'fail' predicate confusion

[everlonggg]

The fail predicate seems a bit unpredictable under certain cases to me. Take the following example.

Code:
a(1).
a(2).
b(3).
b(4).
go:- a(X), write(X).
go:- a(X), write(X), b(Y), write(Y).


Query this with ?- go, fail.

The result is 12134234. How does this make sense? Can somebody explain the logic behind that? Thanks.

 

[joel76]

you wrote "go, fail."

you have 2 rules for go.

the first rule is executed ==> 1
fail
Prolog searches for the last point of choice : fact with a
the first rule is executed ==> 2
fail
the first rule is no longer avalaible (only 2 facts with a).
the second rule is executed ==> 1, 3
fail
the second rule is executed at the last point of choice : fact with b ==> 4
fail
the second rule is executed at the last point of choice : fact with a because all the choices for b have been explored ==> 2
etc, etc.

 

[everlonggg]

the second rule is executed at the last point of choice : fact with b ==> 4
fail"

This step does not make logical sense to me. Why would this not print 2, 4?

 

[joel76]

After 13 you would have 24 ?
No, because Prolog starts from the end of the rule to search back the last point of choice.

 

[everlonggg]

Sorry, 14. That was a mistake on my behalf.

So what you're saying is that the fail predicate doesn't make it start the whole second rule over again, but only checks for a new b(Y)?

 

[joel76]

Yes backtrack search from the last point of choice at b(Y).

 

[everlonggg]

Gotcha! And in the way it works in recursive list cases... let's make up an example.

f([], 0).
f([H|T], X):- f(T, Y), g(H,Z), X is Y+Z.
g(a, 1).
g(X, 0).

inputting f([a,b,a],X), write(X), fail. to this will give you something like...
2110. It will get the first 2, fail. go back and instead of satisfying g(a, 1), satisfy g(X, 0). Then we get a 1. Fail. Now where would it traverse from there?

Thanks! I think I'm starting to catch onto this.