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I am having some trouble with the following problem from Brualdi's Introductory Combinatorics (Chapter 14 Exercise 47).

The nine squares of a $3\times3$ chessboard are to be colored red and blue. The chessboard is free to rotate but cannot be flipped over. Determine the generating function for the number of non-equivalent colorings and the total number of non-equivalent colorings.

I think the problem arises from a generally poor understanding of some the material in the chapter this problem comes from , but also because the solution to the second part of the question given in the back of the book does not match the answer given in this previous question.
My book's solution is

The cycle index of the group of permutations is

$P_G(z_1,z_2,\dots,z_{10}) = \frac{z_1^{10}+2z_1z_4^2+z_1z_2^4}{4}$

Hence the number of nonequivalent colorings is $P_G(2,2,\dots,2) = \frac{2^{10}+2^4+2^5}{4}=2^8+2^3+2^2$

while the solution provided in the question linked above is given as

$\frac{2^9+4\cdot 2^3 + 2 \cdot 2^5 + 4 \cdot 2^3}{4} = 160$

Unfortunately, these two solutions do not coincide. Which one is correct??? I am confused about why my books' solution has 10 variables in its generating function on the left hand side, or why there is a 10 in the exponent of $z_1$ on the right hand side.

I am also unsure of how to apply the counting theorem when the symmetry group is smaller in order than the set we are trying to color. For instance it seems like our symmetry group should be $C_4$ but how do we apply this to the 9 squares of our chessboard?

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FofX
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  • I don't understand; what is the generating function of a number? – Servaes Dec 15 '18 at 03:17
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    There are only nine spaces on a $3\times 3$ board. I do not see where they could have gotten $10$ from in your book. Are you sure you are looking at the correct answer in the book? – JMoravitz Dec 15 '18 at 03:17
  • Just heuristically, I'd expect the number to be closer to $2^9/4 = 128$ since 1 unique board can create 4 different permutations of the $2^9$ variant, whereas the book version is slightly over $2^9/2$ which gives me the sneaking suspicion they accounted for only reflections – wjmccann Dec 15 '18 at 03:22
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    The end result is that you should apply burnside's lemma (or some equivalent formulation). There are $2^9$ ways to color the board which remain unchanged by the identity rotation. There are $2^5$ (top row, one side, and mid) ways to color which remain unchanged by rotation $180^\circ$ clockwise. Finally there are $2^3$ (corner, side, mid) ways to color which remain unchanged by rotation $90^\circ$ clockwise and similarly so for counter-clockwise. By my count I get a total of $\frac{2^9+2^5+2\cdot 2^3}{4}$ – JMoravitz Dec 15 '18 at 03:24
  • @JMoravitz Yes that is the solution to problem 47! Ok that seems very reasonable, although your answer differs from the other answer given in the question I linked to. Let me just try to digest that...EDIT: Also yea we are essentially using burnside's lemma. I don't know if you are familiar with cycle indices but that's what we use. – FofX Dec 15 '18 at 03:28
  • @Servaes The generating function refers to what my book refers to as the cycle index of a group. Then you plug in 2 (in this case) to get the answer for the second half of the question. – FofX Dec 15 '18 at 03:33
  • @JMoravitz After thinking about it I believe your answer is indeed correct. I think the generating function should have read $P_G(z_1,\dots,z_9)=\frac{z_1^{9}+2z_1z_4^2+z_1z_2^4}{4}$. If you plug 2 into all the $z$'s you get your answer, which is I think what the book intended as well, they just got a little carried away... – FofX Dec 15 '18 at 03:49

1 Answers1

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You have 4 element of groups as below:

  • identity - then we have $x_1^9$ that means we have $9$ $1-length$ cycles
  • rotation by $90^o$ - $x_1 x_4^2$
  • rotation by $180^o$ - $x_1 x_2^4$
  • rotation by $270^o$ - you can note that this is just rotation by $90^o$ but in other direction so there is again $x_1 x_4^2$
    From Polya couting theory we have $$F(x_1,...,x_9) = \frac{x_1^9 + 2x_1 x_4^2 + x_1 x_2^4}{4} $$

for generating functions for all coloring in use $2$ colors it will be: $$F(a+b,...,a^9+b^9) $$ for example number of coloring in use of $3$ red and $5$ blue will be a coefficient with $a^3 b^5$: $$[a^3 b^5]F(a+b,...,a^9+b^9) $$ For all possibilities we should put $$a = b = 1 $$ and you got $$ F(2,...,2) = \frac{2^9 + 2^4 + 2^5}{4} = 140 $$

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