coloring 4x4 square by 2 colors
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we have 4x4 square. We can paint 16 small squares white or black color. neighbouring squares can not be black colored (sharig an edge). how many patterns we can get? (if we reach a coloring from rotating another one, then those two are same. if we reach a coloring from reflecting another one, then those two are different)
combinatorics
asked Nov 20 at 9:19
Rasim Abi
11
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we have 4x4 square. We can paint 16 small squares white or black color. neighbouring squares can not be black colored (sharig an edge). how many patterns we can get? (if we reach a coloring from rotating another one, then those two are same. if we reach a coloring from reflecting another one, then those two are different)
combinatorics
asked Nov 20 at 9:19
Rasim Abi
11
1
Come on, downvoter. At least let the OP know what you think is wrong with the question (perhaps you didn't see he's a "New contributer"?). Anyway, since that person won't, let me say this: this website is meant to help you solve problems you get stuck on. As your question is posed now, we have no way to tell whether you even tried the problem. Share your work with us so we can see where exactly you got stuck. Besides that, showing your work simply shows you put effort into it and we like that here :)
– vrugtehagel
Nov 20 at 9:23
yes i prepared the question but till now could not solve it. i checked similar questions but this is different. first time im using this side.
– Rasim Abi
Nov 20 at 9:28
Then please edit your question to include your work, show us what you've tried.
– vrugtehagel
Nov 20 at 9:29
$4times 4$ is a very small grid. If one ignore the issue of overcounting due to rotational symmetry, you can count the number of configuration without nearby black squares with a few line of code. You will find there are $1234$ possible configurations. To handle overcounting due to symmetry, first figure out the number of configurations with $2$-fold symmetry $(34)$ and $4$-fold symmetry $(4)$ and then apply Burnside's lemma.
– achille hui
Nov 20 at 23:43
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
we have 4x4 square. We can paint 16 small squares white or black color. neighbouring squares can not be black colored (sharig an edge). how many patterns we can get? (if we reach a coloring from rotating another one, then those two are same. if we reach a coloring from reflecting another one, then those two are different)
combinatorics
asked Nov 20 at 9:19
Rasim Abi
11
we have 4x4 square. We can paint 16 small squares white or black color. neighbouring squares can not be black colored (sharig an edge). how many patterns we can get? (if we reach a coloring from rotating another one, then those two are same. if we reach a coloring from reflecting another one, then those two are different)
combinatorics
combinatorics
asked Nov 20 at 9:19
Rasim Abi
11
asked Nov 20 at 9:19
Rasim Abi
11
asked Nov 20 at 9:19
Rasim Abi
11
asked Nov 20 at 9:19
Rasim Abi
11
asked Nov 20 at 9:19
Rasim Abi
11
11
1
Come on, downvoter. At least let the OP know what you think is wrong with the question (perhaps you didn't see he's a "New contributer"?). Anyway, since that person won't, let me say this: this website is meant to help you solve problems you get stuck on. As your question is posed now, we have no way to tell whether you even tried the problem. Share your work with us so we can see where exactly you got stuck. Besides that, showing your work simply shows you put effort into it and we like that here :)
– vrugtehagel
Nov 20 at 9:23
yes i prepared the question but till now could not solve it. i checked similar questions but this is different. first time im using this side.
– Rasim Abi
Nov 20 at 9:28
Then please edit your question to include your work, show us what you've tried.
– vrugtehagel
Nov 20 at 9:29
$4times 4$ is a very small grid. If one ignore the issue of overcounting due to rotational symmetry, you can count the number of configuration without nearby black squares with a few line of code. You will find there are $1234$ possible configurations. To handle overcounting due to symmetry, first figure out the number of configurations with $2$-fold symmetry $(34)$ and $4$-fold symmetry $(4)$ and then apply Burnside's lemma.
– achille hui
Nov 20 at 23:43
add a comment |
1
Come on, downvoter. At least let the OP know what you think is wrong with the question (perhaps you didn't see he's a "New contributer"?). Anyway, since that person won't, let me say this: this website is meant to help you solve problems you get stuck on. As your question is posed now, we have no way to tell whether you even tried the problem. Share your work with us so we can see where exactly you got stuck. Besides that, showing your work simply shows you put effort into it and we like that here :)
– vrugtehagel
Nov 20 at 9:23
yes i prepared the question but till now could not solve it. i checked similar questions but this is different. first time im using this side.
