Question

我是Prolog的新手，我想写poppler(Nums, Plate, Tastiness)，其中列出了正好9个数字作为输入，如果可能的话，返回那些形成美味的poppler牌的数字的排列{{1以行主格式读取。

如果三排，两列和两条主对角线中的每一个的Popplers的总和相同，则称Poppler板是美味的。这个共同的总和被称为它的美味。

例如，这是一道美味的Poppler牌，带有美味Plate：

这是我的尝试：

但这不起作用。我该如何解决？

Answer 1

以下是您的代码的固定版本以及一些注释。在SWI-Prolog中测试。

它有效，但它确实很慢（对于你的例子会工作几分钟）。这是因为搜索空间很大，并且没有搜索空间修剪。

你应该真正使用约束编程方法解决这个问题 - 它以一种聪明的方式修剪搜索空间，并且该程序可以立即工作。

% should really just use length/2
size([], 0).
size([Head|T],N) :- size(T,N1), N is N1+1.

% could use simpler version of this like "is_equal([X, Y, Z], Sum)"
is_equal([U, V, W], [X, Y, Z], Sum) :- Sum is U + V + W, Sum is X + Y + Z.

poppler(Nums, Plate, Tastiness) :- 
    size(Nums, 9),
    [A, B, C, D, E, F, G, H, I] = Plate,

    msort(Nums, Sorted),

    member(A, Nums),
    member(B, Nums),
    member(C, Nums),
    member(D, Nums),
    member(E, Nums),
    member(F, Nums),
    member(G, Nums),
    member(H, Nums),
    member(I, Nums),

    % Check if Plate is a permutation of Nums
    msort(Plate, Sorted),

    is_equal([A, B, C], [D, E, F], Tastiness),
    is_equal([A, B, C], [G, H, I], Tastiness),
    is_equal([G, H, I], [D, E, F], Tastiness),
    is_equal([A, D, G], [B, E, H], Tastiness),
    is_equal([A, D, G], [C, F, I], Tastiness),
    is_equal([B, E, H], [C, F, I], Tastiness),
    is_equal([A, E, I], [C, E, G], Tastiness).

Answer 2

通过约束逻辑编程看起来是一个完美的问题。

这是我在ECLiPSe CLP Prolog中的解决方案（可以翻译成其他Prolog系统）：

:- lib(gfd).

poppler(Nums, Plate, S) :-
   [A, B, C, D, E, F, G, H, I] = Plate,
   sorted(Nums, Sorted), sorted(Plate, Sorted),
   % rows
   A + B + C #= S,
   D + E + F #= S,
   G + H + I #= S,
   % colums
   A + D + G #= S,
   B + E + H #= S,
   C + F + I #= S,
   % diagonals
   A + E + I #= S,
   C + E + G #= S,
   labeling(Plate).

试运行：

[eclipse]: poppler([1, 2, 3, 4, 5, 6, 7, 8, 9], Plate, 15).
Plate = [2, 7, 6, 9, 5, 1, 4, 3, 8]

Answer 3

我认为你的主要问题是使用member / 2会产生多次尝试，而不是稍后会丢弃。你可以改为使用置换/ 2：

poppler0(Nums, Plate, Tastiness):-
    Plate = [A, B, C, D, E, F, G, H, I],
    permutation(Nums, Plate),
    is_equal([A, B, C], [D, E, F], Tastiness),
    is_equal([A, B, C], [G, H, I], Tastiness),
    is_equal([G, H, I], [D, E, F], Tastiness),
    is_equal([A, D, G], [B, E, H], Tastiness),
    is_equal([A, D, G], [C, F, I], Tastiness),
    is_equal([B, E, H], [C, F, I], Tastiness),
    is_equal([A, E, I], [C, E, G], Tastiness).

产量

?- numlist(1,9,L),poppler0(L,X,15).
L = [1, 2, 3, 4, 5, 6, 7, 8, 9],
X = [2, 7, 6, 9, 5, 1, 4, 3, 8] ;
L = [1, 2, 3, 4, 5, 6, 7, 8, 9],
X = [2, 9, 4, 7, 5, 3, 6, 1, 8] ;
...

而不是member / 3，select / 3将不复制：

poppler1(Nums, Plate, Tastiness):-
    Plate = [A, B, C, D, E, F, G, H, I],
    %permutation(Nums, Plate),
    select(A, Nums, Num1),
    select(B, Num1, Num2),
    select(C, Num2, Num3),
    select(D, Num3, Num4),
    select(E, Num4, Num5),
    select(F, Num5, Num6),
    select(G, Num6, Num7),
    select(H, Num7, Num8),
    select(I, Num8, []),
    is_equal([A, B, C], [D, E, F], Tastiness),
    is_equal([A, B, C], [G, H, I], Tastiness),
    is_equal([G, H, I], [D, E, F], Tastiness),
    is_equal([A, D, G], [B, E, H], Tastiness),
    is_equal([A, D, G], [C, F, I], Tastiness),
    is_equal([B, E, H], [C, F, I], Tastiness),
    is_equal([A, E, I], [C, E, G], Tastiness).

