\fi
\chapter{Ranking Combinatorial Structures}
+\id{rankchap}
+
+\section{Ranking and unranking}
+
+The techniques for building efficient data structures on the RAM 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 make use of the fact
+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.~in~\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
+\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: $\pi[i\ldots j] = (\pi[i],\ldots,\pi[j])$.
+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, for fixed~$\pi[1]$ 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)ranking
+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 $\<Rank>(\pi,i,n,A)$: Compute the rank of a~permutation $\pi[i\ldots n]$ on~$A$.
+\id{rankalg}
+\algo
+\:If $i\ge n$, return~0.
+\:$a\=R_A(\pi[i])$.
+\:$b\=\<Rank>(\pi,i+1,n,A \setminus \{\pi[i]\})$.
+\:Return $a\cdot(n-i)! + b$.
+\endalgo
+
+\>We can call $\<Rank>(\pi,1,n,[n])$ for ranking on~$[n]$, i.e., to calculate
+$L(\pi,[n])$.
+
+\alg $\<Unrank>(j,i,n,A)$: Return an~array~$\pi$ such that $\pi[i\ldots n]$ is the $j$-th permutation on~$A$.
+\id{unrankalg}
+\algo
+\:If $i>n$, return $(0,\ldots,0)$.
+\:$x\=R^{-1}_A(\lfloor j/(n-i)! \rfloor)$.
+\:$\pi\=\<Unrank>(j\bmod (n-i)!,i+1,n,A\setminus \{x\})$.
+\:$\pi[i]\=x$.
+\:Return~$\pi$.
+\endalgo
+
+\>We can call $\<Unrank>(j,1,n,[n])$ for the unranking problem on~$[n]$, i.e., to get $L^{-1}(j,[n])$.
+
+\para
+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 invokation 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~$W$ 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
+
+\thmn{Lexicographic ranking of permutations \cite{mm:rank}}
+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 order of elements dependent 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 when both ranking and unraking.
+
+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 {\secitfont k\/}-permutations}
+\id{kpranksect}
+
+The technique from the previous section can be also generalized to lexicographic ranking of
+\df{$k$-permutations,} that is of ordered $k$-tuples 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 numbers 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~$H$ 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 \<Rank> from Algorithm \ref{vecops},
+this time on the vector~$\bf h$.
+
+The only operation 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, $i=\<Rank>(r,j+1)-1$. 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 \cite{mm:rank}}
+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 randomly. 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~$i$.
+
+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 $T(\pi)=\{ (i,\pi(i)) ; i\in[n] \}$.
+\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, $N_j := \#\{ \pi\in{\cal P}: \#(H\cap T(\pi)) = j \}$.
+\:$r_k$ 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 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: $H=\{ (i,i) : i\in[n] \}$. 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 $n!/e$.
+
+\obs\id{matchobs}%
+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 in the following lemma:
+
+\lemma\id{permchar}%
+The following sets have the same cardinality:
+
+\itemize\ibull
+\:permutations which obey a~given restriction graph~$G$,
+\:non-attacking placements of rooks on a~$n\times n$ board avoiding holes
+ which 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
+See observations \ref{rooksobs} and~\ref{matchobs}.
+\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 satistying 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.
+
+\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 \cite{dinic:flow} for the flow algorithm, \cite{even:dinic} for the time bound
+and \cite{schrijver} for more references on flows and matchings.)
+\qed
+
+\para
+The bad news is that computing the permanent is known to be~$\#P$-complete even
+for zero-one matrices (as proven by Valiant in \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
+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 $P=\#P$.
+
+\proof
+We will show that a~polynomial-time ranking algorithm would imply a~polynomial-time
+algorithm for computing the permanent of an~arbitrary zero-one matrix, which
+is a~$\#P$-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.
+
+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
+
+\para
+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}.
+
+\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 customary
+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}:
+
+\alg\id{rrankalg}%
+$\<Rank>(\pi,i,n,A,M)$: 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 $M[1,k]=1$.
+ \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 $b + \<Rank>(\pi,i+1,n,A\setminus\{\pi[i]\},M^{1,a+1})$.
+\endalgo
+
+\>To calculate the rank of~$\pi\in{\cal P}_{A,M}$, we call $\<Rank>(\pi,1,\vert A\vert,A,M)$.
+
+\alg\id{runrankalg}%
+$\<Unrank>(j,i,n,A,M)$: 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 above).
+\:$x\=R^{-1}_A(a-1)$.
+\:$\pi\=\<Unrank>(j-C_{a-1}, i+1, n, A\setminus\{x\}, M^{1,a})$.
+\:$\pi[i]\=x$.
+\:Return~$\pi$.
+\endalgo
+
+\>To find the $j$-th permutation in~${\cal P}_{A,M}$, we call $\<Unrank>(j,1,\vert A\vert,A,M)$.
