MAS334 Combinatorics 9 Tabular methods 11 Hall’s marriage theorem

10. Inclusion-exclusion for matching problems

Definition 10.1.

Let $B$ be an $n\times n$ board, with each square coloured black or white as before. We then write $\overline{B}$ for the board with the colours exchanged, so the white squares for $B$ are the black squares for $\overline{B}$ and vice-versa. We call $\overline{B}$ the complement of $B$ .

Example 10.2.

$B$ and $\overline{B}$ could be as follows.

As before, we write $c_{k}(B)$ for the number of ways of placing $k$ non-challenging rooks on $B$ , or equivalently the numbers of partial matchings of size $k$ for the corresponding matching problem. Similarly, we write $c_{k}(\overline{B})$ for the number of ways of placing $k$ non-challenging rooks on $\overline{B}$ . We are particularly interested in $c_{n}(\overline{B})$ , which is the number of full matchings for $\overline{B}$ . It turns out that if we know $c_{0}(B),\dotsc,c_{n}(B)$ then we can work out $c_{n}(\overline{B})$ :

Theorem 10.3.

$c_{n}(\overline{B})=\sum_{k=0}^{n}(-1)^{k}(n-k)!c_{k}(B)$ .

We will prove this after a preliminary result. The Inclusion-Exclusion Principle (from Section 5) will play a central rôle in the proof. We will then use this theorem to solve the snap problem (Problem 7.15) and the ménage problem (Problem 7.16).

Remark 10.4.

Theorem 10.3 covers only the case of a square board of size $n\times n$ , and gives only the top coefficient $c_{n}(\overline{B})$ in the rook polynomial of the complementary board. However, the statement can be generalised as follows: if $B$ is part of an $n\times m$ board, and $0\leq p\leq\min(n,m)$ then

c_{p}(\overline{B})=\sum_{k=0}^{p}(-1)^{k}\binom{n-k}{p-k}\binom{m-k}{p-k}(p-k% )!c_{k}(B).

The proof is quite similar to that of Theorem 10.3, with only a few additional complications; we leave it to the reader to work out the details.

Definition 10.5.

We write $S$ for the set of ways of placing $n$ non-challenging rooks on the full $n\times n$ board (so $|S|=n!$ by Proposition 7.11). Now let $x$ be a position on the board. We define $S_{x}\subseteq S$ to be the set of rook placements that include a rook at $x$ . More generally, given a set $X=\{x_{1},\dotsc,x_{m}\}$ of board positions, we put

	$\displaystyle S_{X}$	$\displaystyle=\bigcap_{i}S_{x_{i}}=S_{x_{1}}\cap\dotsb\cap S_{x_{m}}$
		$\displaystyle=\{\text{ non-challenging rook placements with a rook at each % position }x_{i}\}.$

Lemma 10.6.

Consider a set $X=\{x_{1},\dotsc,x_{m}\}$ as above.

(a)

If there is a row containing two of the positions $x_{i}$ , or a column containing two of the positions $x_{i}$ , then $S_{X}=\emptyset$ and so $|S_{X}|=0$ .
(b)

Otherwise, we have $|S_{X}|=(n-m)!$ .

Proof.

Interactive demo

•

Suppose that some row contains both $x_{i}$ and $x_{j}$ (with $i\neq j$ ). Then if we place two rooks at $x_{i}$ and $x_{j}$ , then they will challenge each other, so the placement does not count as an element of $S_{X}$ . This means that it is impossible to have any elements of $S_{X}$ , so $S_{X}=\emptyset$ and $|S_{X}|=0$ .
•

The same applies if some column contains both $x_{i}$ and $x_{j}$ with $i\neq j$ .
•

Suppose instead that the $x_{i}$ are all in different rows, say $r_{1},\dotsc,r_{m}$ , and also in different columns, say $c_{1},\dotsc,c_{m}$ . Let $A$ be the set of rows other than $r_{1},\dotsc,r_{m}$ , so $|A|=n-m$ . Let $B$ be the set of columns other than $c_{1},\dotsc,c_{m}$ , so $|B|=n-m$ . To get a full rook placement in $S_{X}$ we need to place rooks on $X$ , then place $n-m$ rooks on the board $A\times B$ , which is an $(n-m)\times(n-m)$ board with no black squares. Proposition 7.11 tells us that there are $(n-m)!$ ways to do this, so $|S_{X}|=(n-m)!$ .

∎

Proof of Theorem 10.3.