此外，由于排列现在是“切片”，我们可以“推送”＃39;一些早期的测试，以使整体更快：

poppler2(Nums, Plate, Tastiness):-
    Plate = [A, B, C, D, E, F, G, H, I],
    select(A, Nums, Num1),
    select(B, Num1, Num2),
    select(C, Num2, Num3),
    select(D, Num3, Num4),
    select(E, Num4, Num5),
    select(F, Num5, Num6),
    is_equal([A, B, C], [D, E, F], Tastiness),
    select(G, Num6, Num7),
    select(H, Num7, Num8),
    select(I, Num8, []),
    is_equal([A, B, C], [G, H, I], Tastiness),
    is_equal([G, H, I], [D, E, F], Tastiness),
    is_equal([A, D, G], [B, E, H], Tastiness),
    is_equal([A, D, G], [C, F, I], Tastiness),
    is_equal([B, E, H], [C, F, I], Tastiness),
    is_equal([A, E, I], [C, E, G], Tastiness).

?- numlist(1,9,L),time(poppler0(L,X,15)).
% 642,293 inferences, 0.253 CPU in 0.256 seconds (99% CPU, 2540589 Lips)
L = [1, 2, 3, 4, 5, 6, 7, 8, 9],
X = [2, 7, 6, 9, 5, 1, 4, 3, 8] 
.

8 ?- numlist(1,9,L),time(poppler1(L,X,15)).
% 385,446 inferences, 0.217 CPU in 0.217 seconds (100% CPU, 1777885 Lips)
L = [1, 2, 3, 4, 5, 6, 7, 8, 9],
X = [2, 7, 6, 9, 5, 1, 4, 3, 8] 
.

9 ?- numlist(1,9,L),time(poppler2(L,X,15)).
% 48,409 inferences, 0.029 CPU in 0.029 seconds (100% CPU, 1643812 Lips)
L = [1, 2, 3, 4, 5, 6, 7, 8, 9],
X = [2, 7, 6, 9, 5, 1, 4, 3, 8]

另一个小问题是，某些总和的评估时间超过了一个时间，这可能是由于您选择使用is_equal / 3对测试进行编码。我会改为写

poppler3(Nums, Plate, Tastiness):-
    Plate = [A, B, C, D, E, F, G, H, I],
    select(A, Nums, Num1),
    select(B, Num1, Num2),
    select(C, Num2, Num3),
    sumlist([A, B, C], Tastiness),
    select(D, Num3, Num4),
    select(E, Num4, Num5),
    select(F, Num5, Num6),
    sumlist([D, E, F], Tastiness),
    select(G, Num6, Num7),
    sumlist([A, D, G], Tastiness),
    sumlist([C, E, G], Tastiness),
    select(H, Num7, Num8),
    sumlist([B, E, H], Tastiness),
    select(I, Num8, []),
    sumlist([G, H, I], Tastiness),
    sumlist([C, F, I], Tastiness),
    sumlist([A, E, I], Tastiness).

产生

?- numlist(1,9,L),time(poppler3(L,X,15)).
% 14,371 inferences, 0.004 CPU in 0.004 seconds (99% CPU, 3359784 Lips)
L = [1, 2, 3, 4, 5, 6, 7, 8, 9],
X = [2, 7, 6, 9, 5, 1, 4, 3, 8] 
.

但同样，sumlist / 2比一般要求更为通用，并且可以进一步获得内联总和：

poppler4(Nums, Plate, Tastiness):-
    Plate = [A, B, C, D, E, F, G, H, I],
    select(A, Nums, Num1),
    select(B, Num1, Num2),
    select(C, Num2, Num3),
    A+B+C =:= Tastiness,
    select(D, Num3, Num4),
    select(E, Num4, Num5),
    select(F, Num5, Num6),
    D+E+F =:= Tastiness,
    select(G, Num6, Num7),
    A+D+G =:= Tastiness,
    C+E+G =:= Tastiness,
    select(H, Num7, Num8),
    B+E+H =:= Tastiness,
    select(I, Num8, []),
    G+H+I =:= Tastiness,
    C+F+I =:= Tastiness,
    A+E+I =:= Tastiness.

产量

?- numlist(1,9,L),time(poppler4(L,X,15)).
% 3,394 inferences, 0.002 CPU in 0.002 seconds (100% CPU, 1827856 Lips)
L = [1, 2, 3, 4, 5, 6, 7, 8, 9],
X = [2, 7, 6, 9, 5, 1, 4, 3, 8] 
.

Prolog数字列表的总和

3 个答案: