Using Pattern Matching in Prolog to Find Submultisets

Question

I am new to prolog and I was wondering if anyone could help me with this problem. The problem: given the integers 1,2,3,4, and the predicates mult/2, div/2, div/2, minus/2, and minus/2, and eval/2, I need to write a predicate solutions/1 that, when called like this:

?- solutions(L).

it should terminate with the variable L unified to a list of expressions with value 6. Expressions are of the form:

X, Y, exp/2

But my code is not working. I have two versions. The first freezes up SWI-Prolog, not returning any answer after I type a period, and not letting me evaluate anything else afterward:

eval(1,1.0).
eval(2,2.0).
eval(3,3.0).
eval(4,4.0).

eval(mult(X,Y),Z) :-
    eval(X,A),
    eval(Y,B),
    Z is A*B.

eval(div(X,Y),Z) :-
    eval(X,A),
    eval(Y,B),
    Z is A/B.

eval(minus(X,Y),Z) :-
    eval(X,A),
    eval(Y,B),
    Z is A-B.

solutions(L) :-
    setof(X,eval(X,6),L),
    print(L).

The second version just returns false when I type ?- solutions(L).:

solutions(L) :-
    setof([exp,X,Y],eval(exp(X,Y),6),L),
    print(L).

Thank you so much for taking the time to help!

First version doesn't freeze but got into an infinite loop due to recursion that won't terminate. `eval` calls `eval` endlessly since you're missing the logic for it to terminate. You could try `trace.` and run your query again and you'll see after several steps where it starts to repeat itself. In the second version, in the code you're showing, I see no way for `eval(exp(X,Y), 6)` to succeed since there's no fact or predicate head that matches. `exp` isn't shown anywhere. — lurker, Dec 13 '15 at 03:13
@lurker thanks for the feedback! That makes sense; I was pretty sure that it wouldn't be able to find `exp` anywhere, but I didn't know how else to pattern match to the predicates mult/2, div/2, etc.,. as this is my very first time coding in prolog. If you have any suggestions about how to do that, please lmk! — Maclafel7, Dec 13 '15 at 05:37

Sahil Jain · Answer 1 · 2015-12-13T09:07:07.747

0

The problem is that your code is going in infinite recursion with eval/2 predicate.

You can try this solution:

num(1).
num(2).
num(3).
num(4).

eval(mult(A,B),Z) :-
    num(A),
    num(B),
    Z is A*B.

eval(div(A,B),Z) :-
    num(A),
    num(B),
    Z is A/B.

eval(minus(A,B),Z) :-
    num(A),
    num(B),
    Z is A-B.

test(L) :-
    setof(X,eval(X,6),L),
    print(L).

Which yields:

?- test(L).
[mult(2,3),mult(3,2)]
L = [mult(2, 3), mult(3, 2)].

edited Dec 13 '15 at 09:07

answered Dec 13 '15 at 03:56

Sahil Jain

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Hi @Sahil thanks for the answer! I just tried this, and it won't compile. I think because there is no base case for eval/2... – Maclafel7 Dec 13 '15 at 05:41
@Maclafel7 : Check the updated code. This is what I get on compiling the code. – Sahil Jain Dec 13 '15 at 09:07

score 0 · Accepted Answer · answered Dec 13 '15 at 06:45

Maybe you're looking for something like

solutions(L) :-
    Ns = [1,2,3,4],
    Ex = [*,/,-],
    findall((X,Y,E),
       (member(X,Ns),member(Y,Ns),member(E,Ex),F=..[E,X,Y],6=:=F),
       L).

that yields

?- solutions(L).
L = [(2, 3,  (*)),  (3, 2,  (*))].

Expressions are usually recursive, that is, arguments could be expressions instead of plain numbers. But then, in my opinion your problem is underspecified, as we need criteria to stop the infinite flow of solutions resulting - for instance - by repeated application of operations that don't change the value. Like multiply or divide by 1.

Using Pattern Matching in Prolog to Find Submultisets

2 Answers2