta z)(1-\phi z)}$, where $\theta=3+\sqrt8$ and
$\phi=3-\sqrt8$. Let $w=\theta z$ and $\alpha=\phi/\theta$ so that
$A(z)=B(w)=1/\sqrt{(1-w)(1-\alpha w)}=\sum(a_n/\theta↑n)w↑n$.
We therefore want to find the asymptotics of the coefficients of
the inverse of \eq(4.108). We have $1/\sqrt{@1-\alpha w}=1/\sqrt{@1-\alpha-\alpha(w-1)} =(1-\alpha)↑{-1/2}+(w-1)R(w)$ where $R$ is analytic for $|w|<\alpha↑{-1}$, so the
coefficients $r_n$ of $R$ are $O(\beta↑{-n})$ for some $\beta>1$. Thus
$B(w)=(1-\alpha )↑{-1/2}(1-w)↑{-1/2}+(1-w)↑{1/2}R(w)$, where the latter term is
$\sum{1/2@\choose k}(-w)↑k\sum r_mw↑m$; it follows as in \eq(4.114) that its $n$th
coefficient is $O(n↑{-3/2})$. The $n$th coefficient of the first term is
$(1-\alpha )↑{-1/2}{-1/2@\choose \,\,n}(-1)↑n=(1-\alpha )↑{-1/2}{n-1/2@\choose n}$, which is
of order $n↑{-1/2}$, so $a_n=\theta↑n(1-\alpha )↑{-1/2}{n-1/2@\choose n}+ O(a_n/n)$.

We have ${-1/2@\choose \,\,n}(-1)↑n=2↑{-2n}{2n\choose n}$, and Stirling's
approximation tells us that this is $1/\sqrt{\pi n}+O(n↑{-3/2})$. Thus the
$$a_n={1+\sqrt2\over2↑{5/4}\sqrt{\pi n}}(3+\sqrt8)↑n+O\bigl((3+\sqrt8)↑n n↑{-3/2}\bigr).$$

Incidentally, the numbers $a_{mn}$ arise in surprisingly many contexts.
We have, for example, $a_{mn}=\kern-.3pt\sum{m\choose k}{n+k\choose m}=\kern-.3pt\sum{(m+n-k)!\over (m-k)!(n-k)!k!}=\kern-.3pt\sum{m\choose k}{n\choose k}2↑k=\sum{k\choose m}{k\choose n} 2↑{-1-k}$. Also, $a_{mn}$ is the number of different
$n$-tuples of integers $(x_1,\ldots, x_n)$ such that $|x_1|+\cdots+|x_n|\le m$; this is the volume of a sphere of
radius $m$ in the $n$-dimensional Lee metric.''

\vskip .25in
\ans 3.
The probability that a particular student gets A$↑+$ with exactly $m$ problems
correct is the probability of scoring $m$ $\bigl($namely $p↑m(1-p)\bigr)$
times the probability that each other student missed at least one of the first
$m$ problems $\bigl($namely $(1-p↑m)↑{n-1}\bigr)$. Multiplying by $n$, since
each student has the chance for an A$↑+$, we obtain $A↑+_n=n(1-p)\sum_{m\ge 0} p↑m(1-p↑m)↑{n-1}$. (Similar formulas arise in the analysis of radix exchange
sorting in [Knuth III, 5.2.2], when $p={1\over2}$, and in the more general
treatment of exercise 6.3--19.)

Let $Q_n=nA↑+_{n+1}(n+1)↑{-1}(1-p)↑{-1}=\sum_{m\ge 0}np↑m(1-p↑m)↑n$. Let
$x=np↑m$; the summand is $x(1-x/n)↑n$, which is $xe↑{-x}\bigl(1+O(x↑2/n)\bigr)$
when $x\le n↑\epsilon$. Let $T_n=\sum_{m\ge 0}np↑me↑{-np↑m}$. We have $Q_n-T_n=X_n+Y_n$
where
\baselineskip17pt \eqalign{X_n&=\textstyle\sum_{m\ge 0,np↑m\ge n↑\epsilon }\,np↑m\bigl((1-p↑m)↑n-e↑{-np↑m}\bigr)\cr &=\textstyle\sum_{m\ge 0,np↑m\ge n↑\epsilon }\,np↑mO(e↑{-np↑m})=O(n\log n\,e↑{-n↑\epsilon })\cr}
is exponentially small, since $1-p↑m\le e↑{-p↑m}$ and there are $O(\log n)$ terms.
Also
\baselineskip17pt \eqalign{Y_n&=\textstyle\sum_{m\ge 0,np↑m<n↑\epsilon }\,np↑m\bigl((1-p↑m)↑n-e↑{-np↑m}\bigr)\cr &=\textstyle\sum_{m\ge 0,np↑m<n↑\epsilon }\,np↑mO\bigl( e↑{-np↑m}(np↑m)↑2/n\bigr),\cr}
which is $O(n↑{3\epsilon -1})$ since it reduces to a geometric series after we
use the obvious upper bound $e↑{-np↑m}\le 1$. Applying the Gamma function method,
we have
\eqalign{T_n&=\sum_{m\ge 0}{1\over2\pi i}\int_{1/2-i\infty }↑{1/2+i\infty }np↑m\Gamma(z) (np↑m)↑{-z}dz\cr \noalign{\vskip 5pt} &={n\over2\pi i}\int_{1/2-i\infty }↑{1/2+i\infty }{\Gamma(z)n↑{-z}\over1-p↑{1-z}}dz,\cr}
(cf.\ Eq.\ 5.2.2--45), which can be evaluated as the negative of the sum of
the inte\-grand's residues at its poles in the right half plane. Thus $$T_n= -{1\over\ln p}-{2\over\ln p}\sum_{k\ge 1}\textstyle\Re\bigl(\Gamma(1+2\pi ik/\!\ln p) \exp(-2\pi ik\ln n/\!\ln p)\bigr)+O(n↑{-M})$$ for arbitrary $M$. The quantity
in the sum is bounded since it is periodic in $n$ (note that it has the
same value at $n$ and $pn$). So we can say that $A↑+_n=(1-p)/\!\ln({1\over p}) +f(n)+O({1\over n})$, where $f(n)$ is a certain periodic function. The
absolute value of $f(n)$ is extremely small unless $p$ is extremely small,
since $\Gamma(1+ti)=O(t↑{1/2}e↑{-\pi t/2})$; and each term of $f(n)$ has average
value zero, so $f(n)$ is zero on the average. But $f(n)$ is present and it
is not $o(1)$. One might suspect that $A↑+_n$ would approach 0 or 1 when $n\to \infty$,
so the result is a bit surprising.

Exercise 5.2.2--54 gives another approach to the answer, by which we obtain
the convergent series
$$A↑+_n={1-p\over\ln(1/p)}\biggl(1+2n\sum_{k\ge 1}\Re\bigl( B(n,1+\textstyle{2\pi ik\over\ln p})\bigr)\biggr).$$
The Beta function in this sum has the asymptotic value $$\textstyle n↑{-1-ibk}\Gamma(1+ibk)\bigl(1-{1\over2}(ibk+b↑2k↑2)n↑{-1}+O(n↑{-2})\bigr),$$
where $b=2\pi/\!\ln p$; so we obtain the periodic function mentioned above,
as well as the coefficient of $n↑{-1}$. (It appears that exercise 5.2.2--54
should be mentioned much more prominently in the next edition of [Knuth III].)

\eject
\psmajor{Appendix G: Midterm Exam III and Solutions}
\psminor{Midterm Exam III}

\def\yskip{\vskip 3pt plus 1pt minus 1pt}
\def\prob #1. {\par\yskip\noindent{\bf Problem #1.}\enspace\ignorespaces}

\prob 1.
Let $C_n$ be the $n@$th Catalan number,
$$C_n={2n\choose n}\,{1\over n+1}\,,$$
and let $Q_n=C_n(H_{2n}-H_{n+1})$. Thus we have the following values for small~$n$:
\vcenter{\halign{\hfil#&\hfil\;#\;\hfil&\hfil#\qquad &\hfil#\qquad &\hfil#\hfil\qquad &\hfil#\hfil\qquad &\hfil#\hfil\qquad &\hfil#\hfil&#\hfil\cr n&=&0&1&2&3&4&5\cr \noalign{\smallskip} C_n&=&1&1&2&5&14&42\cr \noalign{\smallskip} Q_n&=&-1&0&{1\over 2}&{11\over 6}&{73\over 12}&{1207\over 60}\cr}}
Prove the amazing identity
$$\sum_{k=0}↑nC_kQ_{n-k}=Q_{n+1}-{2n+1\choose n}\,{1\over n+1}\,.$$
{\sl Hint:\/} Consider the derivative of
$$B(z)↑x=\sum_{n\ge 0}{2n+x\choose n}\,{x\over 2n+x}\,z↑n\,,\qquad B(z)={1-\sqrt{@1-4z\mathstrut}\over 2z}$$
with respect to $x$.

\prob 2.
Given $0\le m<F_{n+2}$, the {\sl Fibonacci representation\/} of~$m$ is defined
to be $(d_n\ldots d_2d_1)_F$ if $m=d_nF_{n+1}+\cdots +d_2F_3+d_1F_2$,
where each $d_k$ is 0 or~1 and $d_kd_{k+1}=0$ for $1\le k<n$. The
{\sl Fibonacci permutation of\/} order~$n$ is the permutation
of $\{0,1,\ldots,F_{n+2}-1\}$ that is obtained by reflecting the representations
of $0,1,\ldots,F_{n+2}-1$
from right to left. For example, here are the representations and their
reflections when $n=4$:
\vcenter{\halign{\hfil#\hfil\qquad&\hfil#\hfil\qquad &\hfil#\hfil\qquad &\hfil#\hfil\qquad &\hfil#\hfil\qquad &\hfil#\hfil\cr m&representation&reflection&permuted value&inversions\cr \noalign{\smallskip} 0&(0000)_F&(0000)_F&0&0\cr 1&(0001)_F&(1000)_F&5&0\cr 2&(0010)_F&(0100)_F&3&1\cr 3&(0100)_F&(0010)_F&2&2\cr 4&(0101)_F&(1010)_F&7&0\cr 5&(1000)_F&(0001)_F&1&4\cr 6&(1001)_F&(1001)_F&6&1\cr 7&(1010)_F&(0101)_F&4&3&.\cr}}
Let $X_n$ be the total number of inversions in the Fibonacci permutation
of order~$n$. (When $1\le n\le 5$ we have $X_n=0,1,3,11,32$, respectively.)
Find a closed form for~$X_n$ in terms of Fibonacci numbers. (Do not use
the number $(1+\sqrt{5}{\thinspace})/2$ explicitly in your answer.)

\prob 3.
The Eikooc Monster is a dual to Paterson's Cookie Monster: The probability
that it {\sl doesn't\/} grow is proportional to its size. More precisely,
if $E$ has eaten $k$~cookies before a new cookie is thrown, it eats the
new cookie with probability $1-pk$. (Monster~$C$ absorbs cookies that fall on~it,
while $E$ eats those that it can see in the rest of the yard.) The differential
operator~$\Theta$ corresponding to~$E$ is
$$x+p(1-x)xD.$$

Find a family of eigenoperators for $\Theta$ corresponding to the book's family
$V_1,V_2,\ldots\,$. Use
your operators to deduce the mean and variance of the number of distinct
coupons collected
after $n$~purchases of random coupons drawn uniformly and
independently from the set $\{1,2,\ldots,m\}$.

\smallskip
Derive asymptotic formulas for the mean and variance of this number when
$n=cm$, for fixed~$c$, correct to $O(1)$ as $m$ and $n\to\infty$.

\prob 4.
Find the mean and variance of the number of comparisons when the following
sorting algorithm is applied to~$n$ distinct inputs $a[1\,.\,.\,n]$:

\smallskip
{\bf procedure} \hbox{\sl pokeysort\/\hskip1pt}($n$: {\sl integer\/});\cr
\qquad {\bf begin if} $n>1$ {\bf then}\cr
\qquad\qquad {\bf repeat} Set $k$ to random element of $\{1,2,\ldots,n\}$;\cr
\qquad\qquad Exchange $a[k]\leftrightarrow a[n]$;\cr
\qquad\qquad \hbox{\sl pokeysort\/\hskip1pt}$(n-1)$;\cr
\qquad\qquad {\bf until} $a[n-1]\le a[n]$;\cr
}

\vskip 10pt
\psminor{Solutions to Midterm Exam III}
\ans 1.
[The class presented a variety of interesting approaches; here is yet
another, which includes several formulas that may be handy in other
investigations.] If we write $C_{-1}=-1$, we
have $B(z)=\sum_{n\ge 0}C_nz↑n$ and $B(z)↑{-1}=-\sum_{n\ge 0}C_{n-1}z↑n$. The
derivative of the hinted formula~is
$$B(z)↑x\ln B(z)=\sum_{n\ge 0}{2n+x\choose n}\,{z↑n\over 2n+x}\,\bigl(1+x (H_{2n+x-1}-H_{n+x})\bigr)\,.$$
The special case $x=0$ gives $\ln B(z)=\sum_{n\ge 1}{2n\choose n}\,{1\over 2n}\,z↑n$;
the special case $x=1$ gives $B(z)\ln B(z)=\sum_{n\ge 1}(C_n+Q_n)z↑n$.
Multiplying by $B(z)↑y$ and equating coefficients of~$z↑n$ gives
$$\let\quad\relax \twoline{\sum_{k=0}↑n{2k{+}x\choose k}\,{1\over 2k{+}x} \bigl(1+x(H_{2k+x-1}-H_{k+x})\bigr){2n{-}2k{+}y\choose n-k}\,{y\over 2n{-}2k{+}y}}{1pt}{={2n{+}x{+}y\choose n}\,{1\over 2n{+}x{+}y}\,\bigl(1+(x{+}y)(H_{2n+x+y-1}-H_{n+x+y})\bigr).}$$
Set $x=1$ and $y=-1$, getting $\sum_{k=0}↑n(C_k+Q_k)(-C_{n-1-k})={2n\choose n}\, {1\over 2n}$. But when $n>0$ we have
$\sum_{k=0}↑nC_kC_{n-1-k}=0$, hence
$\sum_{k=0}↑{n+1}Q_kC_{n-k}=-{2n+2@\choose n+1}\,{1\over 2n+2}=-{2n+1\choose n} {1\over n+1}$.
\par}
\vskip2pt
\ans 2.
\vskip-2pt
[This was in part an exercise in mastering complexity without getting mired
in details.] Everybody solved this problem by deriving a recurrence, usually with
the idea that $X_n=X_{n-1}+X_{n-2}+C_n$ where $C_n$ is the number of
cross inversions'' between the first block of $F_{n+1}$ values and the
last block of~$F_n$ values. The value of~$C_n$ can be written in several ways,
notably $C_n=Y_n+Z_{n-1}+Y_{n-2}+Z_{n-3}+\cdots=C_n+Y_n+Z_{n-1}+C_{n-2}$, where $Y_n= {F_n\choose 2}$, $Z_n=Y_n+F_n$, and $C_0=C_1=0$. It turns out that
$C_n={1\over 2}\,F_{n-1}(F_{n+2}-1)$. Jim Hwang made the interesting observation
that the inversion table entries $B_0 B_1 B_2\ldots$ begin the same for each~$n$;
therefore it would be of interest to study the partial sums
$B_0+B_1+\cdots +B_{m-1}$ as a function of~$m$.

But there's another interesting approach to the solution, based directly
on the binary representations: Each inversion corresponds to strings
$\alpha$, $\beta$, $\beta'$, $\gamma$ of~0s and~1s such that
$$(\alpha\;0\;\beta\;1\;\gamma)_F<(\alpha\;1\;\beta'\;0\;\gamma)_F\,, \qquad (\alpha\;0\;\beta\;1\;\gamma)_F↑R>(\alpha\;1\;\beta'\;0\;\gamma)_F↑R\,.$$
(If $i<j$ and $a_i>a_j$, the Fibonacci forms of~$i$ and~$j$ must differ
first in~0 versus~1, last in~1 versus~0.) The number of such pairs
with $|\alpha|=k$, $|\beta|=n-k-l$, and $|\gamma|=l$, is
$F_{k+1}F↑2_{n-k-l-1}F_{l+1}$; hence $X_n$ is the sum of this quantity
over $0\le k$, $l\le n$.

Let $F(z)=\sum F_{k+1}z↑k=1/(1-z-z↑2)$ and
$$G(z)=\sum_{k\ge 0}F_{k+1}↑2z↑k={1\over 5}\left({3-2z\over 1-3z+z↑2}+{2\over 1+z}\right)\,.$$
Then $X_n$ is $[z↑{n-2}]\,F(z)↑2G(z)$. The partial fraction expansion of
this generating function involves the denominators
$(1-\phi z)↑2$, $(1-\phi z)$, $(1-\hat{\phi}z)↑2$, $(1-\hat{\phi}z)$,
$(1-\phi↑2z)$, $(1-\hat{\phi}↑2z)$, and $1+z$. Hence there must be seven
constants such that
$$X_n=(\alpha n+\beta)F_n+(\gamma n+\delta)F_{n+1}+\epsilon\,F_{2n}+\zeta F_{2n+1}+\eta(-1)↑n\,.$$
{\rh macsyma} quickly determines these constants when given the values
of~$X_n$ for $1\le n\le 7$, using {\tt solve}'. [Incidentally,
{\tt solve}' is much quicker
$$X_n={7F_{2n+1}+4F_{2n}-(4n+15)F_{n+1}+(2n+7)F_n+8(-1)↑n\over 20}\,.$$

Incidentally, a random permutation of this size would have exactly
$I_n={1\over 4}\, F_{n+2}(F_{n+2}-1)={1\over 20}\,\bigl(7F_{2n+1}+4F_{2n}-2(-1)↑n-5F_{n+1} -5F_n\bigr)$ inversions on the average. The Fibonacci permutation is pretty
random'' by the inversion-count criterion, since $X_n-I_n$ is of order $\sqrt{I_n}\log I_n$.

\ans 3.
We have $U_nx=nU_{n-1}+U_n$, hence $U_n(1-x)=-n\,U_{n-1}$.
Let's search for an eigenoperator of the form $U_nx↑{n-a}$: We have
\eqalign{U_nx↑{n-a}\Theta&=U_nx↑{n-a}\bigl(x+p(1-x)xD\bigr)\cr &=U_nx↑{n+1-a}+p\,U_n(1-x)x↑{n+1-a}D\cr &=U_nx↑{n+1-a}-pn\,U_{n-1}\bigl(Dx↑{n+1-a}-(n+1-a)x↑{n-a}\bigr)\cr &=(1-pn)U_nx↑{n+1-a}+pn(n+1-a)U_{n-1}x↑{n-a}\cr &=(1-pn)(U_nx↑{n-a}+n\,U_{n-1}x↑{n-a})+pn(n+1-a)U_{n-1}x↑{n-a}\cr &=(1-pn)U_nx↑{n-a}+(n+pn-pna)U_{n-1}x↑{n-a}\,.\cr}
Therefore we get an eigenoperator with eigenvalue $1-pn$ when $a=1+p↑{-1}$.

The formula $U_nf(x)=\sum_k{n\choose k}f↑{(k)}(1)U_{n-k}$ tells us that
the eigenoperator $U_nx↑{n-a}$ can be written $\sum_k{n\choose k} (n-1-p↑{-1})↑{\underline{k}} U_{n-k}$. It is convenient to normalize it so that the coefficient of~$U_0$
is~$+1$; then $V_nG(z)=1$ when $G(z)=1$. With this normalization
(suggested by Arif Merchant), we have
\eqalignno{V_n&=\sum_k{n\choose k}\,{(-1)↑k\over (p↑{-1})↑{\underline{k}}} \,U_k\cr \noalign{\hbox{and therefore}\nobreak} U_n&=(p↑{-1})↑{\underline{n}}\sum_k{n\choose k}(-1)↑kV_k\,.\cr}
If $G_n(z)=\Theta↑n(1)$, the mean and variance are now easily found
to be respectively $p↑{-1}\bigl(1-(1-p)↑n\bigr)$ and $p↑{-2}(1-p)(1-2p)↑n+p↑{-1}(1-p)↑n -p↑{-2}(1-p)↑{2n}$, in agreement with the answer to [Knuth III;
exercise 5.2.5--5] when $p=1/m$.

When $n=cm$, the mean is $(1-e↑{-c})m+O(1)$; the variance reduces to
$(e↑c-1-c)e↑{-2c}m+O(1)$, fairly small.

\def\mean{\mathop{\rm Mean\mskip1mu}}
\def\var{\mathop{\rm Var\mskip1mu}}

\ans 4.
The probability generating function $G_n(z)$ is defined by the recurrence
$G_1(z)=1$, $G_n(z)=z\,G_{n-1}(z) \bigl({1\over n}+{n-1\over n}\,G_n(z)\bigr)$ for $n>1$. Hence
$$G_n(z)=F_n\bigl(z\,G_{n-1}(z)\bigr)\,,\qquad F_n(z)={z\over n-(n-1)z}\,.$$
Now [Knuth I; exercise 1.2.10--17] tells us that
\eqalign{\mean(G_n)&=\mean(F_n)\bigl(1+\mean(G_{n-1})\bigr) =n\bigl(1+\mean(G_{n-1})\bigr)\cr \noalign{\smallskip} \var(G_n)&=\var(F_n)\bigl(1+\mean(G_{n-1})\bigr)↑2+\mean(F_n) \var(G_{n-1})\cr \noalign{\smallskip} &={n-1\over n}\,\mean(G_n)↑2+n\,\var(G_{n-1})\,.\cr}
Dividing these recurrences by $n!$ leads to sums such as
$$\mean(G_n)=\sum_{1\le k<n}\,{n!\over k!}=S_n-n!-1$$
where $S_n$ has a convenient closed form:
$$S_n=\sum_{0\le k\le n}\,{n!\over k!}=n!\,e-e\gamma(n+1,1)=\lfloor n!\,e\rfloor\,.$$
The variance can also be expressed in terms of $S_n$:
$$\var(G_n)=(S_n-n!-n-1)↑2+3S_n-n!-n↑2-2n-3\,.$$
Incidentally, we can solve'' the recurrence for $G_n(z)$ and write
$$G_n(z)=z↑n/H_n(a)\,,\qquad H_n(z)=n!\,\left(z+\sum_{k=2}↑n\,\left({1\over k!} -{1\over (k-1)!}\right)z↑k\right)\,;$$
then $\var(G_n)=-\var(H_n)$, and the latter can be calculated directly.

\psmajor{Appendix H: Final Exam III and Solutions}
\psminor{Final Exam III}

\prob 1.
A certain gambler starts with \$1 at time~0. If he has$\$k$ at time~$t$,
he makes $k$ independent fair bets of \$1 each, double or nothing; this determines his capital at time$t+1$.$\bigl($Thus, at time~1, he is equally likely to be broke or to have~\$2. At time~2, he has (\$0, \$2, \$4) with probability$({5\over 8}\,,\;{2\over 8}\,,\;{1\over 8})$.$\bigr)$(a)~Find the mean and variance of his capital at time~$t$. (b)~Find the asymptotic probability~$p_t$that he is broke at time~$t$, with absolute error$O(t↑{-2})$. {\sl Hint:\/} Consider the quantity$1/(1-p_{t+1})-1/(1-p_t)-1/2$. \prob 2. Each of$n$parallel processors is executing a random process such that, if the processor is running at time~$t$, it is stopped at time$t+1$with probability~$p$(independent of all other circumstances). Once stopped, a~processor remains stopped. Find the asymptotic value of the expected waiting time until all processors have stopped, assuming that they are all running at time~0. Your answer should be correct to$O(n↑{-1})$. \prob 3. Find the asymptotic value of$S_n=\sum_{1\le k<n/2}e↑{-k↑2\!/n}k↑{-2}$, with absolute error$O(n↑{-10})$. (This is the quantity$r_{-2}(n/2)$in [Knuth~III; Eq.\ 5.2.2--35].) \prob 4. Last year's AA qual included a problem in which it was shown that the number of ways to arrange$n~coins into a so-called fountain'' has the generating function $$\sum_{n\ge 0}f_nz↑n={1\over\displaystyle 1-{\strut z↑{\phantom{1↑{(}}}\over \displaystyle 1-{z↑{2↑{\mathstrut}}\over \displaystyle 1-{z↑{3↑{\mathstrut}}\over\ldots}}}} ={P(z)\over Q(z)}$$ where \eqalign{P(z)&=\sum_{k\ge 0}{(-1)↑kz↑{k(k+1)}\over (1-z)(1-z↑2)\,\ldots\,(1-z↑k)}\;,\cr Q(z)&=\sum_{k\ge 0}{(-1)↑kz↑{k↑2}\over (1-z)(1-z↑2)\,\ldots\,(1-z↑k)}\;.\cr} Prove thatQ(z)$has exactly one root~$\rho$in the disk$|z|\le .6$, and that$f_n=c\rho↑{-n}+O(.6↑{-n})$for some nonzero constant~$c$. Use {\rh macsyma} to evaluate$\rho↑{-1}$and~$c$to several decimal places. {\sl Hints:\/} First express the quantity$(1-z)(1-z↑2)(1-z↑3)Q(z)$in the form$A(z)+R(z)$, where$A(z)$is a polynomial of degree~9 and$|R(z)|$is small when$|z|<1$. Find the roots of$A(z)$using the {\tt allroots} command. Then find a radius~$r$such that$A(z)$has exactly one root for$|z|\le r$and such that$|R(z)|<|A(z)|$for$|z|=r$. Then apply Rouch\'e's theorem. \vskip 10pt \psminor{Solutions to Final Exam III} \ans 1. (a)$G_0(z)=z$;$G_{t+1}(z)=f\bigl(G_t(z)\bigr)$, where$f(z)={1\over 2}+{1\over 2}
z↑2$. Since$f$has mean and variance~1, [Knuth~I; exercise 1.2.10--17] implies that${\rm Mean}(G_t)=1$and${\rm Var}(G_t)=t$. (b) Let$\epsilon_t=(1-p_t)/2$. We have$p_t=G_t(0)$, so$p_{t+1}={1\over 2}+{1\over 2}p_t↑2$. Hence$\epsilon_{t+1}=\epsilon_t(1-\epsilon_t)$, and we have $${1\over\epsilon_{t+1}}={1\over\epsilon_t}+1+{1\over \epsilon_t↑{-1}-1}\,,\qquad \epsilon_0={1\over 2}\;.$$ A bootstrapping argument now shows that$\epsilon_t↑{-1}\ge t+2, hence \eqalignno{\epsilon_t↑{-1}&\le t+2+\sum_{k=0}↑{t-1}{1\over k+1}=t+2+H_t\,;\cr \noalign{\hbox{hence}} \epsilon_t↑{-1}&\ge t+2+\sum_{k=0}↑{t-1}{1\over k+1+H_k}\cr \noalign{\smallskip} &=t+2+\sum_{k=0}↑{t-1}\left({1\over k+1}+O\left({\log k\over k↑2}\right)\right) =t+2+H_t+O(1)\,.\cr} We have proved that\epsilon_t↑{-1}=t+\ln t+O(1)$; hence$p_t=1-2/\bigl(t+\ln t+O(1)\bigr)=1-2t↑{-1}+2t↑{-2}\ln t+O(t↑{-2})$. [Let$C=\lim_{t\to\infty}(\epsilon_t↑{-1}-t-H_t)$. Is it possible to go further and estimate the quantity$\delta_t=\epsilon_t↑{-1}-t-H_t-C$, as$t\to\infty$?] \ans 2. Let$q=1-p$and$Q=1/q$. The probability that the processors are not all stopped at time~$t$is$R_t=1-(1-q↑t)↑n$, so the expected waiting time is$W=\sum_{t\ge 0}R_t=1-\sum_{t\ge 1}\sum_{k\ge 1}{n\choose k}(-1)↑kq↑{tk}=1-\sum_{k\ge 1}
{n\choose k}(-1)↑k/(Q↑k-1)$. We proceed as in [Knuth III; exercise 5.2.2--54] to represent the sum~as $${(-1)↑n\,n!\over 2\pi i}\oint{dz\over z(z-1)\,\ldots\,(z-n)\,(Q↑z-1)}$$ where the contour encircles$\{1,\ldots,n\}$and no other poles. If we increase the contour to a large rectangle whose upper and lower segments have imaginary part$\pm 2\pi(N+{1\over 2})/\ln Q$where$N$is an integer, the contour integral approaches zero, so the sum of the residues inside approaches zero. The residue at~0 is the coefficient of~$z$in $${1\over (1-z)(1-{1\over 2}\,z)\,\ldots\,(1-{1\over n}\,z)(1+{1\over 2}\, z\ln Q+\cdots)\ln Q}\;,$$ namely$(H_n-{1\over 2}\ln Q)/\ln Q$. The sum of residues at$1,\ldots,n$is$1-W$. And the sum of residues at$\ln Q+ibm$and$\ln Q-ibm$, where$b=2\pi/\ln Q$and$m\ge 1$, is$1/\ln Qtimes twice the real part~of \eqalign{{n!\over (ibm)(ibm+1)\,\ldots\,(ibm+n)} &=B(n+1,ibm)=\Gamma(ibm)\,n↑{\underline{ibm}}\cr \noalign{\smallskip} &=\Gamma(ibm)\,n↑{ibm}\bigl(1+O(n↑{-1})\bigr)\,.\cr} (The last estimate comes by expandingn↑{\underline{ibm}}$in terms of generalized Stirling numbers; for example, we have $$n↑{\underline{\alpha}}=n↑{\alpha}\left(\left[{\alpha\atop\alpha}\right] -\left[{\alpha\atop\alpha-1}\right]\,n↑{-1}+\left[{\alpha\atop\alpha-2}\right] \,n↑{-2}+O(n↑{-3})\right)\,.$$ See [GKP; exercise 9.44].) Now$|\Gamma(ibm)n↑{ibm}|=O(e↑{-\pi m/2})$, so we have $$W={H_n\over\ln Q}+{1\over 2}+{2\over\ln Q}\sum_{m\ge 1}\Re\bigl(\Gamma(ibm) n↑{ibm}\bigr)+O(n↑{-1})\,.$$ The sum is a bounded function$f(n)$that is periodic in the sense that$f(n)=f(Qn)$. Tom\'as Feder used Euler's summation formula to deduce the remarkable representation $$f(n)=\int_0↑\infty\left(\!\left({\log u/n\over \log Q}\right)\!\right)\,e↑{-u}\,du$$ where$(\mskip-1mu(x)\mskip-1mu)$is the sawtooth function [Knuth II; \S3.3.3]. \ans 3. Let$g(x)=(e↑{-x↑2}-1)/x↑2$and$f(x)=g(x/\sqrt{n}\thinspace). Then \eqalign{\quad&\hskip-1em nS_n-1-nH_{n-1}↑{(2)}\cr &=\sum_{0\le k<n}f(k)+O(e↑{-n/4})\cr \noalign{\smallskip} &=\int_0↑nf(x)\,dx+\sum_{j=1}↑{18}\,{B_j\over j!}\,f↑{(j-1)}(x) \,\biggr|_0↑n+O(n↑{-9})\cr \noalign{\smallskip} &=\sqrt{n}\int_0↑{\sqrt{n}}g(y)\,dy+\sum_{j=1}↑{18}\,{B_j\over j!}\, n↑{-(j-1)/2}g↑{(j-1)}(y)\,\biggr|_0↑{\sqrt n} +O(n↑{-9})\,.\cr} Consider first \eqalign{\int_0↑{\sqrt{n}}g(y)\,dy&={1-e↑{-y↑2}\over y}\,\biggr|_0↑{\sqrt{n}} -2\int_0↑{\sqrt{n}}e↑{-y↑2}\,dy\cr \noalign{\smallskip} &={1\over\sqrt{n}}-2\int_0↑\infty e↑{-y↑2}\,dy+O(e↑{-n})\,.\cr} We also have $$g↑{(j-1)}(\sqrt{n}\thinspace)=(-1)↑j2↑{@\overline{j-1}}n↑{-(j+1)/2}-O(e↑{-n})$$ andg↑{(j-1)}(0)$is nonzero only when$j$is odd, so we can square it for$j>1$. Thus $$nS_n-1-nH_{n-1}↑{(2)}=1-\sqrt{\pi n}-{1\over 2}\,\left(-{1\over n}+1\right) +\sum_{j=1}↑9\,{B_{2j}\over n↑{2j}}+O(n↑{-9})\,.$$ Also $$H_{n-1}↑{(2)}={\pi↑2\over 6}-{1\over n}-{1\over 2n↑2}-\sum_{j=1}↑9\, {B_{2j}\over n↑{2j+1}}+O(n↑{-10})$$ by Euler's summation formula. Finally, therefore, $$S_n={\pi↑2\over 6}-\sqrt{\pi\over n}+{1\over 2n}+O(n↑{-10})\,.$$ The error is, in fact,$O(n↑{-1000000})$. (Check:$S_{10}=1.134434895$; the approximation yields$1.134434945$.) There are (at least) two other ways to solve this problem: We can use the Gamma-function approach, as pointed out in [Knuth III; exercise 5.2.2--51]; or we can use the Poisson summation formula (found, for example, in [Henrici II]). \ans 4. (The hints are due to A.~Odlyzko.) We have $$A(z)=1-2z-z↑2+z↑3+3z↑4+z↑5-2z↑6-z↑7-z↑9$$ and $$R(z)={z↑{16}\over 1-z↑4}\left(1-{z↑9\over 1-z↑5}+{z↑9\over 1-z↑5}\, {z↑{11}\over 1-z↑6}-\cdots\right)\,.$$ If$|z|=r<1$we have $$|R(z)|\le {r↑{16}\over 1-r↑4}\left(1+{r↑9\over 1-r↑5}+\left({r↑9\over 1-r↑5} \right)↑2+\cdots\right)={r↑{16}\over 1-r↑4}\,{1-r↑5\over 1-r↑5-r↑9}\;.$$ The roots of$A(z)=0are, in increasing order of magnitude: \vcenter{\halign{\hfil#\;&#\hfil\qquad&\hfil#\;&#\hfil\cr r_1&\approx .58\cr r_2,r_3&\approx .75\pm .08i&|r_2|&=|r_3|\approx .75\cr r_4,r_5&\approx -.47\pm .82i&|r_4|&=|r_5|\approx .94\cr r_6,r_7&\approx -1.06\pm .37i&|r_6|&=|r_7|\approx 1.12\cr r_8,r_9&\approx .49\pm 1.58i&|r_8|&=|r_9|\approx 1.66\,.\cr}} To apply Rouch\'e's theorem, we want to find a value ofr$such that$|A(z)|$is relatively large when$|z|=r$but$|R(z)|$is relatively small. The hard part is to show that$|A(z)|=|z-r_1|\ldots|z-z_9|$is relatively large. %When$|z|=r$we have$|z-r_k|\ge \bigl|r-|r_k|\bigr|$. To get the best %bound of this kind on$|z-r_1|\,|z-r_2|\,|z-r_3|$, we choose %$r={2\over 3}r_1+{1\over 3}|r_2|\approx .63$; then$|z-r_1|\,|z-r_2|\,|z-r_3|
%\ge (r-r_1)(|r_2|-r)↑2\approx (0.059)(0.12)↑2=.00081$. %The value of$|R(z)|$for$|z|=r$is at most$.00084$, so %we would like to show that$|z-r_4|\ldots|z-r_9|is a bit greater than~1. One idea is to observe that |z-a+bi|\,|z-a-bi|\ge \cases{(r-|a|)↑2+b↑2,&if a(r↑2+a↑2+b↑2)>2r(a↑2+b↑2);\cr \noalign{\smallskip} \min((r-|a|)↑2+b↑2,\bigl|b(r↑2-a↑2-b↑2)\bigr|/|a+ib|),\hidewidth\cr &otherwise.\cr} (The proof is by settingz=re↑{i\theta}$and taking the derivative with respect to~$\theta$. Extrema occur when$\sin\theta=0$or when we have$\cos\theta=
{a(r↑2+a↑2+b↑2)}/\allowbreak{(2r(a↑2+b↑2))}$.) Unfortunately this idea isn't enough by itself; the product of all these bounds turns out to be less than$r↑{16}$. Better bounds are possible if we use the inequality$|z-r_k|\ge \bigl||r_k-r|-|z-r|\bigr|$. Then if$|r_k-r|>.5$we can conclude that$|z-r_k|\ge |r_k-r|-.5$, whenever$|z-r|\le .5$; similarly if$|r_k-r|<.5$we can conclude that$|z-r_k|\ge .5-|r_k-r|$, whenever$|z-r|\ge .5$. Putting these ideas together yields a rigorous proof that$|A(z)|>|R(z)|$for all$z$on the circle$|z|=r$, for any choice of$r$between$.59$and$.68$. (See the attached {\rh macsyma} transcript. The computed values~$r_1$, \dots,~$r_9$are only approximations to the true roots of$A(z)$; but the fact that the difference$(z-r_1)\ldots (z-r_9)-A(z)$has very small coefficients implies that our calculations are plenty accurate when$|z|\le 1$.) Consequently Rouch\'e's theorem applies, and$Q(z)$has exactly one root$\rho_0$inside$|z|=r$. This root is real, and Newton's method converges quickly to $$\rho_0=0.57614876914275660229786\ldots\;.$$ The contour integral $${1\over 2\pi i}\oint_{|z|=r}{P(z)\,dz\over Q(z)\,z↑{n+1}}$$ is$O(r↑{-n})$, and the sum of residues inside is $$f_n+{P(\rho_0)\over\rho_0↑{n+1}\,Q'(\rho_0)}\,.$$ Hence we have$f_n=c_0\rho_0↑{-n}+O(r↑{-n})$, where$c_0=P(\rho_0)/\bigl(
\rho_0Q'(\rho_0)\bigr); numerically \eqalign{{1\over\rho_0}&=1.7356628245303472565826\ldots\,;\cr c_0&=0.312363324596741453066279\ldots\,.\cr} It turns out that the next largest root ofQ(z)is also real; it is \eqalign{\rho_1&=.81559980\,;\cr c_1&=P(\rho_1)/\bigl(\rho_1Q'(\rho_1)\bigr)=.03795269\,.\cr} The graph ofQ(z)$looks like this for$.5\le z\le .9$: $$\unitlength=4in \beginpicture(.4,.35)(.5,-.16) \put(.5,.1){\line(1,0){.4}} \put(.5,-.1){\line(1,0){.4}} \put(.4,0){\line(1,0){.65}} \multiput(.5,-.12)(.1,0){5}{\line(0,1){.24}} \put(.35,.1){\makebox(0,0){+0.1}} \put(.35,0){\makebox(0,0){0}} \put(.35,-.1){\makebox(0,0){-0.1}} \put(.5,-.16){\makebox(0,0){.5}} \put(.6,-.16){\makebox(0,0){.6}} \put(.7,-.16){\makebox(0,0){.7}} \put(.8,-.16){\makebox(0,0){.8}} \put(.9,-.16){\makebox(0,0){.9}} \put(1,-.03){\makebox(0,0){z}} \put(.25,0){\makebox(0,0){Q(z)}} \put(0,0){\squine(0.5, 0.5236327368284876, 0.5400000000000001, 0.1607637889320886, 0.1077938177855414, 0.07301236593798955)} \put(0,0){\squine(0.5400000000000001, 0.5622130799750084, 0.5800000000000001, 0.07301236593798955, 0.02580818992881538, -0.007216386230973315)} \put(0,0){\squine(0.5800000000000001, 0.6016056422468865, 0.6200000000000001, -0.007216386230973315, -0.04733109843864552, -0.07280004719818223)} \put(0,0){\squine(0.6200000000000001, 0.6411242272954975, 0.6600000000000002, -0.07280004719818223, -0.102048791285633, -0.114783453094367)} \put(0,0){\squine(0.6600000000000002, 0.6804856360064547, 0.7000000000000002, -0.114783453094367, -0.1286042195637996, -0.1240537441044569)} \put(0,0){\squine(0.7000000000000002, 0.719115264339661, 0.7400000000000003, -0.1240537441044569, -0.119596333076086, -0.0963585241403373)} \put(0,0){\squine(0.7400000000000003, 0.7506632423766526, 0.7800000000000003, -0.0963585241403373, -0.0844938592954389, -0.04237635843304233)} \put(0,0){\squine(0.7800000000000003, 0.804765033367961, 0.82, -0.04237635843304233, -0.00682228229334505, 0.003190035175067329)} \put(0,0){\squine(0.82, 0.836420823067818, 0.86, 0.003190035175067329, 0.01398168950674518, 0.006320703647183735)} \put(0,0){\squine(0.86, 0.883689352464209, 0.9, 0.006320703647183735, -0.00137607876267075, -0.0006123112298865863)} \endpicture$$ There is another root between .88 and .89. To check, Odlyzko computed$f_{120}=17002133686539084706594617194$, and found that$f_{120}-c_0/\rho_0↑{120}\approx 1.6\times 10↑9$. If we subtract$c_1/\rho_1↑{120}$the error goes down to$1.3\times 10↑5$. (Odlyzko's work was published in [Odlyzko 88] after this exam was given.) \eject {\parfillskip=0pt plus 1fil minus 4pt \vermode This is MACSYMA 304 (C1) t(k,z):=z↑(k↑2)/prod(1-z↑j,j,1,k); \good (C2) q(n,z):=sum((-1)↑k*t(k,z),k,0,n); \good (C3) a:num(factor(q(3,z))); 9 7 6 5 4 3 2 (D3) Z + Z + 2 Z - Z - 3 Z - Z + Z + 2 Z - 1 \good (C4) allroots(a); \good (C5) for n thru 9 do print(n,r[n]:rhs(part(d4,n)),abs(r[n])); 1 0.575774066 0.575774066 2 0.81792161 %I - 0.469966464 0.94332615 3 - 0.81792161 %I - 0.469966464 0.94332615 4 0.07522564 %I + 0.74832744 0.75209896 5 0.74832744 - 0.07522564 %I 0.75209896 6 0.36716983 %I - 1.05926119 1.1210923 7 - 0.36716983 %I - 1.05926119 1.1210923 8 1.58184962 %I + 0.493013173 1.65689777 9 0.493013173 - 1.58184962 %I 1.65689777 \good (C6) rmax(r):=r↑16/(1-r↑4)/(1-r↑9/(1-r↑5)); \good (C7) bound1(a,b,r):=block([t,s],s:a↑2+b↑2,t:(r-abs(a))↑2+b↑2, if a*(r↑2+s)>2*r*s then t else min(t,abs(b*(r↑2-s))/sqrt(s))); \good (C8) bound2(a,b,r):=block([t,s],s:abs(a+b*%i-r),t:bound1(a,b,r), if s<.5 then t else max(t,(.5-s)↑2)); \good (C9) bound3(a,b,r):=block([t,s],s:abs(a+b*%i-r),t:bound1(a,b,r), if s>.5 then t else max(t,(s-.5)↑2)); \good (C10) amin1(r):=(r-r[1])*prod(bound2(realpart(r[2*k]),imagpart(r[2*k]),r), k,1,4) + 0*"a lower bound for all z such that |z-r|>=.5"; \good (C11) amin2(r):=(r-r[1])*prod(bound3(realpart(r[2*k]),imagpart(r[2*k]),r), k,1,4) + 0*"a lower bound for all z such that |z-r|<=.5"; \good (C12) amin(r):=min(amin1(r),amin2(r)); \good (C13) for n:58 thru 70 do print(n,rmax(n*.01),amin(n*.01)); 58 1.86410865E-4 1.40064462E-4 59 2.4762821E-4 4.7992996E-4 60 3.2769739E-4 8.2895893E-4 61 4.320998E-4 1.18362144E-3 62 5.6784198E-4 1.54014562E-3 63 7.438718E-4 1.89452055E-3 64 9.7160927E-4 2.24249464E-3 65 1.26562865E-3 2.57957187E-3 66 1.6445353E-3 2.90100428E-3 67 2.13209912E-3 3.19357002E-3 68 2.75873208E-3 3.17922997E-3 69 3.56342027E-3 3.1048643E-3 70 4.596277E-3 2.92984536E-3 \good (C14) qprime(n,z):=sum((-1)↑k*t(k,z)*logtprime(k,z),k,0,n); \good (C15) logtprime(k,z):=k↑2/z+sum(j*z↑(j-1)/(1-z↑j),j,1,k); \good (C16) loop(z):=block([zo,zn],zo:0,zn:z, while abs(zo-zn)>10↑-10 do(zo:zn,print(zn:iterate(zo))),zo); \good (C17) t(8,.59)+0*"an upper bound on the alternating sum Q(.59)-Q(8,.59)"; (D17) 1.3545675E-14 \good (C18) iterate(z):=bfloat(z-q(8,z)/qprime(8,z)); \good (C19) loop(5.8B-1); 5.761132798756077B-1 5.761487662923891B-1 5.761487691427566B-1 5.761487691427566B-1 (D19) 5.761487691427566B-1 \good (C20) p(n,z):=sum((-1)↑k*t(k,z)*z↑k,k,0,n); \good (C21) c(rho):=-p(8,rho)/(rho*qprime(8,rho)); \good (C22) c(d19); (D18) 3.123633245967415B-1 \good (C23) expand(prod(z-r[k],k,1,9)-d3); 8 8 7 (D23) - 1.49011612E-8 %I Z - 7.4505806E-9 Z - 1.49011612E-8 %I Z \good 6 6 5 + 1.49011612E-8 %I Z + 8.9406967E-8 Z + 2.98023224E-8 %I Z \good 5 4 4 + 1.63912773E-7 Z + 1.1920929E-7 %I Z - 1.78813934E-7 Z \good 3 3 2 - 4.47034836E-8 %I Z - 2.98023224E-7 Z - 1.04308128E-7 %I Z \good 2 - 2.01165676E-7 Z - 5.2154064E-8 %I Z + 1.49011612E-7 Z + 7.4505806E-9 %I \good (C24) "The sum of the absolute values of those coefficients is an upper bound on the difference between the true A(z) and the polynomial that is bounded by amin"; } \psmajor{Appendix I: A Qualifying Exam Problem and Solution} \psminor{Qual Problem} The result of a recent midterm problem was to analyze LBTs and to show that their average path length is about the same as that of ordinary binary search trees. But shortly after the midterm was graded, our sources discovered that Quick was undaunted by that analysis. According to reliable reports, he has recently decided to try salvaging his idea by including new information in each node. The nodes in Quick's new data structures, which he calls ILBTs (Improved Late Binding Trees), contain a size field that tells how many leaves are in the subtree rooted at that node. Step (4) on page 106 is now replaced by a new step: When a branch node is being split, the insertion continues in whichever subtree is currently smaller. (If the subtree sizes are equal, a random decision is made as before.) The purpose of this problem is to carry out a top level'' analysis of Quick's new algorithm. Let$p_{nk}$be the probability that the root is$(k\\k+1)$after inserting a random permutation of$\{1,\ldots,n\}$. (We assume that all permutations of the$x$'s are equally likely; first$x_1$is made into an ILBT by itself, then$x_2$through$x_n$are inserted one by one.) Let$P_{nk}=n!\,p_{nk}$. Then it can be verified that we have the following values of$P_{nk}$for$1\le k<n$and$1\le n\le 6: \baselineskip13.pt \vbox{\halign{#:&\quad\hfil#&\quad\hfil#&\quad\hfil#&\quad\hfil#&\quad\hfil#\cr n=2&2\cr n=3&3&3\cr n=4&6&12&6\cr n=5&18&42&42&18\cr n=6&72&162&252&162&72\cr}} \yskip\noindent (a) Find a recurrence relation that defines the numbersP_{nk}$. \yskip\noindent (b) Let$Q_{nk}=2P_{nk}\max(k,n-k)/\bigl( k!(n-k)!\bigr), so that we have the following triangle: \baselineskip13.5pt \vbox{\halign{#:&\quad\hfil#&\quad\hfil#&\quad\hfil#&\quad\hfil#&\quad\hfil#\cr n=2&4\cr n=3&6&6\cr n=4&6&12&6\cr n=5&6&21&21&6\cr n=6&6&27&42&27&6\cr}} Show that for most values ofn$and$k$the numbers$Q_{nk}$satisfy the same recurrence as Pascal's triangle, i.e.,$Q_{nk}=Q_{(n-1)k}+Q_{(n-1)(k-1)}$. Find all the exceptions, and state the recurrence obeyed at the exceptional points. \yskip\noindent (c) Let$a_k=Q_{(2k)k}$. Prove that for$k>1$, $$a_k=\sum_{1\le j<k}{2j+1\over j}\,a_jc_{k-j},$$ where$c_n$is the number of binary trees with$n$external nodes. \yskip\noindent (d) Let$B(z)={1\over2}(1+\sqrt{@1-4z}\,)$and$C(z)={1\over2}(1-\sqrt{@1-4z}\,)$, so that$B(z)+C(z)=1$,$B(z)-C(z)=\sqrt{@1-4z}$,$B(z)C(z)=z$, and$C(z)↑2=C(z)-z$; recall that$C(z)$is the generating function$c_1z+c_2z↑2+
c_3z↑3+\cdots$for binary trees. Let$f_k=a_k/k$, and set up the generating function$F(z)=f_1z+f_2z↑2+\cdots\,$. Convert the recurrence in part (c) to a differential equation for$F$, and solve this equation to obtain a closed form'' for$a_k$. [{\sl Possible hint:\/} Show that the derivative of$B(z)F(z)$has a simple form.] \yskip\noindent (e) Apply the recurrence of part (b) to the generating function$Q(w,z)=
\sum_{k,n}Q_{nk}w↑kz↑{n-k}$, and use the values of$a_k$found in part (d) to obtain a formula for$Q(w,z)$as an explicit function of$w$and$z$. \yskip\noindent (f) Find a simple'' expression for the coefficient of$w↑nz↑{n+r}$in the power series for$\sqrt{@1-4wz}/(1-w-z)$, when$r\ge 0$. [{\sl Hint:\/} Consider the problem for fixed$r$and variable$n$. You may wish to use the identity$C(z)↑s/\sqrt{@1-4z}=\sum_n{2n+s\choose n}z↑{n+s}$and the facts about$B(z)$and$C(z)$that are stated in~(d).] \yskip\noindent (g) Show that, therefore, $$p_{nk}={1\over2}\left({k+1\over n-k}-{k\over n-k+1}\right) -{1\over2n}+{2k\over n(n-1)}\qquad\hbox{for 1\le k<{1\over2}n.}$$ Note: Do NOT simply take this formula or an equivalent one and prove it by induction. You should present a scenario that explains how you could have discovered this solution by yourself in a systematic manner without lucky guesses. \vfill\eject \psminor{Qual Solution} (a) If$x_1\ldots x_n$is a permutation of$\{1,\ldots,n\}$, let${\bar x}_1\ldots{\bar x}_{n-1}$be the permutation of$\{1,\ldots,n-1\}$that arises when the elements of$x_1\ldots x_{n-1}$that exceed$x_n$are reduced by 1. The permutation$x_1\ldots x_n$leads to the root$(k\\k+1)$if and only if one of the following happens: (1)$x_n<k$and${\bar x}_1\ldots{\bar x}_{n-1}$leads to the root$(k-1\\k)$. (2)$x_n=k$and${\bar x}_1\ldots{\bar x}_{n-1}$leads to the root$(k-1\\k)$and either$k-1<n-k$or ($k-1=n-k$and a random coin flip comes up heads). (3)~$x_n=k+1$and${\bar x}_1\ldots{\bar x}_{n-1}$leads to the root$(k\\k+1)$and either$k>n-1-k$or ($k=n-1-k$and a random coin flip comes up tails). (4)~$x_n>k+1$and${\bar x}_1\ldots{\bar x}_{n-1}$leads to the root$(k\\k+1)$. Therefore we find, for$1\le k<n$and$n>2$, $$\twoline{P_{nk}= P_{(n-1)(k-1)}\bigl( k-1+[n+1>2k]+\textstyle{1\over2}[n+1=2k]\bigr)}{3pt}{\null+ P_{(n-1)k}\bigl( n-k-1+[n-1<2k]+\textstyle{1\over2}[n-1=2k]\bigr).