Question

I have the following snippet of prolog code:

num(0).
num(X) :- num(X1), X is X1 + 1.

fact(0,1) :-!.
fact(X,Y) :- X1 is X-1, fact(X1,Y1), !, Y is Y1 * X.

fact(X) :- num(Y), fact(Y,X).

Can somebody please explain why the following command causes a stack overflow? Thanks in advance.

fact(6).

Answer 1

First, looking at the rules

  num(0).
  num(X) :- num(X1), X is X1 + 1.

the predicate num(Y) will be immediately valid for Y = 0.

Thus the rule

  fact(X) :- num(Y), fact(Y,X).

can be simplified as

  fact(X) :- fact(0,X).

that will find a match for fact(0,1). For X = 6, what happens instead is, as no rule defines a predicate for fact(0,6), a search is started with fact(-1,V1), followed with fact(-2,V2) etc… until a match occurs for a fact(-value, Var) where the local result would be the Var found.

This cannot happen, and an infinite loop consumes the whole stack, until an error is triggered.