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Handle some confilicts in chapter 1 and 2

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Roope Salmi 2017-02-26 16:50:31 +02:00 committed by Roope Salmi
parent dc4e33512a 65186779bc
commit 57e13ada8b
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book.tex
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@ -1,6 +1,6 @@
\documentclass[twoside,12pt,a4paper,english]{book}
%\includeonly{luku01,kirj}
%\includeonly{chapter01,list}
\usepackage[english]{babel}
\usepackage[utf8]{inputenc}
@ -27,6 +27,8 @@
\usepackage{titlesec}
\usepackage{skak}
\usetikzlibrary{patterns,snakes}
\pagestyle{plain}
@ -48,6 +50,7 @@
\author{\Large Antti Laaksonen}
\makeindex
\usepackage[totoc]{idxlayout}
\titleformat{\subsubsection}
{\normalfont\large\bfseries\sffamily}{\thesubsection}{1em}{}
@ -64,7 +67,7 @@
\setcounter{tocdepth}{1}
\tableofcontents
\include{johdanto}
\include{preface}
\mainmatter
\pagenumbering{arabic}
@ -73,41 +76,45 @@
\newcommand{\key}[1] {\textbf{#1}}
\part{Basic techniques}
\include{luku01}
\include{luku02}
\include{luku03}
\include{luku04}
\include{luku05}
\include{luku06}
\include{luku07}
\include{luku08}
\include{luku09}
\include{luku10}
\include{chapter01}
\include{chapter02}
\include{chapter03}
\include{chapter04}
\include{chapter05}
\include{chapter06}
\include{chapter07}
\include{chapter08}
\include{chapter09}
\include{chapter10}
\part{Graph algorithms}
\include{luku11}
\include{luku12}
\include{luku13}
\include{luku14}
\include{luku15}
\include{luku16}
\include{luku17}
\include{luku18}
\include{luku19}
\include{luku20}
\include{chapter11}
\include{chapter12}
\include{chapter13}
\include{chapter14}
\include{chapter15}
\include{chapter16}
\include{chapter17}
\include{chapter18}
\include{chapter19}
\include{chapter20}
\part{Advanced topics}
\include{luku21}
\include{luku22}
\include{luku23}
\include{luku24}
\include{luku25}
\include{luku26}
\include{luku27}
\include{luku28}
\include{luku29}
\include{luku30}
\include{kirj}
\include{chapter21}
\include{chapter22}
\include{chapter23}
\include{chapter24}
\include{chapter25}
\include{chapter26}
\include{chapter27}
\include{chapter28}
\include{chapter29}
\include{chapter30}
\cleardoublepage
\phantomsection
\addcontentsline{toc}{chapter}{Bibliography}
\include{list}
\cleardoublepage
\printindex
\end{document}la
\end{document}

37
chapter01.tex
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@ -117,10 +117,10 @@ but now it suffices to write \texttt{cout}.
The code can be compiled using the following command:
\begin{lstlisting}
g++ -std=c++11 -O2 -Wall code.cpp -o code
g++ -std=c++11 -O2 -Wall code.cpp -o bin
\end{lstlisting}
This command produces a binary file \texttt{code}
This command produces a binary file \texttt{bin}
from the source code \texttt{code.cpp}.
The compiler follows the C++11 standard
(\texttt{-std=c++11}),
@ -286,7 +286,7 @@ Still, it is good to know that
the \texttt{g++} compiler also provides
a 128-bit type \texttt{\_\_int128\_t}
with a value range of
$-2^{127} \ldots 2^{127}-1$ or $-10^{38} \ldots 10^{38}$.
$-2^{127} \ldots 2^{127}-1$ or about $-10^{38} \ldots 10^{38}$.
However, this type is not available in all contest systems.
\subsubsection{Modular arithmetic}
@ -624,7 +624,7 @@ For example, in the above set
New sets can be constructed using set operations:
\begin{itemize}
\item The \key{intersection} $A \cap B$ consists of elements
that are both in $A$ and $B$.
that are in both $A$ and $B$.
For example, if $A=\{1,2,5\}$ and $B=\{2,4\}$,
then $A \cap B = \{2\}$.
\item The \key{union} $A \cup B$ consists of elements
@ -778,7 +778,9 @@ n! & = & n \cdot (n-1)! \\
\index{Fibonacci number}
The \key{Fibonacci numbers} arise in many situations.
The \key{Fibonacci numbers}
%\footnote{Fibonacci (c. 1175--1250) was an Italian mathematician.}
arise in many situations.
They can be defined recursively as follows:
\[
\begin{array}{lcl}
@ -790,7 +792,8 @@ f(n) & = & f(n-1)+f(n-2) \\
The first Fibonacci numbers are
\[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \ldots\]
There is also a closed-form formula
for calculating Fibonacci numbers:
for calculating Fibonacci numbers\footnote{This formula is sometimes called
\index{Binet's formula} \key{Binet's formula}.}:
\[f(n)=\frac{(1 + \sqrt{5})^n - (1-\sqrt{5})^n}{2^n \sqrt{5}}.\]
\subsubsection{Logarithms}
@ -887,12 +890,13 @@ The International Collegiate Programming Contest (ICPC)
is an annual programming contest for university students.
Each team in the contest consists of three students,
and unlike in the IOI, the students work together;
there is even only one computer available for each team.
there is only one computer available for each team.
The ICPC consists of several stages, and finally the
best teams are invited to the World Finals.
While there are tens of thousands of participants
in the contest, there are only 128 final slots available,
in the contest, there are only a small number\footnote{The exact number of final
slots varies from year to year; in 2016, there were 128 final slots.} of final slots available,
so even advancing to the finals
is a great achievement in some regions.
@ -924,7 +928,7 @@ Google Code Jam and Yandex.Algorithm.
Of course, companies also use those contests for recruiting:
performing well in a contest is a good way to prove one's skills.
\section{Books}
\section{Resources}
\subsubsection{Competitive programming books}
@ -933,12 +937,11 @@ concentrate on competitive programming and algorithmic problem solving:
\begin{itemize}
\item S. Halim and F. Halim:
\emph{Competitive Programming 3: The New Lower Bound of Programming Contests}, 2013
\emph{Competitive Programming 3: The New Lower Bound of Programming Contests} \cite{hal13}
\item S. S. Skiena and M. A. Revilla:
\emph{Programming Challenges: The Programming Contest Training Manual},
Springer, 2003
\item \emph{Looking for a Challenge? The Ultimate Problem Set from
the University of Warsaw Programming Competitions}, 2012
\emph{Programming Challenges: The Programming Contest Training Manual} \cite{ski03}
\item K. Diks et al.: \emph{Looking for a Challenge? The Ultimate Problem Set from
the University of Warsaw Programming Competitions} \cite{dik12}
\end{itemize}
The first two books are intended for beginners,
@ -952,9 +955,9 @@ Some good books are:
\begin{itemize}
\item T. H. Cormen, C. E. Leiserson, R. L. Rivest and C. Stein:
\emph{Introduction to Algorithms}, MIT Press, 2009 (3rd edition)
\emph{Introduction to Algorithms} \cite{cor09}
\item J. Kleinberg and É. Tardos:
\emph{Algorithm Design}, Pearson, 2005
\emph{Algorithm Design} \cite{kle05}
\item S. S. Skiena:
\emph{The Algorithm Design Manual}, Springer, 2008 (2nd edition)
\emph{The Algorithm Design Manual} \cite{ski08}
\end{itemize}

