5

I was having trouble with these two lines:

list_swizzle(L, [], L).
list_swizzle([], L, L).

The problem was that if the both of the first two arguments are the empty list, the first two statements would both be used, returning the same answer. However, if I put a cut in one, it wrecks backtracking. I eventually put in this line above them:

list_swizzle([], [], []):- !.

And it works. But I was wondering if there is a more elegant solution.

false
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Nathan
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  • This cut is again a red cut. You can see this for: `list_swizzle(Xs,Ys,Zs), Xs = [_]` which fails with above cut. It succeeds with an answer as given by @daf. – false Oct 07 '12 at 17:59

1 Answers1

4

Here's my version:

list_swizzle([H|T], [], [H|T]).
list_swizzle([], L, L).

I'm counting on [] not unifying against [H|T] in the first fact. In other words [] has no T because it's the empty list so the first fact doesn't match goals with a [] in the first arg.

I've run this successfully on SWI-Prolog (Multi-threaded, 32 bits, Version 5.8.2)

$ cat tt.pl

s([H|T], [], [H|T]).
s([], L, L).

....

For help, use ?- help(Topic). or ?- apropos(Word).

?- [tt].
% tt compiled 0.00 sec, 920 bytes
true.

?- s(L,[],[]).
L = [].

?- 
% halt
Dafydd Rees
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