I have 2 lists with random number of elemets. Eg A=[1,2,4,5] and B=[1,2,3]. Result should be 2.
Code that I tried:
domains
Numbers1 = integer*
Numbers2 = integer*
int_list=integer*
predicates
nondeterm prinadl(integer, int_list)
clauses
//here going the code that read number that I've entered, and according to entered numer,programm should do something
answer(T):- T=5,
P = 0,
write ("Enter the 1st list"), readterm (int_list, L),
write ("Enter the 2nd list"), readterm (int_list, L2),
L2 = [H|V], prinadl(H, L), P1 = P + 1,
write(L2, P1, V).
prinadl (X, L):- L=[X|_], !.
prinadl (X, L):- L=[_|T], prinadl (X, T).
I'm totally new with prolog. Can you please say me where I'm wrong? All I need is to get number of matches printed to the console.
Thanks in advance.
This answer is based on two things: first, guesswork. second, if_/3 by @false.
Let's define
the meta-predicate count_left_while2/4.
count_left_while2(P_2,Xs,Ys,N)
counts
the number N of corresponding list items in Xs and Ys fulfilling P_2. Proceeding from left to right, count_left_while2 stops at the first two items not satisfying P_2. It also stops when one list is empty, but the other one is not.
:- use_module(library(clpfd)).
:- meta_predicate count_left_while2(2,?,?,?).
count_left_while2(P_2,Xs,Ys,N) :-
N #>= 0,
list_list_countleft_while(Xs,Ys,N,P_2).
nil_or_cons([]).
nil_or_cons([_|_]).
:- meta_predicate list_list_countleft_while(?,?,?,2).
list_list_countleft_while([],Xs,0,_) :-
nil_or_cons(Xs).
list_list_countleft_while([X|Xs],Ys,N,P_2) :-
list_list_prev_countleft_while(Ys,Xs,X,N,P_2).
:- meta_predicate list_list_prev_countleft_while(?,?,?,?,2).
list_list_prev_countleft_while([],_,_,0,_).
list_list_prev_countleft_while([Y|Ys],Xs,X,N,P_2) :-
if_(call(P_2,X,Y),
( N0 #>= 0, N #= N0+1, list_list_countleft_while(Xs,Ys,N0,P_2) ),
N = 0).
Let's use it in combination with reified term equality predicate (=)/3, like this:
:- count_left_while2(=,[1,2,4,5],[1,2,3],N).
N = 2.
