\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
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~$i$.
We will follow the path unthreaded by Kaplansky and Riordan
\cite{kaplansky:rooks} and charted by Stanley in \cite{stanley:econe}.
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 $T(\pi)=\{ (i,\pi(i)) ; i\in[n] \}$.
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, $N_j := \#\{ \pi\in{\cal P}: \#(H\cap T(\pi)) = j \}$.
\:$r_k$ 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.
$$
N(x) = \sum_{j\ge 0} N_j x^j.
$$
-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}}
$$
\proof
-It is sufficient to prove that the equality holds for all integer~$x$.
+If two polynomials of degree~$n$ coincide at more than~$n$ points, they
+are identical, therefore it is sufficient to prove that the equality holds
+for all $x\in{\bb N}^+$.
The $N(x)$ counts the ways of placing~$n$ rooks on the board and labeling
each of them which stands on a~hole with an~element of~$[x]$. The right-hand
side counts the same: We can obtain any such configuration by placing $k$~rooks
on~$H$ first, labeling them with elements of~$\{2,\ldots,x\}$, placing
additional $n-k$ rooks on the remaining rows and columns (there are $(n-k)!$ ways
-how to do this) and labeling the new rooks standing on a~hole with~1.
+how to do this) and labeling those of the the new rooks standing on a~hole with~1.
\qed
\cor
When we substitute~$x=0$ in the above equality, we get a~formula for the
-number of rook placements avoiding~$H$:
+number of rook placements avoiding the holes altogether:
$$N_0 = N(0) = \sum_{k=0}^n (-1)^k \cdot (n-k)! \cdot r_k.$$
\example\id{hatcheck}%
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.
\itemize\ibull
\:permutations which obey a~given restriction graph~$G$,
-\:non-attacking placements of rooks on a~$n\times n$ board avoiding holes,
+\: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$.
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
+determine whether there exists at least one permutation satistying a~given set of restrictions:
+It is sufficient to reduce the particular 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)$.
\thm
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 $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.
+bipartite graph with its 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
\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}$,
-we will consider the set ${\cal P}_A$ of all permutations on~$A$, as usually
+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~$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$.
\:$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})$.
+\: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 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
\rem
This time bound 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
+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, perform local operations on the graph
+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 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 derangements one can achieve linear time complexity.
\examplen{Ranking of hatcheck permutations a.k.a.~derangements}\id{hatrank}%
As we already know, the hatcheck permutations correspond to restriction
we will show that the submatrices of~$D_n$ share a~nice property:
\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.
+As the number of matchings is an~isomorphism invariant, the lemma follows.
\qed
\defn
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
+Instead of maintaining the matrix~$M$ over the course of the algorithm, it is sufficient to remember
+the number~$z$ of zeroes in this 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),
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$).
+$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),$$
+or $n_0(z-1,\vert A\vert-1)$ depending on the set~$Z$. So we get:
+$$C_a = r\cdot n_0(z-1,\vert A\vert-1) + (a-r) \cdot n_0(z,\vert A\vert-1),$$
where $r=R_Z(R^{-1}_A(a))$, that is the number of restricted elements among the $a$~smallest ones in~$A$.
It remains to show how to precompute the table of the $n_0$'s efficiently.
}$$
\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
+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.
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