Number of rotation invariant 6-colorings of a cube
$begingroup$
Consider a cube colored with six distinct colors on its six faces. How many non-equivalent colorings upto rotations are there? That is, how many ways can we color the cube so that we dont get the same colorings by rotating any colorings about any of its canaonical axes?
I think the permutation groups have a role to play here. Typivcally, is it related to automorphisms of $S_6$ under rotations, or do dihedral groups come into sight? Thanks beforehand.
combinatorics coloring
asked Dec 10 '18 at 7:35
vidyarthividyarthi
2,9291832
$endgroup$
|
show 7 more comments
$begingroup$
Consider a cube colored with six distinct colors on its six faces. How many non-equivalent colorings upto rotations are there? That is, how many ways can we color the cube so that we dont get the same colorings by rotating any colorings about any of its canaonical axes?
I think the permutation groups have a role to play here. Typivcally, is it related to automorphisms of $S_6$ under rotations, or do dihedral groups come into sight? Thanks beforehand.
combinatorics coloring
asked Dec 10 '18 at 7:35
vidyarthividyarthi
2,9291832
$endgroup$
$begingroup$
If the colors are distinct on all faces, how can we get the same coloring by rotation?
$endgroup$
– Aravind
Dec 10 '18 at 7:39
$begingroup$
@Aravind so you mean to say that the required number is equal to the number of ways to color the cube using six distinct colors? Is it $6!$?But, I have seen the answer is somewhere to be a two digit number
$endgroup$
– vidyarthi
Dec 10 '18 at 7:41
$begingroup$
@Aravind but, suppose if the top and bottom faces, or right and left faces are colored red and green. Then by rotating $180$ degrees on either top to bottom or right to left, dont we get the same colorings?
$endgroup$
– vidyarthi
Dec 10 '18 at 7:44
$begingroup$
I think you are using the words in a very confusing/confused way. Isn't a "rotationally invariant coloring" one which is unchanged by any rotation? (And since the rotations carry any face to any other face, the only possibility is that the faces are all colored the same.)
$endgroup$
– ancientmathematician
Dec 10 '18 at 7:54
$begingroup$
You might want to say something along the lines of "non-equivalent up to rotations".
$endgroup$
– Ivan Neretin
Dec 10 '18 at 7:56
|
show 7 more comments
$begingroup$
Consider a cube colored with six distinct colors on its six faces. How many non-equivalent colorings upto rotations are there? That is, how many ways can we color the cube so that we dont get the same colorings by rotating any colorings about any of its canaonical axes?
I think the permutation groups have a role to play here. Typivcally, is it related to automorphisms of $S_6$ under rotations, or do dihedral groups come into sight? Thanks beforehand.
combinatorics coloring
asked Dec 10 '18 at 7:35
vidyarthividyarthi
2,9291832
$endgroup$
Consider a cube colored with six distinct colors on its six faces. How many non-equivalent colorings upto rotations are there? That is, how many ways can we color the cube so that we dont get the same colorings by rotating any colorings about any of its canaonical axes?
I think the permutation groups have a role to play here. Typivcally, is it related to automorphisms of $S_6$ under rotations, or do dihedral groups come into sight? Thanks beforehand.
combinatorics coloring
combinatorics coloring
asked Dec 10 '18 at 7:35
vidyarthividyarthi
2,9291832
asked Dec 10 '18 at 7:35
vidyarthividyarthi
2,9291832
edited Dec 10 '18 at 8:45
vidyarthi
asked Dec 10 '18 at 7:35
vidyarthividyarthi
2,9291832
asked Dec 10 '18 at 7:35
vidyarthividyarthi
2,9291832
asked Dec 10 '18 at 7:35
vidyarthividyarthi
2,9291832
2,9291832
$begingroup$
If the colors are distinct on all faces, how can we get the same coloring by rotation?
$endgroup$
– Aravind
Dec 10 '18 at 7:39
$begingroup$
@Aravind so you mean to say that the required number is equal to the number of ways to color the cube using six distinct colors? Is it $6!$?But, I have seen the answer is somewhere to be a two digit number
$endgroup$
– vidyarthi
Dec 10 '18 at 7:41
$begingroup$
@Aravind but, suppose if the top and bottom faces, or right and left faces are colored red and green. Then by rotating $180$ degrees on either top to bottom or right to left, dont we get the same colorings?
$endgroup$
– vidyarthi
Dec 10 '18 at 7:44
$begingroup$
I think you are using the words in a very confusing/confused way. Isn't a "rotationally invariant coloring" one which is unchanged by any rotation? (And since the rotations carry any face to any other face, the only possibility is that the faces are all colored the same.)
$endgroup$
– ancientmathematician
Dec 10 '18 at 7:54
$begingroup$
You might want to say something along the lines of "non-equivalent up to rotations".
$endgroup$
– Ivan Neretin
Dec 10 '18 at 7:56
|
show 7 more comments
$begingroup$
If the colors are distinct on all faces, how can we get the same coloring by rotation?
$endgroup$
– Aravind
Dec 10 '18 at 7:39
$begingroup$
@Aravind so you mean to say that the required number is equal to the number of ways to color the cube using six distinct colors? Is it $6!$?But, I have seen the answer is somewhere to be a two digit number
$endgroup$
– vidyarthi
Dec 10 '18 at 7:41
$begingroup$
@Aravind but, suppose if the top and bottom faces, or right and left faces are colored red and green. Then by rotating $180$ degrees on either top to bottom or right to left, dont we get the same colorings?
$endgroup$
– vidyarthi
Dec 10 '18 at 7:44
$begingroup$
I think you are using the words in a very confusing/confused way. Isn't a "rotationally invariant coloring" one which is unchanged by any rotation? (And since the rotations carry any face to any other face, the only possibility is that the faces are all colored the same.)
$endgroup$
– ancientmathematician
Dec 10 '18 at 7:54
$begingroup$
You might want to say something along the lines of "non-equivalent up to rotations".
$endgroup$
– Ivan Neretin
Dec 10 '18 at 7:56
$begingroup$
If the colors are distinct on all faces, how can we get the same coloring by rotation?
$endgroup$
– Aravind
Dec 10 '18 at 7:39
$begingroup$
If the colors are distinct on all faces, how can we get the same coloring by rotation?
$endgroup$
– Aravind
Dec 10 '18 at 7:39
$begingroup$
@Aravind so you mean to say that the required number is equal to the number of ways to color the cube using six distinct colors? Is it $6!$?But, I have seen the answer is somewhere to be a two digit number
$endgroup$
– vidyarthi
Dec 10 '18 at 7:41
$begingroup$
@Aravind so you mean to say that the required number is equal to the number of ways to color the cube using six distinct colors? Is it $6!$?But, I have seen the answer is somewhere to be a two digit number
$endgroup$
– vidyarthi
Dec 10 '18 at 7:41
$begingroup$
@Aravind but, suppose if the top and bottom faces, or right and left faces are colored red and green. Then by rotating $180$ degrees on either top to bottom or right to left, dont we get the same colorings?
$endgroup$
– vidyarthi
Dec 10 '18 at 7:44
$begingroup$
@Aravind but, suppose if the top and bottom faces, or right and left faces are colored red and green. Then by rotating $180$ degrees on either top to bottom or right to left, dont we get the same colorings?
$endgroup$
– vidyarthi
Dec 10 '18 at 7:44
$begingroup$
I think you are using the words in a very confusing/confused way. Isn't a "rotationally invariant coloring" one which is unchanged by any rotation? (And since the rotations carry any face to any other face, the only possibility is that the faces are all colored the same.)
$endgroup$
– ancientmathematician
Dec 10 '18 at 7:54
$begingroup$
I think you are using the words in a very confusing/confused way. Isn't a "rotationally invariant coloring" one which is unchanged by any rotation? (And since the rotations carry any face to any other face, the only possibility is that the faces are all colored the same.)
$endgroup$
– ancientmathematician
Dec 10 '18 at 7:54
$begingroup$
You might want to say something along the lines of "non-equivalent up to rotations".
$endgroup$
– Ivan Neretin
Dec 10 '18 at 7:56
$begingroup$
You might want to say something along the lines of "non-equivalent up to rotations".
$endgroup$
– Ivan Neretin
Dec 10 '18 at 7:56
|
show 7 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Let the six colors be $C_1,dots, C_6$. Place the cube with the bottom colored $C_1$. There are five different choices for the top face. For each of these the other four colors can be placed on the cycle of 4 vertical faces in $4!/4=6$ ways. Hence in total, 30 colorings.
answered Dec 10 '18 at 8:54
ancientmathematicianancientmathematician
4,4581413
$endgroup$
$begingroup$
fantastic! so no burnside lemma. By the way, how would do it by burnside lemma? The wikipedia link gives way beyond three digit answer!
