问题
I'm developing a game named woodntDie. This game consists in 9 pieces which should look like this:
The aim is to arrange all those pieces and make it look like a dice:
This is the four final possible solutions with all those 9 pieces.
I'm trying to implement some restrictions but those arent enough:
sum([V5, V6, V11, V12, V17, V18, V23, V24, V29, V30, V35, V36, V41, V42, V47, V48, V53, V54], #=, 7),
The sum of the opposite faces should always be 7. In this restriction I'm telling that the sum of the top and bottom face it's 7.
sum([V2, V19, V7, V16, V31, V26], #=, 7),
The sum of the front and back faces is 7 aswell.
sum([V1, V49, V13, V8, V37, V25], #=, 7),
The sum of the left and right face is 7.
sum(Vars, #=, 21),
The sum of all the dots it's 21.
But I'm getting a really weird output, I'm getting repeated pieces and a load of solutions that doesnt make any sense and dont form a dice at all. I think I need few more restrictions and my questions are:
How do I make a restrictions saying that some pieces cant be repeated (notice that I can have two F pieces but all the other ones cant be similar)
How do I make a new restriction with this:
The only other remaining bar to have two spots on its side is C, so C must combine with A to make the 4 or 5 face
The only other bar to have a spot on its side near one end is E, so E must combine with B to make the 2 or 3 face.
I mean, is there any way to say "the sum of VX, VY, VZ it's 2 or 3".
And do you have any other opinion that would make the solution easier?
I'm struggling here and I really dont know what to do, I'd really appreciate all the help I could get.
Here's the whole code:
:-use_module(library(clpfd)).
:-use_module(library(lists)).
dice(Vars):-
Vars=[V1, V2, V3, V4, V5, V6, V7, V8, V9, V10, V11, V12, V13, V14, V15, V16, V17, V18,
V19, V20, V21, V22, V23, V24, V25, V26, V27, V28, V29, V30, V31, V32, V33, V34, V35, V36,
V37, V38, V39, V40, V41, V42, V43, V44, V45, V46, V47, V48, V49, V50, V51, V52, V53, V54],
% pecas dos topos
domain([V5, V6, V11, V12, V17, V18, V23, V24, V29, V30, V35, V36, V41, V42, V47, V48, V53, V54], 0, 1),
% pecas laterais
domain([V1, V2, V3, V4, V7, V8, V9, V10, V13, V14, V15, V16, V19, V20, V21, V22, V25, V26, V27, V28,
V31, V32, V33, V34, V37, V38, V39, V40, V43, V44, V45, V46, V49, V50, V51, V52], 0,2),
table([
[V1, V2, V3, V4, V5, V6], %%peca1
[V7, V8, V9, V10, V11, V12], %%peca2
[V13, V14, V15, V16, V17, V18], %%peca3
[V19, V20, V21, V22, V23, V24], %%peca4
[V25, V26, V27, V28, V29, V30], %%peca5
[V31, V32, V33, V34, V35, V36], %%peca6
[V37, V38, V39, V40, V41, V42], %%peca7
[V43, V44, V45, V46, V47, V48], %%peca8
[V49, V50, V51, V52, V53, V54]], %%peca9
[[2,2,0,0,1,0], %%peca1
[2,1,0,0,1,1], %%peca2
[2,0,0,0,1,1], %%peca3
[2,0,0,0,0,0], %%peca4
[1,0,0,0,1,0], %%peca5
[1,0,0,0,0,0], %%peca6
[1,0,0,0,0,0], %%peca7
[0,0,0,0,1,0], %%peca8
[0,0,0,0,0,0]]), %%peca9
% restricoes para as faces que envolve os topos de cada peça
% Na figura de descriçºao das peças os topos de cima estão guardados
% na posicao 5 de cada lista. Os topos de baixo estoa guardados na
% posicao 6 de cada lista
% Restrições quanto à soma de faces opostas(=7)
V2 + V19 + V7 #= TotalF1,
V8 + V37 + V25 #= TotalF2,
V5 + V23 + V11 + V53 + V47 + V41 + V29 + V35 + V17 #= TotalF3,
V18 + V36 + V30 + V54 + V48 + V42 + V6 + V24 + V12 #= TotalF4,
V1 + V49 + V13 #= TotalF5,
V16+ V31 + V26 #= TotalF6,
Totals = [TotalF1, TotalF2, TotalF3, TotalF4, TotalF5, TotalF6],
domain(Totals, 1,6),
all_different(Totals),
TotalF3 + TotalF4 #= 7,
TotalF1 + TotalF6 #= 7,
TotalF5 + TotalF2 #= 7,
A #= 100000*V1 + 10000*V2+ 1000*V5,
element(P1,Vars,A),
B #= 100000*V7 + 10000*V8 + 10*V11 + V12,
element(P2,Vars,B),
C #= 100000*V13 + 10*V17 + V18,
element(P3,Vars,C),
D #= 100000*V19,
element(P4,Vars,D),
E #= 100000*V25 + 10*V29,
element(P5,Vars,E),
F #= 100000*V31,
element(P6,Vars,F),
G #= 100000*V37,
element(P7,Vars,G),
H #= 100*V47,
element(P8,Vars,H),
I #= 0,
element(P9,Vars,I),
F#=P7,
F#=G,
PiecesIndex = [P1, P2, P3, P4, P5, P6, P8, P9],
all_different(PiecesIndex),
%There are 21 spots on the pieces
%sum(Vars, #=, 21),
labeling([],Vars),
show(Vars).
show([V1, V2, V3, V4, V5, V6, V7, V8, V9, V10, V11, V12, V13, V14, V15, V16, V17, V18,
V19, V20, V21, V22, V23, V24, V25, V26, V27, V28, V29, V30, V31, V32, V33, V34, V35, V36,
V37, V38, V39, V40, V41, V42, V43, V44, V45, V46, V47, V48, V49, V50, V51, V52, V53, V54]) :-
write([V1, V2, V3, V4, V5, V6]), nl,
write([V7, V8, V9, V10, V11, V12]), nl,
write([V13, V14, V15, V16, V17, V18]), nl,
write([V19, V20, V21, V22, V23, V24]), nl,
write([V25, V26, V27, V28, V29, V30]), nl,
write([V31, V32, V33, V34, V35, V36]), nl,
write([V37, V38, V39, V40, V41, V42]), nl,
write([V43, V44, V45, V46, V47, V48]), nl,
write([V49, V50, V51, V52, V53, V54]), nl.
and here's a pic with the representation of the dice and all those V's (each V represents the dots on each face)
ALL THE INFORMATION ABOUT THE GAME
Thank you so much in advance
Any doubt just ask or add me on skype: preserverance
Best regards
回答1:
I mean, is there any way to say "the sum of VX, VY, VZ it's 2 or 3".
If this is the crucial question, try this:
(sum([VX,VY,VZ],#=,2);sum([VX,VY,VZ],#=,3))
来源：https://stackoverflow.com/questions/34257684/prolog-restrictions-puzzle