Colorings of a $3times3$ chessboard
$begingroup$
I am having some trouble with the following problem from Brualdi's Introductory Combinatorics (Chapter 14 Exercise 47).
The nine squares of a $3times3$ 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
$frac{2^9+4cdot 2^3 + 2 cdot 2^5 + 4 cdot 2^3}{4} = 160$
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?
combinatorics generating-functions chessboard polya-counting-theory
edited Dec 15 '18 at 11:59
Alex Ravsky
39.6k32181
asked Dec 15 '18 at 3:13
FofXFofX
637416
$endgroup$
|
show 2 more comments
$begingroup$
I am having some trouble with the following problem from Brualdi's Introductory Combinatorics (Chapter 14 Exercise 47).
The nine squares of a $3times3$ 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
$frac{2^9+4cdot 2^3 + 2 cdot 2^5 + 4 cdot 2^3}{4} = 160$
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?
combinatorics generating-functions chessboard polya-counting-theory
edited Dec 15 '18 at 11:59
Alex Ravsky
39.6k32181
asked Dec 15 '18 at 3:13
FofXFofX
637416
$endgroup$
$begingroup$
I don't understand; what is the generating function of a number?
$endgroup$
– Servaes
Dec 15 '18 at 3:17
2
$begingroup$
There are only nine spaces on a $3times 3$ 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?
$endgroup$
– JMoravitz
Dec 15 '18 at 3:17
$begingroup$
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
$endgroup$
– wjmccann
Dec 15 '18 at 3:22
1
$begingroup$
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 $frac{2^9+2^5+2cdot 2^3}{4}$
$endgroup$
– JMoravitz
Dec 15 '18 at 3:24
$begingroup$
@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.
$endgroup$
– FofX
Dec 15 '18 at 3:28
|
show 2 more comments
$begingroup$
I am having some trouble with the following problem from Brualdi's Introductory Combinatorics (Chapter 14 Exercise 47).
The nine squares of a $3times3$ 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
$frac{2^9+4cdot 2^3 + 2 cdot 2^5 + 4 cdot 2^3}{4} = 160$
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?
combinatorics generating-functions chessboard polya-counting-theory
edited Dec 15 '18 at 11:59
Alex Ravsky
39.6k32181
asked Dec 15 '18 at 3:13
FofXFofX
637416
$endgroup$
I am having some trouble with the following problem from Brualdi's Introductory Combinatorics (Chapter 14 Exercise 47).
The nine squares of a $3times3$ 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
$frac{2^9+4cdot 2^3 + 2 cdot 2^5 + 4 cdot 2^3}{4} = 160$
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?
combinatorics generating-functions chessboard polya-counting-theory
combinatorics generating-functions chessboard polya-counting-theory
edited Dec 15 '18 at 11:59
Alex Ravsky
39.6k32181
asked Dec 15 '18 at 3:13
FofXFofX
637416
edited Dec 15 '18 at 11:59
Alex Ravsky
39.6k32181
asked Dec 15 '18 at 3:13
FofXFofX
637416
edited Dec 15 '18 at 11:59
Alex Ravsky
39.6k32181
edited Dec 15 '18 at 11:59
Alex Ravsky
39.6k32181
edited Dec 15 '18 at 11:59
Alex Ravsky
39.6k32181
39.6k32181
asked Dec 15 '18 at 3:13
FofXFofX
637416
asked Dec 15 '18 at 3:13
FofXFofX
637416
asked Dec 15 '18 at 3:13
FofXFofX
637416
637416
$begingroup$
I don't understand; what is the generating function of a number?
$endgroup$
– Servaes
Dec 15 '18 at 3:17
2
$begingroup$
There are only nine spaces on a $3times 3$ 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?
$endgroup$
– JMoravitz
Dec 15 '18 at 3:17
$begingroup$
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
$endgroup$
– wjmccann
Dec 15 '18 at 3:22
1
$begingroup$
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 $frac{2^9+2^5+2cdot 2^3}{4}$
$endgroup$
– JMoravitz
Dec 15 '18 at 3:24
$begingroup$
@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.
$endgroup$
– FofX
Dec 15 '18 at 3:28
|
show 2 more comments
$begingroup$
I don't understand; what is the generating function of a number?
$endgroup$
– Servaes
Dec 15 '18 at 3:17
2
$begingroup$
There are only nine spaces on a $3times 3$ 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?
$endgroup$
– JMoravitz
Dec 15 '18 at 3:17
$begingroup$
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
$endgroup$
– wjmccann
Dec 15 '18 at 3:22
1
$begingroup$
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 $frac{2^9+2^5+2cdot 2^3}{4}$
$endgroup$
– JMoravitz
Dec 15 '18 at 3:24
$begingroup$
@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.
$endgroup$
– FofX
Dec 15 '18 at 3:28
$begingroup$
I don't understand; what is the generating function of a number?
$endgroup$
– Servaes
Dec 15 '18 at 3:17
$begingroup$
I don't understand; what is the generating function of a number?
$endgroup$
– Servaes
Dec 15 '18 at 3:17
2
2
$begingroup$
There are only nine spaces on a $3times 3$ 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?
$endgroup$
– JMoravitz
Dec 15 '18 at 3:17
$begingroup$
There are only nine spaces on a $3times 3$ 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?
$endgroup$
– JMoravitz
Dec 15 '18 at 3:17
$begingroup$
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
$endgroup$
– wjmccann
Dec 15 '18 at 3:22
$begingroup$
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
$endgroup$
– wjmccann
Dec 15 '18 at 3:22
1
1
$begingroup$
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 $frac{2^9+2^5+2cdot 2^3}{4}$
$endgroup$
– JMoravitz
Dec 15 '18 at 3:24
$begingroup$
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 $frac{2^9+2^5+2cdot 2^3}{4}$
$endgroup$
– JMoravitz
Dec 15 '18 at 3:24
$begingroup$
@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.
$endgroup$
– FofX
Dec 15 '18 at 3:28
$begingroup$
@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.
$endgroup$
– FofX
Dec 15 '18 at 3:28
|
show 2 more comments
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$begingroup$
I don't understand; what is the generating function of a number?
$endgroup$
– Servaes
Dec 15 '18 at 3:17
2
$begingroup$
There are only nine spaces on a $3times 3$ 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?
$endgroup$
– JMoravitz
Dec 15 '18 at 3:17
$begingroup$
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
$endgroup$
– wjmccann
Dec 15 '18 at 3:22
1
$begingroup$
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 $frac{2^9+2^5+2cdot 2^3}{4}$
$endgroup$
– JMoravitz
Dec 15 '18 at 3:24
$begingroup$
@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.
$endgroup$
– FofX
Dec 15 '18 at 3:28