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我在 Haskell 中编写了一个数独求解器。它遍历一个列表，当它找到“0”（一个空单元格）时，它会得到适合的数字并尝试它们：

import Data.List (group, (\\), sort)
import Data.Maybe (fromMaybe)

row :: Int -> [Int] -> [Int]
row y grid = foldl (\acc x -> (grid !! x):acc) [] [y*9 .. y*9+8]
    where y' = y*9
column :: Int -> [Int] -> [Int]
column x grid = foldl (\acc n -> (grid !! n):acc) [] [x,x+9..80]
box :: Int -> Int -> [Int] -> [Int]
box x y grid = foldl (\acc n -> (grid !! n):acc) [] [x+y*9*3+y' | y' <- [0,9,18], x <- [x'..x'+2]]
    where x' = x*3

isValid :: [Int] -> Bool
isValid grid = and [isValidRow, isValidCol, isValidBox]
    where isValidRow = isValidDiv row
          isValidCol = isValidDiv column
          isValidBox = and $ foldl (\acc (x,y) -> isValidList (box x y grid):acc) [] [(x,y) | x <- [0..2], y <- [0..2]]
          isValidDiv f = and $ foldl (\acc x -> isValidList (f x grid):acc) [] [0..8]
          isValidList = all (\x -> length x <= 1) . tail . group . sort -- tail removes entries that are '0'

isComplete :: [Int] -> Bool        
isComplete grid = length (filter (== 0) grid) == 0

solve :: Maybe [Int] -> Maybe [Int]
solve grid' = foldl f Nothing [0..80]
    where grid = fromMaybe [] grid' 
          f acc x
            | isValid grid = if isComplete grid then grid' else f' acc x
            | otherwise    = acc
          f' acc x 
            | (grid !! x) == 0 = case guess x grid of 
                Nothing -> acc
                Just x -> Just x
            | otherwise        = acc

guess :: Int -> [Int] -> Maybe [Int]
guess x grid
    | length valid /= 0 = foldl f Nothing valid
    | otherwise         = Nothing
    where valid = [1..9] \\ (row rowN grid ++ column colN grid ++ box (fst boxN) (snd boxN) grid) -- remove numbers already used in row/collumn/box
          rowN = x `div` 9 -- e.g. 0/9=0 75/9=8
          colN = x - (rowN * 9) -- e.g. 0-0=0 75-72=3
          boxN = (colN `div` 3, rowN `div` 3)
          before x = take x grid
          after x = drop (x+1) grid
          f acc y = case solve $ Just $ before x ++ [y] ++ after x of
            Nothing -> acc
            Just x -> Just x

对于某些谜题，这是可行的，例如这个：

sudoku :: [Int]
sudoku = [5,3,0,6,7,8,0,1,2,
          6,7,0,0,0,0,3,4,8,
          0,0,8,0,0,0,5,0,7,
          8,0,0,0,0,1,0,0,3,
          4,2,6,0,0,3,7,9,0,
          7,0,0,9,0,0,0,5,0,
          9,0,0,5,0,7,0,0,0,
          2,8,7,4,1,9,6,0,5,
          3,0,0,2,8,0,1,0,0]

花了不到一秒钟，但是这个：

sudoku :: [Int]
sudoku = [5,3,0,0,7,0,0,1,2,
          6,7,0,0,0,0,3,4,8,
          0,0,0,0,0,0,5,0,7,
          8,0,0,0,0,1,0,0,3,
          4,2,6,0,0,3,7,9,0,
          7,0,0,9,0,0,0,5,0,
          9,0,0,5,0,7,0,0,0,
          2,8,7,4,1,9,6,0,5,
          3,0,0,2,8,0,1,0,0]

我还没有看到完成。我不认为这是该方法的问题，因为它确实返回了正确的结果。

分析表明大部分时间都花在了“isValid”函数上。该功能是否存在明显低效/缓慢的问题？

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2 回答 2

6

实施当然是可以改进的，但这不是问题。问题是对于第二个网格，简单的猜测和检查算法需要大量的回溯。即使您将每个函数加速 1000 倍，也会有一些网格，它仍然需要宇宙年龄的几倍才能找到（首先，如果网格不是唯一的）解决方案。

你需要一个更好的算法来避免这种情况。避免这种情况的一种相当有效的方法是首先猜测可能性最少的正方形。这并不能避免所有坏情况，但会大大减少它们。

您还应该做的一件事是将length thing == 0支票替换为null thing. 由于此处出现的列表相对较短，因此效果有限，但总的来说它可能是戏剧性的（通常您也不应该使用length list <= 1,null $ drop 1 list而是使用）。

于 2012-02-20T16:04:42.290 回答
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isValidList = all (\x -> length x <= 1) . tail . group . sort -- tail removes entries that are '0'

如果原始列表不包含任何零，tail将删除其他内容，可能是两个 1 的列表。我会替换tail . group. sort为group . sort . filter (/= 0).

我不明白为什么isValidBox并且isValidDiv使用foldlasmap似乎就足够了。我错过了什么/他们在做一些非常聪明的事情吗？

于 2012-02-20T15:55:05.157 回答