Mixing
Coloring a square
$begingroup$
Let $V$ be the set of all four vertices of a square , and ${1,2,3}$ the set of all possible colours. We call a coloring of a square , a function $ f: Vrightarrow$ ${1,2,3}$ a coloring of a square, and $F_H={$$ f: Vrightarrow$ ${1,2,3}$$}$
Also , we say that $f thicksim g$ if there exists a permutation $psi : V rightarrow V $ , such that $ f circ psi = g$ , which preserves the sides of the square , namely , if ab is a side , $psi (a) psi(b)$ is also a side.
I need to come up with the number of elements in $F_H/thicksim$ and the number of elements in each of the classes .
I have found all elements in $F_H/thicksim$ , according to the number of colours and the order we use them , but I cannot find the number of elements in each of those sets.
combinatorics geometry elementary-set-theory
edited Jan 26 at 18:07
Andrés E. Caicedo
65.8k8160251
asked Jan 26 at 17:26
Adolfo GarciaAdolfo Garcia
404
$endgroup$
add a comment |
$begingroup$
Let $V$ be the set of all four vertices of a square , and ${1,2,3}$ the set of all possible colours. We call a coloring of a square , a function $ f: Vrightarrow$ ${1,2,3}$ a coloring of a square, and $F_H={$$ f: Vrightarrow$ ${1,2,3}$$}$
Also , we say that $f thicksim g$ if there exists a permutation $psi : V rightarrow V $ , such that $ f circ psi = g$ , which preserves the sides of the square , namely , if ab is a side , $psi (a) psi(b)$ is also a side.
I need to come up with the number of elements in $F_H/thicksim$ and the number of elements in each of the classes .
I have found all elements in $F_H/thicksim$ , according to the number of colours and the order we use them , but I cannot find the number of elements in each of those sets.
combinatorics geometry elementary-set-theory
edited Jan 26 at 18:07
Andrés E. Caicedo
65.8k8160251
asked Jan 26 at 17:26
Adolfo GarciaAdolfo Garcia
404
$endgroup$
add a comment |
$begingroup$
Let $V$ be the set of all four vertices of a square , and ${1,2,3}$ the set of all possible colours. We call a coloring of a square , a function $ f: Vrightarrow$ ${1,2,3}$ a coloring of a square, and $F_H={$$ f: Vrightarrow$ ${1,2,3}$$}$
Also , we say that $f thicksim g$ if there exists a permutation $psi : V rightarrow V $ , such that $ f circ psi = g$ , which preserves the sides of the square , namely , if ab is a side , $psi (a) psi(b)$ is also a side.
I need to come up with the number of elements in $F_H/thicksim$ and the number of elements in each of the classes .
I have found all elements in $F_H/thicksim$ , according to the number of colours and the order we use them , but I cannot find the number of elements in each of those sets.
combinatorics geometry elementary-set-theory
edited Jan 26 at 18:07
Andrés E. Caicedo
65.8k8160251
asked Jan 26 at 17:26
Adolfo GarciaAdolfo Garcia
404
$endgroup$
Let $V$ be the set of all four vertices of a square , and ${1,2,3}$ the set of all possible colours. We call a coloring of a square , a function $ f: Vrightarrow$ ${1,2,3}$ a coloring of a square, and $F_H={$$ f: Vrightarrow$ ${1,2,3}$$}$
Also , we say that $f thicksim g$ if there exists a permutation $psi : V rightarrow V $ , such that $ f circ psi = g$ , which preserves the sides of the square , namely , if ab is a side , $psi (a) psi(b)$ is also a side.
I need to come up with the number of elements in $F_H/thicksim$ and the number of elements in each of the classes .
