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
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}%
Let $D$ be a~submatrix of~$D_n$ obtained by deletion of rows and columns.
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~$M$ 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),
-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~$k-1$, it 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{(*)}$$
\proof
The base cases of the recurrence are straightforward: $n_0(0,d)$ counts the
stay in effect, so there are~$n_0(z,d-1)$ ways how to do so.
\qed
+\lemma
+The function~$n_0$ also satisfies the following recurrence:
+$$
+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\}$ which satisfy~$M_z^{1,1}$,
+so there are $n_0(z-1,d-1)$ of them.
+\qed
+
+\cor 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
+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),
+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$, it 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.
+
+
\endpart
projects / saga.git / commitdiff