New references etc.

chapter01.tex

@@ -778,7 +778,7 @@ 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 +790,8 @@ 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}

chapter02.tex

@@ -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 this 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}
 
@@ -352,7 +356,7 @@ 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
+\key{maximum subarray sum}\footnote{J. Bentley's
 book \emph{Programming Pearls} \cite{ben86} made this problem popular.}, i.e.,
 the largest possible sum of numbers
 in a contiguous region in the array.
@@ -470,7 +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 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.

chapter03.tex

@@ -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{knu98},
+mergesort was invented by J. von Neumann in 1945.}
 that is based on recursion.
 
 Mergesort sorts a subarray \texttt{t}$[a,b]$ as follows:

chapter04.tex

@@ -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

chapter05.tex

@@ -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
+algorithm tells us that $q(n)=92$.
-there are 92 ways to place 8
-queens to the $8 \times 8$ chessboard.
 When $n$ increases, the search quickly becomes slow,
 because the number of the solutions increases
 exponentially.
-For example, calculating the ways to
+For example, calculating $q(16)=14772512$
-place 16 queens to the $16 \times 16$
+using the above algorithm already takes about a minute
-chessboard already takes about a minute
+on a modern computer\footnote{There is no known way to efficiently
-on a modern computer
+calculate larger values of $q(n)$. The current record is
-(there are 14772512 solutions).
+$q(27)=234907967154122528$, 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 $\sqrt{2^n}$.
 

chapter06.tex

@@ -530,7 +530,9 @@ 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
@@ -671,114 +673,4 @@ character & codeword \\
 \texttt{C} & 10 \\
 \texttt{D} & 111 \\
 \end{tabular}
-\end{center}
+\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:
-% \[\begin{array}{rcl}
-% c(b) & \ge & c(a) \\
-% c(b)\cdot(s(b)-s(a)) & \ge & c(a)\cdot (s(b)-s(a)) \\
-% c(b)\cdot s(b)-c(b)\cdot s(a) & \ge & c(a)\cdot s(b)-c(a)\cdot s(a) \\
-% c(a)\cdot s(a)+c(b)\cdot s(b) & \ge & c(a)\cdot s(b)+c(b)\cdot s(a) \\
-% \end{array}\]

chapter07.tex

@@ -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.

chapter09.tex

@@ -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.

list.tex

@@ -107,7 +107,13 @@
   Computer Science and Computational Biology},
   Cambridge University Press, 1997.
 
+\bibitem{hor74}
+  E. Horowitz and S. Sahni.
+  Computing partitions with applications to the knapsack problem.
+  \emph{Journal of the ACM}, 21(2):277--292, 1974.
+
 \bibitem{huf52}
+  D. A. Huffman.
   A method for the construction of minimum-redundancy codes.
   \emph{Proceedings of the IRE}, 40(9):1098--1101, 1952.
 
@@ -125,11 +131,20 @@
   The statistics of dimers on a lattice: I. The number of dimer arrangements on a quadratic lattice.
   \emph{Physica}, 27(12):1209--1225, 1961.
 
+\bibitem{knu98}
+  D. E. Knuth.
+  \emph{The Art of Computer Programming. Volume 3: Sorting and Searching}, Addison–Wesley, 1998 (2nd edition).
+
 \bibitem{kru56}
   J. B. Kruskal.
   On the shortest spanning subtree of a graph and the traveling salesman problem.
   \emph{Proceedings of the American Mathematical Society}, 7(1):48--50, 1956.
 
+\bibitem{lev66}
+  V. I. Levenshtein.
+  Binary codes capable of correcting deletions, insertions, and reversals.
+  \emph{Soviet physics doklady}, 10(8):707--710, 1966.
+
 \bibitem{mai84}
   M. G. Main and R. J. Lorentz.
   An $O(n \log n)$ algorithm for finding all repetitions in a string.
@@ -145,11 +160,20 @@
   Shortest connection networks and some generalizations.
   \emph{Bell System Technical Journal}, 36(6):1389--1401, 1957.
 
+\bibitem{q27}
+  27-Queens Puzzle: Massively Parallel Enumeration and Solution Counting.
+  \url{https://github.com/preusser/q27}
+
 \bibitem{sha81}
   M. Sharir.
   A strong-connectivity algorithm and its applications in data flow analysis.
   \emph{Computers \& Mathematics with Applications}, 7(1):67--72, 1981.
 
+\bibitem{sta06}
+  P. Stańczyk.
+  \emph{Algorytmika praktyczna w konkursach Informatycznych},
+  MSc thesis, University of Warsaw, 2006.
+
 \bibitem{str69}
   V. Strassen.
   Gaussian elimination is not optimal.