我已经在prolog中构建了一个代码来找到一系列合法的动作,其中骑士只落在棋盘的每个方格(8x8)上一次。
我使用了如下逻辑: 有8种类型的骑士动作:
正确1向下2步:
move(X,Y) :-
C_X is X mod 8,
R_X is X // 8,
C_Y is C_X + 1, % 1 right
C_Y < 8,
R_Y is R_X + 2, % 2 down
R_Y < 8,
Y is R_Y * 8 + C_Y,
Y >= 0,
X >= 0,
X < 64,
Y < 64.
对于所有8种类型的动作都重复这一步
问题是我的代码效率不高,找到正确的路径需要太多步骤。 有谁知道解决这个问题的有效方法?
答案 0 :(得分:7)
能够在可行的时间Warnsdorff's rule内解决8x8 Knight的巡回难题可能是必须的。
我在B-Prolog中创建了一个程序,可以很快地解决这个问题。如果您需要将该程序放在其他Prolog中 - 翻译它或者只是使用它中的一些想法并不难。
knight_moves(X, Y, NewX, NewY) :-
( NewX is X - 1, NewY is Y - 2
; NewX is X - 1, NewY is Y + 2
; NewX is X + 1, NewY is Y - 2
; NewX is X + 1, NewY is Y + 2
; NewX is X - 2, NewY is Y - 1
; NewX is X - 2, NewY is Y + 1
; NewX is X + 2, NewY is Y - 1
; NewX is X + 2, NewY is Y + 1 ).
possible_knight_moves(R, C, X, Y, Visits, NewX, NewY) :-
knight_moves(X, Y, NewX, NewY),
NewX > 0, NewX =< R,
NewY > 0, NewY =< C,
\+ (NewX, NewY) in Visits.
possible_moves_count(R, C, X, Y, Visits, Count) :-
findall(_, possible_knight_moves(R, C, X, Y, Visits, _NewX, _NewY), Moves),
length(Moves, Count).
:- table warnsdorff(+,+,+,+,+,-,-,min).
warnsdorff(R, C, X, Y, Visits, NewX, NewY, Score) :-
possible_knight_moves(R, C, X, Y, Visits, NewX, NewY),
possible_moves_count(R, C, NewX, NewY, [(NewX, NewY) | Visits], Score).
knight(R, C, X, Y, Visits, Path) :-
length(Visits, L),
L =:= R * C - 1,
NewVisits = [(X, Y) | Visits],
reverse(NewVisits, Path).
knight(R, C, X, Y, Visits, Path) :-
length(Visits, L),
L < R * C - 1,
warnsdorff(R, C, X, Y, Visits, NewX, NewY, _Score),
NewVisits = [(X, Y) | Visits],
knight(R, C, NewX, NewY, NewVisits, Path).
| ?- time(knight(8, 8, 1, 1, [], Path)).
CPU time 0.0 seconds.
Path = [(1,1),(2,3),(1,5),(2,7),(4,8),(6,7),(8,8),(7,6),(6,8),(8,7),(7,5),(8,3),(7,1),(5,2),(3,1),(1,2),(2,4),(1,6),(2,8),(3,6),(1,7),(3,8),(5,7),(7,8),(8,6),(7,4),(8,2),(6,1),(7,3),(8,1),(6,2),(4,1),(2,2),(1,4),(2,6),(1,8),(3,7),(5,8),(7,7),(8,5),(6,6),(4,7),(3,5),(5,6),(6,4),(4,3),(5,5),(6,3),(5,1),(7,2),(8,4),(6,5),(4,4),(3,2),(5,3),(4,5),(3,3),(2,1),(1,3),(2,5),(4,6),(3,4),(4,2),(5,4)]
yes
答案 1 :(得分:0)
这是答案集编程(ASP)解决方案。它可以用来在可接受的时间内找到24x24的第一个解决方案,并且可以轻松地适应8x8的情况。它也使用Warnsdorff规则,但是比后向链接解决方案快一点:
向后链接:
?- time(knight_tour((1,1), X)).
% Up 878 ms, GC 32 ms, Thread Cpu 859 ms (Current 10/30/18 20:55:28)
X = [(1,1),(3,2),(5,1),(7,2),(9,1),(11,2),(13,1),(15,2),(17,1), ...
正向链接(使用ASP选择):
?- time(knight_tour((1,1), X)).
% Up 411 ms, GC 0 ms, Thread Cpu 406 ms (Current 10/28/18 20:45:05)
X = [(1,1),(3,2),(5,1),(7,2),(9,1),(11,2),(13,1),(15,2),(17,1), ...
正向链接代码更快,因为它使用正向存储来检查是否已完成移动。这比使用成员谓词进行此检查要快。答案集编程代码为:
:- use_module(library(basic/lists)).
:- use_module(library(minimal/asp)).
knight_tour(Start, Solution) :-
post(go(Start, 1)),
findall(X, go(X,_), Solution).
choose(S) <= posted(go(X,N)), N \== 576,
findall(W-Y, (move(X, Y), weight(Y, X, W)), L),
keysort(L, R),
M is N+1,
strip_and_go(R, M, S).
strip_and_go([_-Y|L], M, [go(Y, M)|R]) :-
strip_and_go(L, M, R).
strip_and_go([], _, []).
weight(X, Z, N) :-
findall(Y, (move(X, Y), Z \== Y), L),
length(L, N).
move(X, Y) :-
knight_move(X, Y),
verify(Y),
\+ clause(go(Y, _), true).
代码使用Jekejeke Prolog中的新模块"asp"。带有谓词knight_move / 2和verify / 1的完整代码位于要旨here上。也可以找到反向链接代码,以便可以并排比较代码。