Trim the tree at the specified depth. In Prolog

问题

Arguments: arbitrary binary tree; necessary depth; result tree.

Result:

?- pred(s(f(b(m,k),a),t(a,g)),2,X). 
X = s(f,t) yes 
?- pred(s(f(b(m,k),a),t(a,g)),3,X). 
X = s(f(b,a),t(a,g)) yes 
?-

Can someone help me with it?

回答1:

Let's rearrange your examples:

?- pred( s( f(b(m,k),a), t(a,g)), 3, X). 
X = s(f(b,a),t(a,g)) yes

?- pred( s( f(b(m,k),a), t(a,g)), 2, X). 
X = s(f,t) yes 

?- pred( s( f(b(m,k),a), t(a,g)), 1, X). 
X = s yes 

?- pred( s( f(b(m,k),a), t(a,g)), 0, X). 
no

Now it's clear what needs to be done, isn't it?

Another piece to this puzzle is the so-called "univ" predicate, =..,

9 ?- s( f(b(m,k),a), t(a,g)) =.. [A, B, C].
A = s,
B = f(b(m, k), a),
C = t(a, g).

10 ?- X =.. [s, f(b(m,k),a), t(a,g)].
X = s(f(b(m, k), a), t(a, g)).

11 ?- X =.. [s, f(b,a), t(a,g)].
X = s(f(b, a), t(a, g)).

12 ?- X =.. [s, f, t].
X = s(f, t).

13 ?- X =.. [s].
X = s.

14 ?- s =.. X.
X = [s].

That's how you can take apart your data, and build it up again.

Lastly, you will need to use recursion:

recursion(       In, Out) :-
  base_relation( In, Out).

recursion(       In, Out) :-
  constituents(  In,     SelfSimilarParts,                 LeftOvers),
  maplist( recursion,     SelfSimilarParts, InterimResults),
  constituents(      Out,                  InterimResults, LeftOvers).

来源：https://stackoverflow.com/questions/50871893/trim-the-tree-at-the-specified-depth-in-prolog