MAS334 Combinatorics 7 Rook polynomials 9 Tabular methods

8. Reduction theorems

We will now introduce two results that allow us to calculate the rook polynomial of a board in terms of rook polynomials of smaller or simpler boards.

Theorem 8.1.

Let $B$ be an $n\times n$ board, in which some squares may be blocked off. Let $s$ be an unblocked square. Let $C$ be the same as $B$ , except that $s$ is blocked off. Let $D$ be the same as $B$ , except that $s$ ’s row and column are removed. Then for $k>0$ we have $c_{k}(B)=c_{k}(C)+c_{k-1}(D)$ , and so

r_{B}(x)=r_{C}(x)+x\,r_{D}(x).

Proof.

Interactive demo

The constant term on both sides is equal to one. Consider instead the coefficients of $x^{k}$ , where $k>0$ . The coefficient in $r_{B}(x)$ is $c_{k}(B)$ , which is the number of ways of placing $k$ non-challenging rooks on $B$ . We can write this as $c_{k}(B)=p+q$ , where $p$ is the number of placements that do not have a rook at $s$ , and $q$ is the number of placements that do have a rook at $s$ . To place $k$ rooks on $B$ without using $s$ is the same as to place $k$ rooks on $C$ , so $p=c_{k}(C)$ . To place $k$ rooks on $B$ including $s$ is the same as to place a rook at $s$ , and then place $k-1$ more rooks on $B$ , avoiding $s$ ’s row and column. This in turn is the same as placing $k-1$ rooks on $D$ , so we have $q=c_{k-1}(D)$ . Note also that $c_{k-1}(D)$ is the coefficient of $x^{k-1}$ in $r_{D}(x)$ , which is the same as the coefficient of $x^{k}$ in $x\,r_{D}(x)$ . The equation $c_{k}(B)=p+q$ now tells us that $c_{k}(B)=c_{k}(C)+c_{k-1}(D)$ , or that the coefficients of $x^{k}$ are the same in $r_{B}(x)$ and $r_{C}(x)+x\,r_{D}(x)$ . As this works for all $k$ , we have $r_{B}(x)=r_{C}(x)+x\,r_{D}(x)$ as claimed. ∎

Remark 8.2.

We say that $C$ is the result of blocking $s$ , and that $D$ is the result of stripping $s$ .

Example 8.3.

Boards $B$ , $C$ and $D$ could be as follows:

As all squares in $D$ are white, Proposition 7.11 gives

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

Next, in $C$ the middle row is fully blacked out, and it is easy to see that this makes it irrelevant, so $C$ is equivalent to a $4\times 5$ board in which all squares are white. We can therefore use Proposition 7.11 again to get

r_{C}(x)=\sum_{k=0}^{4}\binom{4}{k}\binom{5}{k}k!x^{k}=1+20x+120x^{2}+240x^{3}% +120x^{4}.

The theorem now gives

r_{B}(x)=r_{C}(x)+x\,r_{D}(x)=1+21x+136x^{2}+312x^{3}+216x^{4}+120x^{5}.

Example 8.4.

It is interesting to see how Theorem 8.1 works out in some trivial cases. First consider the linear board $L_{n}$ from Example 7.5, so $L_{n}$ just consists of $n$ blank squares in a row. Blocking any square gives $L_{n-1}$ , and stripping any square gives the empty board, so Theorem 8.1 gives $r_{L_{n}}(x)=r_{L_{n-1}}(x)+x\,r_{\emptyset}(x)$ . Even for the empty board, there is a unique way of placing no rooks, so we have $r_{\emptyset}(x)=1$ , so $r_{L_{n}}(x)=r_{L_{n-1}}(x)+x$ . From this it follows inductively that $r_{L_{n}}(x)=1+nx$ (which is already obvious from the definitions, as we remarked in Example 7.5.)

Now consider the diagonal board $D_{n}$ from Example 7.6. Here blocking any square gives $D_{n-1}$ , and stripping any square also gives $D_{n-1}$ . Thus, Theorem 8.1 gives $r_{D_{n}}(x)=(1+x)r_{D_{n-1}}(x)$ . From this it follows inductively that $r_{D_{n}}(x)=(1+x)^{n}$ (which is already obvious from the definitions, as we remarked in Example 7.6.)

