我相信您的问题比示例代码所示的要复杂一些。首先,让我们明确一些定义:
Let pathCount x y
是结束于 (x, y) 的路径数。我们有
pathCount :: Int -> Int -> Integer
pathCount x y
| y == 0 = if x == 0 then 1 else 0
| otherwise = sum [ pathCount (x + d) (y - 1) | d <- [-1..1]]
现在让我们pathSum x y
是以 (x, y) 结尾的所有路径的总和。我们有:
pathSum :: Int -> Int -> Integer
pathSum x y
| y == 0 = if x == 0 then 1 else 0
| otherwise = sum [ pathSum (x + d) (y - 1) + node x y * pathCount (x + d) (y - 1)
| d <- [-1..1] ]
有了这个助手,我们终于可以定义node x y
适当地:
node :: Int -> Int -> Integer
node x y
| y == 0 = if x == 0 then 1 else 0
| otherwise = sum [ pathSum (x + d) (y - 1) | d <- [-1..1]]
该算法目前的形式是指数时间。但是我们可以添加记忆化使加法次数成二次方。这memoizeHackage 上的软件包让这一切变得非常简单。完整示例:
import Control.Monad
import Data.List (intercalate)
import Data.Function.Memoize (memoize2)
node' :: Int -> Int -> Integer
node' x y
| y == 0 = if x == 0 then 1 else 0
| otherwise = sum [ pathSum (x + d) (y - 1) | d <- [-1..1]]
node = memoize2 node'
pathCount' :: Int -> Int -> Integer
pathCount' x y
| y == 0 = if x == 0 then 1 else 0
| otherwise = sum [ pathCount (x + d) (y - 1) | d <- [-1..1]]
pathCount = memoize2 pathCount'
pathSum' :: Int -> Int -> Integer
pathSum' x y
| y == 0 = if x == 0 then 1 else 0
| otherwise = sum [ pathSum (x + d) (y - 1) + node x y * pathCount (x + d) (y - 1)
| d <- [-1..1] ]
pathSum = memoize2 pathSum'
main =
forM_ [0..n] $ \y ->
putStrLn $ intercalate " " $ map (show . flip node y) [-n..n]
where n = 5
Output:
0 0 0 0 0 1 0 0 0 0 0
0 0 0 0 1 1 1 0 0 0 0
0 0 0 2 4 6 4 2 0 0 0
0 0 4 16 40 48 40 16 4 0 0
0 8 72 352 728 944 728 352 72 8 0
16 376 4248 16608 35128 43632 35128 16608 4248 376 16
正如您所看到的算法,数字的大小很快就会失控。所以运行时间是notO(n^2),而算术运算次数为。