# Difference between revisions of "99 questions/Solutions/70"

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Line 173: | Line 173: | ||

treeToString (Node x ts) | treeToString (Node x ts) | ||

= [x] ++ (concat $ intersperse "^" (map treeToString ts)) ++ "^" | = [x] ++ (concat $ intersperse "^" (map treeToString ts)) ++ "^" | ||

+ | </haskell> | ||

+ | |||

+ | ---- | ||

+ | |||

+ | Since the exercise was really designed for Prolog, let's write an implementation at the class level. This is really a parsing/pretty-printing solution, with the names to match. The requested predicate is BuildTree: | ||

+ | |||

+ | <haskell> | ||

+ | class ParseTree str t '[] => BuildTree (str :: [Instr c]) (t :: Tree c) | str -> t, t -> str | ||

+ | instance ParseTree str t '[] => BuildTree str t | ||

+ | </haskell> | ||

+ | |||

+ | We use two helpers to implement it: | ||

+ | |||

+ | <haskell> | ||

+ | class ParseTree (str :: [Instr c]) (t :: Tree c) (rem :: [Instr c]) | str -> t rem, t rem -> str | ||

+ | class ParseForest (str :: [Instr c]) (ts :: [Tree c]) (rem :: [Instr c]) | str -> ts rem, ts rem -> str | ||

+ | |||

+ | instance ParseTree ('Down c ': 'Up ': r) ('Node c '[]) r | ||

+ | instance ParseForest ('Down d ': is) (t ': ts) r => ParseTree ('Down c ': 'Down d ': is) ('Node c (t ': ts)) r | ||

+ | instance ParseForest ('Up ': is) '[] is | ||

+ | instance ( ParseTree ('Down c ': is) t r | ||

+ | , ParseForest r ts r') => ParseForest ('Down c ': is) (t ': ts) r' | ||

</haskell> | </haskell> | ||

[[Category:Programming exercise spoilers]] | [[Category:Programming exercise spoilers]] |

## Revision as of 19:46, 23 September 2017

(**) Tree construction from a node string.

We suppose that the nodes of a multiway tree contain single characters. In the depth-first order sequence of its nodes, a special character ^ has been inserted whenever, during the tree traversal, the move is a backtrack to the previous level.

By this rule, the tree below (`tree5`) is represented as: `afg^^c^bd^e^^^`

Define the syntax of the string and write a predicate tree(String,Tree) to construct the Tree when the String is given. Make your predicate work in both directions.

We could write separate printing and parsing functions, but the problem statement asks for a bidirectional function.

First we need a parser monad, with some primitives:

```
newtype P a = P { runP :: String -> Maybe (a, String) }
instance Monad P where
return x = P $ \ s -> Just (x, s)
P v >>= f = P $ \ s -> do
(x, s') <- v s
runP (f x) s'
instance MonadPlus P where
mzero = P $ \ _ -> Nothing
P u `mplus` P v = P $ \ s -> u s `mplus` v s
charP :: P Char
charP = P view_list
where view_list [] = Nothing
view_list (c:cs) = Just (c, cs)
literalP :: Char -> P ()
literalP c = do { c' <- charP; guard (c == c') }
spaceP :: P ()
spaceP = P (\ s -> Just ((), dropWhile isSpace s))
```

Next a `Syntax` type, combining printing and parsing functions:

```
data Syntax a = Syntax {
display :: a -> String,
parse :: P a
}
```

(We don't use a class, because we want multiple syntaxes for a given type.) Some combinators for building syntaxes:

```
-- concatenation
(<*>) :: Syntax a -> Syntax b -> Syntax (a,b)
a <*> b = Syntax {
display = \ (va,vb) -> display a va ++ display b vb,
parse = liftM2 (,) (parse a) (parse b)
}
-- alternatives
(<|>) :: Syntax a -> Syntax b -> Syntax (Either a b)
a <|> b = Syntax {
display = either (display a) (display b),
parse = liftM Left (parse a) `mplus` liftM Right (parse b)
}
char :: Syntax Char
char = Syntax return charP
literal :: Char -> Syntax ()
literal c = Syntax (const [c]) (literalP c)
space :: Syntax ()
space = Syntax (const " ") spaceP
iso :: (a -> b) -> (b -> a) -> Syntax a -> Syntax b
iso a_to_b b_to_a a = Syntax {
display = display a . b_to_a,
parse = liftM a_to_b (parse a)
}
```