Definition 8.5.

Let $B$ be an $n\times n$ board with some squares blocked off, as before. Suppose that the set of unblocked squares has been split into two parts $C$ and $D$ , so that $B=C\cup D$ and $C\cap D=\emptyset$ . We say that $C$ and $D$ are fully disjoint if

(a)

There is no row that contains a square from $C$ and also a square from $D$ ; and
(b)

There is no column that contains a square from $C$ and also a square from $D$ .

Example 8.6.

Interactive demo

Consider the board $B$ as shown on the left below.

We can divide $B$ into disjoint subboards $C$ and $D$ , as shown by the second picture. Note that:

(a)

Rows $1,3$ and $5$ contain only $C$ ’s, whereas rows $2$ and $4$ contain only $D$ ’s.
(b)

Columns $1,2$ and $5$ contain only $C$ ’s, whereas columns $3$ and $4$ contain only $D$ ’s.

This shows that $C$ and $D$ are fully disjoint.

Theorem 8.7.

Let $B$ be an $n\times n$ board with some squares blocked off, as before. Suppose that $B$ can be split into two subboards $C$ and $D$ , which are fully disjoint. Note that $C$ and $D$ can be regarded as boards in their own right, so they have rook polynomials $r_{C}(x)$ and $r_{D}(x)$ . Then

r_{B}(x)=r_{C}(x)\,r_{D}(x).

(We call this method factoring.)

Proof.

To place $k$ rooks in $B$ , we first need to decide how many of them will be in $C$ , and how many in $D$ . If we place $i$ rooks in $C$ , then there will be $k-i$ in $D$ . There are $c_{i}(C)$ ways of placing $i$ non-challenging rooks in $C$ , and $c_{k-i}(D)$ ways of placing $k-i$ non-challenging rooks in $D$ . Moreover, the rooks in $C$ cannot challenge those in $D$ or vice-versa, because of conditions (b) and (c) in the theorem. Thus, we can put the two sets of rooks together, and we always have a set of $k$ non-challenging rooks in $B$ . From this we see that there are $c_{i}(C)\times c_{k-i}(D)$ ways of placing $k$ non-challenging rooks in $B$ , with precisely $i$ of them in $C$ . By considering all possible values of $i$ , we get $c_{k}(B)=\sum_{i=0}^{k}c_{i}(C)c_{k-i}(D)$ . This in turn gives

	$\displaystyle r_{B}(x)$	$\displaystyle=\sum_{k\geq 0}c_{k}(B)x^{k}=\sum_{k\geq 0}\sum_{i=0}^{k}c_{i}(C)% c_{k-i}(D)x^{k}$
		$\displaystyle=\left(\sum_{i\geq 0}c_{i}(C)x^{i}\right)\left(\sum_{j\geq 0}c_{j% }(D)x^{j}\right)=r_{C}(x)r_{D}(x).$

∎

Example 8.8.

Consider again the board $B$ from example 8.6

Boards $C$ and $D$ are small enough that we can calculate the rook polynomials by inspection:

	$\displaystyle r_{C}(x)$	$\displaystyle=1+8x+14x^{2}+4x^{3}$
	$\displaystyle r_{D}(x)$	$\displaystyle=1+3x+x^{2}.$

Theorem 8.7 therefore gives

	$\displaystyle r_{B}(x)$	$\displaystyle=r_{C}(x)r_{D}(x)=(1+8x+14x^{2}+4x^{3})(1+3x+x^{2})$
		$\displaystyle=1+11x+39x^{2}+54x^{3}+26x^{4}+4x^{5}.$

Problem 8.9.

Consider the following table, in which all the entries are in the range 1-5:

You can check that there are no repeats in any row or column. How many ways are there of adding a fourth row, so that there are still no repeats in any row or column?

Solution.

