Prolog counting using s(0) and p(0)

Question

I am having some issues with a part of my revision for my prolog exam.

I need to create a recursive statement that will be called simplify/2. An example use would be

simplify(s(p(s(0))),Z) 

which would result in Z being s(0). S stands for successor and P predecessor. So: s(0) is 1, s(s(0)) is 2 and p(0) is -1 etc. and p(s(p(p(0)))) would be p(p(0)).

The code I initially had was

check(s(0),s(0)).
check(s(X),s(0)) :- check(X,s(s(0))).
check(p(X),s(0)) :- check(X,0).

But this clearly doesn't work as the second part needs to be kept as a variable that is added on to itself during the recursive call. I'll have another look at it in around 30 minutes because my head is fried at the moment.

Answer 1

My attempt:

simplify(X, Z) :-
    simplify(X, 0, Z).
simplify(0, Z, Z).
simplify(s(X), 0, Z) :- simplify(X, s(0), Z).
simplify(p(X), 0, Z) :- simplify(X, p(0), Z).
simplify(p(X), s(Y), Z) :- simplify(X, Y, Z).
simplify(s(X), p(Y), Z) :- simplify(X, Y, Z).
simplify(s(X), s(Y), Z) :- simplify(X, s(s(Y)), Z).
simplify(p(X), p(Y), Z) :- simplify(X, p(p(Y)), Z).

Update - shorter version:

simplify(X, Z) :-
    simplify(X, 0, Z).
simplify(0, Z, Z).
simplify(p(X), s(Y), Z) :- simplify(X, Y, Z).
simplify(s(X), p(Y), Z) :- simplify(X, Y, Z).
simplify(s(X), Y, Z) :- Y \= p(_), simplify(X, s(Y), Z).
simplify(p(X), Y, Z) :- Y \= s(_), simplify(X, p(Y), Z).

Some tests:

?- simplify(s(p(s(0))), Z).
Z = s(0) 

?- simplify(p(s(p(p(0)))), Z).
Z = p(p(0)) 

?- simplify(p(p(s(s(0)))), Z).
Z = 0 

Answer 2

z(0).
z(s(X)) :-
   z(X).
z(p(X)) :-
   z(X).

z_canonized(Z, C) :-
   z_canonized(Z, 0, C).

z_canonized(0, C,C).
z_canonized(s(N), C0,C) :-
   z_succ(C0,C1),
   z_canonized(N, C1,C).
z_canonized(p(N), C0,C) :-
   z_pred(C0,C1),
   z_canonized(N, C1,C).

z_succ(0,s(0)).
z_succ(s(X),s(s(X))). % was: z_succ(X,s(X)) :- ( X = 0 ; X = s(_) ).
z_succ(p(X),X).

z_pred(0,p(0)).
z_pred(p(X),p(p(X))).
z_pred(s(X),X).