Arithmetic comparison in Prolog

Question

Let's say we have two variables X and Y. X is 53 and Y is 52. What I want to do is compare them by adding 1 to the Y, so that it would be 53 - therefore X would be equal to Y + 1.

I am trying to do it by simply using equal operator and addition for Y variables, like so:

X == Y + 1

Even though this looks simple enough, I am getting false as the result. What am I missing?

Are you sure the variables are grounded? – Willem Van Onsem Sep 23 '15 at 10:42 — Willem Van Onsem, Sep 23 '15 at 10:42

score 2 · Accepted Answer · answered Sep 23 '15 at 11:11

2

?- X = 50+2, Y = 50+1, X =:= Y + 1.

as you can see, (=:=)/2 evaluates both sides, as do (>)/2, etc

answered Sep 23 '15 at 11:11

CapelliC

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score 1 · Answer 2 · answered Sep 23 '15 at 11:09

If you are reasoning over integers, use your Prolog system's CLP(FD) constraints to compare and evaluate arithmetic integer expressions.

For example, in SICStus, SWI and YAP, after use_module(library(clpfd):

?- 53 #= 52 + 1. 
true.

This works in all directions.

Other examples:

?- X #= 52 + 1.
X = 53.

?- 53 #= Y + 1.
Y = 52.

?- 53 #= 52 + 1.
true.

score 0 · Answer 3 · answered Sep 23 '15 at 10:43

In order to evaluate expression trees, the is predicate is used:

X is Y+1.

you must be careful however, this will only work as a test if both X and Y are grounded. And will always error if variables on the right hand side (Y in this case) is not grounded.

swipl demo:

?- X = 53, Y = 52, X is Y+1.
X = 53,
Y = 52.

?- X = 53, Y = 52, X is Y.
false.
?- X = 53, X is Y+1.
ERROR: is/2: Arguments are not sufficiently instantiated
?- Y = 52, X is Y+1.
Y = 52,
X = 53.

Arithmetic comparison in Prolog

3 Answers3