I have found all elements in $F_H/thicksim$ , according to the number of colours and the order we use them , but I cannot find the number of elements in each of those sets.
combinatorics geometry elementary-set-theory
combinatorics geometry elementary-set-theory
edited Jan 26 at 18:07
Andrés E. Caicedo
65.8k8160251
asked Jan 26 at 17:26
Adolfo GarciaAdolfo Garcia
404
edited Jan 26 at 18:07
Andrés E. Caicedo
65.8k8160251
asked Jan 26 at 17:26
Adolfo GarciaAdolfo Garcia
404
edited Jan 26 at 18:07
Andrés E. Caicedo
65.8k8160251
edited Jan 26 at 18:07
Andrés E. Caicedo
65.8k8160251
edited Jan 26 at 18:07
Andrés E. Caicedo
65.8k8160251
65.8k8160251
asked Jan 26 at 17:26
Adolfo GarciaAdolfo Garcia
404
asked Jan 26 at 17:26
Adolfo GarciaAdolfo Garcia
404
asked Jan 26 at 17:26
Adolfo GarciaAdolfo Garcia
404
404
add a comment |
add a comment |
1 Answer
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$begingroup$
Notice that, if we had a colouring with four distinct colours, then there would be $|D_4| = 8$ equivalent colourings. In our actual cases, some of these will give identical colourings, so the number will reduce, but not increase: the cardinality of the equivalence class of a colouring $f$ is precisely the cardinality of the set ${sigma(f) | sigma in D_4 }$ of images of your colouring under $D_4$, so we can just check each of those eight and count which are distinct.
Going through the options:
If our colouring is constant, then clearly its equivalence class has only one element.
If our colouring has three vertices of one colour, and the fourth difference, then its equivalence class has four elements (indexed by where the different colour is).
If our colouring has two vertices each of two different colours, and the same colours are adjacent, then there are again four equivalent colourings (indexed by which side has both vertices one of the colours).
If our colouring has two vertices of each of two different colours, and the same colours are not adjacent, then there are only two equivalent colourings (indexed by which diagonal has both vertices one of the colours).
If our colouring has two vertices of one colour, and one of each other colour, and the two same-colour vertices are adjacent, then all images of our colouring under $D_4$ are distinct, so there are eight equivalent colourings.
If our colouring has two vertices of one colour, and one of each other colour, and the two same-colour vertices are not adjacent, then there are four distinct images (fix which diagonal is same-coloured (2), then swap the positions of the other two colours).
answered Jan 26 at 17:48
user3482749user3482749
4,3111019
$endgroup$
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1 Answer
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$begingroup$
Notice that, if we had a colouring with four distinct colours, then there would be $|D_4| = 8$ equivalent colourings. In our actual cases, some of these will give identical colourings, so the number will reduce, but not increase: the cardinality of the equivalence class of a colouring $f$ is precisely the cardinality of the set ${sigma(f) | sigma in D_4 }$ of images of your colouring under $D_4$, so we can just check each of those eight and count which are distinct.
Going through the options:
If our colouring is constant, then clearly its equivalence class has only one element.
If our colouring has three vertices of one colour, and the fourth difference, then its equivalence class has four elements (indexed by where the different colour is).
If our colouring has two vertices each of two different colours, and the same colours are adjacent, then there are again four equivalent colourings (indexed by which side has both vertices one of the colours).
If our colouring has two vertices of each of two different colours, and the same colours are not adjacent, then there are only two equivalent colourings (indexed by which diagonal has both vertices one of the colours).
If our colouring has two vertices of one colour, and one of each other colour, and the two same-colour vertices are adjacent, then all images of our colouring under $D_4$ are distinct, so there are eight equivalent colourings.
If our colouring has two vertices of one colour, and one of each other colour, and the two same-colour vertices are not adjacent, then there are four distinct images (fix which diagonal is same-coloured (2), then swap the positions of the other two colours).
answered Jan 26 at 17:48
user3482749user3482749
4,3111019
$endgroup$
add a comment |
$begingroup$
Notice that, if we had a colouring with four distinct colours, then there would be $|D_4| = 8$ equivalent colourings. In our actual cases, some of these will give identical colourings, so the number will reduce, but not increase: the cardinality of the equivalence class of a colouring $f$ is precisely the cardinality of the set ${sigma(f) | sigma in D_4 }$ of images of your colouring under $D_4$, so we can just check each of those eight and count which are distinct.
