【发布时间】:2013-11-24 21:09:48
【问题描述】:
我目前正在解决一个特定的 kenken 问题,我注意到大约需要 5 秒才能解决。这是一个较小的谜题。但是,我还需要解决一个更大的问题,我担心找到解决方案需要很长时间。这是我的代码中最相关的部分(我已经删除了多余的部分,它们是预期的形式):
getlarger(First, Second, First) :- First >= Second.
getlarger(First, Second, Second) :- First < Second.
getsmaller(First, Second, Second) :- First >= Second.
getsmaller(First, Second, First) :- First < Second.
subtractsmallerfromlarger(First, Second, Result) :- getlarger(First, Second, Larger),
getsmaller(First, Second, Smaller), Result is Larger - Smaller.
intdividelargerbysmaller(First, Second, Result) :- getlarger(First, Second, Larger),
getsmaller(First, Second, Smaller), Result is Larger // Smaller.
groupof4(List) :- nodups(List).
allrowsof4([X1, X2, X3, X4, X5, X6, X7, X8, X9, X10, X11, X12, X13, X14, X15, X16]) :-
groupof4([X1, X5, X9, X13]), groupof4([X2, X6, X10, X14]), %snip....
allcolumnsof4([X1, X2, X3, X4, X5, X6, X7, X8, X9, X10, X11, X12, X13, X14, X15, X16]) :-
groupof4([X1, X2, X3, X4]), groupof4([X5, X6, X7, X8]), %snip....
validnumbers4([X1, X2, X3, X4, X5, X6, X7, X8, X9, X10, X11, X12, X13, X14, X15, X16]) :-
validnumber4(X1), validnumber4(X2), %.....
kenken([X1, X2, X3, X4, X5, X6, X7, X8, X9, X10, X11, X12, X13, X14, X15, X16]) :-
%snip...
validnumbers4([X1, X2, X3, X4, X5, X6, X7, X8, X9, X10, X11, X12, X13, X14, X15, X16]),
X5 * X1 * X2 =:= 2,
%Additional Arithmetic contraints removed
intdividelargerbysmaller(X11, X15, 2), %Dividend =:= 2,
subtractsmallerfromlarger(X12, X16, 3), %Difference =:= 3,
X13 * X14 =:= 6,
allcolumnsof4([X1, X2, X3, X4, X5, X6, X7, X8, X9, X10, X11, X12, X13, X14, X15, X16]),
allrowsof4([X1, X2, X3, X4, X5, X6, X7, X8, X9, X10, X11, X12, X13, X14, X15, X16]).
对于已经定义的值,我使用语法:X1 is 5,我把它放在我的代码顶部,认为这样可以加快查找速度。
在不必让我的代码变得非常复杂的情况下,有什么地方可以让它更高效一些吗?
此外,我注意到将 intdividelargerbysmaller 和 subsmallerfromlarger “Result is”更改为“Result =”会使查询花费很长时间/无法解决,而如果我在第三个插槽中使用变量,它会同时解决这两个问题。
此外,我注意到如果我将行和列检查移动到接近开头的位置,查询会花费很长时间。
【问题讨论】:
标签: optimization prolog clpfd