Question

SPOILER ：这是Project Euler问题＃18。 自行承担风险

问题是找到＆＃34;最大值＆＃34;从帕斯卡三角形顶部严格向下到底部的所有非确定性路径。我试图通过在三角形的行上进行折叠来计算总和。

这是输入字符串，以及一些基本准备工作：

inputLines = ["75",
              "95 64",
              "17 47 82",
              "18 35 87 10",
              "20 04 82 47 65",
              "19 01 23 75 03 34",
              "88 02 77 73 07 63 67",
              "99 65 04 28 06 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 04 68 89 53 67 30 73 16 69 87 40 31",
              "04 62 98 27 23 09 70 98 73 93 38 53 60 04 23"]

input = map (map (read ::String ->Integer) . words) inputLines

prepareElems :: [[Integer]] -> [[Elem Integer]]
prepareElems = map mkElemsFromTri

这里有两个想法 - 有一排树，有一排三角形。我没有做很多类型级抽象，但这并不重要。这个想法是三角形中的每一行都是一个＆＃34;并行枚举＆＃34;其中的二项式的长度（即：[(4,0), (3,1), (2,2), ... (0,4)]），并且树的每一行都得到了＃34; copy-forked＆＃34;在将标记的三角形行应用于树的行之前，确保在每个机会分叉时保留非确定性的完整性。这就是我的技术：

data BiLabel = BiLabel Integer Integer
  deriving (Show, Eq)

leftLabel (BiLabel x _) = x
rightLabel (BiLabel _ y) = y

parEnumBiLabel :: Integer -> [BiLabel]
parEnumBiLabel n = map ( \x ->BiLabel x $ (n-1)-x ) [0..(n-1)]


forkBiLabel :: BiLabel -> (BiLabel, BiLabel)
forkBiLabel (BiLabel x y) = (BiLabel (x+1) y,BiLabel x (y+1))

data Elem a = Elem {label :: BiLabel, element :: a}
  deriving (Show, Eq)

forkElem :: Elem a -> [Elem a]
forkElem (Elem l a) = [Elem left a,Elem right a]
  where
    (left,right) = forkBiLabel l

-- Does a binomial expansion, but cloning the elements and making new labels
cloneNextLevel :: [Elem Integer] -> [Elem Integer]
cloneNextLevel = concatMap forkElem

我的问题是该代码适用于一行输入行，但在折叠多个行时失败。我的直觉告诉我，我的＆＃34; copy-fork＆＃34;技术是在我的overElems函数可以通过应用它预期有权访问的标签之前创建新标签。以下是我的主要职能：

-- this applys a 1-ary function to all Elems of a list that match the label onLabel :: (a -> a) -> BiLabel -> [Elem a] -> [Elem a] onLabel f (BiLabel la lb) (x:xs) | label x == BiLabel la lb = processHead : processTail | otherwise = x : processTail where processHead = Elem (BiLabel la lb) $ f (element x) processTail = onLabel f (BiLabel la lb) xs onLabel _ _ [] = [] -- this _tries_ to do the equivalent of `onLabel`, but over lists and 2-ary functions overElems :: (a -> a -> a) -> [Elem a] -> [Elem a] -> [Elem a] overElems f (x:xs) ys = overElems f xs ( onLabel (f $ element x) (label x) ys ) overElems _ [] ys = ys -- This starts with the first element of `input` as an accumulator, just because -- then I won't have to copy it over as part of the fold, and can simply -- `cloneNextLevel` before I process it with `overElems`. calculate :: [[Elem Integer]] -> [Elem Integer] calculate = foldr (\z acc -> overElems (+) z (cloneNextLevel acc)) [Elem (BiLabel 0 0) (head $ head input)]

奇怪的是折叠一行工作，但折叠多次无法应用更高阶函数，但仍然将结果扩展到正确的大小。以下是一些示例输入：

\> calculate $ prepareElems [[2,3]] [Elem {label = BiLabel 1 0, element = 78},Elem {label = BiLabel 0 1, element = 77}] -- that's correct, because the first element of the tree is 75 \> calculate $ prepareElems [[2,3],[10,20,30]] [Elem {label = BiLabel 2 0, element = 75},Elem {label = BiLabel 1 1, element = 75}, Elem {label = BiLabel 1 1, element = 75},Elem {label = BiLabel 0 2, element = 75}] -- this is the correct size and labeling, but not the right contents.

我的列表处理中的懒惰会导致这种情况吗？我的感觉是标签枚举是在overWith有机会通过标签应用函数之前发生的。这也是我的代码full page。

Pascal三角形路径求和通过折叠与基于二项式扩展的累加器

0 个答案: