From: Martin Mares Date: Sat, 13 Sep 2008 14:31:02 +0000 (+0200) Subject: Trimmed sections that will not be used in the paper. X-Git-Url: http://mj.ucw.cz/gitweb/?a=commitdiff_plain;h=840079a443f03290a0df52fce00f45834e4e892f;p=saga.git Trimmed sections that will not be used in the paper. --- diff --git a/Makefile b/Makefile index cbd8ba1..7f3bb11 100644 --- a/Makefile +++ b/Makefile @@ -1,6 +1,6 @@ all: saga.ps -CHAPTERS=cover pref mst ram adv opt dyn appl rank epilog notation +CHAPTERS=pref mst ram adv opt dyn appl epilog notation %.dvi: %.tex macros.tex fonts12.tex fonts10.tex biblio.bib mjalpha.bst tex $< && mv $*.toc $*.tok diff --git a/cover.tex b/cover.tex deleted file mode 100644 index cec1c03..0000000 --- a/cover.tex +++ /dev/null @@ -1,84 +0,0 @@ -\ifx\endpart\undefined -\input macros.tex -\fi - -{ -\nopagenumbers -\parindent=0pt - -%%% Title page %%% - -{ - -\vglue 0.7in - -\font\ft=cmr12 at 14pt -\font\xt=cmb17 at 24pt -\font\yt=cmti17 -\font\ct=cmsy17 at 18pt -\ft -\baselineskip=20pt - -\centerline{Charles University in Prague} -\centerline{Faculty of Mathematics and Physics} - -\vfil -\vfil - -\centerline{\xt Doctoral Thesis} - -\vfil -\vfil - -\centerline{\epsfxsize=0.4\hsize\epsfbox{pic/mfflogo.eps}} - -\vfil -\vfil - -\centerline{\xt Graph Algorithms} - -\vfil - -\centerline{\yt {\ct M}\kern-0.13em artin {\ct M}\kern-0.13em are\v{s}} - -\vfil - -\centerline{Department of Applied Mathematics} -\centerline{Malostransk\'e n\'am.~25} -\centerline{Prague, Czech Republic} - -\vfil - -\centerline{Supervisor: Prof.~RNDr.~Jaroslav Ne\v{s}et\v{r}il, DrSc.} -\centerline{Branch I4: Discrete Models and Algorithms} - -\vfil - -\centerline{Updated 13-09-2008} - -\vskip 0.5in -\eject - -} - -%%% Subtitle page %%% - -\vglue 0pt -\vfill - -I hereby declare that I have written this thesis on my own and using exclusively -the cites sources. For any work in the thesis that has been co-published with -other authors, I have the permission of them all to include this work in my thesis. -I~authorize the Charles University to lend this document to other institutions -and individuals for academic or research purposes. - -\bigskip - -\leftline{Martin Mare\v{s}} -\leftline{Prague, May 4th, 2008} - -\eject - -} - -\endpart diff --git a/macros.tex b/macros.tex index 3ab39f7..f6faac0 100644 --- a/macros.tex +++ b/macros.tex @@ -18,7 +18,7 @@ % Parameters for final typesetting (bigger inner margins etc.) \newif\iffinal -\finaltrue +\finalfalse \newdimen\hwobble \hwobble=10mm \advance\hsize by -10mm diff --git a/rank.tex b/rank.tex deleted file mode 100644 index b4b8ffe..0000000 --- a/rank.tex +++ /dev/null @@ -1,756 +0,0 @@ -\ifx\endpart\undefined -\input macros.tex -\fi - -\chapter{Ranking Combinatorial Structures} -\id{rankchap} - -\section{Ranking and unranking}\id{ranksect}% - -The techniques for building efficient data structures on the RAM, which we have described -in Chapter~\ref{ramchap}, can be also used for a~variety of problems related -to ranking of combinatorial structures. Generally, the problems are stated -in the following way: - -\defn\id{rankdef}% -Let~$C$ be a~set of objects and~$\prec$ a~linear order on~$C$. The \df{rank} -$R_{C,\prec}(x)$ of an~element $x\in C$ is the number of elements $y\in C$ such that $y\prec x$. -We will call the function $R_{C,\prec}$ the \df{ranking function} for $C$ ordered by~$\prec$ -and its inverse $R^{-1}_{C,\prec}$ the \df{unranking function} for $C$ and~$\prec$. When the set -and the order are clear from the context, we will use plain~$R(x)$ and $R^{-1}(x)$. -Also, when $\prec$ is defined on a~superset~$C'$ of~$C$, we naturally extend $R_C(x)$ -to elements $x\in C'\setminus C$. - -\example -Let us consider the set $C_k=\{\0,\1\}^k$ of all binary strings of length~$k$ ordered -lexicographically. Then $R^{-1}(i)$ is the $i$-th smallest element of this set, that -is the number~$i$ written in binary and padded to~$k$ digits (i.e., $\(i)_k$ in the -notation of Section~\ref{bitsect}). Obviously, $R(x)$ is the integer whose binary -representation is the string~$x$. - -\para -In this chapter, we will investigate how to compute the ranking and unranking -functions for different sets efficiently. Usually, we will observe -that the ranks (and hence the input and output of our algorithm) are large -numbers, so we can use the integers of a~similar magnitude to represent non-trivial -data structures. - -\para -Until the end of the chapter, we will always assume that our model of computation -is the Random Access Machine (more specifically, the Word-RAM). - -%-------------------------------------------------------------------------------- - -\section{Ranking of permutations} -\id{pranksect} - -One of the most common ranking problems is ranking of permutations on the set~$[n]=\{1,2,\ldots,n\}$. -This is frequently used to create arrays indexed by permutations: for example in Ruskey's algorithm -for finding Hamilton cycles in Cayley graphs (see~\cite{ruskey:ham} and \cite{ruskey:hce}) -or when exploring state spaces of combinatorial puzzles like the Loyd's Fifteen \cite{ss:fifteen}. -Many other applications are surveyed by Critani et al.~\cite{critani:rau} and in -most cases, the time complexity of the whole algorithm is limited by the efficiency -of the (un)ranking functions. - -The permutations are usually ranked according to their lexicographic order. -In fact, an~arbitrary order is often sufficient if the ranks are used solely -for indexing of arrays. The lexicographic order however has an~additional advantage -of a~nice structure, which allows various operations on permutations to be -performed directly on their ranks. - -Na\"\i{}ve algorithms for lexicographic ranking require time $\Theta(n^2)$ in the -worst case \cite{reingold:catp} and even on average~\cite{liehe:raulow}. -This can be easily improved to $O(n\log n)$ by using either a binary search -tree to calculate inversions, or by a divide-and-conquer technique, or by clever -use of modular arithmetic (all three algorithms are described in Knuth -\cite{knuth:sas}). Myrvold and Ruskey \cite{myrvold:rank} mention further -improvements to $O(n\log n/\log \log n)$ by using the RAM data structures of Dietz -\cite{dietz:oal}. - -Linear time complexity was reached by Myrvold and Ruskey \cite{myrvold:rank} -for a~non-lexicographic order, which is defined locally by the history of the -data structure --- in fact, they introduce a linear-time unranking algorithm -first and then they derive an inverse algorithm without describing the order -explicitly. However, they leave the problem of lexicographic ranking open. - -We will describe a~general procedure which, when combined with suitable -RAM data structures, yields a~linear-time algorithm for lexicographic -(un)ranking. - -\nota\id{brackets}% -We will view permutations on a~finite set $A\subseteq {\bb N}$ as ordered $\vert A\vert$-tuples -(in other words, arrays) containing every element of~$A$ exactly once. We will -use square brackets to index these tuples: $\pi=(\pi[1],\ldots,\pi[\vert A\vert])$, -and sub-tuples: $\pi[i\ldots j] = (\pi[i],\pi[i+1],\ldots,\pi[j])$. -The lexicographic ranking and unranking functions for the permutations on~$A$ -will be denoted by~$L(\pi,A)$ and $L^{-1}(i,A)$ respectively. - -\obs\id{permrec}% -Let us first observe that permutations have a simple recursive structure. -If we fix the first element $\pi[1]$ of a~permutation~$\pi$ on the set~$[n]$, the -elements $\pi[2], \ldots, \pi[n]$ form a~permutation on $[n]-\{\pi[1]\} = \{1,\ldots,\pi[1]-1,\pi[1]+1,\ldots,n\}$. -The lexicographic order of two permutations $\pi$ and~$\pi'$ on the original set is then determined -by $\pi[1]$ and $\pi'[1]$ and only if these elements are equal, it is decided -by the lexicographic comparison of permutations $\pi[2\ldots n]$ and $\pi'[2\ldots n]$. -Moreover, when we fix $\pi[1]$, all permutations on the smaller set occur exactly -once, so the rank of $\pi$ is $(\pi[1]-1)\cdot (n-1)!$ plus the rank of -$\pi[2\ldots n]$. - -This gives us a~reduction from (un)ranking of permutations on $[n]$ to (un)rank\-ing -of permutations on a $(n-1)$-element set, which suggests a straightforward -algorithm, but unfortunately this set is different from $[n-1]$ and it even -depends on the value of~$\pi[1]$. We could renumber the elements to get $[n-1]$, -but it would require linear time per iteration. To avoid this, we generalize the -problem to permutations on subsets of $[n]$. For a permutation $\pi$ on a~set -$A\subseteq [n]$ of size~$m$, similar reasoning gives a~simple formula: -$$ -L((\pi[1],\ldots,\pi[m]),A) = R_A(\pi[1]) \cdot (m-1)! + -L((\pi[2],\ldots,\pi[m]), A\setminus\{\pi[1]\}), -$$ -which uses the ranking function~$R_A$ for~$A$. This recursive formula immediately -translates to the following recursive algorithms for both ranking and unranking -(described for example in \cite{knuth:sas}): - -\alg $\(\pi,i,n,A)$: Compute the rank of a~permutation $\pi[i\ldots n]$ on~$A$. -\id{rankalg} -\algo -\:If $i\ge n$, return~0. -\:$a\=R_A(\pi[i])$. -\:$b\=\(\pi,i+1,n,A \setminus \{\pi[i]\})$. -\:Return $a\cdot(n-i)! + b$. -\endalgo - -\>We can call $\(\pi,1,n,[n])$ for ranking on~$[n]$, i.e., to calculate -$L(\pi,[n])$. - -\alg $\(j,i,n,A)$: Return an~array~$\pi$ such that $\pi[i\ldots n]$ is the $j$-th permutation on~$A$. -\id{unrankalg} -\algo -\:If $i>n$, return $(0,\ldots,0)$. -\:$x\=R^{-1}_A(\lfloor j/(n-i)! \rfloor)$. -\:$\pi\=\(j\bmod (n-i)!,i+1,n,A\setminus \{x\})$. -\:$\pi[i]\=x$. -\:Return~$\pi$. -\endalgo - -\>We can call $\(j,1,n,[n])$ for the unranking problem on~$[n]$, i.e., to calculate $L^{-1}(j,[n])$. - -\paran{Representation of sets}% -The most time-consuming parts of the above algorithms are of course operations -on the set~$A$. If we store~$A$ in a~data structure