Prolog count list elements higher than n

Question

I\'m kinda new to Prolog so I have a few problems with a certain task. The task is to write a tail recursive predicate count_elems(List,N,Count) condition

Answer 1

Preserve logical-purity with clpfd!

Here's how:

:- use_module(library(clpfd)).

count_elems([],_,0).
count_elems([X|Xs],Z,Count) :-
   X #=< Z,
   count_elems(Xs,Z,Count).
count_elems([X|Xs],Z,Count) :-
   X #> Z,
   Count #= Count0 + 1,
   count_elems(Xs,Z,Count0).

Let's have a look at how versatile count_elems/3 is:

?- count_elems([1,2,3,4,5,4,3,2],2,Count).
Count = 5 ;                                   % leaves useless choicepoint behind
false.

?- count_elems([1,2,3,4,5,4,3,2],X,3).
X = 3 ;
false.

?- count_elems([1,2,3,4,5,4,3,2],X,Count).
Count = 0, X in 5..sup ;
Count = 1, X = 4       ;
Count = 3, X = Count   ;
Count = 5, X = 2       ;
Count = 7, X = 1       ;
Count = 8, X in inf..0 .

---

Edit 2015-05-05

We could also use meta-predicate tcount/3, in combination with a reified version of (#<)/2:

#<(X,Y,Truth) :- integer(X), integer(Y), !, ( X Truth=true ; Truth=false ).
#<(X,Y,true)  :- X #<  Y.
#<(X,Y,false) :- X #>= Y.

Let's run above queries again!

?- tcount(#<(2),[1,2,3,4,5,4,3,2],Count).
Count = 5.                                           % succeeds deterministically

?- tcount(#<(X),[1,2,3,4,5,4,3,2],3).
X = 3 ;
false.

?- tcount(#<(X),[1,2,3,4,5,4,3,2],Count).
Count = 8, X in inf..0 ;
Count = 7, X = 1       ;
Count = 5, X = 2       ;
Count = 3, X = Count   ;
Count = 1, X = 4       ; 
Count = 0, X in 5..sup .

A note regarding efficiency:

• count_elems([1,2,3,4,5,4,3,2],2,Count) left a useless choicepoint behind.
• tcount(#<(2),[1,2,3,4,5,4,3,2],Count) succeeded deterministically.