列表列表-递归对序言

Question

how can I put all the pairs in a one list?

I have -

pair([H|T], [H,E]) :- member(E, T).
pair([_|T], P) :- pair(T, P).

and I want the answer will be a list of pairs.

so I try -

listPairs([],[Res1]):-
      Res1=[].
listPairs(L,[Res2]):-
    L=[H|T],
    append(pair([H|T],[Res2]),listPairs(T,[Res2])).

but I miss something of the lists.. because it is not compile.

Answer 1

I don't really see what you aim to do with append/2 . You can definitely not put goals that should be called in the append/3 because in that case, they will be seen as functors.

You can however easily use the findall/3 predicate which is implemented in nearly all Prolog systems (well at least all I have seen that are actively used):

listPairs(L,Pairs) :-
    findall(Pair,pair(L,Pair),Pairs).

findall/3 works as follows:

findall(Format,Goal,List).

Here you specify a Goal (here pair(L,Pair) ) and Prolog will call the Goal . Each time the Goal succeed, Prolog will suspend the interpreter and add en element formatted as Format to the List . When the Goal fails, the list of all Format s is returned.

If I run the query, I get:

?- listPairs([1,a,2,'5'],Pairs).
Pairs = [[1, a], [1, 2], [1, '5'], [a, 2], [a, '5'], [2, '5']].

Note that L must be grounded (as list, meaning without uninstantiated tail, an uninstantiated element is fine), since otherwise the number of pair/2 elements that you will generate is infinite and thus you will run out of global stack (or memory in general).