How to make the program give the other alternate outputs

Help!
1

I’m using: SWI-Prolog Multi-threaded, 64 bits, Version 7.2.3

I want the code to:
find all possible permutations of the given list. It must run select2() function for different possible values of H, thereby giving the different permutations. However each time by base case is fixed and therefore for empty list ([]) it must always return empty list.

But what I’m getting is:
When I am running the Prolog Program , first time it gives the list itself which is correct, then when I try to find other possible permutations by using ‘;’ operator, for finding other alternatives, it tries to run the rule for the base case instead of the fact. I understand that maybe thats what ‘;’ does. However my requirement is to run the fact always for base case and every time the only variations should result from the select2() function whose output decides the working of the following permutation() function

My code looks like this:

permutation([],[]).
permutation(L,[H|T]):-
        select2(H,L,R),
        permutation(R,T).

select2(_,[],[]).
select2(X,[X|T],T):-!.
select2(X,[H|T],[H|L]):-
	select2(X,T,L).

My trace looks like this
?- trace,permutation([3,2,1],L).
Call: (8) permutation([3, 2, 1], _G1620) ? creep
Call: (9) select2(_G1738, [3, 2, 1], _G1750) ? creep
Exit: (9) select2(3, [3, 2, 1], [2, 1]) ? creep
Call: (9) permutation([2, 1], _G1739) ? creep
Call: (10) select2(_G1741, [2, 1], _G1753) ? creep
Exit: (10) select2(2, [2, 1], [1]) ? creep
Call: (10) permutation([1], _G1742) ? creep
Call: (11) select2(_G1744, [1], _G1756) ? creep
Exit: (11) select2(1, [1], []) ? creep
Call: (11) permutation([], _G1745) ? creep
Exit: (11) permutation([], []) ? creep
Exit: (10) permutation([1], [1]) ? creep
Exit: (9) permutation([2, 1], [2, 1]) ? creep
Exit: (8) permutation([3, 2, 1], [3, 2, 1]) ? creep
L = [3, 2, 1] ;
Redo: (11) permutation([], _G1745) ? creep
Call: (12) select2(_G1747, [], _G1759) ? creep
Exit: (12) select2(_G1747, [], []) ? creep
Call: (12) permutation([], _G1748) ? creep
Exit: (12) permutation([], []) ? creep
Exit: (11) permutation([], [_G1747]) ? creep
Exit: (10) permutation([1], [1, _G1747]) ? creep
Exit: (9) permutation([2, 1], [2, 1, _G1747]) ?
…
This is a part of the trace. As can be clearly seen, from the alternative output case, it goes into a computation which is yielding the results I want([2,1,_G1747]).

I am a beginner here so please pardon any mistakes

2

What is the code for select2?
(And it’s not a “function”, which usually means that it returns a single result; it’s a “predicate”.)

Anyway, if you look at your output, you’ll see that the last “Call” was (11) permutation([], _G1745). This is what “Redo” tries to do. (I’m guessing this is your question about “tries to run the rule for the base case”). The second clause also matches this, but without further information, I can’t tell what your problem is.

3

Oh, I am very sorry. I forgot to include the code for select2. Anyway select2 selects one element from the list and gives the remaining list as output. So as we are giving a variable as the element to remove, I want it to work such that every time it takes one of the elements out and gives remaining list which is permuted on.

4

I think the cut (! in the second clause of select/3) is likely the problem; that removes the choice point there & prevents finding subsequent variations in the third clause.

5

Also, the first clause of select2 shouldn’t be there, unless you think that select2(5, [], []) is correct.