Fixed points free permutations [duplicate] The 2019 Stack Overflow Developer Survey Results Are InFaster derangements?How to remove repeated permutations?Find all permutations with reversals / cyclic permutations removedGenerating permutations of at most 'n' elements, and where a specific subset of elements always appearsNested permutationsCreating all permutations of list with “alignment”Duplicate-free results of permutations + constant listCyclic and Non-cyclic PermutationsPermutations of lists of fixed even numbersHow to generate all involutive permutations?How do I iterate through all permutations of a list?
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Fixed points free permutations [duplicate]
The 2019 Stack Overflow Developer Survey Results Are InFaster derangements?How to remove repeated permutations?Find all permutations with reversals / cyclic permutations removedGenerating permutations of at most 'n' elements, and where a specific subset of elements always appearsNested permutationsCreating all permutations of list with “alignment”Duplicate-free results of permutations + constant listCyclic and Non-cyclic PermutationsPermutations of lists of fixed even numbersHow to generate all involutive permutations?How do I iterate through all permutations of a list?
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Faster derangements?
9 answers
How to generate a list of fixpoint free permutations of n elements in mathematica?
list-manipulation permutation
asked Apr 5 at 10:30
Darwin1871Darwin1871
33717
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marked as duplicate by corey979, Carl Woll list-manipulation
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Faster derangements?
9 answers
How to generate a list of fixpoint free permutations of n elements in mathematica?
list-manipulation permutation
asked Apr 5 at 10:30
Darwin1871Darwin1871
33717
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This question already has an answer here:
Faster derangements?
9 answers
How to generate a list of fixpoint free permutations of n elements in mathematica?
list-manipulation permutation
asked Apr 5 at 10:30
Darwin1871Darwin1871
33717
$endgroup$
This question already has an answer here:
Faster derangements?
9 answers
How to generate a list of fixpoint free permutations of n elements in mathematica?
This question already has an answer here:
Faster derangements?
9 answers
list-manipulation permutation
list-manipulation permutation
asked Apr 5 at 10:30
Darwin1871Darwin1871
33717
asked Apr 5 at 10:30
Darwin1871Darwin1871
33717
asked Apr 5 at 10:30
Darwin1871Darwin1871
33717
asked Apr 5 at 10:30
Darwin1871Darwin1871
33717
asked Apr 5 at 10:30
Darwin1871Darwin1871
33717
33717
marked as duplicate by corey979, Carl Woll list-manipulation
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2 Answers
2
active
oldest
votes
$begingroup$
Here a brute force method.
n = 4;
perms = Permutations[Range[n]];
Pick[perms, Unitize[Min[Abs[# - Range[n]]] & /@ perms], 1]
2, 1, 4, 3, 2, 3, 4, 1, 2, 4, 1, 3, 3, 1, 4, 2, 3, 4, 1,
2, 3, 4, 2, 1, 4, 1, 2, 3, 4, 3, 1, 2, 4, 3, 2, 1
answered Apr 5 at 10:41
Henrik SchumacherHenrik Schumacher
59.7k582166
$endgroup$
add a comment |
$begingroup$
With[n = 4,
Select[Permutations[Range[n]], Length[PermutationSupport[#]] == n &]]
2, 1, 4, 3, 2, 3, 4, 1, 2, 4, 1, 3, 3, 1, 4, 2, 3, 4, 1, 2, 3, 4, 2, 1, 4, 1, 2, 3, 4, 3, 1, 2, 4, 3, 2, 1
The fraction of permutations satisfying this condition is $1/e$ as $ntoinfty$, so the above code is not very wasteful.
answered Apr 5 at 11:30
RomanRoman
4,96511130
$endgroup$
$begingroup$
Thank you all very much for the hints.
$endgroup$
– Darwin1871
Apr 6 at 10:28
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here a brute force method.
n = 4;
perms = Permutations[Range[n]];
Pick[perms, Unitize[Min[Abs[# - Range[n]]] & /@ perms], 1]
2, 1, 4, 3, 2, 3, 4, 1, 2, 4, 1, 3, 3, 1, 4, 2, 3, 4, 1,
2, 3, 4, 2, 1, 4, 1, 2, 3, 4, 3, 1, 2, 4, 3, 2, 1
answered Apr 5 at 10:41
Henrik SchumacherHenrik Schumacher
59.7k582166
$endgroup$
add a comment |
$begingroup$
Here a brute force method.
n = 4;
perms = Permutations[Range[n]];
Pick[perms, Unitize[Min[Abs[# - Range[n]]] & /@ perms], 1]
2, 1, 4, 3, 2, 3, 4, 1, 2, 4, 1, 3, 3, 1, 4, 2, 3, 4, 1,
2, 3, 4, 2, 1, 4, 1, 2, 3, 4, 3, 1, 2, 4, 3, 2, 1
answered Apr 5 at 10:41
Henrik SchumacherHenrik Schumacher
59.7k582166
$endgroup$
add a comment |
$begingroup$
Here a brute force method.
