I have currently started with PROLOG and I want to write a predicate which checks if a given object is in this list or not. If the object is in the list the predicate should return the index of the element. If the element is not found it should return 0.
It should work like this: find(3,[1,4,5,3,2,3],N). -> yes. N / 4
find(2,[1,3,4,5,6,7],N). -> yes. N / 0
But I have problems with counting up the index N and maybe someone here can help. I've seen many different ways on how to traverse a list but I found so many different ways and I wasn't able to understand how they work. I would be really happy if someone can provide a solution and explain how it works and why.
This is what I wrote so far:
find(X, [X|TAIL], N) :- N is 1, write(N).
find(X, [], N) :- N is 0, write(N).
find(X, [_|TAIL], N) :- find(X, TAIL, N + 1).
It is working for the basecases:
find(a, [a, b, c, d, e, f, g], N) -> yes. N / 1.
find(j, [a, b, c, d, e, f, g], N) -> yes. N / 0.
But when it is starting with recursion It is not working anymore and I don't understand what's going wrong.
For the recursion case it gives me this: find(b, [a, b, c, d, e, f, g], N) -> no.
I am thankful for every answer and every comment!
yes. N = 0. Examples: ?- find(a, [], N) -> yes. N = 0. and ?- find(a, [b,c,d], N) -> yes. N = 0.?- find(a, [a, b, c], N) -> yes. N = 1 and ?- find(a, [b,c,a,d], N) -> yes. N = 3.?- find(a, [a,b,c,a], N) -> yes. N = 1.?- find(a, [a, b,c], 0) and ?- find(a, [b, a, c], 0) and every other query where a is in the list should be answered with no.?- find(a, [a, b, c, a], 4) should be answered with no because the a with index 4 is not the first apperance of a.Using the argument order of nth1/3
nth1_once_else_0(Nth1, Lst, Elem) :-
nth1_once_else_0_when_(Lst, Elem, 1, Nth1).
nth1_once_else_0_thaw_([H|_], Elem, Upto, Nth1) :-
Upto == Nth1,
!,
H = Elem.
nth1_once_else_0_thaw_(L, _Elem, _Upto, Nth1) :-
L == [],
!,
Nth1 = 0.
nth1_once_else_0_thaw_(L, Elem, _Upto, Nth1) :-
Nth1 == 0,
!,
nth1_once_else_0_thaw_0_(L, Elem).
nth1_once_else_0_thaw_(_L, _Elem, Upto, Nth1) :-
% If gone past Nth1, fail
nonvar(Nth1),
Nth1 =< Upto,
!,
fail.
nth1_once_else_0_thaw_([H|T], Elem, Upto, Nth1) :-
( nonvar(Nth1),
Nth1 > Upto -> true
; H \= Elem
),
!,
% Elements must be different
dif(H, Elem),
Upto1 is Upto + 1,
nth1_once_else_0_when_(T, Elem, Upto1, Nth1).
nth1_once_else_0_thaw_([H|_], Elem, Upto, Nth1) :-
?=(H, Elem),
% Able to compare
H = Elem,
!,
% Elements match
Upto = Nth1.
nth1_once_else_0_thaw_([H|T], Elem, Upto, Nth1) :-
% Wait for possible comparison
when(
(?=(H, Elem) ; nonvar(Nth1)),
nth1_once_else_0_thaw_([H|T], Elem, Upto, Nth1)
).
nth1_once_else_0_when_(L, Elem, Upto, Nth1) :-
var(L),
!,
when(
(nonvar(L), nonvar(Elem) ; nonvar(Nth1)),
nth1_once_else_0_thaw_(L, Elem, Upto, Nth1)
).
nth1_once_else_0_when_([], _Elem, _Upto, 0).
nth1_once_else_0_when_([H|T], Elem, Upto, Nth1) :-
when(
(nonvar(H), nonvar(Elem) ; nonvar(Nth1)),
nth1_once_else_0_thaw_([H|T], Elem, Upto, Nth1)
).