of a~known time complexity, the complexity -of the whole algorithm is easy to calculate: - -\lemma\id{ranklemma}% -Suppose that there is a~data structure maintaining a~subset of~$[n]$ under a~sequence -of deletions, which supports ranking and unranking of elements, and that -the time complexity of a~single operation is at most~$t(n)$. -Then lexicographic ranking and unranking of permutations can be performed in time $\O(n\cdot t(n))$. - -\proof -Let us analyse the above algorithms. The depth of the recursion is~$n$ and in each -nested invocation of the recursive procedure we perform a~constant number of operations. -All of them are either trivial, or calculations of factorials (which can be precomputed in~$\O(n)$ time), -or operations on the data structure. -\qed - -\example -If we store~$A$ in an~ordinary array, we have insertion and deletion in constant time, -but ranking and unranking in~$\O(n)$, so $t(n)=\O(n)$ and the algorithm is quadratic. -Binary search trees give $t(n)=\O(\log n)$. The data structure of Dietz \cite{dietz:oal} -improves it to $t(n)=O(\log n/\log \log n)$. In fact, all these variants are equivalent -to the classical algorithms based on inversion vectors, because at the time of processing~$\pi[i]$, -the value of $R_A(\pi[i])$ is exactly the number of elements forming inversions with~$\pi[i]$. - -\para -To obtain linear time complexity, we will make use of the representation of -vectors by integers on the RAM as developed in Section~\ref{bitsect}, but first -of all, we will make sure that the ranks are large numbers, so the word size of the -machine has to be large as well: - -\obs -$\log n! = \Theta(n\log n)$, therefore the word size must be~$\Omega(n\log n)$. - -\proof -We have $n^n \ge n! \ge \lfloor n/2\rfloor^{\lfloor n/2\rfloor}$, so $n\log n \ge \log n! \ge \lfloor n/2\rfloor\cdot\log \lfloor n/2\rfloor$. -\qed - -\>Thus we get the following theorem: - -\thmn{Lexicographic ranking of permutations} -When we order the permutations on the set~$[n]$ lexicographically, both ranking -and unranking can be performed on the RAM in time~$\O(n)$. - -\proof -We will store the elements of the set~$A$ in a~sorted vector. Each element has -$\O(\log n)$ bits, so the whole vector takes $\O(n\log n)$ bits, which by the -above observation fits in a~constant number of machine words. We know from -Algorithm~\ref{vecops} that ranks can be calculated in constant time in such -vectors and that insertions and deletions can be translated to ranks and -masking. Unranking, that is indexing of the vector, is masking alone. -So we can apply the previous Lemma \ref{ranklemma} with $t(n)=\O(1)$. -\qed - -\rem -We can also easily derive the non-lexicographic linear-time algorithm of Myrvold -and Ruskey~\cite{myrvold:rank} from our algorithm. We will relax the requirements -on the data structure to allow the order of elements to depend on the history of the -structure (i.e., on the sequence of deletes performed so far). We can observe that -although the algorithm no longer gives the lexicographic ranks, the unranking function -is still an~inverse of the ranking function, because the sequence of deletes -from~$A$ is the same during both ranking and unranking. - -The implementation of the relaxed structure is straightforward. We store the set~$A$ -in an~array~$\alpha$ and use the order of the elements in~$\alpha$ determine the -order on~$A$. We will also maintain an~``inverse'' array $\alpha^{-1}$ such that -$\alpha[\alpha^{-1}[x]]=x$ for every~$x\in A$. Ranking and unranking can be performed -by a~simple lookup in these arrays: $R_A(x)=\alpha^{-1}[x]$, $R^{-1}(i)=\alpha[i]$. -When we want to delete an~element, we exchange it with the last element in the -array~$\alpha$ and update~$\alpha^{-1}$ accordingly. - - -%-------------------------------------------------------------------------------- - -\section{Ranking of \iftoc $k$\else{\secitfont k\/}\fi-permutations} -\id{kpranksect} - -The ideas from the previous section can be also generalized to lexicographic ranking of -\df{$k$-permutations,} that is of ordered $k$-tuples of distinct elements drawn from the set~$[n]$. -There are $n^{\underline k} = n\cdot(n-1)\cdot\ldots\cdot(n-k+1)$ -such $k$-permutations and they have a~recursive structure similar to the one of -the permutations. We will therefore use the same recursive scheme as before -(algorithms \ref{rankalg} and \ref{unrankalg}), but we will modify the first step of both algorithms -to stop after the first~$k$ iterations. We will also replace the number $(n-i)!