Hello and thank you for taking the time to read this question. I have the following problem:
Given a list that returned the number of numbers greater than X Example:
greater (4, [1,2,3,4,5,6], N) Result. N = 2 My code is:
greater(0,[],0):-!, fail.
greater(N,[N],1).
greater(N,[H|Q],X):-H>N,greater(Q,N,X),X is X+1.
The problem is that PROLOG only returns False but not the value of X.
I hope you can explain to me what I am doing wrong, I thank you in advance for your
Since a predicate call fails in Prolog if there are no success paths for that call, then the following predicate clause serves no purpose. You can remove it.
greater(0,[],0):-!, fail.
Your next clause is your recursive base case and is incorrectly formulated:
greater(N,[N],1).
This succeeds even though it violates your condition that you want to count elements in the list that are greater than N. N is not greater than N. What should this clause look like if you want greater(N, [X], 1). to succeed?
In your recursive clause, you have a problem:
greater(N,[H|Q],X):-H>N,greater(Q,N,X),X is X+1.
X is X+1 will always fail because the value of X cannot possibly ever be the same as the value X+1. That is, there is no number that is equal to itself plus one. You need to use an auxiliary variable:
greater(N,[H|Q],X):-H>N,greater(Q,N,X1),X is X1+1.
Finally, you're missing the case when H =< N:
greater(N,[H|Q],X):-H=<N, ....
What should this clause look like?