Prolog performance and recursion type

Question

I was playing with permutation in a couple of programs and stumbled upon this little experiment:

Permutation method 1:

permute([], []).
         


        

Answer 1

Really nice question.

Waiting for someone to post a time/space analysis, the only caveat I can offer is that method 1 & 2 don't terminate when first argument is free, while method 3 does.

Anyway, method 1 seems really much more efficient than the builtin. Good to know.

edit: and given that the library implementation merely adjust instantiation of arguments and calls method 1, I'm going to discuss on SWI-Prolog mailing list your method 2 as alternative (or, you prefer to do it yourself, let me know).

more edit: I forgot previously to point out that permutation/3 (let's say, method 2), gives lexicographically ordered solutions, while method 1 doesn't. I think that could be a strong preferential requirement, but should be expressed as an option, given the performance gain that method 1 allows.

?- time(call_nth(permute1([0,1,2,3,4,5,6,7,8,9],P),1000000)).
% 3,112,758 inferences, 3,160 CPU in 3,162 seconds (100% CPU, 984974 Lips)
P = [1, 4, 8, 3, 7, 6, 5, 9, 2|...] .

?- time(call_nth(permute2([0,1,2,3,4,5,6,7,8,9],P),1000000)).
% 10,154,843 inferences, 9,779 CPU in 9,806 seconds (100% CPU, 1038398 Lips)
P = [2, 7, 8, 3, 9, 1, 5, 4, 6|...] .

YAP gives still more gain!

?- time(call_nth(permute1([0,1,2,3,4,5,6,7,8,9],P),1000000)).
% 0.716 CPU in 0.719 seconds ( 99% CPU)
P = [1,4,8,3,7,6,5,9,2,0]

?- time(call_nth(permute2([0,1,2,3,4,5,6,7,8,9],P),1000000)).
% 8.357 CPU in 8.368 seconds ( 99% CPU)
P = [2,7,8,3,9,1,5,4,6,0]

edit: I've posted a comment on SWI-Prolog doc page about this theme.