CODEWITHSUNDEEP
prolog
JMFS
JMFS

Reputation: 297

Not in Swi-Prolog 6.2

I have this facts:

morpheme(a,82).
morpheme(r,83).

And i have this rule:

foo(er,Start) :-
    morpheme(a,Start), morpheme(r,I), I is Start+1,
    not(morfema(_,J)), J is I+1.

When I ask the query:

foo(er,82).

I got "false" (wrong answer) instead of "true" (correct answer).

In the query I've tried to say that: "if there is a morpheme AR in the start position "Start" and there are no more morphemes in higher positions (higher than Start+1), then fires the rule".

I tried using \+ and cut-fail (http://stackoverflow.com/questions/3850563/writing-prolog-statement-with-not-operator) but no success :(

I think that the problem is located in the way i wrote the rule.

Thanks in advance!!!

Upvotes: 1

Views: 172

Answers (2)

JMFS
JMFS

Reputation: 297

Thanks False! That solved the problem. The modified code is below:

morpheme(a,82).
morpheme(r,83).

not(X) :- \+ X.

foo(er,Start) :-
    morpheme(a,Start), I is Start+1, morpheme(r,I),
    J is I+1, not(morpheme(_,J)).

Upvotes: 0

false
false

Reputation: 10102

Typo: morfema / morpheme.

But the deeper problem is the variable J: At the point in time when the negation is tried, J is an uninstantiated variable. Only afterwards, it gets the value you expect.

So exchange the two goals - and better use \+ in place of not!

The \+ means: not provable at this point in time. It therefore reveals a bit the way how Prolog programs are executed.

There is another problem in your program: You may move I is Start+1 one to the left. In this manner the morpheme(r,I) will be a ground goal. It might be executed faster.

Upvotes: 1

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