\fi
\chapter{Ranking Combinatorial Structures}
+\id{rankchap}
-\section{Ranking and unranking}
+\section{Ranking and unranking}\id{ranksect}%
-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
+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:
\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
+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.
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
+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.
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
+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}.
\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: $\pi=(\pi[1],\ldots,\pi[\vert A\vert])$.
-The corresponding lexicographic ranking and unranking functions will be denoted by~$L(\pi,A)$
-and $L^{-1}(i,A)$ respectively.
+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.
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 $(\pi[2],\ldots,\pi[n])$ and
-$(\pi'[2],\ldots,\pi'[n])$. 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 $(\pi[2],\ldots,\pi[n])$.
+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)ranking
+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 $[n-1]$,
\>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~$A$.
+\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 $(0,\ldots,0)$.
-\:$a\=R^{-1}_A(\lfloor j/(n-i)! \rfloor)$.
-\:$\pi\=\<Unrank>(j\bmod (n-i)!,i+1,n,A\setminus \{a\})$.
-\:$\pi[i]\=a$.
+\:$x\=R^{-1}_A(\lfloor j/(n-i)! \rfloor)$.
+\:$\pi\=\<Unrank>(j\bmod (n-i)!,i+1,n,A\setminus \{x\})$.
+\:$\pi[i]\=x$.
\:Return~$\pi$.
\endalgo
-\>We can call $\<Unrank>(j,1,n,[n])$ for the unranking problem on~$[n]$, i.e., to get $L^{-1}(j,[n])$.
+\>We can call $\<Unrank>(j,1,n,[n])$ for the unranking problem on~$[n]$, i.e., to calculate $L^{-1}(j,[n])$.
-\para
+\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:
\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.
+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
machine has to be large as well:
\obs
-$\log n! = \Theta(n\log n)$, therefore the word size~$W$ must be~$\Omega(n\log n)$.
+$\log n! = \Theta(n\log n)$, 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
-\thmn{Lexicographic ranking of permutations \cite{mm:rank}}
+\>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~$\O(n)$.
\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
+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 when both ranking and unraking.
+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
%--------------------------------------------------------------------------------
-\section{Ranking of {\secitfont k\/}-permutations}
+\section{Ranking of \iftoc $k$\else{\secitfont k\/}\fi-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~$[n]$.
+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
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.
+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.
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
+are needed to represent the rank, so the vector representation of~$H$ certainly fits in
a~constant number of words.
\lemma
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 which are smaller than~$x$, therefore $R_A(x)=x-R_H(x)$.
+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~$\bf h$
+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
we have just proven the following theorem, which brings this section to
a~happy ending:
-\thmn{Lexicographic ranking of $k$-permutations \cite{mm:rank}}
+\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~$\O(k)$.
%--------------------------------------------------------------------------------
-\section{Hatcheck lady and other derangements}
+\section{Restricted permutations}
-Another interesting class of combinatorial objects which can be counted and
+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
+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.
-\nota\id{permnota}
+\defn\id{permnota}%
We will fix a~non-negative integer~$n$ and use ${\cal P}$ for the set of
all~permutations on~$[n]$.
-
-\defn
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 for every~$i$, $(i,\pi(i))$ is an~edge of~$G$.
+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
-Let~$n$ be a~non-negative integer. Then:
\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 $T(\pi)=\{ (i,\pi(i)) ; i\in[n] \}$.
+\: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 in
+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.
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, $N_j := \#\{ \pi\in{\cal P}: \#(H\cup T(\pi)) = j \}$.
+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.
$$
N(x) = \sum_{j\ge 0} N_j x^j.
$$
-As $N_j=0$ for $j>n$, the function is in fact a~finite polynomial.
+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}}
$$
\proof
-It is sufficient to prove that the equality holds for all integer~$x$.
+If two polynomials of degree~$n$ coincide at more than~$n$ points, they
+are identical, therefore it is sufficient to prove that the equality holds
+for all $x\in{\bb N}^+$.
The $N(x)$ counts the ways of placing~$n$ rooks on the board and labeling
each of them which stands on a~hole with an~element of~$[x]$. The right-hand
side counts the same: We can obtain any such configuration by placing $k$~rooks
on~$H$ first, labeling them with elements of~$\{2,\ldots,x\}$, placing
additional $n-k$ rooks on the remaining rows and columns (there are $(n-k)!$ ways
-how to do this) and labeling the new rooks standing on a~hole with~1.
+how to do this) and labeling those of the new rooks standing on a~hole with~1.
\qed
\cor
When we substitute~$x=0$ in the above equality, we get a~formula for the
-number of rook placements avoiding~$H$:
+number of rook placements avoiding the holes altogether:
$$N_0 = N(0) = \sum_{k=0}^n (-1)^k \cdot (n-k)! \cdot r_k.$$
\example\id{hatcheck}%
Let us apply this theory to the hatcheck lady problem. The set~$H$ of holes is the main diagonal
-of the board: $H=\{ (i,i) : i\in[n] \}$. When we want to place $k$~rooks on the holes,
+of the board: $H=\{ (i,i) \mid i\in[n] \}$. When we want to place $k$~rooks on the holes,
we can do that in $r_k={n\choose k}$ ways. By the previous corollary, the number of
derangements is:
$$
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}%
+\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 or column translates to the edges not touching
+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.
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:
+We will summarize our observations by 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~$G$,
+\: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
-See observations \ref{rooksobs} and~\ref{matchobs}.
+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 a~permutation satistying a~given set of restrictions exists:
-It is sufficient to reduce the corresponding matching problem to finding a~maximum flow in a~suitable
-unit-capacity network. The flow can be then found using the Dinic's algorithm
-in time $\O(\sqrt{n}\cdot m)$. Here $m$~is the number of edges in the network, which
-is linear in the size of the graph~$G$, therefore at worst $\Theta(n^2)$.
-(See \cite{dinic:flow} for the algorithm, \cite{even:dinic} for the time bound
-and \cite{schrijver} for more references on flows and matchings.)
-
-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}).
+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
+\thm\id{pcomplete}%
If there is a~polynomial-time algorithm for lexicographic ranking of permutations with
-an~arbitrary set of restrictions (which is a~part of the input), then $P=\#P$.
+a~set of restrictions which is a~part of the input, then $\rm P=\#P$.
\proof
-We will show that such a~ranking algorithm would enable us to compute the permanent
-of an~arbitrary zero-one matrix, which is a~$\#P$-complete problem. Let~$G$ be the
-bipartite graph with the bipartite adjacency matrix equal to the given matrix.
-The permanent of the matrix is then equal to the number of perfect matchings
-in~$G$, which is one more than the rank of the lexicographically maximal perfect
-matching in~$G$. As ranking of perfect matchings in~$G$ corresponds to ranking
-of permutations restricted by~$G$, it remains to show that we can find the
-lexicographically maximal permitted permutation in polynomial time.
+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. We can do this for example by using the
-Dinic's algorithm as described above on~the graph of remaining restrictions
-(i.e., $G$ with the vertices 1 and~$\pi[1]$ and removed together with the corresponding
-edges). Once we have~$\pi[1]$, we can fix it and proceed by finding $\pi[2]$
-in the same way, using the reduced graph. This way we construct the whole
-maximal permutation~$\pi$ in~$\O(n^2)$ calls to the Dinic's algorithm.
+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
+\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 slightly. For every finite set of elements $A\subset{\bb N}$,
+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
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~$A$) by $N_0(M) = {\per M}$.
+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 $i$-th row and $j$-th column removed.
+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 $\pi'=(\pi[2],\ldots,\pi[n])$ on the set~$A'=A\setminus\{\pi[1]\}$.
+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}%
$\<Rank>(\pi,i,n,A,M)$: 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 $M[1,k]=1$.
