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列表中的 Prolog 递归,最后一个元素

更新时间:2023-11-10 10:37:34

您的原始版本更易于阅读.特别是,递归规则读取 - 从右到左读取它

Your original version is much simpler to read. In particular, the recursive rule reads - reading it right-to-left

last_but_one(X, [_|T]) :- last_but_one(X, T).
                          ^^^^^^^^^^
                              provided X is the lbo-element in T

                       ^^  then, it follows, that (that's an arrow!)
^^^^^^^^^^^^^^^^^^^^^^
      X is also the lbo-element of T with one more element

换句话说:如果您在给定列表T中已经有一个lbo元素,那么您可以构造新列表,其中包含前面也具有相同lbo元素的任何其他元素.

In other words: If you have already an lbo-element in a given list T, then you can construct new lists with any further elements in front that also have the very same lbo-element.

人们可能会争论哪个版本在效率方面更可取.如果你真的很喜欢,那就去吧:

One might debate which version is preferable as to efficiency. If you are really into that, rather take:

last_but_one_f1(E, Es) :-
   Es = [_,_|Xs],
   xs_es_lbo(Xs, Es, E).

xs_es_lbo([], [E|_], E).
xs_es_lbo([_|Xs], [_|Es], E) :-
   xs_es_lbo(Xs, Es, E).

甚至:

last_but_one_f2(E, [F,G|Es]) :-
    es_f_g(Es, F, G, E).

es_f_g([], E, _, E).
es_f_g([G|Es], _, F, E) :-
   es_f_g(Es, F, G, E).

永远不要忘记一般测试:

Never forget general testing:

| ?- last_but_one(X, Es).
Es = [X,_A] ? ;
Es = [_A,X,_B] ? ;
Es = [_A,_B,X,_C] ? ;
Es = [_A,_B,_C,X,_D] ? ;
Es = [_A,_B,_C,_D,X,_E] ? ;
Es = [_A,_B,_C,_D,_E,X,_F] ? ...

以下是我的 olde labtop 上的一些基准测试:

And here are some benchmarks on my olde labtop:

          SICStus     SWI
          4.3.2     7.3.20-1
    --------------+----------+--------
    you   0.850s  |   3.616s |  4.25×
    they  0.900s  |  16.481s | 18.31×
    f1    0.160s  |   1.625s | 10.16×
    f2    0.090s  |   1.449s | 16.10×
    mat   0.880s  |   4.390s |  4.99×
    dcg   3.670s  |   7.896s |  2.15×
    dcgx  1.000s  |   7.885s |  7.89×
    ap    1.200s  |   4.669s |  3.89×

差异很大的原因是 f1f2 都完全确定地运行,没有创建任何选择点.

The reason for the big difference is that both f1 and f2 run purely determinate without any creation of a choicepoint.

使用

bench_last :-
   + ( length(Ls, 10000000),
        member(M, [you,they,f1,f2,mat,dcg,dcgx,ap]), write(M), write(' '),
        atom_concat(last_but_one_,M,P), + time(call(P,L,Ls))
   ).