prolog , Avoid recursion in predicate

Avoid recursion in predicate


Tag: prolog

I have the following question for the following predicate, how can i drop the recursive call f(T,S1) from both predicates.

Flow model: (i,o)

    S1 > 2,!,
    S is S1 + H.
    S is S1 + 1.

This is a trick question, and I am not that good with prolog. the second predicate fails always.= > can be droped, but what about the second. As I see this method is a list elements counter. Thanks.

Is that even possible?


OK, this is a complex issue. You assume it is a trick question, but is it really one? How can we be sure? I will let library(clpfd) do the thinking for me. First I will rewrite your program:

:- use_module(library(clpfd)).

    S1 #> 2,
    S #= S1 + H.
    S1 #=< 2,
    S #= S1 + 1.

(And just a remark: putting these tests after the recursion twice will make this program require exponentially many inferences, but let's stick to it...)

So I would like to reason about that program in the most general manner. Therefore, I will not take concrete values and then try to figure out a theory. Instead I will ask very general questions (using SICStus):

| ?- assert(clpfd:full_answer).
| ?- length(L,N), fx(L,S).
   L = [],
   N = 0,
   S = 0 ?
   L = [_A],
   N = 1,
   S = 1 ?
   L = [_A,_B],
   N = 2,
   S = 2 ?
   L = [_A,_B,_C],
   N = 3,
   S = 3 ?
   L = [_A,_B,_C,_D],
   N = 4,
   _A in inf..sup,
   S in inf..sup ?
   L = [_A,_B,_C,_D,_E],
   N = 5, ...

So please look at the answers N = 0 up to N = 3: There no constraints are involved, effectively all the elements of the list [_A,_B,...] are ignored. However, starting with N = 4 the first element _A now influences the "sum" S since the equation S #= _A+3 holds! With larger values things become more and more complex.

In any case, I cannot see how this could be a trick question. The last three elements are ignored. Well, that's kind of a trick. But otherwise elements (or at least some of them) influence the outcome!


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