Factorize a permutation with elements from two permutation groups
$begingroup$
Given two permutation groups $G,H$ on the same set $M$, and a permutation $sigma$ on $M$, how to find a factorization $sigma = gh$ (but not $sigma = hg$), where $g in G$ and $h in H$?
This problem arises from modeling variations of Rubik's Cube with permutation groups. One problem with this approach is that these puzzles often contain indistinguishable stickers. For example, in the $4times4times4$ puzzle the four center stickers on each face are indistinguishable, giving a total of $24^6$ potential "solved" configurations. But not all of these are actually possible. Calculations with Mathematica show that for each face, only $12$ permutations of its center stickers are achievable, without disrupting the configuration of other stickers. Therefore in a scrambled state we cannot randomly assign to each center sticker a target position, otherwise the permutation may not be solvable.
group-theory finite-groups permutations rubiks-cube
edited Dec 24 '18 at 13:43
Shaun
9,338113684
asked Oct 25 '17 at 20:30
user2249675user2249675
38718
$endgroup$
add a comment |
$begingroup$
Given two permutation groups $G,H$ on the same set $M$, and a permutation $sigma$ on $M$, how to find a factorization $sigma = gh$ (but not $sigma = hg$), where $g in G$ and $h in H$?
This problem arises from modeling variations of Rubik's Cube with permutation groups. One problem with this approach is that these puzzles often contain indistinguishable stickers. For example, in the $4times4times4$ puzzle the four center stickers on each face are indistinguishable, giving a total of $24^6$ potential "solved" configurations. But not all of these are actually possible. Calculations with Mathematica show that for each face, only $12$ permutations of its center stickers are achievable, without disrupting the configuration of other stickers. Therefore in a scrambled state we cannot randomly assign to each center sticker a target position, otherwise the permutation may not be solvable.
group-theory finite-groups permutations rubiks-cube
edited Dec 24 '18 at 13:43
Shaun
9,338113684
asked Oct 25 '17 at 20:30
user2249675user2249675
38718
$endgroup$
$begingroup$
This is a coset intersection problem - you are looking for $Gsigma cap H$. There is a discussion about it here
$endgroup$
– Derek Holt
Oct 25 '17 at 20:40
add a comment |
$begingroup$
Given two permutation groups $G,H$ on the same set $M$, and a permutation $sigma$ on $M$, how to find a factorization $sigma = gh$ (but not $sigma = hg$), where $g in G$ and $h in H$?
This problem arises from modeling variations of Rubik's Cube with permutation groups. One problem with this approach is that these puzzles often contain indistinguishable stickers. For example, in the $4times4times4$ puzzle the four center stickers on each face are indistinguishable, giving a total of $24^6$ potential "solved" configurations. But not all of these are actually possible. Calculations with Mathematica show that for each face, only $12$ permutations of its center stickers are achievable, without disrupting the configuration of other stickers. Therefore in a scrambled state we cannot randomly assign to each center sticker a target position, otherwise the permutation may not be solvable.
group-theory finite-groups permutations rubiks-cube
edited Dec 24 '18 at 13:43
Shaun
9,338113684
asked Oct 25 '17 at 20:30
user2249675user2249675
38718
$endgroup$
Given two permutation groups $G,H$ on the same set $M$, and a permutation $sigma$ on $M$, how to find a factorization $sigma = gh$ (but not $sigma = hg$), where $g in G$ and $h in H$?
This problem arises from modeling variations of Rubik's Cube with permutation groups. One problem with this approach is that these puzzles often contain indistinguishable stickers. For example, in the $4times4times4$ puzzle the four center stickers on each face are indistinguishable, giving a total of $24^6$ potential "solved" configurations. But not all of these are actually possible. Calculations with Mathematica show that for each face, only $12$ permutations of its center stickers are achievable, without disrupting the configuration of other stickers. Therefore in a scrambled state we cannot randomly assign to each center sticker a target position, otherwise the permutation may not be solvable.
group-theory finite-groups permutations rubiks-cube
group-theory finite-groups permutations rubiks-cube
edited Dec 24 '18 at 13:43
Shaun
9,338113684
asked Oct 25 '17 at 20:30
user2249675user2249675
38718
edited Dec 24 '18 at 13:43
Shaun
9,338113684
asked Oct 25 '17 at 20:30
user2249675user2249675
38718
edited Dec 24 '18 at 13:43
Shaun
9,338113684
edited Dec 24 '18 at 13:43
Shaun
9,338113684
edited Dec 24 '18 at 13:43
Shaun
9,338113684
9,338113684
asked Oct 25 '17 at 20:30
user2249675user2249675
38718
asked Oct 25 '17 at 20:30
user2249675user2249675
38718
asked Oct 25 '17 at 20:30
user2249675user2249675
38718
38718
$begingroup$
This is a coset intersection problem - you are looking for $Gsigma cap H$. There is a discussion about it here
$endgroup$
– Derek Holt
Oct 25 '17 at 20:40
add a comment |
$begingroup$
This is a coset intersection problem - you are looking for $Gsigma cap H$. There is a discussion about it here
$endgroup$
– Derek Holt
Oct 25 '17 at 20:40
$begingroup$
This is a coset intersection problem - you are looking for $Gsigma cap H$. There is a discussion about it here
$endgroup$
– Derek Holt
Oct 25 '17 at 20:40
$begingroup$
This is a coset intersection problem - you are looking for $Gsigma cap H$. There is a discussion about it here
$endgroup$
– Derek Holt
Oct 25 '17 at 20:40
add a comment |
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$begingroup$
This is a coset intersection problem - you are looking for $Gsigma cap H$. There is a discussion about it here
$endgroup$
– Derek Holt
Oct 25 '17 at 20:40