prolog , Avoid recursion in predicate
Avoid recursion in predicate
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:
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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