$12\cdots t$-Permutation Avoiding (0, 1)-Matrices

Abstract

Let n and t be integers with $1<t\le n$. Let A be an $n\times n$ (0, 1)-matrix with a positive permanent, that is, for which there exists a permutation matrix $P\le A$ (entrywise order). We investigate the minimum number $\alpha (n,t)$ of zeros possible in such a matrix A which avoids a P with a $12\cdots t$-pattern, that is, for which there does not exist a permutation matrix $P\le A$ containing the identity matrix $I_t$ as a submatrix. We conjecture that $\alpha (n,t)={{k+1}\atopwithdelims ()2}$ where $k=n-t+1$. We prove this conjecture is correct when $t=2 \text{ or } 3$ and we consider for which matrices equality holds. We also prove the conjecture is correct for all t if $n\ge 2k-3$. Finally, we investigate which $12\cdots t$-permutation avoiding matrices have the maximum permanent.

Appendix

In this appendix we show that Conjecture 1.1 is also true for $n=2k-2$ and $n=2k-3$, that is, in addition to $n\ge 2k-1$, Theorem 3.6 is also true for $n=2k-2$ and $n=2k-3$. The following lemma is used in the proof.

Lemma 5.1

Suppose there exist integers n and t with $2 <t \le n$ such that there exists an $n \times n$ (0, 1)-matrix with a nonzero permanent which is $123\cdots t$-permutation avoiding, and which has fewer than $f(n,t) = {{n-t+2}\atopwithdelims ()2}$ 0’s. If A is such a matrix with n minimal, then A is fully indecomposable.

Proof

Let $k = n-t+1$. Assume to the contrary that A is such a matrix of smallest possible order n with A partly decomposable. (We know that $n > 5$, since if $n \le 5$ then, by Theorem 3.6, A has at least $k(k+1)/2 = {{n-t+2}\atopwithdelims ()2} = f(n,t)$ 0’s.) Hence there exist permutation matrices P and Q such that $A' = PAQ$ has form

$$\begin{aligned} A'=\left[ \begin{array}{c|c} O_{n_1,n_2}& A_1\\ \hline A_2& X\end{array}\right] , \end{aligned}$$

where $A_1$ is $n_1 \times n_1$, $A_2$ is $n_2\times n_2$, and $n_1 + n_2 = n$. By taking the rows and columns of $A_1$ and $A_2$ in the same relative order as the rows and columns of A, the matrices $A_1, A_2$, and $A'$ will all be $123\cdots t$-permutation avoiding. Since $A'$ has at least $\left( f(n_1,t) + f(n_2,t) + n_1n_2\right)$ 0’s and, as can be shown by direct calculation, this is equal to $f(n,t) + {{t-1}\atopwithdelims ()2}$, $A'$ has more than f(n, t) 0’s, a contraction. $\square$

Note that Conjecture 1.1 says that no matrix satisfies the hypotheses of Lemma 5.1.

Case of Theorem  3.6 with $n=2k-2$:

Now A is $(k-1)$-permutation avoiding and we choose A to be a counterexample with n smallest. By Lemma 5.1A is fully indecomposable. We assume that no row or column has more than $\frac{n}{2}=k-1$ 0’s; otherwise, as in Theorem 3.6, we are done by induction.

By the argument in the proof of Theorem 3.6, for each $j\in [k-1]$, $H_j$ and $T_j$ each have at least $\big \lceil \frac{k-j+1}{2} \big \rceil$ 0’s with no overlap. This gives a total of at least

$$\begin{aligned} \left( 2\sum _{j=2}^{k-1} \bigg \lceil \frac{k-j+1}{2} \bigg \rceil \right) +\bigg \lceil \frac{k}{2}\bigg \rceil =\frac{k(k+1)}{2}-2 \end{aligned}$$

0’s. Hence if A has only $\left( \frac{k(k+1)}{2}-1\right)$ 0’s, then each of $H_j$ and $T_j$ has exactly $\big \lceil \frac{k-j+1}{2} \big \rceil$ 0’s for all $j\in [k-1]$, and $H_k= H_k^* \cup T_k^*$ has precisely one 0. Both $H_k^*$ and $T_k^*$ have $k-1$ elements. Without loss of generality, assume the 0 in $H_k=T_k$ is in $T_k^*.$

Claim: Let j and t be positive integers such that $j=k+2t-1\le n$. Then $T_j$ has precisely t 0’s none of which are in $T_j^*.$

Proof of the Claim. Since $j=k+2t-1>k,$ $T_j=H_{(n+2-j)}=H_{2k-j}=H_{k-2t+1}$ has precisely $t=\big \lceil \frac{k-(k-2t+1)+1}{2}\big \rceil$ 0’s.

