Better termination for s(X)-sum

Question

(Let me sneak that in within the wave of midterm questions.)

A common definition for the sum of two natural numbers is nat_nat_sum/3:

na         


        

Answer 1

Ok, seems its over. The solution I was thinking of was:

nat_nat_sum2(0, N,N).
nat_nat_sum2(s(N), 0, s(N)).
nat_nat_sum2(s(N), s(M), s(s(O))) :-
   nat_nat_sum2(N, M, O).

But as I realize, that's just the same as @mat's one which is almost the same as @WillNess'es.

Is this really the better nat_nat_sum/3? The original's runtime is independent of B (if we ignore one (1) occurs check for the moment).

There is another downside of my solution compared to @mat's solution which naturally extends to nat_nat_nat_sum/3

nat_nat_nat_sum(0, B, C, D) :-
   nat_nat_sum(B, C, D).
nat_nat_nat_sum(s(A), B, C, s(D)) :-
   nat_nat_nat_sum2(B, C, A, D).

Which gives

nat_nat_nat_sum(A,B,C,D)terminates_if b(A),b(B);b(A),b(C);b(B),b(C);b(D).

(provable in the unfolded version with cTI)