The following Prolog program defines a predicate fact/2
for computing the factorial of an integer in successor arithmetics:
fact(0, s(0)).
fact(s(X), Y) :-
fact(X, Z),
prod(s(X), Z, Y).
prod(0, _, 0).
prod(s(U), V, W) :-
sum(V, X, W),
prod(V, U, X).
sum(0, Y, Y).
sum(s(X), Y, s(Z)) :-
sum(X, Y, Z).
It works with queries in this argument mode:
?- fact(s(0), s(0)).
true
; false.
It also works with queries in this argument mode:
?- fact(s(0), Y).
Y = s(0)
; false.
It also works with queries in this argument mode:
?- fact(X, Y).
X = 0, Y = s(0)
; X = Y, Y = s(0)
; X = Y, Y = s(s(0))
; X = s(s(s(0))), Y = s(s(s(s(s(s(0))))))
; …
But it exhausts resources with queries in this argument mode:
?- fact(X, s(0)).
X = 0
; X = s(0)
;
Stack limit (0.2Gb) exceeded
Stack sizes: local: 4Kb, global: 0.2Gb, trail: 0Kb
Stack depth: 2,503,730, last-call: 100%, Choice points: 13
In:
[2,503,730] sum('<garbage_collected>', _1328, _1330)
[38] prod('<garbage_collected>', <compound s/1>, '<garbage_collected>')
[33] fact('<garbage_collected>', <compound s/1>)
[32] fact('<garbage_collected>', <compound s/1>)
[31] swish_trace:swish_call('<garbage_collected>')
How to implement the factorial sequence in successor arithmetics for all argument modes?
The first question must be why? A failure-slice helps to understand the problem:
fact(0, s(0)) :- false. fact(s(X), Y) :- fact(X, Z), false,prod(s(X), Z, Y).
This fragment alone terminates only if the first argument is given. If it is not, then there is no way to prevent non-termination, as Y
is not restricted in any way in the visible part. So we have to change that part. A simple way is to observe that the second argument continually increases. In fact it grows quite fast, but for the sake of termination, one is enough:
fact2(N, F) :-
fact2(N, F, F).
fact2(0, s(0), _).
fact2(s(X), Y, s(B)) :- fact2(X, Z, B), prod(s(X), Z, Y).
And, should I add, this can be even proved.
fact2(A,B)terminates_if b(A);b(B).
% optimal. loops found: [fact2(s(_),s(_))]. NTI took 0ms,73i,73i
But, there is a caveat...
If only
F
is known, the program will now require temporally space proprotional to |F
|! That is not an exclamation point but a factorial sign...