Prolog recursive subtraction

Question

I picked up prolog a couple of days ago and I 'm kind of stuck to this question. I want to subtract a number recursively until that number becomes less than 0. In pseudocode that would be like:

N:=0
while(Y>=X)
{
    Y := Y-X
    N := N+1
    Y := Y+2
}

So for example if I have Y=20 and X=10 then we would get N=2 and Y=4.

Any ideas? Thanks in advance. Any help appreciated. I'm using SWI Prolog.

EDIT 1 What I've accomplished so far is(although I'm not sure even if its correct):

sufficient(X, Y, M, N, F) :- 
    F is Y-X,
    Y>=X,
    plus(M, 1, N),
    sufficient(X, F, N, N, F).

I have problem finding my base case, I'm confused on how to implement it. Also, in the sufficient I have implemented, obviously when Y<X it terminates returning false. Is there a way to get the N and F before terminating? I am feeling that I am not thinking the "prolog" way, since I am mostly used on C and that vagues my thinking. Thanks.

EDIT 2

I have found my base case and I can stop recursion however, I can't manage to ge the correct values. My code:

sufficient(X, Y, M, N, F) :- Y<X.
sufficient(X, Y, M, N, F) :- 
   F is Y-X,
   plus(M, 1, N),
   sufficient(X, F, N, D, E).

Thing is after the first recursion, if for example I call sufficient as sufficient(10,21,0,N,F). from the swi prolog command prompt, I 'll get N=1 and F=11. That happens because I make 2 new variables D and E. If I don't make those 2 new variables(D and E), and at the 3rd sufficient in the code I call N and F instead of D and E at the line F is Y-X, I get a false, because F is 11 and Y-X is 1. Do I have to set the a subtraction function myself, since F is Y-X is not exactly a subtraction? Any ideas on how to do it?

Try successor-arithmetics first! You can define addition there, which is the very same as subtraction! — false
Take nat_nat_sum/3: nat_nat_dif(A, B, D) :- nat_nat_sum(B, D, A). — false
If X and Y are known, successor arithmetic won't be needed; is is sufficient. — Topological Sort

Topological Sort Topological Sort · Accepted Answer · 2016-12-02T17:41:22

All recursive functions need at least one base case. In what circumstance should your function say, OK, I have the answer, no need to recurse?

It would be the case in which your pseudocode loop is done, right?

Usually we write it in this format:

factorial(0,1).                          % The factorial of 0 is 1.
factorial(N,Factorial) :-
   N>0,                                   % You may need to test applicability 
                                          %    of this recursive clause
   NMinus1 is N-1,                        % maybe some setup
   factorial(NMinus1,FactorialOfNMinus1), %recursive call
   Factorial is N*FactorialOfNMinus1).    %and maybe some code after

I wouldn't want to do your homework for you, but this should get you going:

sufficient(X,Y,M,N,F) :- %whatever condition means you're done, 
                         % and results = whatever they should
sufficient(X,Y,M,N,F) :- %whatever condition means you aren't done
                         % and setting up results w/ a recursive call

One more hint: looks like M is a temporary variable and need not be a parameter.

Prolog recursive subtraction

1 Answers