獲取位置

Question

Rellevant類型：

Data Sudoku = Sudoku [[Maybe Int]] 
type Block = [Maybe Int] 
rows :: Sudoku -> [[Maybe Int]] 
rows (Sudoku rs) = rs 

我有一個列表的列表，[也許INT]]，我已經說

type Block = [Maybe Int] 

我是分配工作是創建一個數獨求解器，所以我需要一個函數來返回一個'空白'單元格的位置，空白爲Nothing。 數獨的實際佈局將

Sudoku [ 
     , list1 
     , list2 
     .... ] 

，直到我有九個列表，每個包含九個元素，沒有或只是詮釋。

type Pos = (Int, Int) 

第一個Int in Pos應指示哪個'行'包含空白單元格。第二個應該指出那個Row是哪個Element是空白的。 我寫

whichRow :: Sudoku -> Int 
whichRow (Sudoku (x:xs)) = 
    if isNothingPresent x == False then 1 + whichRow (Sudoku xs) else 1 

whereIsNothing :: Block -> Int 
whereIsNothing (x:xs) = if x == Nothing then 1 else 1 + whereIsNothing xs 

isNothingPresent :: Block -> Bool 
isNothingPresent b 
    | Nothing `notElem` b = False 
    | otherwise   = True 

現在我需要這些功能結合起來，創造一個功能空白，這將檢查一個數獨，並返回一個空白單元格的位置。我打得四處，並試圖：

blank :: Sudoku -> Pos 
blank sud = do k <- (whichRow sud) 
       n <- (whereIsNothing (head (drop (k-1) (rows sud)))) 
       (return (k, n)) 

這顯然是錯誤的，我得到錯誤類型和各種東西。它看起來很奇怪，那裏也有一個do block。任何人都可以將我指向正確的方向嗎？謝謝！

我得到這些錯誤：

Sudoku.hs:159:22: 
    Couldn't match expected type `(Int, Int)' with actual type `Int' 
    In a stmt of a 'do' block: k <- (whichRow sud) 
    In the expression: 
     do { k <- (whichRow sud); 
      n <- (whereIsNothing (head (drop (k - 1) (rows sud)))); 
      (return (k, n)) } 

Sudoku.hs:160:22: 
    Couldn't match expected type `(Int, t0)' with actual type `Int' 
    In a stmt of a 'do' block: 
     n <- (whereIsNothing (head (drop (k - 1) (rows sud)))) 
    In the expression: 
     do { k <- (whichRow sud); 
      n <- (whereIsNothing (head (drop (k - 1) (rows sud)))); 
      (return (k, n)) } 

Sudoku.hs:161:24: 
    Couldn't match expected type `Int' with actual type `(Int, t0)' 
    Relevant bindings include n :: t0 (bound at Sudoku.hs:160:16) 
    In the first argument of `return', namely `(k, n)' 
    In a stmt of a 'do' block: (return (k, n)) 
    In the expression: 
     do { k <- (whichRow sud); 
      n <- (whereIsNothing (head (drop (k - 1) (rows sud)))); 
      (return (k, n)) } 

再次編輯：此外，我敢肯定，這個功能會表現得可笑，如果沒有空白單元格..

Answer 1

首先，使用尾遞歸，但不是遞歸函數。

whichRow :: Sudoku -> Int 
whichRow s = whichRow' s 0 -- tail-recursive 
    where 
    whichRow' (Sudoku (x:xs)) i = if isNothingPresent x 
     then 1 
     else whichRow' (Sudoku xs) (i+1) 

二，不要忘了看看所有情況下，模式匹配：

whichRow :: Sudoku -> Int 
whichRow s = whichRow' s 0 
    where 
    whichRow' (Sudoku [])  i = i  -- missing case 
    whichRow' (Sudoku (x:xs)) i = if isNothingPresent x 
     then 1 
     else whichRow' (Sudoku xs) (i+1) 

三，你爲什麼用這麼多括號？

blank :: Sudoku -> Pos 
blank sud = do k <- whichRow sud 
       n <- whereIsNothing (head (drop (k-1) (rows sud))) 
       return (k, n) 

但主要 - 這個代碼是無效的。 do是一個monad，但只使用Int。使用let代替

blank :: Sudoku -> Pos 
blank sud = let k = whichRow sud in 
      let n = whereIsNothing $ head $ drop (k-1) (rows sud) in 
      (k, n) 

或where

blank :: Sudoku -> Pos 
blank sud = (k, n) 
    where 
    k = whichRow sud 
    n = whereIsNothing $ head $ drop (k-1) (rows sud) 

四，不要重複使用的Bool了很多，比如if smth == True then ....，寫，而不是if smth then ...。讓我們來看看

isNothingPresent :: Block -> Bool 
isNothingPresent b 
    | Nothing `notElem` b = False 
    | otherwise   = True 

，我們可以把它

isNothingPresent :: Block -> Bool 
isNothingPresent b = not $ Nothing `notElem` b 

，或者更簡單地

isNothingPresent :: Block -> Bool 
isNothingPresent b = Nothing `elem` b