31
chapter02.tex
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@ -281,7 +281,11 @@ Still, there are many important problems for which
no polynomial algorithm is known, i.e.,
nobody knows how to solve them efficiently.
\key{NP-hard} problems are an important set
of problems, for which no polynomial algorithm is known \cite{gar79}.
of problems, for which no polynomial algorithm
is known\footnote{A classic book on the topic is
M. R. Garey's and D. S. Johnson's
\emph{Computers and Intractability: A Guide to the Theory
of NP-Completeness} \cite{gar79}.}.
\section{Estimating efficiency}
@ -309,15 +313,14 @@ assuming a time limit of one second.
\begin{center}
\begin{tabular}{ll}
input size ($n$) & required time complexity \\
typical input size & required time complexity \\
\hline
$n \le 10^{18}$ & $O(1)$ or $O(\log n)$ \\
$n \le 10^{12}$ & $O(\sqrt n)$ \\
$n \le 10^6$ & $O(n)$ or $O(n \log n)$ \\
$n \le 5000$ & $O(n^2)$ \\
$n \le 500$ & $O(n^3)$ \\
$n \le 25$ & $O(2^n)$ \\
$n \le 10$ & $O(n!)$ \\
$n \le 20$ & $O(2^n)$ \\
$n \le 500$ & $O(n^3)$ \\
$n \le 5000$ & $O(n^2)$ \\
$n \le 10^6$ & $O(n \log n)$ or $O(n)$ \\
$n$ is large & $O(1)$ or $O(\log n)$ \\
\end{tabular}
\end{center}
@ -353,8 +356,8 @@ time and even in $O(n)$ time.
Given an array of $n$ integers $x_1,x_2,\ldots,x_n$,
our task is to find the
\key{maximum subarray sum}\footnote{Bentley's
book \emph{Programming Pearls} \cite{ben86} made this problem popular.}, i.e.,
\key{maximum subarray sum}\footnote{J. Bentley's
book \emph{Programming Pearls} \cite{ben86} made the problem popular.}, i.e.,
the largest possible sum of numbers
in a contiguous region in the array.
The problem is interesting when there may be
@ -444,8 +447,8 @@ and the sum of the numbers is calculated to the variable $s$.
The variable $p$ contains the maximum sum found during the search.
The time complexity of the algorithm is $O(n^3)$,
because it consists of three nested loops and
each loop contains $O(n)$ steps.
because it consists of three nested loops
that go through the input.
\subsubsection{Algorithm 2}
@ -471,7 +474,9 @@ After this change, the time complexity is $O(n^2)$.
\subsubsection{Algorithm 3}
Surprisingly, it is possible to solve the problem
in $O(n)$ time, which means that we can remove
in $O(n)$ time\footnote{In \cite{ben86}, this linear-time algorithm
is attributed to J. B. Kadene, and the algorithm is sometimes
called \index{Kadene's algorithm} \key{Kadene's algorithm}.}, which means that we can remove
one more loop.
The idea is to calculate, for each array position,
the maximum sum of a subarray that ends at that position.

6
chapter03.tex
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@ -326,7 +326,8 @@ of the algorithm is at least $O(n^2)$.
It is possible to sort an array efficiently
in $O(n \log n)$ time using algorithms
that are not limited to swapping consecutive elements.
One such algorithm is \key{mergesort}
One such algorithm is \key{mergesort}\footnote{According to \cite{knu983},
mergesort was invented by J. von Neumann in 1945.}
that is based on recursion.
Mergesort sorts a subarray \texttt{t}$[a,b]$ as follows:
@ -538,8 +539,7 @@ but use some other information.
An example of such an algorithm is
\key{counting sort} that sorts an array in
$O(n)$ time assuming that every element in the array
is an integer between $0 \ldots c$ where $c$
is a small constant.
is an integer between $0 \ldots c$ and $c=O(n)$.
The algorithm creates a \emph{bookkeeping} array
whose indices are elements in the original array.

4
chapter04.tex
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@ -196,7 +196,7 @@ for (auto x : s) {
}
\end{lstlisting}
An important property of sets
An important property of sets is
that all the elements are \emph{distinct}.
Thus, the function \texttt{count} always returns
either 0 (the element is not in the set)
@ -723,7 +723,7 @@ $5 \cdot 10^6$ & $10{,}0$ s & $2{,}3$ s & $0{,}9$ s \\
\end{tabular}
\end{center}
Algorithm 1 and 2 are equal except that
Algorithms 1 and 2 are equal except that
they use different set structures.
In this problem, this choice has an important effect on
the running time, because algorithm 2

19
chapter05.tex
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@ -436,18 +436,18 @@ the $4 \times 4$ board are numbered as follows:
\end{tikzpicture}
\end{center}
Let $q(n)$ denote the number of ways
to place $n$ queens to te $n \times n$ chessboard.
The above backtracking
algorithm tells us that
there are 92 ways to place 8
queens to the $8 \times 8$ chessboard.
algorithm tells us that $q(n)=92$.
When $n$ increases, the search quickly becomes slow,
because the number of the solutions increases
exponentially.
For example, calculating the ways to
place 16 queens to the $16 \times 16$
chessboard already takes about a minute
on a modern computer
(there are 14772512 solutions).
For example, calculating $q(16)=14772512$
using the above algorithm already takes about a minute
on a modern computer\footnote{There is no known way to efficiently
calculate larger values of $q(n)$. The current record is
$q(27)=234907967154122528$, calculated in 2016 \cite{q27}.}.
\section{Pruning the search}
@ -716,7 +716,8 @@ check if the sum of any of the subsets is $x$.
The running time of such a solution is $O(2^n)$,
because there are $2^n$ subsets.
However, using the meet in the middle technique,
we can achieve a more efficient $O(2^{n/2})$ time solution.
we can achieve a more efficient $O(2^{n/2})$ time solution\footnote{This
technique was introduced in 1974 by E. Horowitz and S. Sahni \cite{hor74}.}.
Note that $O(2^n)$ and $O(2^{n/2})$ are different
complexities because $2^{n/2}$ equals $\sqrt{2^n}$.

120
chapter06.tex
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@ -103,8 +103,12 @@ is 6, the greedy algorithm produces the solution
$4+1+1$ while the optimal solution is $3+3$.
It is not known if the general coin problem
can be solved using any greedy algorithm.
can be solved using any greedy algorithm\footnote{However, it is possible
to \emph{check} in polynomial time
if the greedy algorithm presented in this chapter works for
a given set of coins \cite{pea05}.}.
However, as we will see in Chapter 7,
in some cases,
the general problem can be efficiently
solved using a dynamic
programming algorithm that always gives the
@ -530,7 +534,9 @@ the string \texttt{AB} or the string \texttt{C}.
\subsubsection{Huffman coding}
\key{Huffman coding} \cite{huf52} is a greedy algorithm
\key{Huffman coding}\footnote{D. A. Huffman discovered this method
when solving a university course assignment
and published the algorithm in 1952 \cite{huf52}.} is a greedy algorithm
that constructs an optimal code for
compressing a given string.
The algorithm builds a binary tree
@ -672,113 +678,3 @@ character & codeword \\
\texttt{D} & 111 \\
\end{tabular}
\end{center}
% \subsubsection{Miksi algoritmi toimii?}
%
% Huffmanin koodaus on ahne algoritmi, koska se
% yhdistää aina kaksi solmua, joiden painot ovat
% pienimmät.
% Miksi on varmaa, että tämä menetelmä tuottaa
% aina optimaalisen koodin?
%
% Merkitään $c(x)$ merkin $x$ esiintymiskertojen
% määrää merkkijonossa sekä $s(x)$
% merkkiä $x$ vastaavan koodisanan pituutta.
% Näitä merkintöjä käyttäen merkkijonon
% bittiesityksen pituus on
% \[\sum_x c(x) \cdot s(x),\]
% missä summa käy läpi kaikki merkkijonon merkit.
% Esimerkiksi äskeisessä esimerkissä
% bittiesityksen pituus on
% \[5 \cdot 1 + 1 \cdot 3 + 2 \cdot 2 + 1 \cdot 3 = 15.\]
% Hyödyllinen havainto on, että $s(x)$ on yhtä suuri kuin
% merkkiä $x$ vastaavan solmun \emph{syvyys} puussa
% eli matka puun huipulta solmuun.
%
% Perustellaan ensin, miksi optimaalista koodia vastaa
% aina binääripuu, jossa jokaisesta solmusta lähtee
% alaspäin joko kaksi haaraa tai ei yhtään haaraa.
% Tehdään vastaoletus, että jostain solmusta lähtisi
% alaspäin vain yksi haara.
% Esimerkiksi seuraavassa puussa tällainen tilanne on solmussa $a$:
% \begin{center}
% \begin{tikzpicture}[scale=0.9]
% \node[draw, circle, minimum size=20pt] (3) at (3,1) {\phantom{$a$}};
% \node[draw, circle, minimum size=20pt] (2) at (4,0) {$b$};
% \node[draw, circle, minimum size=20pt] (5) at (5,1) {$a$};
% \node[draw, circle, minimum size=20pt] (6) at (4,2) {\phantom{$a$}};
%
% \path[draw,thick,-] (2) -- (5);
% \path[draw,thick,-] (3) -- (6);
% \path[draw,thick,-] (5) -- (6);
% \end{tikzpicture}
% \end{center}
% Tällainen solmu $a$ on kuitenkin aina turha, koska se
% tuo vain yhden bitin lisää polkuihin, jotka kulkevat
% solmun kautta, eikä sen avulla voi erottaa kahta
% koodisanaa toisistaan. Niinpä kyseisen solmun voi poistaa
% puusta, minkä seurauksena syntyy parempi koodi,
% eli optimaalista koodia vastaavassa puussa ei voi olla
% solmua, josta lähtee vain yksi haara.
%
% Perustellaan sitten, miksi on joka vaiheessa optimaalista
% yhdistää kaksi solmua, joiden painot ovat pienimmät.
% Tehdään vastaoletus, että solmun $a$ paino on pienin,
% mutta sitä ei saisi yhdistää aluksi toiseen solmuun,
% vaan sen sijasta tulisi yhdistää solmu $b$
% ja jokin toinen solmu:
% \begin{center}
% \begin{tikzpicture}[scale=0.9]
% \node[draw, circle, minimum size=20pt] (1) at (0,0) {\phantom{$a$}};
% \node[draw, circle, minimum size=20pt] (2) at (-2,-1) {\phantom{$a$}};
% \node[draw, circle, minimum size=20pt] (3) at (2,-1) {$a$};
% \node[draw, circle, minimum size=20pt] (4) at (-3,-2) {\phantom{$a$}};
% \node[draw, circle, minimum size=20pt] (5) at (-1,-2) {\phantom{$a$}};
% \node[draw, circle, minimum size=20pt] (8) at (-2,-3) {$b$};
% \node[draw, circle, minimum size=20pt] (9) at (0,-3) {\phantom{$a$}};
%
% \path[draw,thick,-] (1) -- (2);
% \path[draw,thick,-] (1) -- (3);
% \path[draw,thick,-] (2) -- (4);
% \path[draw,thick,-] (2) -- (5);
% \path[draw,thick,-] (5) -- (8);
% \path[draw,thick,-] (5) -- (9);
% \end{tikzpicture}
% \end{center}
% Solmuille $a$ ja $b$ pätee
% $c(a) \le c(b)$ ja $s(a) \le s(b)$.
% Solmut aiheuttavat bittiesityksen pituuteen lisäyksen
% \[c(a) \cdot s(a) + c(b) \cdot s(b).\]
% Tarkastellaan sitten toista tilannetta,
% joka on muuten samanlainen kuin ennen,
% mutta solmut $a$ ja $b$ on vaihdettu keskenään:
% \begin{center}
% \begin{tikzpicture}[scale=0.9]
% \node[draw, circle, minimum size=20pt] (1) at (0,0) {\phantom{$a$}};
% \node[draw, circle, minimum size=20pt] (2) at (-2,-1) {\phantom{$a$}};
% \node[draw, circle, minimum size=20pt] (3) at (2,-1) {$b$};
% \node[draw, circle, minimum size=20pt] (4) at (-3,-2) {\phantom{$a$}};
% \node[draw, circle, minimum size=20pt] (5) at (-1,-2) {\phantom{$a$}};
% \node[draw, circle, minimum size=20pt] (8) at (-2,-3) {$a$};
% \node[draw, circle, minimum size=20pt] (9) at (0,-3) {\phantom{$a$}};
%
% \path[draw,thick,-] (1) -- (2);
% \path[draw,thick,-] (1) -- (3);
% \path[draw,thick,-] (2) -- (4);
% \path[draw,thick,-] (2) -- (5);
% \path[draw,thick,-] (5) -- (8);
% \path[draw,thick,-] (5) -- (9);
% \end{tikzpicture}
% \end{center}
% Osoittautuu, että tätä puuta vastaava koodi on
% \emph{yhtä hyvä tai parempi} kuin alkuperäinen koodi, joten vastaoletus
% on väärin ja Huffmanin koodaus
% toimiikin oikein, jos se yhdistää aluksi solmun $a$
% jonkin solmun kanssa.
% Tämän perustelee seuraava epäyhtälöketju:
% \[\begin{array}{rcl}
% c(b) & \ge & c(a) \\
% c(b)\cdot(s(b)-s(a)) & \ge & c(a)\cdot (s(b)-s(a)) \\
% c(b)\cdot s(b)-c(b)\cdot s(a) & \ge & c(a)\cdot s(b)-c(a)\cdot s(a) \\
% c(a)\cdot s(a)+c(b)\cdot s(b) & \ge & c(a)\cdot s(b)+c(b)\cdot s(a) \\
% \end{array}\]