+\:$b\=C_a=\sum_k N_0(M^{1,k})$ 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 $b + \<Rank>(\pi,i+1,n,A\setminus\{\pi[i]\},M^{1,a})$.
+\: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 $\<Rank>(\pi,1,\vert A\vert,A,M)$.
\algo
\:If $i>n$, return $(0,\ldots,0)$.
-\:Find minimum $\ell$ such that $C_\ell > j$ (where $C_\ell$ is as above)
-\:$a\=R^{-1}_A(\ell-1)$.
-\:$\pi\=\<Unrank>(j-C_{\ell-1}, i+1, n, A\setminus\{a\}, M^{1,\ell})$.
-\:$\pi[i]\=a$.
+\:Find minimum $a$ such that $C_a > j$ (where $C_a$ is as in \<Rank> 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
\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
+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.)
and $\O(n^2\cdot t(n))$ by the computations of the~$N_0$'s.
\qed
-\rem
-This time bound is obviously very coarse, its main purpose was to demonstrate that
+\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 such improvement
-is to replace the matrix~$M$ by the corresponding restriction graph and instead of
-copying the matrix at every level of recursion, perform local operations on the graph
+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 large matrices.
+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
-and we will show that for derangements one can achieve linear time complexity.
+instead. We will show that for the derangements one can achieve linear time complexity.
-\examplen{Ranking of hatcheck permutations a.k.a.~derangements}\id{hatrank}%
+\nota\id{hatrank}%
As we already know, the hatcheck permutations correspond to restriction
-matrices which contain zeroes only on the main diagonal and graphs which are
-complete bipartite with the matching $\{(i,i) : i\in[n]\}$ deleted. For
+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 a~nice property:
+we will show that the submatrices of~$D_n$ share several nice properties:
\lemma\id{submatrixlemma}%
-If $D$ is a~submatrix of~$D_n$ obtained by deletion of rows and columns.
+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 the number of zero entries in it.
+and on the number of zero entries in~$D$.
\proof
We know from Lemma~\ref{permchar} that the permanent counts matchings in 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 matching,
+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 (but this matching is not necessarily perfect). 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 isomorphism invariant, the lemma follows.
+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.
-\para
-Instead of maintaining the matrix~$M$ in the algorithm, it is sufficient to keep
-the number~$z$ of zeroes in the matrix and the set~$Z$ which 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.
-
-The submatrix $M'=M^{1,k}$ for an~element $x$ of~rank~$k-1$ is described by either
-$z'=z-1$ 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 a~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~$Z$. So 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~$A$.
-
-It remains to show how to precompute the table of the $n_0$'s efficiently.
-
\lemma
The function~$n_0$ satisfies the following recurrence:
$$\eqalign{
n_0(d,d) &= d! \cdot \sum\nolimits_{k=0}^d {(-1)^k \over k!}, \cr
\noalign{\medskip}
n_0(z,d) &= z\cdot n_0(z-1,d-1) + (d-z)\cdot n_0(z,d-1) \quad\hbox{for $z<d$.} \cr
-}$$
+}\eqno{(*)}$$
\proof
-The base cases of the recurrence are simple: $n_0(0,d)$ counts the number of
-unrestricted permutations on~$[d]$ and $n_0(d,d)$ is equal to the number of derangements
-on~$[d]$, which we have already computed in Observation \ref{hatcheck}. Let us
+The base cases of the recurrence are straightforward: $n_0(0,d)$ counts the
+unrestricted permutations on~$[d]$, and $n_0(d,d)$ is equal to the number of derangements
+on~$[d]$, which we have already computed in Example \ref{hatcheck}. Let us
prove the third formula.
We will count the permutations~$\pi$ restricted by a~matrix~$M$ of the given parameters
stay in effect, so there are~$n_0(z,d-1)$ ways how to do so.
\qed
+\lemma
+The function~$n_0$ also satisfies the following recurrence:
+$$
+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
+
+\paran{The algorithm}\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 remembering 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$ whose rank is~$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 be therefore 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