Going through the options:
If our colouring is constant, then clearly its equivalence class has only one element.
If our colouring has three vertices of one colour, and the fourth difference, then its equivalence class has four elements (indexed by where the different colour is).
If our colouring has two vertices each of two different colours, and the same colours are adjacent, then there are again four equivalent colourings (indexed by which side has both vertices one of the colours).
If our colouring has two vertices of each of two different colours, and the same colours are not adjacent, then there are only two equivalent colourings (indexed by which diagonal has both vertices one of the colours).
If our colouring has two vertices of one colour, and one of each other colour, and the two same-colour vertices are adjacent, then all images of our colouring under $D_4$ are distinct, so there are eight equivalent colourings.
If our colouring has two vertices of one colour, and one of each other colour, and the two same-colour vertices are not adjacent, then there are four distinct images (fix which diagonal is same-coloured (2), then swap the positions of the other two colours).
answered Jan 26 at 17:48
user3482749user3482749
4,3111019
$endgroup$
add a comment |
$begingroup$
Notice that, if we had a colouring with four distinct colours, then there would be $|D_4| = 8$ equivalent colourings. In our actual cases, some of these will give identical colourings, so the number will reduce, but not increase: the cardinality of the equivalence class of a colouring $f$ is precisely the cardinality of the set ${sigma(f) | sigma in D_4 }$ of images of your colouring under $D_4$, so we can just check each of those eight and count which are distinct.
Going through the options:
If our colouring is constant, then clearly its equivalence class has only one element.
If our colouring has three vertices of one colour, and the fourth difference, then its equivalence class has four elements (indexed by where the different colour is).
If our colouring has two vertices each of two different colours, and the same colours are adjacent, then there are again four equivalent colourings (indexed by which side has both vertices one of the colours).
If our colouring has two vertices of each of two different colours, and the same colours are not adjacent, then there are only two equivalent colourings (indexed by which diagonal has both vertices one of the colours).
If our colouring has two vertices of one colour, and one of each other colour, and the two same-colour vertices are adjacent, then all images of our colouring under $D_4$ are distinct, so there are eight equivalent colourings.
If our colouring has two vertices of one colour, and one of each other colour, and the two same-colour vertices are not adjacent, then there are four distinct images (fix which diagonal is same-coloured (2), then swap the positions of the other two colours).
answered Jan 26 at 17:48
user3482749user3482749
4,3111019
$endgroup$
Notice that, if we had a colouring with four distinct colours, then there would be $|D_4| = 8$ equivalent colourings. In our actual cases, some of these will give identical colourings, so the number will reduce, but not increase: the cardinality of the equivalence class of a colouring $f$ is precisely the cardinality of the set ${sigma(f) | sigma in D_4 }$ of images of your colouring under $D_4$, so we can just check each of those eight and count which are distinct.
Going through the options:
If our colouring is constant, then clearly its equivalence class has only one element.
If our colouring has three vertices of one colour, and the fourth difference, then its equivalence class has four elements (indexed by where the different colour is).
If our colouring has two vertices each of two different colours, and the same colours are adjacent, then there are again four equivalent colourings (indexed by which side has both vertices one of the colours).
If our colouring has two vertices of each of two different colours, and the same colours are not adjacent, then there are only two equivalent colourings (indexed by which diagonal has both vertices one of the colours).
If our colouring has two vertices of one colour, and one of each other colour, and the two same-colour vertices are adjacent, then all images of our colouring under $D_4$ are distinct, so there are eight equivalent colourings.
If our colouring has two vertices of one colour, and one of each other colour, and the two same-colour vertices are not adjacent, then there are four distinct images (fix which diagonal is same-coloured (2), then swap the positions of the other two colours).
answered Jan 26 at 17:48
user3482749user3482749
4,3111019
answered Jan 26 at 17:48
user3482749user3482749
4,3111019
answered Jan 26 at 17:48
user3482749user3482749
4,3111019
answered Jan 26 at 17:48
user3482749user3482749
4,3111019
4,3111019
add a comment |
add a comment |
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