n = 4;
perms = Permutations[Range[n]];
Pick[perms, Unitize[Min[Abs[# - Range[n]]] & /@ perms], 1]
2, 1, 4, 3, 2, 3, 4, 1, 2, 4, 1, 3, 3, 1, 4, 2, 3, 4, 1,
2, 3, 4, 2, 1, 4, 1, 2, 3, 4, 3, 1, 2, 4, 3, 2, 1
answered Apr 5 at 10:41
Henrik SchumacherHenrik Schumacher
59.7k582166
$endgroup$
Here a brute force method.
n = 4;
perms = Permutations[Range[n]];
Pick[perms, Unitize[Min[Abs[# - Range[n]]] & /@ perms], 1]
2, 1, 4, 3, 2, 3, 4, 1, 2, 4, 1, 3, 3, 1, 4, 2, 3, 4, 1,
2, 3, 4, 2, 1, 4, 1, 2, 3, 4, 3, 1, 2, 4, 3, 2, 1
answered Apr 5 at 10:41
Henrik SchumacherHenrik Schumacher
59.7k582166
answered Apr 5 at 10:41
Henrik SchumacherHenrik Schumacher
59.7k582166
answered Apr 5 at 10:41
Henrik SchumacherHenrik Schumacher
59.7k582166
answered Apr 5 at 10:41
Henrik SchumacherHenrik Schumacher
59.7k582166
59.7k582166
add a comment |
add a comment |
$begingroup$
With[n = 4,
Select[Permutations[Range[n]], Length[PermutationSupport[#]] == n &]]
2, 1, 4, 3, 2, 3, 4, 1, 2, 4, 1, 3, 3, 1, 4, 2, 3, 4, 1, 2, 3, 4, 2, 1, 4, 1, 2, 3, 4, 3, 1, 2, 4, 3, 2, 1
The fraction of permutations satisfying this condition is $1/e$ as $ntoinfty$, so the above code is not very wasteful.
answered Apr 5 at 11:30
RomanRoman
4,96511130
$endgroup$
$begingroup$
Thank you all very much for the hints.
$endgroup$
– Darwin1871
Apr 6 at 10:28
add a comment |
$begingroup$
With[n = 4,
Select[Permutations[Range[n]], Length[PermutationSupport[#]] == n &]]
2, 1, 4, 3, 2, 3, 4, 1, 2, 4, 1, 3, 3, 1, 4, 2, 3, 4, 1, 2, 3, 4, 2, 1, 4, 1, 2, 3, 4, 3, 1, 2, 4, 3, 2, 1
The fraction of permutations satisfying this condition is $1/e$ as $ntoinfty$, so the above code is not very wasteful.
answered Apr 5 at 11:30
RomanRoman
4,96511130
$endgroup$
$begingroup$
Thank you all very much for the hints.
$endgroup$
– Darwin1871
Apr 6 at 10:28
add a comment |
$begingroup$
With[n = 4,
Select[Permutations[Range[n]], Length[PermutationSupport[#]] == n &]]
2, 1, 4, 3, 2, 3, 4, 1, 2, 4, 1, 3, 3, 1, 4, 2, 3, 4, 1, 2, 3, 4, 2, 1, 4, 1, 2, 3, 4, 3, 1, 2, 4, 3, 2, 1
The fraction of permutations satisfying this condition is $1/e$ as $ntoinfty$, so the above code is not very wasteful.
answered Apr 5 at 11:30
RomanRoman
4,96511130
$endgroup$
With[n = 4,
Select[Permutations[Range[n]], Length[PermutationSupport[#]] == n &]]
2, 1, 4, 3, 2, 3, 4, 1, 2, 4, 1, 3, 3, 1, 4, 2, 3, 4, 1, 2, 3, 4, 2, 1, 4, 1, 2, 3, 4, 3, 1, 2, 4, 3, 2, 1
The fraction of permutations satisfying this condition is $1/e$ as $ntoinfty$, so the above code is not very wasteful.
answered Apr 5 at 11:30
RomanRoman
4,96511130
edited Apr 5 at 11:40
answered Apr 5 at 11:30
RomanRoman
4,96511130
answered Apr 5 at 11:30
RomanRoman
4,96511130
answered Apr 5 at 11:30
RomanRoman
4,96511130
4,96511130
$begingroup$
Thank you all very much for the hints.
$endgroup$
– Darwin1871
Apr 6 at 10:28
add a comment |
$begingroup$
Thank you all very much for the hints.
$endgroup$
– Darwin1871
Apr 6 at 10:28
$begingroup$
Thank you all very much for the hints.
$endgroup$
– Darwin1871
Apr 6 at 10:28
$begingroup$
Thank you all very much for the hints.
$endgroup$
– Darwin1871
Apr 6 at 10:28
add a comment |
