Prolog - count occurrence of number

Question

I want to write predicate which can count all encountered number:

count(1, [1,0,0,1,0], X).
X = 2.

I tried to write it like:

count(_, [], 0).
count(Num, [H|T], X) :- count(Num, T, X1), Num = H, X is X1 + 1.

Why doesn't work it?

Answer 1

Why doesn't work it?

Prolog is a programming language that often can answer such question directly. Look how I tried out your definition starting with your failing query:

?- count(1, [1,0,0,1,0], X).
   false.
?- count(1, Xs, X).
   Xs = [], X = 0
;  Xs = [1], X = 1
;  Xs = [1,1], X = 2
;  Xs = [1,1,1], X = 3
;  ... .
?- Xs = [_,_,_], count(1, Xs, X).
   Xs = [1,1,1], X = 3.

So first I realized that the query does not work at all, then I generalized the query. I replaced the big list by a variable Xs and said: Prolog, fill in the blanks for me! And Prolog did this and reveals us precisely the cases when it will succeed.

In fact, it only succeeds with lists of 1s only. That is odd. Your definition is too restricted - it correctly counts the 1s in lists where there are only ones, but all other lists are rejected. @coder showed you how to extend your definition.

Here is another one using library(reif) for SICStus|SWI. Alternatively, see tfilter/3.

count(X, Xs, N) :-
   tfilter(=(X), Xs, Ys),
   length(Ys, N).

A definition more in the style of the other definitions:

count(_, [], 0).
count(E, [X|Xs], N0) :-
   if_(E = X, C = 1, C = 0),
   count(E, Xs, N1),
   N0 is N1+C.

And now for some more general uses:

How does a four element list look like that has 3 times a 1 in it?

?- length(L, 4), count(1, L, 3).
   L = [1,1,1,_A], dif(1,_A)
;  L = [1,1,_A,1], dif(1,_A)
;  L = [1,_A,1,1], dif(1,_A)
;  L = [_A,1,1,1], dif(1,_A)
;  false.

So the remaining element must be something different from 1.

That's the fine generality Prolog offers us.

Answer 2

I have decided to add my solution to the list here.

Other solutions here use either explicit unification/failure to unify, or libraries/other functions, but mine uses cuts and implicit unification instead. Note my solution is similar to Ilario's solution but simplifies this using cuts.

count(_, [], 0) :- !.

count(Value, [Value|Tail],Occurrences) :- !, 
    count(Value,Tail,TailOcc), 
    Occurrences is TailOcc+1.

count(Value, [_|Tail], Occurrences) :- count(Value,Tail,Occurrences).

How does this work? And how did you code it?

It is often useful to equate solving a problem like this to solving a proof by induction, with a base case, and then a inductive step which shows how to reduce the problem down.

Line 1 - base case

Line 1 (count(_, [], 0) :- !.) handles the "base case". As we are working on a list, and have to look at each element, the simplest case is zero elements ([]). Therefore, we want a list with zero elements to have no instances of the Value we are looking for.

Note I have replaced Value in the final code with _ - this is because we do not care what value we are looking for if there are no values in the list anyway! Therefore, to avoid a singleton variable we negate it here.

I also added a ! (a cut) after this - as there is only one correct value for the number of occurrences we do not want Prolog to backtrack and fail - therefore we tell Prolog we found the correct value by adding this cut.

Lines 2/3 - inductive step

Lines 2 and 3 handle the "inductive step". This should handle if we have one or more elements in the list we are given. In Prolog we can only directly look at the head of the list, therefore let us look at one element at a time. Therefore, we have two cases - either the value at the head of the list is the Value we are looking for, or it is not.

Line 2

Line 2 (count(Value, [Value|Tail],Occurrences) :- !, count(Value,Tail,TailOcc), Occurrences is TailOcc+1.) handles if the head of our list and the value we are looking for match. Therefore, we simply use the same variable name so Prolog will unify them.

A cut is used as the first step in our solution (which makes each case mutually exclusive, and makes our solution last-call-optimised, by telling Prolog not to try any other rules).

Then, we find out how many instances of our term there are in the rest of the list (call it TailOcc). We don't know how many terms there are in the list we have at the moment, but we know it is one more than there are in the rest of the list (as we have a match).

Once we know how many instances there are in the rest of the list (call this Tail), we can take this value and add 1 to it, then return this as the last value in our count function (call this Occurences).

Line 3

Line 3 (count(Value, [_|Tail], Occurrences) :- count(Value,Tail,Occurrences).) handles if the head of our list and the value we are looking for do not match.

As we used a cut in line 2, this line will only be tried if line 2 fails (i.e. there is no match).

We simply take the number of instances in the rest of the list (the tail) and return this same value without editing it.

Answer 3

You simply let the predicate fail at the unification Num = X. Basically, it's like you don't accept terms which are different from the only one you are counting.

I propose to you this simple solution which uses tail recursion and scans the list in linear time. Despite the length, it's very efficient and elegant, it exploits declarative programming techniques and the backtracking of the Prolog engine.

count(C, L, R) :-
    count(C, L, 0, R).

count(_, [], Acc, Acc).
count(C, [C|Xr], Acc, R) :-
    IncAcc is Acc + 1,
    count(C, Xr, IncAcc, R).
count(C, [X|Xr], Acc, R) :-
    dif(X, C),
    count(C, Xr, Acc, R).

count/3 is the launcher predicate. It takes the term to count, the list and gives to you the result value. The first count/4 is the basic case of the recursion. The second count/4 is executed when the head of the list is unified with the term you are looking for. The third count/4 is reached upon backtracking: If the term doesn’t match, the unification fails, you won't need to increment the accumulator.

Acc allows you to scan the entire list propagating the partial result of the recursive processing. At the end you simply have to return it.

Answer 4

The problem is that as stated by @lurker if condition (or better unification) fails then the predicate will fail. You could make another clause for this purpose, using dif/2 which is pure and defined in the iso:

count(_, [], 0).
count(Num, [H|T], X) :- dif(Num,H), count(Num, T, X).
count(Num, [H|T], X) :- Num = H, count(Num, T, X1), X is X1 + 1.

The above is not the most efficient solution since it leaves many choice points but it is a quick and correct solution.

Answer 5

I solved it myself:

count(_, [], 0).
count(Num, [H|T], X) :- Num \= H, count(Num, T, X).
count(Num, [H|T], X) :- Num = H, count(Num, T, X1), X is X1 + 1.