Question

我在理解Haskell中Knuth-Morris-Pratt算法的实现时遇到了麻烦。

http://twanvl.nl/blog/haskell/Knuth-Morris-Pratt-in-Haskell

特别是我不了解自动机的结构。我知道它使用“绑结”方法来构建它，但我不清楚，我也不知道为什么它应该具有正确的复杂性。

我想知道的另一件事是你是否认为这个实现可以很容易地推广到实现Aho-Corasick算法。

感谢您的回答！

Answer 1

所以这是算法：

makeTable :: Eq a => [a] -> KMP a
makeTable xs = table
   where table = makeTable' xs (const table)

makeTable' []     failure = KMP True failure
makeTable' (x:xs) failure = KMP False test
   where  test  c = if c == x then success else failure c
          success = makeTable' xs (next (failure x))

使用它，让我们看一下为“shoeshop”构建的表：

makeTable "shoeshop" = table0

table0 = makeTable' "shoeshop" (const table0)
       = KMP False test0

test0 c = if c == 's' then success1 else const table0 c
        = if c == 's' then success1 else table0

success1 = makeTable' "hoeshop" (next (const table0 's'))
         = makeTable' "hoeshop" (next table0)
         = makeTable' "hoeshop" test0
         = KMP False test1

test1 c = if c == 'h' then success2 else test0 c

success2 = makeTable' "oeshop" (next (test0 'h'))
         = makeTable' "oeshop" (next table0)
         = makeTable' "oeshop" test0
         = makeTable' "oeshop" test0
         = KMP False test2

test2 c = if c == 'o' then success3 else test0 c

success3 = makeTable' "eshop" (next (test0 'o'))
         = makeTable' "eshop" (next table0)
         = makeTable' "eshop" test0
         = KMP False test3

test3 c = if c == 'e' then success4 else test0 c

success4 = makeTable' "shop" (next (test0 'e'))
         = makeTable' "shop" (next table0)
         = makeTable' "shop" test0
         = KMP False test4

test4 c = if c == 's' then success5 else test0 c

success5 = makeTable' "hop" (next (test0 's'))
         = makeTable' "hop" (next success1)
         = makeTable' "hop" test1
         = KMP False test5

test5 c = if c == 'h' then success6 else test1 c

success6 = makeTable' "op" (next (test1 'h'))
         = makeTable' "op" (next success2)
         = makeTable' "op" test2
         = KMP False test6

test6 c = if c == 'o' then success7 else test2 c

success7 = makeTable' "p" (next (test2 'o'))
         = makeTable' "p" (next success3)
         = makeTable' "p" test3
         = KMP False test7

test7 c = if c == 'p' then success8 else test3 c

success8 = makeTable' "" (next (test3 'p'))
         = makeTable' "" (next (test0 'p'))
         = makeTable' "" (next table0)
         = makeTable' "" test0
         = KMP True test0

注意success5如何使用消耗的's'来回溯模式的初始's'。

现在逐步了解isSubstringOf2 "shoeshop" $ cycle "shoe"时发生的事情。

当test7无法与'p'匹配时，会发现它会回到test3以尝试匹配'e'，以便我们循环显示success4，{{1}无限期地{}，success5和success6。

Haskell中的Knuth-Morris-Pratt算法

1 个答案: