简化CLP难题中的约束

Question

在当地的比赛晚上，四个小伙子参加拼字游戏和国际象棋比赛。利亚姆在国际象棋中击败了马克，詹姆斯获得第三名，16岁的他赢了。利亚姆在拼字游戏中获得第二名，15岁的利亚姆获胜，詹姆斯击败18岁的利亚姆·利亚姆，19岁的利亚姆利亚姆获得第三名。凯文比马克小3岁。国际象棋最后一名，拼字游戏第三名，只有一个小伙子在两场比赛中都获得了相同的位置。

我觉得我的解决方案比需要的更笨重：

:- use_module(library(clpfd)).
check(AJ, AK, AL, AM, CJ, CK, CL, CM, SJ, SK, SL, SM) :-
AJ in 15..16 / 18..19,
AK in 15..16 / 18..19,
AL in 15..16 / 18..19,
AK in 15..16 / 18..19,
all_different([AJ, AK, AL, AM]),
CJ in 1..4, CK in 1..4, CL in 1..4, CM in 1..4,
SJ in 1..4, SK in 1..4, SL in 1..4, SM in 1..4,
all_different([CJ, CK, CL, CM]),
all_different([SJ, SK, SL, SM]),
CL #< CM,
CJ #= 3,
( AJ #= 16, CJ #= 1 ;
AK #= 16, CK #= 1 ;
AL #= 16, CL #= 1 ;
AM #= 16, CM #= 1 ),
SL #= 2,
( AJ #= 15, SJ #= 1 ;
AK #= 15, SK #= 1 ;
AL #= 15, SL #= 1 ;
AM #= 15, SM #= 1 ),
AK #= AM - 3,
( CJ #= 4, SJ #= 3 ;
CK #= 4, SK #= 3 ;
CL #= 5, SL #= 3 ;
CM #= 4, SM #= 3 ),
( CJ #=  SJ, CK #= SK, CL #= SL, CM #= SM ;
CJ #= SJ, CK #=  SK, CL #= SL, CM #= SM ;
CJ #= SJ, CK #= SK, CL #=  SL, CM #= SM ;
CJ #= SJ, CK #= SK, CL #= SL, CM #=  SM ).

有没有更好的方式来表达约束？

---

建议后的改进版本：

:- use_module(library(clpfd)).
check(AJ, AK, AL, AM, CJ, CK, CL, CM, SJ, SK, SL, SM) :-
permutation([AJ, AK, AL, AM], [15, 16, 18, 19]),
permutation([CJ, CK, CL, CM], [1, 2, 3, 4]),
permutation([SJ, SK, SL, SM], [1, 2, 3, 4]),
CL #< CM,
CJ #= 3,
( AJ #= 16, CJ #= 1 ;
AK #= 16, CK #= 1 ;
AL #= 16, CL #= 1 ;
AM #= 16, CM #= 1 ),
SL #= 2,
( AJ #= 15, SJ #= 1 ;
AK #= 15, SK #= 1 ;
AL #= 15, SL #= 1 ;
AM #= 15, SM #= 1 ),
AK #= AM - 3,
( CJ #= 4, SJ #= 3 ;
CK #= 4, SK #= 3 ;
CL #= 5, SL #= 3 ;
CM #= 4, SM #= 3 ),
( CJ #=  SJ, CK #= SK, CL #= SL, CM #= SM ;
CJ #= SJ, CK #=  SK, CL #= SL, CM #= SM ;
CJ #= SJ, CK #= SK, CL #=  SL, CM #= SM ;
CJ #= SJ, CK #= SK, CL #= SL, CM #=  SM ).

Answer 1

这里有一个SWI-Prolog版本，使用全局约束element/3来访问/查找不同列表中的索引/值。(可以将此模型视为element/3:-中的练习。(

:- use_module(library(clpfd)).
go :-
% At the local games evening, four lads were competing in the Scrabble and 
% chess competitions.
N = 4,
length(Lads,N),
Lads = [James,Kevin,Liam,Mark],
Lads = [1,2,3,4],
Lads ins 1..N,
LadsS = ['James','Kevin','Liam','Mark'],

length(Chess,N),
Chess ins 1..N,

length(Scrabble,N),
Scrabble ins 1..N,
length(Ages,N),
Ages ins 15..16 / 18..19,
all_different(Chess),
all_different(Scrabble),
all_different(Ages),
element(Ages15,Ages,15),
element(Ages16,Ages,16),
element(Ages18,Ages,18),
element(Ages19,Ages,19),

% Liam beat Mark in chess, James came third and the 16 year old won.
element(Liam,Chess,ChessLiam),
element(Mark,Chess,ChessMark),
ChessLiam #< ChessMark,
element(James,Chess,3),
element(Ages16,Chess,1),
% Liam came second in Scrabble, the 15 year old won; James beat the 18 year old 
% and the 19 year old came third.
element(Liam,Scrabble,2),
element(Ages15,Scrabble,1),
element(Ages18,Scrabble,ScrabbleAges18),
element(James,Scrabble,ScrabbleJames),
ScrabbleJames #< ScrabbleAges18,
element(Ages19,Scrabble,3),
% Kevin is 3 years younger than Mark.
element(Kevin,Ages,AgesKevin),
element(Mark,Ages,AgesMark),  
AgesKevin + 3 #= AgesMark,

% The person who came last in chess, came third in Scrabble and only one lad
% got the same position in both games.
element(ChessPlace4,Chess,4),
element(ChessPlace4,Scrabble,3),
sums(Chess,Scrabble,0,Sums),
Sums #= 1,

% Can you determine the ages of the lads and the positions in the two games?
flatten([Chess,Scrabble,Ages,Sums],Vars),
label(Vars),

writeln(chess=Chess),
writeln(scrabble=Scrabble),
writeln(ages=Ages),
sol(LadsS,Ages,Scrabble,Chess),
nl,

fail,
nl.
go.

sums([],[],Total,Total).
sums([C|Chess],[S|Scrabble],Total0,Total) :-
B in 0..1,
C #= S #<==> B #= 1,
Total1 #= Total0 + B,         
sums(Chess,Scrabble,Total1,Total).
sol([L|Lads],[A|Ages],[S|Scrabble],[C|Chess]) :-
format("~w (~d) Scrabble: ~d Chess: ~d~n",[L,A,S,C]),
sol(Lads,Ages,Scrabble,Chess).

(完整程序可在此处获取：http://hakank.org/swi_prolog/scrabble_contest.pl(