在指定深度修剪树.在序言中 [英] Trim the tree at the specified depth. In Prolog

问题描述

参数:任意二叉树；必要的深度；结果树.

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

结果:

?- 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 
?-

有人可以帮我吗?

推荐答案

让我们重新排列您的示例:

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

现在很清楚需要做什么，不是吗?

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

此难题的另一个组成部分是所谓的"univ"谓词， =.. ，

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].

这就是您可以拆分数据并再次构建的方法.

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

最后，您将需要使用递归:

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).

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