10
chapter07.tex
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@ -708,7 +708,8 @@ depends on the values of the objects.
\index{edit distance}
\index{Levenshtein distance}
The \key{edit distance} or \key{Levenshtein distance}
The \key{edit distance} or \key{Levenshtein distance}\footnote{The distance
is named after V. I. Levenshtein who discussed it in connection with binary codes \cite{lev66}.}
is the minimum number of editing operations
needed to transform a string
into another string.
@ -983,9 +984,10 @@ $2^m$ distinct rows and the time complexity is
$O(n 2^{2m})$.
As a final note, there is also a surprising direct formula
for calculating the number of tilings\footnote{Surprisingly,
this formula was discovered independently
by \cite{kas61} and \cite{tem61} in 1961.}:
for calculating the number of tilings:
% \footnote{Surprisingly,
% this formula was discovered independently
% by \cite{kas61} and \cite{tem61} in 1961.}:
\[ \prod_{a=1}^{\lceil n/2 \rceil} \prod_{b=1}^{\lceil m/2 \rceil} 4 \cdot (\cos^2 \frac{\pi a}{n + 1} + \cos^2 \frac{\pi b}{m+1})\]
This formula is very efficient, because it calculates
the number of tilings in $O(nm)$ time,

7
chapter09.tex
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@ -440,7 +440,8 @@ we can conclude that $\textrm{rmq}(2,7)=1$.
\index{binary indexed tree}
\index{Fenwick tree}
A \key{binary indexed tree} or \key{Fenwick tree} \cite{fen94}
A \key{binary indexed tree} or \key{Fenwick tree}\footnote{The
binary indexed tree structure was presented by P. M. Fenwick in 1994 \cite{fen94}.}
can be seen as a dynamic version of a prefix sum array.
This data structure supports two $O(\log n)$ time operations:
calculating the sum of elements in a range
@ -738,7 +739,9 @@ takes $O(1)$ time using bit operations.
\index{segment tree}
A \key{segment tree} is a data structure
A \key{segment tree}\footnote{The origin of this structure is unknown.
The bottom-up-implementation in this chapter corresponds to
the implementation in \cite{sta06}.} is a data structure
that supports two operations:
processing a range query and
modifying an element in the array.

3
chapter10.tex
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@ -391,7 +391,8 @@ to change an iteration over permutations into
an iteration over subsets, so that
the dynamic programming state
contains a subset of a set and possibly
some additional information.
some additional information\footnote{This technique was introduced in 1962
by M. Held and R. M. Karp \cite{hel62}.}.
The benefit in this is that
$n!$, the number of permutations of an $n$ element set,

12
chapter13.tex
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@ -24,7 +24,9 @@ for finding shortest paths.
\index{Bellman–Ford algorithm}
The \key{Bellman–Ford algorithm} \cite{bel58} finds the
The \key{Bellman–Ford algorithm}\footnote{The algorithm is named after
R. E. Bellman and L. R. Ford who published it independently
in 1958 and 1956, respectively \cite{bel58,for56a}.} finds the
shortest paths from a starting node to all
other nodes in the graph.
The algorithm can process all kinds of graphs,
@ -331,7 +333,9 @@ original Bellman–Ford algorithm.
\index{Dijkstra's algorithm}
\key{Dijkstra's algorithm} \cite{dij59} finds the shortest
\key{Dijkstra's algorithm}\footnote{E. W. Dijkstra published the algorithm in 1959 \cite{dij59};
however, his original paper does not mention how to implement the algorithm efficiently.}
finds the shortest
paths from the starting node to all other nodes,
like the Bellman–Ford algorithm.
The benefit in Dijsktra's algorithm is that
@ -594,7 +598,9 @@ at most one distance to the priority queue.
\index{Floyd–Warshall algorithm}
The \key{Floyd–Warshall algorithm} \cite{flo62}
The \key{Floyd–Warshall algorithm}\footnote{The algorithm
is named after R. W. Floyd and S. Warshall
who published it independently in 1962 \cite{flo62,war62}.}
is an alternative way to approach the problem
of finding shortest paths.
Unlike the other algorihms in this chapter,

14
chapter15.tex
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@ -123,7 +123,8 @@ maximum spanning trees by processing the edges in reverse order.
\index{Kruskal's algorithm}
In \key{Kruskal's algorithm} \cite{kru56}, the initial spanning tree
In \key{Kruskal's algorithm}\footnote{The algorithm was published in 1956
by J. B. Kruskal \cite{kru56}.}, the initial spanning tree
only contains the nodes of the graph
and does not contain any edges.
Then the algorithm goes through the edges
@ -409,7 +410,11 @@ belongs to more than one set.
Two $O(\log n)$ time operations are supported:
the \texttt{union} operation joins two sets,
and the \texttt{find} operation finds the representative
of the set that contains a given element.
of the set that contains a given element\footnote{The structure presented here
was introduced in 1971 by J. D. Hopcroft and J. D. Ullman \cite{hop71}.
Later, in 1975, R. E. Tarjan studied a more sophisticated variant
of the structure \cite{tar75} that is discussed in many algorithm
textbooks nowadays.}.
\subsubsection{Structure}
@ -567,7 +572,10 @@ the smaller set to the larger set.
\index{Prim's algorithm}
\key{Prim's algorithm} \cite{pri57} is an alternative method
\key{Prim's algorithm}\footnote{The algorithm is
named after R. C. Prim who published it in 1957 \cite{pri57}.
However, the same algorithm was discovered already in 1930
by V. Jarník.} is an alternative method
for finding a minimum spanning tree.
The algorithm first adds an arbitrary node
to the tree.

4
chapter16.tex
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@ -657,7 +657,9 @@ achieves these properties.
\index{Floyd's algorithm}
\key{Floyd's algorithm} walks forward
\key{Floyd's algorithm}\footnote{The idea of the algorithm is mentioned in \cite{knu982}
and attributed to R. W. Floyd; however, it is not known if Floyd was the first
who discovered the algorithm.} walks forward
in the graph using two pointers $a$ and $b$.
Both pointers begin at a node $x$ that
is the starting node of the graph.