If at least one of these 0’s is in $T_j^*,$ then

$$\begin{aligned} T_j\backslash T_j^*=H_{n+2-j}^*=H_{k-2t+1}^* \end{aligned}$$

has $(n-(k-2t+1)+1)=(k-2+2t)$ entries and at most $(t-1)$ 0’s, and so at least $(k-1+t)$ $1'$s. Hence the transversal obtained by switching out the t 0’s in $T_j$ has at least $(k-1)$ $1'$s in $H_{k-2t+1}^*,$ contradicting that A is $(k-1)$-permutation avoiding. $\Box$

If the 0 in $T_k^*$ can be switched out with a 1 in $T_k^*,$ then we get a permutation matrix with $(k-1)$ $1'$s in $H_k^*,$ contradicting that A is $(k-1)$-permutation avoiding. Hence the 0 in $T_k^*$ cannot be switched out with a 1 in $T_k^*.$ There are two cases to consider depending on whether or not the 0 in $T_k^*$ is in the first or last column of A.

• (i) The 0 in $T_k^*$ is not in position (1, k) or $(k-1,n).$ Suppose it is in position $(i,i+k-1)$ where $i\in [2,k-2].$ Since the entries in position $(i-1,i+k-1)$ and $(i,i+k)$ are both in $T_{k+1}^*,$ by the Claim, they are both equal to 1. To prevent switching out the 0 in $T_k^*,$ the entries in positions $(i,i+k-2)$ and $(i+1,i+k-1)$ must both equal to 0 (see Eq. (2)(ii) for example with $k=8$ where the two d’s must be 0). However, both these entries are in $T_{k-1},$ contradicting that $T_{k-1}$ has only one 0.

• (ii) The 0 in $T_k^*$ is in position (1, k) or $(k-1,n).$ Assume the 0 is in position (1, k). For each t such that $2t\le k-1,$ by the Claim there is a 1 in position $(1,k+2t-1)$, and so to avoid switching out the 0 in position (1, k),  there must be a 0 in position (2t, k). For each t such that $2t\le k-2$ there are $1'$s in positions $(1,k+2t-1)$ and $(2t, k+2t)$ (they are in $T_{k+2t-1}^*$ and $T_{k+1}^*$ respectively). Hence there must be a 0 in position $(2t+1,k)$ or else we could do a 3-way switch of those three entries in place of the entries in positions $(1,k), (2t, k+2t-1),$ and $(2t+1, k+2t)$ in $T_k^*$ to get a transversal of all 1’s including the $(k-1)$ 1’s in $H_k*,$ a contradiction. Hence position (i, k) contains a 0 for all $1\le i\le k-1$. To avoid the existence of a 1 whose row and column contain a total of k 0’s (so that the result then follows by induction), the last $(k-1)$ rows of A must contain all 1’s. For any transversal of all 1’s in A, the 1’s in the last $(k-1)$ rows can be permuted to produce a forbidden transversal containing all 1’s. The argument if the 0 is in position $(k-1,n)$ is similar. See Eq. (9)(i) which shows the top right $7\times 7$ submatrix when $k=8$ and $n=14$. The three c’s in column 1 must be 0’s because d, e, f in row 1 are all 1’s. The three d’s, three e’s, and three f’s cannot be all 1’s, so the d, e, f in the first column must be 0’s.

  $$\begin{aligned} (i) \left[ \begin{array}{c|c|c|c|c|c|c} 0& d& & e& & f& \\ \hline c& 1& d& & 1& & 1\\ \hline d& & 1& 1& & 1& \\ \hline c& & & 1& e& & 1\\ \hline e& & & & 1& 1& \\ \hline c& & & & & 1& f\\ \hline f& & & & & & 1\end{array}\right] \text{ and } (ii) \left[ \begin{array}{c|c|c|c|c|c|c} 1& 1& & 1& & 1& \\ \hline & 1& 1& & 1& & 1\\ \hline & d& 0& 1& & 1& \\ \hline & & d& 1& 1& & 1\\ \hline & & & & 1& 1& \\ \hline & & & & & 1& 1\\ \hline & & & & & & 1\end{array}\right] . \end{aligned}$$

  (9)

Case of Theorem  3.6 with $n=2k-3$:

If some row or column has k 0’s, we are done by induction as in the proof of Theorem 3.6. If some row (or column) has $(k-1)$ 0’s then, as in the proof of Theorem 3.6, to avoid a 1 with a total of at least k 0’s in its row and column, A contains $(n-k+1)$ columns with all 1’s and hence a transversal with the forbidden pattern. Hence we can assume that each row and column has at most $(k-2)<n/2$ 0’s, so by Theorem 3.5, A has at least ${{k+1}\atopwithdelims ()2}$ 0’s. $\Box$