We need to understand the set $C_{n}(\overline{B})\subseteq S$ . Consider a rook placement $P$ for the full board, so $P\in S$ . We claim that $P\in\bigcup_{x\in B}S_{x}$ iff $P\not\in C_{n}(\overline{B})$ . Indeed, we have $P\in\bigcup_{x\in B}S_{x}$ iff there exists a square $x\in B$ such that $P\in S_{x}$ . However, we have $P\in S_{x}$ iff the placement $P$ has a rook at $x$ . Here $x$ is a white square of $B$ so it is a black square of $\overline{B}$ , so if the placement has a rook at $x$ , then it does not count as a placement on $\overline{B}$ , so $P\not\in C_{n}(\overline{B})$ . This argument is reversible, so $P\in\bigcup_{x\in B}S_{x}$ iff $P\not\in C_{n}(\overline{B})$ . By the contrapositive, we have $P\in C_{n}(\overline{B})$ iff $P\not\in\bigcup_{x\in B}S_{x}$ , or in other words $P\in S\setminus\bigcup_{x}S_{x}$ . The negative form of the IEP now tells us that

c_{n}(\overline{B})=|C_{n}(\overline{B})|=\sum_{X\subseteq B}(-1)^{|X|}|S_{X}|.

If $X$ is not a non-challenging rook placement on $B$ , then $|S_{X}|=0$ by Lemma 10.6(a), so we can ignore these terms. On the other hand, if $X$ is a placement of $k$ non-challenging rooks on $B$ , then $|S_{X}|=(n-k)!$ by Lemma 10.6(a). The number of terms of this type is $c_{k}(B)$ , so the sum of all terms of this type is $(-1)^{k}(n-k)!c_{k}(B)$ . Putting this together, we get $c_{n}(\overline{B})=\sum_{k=0}^{n}(-1)^{k}(n-k)!c_{k}(B)$ as claimed. ∎

Remark 10.7.

We can apply Theorem 10.3 to $\overline{B}$ and note that $\overline{\overline{B}}=B$ ; this gives

c_{n}(B)=\sum_{k=0}^{n}(-1)^{k}(n-k)!c_{k}(\overline{B}).

Example 10.8.

As a trivial example, we have $c_{n}(F_{n})=\sum_{k=0}^{n}(-1)^{k}(n-k)!c_{k}(\overline{F_{n}})$ . However, all squares in $\overline{F_{n}}$ are black, so we cannot place any rooks there, so $c_{k}(\overline{F_{n}})=0$ for $k>0$ . On the other hand, we always have $c_{0}=1$ , so our equation becomes

c_{n}(F_{n})=(-1)^{0}n!c_{0}(\overline{F_{n}})=n!,

and we knew this already from Proposition 7.10.

Example 10.9.

Recall from Example 7.6 that $D_{n}$ is the $n\times n$ board with only the diagonal squares coloured white, and that $c_{k}(D_{n})=\binom{n}{k}=n!/(k!(n-k)!)$ . We thus have

c_{n}(\overline{D_{n}})=\sum_{k=0}^{n}(-1)^{k}(n-k)!c_{k}(D_{n})=n!\sum_{k=0}^% {n-1}(-1)^{k}/k!.

On the other hand, Problem 7.13 shows that $C_{n}(\overline{D_{n}})$ is the set of derangements of $\{1,\dotsc,n\}$ . We saw in Proposition 5.11 that the number of derangements is $n!\sum_{k=0}^{n-1}(-1)^{k}/k!$ , so everything is consistent.

Definition 10.10.

The board $Q_{n}$ consists of $2n-1$ white squares in an $n\times n$ board arranged in a zigzag pattern as illustrated on the left below. (To see that there are $2n-1$ white squares, note that there are $n$ rows, and each row contains two white squares, except for the first row, which contains only one white square.) We also consider the board $Q^{\prime}_{n}$ , which has an extra white square at the top right, making $2n$ white squares in total.

We call a board of type $Q_{n}$ a staircase.

Proposition 10.11.

For $0\leq k\leq n$ we have $c_{k}(Q_{n})=\binom{2n-k}{k}$ . We also have

c_{k}(Q^{\prime}_{n})=c_{k}(Q_{n})+c_{k-1}(Q_{n-1})=\binom{2n-k}{k}+\binom{2n-% 1-k}{k-1}.

Proof.