5
chapter17.tex
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@ -368,7 +368,10 @@ performs two depth-first searches.
\index{2SAT problem}
Strongly connectivity is also linked with the
\key{2SAT problem} \cite{asp79}.
\key{2SAT problem}\footnote{The algorithm presented here was
introduced in \cite{asp79}.
There is also another well-known linear-time algorithm \cite{eve75}
that is based on backtracking.}.
In this problem, we are given a logical formula
\[
(a_1 \lor b_1) \land (a_2 \lor b_2) \land \cdots \land (a_m \lor b_m),

97
chapter18.tex
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@ -266,14 +266,18 @@ is $3+4+3+1=11$.
\end{center}
The idea is to construct a tree traversal array that contains
three values for each node: (1) the identifier of the node,
(2) the size of the subtree, and (3) the value of the node.
three values for each node: the identifier of the node,
the size of the subtree, and the value of the node.
For example, the array for the above tree is as follows:
\begin{center}
\begin{tikzpicture}[scale=0.7]
\draw (0,1) grid (9,-2);
\node[left] at (-1,0.5) {node id};
\node[left] at (-1,-0.5) {subtree size};
\node[left] at (-1,-1.5) {node value};
\node at (0.5,0.5) {$1$};
\node at (1.5,0.5) {$2$};
\node at (2.5,0.5) {$6$};
@ -330,6 +334,10 @@ can be found as follows:
\fill[color=lightgray] (4,-1) rectangle (8,-2);
\draw (0,1) grid (9,-2);
\node[left] at (-1,0.5) {node id};
\node[left] at (-1,-0.5) {subtree size};
\node[left] at (-1,-1.5) {node value};
\node at (0.5,0.5) {$1$};
\node at (1.5,0.5) {$2$};
\node at (2.5,0.5) {$6$};
@ -438,6 +446,10 @@ For example, the following array corresponds to the above tree:
\begin{tikzpicture}[scale=0.7]
\draw (0,1) grid (9,-2);
\node[left] at (-1,0.5) {node id};
\node[left] at (-1,-0.5) {subtree size};
\node[left] at (-1,-1.5) {path sum};
\node at (0.5,0.5) {$1$};
\node at (1.5,0.5) {$2$};
\node at (2.5,0.5) {$6$};
@ -491,6 +503,10 @@ the array changes as follows:
\fill[color=lightgray] (4,-1) rectangle (8,-2);
\draw (0,1) grid (9,-2);
\node[left] at (-1,0.5) {node id};
\node[left] at (-1,-0.5) {subtree size};
\node[left] at (-1,-1.5) {path sum};
\node at (0.5,0.5) {$1$};
\node at (1.5,0.5) {$2$};
\node at (2.5,0.5) {$6$};
@ -562,9 +578,9 @@ is node 2:
\node[draw, circle] (3) at (-2,1) {$2$};
\node[draw, circle] (4) at (0,1) {$3$};
\node[draw, circle] (5) at (2,-1) {$7$};
\node[draw, circle] (6) at (-3,-1) {$5$};
\node[draw, circle, fill=lightgray] (6) at (-3,-1) {$5$};
\node[draw, circle] (7) at (-1,-1) {$6$};
\node[draw, circle] (8) at (-1,-3) {$8$};
\node[draw, circle, fill=lightgray] (8) at (-1,-3) {$8$};
\path[draw,thick,-] (1) -- (2);
\path[draw,thick,-] (1) -- (3);
\path[draw,thick,-] (1) -- (4);
@ -572,6 +588,9 @@ is node 2:
\path[draw,thick,-] (3) -- (6);
\path[draw,thick,-] (3) -- (7);
\path[draw,thick,-] (7) -- (8);
\path[draw=red,thick,->,line width=2pt] (6) edge [bend left] (3);
\path[draw=red,thick,->,line width=2pt] (8) edge [bend right=40] (3);
\end{tikzpicture}
\end{center}
@ -583,13 +602,17 @@ finding the lowest common ancestor of two nodes.
One way to solve the problem is to use the fact
that we can efficiently find the $k$th
ancestor of any node in the tree.
Thus, we can first make sure that
both nodes are at the same level in the tree,
and then find the smallest value of $k$
such that the $k$th ancestor of both nodes is the same.
Using this, we can divide the problem of
finding the lowest common ancestor into two parts.
As an example, let us find the lowest common
ancestor of nodes $5$ and $8$:
We use two pointers that initially point to the
two nodes for which we should find the
lowest common ancestor.
First, we move one of the pointers upwards
so that both nodes are at the same level in the tree.
In the example case, we move from node 8 to node 6,
after which both nodes are at the same level:
\begin{center}
\begin{tikzpicture}[scale=0.9]
@ -599,8 +622,8 @@ ancestor of nodes $5$ and $8$:
\node[draw, circle] (4) at (0,1) {$3$};
\node[draw, circle] (5) at (2,-1) {$7$};
\node[draw, circle,fill=lightgray] (6) at (-3,-1) {$5$};
\node[draw, circle] (7) at (-1,-1) {$6$};
\node[draw, circle,fill=lightgray] (8) at (-1,-3) {$8$};
\node[draw, circle,fill=lightgray] (7) at (-1,-1) {$6$};
\node[draw, circle] (8) at (-1,-3) {$8$};
\path[draw,thick,-] (1) -- (2);
\path[draw,thick,-] (1) -- (3);
\path[draw,thick,-] (1) -- (4);
@ -608,26 +631,30 @@ ancestor of nodes $5$ and $8$:
\path[draw,thick,-] (3) -- (6);
\path[draw,thick,-] (3) -- (7);
\path[draw,thick,-] (7) -- (8);
\path[draw=red,thick,->,line width=2pt] (8) edge [bend right] (7);
\end{tikzpicture}
\end{center}
Node $5$ is at level $3$, while node $8$ is at level $4$.
Thus, we first move one step upwards from node $8$ to node $6$.
After this, it turns out that the parent of both nodes $5$
and $6$ is node $2$, so we have found the lowest common ancestor.
After this, we determine the minimum number of steps
needed to move both pointers upwards so that
they will point to the same node.
This node is the lowest common ancestor of the nodes.
The following picture shows how we move in the tree:
In the example case, it suffices to move both pointers
one step upwards to node 2,
which is the lowest common ancestor:
\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (0,3) {$1$};
\node[draw, circle] (2) at (2,1) {$4$};
\node[draw, circle] (3) at (-2,1) {$2$};
\node[draw, circle,fill=lightgray] (3) at (-2,1) {$2$};
\node[draw, circle] (4) at (0,1) {$3$};
\node[draw, circle] (5) at (2,-1) {$7$};
\node[draw, circle,fill=lightgray] (6) at (-3,-1) {$5$};
\node[draw, circle] (6) at (-3,-1) {$5$};
\node[draw, circle] (7) at (-1,-1) {$6$};
\node[draw, circle,fill=lightgray] (8) at (-1,-3) {$8$};
\node[draw, circle] (8) at (-1,-3) {$8$};
\path[draw,thick,-] (1) -- (2);
\path[draw,thick,-] (1) -- (3);
\path[draw,thick,-] (1) -- (4);
@ -637,20 +664,21 @@ The following picture shows how we move in the tree:
\path[draw,thick,-] (7) -- (8);
\path[draw=red,thick,->,line width=2pt] (6) edge [bend left] (3);
\path[draw=red,thick,->,line width=2pt] (8) edge [bend right] (7);
\path[draw=red,thick,->,line width=2pt] (7) edge [bend right] (3);
\end{tikzpicture}
\end{center}
Using this method, we can find the lowest common ancestor
of any two nodes in $O(\log n)$ time after an $O(n \log n)$ time
preprocessing, because both steps can be
performed in $O(\log n)$ time.
Since both parts of the algorithm can be performed in
$O(\log n)$ time using precomputed information,
we can find the lowest common ancestor of any two
nodes in $O(\log n)$ time using this technique.
\subsubsection{Method 2}
Another way to solve the problem is based on
a tree traversal array \cite{ben00}.
a tree traversal array\footnote{This lowest common ancestor algorithm is based on \cite{ben00}.
This technique is sometimes called the \index{Euler tour technique}
\key{Euler tour technique} \cite{tar84}.}.
Once again, the idea is to traverse the nodes
using a depth-first search:
@ -689,23 +717,26 @@ using a depth-first search:
\end{tikzpicture}
\end{center}
However, we use a bit different variant of
the tree traversal array where
However, we use a bit different tree
traversal array than before:
we add each node to the array \emph{always}
when the depth-first search visits the node,
when the depth-first search walks through the node,
and not only at the first visit.
Hence, a node that has $k$ children appears $k+1$ times
in the array, and there are a total of $2n-1$
in the array and there are a total of $2n-1$
nodes in the array.
We store two values in the array:
(1) the identifier of the node, and (2) the level of the
the identifier of the node and the level of the
node in the tree.
The following array corresponds to the above tree:
\begin{center}
\begin{tikzpicture}[scale=0.7]
\node[left] at (-1,1.5) {node id};
\node[left] at (-1,0.5) {level};
\draw (0,1) grid (15,2);
%\node at (-1.1,1.5) {\texttt{node}};
\node at (0.5,1.5) {$1$};
@ -770,6 +801,10 @@ can be found as follows:
\begin{center}
\begin{tikzpicture}[scale=0.7]
\node[left] at (-1,1.5) {node id};
\node[left] at (-1,0.5) {level};
\fill[color=lightgray] (2,1) rectangle (3,2);
\fill[color=lightgray] (5,1) rectangle (6,2);
\fill[color=lightgray] (2,0) rectangle (6,1);

30
chapter19.tex
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@ -22,7 +22,9 @@ problem and no efficient algorithm is known for solving the problem.
\index{Eulerian path}
An \key{Eulerian path} is a path
An \key{Eulerian path}\footnote{L. Euler (1707--1783) studied such paths in 1736
when he solved the famous Königsberg bridge problem.
This was the birth of graph theory.} is a path
that goes exactly once through each edge in the graph.
For example, the graph
\begin{center}
@ -222,7 +224,8 @@ from node 2 to node 5:
\index{Hierholzer's algorithm}
\key{Hierholzer's algorithm} is an efficient
\key{Hierholzer's algorithm}\footnote{The algorithm was published
in 1873 after Hierholzer's death \cite{hie73}.} is an efficient
method for constructing
an Eulerian circuit.
The algorithm consists of several rounds,
@ -395,7 +398,9 @@ so we have successfully constructed an Eulerian circuit.
\index{Hamiltonian path}
A \key{Hamiltonian path} is a path
A \key{Hamiltonian path}
%\footnote{W. R. Hamilton (1805--1865) was an Irish mathematician.}
is a path
that visits each node in the graph exactly once.
For example, the graph
\begin{center}
@ -481,12 +486,12 @@ Also stronger results have been achieved:
\begin{itemize}
\item
\index{Dirac's theorem}
\key{Dirac's theorem}:
\key{Dirac's theorem}: %\cite{dir52}
If the degree of each node is at least $n/2$,
the graph contains a Hamiltonian path.
\item
\index{Ore's theorem}
\key{Ore's theorem}:
\key{Ore's theorem}: %\cite{ore60}
If the sum of degrees of each non-adjacent pair of nodes
is at least $n$,
the graph contains a Hamiltonian path.
@ -525,7 +530,9 @@ It is possible to implement this solution in $O(2^n n^2)$ time.
\index{De Bruijn sequence}
A \key{De Bruijn sequence} is a string that contains
A \key{De Bruijn sequence}
%\footnote{N. G. de Bruijn (1918--2012) was a Dutch mathematician.}
is a string that contains
every string of length $n$
exactly once as a substring, for a fixed
alphabet of $k$ characters.
@ -546,7 +553,7 @@ and each edge adds one character to the string.
The following graph corresponds to the above example:
\begin{center}
\begin{tikzpicture}
\begin{tikzpicture}[scale=0.8]
\node[draw, circle] (00) at (-3,0) {00};
\node[draw, circle] (11) at (3,0) {11};
\node[draw, circle] (01) at (0,2) {01};
@ -628,12 +635,13 @@ The search can be made more efficient by using
\key{heuristics} that attempt to guide the knight so that
a complete tour will be found quickly.
\subsubsection{Warnsdorff's rule}
\subsubsection{Warnsdorf's rule}
\index{heuristic}
\index{Warnsdorff's rule}
\index{Warnsdorf's rule}
\key{Warnsdorff's rule} is a simple and effective heuristic
\key{Warnsdorf's rule}\footnote{This heuristic was proposed
in Warnsdorf's book \cite{war23} in 1823.} is a simple and effective heuristic
for finding a knight's tour.
Using the rule, it is possible to efficiently construct a tour
even on a large board.
@ -655,7 +663,7 @@ possible squares to which the knight can move:
\node at (3.5,1.5) {$d$};
\end{tikzpicture}
\end{center}
In this situation, Warnsdorff's rule moves the knight to square $a$,
In this situation, Warnsdorf's rule moves the knight to square $a$,
because after this choice, there is only a single possible move.
The other choices would move the knight to squares where
there would be three moves available.

34
chapter21.tex
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@ -24,7 +24,7 @@ z = \sqrt[3]{3}.\\
However, nobody knows if there are any three
\emph{integers} $x$, $y$ and $z$
that would satisfy the equation, but this
is an open problem in number theory.
is an open problem in number theory \cite{bec07}.
In this chapter, we will focus on basic concepts
and algorithms in number theory.
@ -205,7 +205,9 @@ so the result of the function is $[2,2,2,3]$.
\index{sieve of Eratosthenes}
The \key{sieve of Eratosthenes} is a preprocessing
The \key{sieve of Eratosthenes}
%\footnote{Eratosthenes (c. 276 BC -- c. 194 BC) was a Greek mathematician.}
is a preprocessing
algorithm that builds an array using which we
can efficiently check if a given number between $2 \ldots n$
is prime and, if it is not, find one prime factor of the number.
@ -327,7 +329,8 @@ The greatest common divisor and the least common multiple
are connected as follows:
\[\textrm{lcm}(a,b)=\frac{ab}{\textrm{gcd}(a,b)}\]
\key{Euclid's algorithm} provides an efficient way
\key{Euclid's algorithm}\footnote{Euclid was a Greek mathematician who
lived in about 300 BC. This is perhaps the first known algorithm in history.} provides an efficient way
to find the greatest common divisor of two numbers.
The algorithm is based on the following formula:
\begin{equation*}
@ -355,6 +358,7 @@ For example,
Numbers $a$ and $b$ are \key{coprime}
if $\textrm{gcd}(a,b)=1$.
\key{Euler's totient function} $\varphi(n)$
%\footnote{Euler presented this function in 1763.}
gives the number of coprime numbers to $n$
between $1$ and $n$.
For example, $\varphi(12)=4$,
@ -432,12 +436,16 @@ int modpow(int x, int n, int m) {
\index{Fermat's theorem}
\index{Euler's theorem}
\key{Fermat's theorem} states that
\key{Fermat's theorem}
%\footnote{Fermat discovered this theorem in 1640.}
states that
\[x^{m-1} \bmod m = 1\]
when $m$ is prime and $x$ and $m$ are coprime.
This also yields
\[x^k \bmod m = x^{k \bmod (m-1)} \bmod m.\]
More generally, \key{Euler's theorem} states that
More generally, \key{Euler's theorem}
%\footnote{Euler published this theorem in 1763.}
states that
\[x^{\varphi(m)} \bmod m = 1\]
when $x$ and $m$ are coprime.
Fermat's theorem follows from Euler's theorem,
@ -517,7 +525,9 @@ cout << x*x << "\n"; // 2537071545
\index{Diophantine equation}
A \key{Diophantine equation} is an equation of the form
A \key{Diophantine equation}
%\footnote{Diophantus of Alexandria was a Greek mathematician who lived in the 3th century.}
is an equation of the form
\[ ax + by = c, \]
where $a$, $b$ and $c$ are constants
and we should find the values of $x$ and $y$.
@ -637,7 +647,9 @@ are solutions.
\index{Lagrange's theorem}
\key{Lagrange's theorem} states that every positive integer
\key{Lagrange's theorem}
%\footnote{J.-L. Lagrange (1736--1813) was an Italian mathematician.}
states that every positive integer
can be represented as a sum of four squares, i.e.,
$a^2+b^2+c^2+d^2$.
For example, the number 123 can be represented
@ -648,7 +660,9 @@ as the sum $8^2+5^2+5^2+3^2$.
\index{Zeckendorf's theorem}
\index{Fibonacci number}
\key{Zeckendorf's theorem} states that every
\key{Zeckendorf's theorem}
%\footnote{E. Zeckendorf published the theorem in 1972 \cite{zec72}; however, this was not a new result.}
states that every
positive integer has a unique representation
as a sum of Fibonacci numbers such that
no two numbers are equal or consecutive
@ -689,7 +703,9 @@ produces the smallest Pythagorean triple
\index{Wilson's theorem}
\key{Wilson's theorem} states that a number $n$
\key{Wilson's theorem}
%\footnote{J. Wilson (1741--1793) was an English mathematician.}
states that a number $n$
is prime exactly when
\[(n-1)! \bmod n = n-1.\]
For example, the number 11 is prime, because

17
chapter22.tex
View File

@ -342,7 +342,9 @@ corresponds to the binomial coefficient formula.
\index{Catalan number}
The \key{Catalan number} $C_n$ equals the
The \key{Catalan number}
%\footnote{E. C. Catalan (1814--1894) was a Belgian mathematician.}
$C_n$ equals the
number of valid
parenthesis expressions that consist of
$n$ left parentheses and $n$ right parentheses.
@ -678,7 +680,9 @@ elements should be changed.
\index{Burnside's lemma}
\key{Burnside's lemma} can be used to count
\key{Burnside's lemma}
%\footnote{Actually, Burnside did not discover this lemma; he only mentioned it in his book \cite{bur97}.}
can be used to count
the number of combinations so that
only one representative is counted
for each group of symmetric combinations.
@ -764,7 +768,10 @@ with 3 colors is
\index{Cayley's formula}
\key{Cayley's formula} states that
\key{Cayley's formula}
% \footnote{While the formula is named after A. Cayley,
% who studied it in 1889, it was discovered earlier by C. W. Borchardt in 1860.}
states that
there are $n^{n-2}$ labeled trees
that contain $n$ nodes.
The nodes are labeled $1,2,\ldots,n$,
@ -827,7 +834,9 @@ be derived using Prüfer codes.
\index{Prüfer code}
A \key{Prüfer code} is a sequence of
A \key{Prüfer code}
%\footnote{In 1918, H. Prüfer proved Cayley's theorem using Prüfer codes \cite{pru18}.}
is a sequence of
$n-2$ numbers that describes a labeled tree.
The code is constructed by following a process
that removes $n-2$ leaves from the tree.

9
chapter23.tex
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@ -245,8 +245,9 @@ two $n \times n$ matrices
in $O(n^3)$ time.
There are also more efficient algorithms
for matrix multiplication\footnote{The first such
algorithm, with time complexity $O(n^{2.80735})$,
was published in 1969 \cite{str69}, and
algorithm was Strassen's algorithm,
published in 1969 \cite{str69},
whose time complexity is $O(n^{2.80735})$;
the best current algorithm
works in $O(n^{2.37286})$ time \cite{gal14}.},
but they are mostly of theoretical interest
@ -749,7 +750,9 @@ $2 \rightarrow 1 \rightarrow 4 \rightarrow 2 \rightarrow 5$.
\index{Kirchhoff's theorem}
\index{spanning tree}
\key{Kirchhoff's theorem} provides a way
\key{Kirchhoff's theorem}
%\footnote{G. R. Kirchhoff (1824--1887) was a German physicist.}
provides a way
to calculate the number of spanning trees
of a graph as a determinant of a special matrix.
For example, the graph

