CODEWITHSUNDEEP
numbersprologcompare
Noisie
Noisie

Reputation: 71

How can i compare natural numbers in prolog?

I have a function that takes two arguments and compares if they are natural numbers in their unit form and if the first arg is bigger than the second!

So here is the code I've written but every time it gets me "no".

nat(0).
nat(s(X)) :- nat(X).
sum(X,0,X) :- nat(X).
sum(X,s(Y),s(Z)) :- sum(X,Y,Z).
gr(X,Y) :- nat(s(X)), nat(s(Y)), X>Y.

What goes wrong? Everything is in Prolog . the function is the gr() .

Upvotes: 1

Views: 559

Answers (1)

user1812457
user1812457

Reputation:

First, you probably want for sum rather this:

sum(0, Y, Y) :-
    nat(Y).
sum(s(X), Y, s(Z)) :-
    sum(X, Y, Z).

This is so that Prolog can recognize that the two clauses are exclusive by only looking at the first argument.

Now to your greater than:

% gr(X, Y) is true if X is greater than Y
gr(X, Y) :- sm(Y, X).
% sm(X, Y) is true if X is smaller than Y
sm(0, s(Y)) :-
    nat(Y).
sm(s(X), s(Y)) :-
    sm(X, Y).

To answer your actual question: what goes wrong is that the operator > works on integers (like 1 or 0 or -19), not on compound terms. The operator @> will work (see the documentation of the implementation you are using), but I have the feeling you might actually want to be explicit about it.

Upvotes: 2

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