Question

I am a beginner in Prolog and I figured I could try to write a simple test to check if a set of integers mod K (L) is a group. I started by trying to find if the said set is additive, ie. if the sum of every two elements of L is also an element of L.

I wrote the following:

group(A,K):-member(B,A),member(C,A),As is B+C, Bs is mod(As,K), member(Bs,A).

I then tried it with this:

trace. group([0,1,2],3).

This, naturally, produces all possible sums and, correctly, answers that the answer is true for all of them.

But after all these cases it finally prints the following:

  1    1  Redo: group([0,1,2],3) ? 
  6    2  Redo: member(1,[0,1,2]) ? 
  6    2  Fail: member(1,[0,1,2]) ? 
  1    1  Fail: group([0,1,2],3) ?
  no

Why the program checks this final case, that, to me, seems nonsensical?

The last case before this is:

  1    1  Redo: group([0,1,2],3) ? 
  6    2  Redo: member(0,[0,1,2]) ? 
  6    2  Fail: member(0,[0,1,2]) ? 
  3    2  Redo: member(1,[0,1,2]) ? 
  3    2  Exit: member(2,[0,1,2]) ? 
  4    2  Call: _158 is 2+2 ? 
  4    2  Exit: 4 is 2+2 ? 
  5    2  Call: _186 is 4 mod 3 ? 
  5    2  Exit: 1 is 4 mod 3 ? 
  6    2  Call: member(1,[0,1,2]) ? 
  6    2  Exit: member(1,[0,1,2]) ? 
  1    1  Exit: group([0,1,2],3) ? 

  true

which does what it should.

Answer 1

Using a tracer for such a purpose is not very helpful. It shows you lots of detail that is irrelevant. Instead, concentrate on a good formulation of the problem. Concentrate on meaningful names. You use A for the set, and B and C for elements. That can be improved!

What you currently test is this:

There exist two elements of a set whose sum modulo K is in the same set.

What you want to test is that :

For all X, Y in S: ( (X+Y) mod K ) in S.

The operation itself can be written as:Z is (X+Y) mod K

How can a tracer explain that to you?

group(S, K) :-
   \+ (
         member(X,S), member(Y,S), Z is (X+Y) mod K,
         \+ member(Z,S)
   ).

?- group([0,1,2],3).
   true.
?- group([0,2],3).
   false.