Interactive demo

By inspection, two rooks on the staircase challenge each other iff they are adjacent. Thus, the allowable rook placements are just subsets of the staircase that have no adjacent elements; in other words, they must be gappy, as in Definition 1.21. Thus, $c_{k}(Q_{n})$ is the number of gappy subsets of size $k$ in the staircase. Proposition 1.23 therefore gives $c_{k}(Q_{n})=\binom{m-k+1}{k}$ , where $m$ is the number of cells in the staircase. This is just $m=2n-1$ , so we get $c_{k}(Q_{n})=\binom{2n-k}{k}$ .

Now consider $Q^{\prime}_{n}$ . Blocking the top right square gives $Q_{n}$ , and stripping it gives a reflected copy of $Q_{n-1}$ . Thus, Theorem 8.1 gives $c_{k}(Q^{\prime}_{n})=c_{k}(Q_{n})+c_{k-1}(Q_{n-1})$ , which is $\binom{2n-k}{k}+\binom{2n-1-k}{k-1}$ by the previous paragraph. ∎

We can now complete our analysis of the ménage problem.

Example 10.12.

Using Proposition 10.11, we see that the rook polynomial coefficients of $Q^{\prime}_{5}$ are as follows:

	$\displaystyle c_{0}(Q^{\prime}_{5})$	$\displaystyle=1$
	$\displaystyle c_{1}(Q^{\prime}_{5})$	$\displaystyle=\binom{9}{1}+\binom{8}{0}=10$
	$\displaystyle c_{2}(Q^{\prime}_{5})$	$\displaystyle=\binom{8}{2}+\binom{7}{1}=35$
	$\displaystyle c_{3}(Q^{\prime}_{5})$	$\displaystyle=\binom{7}{3}+\binom{6}{2}=50$
	$\displaystyle c_{4}(Q^{\prime}_{5})$	$\displaystyle=\binom{6}{4}+\binom{5}{3}=25$
	$\displaystyle c_{5}(Q^{\prime}_{5})$	$\displaystyle=\binom{5}{5}+\binom{4}{4}=2.$

Now note that the board $B$ considered in Problem 7.16 is just the complement of $Q^{\prime}_{5}$ . It follows that

	$\displaystyle c_{5}(B)$	$\displaystyle=5!c_{0}(Q^{\prime}_{5})-4!c_{1}(Q^{\prime}_{5})+3!c_{2}(Q^{% \prime}_{5})-2!c_{3}(Q^{\prime}_{5})+1!c_{4}(Q^{\prime}_{5})-0!c_{5}(Q^{\prime% }_{5})$
		$\displaystyle=120\times 1-24\times 10+6\times 35-2\times 50+1\times 25-1\times 0$
		$\displaystyle=13.$

Thus, there are precisely $13$ full matchings for $B$ . As explained in Problem 7.16, it follows that there are $2\times 120\times 13=3120$ ways to solve the seating problem described there.

We can also now complete our analysis of the snap problem.

Example 10.13.

Let $B$ denote the board shown in Problem 7.15, which is a $16\times 16$ board with four black blocks of size $4\times 4$ on the diagonal. This means that $\overline{B}$ consists of four fully disjoint copies of $F_{4}$ . Proposition 7.11) tells us that

r_{F_{4}}(x)=\sum_{k=0}^{4}\binom{4}{k}^{2}k!x^{k}=1+16x+72x^{2}+96x^{3}+24x^{% 4}.

Moreover, we can use Theorem 8.7 to show that

r_{\overline{B}}(x)=r_{F_{4}}(x)^{4}.

It would be a lot of work to expand this by hand, but it can easily be done with a computer. If $r_{F_{4}}(x)^{4}=\sum_{i=0}^{16}a_{i}x^{i}$ , we then find that $c_{16}(B)=\sum_{i}(-1)^{i}(16-i)!a_{i}$ . Using a computer again, we find that $c_{16}(B)\simeq 2.483\times 10^{11}$ . This is the number of possible games with no snap. On the other hand, the total number of possible games is $16!\simeq 2.092\times 10^{13}$ . Thus, the probability of not getting a snap is

c_{16}(B)/52!\simeq\frac{2.483\times 10^{11}}{2.092\times 10^{13}}\simeq 0.012.

In other words, only about one in a hundred games will end without a snap. This was all for the restricted game where we only use aces, kings, queens and jacks. If we want to use the full pack, we need to expand $r_{F_{4}}(x)^{13}$ as $\sum_{i=0}^{52}b_{i}x^{i}$ , and then calculate $c_{52}(B)=\sum_{i}(-1)^{i}(52-i)!b_{i}$ , which works out to $1.309\times 10^{66}$ . We then divide by $52!\simeq 8.066\times 10^{67}$ to give a probability of approximately $0.016$ .