Replacement rule with all possible permutations

Question

I have a list with expressions like

{1,3,5,a,c,b,3,d}

I want to generate all outputs with permutations of {a,b,c,d}, so for example a rule like

/. {a->a,b->d, c->c, d->b)

In total shere should be k! new lists. For the above example I also do not know how to make a replacement like b<->d which would save some time.

I found the Function PermutationReplace, and something like

PermutationReplace[{1,2,3},SymmetricGroup[3]]

would give me all 6 Permutations of 123. But I dont know how I can apply something like that were the elements I want to permute are arbitrary numbers/characters/variables.

I appreciate any help, and thanks for reading.

Answer 1

f[L_, s_] := With[{pos = Union @@ Lookup[PositionIndex[L], s, {}]},
               Permute[L, Cycles[{pos[[#]]}]] & /@ Permutations[Range[Length[pos]]]]

f[{1, 3, 5, a, c, b, 3, d}, {a, b, c, d}]
(* {{1, 3, 5, d, a, c, 3, b}, {1, 3, 5, b, a, d, 3, c},
    {1, 3, 5, d, b, a, 3, c}, {1, 3, 5, c, d, a, 3, b},
    {1, 3, 5, b, d, c, 3, a}, {1, 3, 5, c, b, d, 3, a},
    {1, 3, 5, c, d, a, 3, b}, {1, 3, 5, c, b, d, 3, a},
    {1, 3, 5, b, d, c, 3, a}, {1, 3, 5, d, a, c, 3, b},
    {1, 3, 5, d, b, a, 3, c}, {1, 3, 5, b, a, d, 3, c},
    {1, 3, 5, b, a, d, 3, c}, {1, 3, 5, b, d, c, 3, a},
    {1, 3, 5, c, b, d, 3, a}, {1, 3, 5, d, b, a, 3, c},
    {1, 3, 5, d, a, c, 3, b}, {1, 3, 5, c, d, a, 3, b},
    {1, 3, 5, d, a, c, 3, b}, {1, 3, 5, d, b, a, 3, c},
    {1, 3, 5, c, d, a, 3, b}, {1, 3, 5, b, d, c, 3, a},
    {1, 3, 5, b, a, d, 3, c}, {1, 3, 5, c, b, d, 3, a}} *)

If L has duplicates then the following can be used to avoid duplicates in the result

f[L_, s_] := Module[{res = L,
   pos = Union @@ Lookup[PositionIndex[L], s, {}]},
  (res[[pos]] = #; res) & /@ Permutations[L[[pos]]]]

f[{5, a, d, 3, d}, {a, b, c, d}]
(* {{5, a, d, 3, d}, {5, d, a, 3, d}, {5, d, d, 3, a}} *)

Answer 2

Here is a simple solution based on Inner product:

rules[vec_] := Inner[Rule, Permutations[vec], vec, List];

Examples:

rules[{a, b, c}]

{{a->a,b->b,c->c},{a->a,c->b,b->c},{b->a,a->b,c->c},{b->a,c->b,a->c},{c->a,a->b,b->c},{c->a,b->b,a->c}}

{1, 3, 5, a, c, b, 3, d} /. rules[{a, b, c, d}]

{{1,3,5,a,c,b,3,d},{1,3,5,a,d,b,3,c},{1,3,5,a,b,c,3,d},{1,3,5,a,b,d,3,c},{1,3,5,a,d,c,3,b},{1,3,5,a,c,d,3,b},{1,3,5,b,c,a,3,d},{1,3,5,b,d,a,3,c},{1,3,5,c,b,a,3,d},{1,3,5,d,b,a,3,c},{1,3,5,c,d,a,3,b},{1,3,5,d,c,a,3,b},{1,3,5,b,a,c,3,d},{1,3,5,b,a,d,3,c},{1,3,5,c,a,b,3,d},{1,3,5,d,a,b,3,c},{1,3,5,c,a,d,3,b},{1,3,5,d,a,c,3,b},{1,3,5,b,d,c,3,a},{1,3,5,b,c,d,3,a},{1,3,5,c,d,b,3,a},{1,3,5,d,c,b,3,a},{1,3,5,c,b,d,3,a},{1,3,5,d,b,c,3,a}}

Answer 3

ClearAll[g]
g[list_, sublist_]:= list /. AssociationThread[sublist, #]& /@ Permutations[sublist]

g[{1, 3, 5, a, c, b, 3, d}, {a, b, c, d}]

{{1, 3, 5, a, c, b, 3, d}, {1, 3, 5, a, d, b, 3, c}, {1, 3, 5, a, b, c, 3, d},
{1, 3, 5, a, d, c, 3, b}, {1, 3, 5, a, b, d, 3, c}, {1, 3, 5, a, c, d, 3, b},
{1, 3, 5, b, c, a, 3, d}, {1, 3, 5, b, d, a, 3, c}, {1, 3, 5, b, a, c, 3, d},
{1, 3, 5, b, d, c, 3, a}, {1, 3, 5, b, a, d, 3, c}, {1, 3, 5, b, c, d, 3, a},
{1, 3, 5, c, b, a, 3, d}, {1, 3, 5, c, d, a, 3, b}, {1, 3, 5, c, a, b, 3, d},
{1, 3, 5, c, d, b, 3, a}, {1, 3, 5, c, a, d, 3, b}, {1, 3, 5, c, b, d, 3, a},
{1, 3, 5, d, b, a, 3, c}, {1, 3, 5, d, c, a, 3, b}, {1, 3, 5, d, a, b, 3, c},
{1, 3, 5, d, c, b, 3, a}, {1, 3, 5, d, a, c, 3, b}, {1, 3, 5, d, b, c, 3, a}}