优化Haskell上的数独求解器

Question

我在Haskell写了一个数独求解器.它通过一个列表,当它找到'0'(一个空单元格)时,它将获得可以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

对于一些谜题,这是有效的,例如这一个:

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]

不到一秒,但是这一个:

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]

我还没有看完.我不认为这是该方法的问题,因为它确实返回正确的结果.

分析表明大部分时间都花在"isValid"函数上.该功能是否存在效率明显低效/缓慢的问题？

Answer 1

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.即使你将每个函数加速1000倍,也会有网格需要几倍于宇宙年龄才能找到(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.避免此类情况的一种相当有效的方法是首先猜测具有最少可能性的平方.这并不能避免所有不良情况,但会减少很多.

您还应该做的一件事是用length thing == 0支票替换支票null thing.由于此处出现相对较短的列表,效果有限,但一般来说它可能是戏剧性的(通常你也不应该使用length list <= 1,null $ drop 1 list而是使用).