Howto remove a word from a Trie structure? - data-structures

Question

maybe I'm not smart enough to learn Haskell, but I'd give it the last chance.
I've got stuck at implementation of entry removal from a tree, Trie like structure to be a more specific (http://en.wikipedia.org/wiki/Trie).
I'm looking for any advices (not a solution !) how to implement such pure function.
I've had an idea about one algorithm. Recreate a new tree by traversing whole tree "skipping" values equal to each character of the word, with edge condition return original tree if next character won't be found. But there arises a problem when a character also belongs to another word.
data Trie = Trie { commonness :: Maybe Int
, children :: [(Char, Trie)]
} deriving (Eq, Read, Show)
-- Creates an empty "dictionary"
trie :: Trie
trie = Trie { commonness = Nothing, children = [] }
-- Inserts a word with given commonness into dictionary
add :: String -> Int -> Trie -> Trie
add [] freq tree
| (0 <= freq) && (freq <= 16) = tree { commonness = Just freq }
| otherwise = error $ "Commonness out of bounds: " ++ (show freq)
add word freq tree = tree { children = traverse word (children tree) }
where
traverse [] tree = error $ "traverse called with [] " ++ (show tree)
traverse (x:xs) [] = [(x, add xs freq trie)]
traverse str#(x:xs) (t:ts)
| x == fst t = (x, add xs freq $ snd t):ts
| otherwise = t:(traverse str ts)
remove :: String -> Trie -> Trie
???
And the data looks like:
GHCi> putStrLn $ groom $ add "learn" 16 $ add "leap" 5 $ add "sing" 7 $ add "lift" 10 trie
Trie{commonness = Nothing,
children =
[('l',
Trie{commonness = Nothing,
children =
[('i',
Trie{commonness = Nothing,
children =
[('f',
Trie{commonness = Nothing,
children = [('t', Trie{commonness = Just 10, children = []})]})]}),
('e',
Trie{commonness = Nothing,
children =
[('a',
Trie{commonness = Nothing,
children =
[('p', Trie{commonness = Just 5, children = []}),
('r',
Trie{commonness = Nothing,
children =
[('n',
Trie{commonness = Just 16, children = []})]})]})]})]}),
('s',
Trie{commonness = Nothing,
children =
[('i',
Trie{commonness = Nothing,
children =
[('n',
Trie{commonness = Nothing,
children =
[('g', Trie{commonness = Just 7, children = []})]})]})]})]}

Answer 1

This is going to be easier if you use a Map Char Trie instead of [(Char,Trie)] for your child table. That is what I'm going to assume for this answer. I'll get you started with the inductive case:
import qualified Data.Map as Map
remove :: String -> Trie -> Trie
remove (c:cs) t = t { children = Map.alter remove' c (children t) }
where
remove' (Just t) = Just (remove cs t)
remove' Nothing = Nothing
remove [] t = ...
I'll leave the base case to you. Here are the docs for the Map function I used, alter. You could get this same solution without using Map if you implemented alter for [(Char,a)].
Exercise: remove' is pretty wordy. See if you can shorten it using fmap.

Answer 2

In Python you can do something like this
def remove_string_helper(self, string, pnode, index):
if pnode:
flag = False
if index < len(string):
flag = self.remove_string_helper(string, pnode.childs.get(string[index]), index + 1)
if index == len(string) and pnode.is_complete_word:
pnode.is_complete_word = False
return len(pnode.childs) == 0
if flag:
pnode.childs.pop(string[index])
return len(self.childs) == 0
return False
def remove_string(self, string):
self.remove_string_helper(string, self.childs.get(string[0]), 1)