Revision history [back]

Here is a possible approach. It first finds one permutation p that sorts a given list, then constructs a subgroup a permutations that fix the sorted list, and then reports the product of p and the subgroup elements:

import itertools import sage.combinat.permutation as permutation

import itertools
import sage.combinat.permutation as permutation

dP = [3,4,3,2,4]
n = len(dP)
S = SymmetricGroup(n)

s_dP = sorted(dP)     # sorted list
p = S( Word(s_dP).standard_permutation() / Word(dP).standard_permutation() )

P = [ permutation.from_cycles(n,[[i+1 for i,_ in g]]) for _,g in itertools.groupby(enumerate(s_dP),key=lambda t:t[1]) ]
G = S.subgroup(P)    # subgroup that fix s_dP
for q in G:
    r = Permutation( q * p )
    assert [dP[i-1] for i in r] == s_dP     # verification, just in case
    print(r)

Here is a possible approach. It first finds one permutation p that sorts a given list, then constructs a subgroup a permutations that fix the sorted list, and then reports the product of p and the subgroup elements:

import itertools import sage.combinat.permutation as permutation

import itertools
import sage.combinat.permutation as permutation

dP = [3,4,3,2,4]
n = len(dP)
S = SymmetricGroup(n)

s_dP = sorted(dP)     # sorted list
p = S( Word(s_dP).standard_permutation() / Word(dP).standard_permutation() )

P = [ permutation.from_cycles(n,[[i+1 for i,_ in g]]) permutation.from_cycles(n,[list(g)]) for _,g in itertools.groupby(enumerate(s_dP),key=lambda t:t[1]) itertools.groupby(range(1,n+1),key=lambda t:s_dP[t-1]) ]
G = S.subgroup(P)    # subgroup that fix s_dP
for q in G:
    r = Permutation( q * p )
    assert [dP[i-1] for i in r] == s_dP     # verification, just in case
    print(r)

Here is a possible approach. It first finds one permutation p that sorts a given list, then constructs a subgroup a permutations that fix the sorted list, and then reports the product of p and the subgroup elements:

import itertools import sage.combinat.permutation as permutation

import itertools
import sage.combinat.permutation as permutation

dP = [3,4,3,2,4]
n = len(dP)
S = SymmetricGroup(n)

s_dP = sorted(dP)     # sorted list
p = S( Word(s_dP).standard_permutation() / Word(dP).standard_permutation() )

P = [ permutation.from_cycles(n,[list(g)]) for _,g in itertools.groupby(range(1,n+1),key=lambda t:s_dP[t-1]) ]
G = S.subgroup(P)    # subgroup that fix s_dP
for q in G:
    r = Permutation( q * p )
    assert [dP[i-1] for i in r] == s_dP     # verification, just in case
    print(r)

Here is a possible approach. It first finds one permutation p that sorts a given list, then constructs a subgroup a permutations that fix the sorted list, and then reports the product of p and the subgroup elements:

import itertools
import sage.combinat.permutation as permutation

dP = [3,4,3,2,4]
n = len(dP)
S = SymmetricGroup(n)

s_dP = sorted(dP)     # sorted list
p = S( Word(s_dP).standard_permutation() / Word(dP).standard_permutation() )

P = [ permutation.from_cycles(n,[list(g)]) permutation.from_cycles(n,[(i,i+1)]) for _,g in itertools.groupby(range(1,n+1),key=lambda t:s_dP[t-1]) for i in list(g)[:-1] ]
G = S.subgroup(P)    # subgroup that fix s_dP
for q in G:
    r = Permutation( q * p )
    assert [dP[i-1] for i in r] == s_dP     # verification, just in case
    print(r)

Here is a possible approach. It first finds one permutation p that sorts a given list, then constructs a subgroup a permutations that fix the sorted list, and then reports the product of p and the subgroup elements:

import itertools
import sage.combinat.permutation as permutation

dP = [3,4,3,2,4]
n = len(dP)
S = SymmetricGroup(n)

s_dP = sorted(dP)     # sorted list
p = S( Word(s_dP).standard_permutation() / Word(dP).standard_permutation() )

P = [ permutation.from_cycles(n,[(i,i+1)]) for _,g in itertools.groupby(range(1,n+1),key=lambda t:s_dP[t-1]) itertools.groupby(range(n),key=lambda t:s_dP[t]) for i in list(g)[:-1] itertools.islice(g,1,None) ]
G = S.subgroup(P)    # subgroup that fix s_dP
for q in G:
    r = Permutation( q * p )
    assert [dP[i-1] for i in r] == s_dP     # verification, just in case
    print(r)