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

Answer 1

So here's the algorithm:

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))

Using that, let's look at the table constructed for "shoeshop":

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

Note how success5 uses the consumed 's' to retrace the initial 's' of the pattern.

Now walk through what happens when you do isSubstringOf2 "shoeshop" $ cycle "shoe".

See that when test7 fails to match 'p', it falls back to test3 to try to match 'e', so that we cycle through success4, success5, success6 and success7 ad infinitum.