$endgroup$
– vidyarthi
Dec 10 '18 at 9:03
1
$begingroup$
The groups has order 24. There are $6!$ colorings in total; the identity fixes each of these; the other 23 rotations don't fix any of them. So the number up to rotation is $frac{1}{24}(6!+0+dots+0)=30$.
$endgroup$
– ancientmathematician
Dec 10 '18 at 9:10
add a comment |
$begingroup$
Paint one surface white. Choose one of $5$ remaining colors for the opposite face. The four remaining colors can be split in $3$ ways into pairs and then put in $2$ ways. This gives a total of $5cdot3cdot2=30$ possibilities.
answered Dec 10 '18 at 8:55
Christian BlatterChristian Blatter
173k7113326
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
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active
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$begingroup$
Let the six colors be $C_1,dots, C_6$. Place the cube with the bottom colored $C_1$. There are five different choices for the top face. For each of these the other four colors can be placed on the cycle of 4 vertical faces in $4!/4=6$ ways. Hence in total, 30 colorings.
answered Dec 10 '18 at 8:54
ancientmathematicianancientmathematician
4,4581413
$endgroup$
$begingroup$
fantastic! so no burnside lemma. By the way, how would do it by burnside lemma? The wikipedia link gives way beyond three digit answer!
$endgroup$
– vidyarthi
Dec 10 '18 at 9:03
1
$begingroup$
The groups has order 24. There are $6!$ colorings in total; the identity fixes each of these; the other 23 rotations don't fix any of them. So the number up to rotation is $frac{1}{24}(6!+0+dots+0)=30$.
$endgroup$
– ancientmathematician
Dec 10 '18 at 9:10
add a comment |
$begingroup$
Let the six colors be $C_1,dots, C_6$. Place the cube with the bottom colored $C_1$. There are five different choices for the top face. For each of these the other four colors can be placed on the cycle of 4 vertical faces in $4!/4=6$ ways. Hence in total, 30 colorings.
answered Dec 10 '18 at 8:54
ancientmathematicianancientmathematician
4,4581413
$endgroup$
$begingroup$
fantastic! so no burnside lemma. By the way, how would do it by burnside lemma? The wikipedia link gives way beyond three digit answer!
$endgroup$
– vidyarthi
Dec 10 '18 at 9:03
1
$begingroup$
The groups has order 24. There are $6!$ colorings in total; the identity fixes each of these; the other 23 rotations don't fix any of them. So the number up to rotation is $frac{1}{24}(6!+0+dots+0)=30$.
$endgroup$
– ancientmathematician
Dec 10 '18 at 9:10
add a comment |
$begingroup$
Let the six colors be $C_1,dots, C_6$. Place the cube with the bottom colored $C_1$. There are five different choices for the top face. For each of these the other four colors can be placed on the cycle of 4 vertical faces in $4!/4=6$ ways. Hence in total, 30 colorings.
answered Dec 10 '18 at 8:54
ancientmathematicianancientmathematician
4,4581413
$endgroup$
Let the six colors be $C_1,dots, C_6$. Place the cube with the bottom colored $C_1$. There are five different choices for the top face. For each of these the other four colors can be placed on the cycle of 4 vertical faces in $4!/4=6$ ways. Hence in total, 30 colorings.
answered Dec 10 '18 at 8:54
ancientmathematicianancientmathematician
4,4581413
answered Dec 10 '18 at 8:54
ancientmathematicianancientmathematician
4,4581413
answered Dec 10 '18 at 8:54
ancientmathematicianancientmathematician
4,4581413
answered Dec 10 '18 at 8:54
ancientmathematicianancientmathematician
4,4581413
4,4581413
$begingroup$
fantastic! so no burnside lemma. By the way, how would do it by burnside lemma? The wikipedia link gives way beyond three digit answer!
$endgroup$
– vidyarthi
Dec 10 '18 at 9:03
1
$begingroup$
The groups has order 24. There are $6!$ colorings in total; the identity fixes each of these; the other 23 rotations don't fix any of them. So the number up to rotation is $frac{1}{24}(6!+0+dots+0)=30$.
$endgroup$
– ancientmathematician
Dec 10 '18 at 9:10
add a comment |
$begingroup$
fantastic! so no burnside lemma. By the way, how would do it by burnside lemma? The wikipedia link gives way beyond three digit answer!
