This is the question for one of my assignments:
Write
repCount(L, X, N)
which is true whenN
is the number of occurrences ofX
in listL
.
Here's my code where I try to tackle the problem recursively:
repCount([], X, N) :-
N is 0.
repCount([H|T], X, N) :-
count([H|T], X, N).
count([], X, 0).
count([H|T], X, N) :-
count(T, X, N1),
X =:= H,
N is N1 + 1.
And it works when I supply a list full of identical numbers like this:
?- repCount([2,2,2], 2, N).
N = 3.
But if I supply a list with at least one different value:
?- repCount([2,2,22], 2, N).
false.
It returns false
. I cannot figure out why this happens or how to change it to 'skip' the non-matching value, rather than declare the whole thing false. Any input is appreciated.
count([H|T], X, N):- count(T, X, N1), X=:=H, N is N1 + 1.
here you declare that N should be N1+1 if X is H; however you do not define what should happen if X is not H (basically missing an else clause) this should work:
count([H|T], X, N):-
count(T, X, N1),
(X=:=H->
N is N1 + 1
; N is N1).
another way would be:
count([H|T], X, N):- count(T, X, N1), X=:=H, N is N1 + 1.
count([H|T], X, N):- X=\=H, count(T, X, N1), N is N1.
but this is inefficient since count(T,X,N1) will be called twice if X is not H. we can fix this by doing the check in the head of the clause:
count([H|T], H, N):- count(T, X, N1), N is N1 + 1.
count([H|T], X, N):- count(T, X, N1), N is N1.
or simply: count([H|T], H, N):- count(T, X, N1), N is N1 + 1.
count([H|T], X, N1):- X=\=H, count(T, X, N1).
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