The last one maps a syntax using an isomorphism between types. Some uses of this function:

```
-- concatenation, with no value in the first part
(*>) :: Syntax () -> Syntax a -> Syntax a
p *> q = iso snd ((,) ()) (p <*> q)
-- list of a's, followed by finish
list :: Syntax a -> Syntax () -> Syntax [a]
list a finish = iso toList fromList (finish <|> (a <*> list a finish))
where toList (Left _) = []
toList (Right (x, xs)) = x:xs
fromList [] = Left ()
fromList (x:xs) = Right (x, xs)
```

Now we can define the syntax of depth-first presentations:

```
df :: Syntax (Tree Char)
df = iso toTree fromTree (char <*> list df (literal '^'))
where toTree (x, ts) = Node x ts
fromTree (Node x ts) = (x, ts)
```

We are using the isomorphism between `Tree a` and `(a, [Tree a])`.
Some examples:

```
Tree> display df tree5
"afg^^c^bd^e^^^"
Tree> runP (parse df) "afg^^c^bd^e^^^"
Just (Node 'a' [Node 'f' [Node 'g' []],Node 'c' [],Node 'b' [Node 'd' [],Node 'e' []]],"")
```

A more naive solution, trying to split the string with stack

```
stringToTree :: String -> Tree Char
stringToTree (x:xs@(y:ys))
| y == '^' = Node x []
| otherwise = Node x (map stringToTree subs)
where subs = snd $ foldl parse ([],[]) $ init xs
parse ([],[]) z = ([z], [[z]])
parse (stack, acc) z = (stack', acc')
where stack'
| z == '^' = init stack
| otherwise = stack ++ [z]
acc' = if stack == []
then acc ++ [[z]]
else (init acc) ++ [(last acc) ++ [z]]
```

A simple solution that uses Standard Prelude functions:

```
stringToTree :: String -> Tree Char
stringToTree (x:'^':"") = Node x []
stringToTree (x:xs) = Node x ys
where
z = map fst $ filter ((==) 0 . snd) $ zip [0..] $
scanl (+) 0 $ map (\x -> if x == '^' then -1 else 1) xs
ys = map (stringToTree . uncurry (sub xs)) $ zip (init z) (tail z)
sub s a b = take (b - a) $ drop a s
```

A solution similar to that of problem 69, special casing the root node:

```
stringToTree :: String -> Tree Char
stringToTree (x:xs) = Node x (fst (stringToTrees xs))
where stringToTrees (x:xs)
| x == '^' = ([], xs)
| otherwise = ([Node x trees0] ++ trees1, rest1)
where (trees0, rest0) = stringToTrees xs
(trees1, rest1) = stringToTrees rest0
```

It's more direct to convert Tree back to string

```
import Data.List
treeToString :: Tree Char -> String
treeToString (Node x ts)
= [x] ++ (concat $ intersperse "^" (map treeToString ts)) ++ "^"
```

Since the exercise was really designed for Prolog, let's write an implementation at the class level. This is really a parsing/pretty-printing solution, with the names to match. The requested predicate is BuildTree:

```
class ParseTree str t '[] => BuildTree (str :: [Instr c]) (t :: Tree c) | str -> t, t -> str
instance ParseTree str t '[] => BuildTree str t
```

We use two helpers to implement it:

```
class ParseTree (str :: [Instr c]) (t :: Tree c) (rem :: [Instr c]) | str -> t rem, t rem -> str
class ParseForest (str :: [Instr c]) (ts :: [Tree c]) (rem :: [Instr c]) | str -> ts rem, ts rem -> str
instance ParseTree ('Down c ': 'Up ': r) ('Node c '[]) r
instance ParseForest ('Down d ': is) (t ': ts) r => ParseTree ('Down c ': 'Down d ': is) ('Node c (t ': ts)) r
instance ParseForest ('Up ': is) '[] is
instance ( ParseTree ('Down c ': is) t r
, ParseForest r ts r') => ParseForest ('Down c ': is) (t ': ts) r'
```