Re: [newbie question] How to compare boolean expression/results?

[Daniel Diaz]

Hi Carlos,

You can simplify your code as follows:

:- initialization(main).

commutative(X, Y) :-
    X =:= Y.

main :-
    A = 1,
    B = 1 / (1-2.220446049250313e-16),
    (   commutative(A, B) ->
        write(true)
    ;
        write(false)
    ).


Since Prolog works with predicates (which can then be true or false, 
i.e. succeed or fail) it is better to avoid the 3rd argument to equal 
and commutative and to simply state commutative succeeds iff X and Y are 
commutative. NB: there is no need to check X =:= Y and Y =:= X (=:= is a 
symetric relation in Prolog). So there is no more need for the predicate 
equal (use =:= directly).

BTW on my machine with e-16 it is NOT commutative while with e-17 it is.

Daniel

Le 04/06/2013 00:52, Carlos Nepomuceno a écrit :
> Hi guys! This is my first post to this list.
>
> I'm not really learning Prolog at the moment but I started to compare how 
> different programming languages treat the following floating point comparison 
> case:
>
>
> a=1
> b=1/(1-epsilon)
> (a==b) == (b==a)
>
> That's basic a test to check if the equality operator ('==') is commutative.
>
> I found it really hard to implement that simple test in Prolog. I've read a lot 
> of stuff and the only thing I've found to work (AFAIK) was the following:
>
> %works in GNU Prolog 1.4.0 
> (http://www.compileonline.com/execute_prolog_online.php)
> :- initialization(main).
> equal(X,Y,Z) :- (X=:=Y -> Z=true; Z=false).
> commutative(X,Y,Z) :- ((equal(X,Y,W), equal(Y,X,W)) -> Z=true; Z=false).
> main :- A = 1,
>          B = 1/(1-2.220446049250313e-16),
>          commutative(A,B,T),
>          write(T).
> %output: true
>
> That seems to be very verbose and inefficient solution.
>
> How can I do it simpler? Is it possible to make it faster?                      
>                 
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