Stuck on an infinite loop in prolog

Question

I'd like my program to find me all sub-sets of size K of the integers 1,2,...,N.

For this, I wrote the following subs(N,X,Y) means that X is a sub-set of size N of the set Y. I defined the following:

subs(0,[],X).
subs(N,[A|R1],[A|R2]):-N>0, N1 is N-1, subs(N1,R1,R2).
subs(N,[A|R1],[B|R2]):-subs(N,[A|R1],R2).
subs(N,[A|R1],[B|R2]):-subs(N,R1,[B|R2]).

And then as a check I ran subs(2,X,[1,2,3,4]).

I got the first answer [1,2], but never did it give a second answer, as it got stuck in an infinite loop. I tried to trace it, and seems that after finding the first answer it does:

   Redo: (8) subs(0, _G613, [3, 4]) ? creep
^  Call: (9) 0>0 ? creep
^  Fail: (9) 0>0 ? creep
   Redo: (8) subs(0, _G613, [3, 4]) ? creep
   Call: (9) subs(0, [_G618|_G619], [4]) ? creep
^  Call: (10) 0>0 ? creep
^  Fail: (10) 0>0 ? creep
   Redo: (9) subs(0, [_G618|_G619], [4]) ? creep
   Call: (10) subs(0, [_G618|_G619], []) ? creep
   Fail: (10) subs(0, [_G618|_G619], []) ? creep
   Redo: (9) subs(0, [_G618|_G619], [4]) ? creep
   Call: (10) subs(0, _G619, [4]) ? creep
   Exit: (10) subs(0, [], [4]) ? creep
   Exit: (9) subs(0, [_G618], [4]) ? creep
   Exit: (8) subs(0, [_G618], [3, 4]) ? creep
   Exit: (7) subs(1, [2, _G618], [2, 3, 4]) ?

So I see that I get stuck with subs(0, _G619, [4]). Does someone have an idea of how to overcome this problem?

Thanks

Your 4th clause has a flaw. The head of the second argument (the subset) has a variable `A` which is singleton. The clause basically reads, *`[A|R1]` is a subset of `N` values from `[B|R2]` if `R1` is a subset of `N` values from `[B|R2]`, for any variable `A`*. That wouldn't be a correct rule for a subset. It's not clear what the purpose of this rule is. — lurker, Mar 17 '15 at 14:27

lurker · Accepted Answer · 2015-03-17T14:51:00.337

4

Your 4th clause has a flaw. The head of the second argument (the subset) has a variable A which is singleton. The clause basically reads, [A|R1] is a subset of N values from [B|R2] if R1 is a subset of N values from [B|R2], for any variable A. That wouldn't be a correct rule for a subset, and results in the infinite loop since it doesn't ultimately reduce to the base case. It's not clear what the purpose of this rule is. You can probably just remove it as the first 3 adequately define the subset.

You also should constrain N in the 3rd clause to avoid duplicate overlap of rule matching.

That plus a little variable clean-up makes your predicate into:

subs(0, [], _).
subs(N, [A|R1], [A|R2]) :- N > 0, N1 is N-1, subs(N1, R1, R2).
subs(N, R1, [_|R2]) :- N > 0, subs(N, R1, R2).

edited Mar 17 '15 at 14:51

answered Mar 17 '15 at 14:32

lurker

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Thanks! But now that it works I noticed something: if I ask for subs(K,[1,2],[1,2,3]) I get the following error: ERROR: Unhandled exception: >/2: Arguments are not sufficiently instantiated , yet if I ask for subs(2,X,[1,2,3]) all works OK ( I get X=[1,2] , X=[1,3] and X=[2,3]). Seems like a weird error? – TheEmeritus Mar 17 '15 at 14:58
1

@TheEmeritus that's because the numeric relational operators, such as `>/2`, require that all of the arguments be fully instantiated. There can be no variables for it to be able to work. If you want your `subs` predicate to be fully relational and allow the case of the first argument being a variable, it will need to be refactored somewhat to handle that case. In Prolog it's often easy to make a predicate work for a subset of variable argument cases, but to make it fully relational can be a substantial challenge. – lurker Mar 17 '15 at 15:03

score 2 · Answer 2 · edited May 23 '17 at 12:31

@lurker's answer is on the semantic level of the predicate. That's fine. But there is an even simpler way of identifying the problem - simply by using at the following failure slice.

subs(0,[],_X) :- false.
subs(N,[A|R1],[A|R2]):- false, N>0, N1 is N-1, subs(N1,R1,R2).
subs(N,[A|R1],[_B|R2]):- false, subs(N,[A|R1],R2).
subs(N,[_A|R1],[B|R2]):- subs(N,R1,[B|R2]), false.

Already this fragment does not terminate for subs(2,X,[1,2,3,4]). However, it should do so. So in the remaining visible part there is a problem you need to address.

There is a single criterion to find this failure slice: The (universal) non-termination of your query. No further information about the actual intended meaning is used to determine this slice.

Stuck on an infinite loop in prolog

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