I want to write the equivalent psudo-function in prolog:
function underhundred(X){
if (X >= 0 && X <=100) return 1;
else return 0;
}
I tried writing this but it does not compile:
underhundred(X,L) :- L is (X => 0, X =< 100 -> L = 1; L = 0) .
What would be the proper way of writing this without using prolog between predicate?
If you indeed want to use the goals L=1 and L=0 and X is an integer, use clpfd!
:- use_module(library(clpfd)).
Reification works like a charm!
?- X in 0..100 #<==> L.
L in 0..1, X in 0..100#<==>L.
What if X gets instantiated?
?- X in 0..100 #<==> L, X=(-7).
L = 0, X = -7. % out of bounds: -7 < 0 =< 100
?- X in 0..100 #<==> L, X=55.
L = 1, X = 55. % within bounds: 0 =< 55 =< 100
?- X in 0..100 #<==> L, X=111.
L = 0, X = 111. % out of bounds: 0 =< 100 < 111
A Prolog query succeeds or fails. If it succeeds it will return the bindings it made to be true.
You can write this predicate using clpfd as:
:-use_module(library(clpfd)).
under_hundred_clpfd(X):-
X in 0..100.
(You might prefer a name such as between_0_100?, if you literally want under 100 then you can use X in inf..99). Some queries:
?-under_hundred_clpfd(5).
true.
?-under_hundred_clpfd(101).
false.
?-under_hundred_clpfd(X).
X in 0..100.
A traditional way to write this is:
under_hundred(X):-
X>=0,
X=<100.
But this way does not work for uninstantiated variables.
?-under_hundred(X).
ERROR: >/2: Arguments are not sufficiently instantiated
So like you say you might have to put a between/3 or length/2 goal to get a solution or similar construct.
underhundred(X):-
length(_,X),
X>=0,
X=<100.
This is not a very good solution as on back tracking it will get into an infinite loop. between/3 behaves better but you don't want it :).
If the main point of the question is how to write an if-then-else construct in Prolog, then a reasonable answer along the lines of the proposed definition is:
underhundred(X,L) :-
( X >= 0, X =< 100 )
-> L = 1
; L = 0.
This will only be useful if underhundred(X,L) is called after X has been sufficiently instantiated.