$ -of permutations on the remaining elements by the number of $(k-i)$-permutations on the same elements, -i.e., by $(n-i)^{\underline{k-i}}$. As $(n-i)^{\underline{k-i}} = (n-i) \cdot (n-i-1)^{\underline{k-i-1}}$, -we can precalculate all these values in linear time. - -Unfortunately, the ranks of $k$-permutations can be much smaller, so we can no -longer rely on the same data structure fitting in a constant number of word-sized integers. -For example, if $k=1$, the ranks are $\O(\log n)$-bit numbers, but the data -structure still requires $\Theta(n\log n)$ bits. - -We do a minor side step by remembering the complement of~$A$ instead, that is -the set of the at most~$k$ elements we have already seen. We will call this set~$H$ -(because it describes the ``holes'' in~$A$). Let us prove that $\Omega(k\log n)$ bits -are needed to represent the rank, so the vector representation of~$H$ certainly fits in -a~constant number of words. - -\lemma -The number of $k$-permutations on~$[n]$ is $2^{\Omega(k\log n)}$. - -\proof -We already know that there $n^{\underline k}$ such $k$-permutations. If $k\le n/2$, -then every term in the product is $n/2$ or more, so $\log n^{\underline k} \ge -k\cdot (\log n - 1)$. If $k\ge n/2$, then $n^{\underline k} \ge n^{\underline{\smash{n/2}}}$ -and $\log n^{\underline k} \ge (n/2)(\log n - 1) \ge (k/2)(\log n - 1)$. -\qed - -\para -It remains to show how to translate the operations on~$A$ to operations on~$H$, -again stored as a~sorted vector~${\bf h}$. Insertion to~$A$ correspond to -deletion from~$H$ and vice versa. The rank of any~$x\in[n]$ in~$A$ is $x$ minus -the number of holes that are smaller than~$x$, therefore $R_A(x)=x-R_H(x)$. -To calculate $R_H(x)$, we can again use the vector operation \ from Algorithm \ref{vecops}, -this time on the vector~$\bf h$. - -The only operation, which we cannot translate directly, is unranking in~$A$. We will -therefore define an~auxiliary vector~$\bf r$ of the same size as~$\bf h$, -containing the ranks of the holes: $r_i=R_A(h_i)=h_i-R_H(h_i)=h_i-i$. -To find the $j$-th smallest element of~$A$, we locate the interval between -holes to which this element belongs: the interval is bordered from below by -a~hole~$h_i$ such that $i$ is the largest index satisfying~$r_i \le j$. -In other words, $i=\(r,j+1)-1$. Finding the right element in the interval -is then easy: $R^{-1}_A(j) = h_i + 1 + j - r_i$. - -\example -If $A=\{2,5,6\}$ and $n=8$, then ${\bf h}=(1,3,4,7,8)$ and ${\bf r} -= (0,1,1,3,3)$. When we want to calculate $R^{-1}_A(2)$, we find $i=2$ and -the wanted element is $h_2+1+2-r_2 = 4+1+2-1 = 6$. - -\para -The vector~$\bf r$ can be updated in constant time whenever an~element is -inserted to~$\bf h$. It is sufficient to shift the fields apart (we know -that the position of the new element in~$\bf r$ is the same as in~$\bf h$), -insert the new value using masking operations and decrease all higher fields -by one in parallel by using a~single subtraction. Updates after deletions -from~$\bf h$ are analogous. - -We have replaced all operations on~$A$ by the corresponding operations on the -modified data structure, each of which works again in constant time. Therefore -we have just proven the following theorem, which brings this section to -a~happy ending: - -\thmn{Lexicographic ranking of $k$-permutations} -When we order the $k$-per\-mu\-ta\-tions on the set~$[n]$ lexicographically, both -ranking and unranking can be performed on the RAM in time~$\O(k)$. - -\proof -We modify algorithms \ref{rankalg} and \ref{unrankalg} for $k$-permutations as -shown at the beginning of this section. We use the vectors $\bf h$ and~$\bf r$ -described above as an~implicit representation of the set~$A$. The modified -algorithm uses recursion $k$~levels deep and as each operation on~$A$ can be -performed in~$\O(1)$ time using $\bf h$ and~$\bf r$, every level takes only -constant time. The time bound follows. \qed - -%-------------------------------------------------------------------------------- - -\section{Restricted permutations} - -Another interesting class of combinatorial objects that can be counted and -ranked are restricted permutations. An~archetypal member of this class are -permutations without a~fixed point, i.e., permutations~$\pi$ such that $\pi(i)\ne i$ -for all~$i$. These are also called \df{derangements} or \df{hatcheck permutations.}\foot{% -As the story in~\cite{matnes:idm} goes, once upon a~time there was a~hatcheck lady who -was so confused that she was giving out the hats completely at random. What is -the probability that none of the gentlemen receives his own hat?} We will present -a~general (un)ranking method for any class of restricted permutations and -derive a~linear-time algorithm for the derangements from it. - -\defn\id{permnota}% -We will fix a~non-negative integer~$n$ and use ${\cal P}$ for the set of -all~permutations on~$[n]$. -A~\df{restriction graph} is a~bipartite graph~$G$ whose parts are two copies -of the set~$[n]$. A~permutation $\pi\in{\cal P}$ satisfies the restrictions -if $(i,\pi(i))$ is an~edge of~$G$ for every~$i$. - -\paran{Boards and rooks}% -We will follow the path unthreaded by Kaplansky and Riordan -\cite{kaplansky:rooks} and charted by Stanley in \cite{stanley:econe}. -We will relate restricted permutations to placements of non-attacking -rooks on a~hollow chessboard. - -\defn -\itemize\ibull -\:A~\df{board} is the grid $B=[n]\times [n]$. It consists of $n^2$ \df{squares.