15
chapter24.tex
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@ -359,7 +359,10 @@ The expected value for $X$ in a geometric distribution is
\index{Markov chain}
A \key{Markov chain} is a random process
A \key{Markov chain}
% \footnote{A. A. Markov (1856--1922)
% was a Russian mathematician.}
is a random process
that consists of states and transitions between them.
For each state, we know the probabilities
for moving to other states.
@ -514,7 +517,11 @@ just to find one element?
It turns out that we can find order statistics
using a randomized algorithm without sorting the array.
The algorithm is a Las Vegas algorithm:
The algorithm, called \key{quickselect}\footnote{In 1961,
C. A. R. Hoare published two algorithms that
are efficient on average: \index{quicksort} \index{quickselect}
\key{quicksort} \cite{hoa61a} for sorting arrays and
\key{quickselect} \cite{hoa61b} for finding order statistics.}, is a Las Vegas algorithm:
its running time is usually $O(n)$
but $O(n^2)$ in the worst case.
@ -560,7 +567,9 @@ but one could hope that verifying the
answer would by easier than to calculate it from scratch.
It turns out that we can solve the problem
using a Monte Carlo algorithm whose
using a Monte Carlo algorithm\footnote{R. M. Freivalds published
this algorithm in 1977 \cite{fre77}, and it is sometimes
called \index{Freivalds' algoritm} \key{Freivalds' algorithm}.} whose
time complexity is only $O(n^2)$.
The idea is simple: we choose a random vector
$X$ of $n$ elements, and calculate the matrices

6
chapter25.tex
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@ -248,7 +248,8 @@ and this is always the final state.
It turns out that we can easily classify
any nim state by calculating
the \key{nim sum} $x_1 \oplus x_2 \oplus \cdots \oplus x_n$,
where $\oplus$ is the xor operation.
where $\oplus$ is the xor operation\footnote{The optimal strategy
for nim was published in 1901 by C. L. Bouton \cite{bou01}.}.
The states whose nim sum is 0 are losing states,
and all other states are winning states.
For example, the nim sum for
@ -367,7 +368,8 @@ so the nim sum is not 0.
\index{Sprague–Grundy theorem}
The \key{Sprague–Grundy theorem} generalizes the
The \key{Sprague–Grundy theorem}\footnote{The theorem was discovered
independently by R. Sprague \cite{spr35} and P. M. Grundy \cite{gru39}.} generalizes the
strategy used in nim to all games that fulfil
the following requirements:

4
chapter29.tex
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@ -42,6 +42,7 @@ After this, it suffices to sum the areas
of the triangles.
The area of a triangle can be calculated,
for example, using \key{Heron's formula}
%\footnote{Heron of Alexandria (c. 10--70) was a Greek mathematician.}
\[ \sqrt{s (s-a) (s-b) (s-c)},\]
where $a$, $b$ and $c$ are the lengths
of the triangle's sides and
@ -500,7 +501,8 @@ so $b$ is outside the polygon.
\section{Polygon area}
A general formula for calculating the area
of a polygon is
of a polygon\footnote{This formula is sometimes called the
\index{shoelace formula} \key{shoelace formula}.} is
\[\frac{1}{2} |\sum_{i=1}^{n-1} (p_i \times p_{i+1})| =
\frac{1}{2} |\sum_{i=1}^{n-1} (x_i y_{i+1} - x_{i+1} y_i)|, \]
where the vertices are

53
chapter30.tex
View File

@ -27,10 +27,10 @@ For example, the table
\begin{tabular}{ccc}
person & arrival time & leaving time \\
\hline
Uolevi & 10 & 15 \\
Maija & 6 & 12 \\
Kaaleppi & 14 & 16 \\
Liisa & 5 & 13 \\
John & 10 & 15 \\
Maria & 6 & 12 \\
Peter & 14 & 16 \\
Lisa & 5 & 13 \\
\end{tabular}
\end{center}
corresponds to the following events:
@ -51,10 +51,10 @@ corresponds to the following events:
\draw[fill] (5,-5.5) circle [radius=0.05];
\draw[fill] (13,-5.5) circle [radius=0.05];
\node at (2,-1) {Uolevi};
\node at (2,-2.5) {Maija};
\node at (2,-4) {Kaaleppi};
\node at (2,-5.5) {Liisa};
\node at (2,-1) {John};
\node at (2,-2.5) {Maria};
\node at (2,-4) {Peter};
\node at (2,-5.5) {Lisa};
\end{tikzpicture}
\end{center}
We go through the events from left to right
@ -85,10 +85,10 @@ In the example, the events are processed as follows:
\draw[fill] (5,-5.5) circle [radius=0.05];
\draw[fill] (13,-5.5) circle [radius=0.05];
\node at (2,-1) {Uolevi};
\node at (2,-2.5) {Maija};
\node at (2,-4) {Kaaleppi};
\node at (2,-5.5) {Liisa};
\node at (2,-1) {John};
\node at (2,-2.5) {Maria};
\node at (2,-4) {Peter};
\node at (2,-5.5) {Lisa};
\path[draw,dashed] (10,0)--(10,-6.5);
\path[draw,dashed] (15,0)--(15,-6.5);
@ -122,7 +122,7 @@ The symbols $+$ and $-$ indicate whether the
value of the counter increases or decreases,
and the value of the counter is shown below.
The maximum value of the counter is 3
between Uolevi's arrival time and Maija's leaving time.
between John's arrival time and Maria's leaving time.
The running time of the algorithm is $O(n \log n)$,
because sorting the events takes $O(n \log n)$ time
@ -270,7 +270,11 @@ we should find the following points:
This is another example of a problem
that can be solved in $O(n \log n)$ time
using a sweep line algorithm.
using a sweep line algorithm\footnote{Besides this approach,
there is also an
$O(n \log n)$ time divide-and-conquer algorithm \cite{sha75}
that divides the points into two sets and recursively
solves the problem for both sets.}.
We go through the points from left to right
and maintain a value $d$: the minimum distance
between two points seen so far.
@ -396,21 +400,20 @@ an easy way to
construct the convex hull for a set of points
in $O(n \log n)$ time.
The algorithm constructs the convex hull
in two steps:
in two parts:
first the upper hull and then the lower hull.
Both steps are similar, so we can focus on
Both parts are similar, so we can focus on
constructing the upper hull.
We sort the points primarily according to
First, we sort the points primarily according to
x coordinates and secondarily according to y coordinates.
After this, we go through the points and always
add the new point to the hull.
After adding a point we check using cross products
whether the tree last point in the hull turn left.
If this holds, we remove the middle point from the hull.
After this we keep checking the three last points
and removing points, until the three last points
do not turn left.
After this, we go through the points and
add each point to the hull.
Always after adding a point to the hull,
we make sure that the last line segment
in the hull does not turn left.
As long as this holds, we repeatedly remove the
second last point from the hull.
The following pictures show how
Andrew's algorithm works:

215
list.tex
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2
preface.tex
View File

@ -12,7 +12,7 @@ The book is especially intended for
students who want to learn algorithms and
possibly participate in
the International Olympiad in Informatics (IOI) or
the International Collegiate Programming Contest (ICPC).
in the International Collegiate Programming Contest (ICPC).
Of course, the book is also suitable for
anybody else interested in competitive programming.