haskell - Knuth-Morris-Pratt algorithm in Haskell

Question

I have a trouble with understanding this implementation of the Knuth-Morris-Pratt algorithm in Haskell.

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

In particular I don't understand the construction of the automaton. I know that it uses the "Tying the Knot" method to construct it, but it isn't clear to me and I also don't know why it should have the right complexity.

Another thing I would like to know is whether you think that this implementation could be easily generalized to implement the Aho-Corasick algorithm.

Thanks for your answers!

Answer 1

所以这里的算法：

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，让我们看看为“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

请注意如何success5使用消耗的“s”来回溯模式的初始“s”。

现在看看当你这样做时会发生什么isSubstringOf2 "shoeshop" $ cycle "shoe"。

看到当test7匹配 'p' 失败时，它会回退到test3尝试匹配 'e'，这样我们就可以循环遍历success4, success5,success6和success7ad infinitum。