Declarative uses of memberchk/2

Question

memberchk/2 is a commonly defined predicate that is defined in terms of member/2 like so:

memberchk(X, Xs) :-
   once(member(X, Xs)         


        

Answer 1

Here is a well-known example use of member/2 that cannot be represented by memberd/2: bridge.pl the bridge scheduling problem given by Pascal Van Hentenryck.

In the setup phase member/2 is used:

setup(K,Ende,Disj):-
    jobs(L),
    make_vars(L,K),
    member([stop,_,Ende],K),
    ....

So here, effectively the first element in the three-element list is used to select a particular task whereas memberd/2 uses the entire element for comparison. As a consequence this setup/3 leaves open a lot of choicepoints (actually, 219). Some (like SICStus) use memberchk/2 in that situation, thereby risking non-monotonicity.

Using the following pure replacement, all choicepoints are avoided.

member3l([N,D,A], Plan) :-
   tmember(l3_t(N,D,A),  Plan).

l3_t(N,D,A, X, T) :-
   X = [Ni|_],
   if_(N = Ni, ( X=[N,D,A], T = true ), T = false ).

tmember(P_2, [X|Xs]) :-
   if_( call(P_2, X), true, tmember(P_2, Xs) ).

Alternatively using library(lambda):

member3li([N,Nd,Na], Plan) :-
   tmember([N,Nd,Na]+\X^T^
       (  X=[Nk|_],
          if_( Nk = N, ( X=[N,Nd,Na], T = true ), T = false ) ),
      Plan).

Other uses of tmember/2:

old_member(X, Xs) :-
   tmember( X+\E^T^( X = E, T = true ; T = false ), Xs).

old_memberd(X, Xs) :-
   tmember(=(X), Xs).

Here is a more compact representation:

member3l([N,D,A], Plan) :-
   tmember({N,D,A}+\[Ni,Di,Ai]^cond_t(N = Ni, [D,A] = [Di,Ai] ), Plan).

Using library(lambda)and cond_t/3:

cond_t(If_1, Then_0, T) :-
   if_(If_1, ( Then_0, T = true ), T = false ).