Interactive demo

We start by extending the table a little:

We want to add new row, in such a way that there are no repeats in any row or column. Column $a$ already contains $1$ , $2$ and $4$ , so the new entry cannot be any of those, so it must be in the set $\{3,5\}$ . Column $b$ already contains $2$ , $4$ and $3$ , so the new entry cannot be any of those, so it must be in the set $\{1,5\}$ . In the same way, we see that the new entries in columns $c$ , $d$ and $e$ must be taken from the sets $\{2,4\}$ , $\{1,2\}$ and $\{3,4\}$ , as indicated in the diagram. This gives us a matching problem, with chessboard diagram as shown on the left below. Adding a new row is equivalent to solving the rook problem for this board. For example, one possible solution is to place rooks at $a3,\;b5,\;c2,\;d1,\;e4$ , as shown in the middle picture. This indicates that we can legitimately add a new row to our table, with column $a$ containing $3$ , column $b$ containing $5$ , column $c$ containing $2$ , column $d$ containing $1$ and column $e$ containing $4$ . The result is shown in the right-hand picture.

We want to find the number of ways to add a new row, which is the same as the number of ways of placing $5$ non-challenging rooks on the board, which is the coefficient of $x^{5}$ in the rook polynomial. As an illustration of the relevant methods, we will in fact calculate the full rook polynomial, by repeatedly using Theorems 8.1 and 8.7.

The following picture shows boards $B_{1},\dotsc,B_{15}$ . We write $r_{k}(x)$ for the rook polynomial of board $B_{k}$ . Our problem is to calculate $r_{1}(x)$ .

By inspection, we have

	$\displaystyle r_{4}(x)$	$\displaystyle=r_{5}(x)=1+2x$
	$\displaystyle r_{7}(x)$	$\displaystyle=r_{8}(x)=r_{11}(x)=r_{12}(x)=1+3x+x^{2}$
	$\displaystyle r_{14}(x)$	$\displaystyle=r_{15}(x)=1+4x+3x^{2}.$

Now consider board $B_{3}$ . We have divided the empty cells into two groups, marked $P$ and $Q$ respectively. There is no row that contains both $P$ and $Q$ , and there is no column that contains both $P$ and $Q$ , so $P$ and $Q$ are fully disjoint. Thus, Theorem 8.7 is applicable, and we get $r_{3}(x)=r_{4}(x)r_{5}(x)$ . We can also factor boards $B_{6}$ , $B_{10}$ and $B_{13}$ in a similar way. We find that

	$\displaystyle r_{3}(x)$	$\displaystyle=r_{4}(x)r_{5}(x)=(1+2x)^{2}=1+4x+4x^{2}$
	$\displaystyle r_{6}(x)$	$\displaystyle=r_{7}(x)r_{8}(x)=(1+3x+x^{2})^{2}=1+6x+11x^{2}+6x^{3}+x^{4}$
	$\displaystyle r_{10}(x)$	$\displaystyle=r_{11}(x)r_{12}(x)=(1+3x+x^{2})^{2}=1+6x+11x^{2}+6x^{3}+x^{4}$
	$\displaystyle r_{13}(x)$	$\displaystyle=r_{14}(x)r_{15}(x)=(1+4x+3x^{2})^{2}=1+8x+22x^{2}+24x^{3}+9x^{4}.$

Now consider board $B_{9}$ , in which we have marked one square with a red dot. Blocking that square gives $B_{13}$ , and stripping it gives $B_{10}$ . Theorem 8.1 therefore tells us that $r_{9}(x)=r_{13}(x)+x\,r_{10}(x)$ . We can also block and strip the marked square in $B_{2}$ to get $B_{6}$ and $B_{3}$ , or we can block and strip the marked square in $B_{1}$ to get $B_{9}$ and $B_{2}$ . We therefore have

	$\displaystyle r_{9}(x)$	$\displaystyle=r_{13}(x)+x\,r_{10}(x)=(1+8x+22x^{2}+24x^{3}+9x^{4})+x\,(1+6x+11% x^{2}+6x^{3}+x^{4})$
		$\displaystyle=1+9x+28x^{2}+35x^{3}+15x^{4}+x^{5}$
	$\displaystyle r_{2}(x)$	$\displaystyle=r_{6}(x)+x\,r_{3}(x)=(1+6x+11x^{2}+6x^{3}+x^{4})+x\,(1+4x+4x^{2})$
		$\displaystyle=1+7x+15x^{2}+10x^{3}+x^{4}$
	$\displaystyle r_{1}(x)$	$\displaystyle=r_{9}(x)+x\,r_{2}(x)=(1+9x+28x^{2}+35x^{3}+15x^{4}+x^{5})+x\,(1+% 7x+15x^{2}+10x^{3}+x^{4})$
		$\displaystyle=1+10x+35x^{2}+50x^{3}+25x^{4}+11x^{4}+2x^{5}.$

