Haskell: 从元组中删除一个成员

Question

我正在尝试在 Haskell 中制作数独求解器，但在制作 deleteV 函数时遇到问题，如果相邻单元格具有相同的值，该函数将从候选列表中删除一个值。电路板和电池都是我定义的，下面是我目前的情况：

type Cell = (Int, [Int])
type Board = [Cell]

rowNo  sq =  sq `div` 9
colNo  sq =  sq `mod` 9
boxNo  sq =  (sq `div` 9 `div` 3 ) * 3
    + (sq `div` 3) `mod` 3

-- Two squares sq1 and sq2 are in the same neighborhood if their row,
-- column, or box numbers are the same.

isNeighbor sq1 sq2 =
    ((rowNo sq1) == (rowNo sq2)) ||
    ((colNo sq1) == (colNo sq2)) ||
    (((boxNo sq1) == (boxNo sq2)))

delV val sq board =
    if board == [] then []
    else if isNeighbor (fst(head board)) sq -- then delV val sq (tail     board)
    then (val `delete` snd(head board)):(delV val sq (tail board))\

我可以让 delV 做它应该做的一部分，但不是全部。当我用“delV 5 2 [(2, [1,5,7]), (26, [1,5,8])]”调用 delV 时，我得到 “[[1,7],[1,5,8]]”，但所需的输出是“[(2, [1,7]), (26, [1,5,8])]”。虽然唯一的区别是在输出中包含索引，但我需要这种格式的索引。 P.S。当前版本的 delV 无法编译，在我尝试修复它的过程中，我又破坏了它。

已编辑：这是 delV 的一个版本，returns“[[1,7],[1,5,8]]”是正确的，但不包括索引：

  delV val sq board =
    if board == [] then []
    else if isNeighbor (fst(head board)) sq then val `delete` snd(head board):delV val sq (tail board)
    else (snd(head board)):delV val sq (tail board)

Answer 1

您有一些格式错误的代码，您想要的输出有一个 7，看起来您想要一个 8。我假设这就是你的意思：

import Data.List

delV val sq board =
  if board == [] then []                                         
  else if isNeighbor (fst(head board)) sq 
       then val `delete` snd(head board):delV val sq (tail board)
       else (snd(head board)):delV val sq (tail board)

哪里

delV :: Int -> Int -> Board -> [[Int]]
delV 5 2 [(2, [1,5,7]), (26, [1,5,8])] == [[1,7],[1,5,8]]

如你所愿

delV :: Int -> Int -> Board -> Board
delV 5 2 [(2, [1,5,7]), (26, [1,5,8])] == [(2, [1,7]), (26, [1,5,8])]

您只需将索引加回去即可。就是这样，继续你的风格：

delV :: Int -> Int -> Board -> Board    
delV val sq board =
  if board == [] then []
  else if isNeighbor (fst(head board)) sq
       then (fst(head board), val `delete` snd(head board)):
               delV val sq (tail board)
       else head board:delV val sq (tail board)

这里是地图方面的：

delV2 :: Int -> Int -> Board -> Board
delV2 value square = map f
  where
    f (index, candidates) | isNeighbor index square =
        (index, delete value candidates)
    f t = t

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