– Rasim Abi
Nov 20 at 9:28
Then please edit your question to include your work, show us what you've tried.
– vrugtehagel
Nov 20 at 9:29
$4times 4$ is a very small grid. If one ignore the issue of overcounting due to rotational symmetry, you can count the number of configuration without nearby black squares with a few line of code. You will find there are $1234$ possible configurations. To handle overcounting due to symmetry, first figure out the number of configurations with $2$-fold symmetry $(34)$ and $4$-fold symmetry $(4)$ and then apply Burnside's lemma.
– achille hui
Nov 20 at 23:43
1
1
Come on, downvoter. At least let the OP know what you think is wrong with the question (perhaps you didn't see he's a "New contributer"?). Anyway, since that person won't, let me say this: this website is meant to help you solve problems you get stuck on. As your question is posed now, we have no way to tell whether you even tried the problem. Share your work with us so we can see where exactly you got stuck. Besides that, showing your work simply shows you put effort into it and we like that here :)
– vrugtehagel
Nov 20 at 9:23
Come on, downvoter. At least let the OP know what you think is wrong with the question (perhaps you didn't see he's a "New contributer"?). Anyway, since that person won't, let me say this: this website is meant to help you solve problems you get stuck on. As your question is posed now, we have no way to tell whether you even tried the problem. Share your work with us so we can see where exactly you got stuck. Besides that, showing your work simply shows you put effort into it and we like that here :)
– vrugtehagel
Nov 20 at 9:23
yes i prepared the question but till now could not solve it. i checked similar questions but this is different. first time im using this side.
– Rasim Abi
Nov 20 at 9:28
yes i prepared the question but till now could not solve it. i checked similar questions but this is different. first time im using this side.
– Rasim Abi
Nov 20 at 9:28
Then please edit your question to include your work, show us what you've tried.
– vrugtehagel
Nov 20 at 9:29
Then please edit your question to include your work, show us what you've tried.
– vrugtehagel
Nov 20 at 9:29
$4times 4$ is a very small grid. If one ignore the issue of overcounting due to rotational symmetry, you can count the number of configuration without nearby black squares with a few line of code. You will find there are $1234$ possible configurations. To handle overcounting due to symmetry, first figure out the number of configurations with $2$-fold symmetry $(34)$ and $4$-fold symmetry $(4)$ and then apply Burnside's lemma.
– achille hui
Nov 20 at 23:43
$4times 4$ is a very small grid. If one ignore the issue of overcounting due to rotational symmetry, you can count the number of configuration without nearby black squares with a few line of code. You will find there are $1234$ possible configurations. To handle overcounting due to symmetry, first figure out the number of configurations with $2$-fold symmetry $(34)$ and $4$-fold symmetry $(4)$ and then apply Burnside's lemma.
– achille hui
Nov 20 at 23:43
add a comment |
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1
Come on, downvoter. At least let the OP know what you think is wrong with the question (perhaps you didn't see he's a "New contributer"?). Anyway, since that person won't, let me say this: this website is meant to help you solve problems you get stuck on. As your question is posed now, we have no way to tell whether you even tried the problem. Share your work with us so we can see where exactly you got stuck. Besides that, showing your work simply shows you put effort into it and we like that here :)
– vrugtehagel
Nov 20 at 9:23
yes i prepared the question but till now could not solve it. i checked similar questions but this is different. first time im using this side.
– Rasim Abi
Nov 20 at 9:28
Then please edit your question to include your work, show us what you've tried.
– vrugtehagel
Nov 20 at 9:29
$4times 4$ is a very small grid. If one ignore the issue of overcounting due to rotational symmetry, you can count the number of configuration without nearby black squares with a few line of code. You will find there are $1234$ possible configurations. To handle overcounting due to symmetry, first figure out the number of configurations with $2$-fold symmetry $(34)$ and $4$-fold symmetry $(4)$ and then apply Burnside's lemma.
– achille hui
Nov 20 at 23:43