$endgroup$
– vidyarthi
Dec 10 '18 at 9:03
1
$begingroup$
The groups has order 24. There are $6!$ colorings in total; the identity fixes each of these; the other 23 rotations don't fix any of them. So the number up to rotation is $frac{1}{24}(6!+0+dots+0)=30$.
$endgroup$
– ancientmathematician
Dec 10 '18 at 9:10
$begingroup$
fantastic! so no burnside lemma. By the way, how would do it by burnside lemma? The wikipedia link gives way beyond three digit answer!
$endgroup$
– vidyarthi
Dec 10 '18 at 9:03
$begingroup$
fantastic! so no burnside lemma. By the way, how would do it by burnside lemma? The wikipedia link gives way beyond three digit answer!
$endgroup$
– vidyarthi
Dec 10 '18 at 9:03
1
1
$begingroup$
The groups has order 24. There are $6!$ colorings in total; the identity fixes each of these; the other 23 rotations don't fix any of them. So the number up to rotation is $frac{1}{24}(6!+0+dots+0)=30$.
$endgroup$
– ancientmathematician
Dec 10 '18 at 9:10
$begingroup$
The groups has order 24. There are $6!$ colorings in total; the identity fixes each of these; the other 23 rotations don't fix any of them. So the number up to rotation is $frac{1}{24}(6!+0+dots+0)=30$.
$endgroup$
– ancientmathematician
Dec 10 '18 at 9:10
add a comment |
$begingroup$
Paint one surface white. Choose one of $5$ remaining colors for the opposite face. The four remaining colors can be split in $3$ ways into pairs and then put in $2$ ways. This gives a total of $5cdot3cdot2=30$ possibilities.
answered Dec 10 '18 at 8:55
Christian BlatterChristian Blatter
173k7113326
$endgroup$
add a comment |
$begingroup$
Paint one surface white. Choose one of $5$ remaining colors for the opposite face. The four remaining colors can be split in $3$ ways into pairs and then put in $2$ ways. This gives a total of $5cdot3cdot2=30$ possibilities.
answered Dec 10 '18 at 8:55
Christian BlatterChristian Blatter
173k7113326
$endgroup$
add a comment |
$begingroup$
Paint one surface white. Choose one of $5$ remaining colors for the opposite face. The four remaining colors can be split in $3$ ways into pairs and then put in $2$ ways. This gives a total of $5cdot3cdot2=30$ possibilities.
answered Dec 10 '18 at 8:55
Christian BlatterChristian Blatter
173k7113326
$endgroup$
Paint one surface white. Choose one of $5$ remaining colors for the opposite face. The four remaining colors can be split in $3$ ways into pairs and then put in $2$ ways. This gives a total of $5cdot3cdot2=30$ possibilities.
answered Dec 10 '18 at 8:55
Christian BlatterChristian Blatter
173k7113326
answered Dec 10 '18 at 8:55
Christian BlatterChristian Blatter
173k7113326
answered Dec 10 '18 at 8:55
Christian BlatterChristian Blatter
173k7113326
answered Dec 10 '18 at 8:55
Christian BlatterChristian Blatter
173k7113326
173k7113326
add a comment |
add a comment |
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$begingroup$
If the colors are distinct on all faces, how can we get the same coloring by rotation?
$endgroup$
– Aravind
Dec 10 '18 at 7:39
$begingroup$
@Aravind so you mean to say that the required number is equal to the number of ways to color the cube using six distinct colors? Is it $6!$?But, I have seen the answer is somewhere to be a two digit number
$endgroup$
– vidyarthi
Dec 10 '18 at 7:41
$begingroup$
@Aravind but, suppose if the top and bottom faces, or right and left faces are colored red and green. Then by rotating $180$ degrees on either top to bottom or right to left, dont we get the same colorings?
$endgroup$
– vidyarthi
Dec 10 '18 at 7:44
$begingroup$
I think you are using the words in a very confusing/confused way. Isn't a "rotationally invariant coloring" one which is unchanged by any rotation? (And since the rotations carry any face to any other face, the only possibility is that the faces are all colored the same.)
$endgroup$
– ancientmathematician
Dec 10 '18 at 7:54
$begingroup$
You might want to say something along the lines of "non-equivalent up to rotations".
$endgroup$
– Ivan Neretin
Dec 10 '18 at 7:56