}$$ (b) It is easy to see that$P_{nk}=P_{n(n-k)}$, so$Q_{nk}=Q_{n(n-k)}$. Thus it suffices to consider$k\le n-k$. If$k<n-k-1$, the above recurrence reads $$\twoline{{Q_{nk}k!(n-k)!\over2(n-k)}= {Q_{(n-1)(k-1)}(k-1)!(n-k)!\over2(n-k)}(k-1+1)}{3pt}{\null+ {Q_{(n-1)k}k!(n-1-k)!\over2(n-k-1)}(n-k-1),}$$ i.e.,$Q_{nk}=Q_{(n-1)(k-1)}+Q_{(n-1)k}$. If$k=n-k$, it reads $$\twoline{{Q_{nk}k!\,k!\over2k}={Q_{(n-1)(k-1)}(k-1)!\,k!\over2k}(k-1+1)}{3pt}{\null+ {Q_{(n-1)k}k!(k-1)!\over2k}(k-1+1),}$$ so Pascal's relation holds again. But if$k=n-k-1$, we have $$\twoline{{Q_{nk}k!(k+1)!\over2(k+1)}=}{3pt }{{Q_{(n-1)(k-1)}(k-1)!(k+1)!\over2(k+1)}(k-1+1)+ {Q_{(n-1)k}k!\,k!\over2k}(k+\textstyle{1\over2}),}$$ hence$Q_{nk}=Q_{(n-1)(k-1)}+Q_{(n-1)k}+{1\over n-1}Q_{(n-1)k}$. By symmetry, if$k=n-k+1$we have$Q_{nk}=Q_{(n-1)(k-1)}+Q_{(n-1)k}+{1\over n-1}Q_{(n-1)(k-1)}$. Pascal's relation therefore holds except when$(n,k)=(2,1)$or when$|n-2k|=1$. (c) It is convenient to tip the triangle sideways and to associate$Q_{nk}$with the point$(k,n-k)$in a grid. We can interpret$Q_{nk}$as 4 times the sum, over all paths from$(1,1)$to$(k,n-k)$, of the products of the weights of the edges, where edges run from$(i,j)$to$(i+1,j)$and to$(i,j+1)$; the weight of such an edge is~1, except when$i=j$it is$1+1/(2j)$. Now$a_k$is 4 times the sum over paths from$(1,1)$to$(k,k)$, so we can break the sum into various sub-sums depending on the greatest diagonal point$(j,j)$on the path, for$j<k$. The$j$th sub-sum is$a_j$times$1+1/(2j)$times the number of subpaths from$(j,j)$to$(k,k)$that do not touch the diagonal, since all edge weights but the first are 1 on such subpaths. There are$2c_{k-j}$such subpaths. (d) Since$kf_k=\sum_j(2j+1)f_jc_{k-j}+4[k=1]$, we have$zF↑\prime(z)=4z+C(z)
\bigl(2zF↑\prime(z)+F(z)\bigr), and this simplifies to $$z\sqrt{@1-4z}\,F↑\prime(z)=4z+C(z)F(z).$$ Following the hint, which follows from the general method of finding an integrating factor for first-order differential equations, we find \eqalign{\bigl( B(z)F(z)\bigr)↑\prime &=B(z)F↑\prime(z)-F(z)/\sqrt{@1-4z}\cr \noalign{\vskip3pt} &={B(z)\over z\sqrt{@1-4z}}\bigl( z\sqrt{@1-4z}\,F↑\prime(z)-C(z)F(z)\bigr)\cr \noalign{\vskip3pt} &=4B(z)/\sqrt{@1-4z}=2/\sqrt{@1-4z}+2.\cr} ThusB(z)F(z)=2C(z)+2z$, and in a few more steps we find the solution$a_n=2n(c_n+c_{n+1})
=n{2n\choose n}\bigl({1\over2n-1}+{2\over n+1}\bigr)$, for$n\ge 1$. (e)$(1-w-z)Q(z)=\sum w↑kz↑{n-k}(Q_{nk}-Q_{(n-1)k}-Q_{(n-1)(k-1)})=
4wz+{1\over2}(w+z)(f_1wz+f_2w↑2z↑2+f_3w↑3z↑3+\cdots\,)$, hence we have $$Q(w,z)={4wz+\textstyle{1\over2}(w↑{-1}{-}w{+}z↑{-1}{-}z -(w↑{-1}{+}w{+}z↑{-1}{+}z)\sqrt{@1{-}4wz}\,)\over 1-w-z}.$$ (f) The coefficient of$w↑nz↑{n+r}$in$g(wz)h(w,z)$is the coefficient of$x↑n$in$g(x)h_r(x)$, if$h_r(x)=\sum a_{m(m+r)}x↑m$and$h(w,z)=
\sum a_{mn}w↑mz↑n$, since multiplication by$g(wz)$affects only the coefficients having the same exponent offset. Hence the coefficient of$w↑nz↑{n+r}$in$\sqrt{@1-4wz}/(1-w-z)$is the coefficient of$x↑n$in$\sqrt{@1-4x}\sum
{2n+r\choose n}x↑n=\bigl( C(x)/x\bigr)↑r=C(x)↑r\bigl( B(x)-C(x)\bigr) x↑{-r}/
\sqrt{@1-4x}=\bigl( C(x)/x\bigr)↑{r-1}/\sqrt{@1-4x}-
x\bigl( C(x)/x\bigr)↑{r+1}/\sqrt{@1-4x}=\sum\bigl({2n+r-1\choose n}-
{2n+r-1\choose n-1}\bigr) x↑n$. (g) For$r>0$, the coefficient of$w↑nz↑{n+r}$in$Q(w,z)$can now be computed by considering the various terms in part (e). Let$b_r={2n+r\choose n}$. Then$Q_{(2n+r)n}=4(b_{r-1}-b_{r-2})+{1\over2}\bigl({2n+r+1\choose n+1}-b_r+b_{r-1}
+b_{r+1}-b_{r-1}-{2n+r\choose n+1}+\break
b_r-b_{r-1}+b_{r-2}+b_r-2b_{r-1}+b_{r-2}
-b_r+b_{r+1}-b_r-b_{r-2}+b_{r-1}-b_{r-2}\bigr)=b_{r+1}+3b_{r-1}-4b_{r-2}$. Multiply by${1\over2}(n+r-1)!\,n!/(2n+r)!$to get$p_{(2n+r)n}$. A dif-\break ferent formula applies when$r=0$, because of the$w↑{-1}$and$z↑{-1}$terms. \yskip\noindent {\sl Final comment: A note to J. H. Quick.}\quad When$x=k/n<\smash{1\over2}$we have$p_{nk}\approx{1\over2}\bigl((1-x)↑{-2}-1\bigr)
+2x$, hence ILBT's do a reasonably good job of partitioning. The distribution of permutations in the left and right subtrees is not random, and we could perhaps pursue the analysis to find the average path length of ILBT's. But really, Mr.\ Quick, your algorithm still does not deserve to be implemented. The average path length will be somewhere between$2n\ln n$and$(1/\!\ln2)\,n\ln n$; the extra time your method takes at each node slows the program down so much that the slightly smaller path length is pointless. It was clear from the start that ILBT's would lose out to other methods in all respects (space, time, ease of implementation, and so on). The only saving feature was that your algorithms lead to instructive mathematics that we might someday be able to apply to the analysis of a really useful method. You undoubtedly knew that too, so thanks for the ideas.'' \vfill\eject % Index \def\rhead{index} \mark{index} \hsize 2.10 in \vsize 5.5 in \newif\ifleft \lefttrue \newbox\leftcol \output{\ifleft \global\setbox\leftcol=\box255 \global\leftfalse \else\twocol\fi} \def\twocol{{\hsize=4.5in \setbox255=\vbox{\line{\box\leftcol \hfill \box255}} \ifdim\vsize<6in \global\vsize=7in \setbox255=\vbox to \vsize{\null \vskip 1in \hbox{\tit Index} \vfill \box255}\fi \plainoutput} \global\lefttrue} \def\par{\endgraf\hangindent 20pt} \parindent 0pt \def\abreak{\penalty-5\vskip 10pt} \parskip 1pt plus 1pt minus.5pt \rightskip 0pt plus 10em \spaceskip .3333em plus .5em \interlinepenalty=5 \hyphenpenalty=10000 % avoid break at hyphen \exhyphenpenalty=10000 % avoid break at dash Abel-Plana formula, 100. Abelian theorem, 49. Aho, Alfred Vaino, 31, 34, 81. algebraic singularities, 70--71. Amble, Ole, 81, 85. ape, 89--90. Apostol, Tom Mike, 59, 81. asymptotic analysis, 46--80, 99--104, 111--114. \abreak Bailey, Wilfred Norman, 13, 81. balanced trees, 34. banana, 89. basic polynomials, 12--13. Bell polynomials, 78. Bender, Edward Anton, 71, 81. Bent, Samuel Watkins, 3. Bernoulli numbers, 63. Bernoulli polynomials, 62--63. Beta function, 114, 122. binary trees, 34, 91, 99, 102--104; {\sl see also\/} late binding trees. binomial identities, 5--14. bootstrapping, 47, 54, 56, 