% Remainder of list does not match
nth1_once_else_0_thaw_0_(L, Elem) :-
var(L),
!,
freeze(L, nth1_once_else_0_thaw_0_(L, Elem)).
nth1_once_else_0_thaw_0_([], _Elem).
nth1_once_else_0_thaw_0_([H|T], Elem) :-
dif(H, Elem),
freeze(T, nth1_once_else_0_thaw_0_(T, Elem)).
Using swi-prolog's unit test ability:
:- begin_tests(nth1_once_else_0).
test(1) :-
nth1_once_else_0(2, [a, b], B), B == b.
test(2) :-
nth1_once_else_0(N, [a, b, c, d, e, f, g], a), N == 1.
test(3) :-
nth1_once_else_0(N, [a, b, c, d, e, f, g], j), N == 0.
test(4) :-
nth1_once_else_0(N, [a, b, c, d, e, f, g], b), N == 2.
test(5) :-
nth1_once_else_0(N, [], a), N == 0.
test(6) :-
nth1_once_else_0(N, [], _), N == 0.
test(7) :-
nth1_once_else_0(N, [a, b, c, a, b, c], a), N == 1.
test(8) :-
\+ nth1_once_else_0(6, [a, b, c, a, b, c], c).
test(9) :-
nth1_once_else_0(N, L, b), L = [a, b], N == 2.
test(10) :-
\+ nth1_once_else_0(0, [a, b], b).
test(11) :-
nth1_once_else_0(N, [a, b, c, a], a), N == 1.
test(12) :-
nth1_once_else_0(N, [a, b, c], d), N == 0.
test(13) :-
nth1_once_else_0(N, [a, b], X), X = b, N == 2.
test(14) :-
nth1_once_else_0(N, L, X), L = [a, b|_], X = b, N == 2.
test(15) :-
nth1_once_else_0(N, L, X), L = [a, b|_], N = 2, X == b.
test(16) :-
nth1_once_else_0(N, L, X), L = [a, b|_], X = z, N = 0.
test(17) :-
L = [a, a], nth1_once_else_0(1, L, a), \+ nth1_once_else_0(2, L, a).
test(18) :-
nth1_once_else_0(N, L, X), L = [a, b|_], L = [a,b,c], N = 0, \+ X = a.
test(19) :-
nth1_once_else_0(N, L, X), L = [a, b|_], X = z, L = [a, b], N == 0.
test(20) :-
nth1_once_else_0(N, L, X), nth1(2, L, b), nth1(1, L, a), X = z, L = [a, b], N == 0.
test(21) :-
nth1_once_else_0(N, L, X), nth1(2, L, b), nth1(1, L, a), L = [a, b], X = a, N == 1.
test(22) :-
nth1_once_else_0(N, L, X), X = a, nth1(2, L, b), nth1(1, L, a), N == 1.
test(23) :-
nth1_once_else_0(N, L, X), X = b, nth1(2, L, b), nth1(1, L, a), N == 2.
test(24) :-
nth1_once_else_0(3, L, c), nth1(3, L, C), C == c.
test(25) :-
nth1_once_else_0(3, L, C), C = c, nth1(3, L, Z), Z == c.
test(26) :-
nth1_once_else_0(N, L, c), N = 3, nth1(N, L, C), C == c.
test(27) :-
nth1_once_else_0(N, L, C), C = c, N = 3, nth1(N, L, Z), Z == c.
test(28) :-
nth1_once_else_0(N, [a, b, c|_], c), N == 3.
test(29) :-
\+ nth1_once_else_0(3, [_, _], z).
test(30) :-
nth1_once_else_0(N, [-_, -_], +_), N == 0.
test(31) :-
nth1_once_else_0(N, L, X), nth1(2, L, b), nth1(1, L, a), X = a, N == 1.
test(32) :-
nth1_once_else_0(N, L, f(b)), L = [f(Z)|T], Z = g(_), T = [f(B)|_], B = b, N == 2.
test(33) :-
nth1_once_else_0(N, L, f(a)), L = [f(A)|_], A = a, N == 1.
test(34) :-
nth1_once_else_0(N, L, f(b)), L = [f(_)|_], N = 2, nth1(2, L, B), B == f(b).
:- end_tests(nth1_once_else_0).
Result:
?- run_tests.
% All 34 tests passed