Prolog数字列表的总和

Question

I am new to Prolog and I want to write poppler(Nums, Plate, Tastiness) that takes a list of exactly 9 numbers as input, and, if possible, returns a permutation of those numbers that forms a delicious poppler plate when Plate is read in row-major format.

A Poppler plate is said to be delicious if the sum of the Popplers in each of the three rows, columns, and two main diagonals is the same. This common sum is called its tastiness.

For example, this is a delicious Poppler plate with tastiness 15 :

2 7 6

9 5 1

4 3 8

Here is my try:

size([], 0).
size([Head|T], N) :-
   size(T, N1),
   N is N1+1.

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

But this doesn't work. How can I fix it?

Answer 1

Here is a fixed version of your code with some comments. Tested in SWI-Prolog.

It works, but it's really slow (will work for minutes for your example). This is because the search space is large, and there is no search space pruning.

Your should really use constraint programming approach for this problem - it prunes search space in a clever way, and that program works instantly.

% 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

Looks like a perfect problem to solve with constraint logic programming.

Here is my solution in ECLiPSe CLP Prolog (can be translated to other Prolog systems):

:- 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).

Test run:

[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

I think your main problem is that using member/2 you are generating much more attempts than will be discarded later. You could instead use permutation/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).

yields

?- 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] ;
...

Instead of member/3, select/3 would get non duplicated:

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).

Also, since the permutation now is 'sliced', we could 'push' some of the tests early, to make the whole faster:

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] 

Another minor problem is that some sum is evaluated more than a time, that's probably due to your choice to code the test with is_equal/3. I would write instead

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).

that yields

?- 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] 
.

but again, sumlist/2 is more general than required, and there is further gain to inline the sum:

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.

yields

?- 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] 
.