Hello! After reading so many useful posts on Brilliant, I have finally set my mind to do one (my first!) on Rook's Polynomial, a useful technique for many combinatorics question (those with certain restrictions as you will see below).

To elaborate further on what exactly is the Rook's Polynomial, it is the generating function for the numbers of arrangement of \(k\) non-attacking rooks on a board \(B\). For those who are new to chess, rooks are chess piece that move and attack horizontally and vertically. Hence, for rooks to be non-attacking to each other, they must not be placed on the same row or column of the board.

In general, the rook polynomial of a \(m \times n\) board \(B_{m,n}\) (denoted as \(R(x, B_{m,n})\)) is: \[R(x,B_{m,n}) = \displaystyle \sum_{k=0}^\infty r_k(B_{m,n})x^k\]

where, \(r_k\) is the number of ways to arrange \(k\) non attacking rooks on board \(B_{m,n}\). For example \( R(x, B_{1,1}) = x +1\) since there are \(1\) way (coefficient of \(x^0 = 1\)) to place \(0\) non-attacking rook and \(1\) way (coefficient of \(x^1\)) to place \(1\) non-attacking rook.

So, how do we deduce a generalisation for the Rook's polynomial for \(2\) -Dimensional board \(B_{m,n}\)? Letting \(i = \min \{m,n\}\), we get \[R(x,B_{m,n}) = \displaystyle \sum_{k=0}^i \binom{m}{k} P(n,k) \text{ } x^k\]

The above can be derived by first choosing \(k\) rows out of the board's \(m\) rows; then permutating the \(k\) out of the \(n\) columns (denoted by \(P(n,k)\)). Recall that \(k\) refers to the numbers of non attacking rooks placed on the board.

One way we can apply the Rook's Polynomial to combinatorics question is by labelling values/variables to the rows and columns of the board. A simple example question is as follows:

- David, Mark, Jerry and Kai joined a card game club. They are required to choose \(1\) out of the \(5\) card games available in the club ---- namely, Monopoly Deal, Poker, Blackjacks, Uno and Cheat. How many ways are there for each of David, Mark, Jerry and Kai to choose a different card game from each other?

Solution: Label the rows of board \(B_{4 \times 5}\) with the names (David, Mark, Jerry and Kai), and the columns with the card games (Monopoly Deal, Poker, Blackjacks, Uno and Cheat). Since each of the members must choose only \(1\) card game and each card game can only be chosen by at most \(1\) member, we can use the rook polynomial to find the ways to arranging \(4\) non-attacking rooks on board \(B_{4 \times 5}\). The answer to the question is hence, \[\binom{4}{4} P(5,4) = 120\]

Rook's Polynomial is a really interesting and (for me) quite a rare topic. Hope you have enjoyed and learnt something! Since there are much more about Rook's Polynomial, you can read it up here! :) Thanks!

- After you have read the above link: You can try this problem which can be solved by Rook's Polynomial.

Image from: Google Search (chess.com)

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestWow, I have never heard of this concept before. Thank you for sharing, very nice first post!

Log in to reply

Thanks for the encouragement :) have just read about it not long ago too

Log in to reply

Really brilliant, I've never seen such a good way of translating such ugly combinatorics questions into something with a fairly nice solution

Log in to reply

The link is also very informative....totally worth checking out

Log in to reply

Very nice first post!

Log in to reply

Thanks!

Log in to reply

Superb.

Log in to reply

Nice explanation :) Thanks for the sample problems too!

Log in to reply

Thanks! :) the link above has more details and sample questions for you to explore!

Log in to reply