CODEWITHSUNDEEP
prolog
Jack R.
Jack R.

Reputation: 11

Simple Prolog predicate - statement is always evaluated as false

I've just started Prolog and truly wonder why the following lines, specifically the 'is' part, always produces false:

highest(L) :-
    path(_,_,Z),
    Z >= L,
    L is Z.
highestWrap :-
    highest(0).

highestWrap is called.

Thanks in advance and have a beautiful day!

Upvotes: 1

Views: 272

Answers (1)

willeM_ Van Onsem
willeM_ Van Onsem

Reputation: 476527

Unless there is a path with length 0, this will not work, and even then, it will likely not yield what you want: it will just say true.

In Prolog variables can only be set once, that means that if L is set to 0, then it remains 0, unless you backtrack over that assignment.

Here it thus means that you call highest(0), next you instruct Prolog to call path(_, _, Z) and this can result in zero, one or more solutions. In case there are no path(_, _, Z)s, then the call will fail. But in case there are, then Z will (if I make correct assumptions about the predicate), have a numerical value, for example 7.

Now the condition Z >= L of course holds in that case (if Z is 7), so that is nog the problem. But now you specify L is Z. That thus means that you call 0 is 7. The is/2 [swi-doc] predicate aims to solve the expression of the second argument (that expression is 7 in the example, so there is not much to solve), and then aims to unify it with the term on the left side. But since 0 is not equal to 7 that fails.

If you want to obtain the highest path, you can for example make use of the aggregate [swi-doc] library:

:- use_module(library(aggregate)).

highest(L) :-
    aggregate(Max(Z), path(_,_,Z), Max(L)).

You can then call it with higest(X) to unify X with the highest value for Z in a call to path(_, _, Z).

Upvotes: 1

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