} -\:A~\df{trace} of a~permutation $\pi\in{\cal P}$ is the set of squares \hbox{$T(\pi)=\{ (i,\pi(i)) ; i\in[n] \}$. \hskip-4em} %%HACK -\endlist - -\obs\id{rooksobs}% -The traces of permutations (and thus the permutations themselves) correspond -exactly to placements of $n$ rooks at the board in a~way such that the rooks do -not attack each other (i.e., there is at most one rook in every row and -likewise in every column; as there are $n$~rooks, there must be exactly one of them in -every row and column). When speaking about \df{rook placements,} we will always -mean non-attacking placements. - -Restricted permutations then correspond to placements of rooks on a~board with -some of the squares removed. The \df{holes} (missing squares) correspond to the -non-edges of~$G$, so $\pi\in{\cal P}$ satisfies the restrictions iff -$T(\pi)$ avoids the holes. - -\defn -Let~$H\subseteq B$ be any set of holes in the board. Then: -\itemize\ibull -\:$N_j$ denotes the number of placements of $n$~rooks on the board such that exactly~$j$ of the rooks -stand on holes. That is: -$$N_j := \bigl\vert\bigl\{ \pi\in{\cal P} \mathbin{\bigl\vert} \vert H\cap T(\pi) \vert = j \bigr\}\bigr\vert.$$ -\:$r_k$ is the number of ways how to place $k$~rooks on the holes. In other words, -this is the number of $k$-element subsets of~$H$ such that no two elements share -a~common row or column. -\:$N$ is the generating function for the~$N_j$'s: -$$ -N(x) = \sum_{j\ge 0} N_j x^j. -$$ -As $N_j=0$ for $j>n$, this function is in fact a~finite polynomial. -\endlist - -\thmn{The number of restricted permutations, Stanley \cite{stanley:econe}} -The function~$N$ can be expressed in terms of the numbers~$r_k$ as: -$$ -N(x) = \sum_{k=0}^n r_k \cdot (n-k)! \cdot (x-1)^k. -$$ - -\proof -If two polynomials of degree~$n$ coincide at more than~$n$ points, they -are identical, therefore it is sufficient to prove that the equality holds -for all $x\in{\bb N}^+$. -The $N(x)$ counts the ways of placing~$n$ rooks on the board and labeling -each of them which stands on a~hole with an~element of~$[x]$. The right-hand -side counts the same: We can obtain any such configuration by placing $k$~rooks -on~$H$ first, labeling them with elements of~$\{2,\ldots,x\}$, placing -additional $n-k$ rooks on the remaining rows and columns (there are $(n-k)!$ ways -how to do this) and labeling those of the new rooks standing on a~hole with~1. -\qed - -\cor -When we substitute~$x=0$ in the above equality, we get a~formula for the -number of rook placements avoiding the holes altogether: -$$N_0 = N(0) = \sum_{k=0}^n (-1)^k \cdot (n-k)! \cdot r_k.$$ - -\example\id{hatcheck}% -Let us apply this theory to the hatcheck lady problem. The set~$H$ of holes is the main diagonal -of the board: $H=\{ (i,i) \mid i\in[n] \}$. When we want to place $k$~rooks on the holes, -we can do that in $r_k={n\choose k}$ ways. By the previous corollary, the number of -derangements is: -$$ -N_0 = \sum_{k=0}^n (-1)^k \cdot (n-k)! \cdot {n\choose k} - = \sum_{k=0}^n (-1)^k \cdot {n!\over k!} - = n! \cdot \sum_{k=0}^n {(-1)^k\over k!}. -$$ -As the sum converges to~$1/e$ when $n$~approaches infinity, we know that the number -of derangements is asymptotically $n!/e$. - -\paran{Matchings and permanents}\id{matchper}% -Placements of~$n$ rooks (and therefore also restricted permutations) can be -also equated with perfect matchings in the restriction graph~$G$. The edges -of the matching correspond to the squares occupied by the rooks, the condition -that no two rooks share a~row nor column translates to the edges not touching -each other, and the use of exactly~$n$ rooks is equivalent to the matching -being perfect. - -There is also a~well-known correspondence between the perfect matchings -in a~bipartite graph and non-zero summands in the formula for the permanent -of the bipartite adjacency matrix~$M$ of the graph. This holds because the -non-zero summands are in one-to-one correspondence with the placements -of~$n$ rooks on the corresponding board. The number $N_0$ is therefore -equal to the permanent of the matrix~$M$. - -We will summarize our observations by the following lemma: - -\lemma\id{permchar}% -The following sets have the same cardinality: - -\itemize\ibull -\:permutations that obey a~given restriction graph~$G$, -\:non-attacking placements of rooks on a~$n\times n$ board avoiding holes - that correspond to non-edges of~$G$, -\:perfect matchings in the graph~$G$, -\:non-zero summands in the permanent of the adjacency matrix of~$G$. -\endlist - -\proof -Follows from \ref{rooksobs} and~\ref{matchper}. -\qed - -\para -The diversity of the characterizations of restricted permutations brings -both good and bad news. The good news is that we can use the -plethora of known results on bipartite matchings. Most importantly, we can efficiently -determine whether there exists at least one permutation satisfying a~given set of restrictions: - -\thm -There is an~algorithm which decides in time $\O(n^{1/2}\cdot m)$ whether there exists -a~permutation satisfying a~given restriction graph. The $n$ and~$m$ are the number -of vertices and edges of the restriction graph. - -\proof -It is sufficient to verify that there exists a~perfect matching in the -given graph. By a~standard technique, this can be reduced in linear time to finding a~maximum -flow in a~suitable unit-capacity network. This flow can be then found using the Dinic's -algorithm in time $\O(\sqrt{n}\cdot m)$. -(See Dinic \cite{dinic:flow} for the flow algorithm, Even and Tarjan \cite{even:dinic} for the time bound -and Schrijver \cite{schrijver} for more references on flows and matchings.) -\qed - -\para -The bad news is that computing the permanent is known to be~$\#\rm P$-complete even -for zero-one matrices (as proven by Valiant \cite{valiant:permanent}). -As a~ranking function for a~set of~matchings can be used to count all such -matchings, we obtain the following theorem: - -\thm\id{pcomplete}% -If there is a~polynomial-time algorithm for lexicographic ranking of permutations with -a~set of restrictions which is a~part of the input, then $\rm P=\#P$. - -\proof -We will show that a~polynomial-time ranking algorithm would imply a~po\-ly\-nom\-ial-time -algorithm for computing the permanent of an~arbitrary zero-one matrix, which -is a~$\#\rm P$-complete problem. - -We know from Lemma \ref{permchar} that non-zero -summands in the permanent of a~zero-one matrix~$M$ correspond to permutations restricted -by a~graph~$G$ whose bipartite adjacency matrix is~$M$. The permanent is -therefore equal to the number of such permutations, which is one more than the -rank of the lexicographically maximum such permutation. -It therefore remains to show that we can find the lexicographically maximum -permutation permitted by~$G$ in polynomial time. -\looseness=1 %%HACK%% - -We can determine $\pi[1]$ by trying all the possible values permitted by~$G$ -in decreasing order and stopping as soon as we find~$\pi[1]$ which can be -extended to a~complete permutation. This can be verified using the previous -theorem on~the graph of the remaining restrictions, i.e., on~$G$ with the vertices -1~on one side and~$\pi[1]$ on the other side removed. -Once we have~$\pi[1]$, proceed by finding $\pi[2]$ in the same way, using the reduced -graph. This way we construct the whole maximum permutation~$\pi$ -in~$\O(n^2)$ calls to the verification algorithm. -\qed - -\paran{Recursive structure}% -However, the hardness of computing the permanent is the only obstacle. -We will show that whenever we are given a~set of restrictions for which -the counting problem is easy (and it is also easy for subgraphs obtained -by deleting vertices), ranking is easy as well. The key will be once again -a~recursive structure, similar to the one we have seen in the case of plain -permutations in \ref{permrec}. -\looseness=1 %%HACK%% - -\nota\id{restnota}% -As we will work with permutations on different sets simultaneously, we have -to extend our notation accordingly. For every finite set of elements $A\subset{\bb N}$, -we will consider the set ${\cal P}_A$ of all permutations on~$A$, as usually -viewed as ordered $\vert A\vert$-tuples. The restriction graph will be represented -by its adjacency matrix~$M\in \{0,1\}^{\vert A\vert\times \vert A\vert}$ and -a~permutation $\pi\in{\cal P}_A$ satisfies~$M$ (conforms to the restrictions) -iff $M[i,j]=1$ whenever $j=R_A(\pi[i])+1$.\foot{The $+1$ is added because -matrices are indexed from~1 while the lowest rank is~0.} -The set of all such~$\pi$ will be denoted by~${\cal P}_{A,M}$ -and their number (which obviously does not depend on the choice of~$A$) by $N_0(M) = {\per M}$. - -We will also frequently need to delete a~row and a~column simultaneously -from~$M$. This operation corresponds to deletion of one vertex from each -part of the restriction graph. We will write $M^{i,j}$ for the matrix~$M$ -with its $i$-th row and $j$-th column removed. - -\obs -Let us consider a~permutation $\pi\in{\cal P}_A$ and $n=\vert A\vert$. -When we fix the value of the element $\pi[1]$, the remaining elements form -a~permutation $\pi'=\pi[2\ldots n]$ on the set~$A'=A\setminus\{\pi[1]\}$. -The permutation~$\pi$ satisfies the restriction matrix~$M$ if and only if -$M[1,a]=1$ for $a=R_A(\pi[1])$ and $\pi'$ satisfies a~restriction matrix~$M'=M^{1,a}$. -This translates to the following counterparts of algorithms \ref{rankalg} -and \ref{unrankalg}: - -\goodbreak %%HACK%% -\alg\id{rrankalg}% -$\(\pi,i,n,A,M)$: Compute the lexicographic rank of a~permutation $\pi[i\ldots n]\in{\cal P}_{A,M}$. - -\algo -\:If $i\ge n$, return 0. -\:$a\=R_A(\pi[i])$. -\:$b\=C_a=\sum_k N_0(M^{1,k})$ over all $k$ such that $1\le k\le a$ and \hbox{$M[1,k]=1$.