121. bounded variation, 60. Boyce, William Edward, 19, 81. Broder, Andrei, 3. de Bruijn, Nicolaas Govert, 30, 47, 50, 81, 82, 100, 103. \abreak Catalan numbers, 115--117. central limit theorem, 75--78. Chebyshev's inequality, 51. Chebyshev's inverse relation, 10. Clarkson, Kenneth Lee, 3, 109. Computer Science 255 aka 360, 3, 85--87. Comtet, Louis, 72, 81. context-free grammar, unambiguous, 92--93. continued fractions, 29--31. contour integrals, 69, 72--74, 76. convergent to a continued fraction,~30. cookie monster, 35--38, 41--44, 116. coupon collecting, 116, 119. cycle leaders, 25. \abreak Darboux's method, 69--72, 74. de Bruijn, Nicolaas Govert, 30, 47, 50, 81, 82, 100, 103. Delange, Hubert, 27, 82. diagonalization of series, 73--74, 112. differencing recurrence relations, 21. differential equations, 19, 24, 92, 131. digital sums, 26. DiPrima, Richard Clyde, 81. disguised linear recurrences, 29--31, 119. dissecting a sum, 48, 55--58. divide and conquer, 31. Doubilet, Peter, 84. Doubly exponential sequences, \hbox{31--34}. Drysdale, Robert Lewis (Scot), III, 3. \abreak Egorychev, Georgi\u\i\ Petrovich, 7; method of coefficients, 7--8. eigenoperators, 35--45, 118. eikooc monster, 116. Erd\H os, P\'al, 82, 86. Euler's constant$\gamma$, 51, 54, 101, 102. Euler's summation formula, 51, 57, 62--63, 100, 122, 123. exponential integral, 67. \abreak factorial powers, 11--13. factoring algorithms, 52. factorization, distinct degree, 52. Feder, Tom\'as, 122. Ferguson, David Elton, 88--89. Fibonacci numbers, 34, 117--118. Fibonacci permutations, 115--118. finite history reccurence relations, 15, 16--20. fountain, 120. Fredman, Michael Lawrence, 27--29, 82. full history reccurence relations, 15, 21--24, 90. \abreak Gamma function, 78, 114, 122. Gamma-function method, 114, 123. generating functions, 8, 11, 16, 18--19, 20, 24, 29, 35--45, 69--80, 89--98, 102--103, 108--110, 121, 129. Gessel, Ira Martin, 7. Glaisher's constant, 100. golden ratio, 34. Golomb, Solomon Wolf, 33. Gould, Henry Wadsworth, 10, 11. grading problem 111--114. Graham, Ronald Lewis, 6, 13, 16, 63, 82, 109, 122. grammar, context free, unambiguous, 92--93. greatest common divisor, 75. Greene, Curtis, 84. Greene, Daniel Hill, 0, 3, 85, 86. grid paths, 7, 111, 112, 131. Guibas, Leonidas Ioannis, 3, 86. \abreak Hadamard product, 74. Hardy, Godfrey Harold, 30, 50, 66, 82. harmonic numbers, identities, 14;\break examples, 23--24, 54, 107, 108, 110, 112; asymp\-totics, 51, 54. hashing, coalesced, 38--41; uniform, 42; secondary clustering, 43--45, 105, 107--108. Henrici, Peter, 13, 82, 123. Hertz Foundation, 3. hidden linear recurrences, 29--31, 119. Hobby, John Douglas, 109. l'Hospital's rule, 72. Hsu, Lee-Tsch Ching-Siur, 10, 82. Hwang, Lucius James, 117. hypergeometric series, 13. \abreak IBM, 122. implicit equations, 47. {\sl in situ} permutation, 25, 85. inclusion and exclusion, 10. induction from the other end, 35, 41--44, 91. inverse relations, 9--11, 65--66. inversion table, 90, 109. inversions, 115, 117. irreducible polynomials, 52. IRT, 76--78. \abreak Jonassen, Arne Tormod, 7, 82. Jordan, Camille, 15, 83. \abreak Kahaner, David Kenneth, 84. Knuth, Donald Ervin, 0, 3, 6, 7, 11, 13, 15, 16, 18, 21, 25, 27, 28, 34, 41, 45, 47, 52, 63, 64, 67, 70, 72, 81--83, 85--87, 90, 99, 100, 102--103, 109, 112--114, 119--123. \abreak Laplace's method, 74--77, 80. late binding trees, 105--110, 128--132. lattice paths, 7, 111, 112, 131. Laurent expansion, 69, 72. lectures, 85--86. Lee metric, 113. left-to-right extrema, 106--109. linear recurrences, 15--24, 29--31. Lueker, George Schick, 18, 83. \abreak {\rh macsyma}, 94--98, 118, 126--127. Mairson, Harry George, 3. median-of-three quicksort, 22--24. Mellin transform, 86, 114. Merchant, Arif Abdulhussein, 118. merging sequences, 28. Mertens, Franz Carl Josef, 67,~83. Milne-Thomson, Louis Melville, 15,~84. minvolution, 28. M\"obius inversion, 65--66. \abreak National Science Foundation, 3. Newton's expansion, 13. Newton's method, 125, 127. nonuniform distribution, 106--109. \abreak$O$-notation, 47. Odlyzko, Andrew Michael, 84, 124--125. Office of Naval Research, 3. Olver, Frank William John, 100. open addressing, {\sl see\/} hashing. operator methods, 12--13, 35--45, 89,~91. orthogonal relation, 9. \abreak Page, Ewan Stafford, 15, 84. parallel processing, 120. partial fractions, 16--17, 55. partitions, 52. Pascal's triangle, 128, 130. Patashnik, Oren, 6, 13, 16, 63, 82, 109, 122. Paterson, Michael Stewart, 35--45, 116. permutations, bubble sort, 88, 90; cocktail shaker, 88, 90; Fibonacci, 115--117; input model, 106, 108--110, 117; obtainable with a stack, 87. Plass, Michael Frederick, 28. Poisson summation formula, 123. pokeysort, 116, 119. polynomials, basic, 12--13; irreducible, 52. prime factors, distinct, 64--68; unusual, 99--102. prime numbers, asymptotically, 64, 101. \abreak Quick, Jonathan Horatio, 105, 128, 132. \abreak radix exchange sort, 11, 113. Ramshaw, Lyle Harold, 86. Read, Ronald Cedric, 81. recurrence relations, linear, 15--24; nonlinear, 25--34. R\'enyi, Alfr\'ed, 82. repertoire approach, 21--24, 107. residue theorem, 72, 76, 103, 112--114, 122, 125. Rice, Stephan Oswald, 82. Riemann zeta function, 55, 65--66. Riordan, John, 9--11, 84. roots of polynomial, 124--127. Rota, Gian-Carlo, 12, 13, 84. Rouch\'e's theorem, 124--125. Rousseau, Cecil Clyde, 8. \abreak saddle point method, 69, 74--80. Sch\"onhage, Arnold, 15, 83. secondary clustering, 43--45, 105. Sedgewick, Robert, 84, 85, 86. semi-invariants, 75, 80. shifting the mean, 78--80. sliding operators, 43--45, 105, 107--108. Sloane, Neal James Alexander, 31, 34, 81. Spiegel, Murray R., 15, 17, 20, 84. Stanley, Richard Peter, 84. Stieltjes constants$\gamma_r\$, 102.

Stieltjes integral, 59--68.

Stirling numbers, 11, 80, 122.

Stirling's approximation, 51, 113.

Stolarsky, Kenneth Barry, 27, 84.

Stolfi, Jorge, 94.

summation by parts 60, 112.

summation factors, 18, 20, 91.

\abreak
tangent, 92.

Tauberian theorem, 50, 53--54, 58.

Taylor's expansion, general, 12.

Thiele expansion, 75--78.

Trabb Pardo, Luis Isidoro, 83.

trees, balanced binary, 34;
binary search, 105--107, 108--110, 128--132;
external path length, 106--107;
late binding, 105--110, 128--132;
ordered oriented, 29;
representing binary, 88--89, 91;
total path length, 109;
traversing binary, 99, 102--104.

\abreak

undetermined coefficients, 17.

\abreak

de la Vall\'ee Poussin, Charles Louis Xavier Joseph, 64.

Vandermonde's theorem, 13.

variance, 37, 51, 91, 119, 121.

Vitter, Jeffrey Scott, 3.

\abreak

Watson, George Neville, 84.

Whittaker, Sir Edmund Taylor, 84.

Wilson, Leslie Blackett, 15, 83.

Winkler, Phyllis Astrid Benson, 3.

Wrig`