+
+\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 the recursion, and by the manipulation with matrices. Because of this,
+we do not any special 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
+
+\rem
+In cases where the 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 in \cite{jerrum:permanent}.
+We then get an~approximation scheme for the ranks.
+
+\rem
+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 to replace the matrix~$M$ by the corresponding restriction graph and instead of
+copying the matrix at every level of recursion, we 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 graphs that are
+complete bipartite with the matching $\{(i,i) : i\in[n]\}$ 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:
+$$\eqalign{
+n_0(0,d) &= d!, \cr
+n_0(d,d) &= d! \cdot \sum\nolimits_{k=0}^d {(-1)^k \over k!}, \cr
+\noalign{\medskip}
+n_0(z,d) &= z\cdot n_0(z-1,d-1) + (d-z)\cdot n_0(z,d-1) \quad\hbox{for $z<d$.} \cr
+}\eqno{(*)}$$
+
+\proof
+The base cases of the recurrence are straightforward: $n_0(0,d)$ counts the
+unrestricted permutations on~$[d]$, and $n_0(d,d)$ is equal to the number of derangements
+on~$[d]$, which we have already computed in Observation \ref{hatcheck}. Let us
+prove the third formula.
+
+We will count the permutations~$\pi$ restricted by a~matrix~$M$ of the given parameters
+$z$ and~$d$. As $z<d$, there is at least one position in the permutation for which
+no restriction applies and by Lemma~\ref{submatrixlemma} we can choose without
+loss of generality that it is the first position.
+
+If we select $\pi[1]$ from the~$z$ restricted elements, the rest of~$\pi$ is a~permutation
+on the remaining elements with one restriction less and there are $n_0(z-1,d-1)$ such
+permutations. On the other hand, if we use an~unrestricted element, all restrictions
+stay in effect, so there are~$n_0(z,d-1)$ ways how to do so.
+\qed
+
+\lemma
+The function~$n_0$ also satisfies the following recurrence:
+$$
+n_0(z-1,d) = n_0(z,d) + n_0(z-1,d-1) \quad\hbox{for $z>0$, $d>0$.} \eqno{(\maltese)}
+$$
+
+\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 $0\le z<d$ we have $n_0(z,d) - n_0(z+1,d) \le n_0(z,d)/d$.
+
+\proof
+According to the recurrence $(\maltese)$, the difference $n_0(z,d) - n_0(z+1,d)$ is
+equal to $n_0(z,d-1)$. We can bound this by plugging the trivial inequality $n_0(z,d-1) \le n_0(z-1,d-1)$
+to~$(*)$, from which we obtain $n_0(z,d) \ge d\cdot n_0(z,d-1)$.
+\qed
+
+\para\id{rrankmod}%
+Let us show how to modify the ranking algorithm (\ref{rrankalg}) using the insight
+we have gained into the structure of derangements.
+
+The algorithm uses the matrix~$M$ only for computing~$N_0$ of its submatrices
+and we have shown that this value depends only on the order of the matrix and
+the number of zeroes in it. We will therefore replace maintenance of the matrix
+by remember the number~$z$ of its zeroes and the set~$Z$ that contains the elements
+$x\in A$ whose locations are restricted (there is a~zero anywhere in the $(R_A(x)+1)$-th
+column of~$M$). In other words, every $x\in Z$ can appear at all positions in the
+permutation except one (and these forbidden positions are different for different~$x$'s),
+while the remaining elements of~$A$ can appear anywhere.
+
+As we already observed (\ref{hatcheck}) that the number of derangements on~$[n]$ is $\Theta(n!)$,
+we can again use word-sized vectors to represent the sets~$A$ and~$Z$ with insertion,
+deletion, ranking and unranking on them in constant time.
+
+When the algorithm selects a~submatrix $M'=M^{1,k}$ for an~element $x$ of~rank~$k-1$, this
+matrix it is described by either by the choice of $z'=z-1$ and~$Z'=Z\setminus\{x\}$ (if $x\in Z$)
+or $z'=z$ and $Z'=Z$ (if $x\not\in Z$).
+All computations of~$N_0$ in the algorithm can therefore be replaced by looking
+up the appropriate $n_0(z',\vert A\vert-1)$ in the precomputed table. Moreover, we can
+calculate a~single~$C_a$ in constant time, because all summands are either $n_0(z,\vert A\vert-1)$
+or $n_0(z-1,\vert A\vert-1)$ depending on the set~$Z$. We get:
+$$C_a = r\cdot n_0(z-1,\vert A\vert-1) + (a-r) \cdot n_0(z,\vert A\vert-1),$$
+where $r=R_Z(R^{-1}_A(a))$, that is the number of restricted elements among the $a$~smallest ones in~$A$.
+
+All operations at a~single level of the \<Rank> function now run in constant time,
+but \<Unrank> 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
projects / saga.git / blobdiff