用haskell实现的八皇后程序
用Haskell解八皇后问题,Haskell才是最精练的程序:
main = print $ queens 8 boardSize = 8 queens 0 = [[]] queens n = [ x : y | y <- queens (n-1), x <- [1..boardSize], safe x y 1] where safe x [] n = True safe x (c:y) n = and [ x /= c , x /= c + n , x /= c - n , safe x y (n+1)]
运行结果:
[[4,2,7,3,6,8,5,1],[5,2,4,7,3,8,6,1],[3,5,2,8,6,4,7,1],[3,6,4,2,8,5,7,1],[5,7,1,3,8,6,4,2],[4,6,8,3,1,7,5,2] ... ... ... ... 5,7,2,6,3,1,4,8]]
另外一种用到了monad的源程序:
import Control.Monad queens n = foldM (\y _ -> [ x : y | x <- [1..n], safe x y 1]) [] [1..n] safe x [] n = True safe x (c:y) n = and [ x /= c , x /= c + n , x /= c - n , safe x y (n+1)] main = mapM_ print $ queens 8
运行结果:
[4,2,7,3,6,8,5,1]
[5,2,4,7,3,8,6,1]
[3,5,2,8,6,4,7,1]
… … … …
[5,7,2,6,3,1,4,8]
albertlee在这篇用 Python 秒掉八皇后问题!中给出的haskell程序:
import Control.Monad import Control.Monad.Writer import Data.List diagonal (x1,y1) (x2,y2) = x1 + y1 == x2 + y2 || x1 - y1 == x2 - y2 nqueens n = execWriter $ f [1..n] 1 [] where f [] _ ps = tell [ps] f cs r ps = forM_ cs $ \c ->; unless (any (diagonal (r,c)) ps) $ f (delete c cs) (r + 1) ((r,c):ps) main = print $ nqueens 4
http://fleurer-lee.com/2009/04/03/haskellqiu-jie-nhuang-hou-wen-ti.html给出的代码:
module Main where import Data.List queue :: Int -> Int -> [[Int]] queue m 1 = [ [x] | x <- [1..m] ] queue m n = concatMap put $ filter putable $ queue m (n-1) where putable xs = (safe_places xs /= []) put xs = map (:xs) $ safe_places xs safe_places xs = [1..m] \\ (concatMap (\(x,y) -> [x-y,x,x+y]) $ zip xs [1..])
下面这个网址给出了另外一种解法,并且有详细的解释:
http://blog.chinaunix.net/uid-8582194-id-374620.html
main = do putStrLn $ "There are " ++ show(length result) ++ " solutions in total" mapM_ (print.zipWith (\x y -> x:show y) ['A'..'H']) result where result = queens 8 queens 0 = [[]] queens n = [ q:b | b <- queens (n-1), q <- [1..8], safe q b ] safe q b = and [ not (checks q b i) | i <- [0..(length b-1)] ] checks q b i = q==b!!i || abs(q - b!!i)==i+1
Haskell了解得还太少,暂时这些代码都看不懂,先收集在这里,以后慢慢理解。
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