\kern-3pt} %%HACK - \cmt{$C_a$ is the number of permutations in ${\cal P}_{A,M}$ whose first element lies - among the first $a$ elements of~$A$.} -\:Return $b + \(\pi,i+1,n,A\setminus\{\pi[i]\},M^{1,a+1})$. -\endalgo - -\>To calculate the rank of~$\pi\in{\cal P}_{A,M}$, we call $\(\pi,1,\vert A\vert,A,M)$. - -\alg\id{runrankalg}% -$\(j,i,n,A,M)$: Return an~array~$\pi$ such that $\pi[i,\ldots,n]$ is the $j$-th -permutation in~${\cal P}_{A,M}$. - -\algo -\:If $i>n$, return $(0,\ldots,0)$. -\:Find minimum $a$ such that $C_a > j$ (where $C_a$ is as in \ above). -\:$x\=R^{-1}_A(a-1)$. -\:$\pi\=\(j-C_{a-1}, i+1, n, A\setminus\{x\}, M^{1,a})$. -\:$\pi[i]\=x$. -\:Return~$\pi$. -\endalgo - -\>To find the $j$-th permutation in~${\cal P}_{A,M}$, we call $\(j,1,\vert A\vert,A,M)$. - -\para -The time complexity of these algorithms will be dominated by the computation of -the numbers $C_a$, which requires a~linear amount of calls to~$N_0$ on every -level of recursion, and by the manipulation with matrices. Because of this, -we do not need any sophisticated data structure for the set~$A$, an~ordinary sorted array -will suffice. (Also, we cannot use the vector representation blindly, because -we have no guarantee that the word size is large enough.) - -\thmn{Lexicographic ranking of restricted permutations} -Suppose that we have a~family of matrices ${\cal M}=\{M_1,M_2,\ldots\}$ such that $M_n\in \{0,1\}^{n\times n}$ -and it is possible to calculate the permanent of~$M'$ in time $\O(t(n))$ for every matrix $M'$ -obtained by deletion of rows and columns from~$M_n$. Then there exist algorithms -for ranking and unranking in ${\cal P}_{A,M_n}$ in time $\O(n^4 + n^2\cdot t(n))$ -if $M_n$ and an~$n$-element set~$A$ are given as a~part of the input. - -\proof -We will combine the algorithms \ref{rrankalg} and \ref{runrankalg} with the supplied -function for computing the permanent. All matrices constructed by the algorithm -are submatrices of~$M_n$ of the required type, so all computations of the function~$N_0$ -can be performed in time $\O(t(n))$ each. - -The recursion is $n$~levels deep. Every level involves -a~constant number of (un)ranking operations on~$A$ and computation of at most~$n$ -of the $C_a$'s. Each such $C_a$ can be derived from~$C_{a-1}$ by constructing -a~submatrix of~$M$ (which takes $\O(n^2)$ time) and computing its $N_0$. We therefore -spend time $\O(n^2)$ on operations with the set~$A$, $\O(n^4)$ on matrix manipulations -and $\O(n^2\cdot t(n))$ by the computations of the~$N_0$'s. -\qed - -\paran{Approximation}% -In cases where efficient evaluation of the permanent is out of our reach, -we can consider using the fully-polynomial randomized approximation scheme -for the permanent described by Jerrum, Sinclair and Vigoda \cite{jerrum:permanent}. -They have described a~randomized algorithm that for every $\varepsilon>0$ -approximates the value of the permanent of an~$n\times n$ matrix with non-negative -entries. The output is within relative error~$\varepsilon$ from the correct value with -probability at least~$1/2$ and the algorithm runs in time polynomial in~$n$ and~$1/\varepsilon$. -From this, we can get a~similar approximation scheme for the ranks. - -\paran{Special restriction graphs}% -There are also deterministic algorithms for computing the number of perfect matchings -in various special graph families (which imply polynomial-time ranking algorithms for -the corresponding families of permutations). If the graph is planar, we can -use the Kasteleyn's algorithm \cite{kasteleyn:crystals} based on Pfaffian -orientations which runs in time $\O(n^3)$. -It has been recently extended to arbitrary surfaces by Yuster and Zwick -\cite{yuster:matching} and sped up to $\O(n^{2.19})$. The counting problem -for arbitrary minor-closed classes (cf.~Section \ref{minorclosed}) is still -open. - -%-------------------------------------------------------------------------------- - -\section{Hatcheck lady and other derangements} - -The time bound for ranking of general restricted permutations shown in the previous -section is obviously very coarse. Its main purpose was to demonstrate that -many special cases of the ranking problem can be indeed computed in polynomial time. -For most families of restriction matrices, we can do much better. One of the possible improvements -is replacing the matrix~$M$ by the corresponding restriction graph and instead of -copying the matrix at every level of recursion, we can perform local operations on the graph -and undo them later. Another useful trick is to calculate the $N_0$'s of the smaller -matrices using information already computed for the larger matrices. - -These speedups are hard to state formally in general (they depend on the -structure of the matrices), so we will concentrate on a~specific example -instead. We will show that for the derangements one can achieve linear time complexity. - -\nota\id{hatrank}% -As we already know, the hatcheck permutations correspond to restriction -matrices that contain zeroes only on the main diagonal, and to graphs that are -complete bipartite with the matching $\{(i,i) \mid i\in[n]\}$ deleted. For -a~given order~$n$, we will call this matrix~$D_n$ and the graph~$G_n$ and -we will show that the submatrices of~$D_n$ share several nice properties: - -\lemma\id{submatrixlemma}% -Let $D$ be a~submatrix of~$D_n$ obtained by deletion of rows and columns. -Then the value of the permanent of~$D$ depends only on the size of~$D$ -and on the number of zero entries in~$D$. - -\proof -We know from Lemma~\ref{permchar} that the permanent counts matchings in the -graph~$G$ obtained from~$G_n$ by removing the vertices corresponding to the -deleted rows and columns of~$D_n$. Therefore we can prove the lemma for -the number of matchings instead. - -As~$G_n$ is a~complete bipartite graph without edges of a~single perfect matching, -the graph~$G$ must be also complete bipartite with some non-touching edges -missing. Two such graphs $G$ and~$G'$ are therefore isomorphic if and only if they have the -same number of vertices and also the same number of missing edges. As the -number of matchings is an~isomorphism invariant, the lemma follows. -\qed - -\rem -There is a~clear combinatorial intuition behind this lemma: if we are -to count permutations with restrictions placed on~$z$ elements and these -restrictions are independent, it does not matter how exactly they look like. - -\defn -Let $n_0(z,d)$ be the permanent shared by all submatrices as described -by the above lemma, which have $d\times d$ entries and exactly~$z$ zeroes. - -\lemma -The function~$n_0$ satisfies the following recurrence: -$$\eqalign{ -n_0(0,d) &= d!, \cr -n_0(d,d) &= d! \cdot \sum\nolimits_{k=0}^d {(-1)^k \over k!}, \cr -\noalign{\medskip} -n_0(z,d) &= z\cdot n_0(z-1,d-1) + (d-z)\cdot n_0(z,d-1) \quad\hbox{for $z0$, $d>0$.} \eqno{(\maltese)} -$$ - -\proof -We will again take advantage of having proven Lemma~\ref{submatrixlemma}, which -allows us to choose arbitrary matrices with the given parameters. Let us pick a~matrix~$M_z$ -containing $z$~zeroes such that $M_z[1,1]=0$. Then define~$M_{z-1}$ which is equal to~$M_z$ -everywhere except $M_{z-1}[1,1]=1$. - -We will count the permutations $\pi\in {\cal P}_d$ satisfying~$M_{z-1}$ in two ways. -First, there are $n_0(z-1,d)$ such permutations. On the other hand, we can divide -the them to two types depending on whether $\pi[1]=1$. Those having $\pi[1]\ne 1$ -are exactly the $n_0(z,d)$ permutations satisfying~$M_z$. The others correspond to -permutations $(\pi[2],\ldots,\pi[d])$ on $\{2,\ldots,d\}$ that satisfy~$M_z^{1,1}$, -so there are $n_0(z-1,d-1)$ of them. -\qed - -\cor\id{nzeroprecalc}% -For a~given~$n$, a~table of the values $n_0(z,d)$ for all $0\le z\le d\le n$ -can be precomputed in time~$\O(n^2)$. - -\proof -Use either recurrence and induction on~$z+d$. -\qed - -\cor\id{smalldiff}% -For every $0\le z function now run in constant time, -but \ needs to search among the~$C_a$'s to find the first of them which -exceeds the given rank. We could use binary search, but that would take $\Theta(\log n)$ -time. There is however a~clever trick: the value of~$C_a$ does not vary too much with -the set~$Z$. Specifically, by Corollary~\ref{smalldiff} the difference between the values -for $Z=\emptyset$ and $Z=A$ is at most $n_0(z-1,\vert A\vert -1)$. It is therefore -sufficient to just divide the rank by $n_0(z-1,\vert A\vert-1)$ and we get either -the correct value of~$a$ or one more. Both possibilities can be checked in constant time. - -We can therefore conclude this section by the following theorem: - -\thmn{Ranking of derangements}% -For every~$n$, the derangements on the set~$[n]$ can be ranked and unranked according to the -lexicographic order in time~$\O(n)$ after spending $\O(n^2)$ on initialization of auxiliary tables. - -\proof -We modify the general algorithms for (un)ranking of restricted permutations (\ref{rrankalg} and \ref{runrankalg}) -as described above (\ref{rrankmod}). Each of the $n$~levels of recursion will then run in constant time. The values~$n_0$ will -be looked up in a~table precalculated in quadratic time as shown in Corollary~\ref{nzeroprecalc}. -\qed - -\endpart diff --git a/saga.tex b/saga.tex index e517459..8724eeb 100644 --- a/saga.tex +++ b/saga.tex @@ -2,7 +2,6 @@ \input fonts12.tex \let\endpart=\relax -\input cover.tex \input pref.tex \unchapter{Table of contents} @@ -14,7 +13,6 @@ \input opt.tex \input dyn.tex \input appl.tex -\input rank.tex \input epilog.tex \appendices @@ -25,6 +23,4 @@ \dumpbib \vfill\eject -\ifodd\pageno\else\hbox{}\fi - \bye