(If we wanted to save writing, it would not be hard to do this calculation without explicitly drawing any of the boards to the right of the dotted line.) In particular, the number of ways of adding an admissible row to our original table is the number of ways of placing $5$ non-challenging rooks on $B_{1}$ , which is the coefficient of $x^{5}$ in $r_{1}(x)$ , which is $2$ . In fact, the two possible rook placements are $a3b5c2d1e4$ and $a5b1c4d2e3$ , so the two possibilities for the extra row are $(3,5,2,1,4)$ and $(5,1,4,2,3)$ .

Problem 8.10.

Calculate the full rook polynomial for the board in Problem 7.14.

Solution.

We use the same method as in the previous example, but written in a slightly more efficient way.

We have boards $B_{1},\dotsc,B_{13}$ . However, $B_{5}$ and $B_{6}$ have not been drawn separately, they are just marked as subsets of $B_{4}$ . Similarly, $B_{9}$ and $B_{10}$ are just marked as subsets of $B_{8}$ , and $B_{12}$ and $B_{13}$ are subsets of $B_{11}$ . By inspection, we have

	$\displaystyle r_{3}(x)$	$\displaystyle=r_{9}(x)=1+2x$
	$\displaystyle r_{5}(x)$	$\displaystyle=r_{12}(x)=r_{13}(x)=1+3x+x^{2}$
	$\displaystyle r_{6}(x)$	$\displaystyle=r_{10}(x)=1+x.$

We next want to use the factoring theorem (Theorem 8.7) to calculate $r_{4}(x)$ , $r_{8}(x)$ and $r_{11}(x)$ . For that result to be applicable, we must check that the relevant subboards are fully disjoint. This is equivalent to the following claim: in boards $B_{4}$ , $B_{8}$ and $B_{11}$ , any two squares in the same row have the same number, and any two squares in the same column have the same number. This is clear by inspection. We therefore have

	$\displaystyle r_{4}(x)$	$\displaystyle=r_{5}(x)r_{6}(x)=(1+3x+x^{2})(1+x)=1+4x+4x^{2}+x^{3}$
	$\displaystyle r_{8}(x)$	$\displaystyle=r_{9}(x)r_{10}(x)=(1+2x)(1+x)=1+3x+2x^{2}$
	$\displaystyle r_{11}(x)$	$\displaystyle=r_{12}(x)r_{13}(x)=(1+3x+x^{2})^{2}=1+6x+11x^{2}+6x^{3}+x^{4}.$

Next, blocking and stripping the marked square in $B_{7}$ gives $B_{11}$ and $B_{8}$ . Similarly, $B_{2}$ gives $B_{4}$ and $B_{3}$ , and $B_{1}$ gives $B_{7}$ and $B_{2}$ . We therefore have

	$\displaystyle r_{7}(x)$	$\displaystyle=r_{11}(x)+x\,r_{8}(x)=1+6x+11x^{2}+6x^{3}+x^{4}+x\,(1+3x+2x^{2})$
		$\displaystyle=1+7x+14x^{2}+8x^{3}+x^{4}$
	$\displaystyle r_{2}(x)$	$\displaystyle=r_{4}(x)+x\,r_{3}(x)=1+4x+4x^{2}+x^{3}+x\,(1+2x)$
		$\displaystyle=1+5x+6x^{2}+x^{3}$
	$\displaystyle r_{1}(x)$	$\displaystyle=r_{7}(x)+x\,r_{2}(x)=1+7x+14x^{2}+8x^{3}+x^{4}+x\,(1+5x+6x^{2}+x% ^{3})$
		$\displaystyle=1+8x+19x^{2}+14x^{3}+2x^{4}.$

In particular, the number of ways of allocating the jobs in Problem 7.14 is the same as the number of ways of placing $4$ non-challenging rooks on $B_{1}$ , which is the coefficient of $x^{4}$ in $r_{1}(x)$ , which is $2$ . This is the same answer as we found previously in Problem 7.14.