Prolog Cut operator in the first predicate

Question

I have the following base in Prolog:

holiday(friday,may1).
weather(friday,fair).
weather(saturday,fair).
weather(sunday,fair).
weekend(saturday).
weekend(sunday).

picnic(Day) :- !,weather(Day,fair), weekend(Day).
picnic(Day) :- holiday(Day,may1).

When I run picnic(When). I get the following trace:

[trace]  ?- picnic(When).
   Call: (6) picnic(_G716) ? creep
   Call: (7) weather(_G716, fair) ? creep
   Exit: (7) weather(friday, fair) ? creep
   Call: (7) weekend(friday) ? creep
   Fail: (7) weekend(friday) ? creep
   Redo: (7) weather(_G716, fair) ? creep
   Exit: (7) weather(saturday, fair) ? creep
   Call: (7) weekend(saturday) ? creep
   Exit: (7) weekend(saturday) ? creep
   Exit: (6) picnic(saturday) ? creep
When = saturday ;
   Redo: (7) weather(_G716, fair) ? creep
   Exit: (7) weather(sunday, fair) ? creep
   Call: (7) weekend(sunday) ? creep
   Exit: (7) weekend(sunday) ? creep
   Exit: (6) picnic(sunday) ? creep
When = sunday.

My doubt is: the cut operator, as I know, should stop to search alternatives when the predicates to the left of the ! signal are true. What's the meaning of the signal in the first position? Why does the interpreter keep searching for another values that can turn the other predicates true?

CapelliC CapelliC · Accepted Answer · 2013-04-02T02:53:23

The effect of that cut is that your second picnic/1 rule will be totally ignored. It will never have a chance to fire during your program lifetime.

But backtracking is still at work among available alternatives (goals to the right of the cut), and you can clearly observe them in your trace.

Prolog Cut operator in the first predicate

2 Answers