我能以某种方式看到haskell尝试使用的模式吗? 它适用于较小的例子,但在较大的例子上崩溃,我很难找到它可能的情况。
euler18 :: [[Integer]]
euler18 = pathPyr minipyr
pathPyr xss = path [(head xss)] (tail xss)
minipyr = [[3],[7,4],[2,4,6]]
pyramid = [[75], [95,64], [17,47,82], [18,35,87,10], [20,4,82,47,65], [19,1,23,75,3,34], [88,2,77,73,7,63,67], [99,65,4,28,6,16,70,92], [41,41,26,56,83,40,80,70,33], [41,48,72,33,47,32,37,16,94,29], [53,71,44,65,25,43,91,52,97,51,14], [70,11,33,28,77,73,17,78,39,68,17,57], [91,71,52,38,17,14,91,43,58,50,27,29,48], [63,66,4,68,89,53,67,30,73,16,69,87,40,31], [4,62,98,27,23,9,70,98,73,93,38,53,60,4,23]]
path :: [[Integer]] -> [[Integer]] -> [[Integer]]
path xss (ys:[]) = extendpath xss ys
path xss (ys:yss) = path (extendpath xss ys) yss
extendpath :: [[Integer]] -> [Integer] -> [[Integer]]
extendpath (xs:[]) (y1:y2:[]) = [y1:xs,y2:xs]
extendpath (xs:xss) (y1:y2:ys) = [y1:xs,y2:xs] ++ extendpath xss (y2:ys)
输出:
*Main> pathPyr pyramid
[[4,63,91,70,53,41,41,99,88,19,20,18,17,95,75],[62,63,91,70,53,41,41,99,88,19,20,18,17,95,75],[62,66,91,70,53,41,41,99,88,19,20,18,17,95,75],[98,66,91,70,53,41,41,99,88,19,20,18,17,95,75],[98,66,71,70,53,41,41,99,88,19,20,18,17,95,75],[27,66,71,70,53,41,41,99,88,19,20,18,17,95,75],[27,4,71,70,53,41,41,99,88,19,20,18,17,95,75],[23,4,71,70,53,41,41,99,88,19,20,18,17,95,75],[23,4,71,11,53,41,41,99,88,19,20,18,17,95,75],[9,4,71,11,53,41,41,99,88,19,20,18,17,95,75],[9,68,71,11,53,41,41,99,88,19,20,18,17,95,75],[70,68,71,11,53,41,41,99,88,19,20,18,17,95,75],[70,68,52,11,53,41,41,99,88,19,20,18,17,95,75],[98,68,52,11,53,41,41,99,88,19,20,18,17,95,75],[98,89,52,11,53,41,41,99,88,19,20,18,17,95,75],[73,89,52,11,53,41,41,99,88,19,20,18,17,95,75],[73,89,52,11,71,41,41,99,88,19,20,18,17,95,75],[93,89,52,11,71,41,41,99,88,19,20,18,17,95,75],[93,53,52,11,71,41,41,99,88,19,20,18,17,95,75],[38,53,52,11,71,41,41,99,88,19,20,18,17,95,75],[38,53,38,11,71,41,41,99,88,19,20,18,17,95,75],[53,53,38,11,71,41,41,99,88,19,20,18,17,95,75],[53,67,38,11,71,41,41,99,88,19,20,18,17,95,75],[60,67,38,11,71,41,41,99,88,19,20,18,17,95,75],[60,67,38,33,71,41,41,99,88,19,20,18,17,95,75],[4,67,38,33,71,41,41,99,88,19,20,18,17,95,75],[4,30,38,33,71,41,41,99,88,19,20,18,17,95,75],[23,30,38,33,71,41,41,99,88,19,20,18,17,95,75]*** Exception: euler18.hs:(16,1)-(17,72): Non-exhaustive patterns in function Main.extendpath
答案 0 :(得分:5)
extendpath :: [[Integer]] -> [Integer] -> [[Integer]]
extendpath (xs:[]) (y1:y2:[]) = [y1:xs,y2:xs]
extendpath (xs:xss) (y1:y2:ys) = [y1:xs,y2:xs] ++ extendpath xss (y2:ys)
虽然它有效,extendpath
每次调用时路径的数量都会加倍。但是行的长度仅增加1.因此,只要您有一个三行以上的行,就会在处理第四行(以及后来的每一行)时遇到模式匹配失败。
在第二行(从零开始,因此具有三个元素的行)行之后,你有
-- LL LR RL RR
extendpath [p_1, p_2, p_3, p_4] [v_1, v_2, v_3, v_4]
~> [v_1:p_1, v_2:p_1] ++ extendpath [p_2, p_3, p_4] [v_2, v_3, v_4]
~> [v_1:p_1, v_2:p_1] ++ [v_2:p_2, v_3:p_2] ++ extendpath [p_3, p_4] [v_3, v_4]
~> [v_1:p_1, v_2:p_1] ++ [v_2:p_2, v_3:p_2] ++ [v_3:p_3, v_4:p_3] ++ extendpath [p_4] [v_4]
并且最后一次调用没有匹配的模式。
现在,路径p_2
和p_3
在同一个地方结束,因此两个可以在下一行中使用相同的两个值进行扩展,但extendpath
不会。 t考虑路径的终点。通常,在n
行上有
n `choose` k
路径以k
点结束[这就是为什么我在这里选择从零开始的计数]。扩展路径时,可能的扩展取决于路径的当前终点。
您需要一种不同的方法。特别是问题67。
答案 1 :(得分:4)
你有大量的无法覆盖的病例。这意味着您有部分功能,错误。尽量避免使用它们。通常,使用Maybe
类型构造函数将它们转换为总函数很容易。然后,您可以使用模式匹配来确保涵盖所有案例。另请注意,head
和tail
是部分功能。
作为诊断这个的黑客,我只想添加类似的内容:
extendPath xs ys = error ("oops, pattern-match failed, arguments were: " ++ show xs ++ show ys)
确保那个案例在之后,因为它是一个全能的。
您也必须为path
和pathPyr
执行此操作。您可能希望通过启用警告来招募Haskell的帮助。
答案 2 :(得分:2)
通过将-W
作为GHC的参数,或者在GHCi中输入:set -W
(并重新编译/重新加载文件!),可以使GHC / GHCi在模式不完整时给出警告。 / p>
答案 3 :(得分:2)
良好的错误警告级别将为您提供所需的所有信息。要触发它,请放入文件的头部:
{-# OPTIONS_GHC -Wall #-}
这也会警告你一些其他东西,很可能。哪个好。
答案 4 :(得分:0)
捕捉缺失模式的另一种选择: