# 99 questions/70B to 73

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(moved solutions to subpages of 99 questions/Solutions) |
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Example in Prolog: |
Example in Prolog: |
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<pre> |
<pre> |
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?- istree(t(a,[t(f,[t(g,[])]),t(c,[]),t(b,[t(d,[]),t(e,[])])])). |
?- istree(t(a,[t(f,[t(g,[])]),t(c,[]),t(b,[t(d,[]),t(e,[])])])). |
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In Haskell, we define multiway trees as a datatype, as in the module [http://www.haskell.org/ghc/docs/latest/html/libraries/base/Data-Tree.html Data.Tree]: |
In Haskell, we define multiway trees as a datatype, as in the module [http://www.haskell.org/ghc/docs/latest/html/libraries/base/Data-Tree.html Data.Tree]: |
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<haskell> |
<haskell> |
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data Tree a = Node a [Tree a] |
data Tree a = Node a [Tree a] |
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deriving (Eq, Show) |
deriving (Eq, Show) |
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</haskell> |
</haskell> |
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Some example trees: |
Some example trees: |
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<haskell> |
<haskell> |
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tree1 = Node 'a' [] |
tree1 = Node 'a' [] |
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] |
] |
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</haskell> |
</haskell> |
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The last is the tree illustrated above. |
The last is the tree illustrated above. |
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− | As in problem 54A, all members of this type are multiway trees; |
+ | As in problem 54A, all members of this type are multiway trees; there is no use for a predicate to test them. |

− | there is no use for a predicate to test them. |
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== Problem 70C == |
== Problem 70C == |
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Example in Haskell: |
Example in Haskell: |
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− | <pre> |
+ | |

+ | <haskell> |
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Tree> nnodes tree2 |
Tree> nnodes tree2 |
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2 |
2 |
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− | </pre> |
+ | </haskell> |

− | Solution: |
+ | [[99 questions/Solutions/70C | Solutions]] |

− | <haskell> |
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− | nnodes :: Tree a -> Int |
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− | nnodes (Node _ ts) = 1 + sum (map nnodes ts) |
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− | </haskell> |
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== Problem 70 == |
== Problem 70 == |
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Make your predicate work in both directions. |
Make your predicate work in both directions. |
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− | Solution: |
+ | [[99 questions/Solutions/70 | Solutions]] |

− | We could write separate printing and parsing functions, but the problem statement asks for a bidirectional function. |
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− | First we need a parser monad, with some primitives: |
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− | <haskell> |
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− | newtype P a = P { runP :: String -> Maybe (a, String) } |
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− | |||

− | instance Monad P where |
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− | return x = P $ \ s -> Just (x, s) |
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− | P v >>= f = P $ \ s -> do |
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− | (x, s') <- v s |
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− | runP (f x) s' |
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− | |||

− | instance MonadPlus P where |
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− | mzero = P $ \ _ -> Nothing |
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− | P u `mplus` P v = P $ \ s -> u s `mplus` v s |
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− | |||

− | charP :: P Char |
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− | charP = P view_list |
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− | where view_list [] = Nothing |
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− | view_list (c:cs) = Just (c, cs) |
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− | |||

− | literalP :: Char -> P () |
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− | literalP c = do { c' <- charP; guard (c == c') } |
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− | |||

− | spaceP :: P () |
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− | spaceP = P (\ s -> Just ((), dropWhile isSpace s)) |
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− | </haskell> |
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− | Next a <tt>Syntax</tt> type, combining printing and parsing functions: |
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− | <haskell> |
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− | data Syntax a = Syntax { |
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− | display :: a -> String, |
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− | parse :: P a |
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− | } |
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− | </haskell> |
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− | (We don't use a class, because we want multiple syntaxes for a given type.) |
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− | Some combinators for building syntaxes: |
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− | <haskell> |
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− | -- concatenation |
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− | (<*>) :: Syntax a -> Syntax b -> Syntax (a,b) |
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− | a <*> b = Syntax { |
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− | display = \ (va,vb) -> display a va ++ display b vb, |
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− | parse = liftM2 (,) (parse a) (parse b) |
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− | } |
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− | |||

− | -- alternatives |
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− | (<|>) :: Syntax a -> Syntax b -> Syntax (Either a b) |
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− | a <|> b = Syntax { |
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− | display = either (display a) (display b), |
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− | parse = liftM Left (parse a) `mplus` liftM Right (parse b) |
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− | } |
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− | |||

− | char :: Syntax Char |
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− | char = Syntax return charP |
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− | |||

− | literal :: Char -> Syntax () |
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− | literal c = Syntax (const [c]) (literalP c) |
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− | |||

− | space :: Syntax () |
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− | space = Syntax (const " ") spaceP |
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− | |||

− | iso :: (a -> b) -> (b -> a) -> Syntax a -> Syntax b |
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− | iso a_to_b b_to_a a = Syntax { |
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− | display = display a . b_to_a, |
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− | parse = liftM a_to_b (parse a) |
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− | } |
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− | </haskell> |
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− | The last one maps a syntax using an isomorphism between types. |
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− | Some uses of this function: |
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− | <haskell> |
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− | -- concatenation, with no value in the first part |
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− | (*>) :: Syntax () -> Syntax a -> Syntax a |
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− | p *> q = iso snd ((,) ()) (p <*> q) |
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− | |||

− | -- list of a's, followed by finish |
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− | list :: Syntax a -> Syntax () -> Syntax [a] |
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− | list a finish = iso toList fromList (finish <|> (a <*> list a finish)) |
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− | where toList (Left _) = [] |
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− | toList (Right (x, xs)) = x:xs |
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− | fromList [] = Left () |
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− | fromList (x:xs) = Right (x, xs) |
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− | </haskell> |
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− | Now we can define the syntax of depth-first presentations: |
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− | <haskell> |
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− | df :: Syntax (Tree Char) |
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− | df = iso toTree fromTree (char <*> list df (literal '^')) |
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− | where toTree (x, ts) = Node x ts |
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− | fromTree (Node x ts) = (x, ts) |
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− | </haskell> |
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− | We are using the isomorphism between <tt>Tree a</tt> and <tt>(a, [Tree a])</tt>. |
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− | Some examples: |
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− | <pre> |
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− | Tree> display df tree5 |
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− | "afg^^c^bd^e^^^" |
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− | Tree> runP (parse df) "afg^^c^bd^e^^^" |
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− | Just (Node 'a' [Node 'f' [Node 'g' []],Node 'c' [],Node 'b' [Node 'd' [],Node 'e' []]],"") |
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− | </pre> |
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== Problem 71 == |
== Problem 71 == |
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Example in Haskell: |
Example in Haskell: |
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− | <pre> |
+ | |

+ | <haskell> |
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Tree> ipl tree5 |
Tree> ipl tree5 |
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9 |
9 |
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Tree> ipl tree4 |
Tree> ipl tree4 |
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2 |
2 |
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− | </pre> |
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− | |||

− | Solution: |
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− | <haskell> |
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− | ipl :: Tree a -> Int |
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− | ipl = ipl' 0 |
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− | where ipl' d (Node _ ts) = d + sum (map (ipl' (d+1)) ts) |
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</haskell> |
</haskell> |
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+ | |||

+ | [[99 questions/Solutions/71 | Solutions]] |
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+ | |||

== Problem 72 == |
== Problem 72 == |
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Example in Haskell: |
Example in Haskell: |
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− | <pre> |
+ | |

+ | <haskell> |
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Tree> bottom_up tree5 |
Tree> bottom_up tree5 |
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"gfcdeba" |
"gfcdeba" |
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− | </pre> |
+ | </haskell> |

+ | |||

+ | [[99 questions/Solutions/72 | Solutions]] |
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− | Solution: |
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− | <haskell> |
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− | bottom_up :: Tree a -> [a] |
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− | bottom_up (Node x ts) = concatMap bottom_up ts ++ [x] |
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− | </haskell> |
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− | A more efficient version using an accumulator: |
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− | <haskell> |
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− | bottom_up :: Tree a -> [a] |
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− | bottom_up t = bottom_up_aux t [] |
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− | where bottom_up_aux :: Tree a -> [a] -> [a] |
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− | bottom_up_aux (Node x ts) xs = foldr bottom_up_aux (x:xs) ts |
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− | </haskell> |
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== Problem 73 == |
== Problem 73 == |
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Example in Haskell: |
Example in Haskell: |
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− | <pre> |
+ | |

+ | <haskell> |
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Tree> display lisp tree1 |
Tree> display lisp tree1 |
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"a" |
"a" |
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Tree> display lisp tree5 |
Tree> display lisp tree5 |
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"(a (f g) c (b d e))" |
"(a (f g) c (b d e))" |
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− | </pre> |
+ | </haskell> |

As a second, even more interesting exercise try to rewrite tree_ltl/2 in a way that the inverse conversion is also possible. |
As a second, even more interesting exercise try to rewrite tree_ltl/2 in a way that the inverse conversion is also possible. |
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− | Solution: |
+ | [[99 questions/Solutions/73 | Solutions]] |

− | using the <tt>Syntax</tt> type used in problem 70 above: |
+ | |

− | <haskell> |
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− | lisp :: Syntax (Tree Char) |
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− | lisp = iso toTree fromTree |
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− | (literal '(' *> (char <*> list (space *> lisp) (literal ')')) <|> char) |
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− | where toTree (Left (x, ts)) = Node x ts |
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− | toTree (Right x) = Node x [] |
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− | fromTree (Node x []) = Right x |
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− | fromTree (Node x ts) = Left (x, ts) |
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− | </haskell> |
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− | Here we use the isomorphism between <tt>Tree a</tt> and <tt>Either (a, [Tree a]) a</tt>, where the list is non-empty. |
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[[Category:Tutorials]] |
[[Category:Tutorials]] |

## Revision as of 15:45, 15 July 2010

This is part of Ninety-Nine Haskell Problems, based on Ninety-Nine Prolog Problems.

## 1 Multiway Trees

A multiway tree is composed of a root element and a (possibly empty) set of successors which are multiway trees themselves. A multiway tree is never empty. The set of successor trees is sometimes called a forest.

## 2 Problem 70B

(*) Check whether a given term represents a multiway tree.

In Prolog or Lisp, one writes a predicate to check this.

Example in Prolog:

?- istree(t(a,[t(f,[t(g,[])]),t(c,[]),t(b,[t(d,[]),t(e,[])])])). Yes

In Haskell, we define multiway trees as a datatype, as in the module Data.Tree:

data Tree a = Node a [Tree a] deriving (Eq, Show)

Some example trees:

tree1 = Node 'a' [] tree2 = Node 'a' [Node 'b' []] tree3 = Node 'a' [Node 'b' [Node 'c' []]] tree4 = Node 'b' [Node 'd' [], Node 'e' []] tree5 = Node 'a' [ Node 'f' [Node 'g' []], Node 'c' [], Node 'b' [Node 'd' [], Node 'e' []] ]

The last is the tree illustrated above.

As in problem 54A, all members of this type are multiway trees; there is no use for a predicate to test them.

## 3 Problem 70C

(*) Count the nodes of a multiway tree.

Example in Haskell:

Tree> nnodes tree2 2

## 4 Problem 70

(**) 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.

## 5 Problem 71

(*) Determine the internal path length of a tree.

We define the internal path length of a multiway tree as the total sum of the path lengths from the root to all nodes of the tree. By this definition, `tree5` has an internal path length of 9.

Example in Haskell:

Tree> ipl tree5 9 Tree> ipl tree4 2

## 6 Problem 72

(*) Construct the bottom-up order sequence of the tree nodes.

Write a predicate bottom_up(Tree,Seq) which constructs the bottom-up sequence of the nodes of the multiway tree Tree.

Example in Haskell:

Tree> bottom_up tree5 "gfcdeba"

## 7 Problem 73

(**) Lisp-like tree representation.

There is a particular notation for multiway trees in Lisp. Lisp is a prominent functional programming language, which is used primarily for artificial intelligence problems. As such it is one of the main competitors of Prolog. In Lisp almost everything is a list, just as in Prolog everything is a term.

The following pictures show how multiway tree structures are represented in Lisp.

Note that in the "lispy" notation a node with successors (children) in the tree is always the first element in a list, followed by its children. The "lispy" representation of a multiway tree is a sequence of atoms and parentheses '(' and ')', which we shall collectively call "tokens". We can represent this sequence of tokens as a Prolog list; e.g. the lispy expression (a (b c)) could be represented as the Prolog list ['(', a, '(', b, c, ')', ')']. Write a predicate tree_ltl(T,LTL) which constructs the "lispy token list" LTL if the tree is given as term T in the usual Prolog notation.

(The Prolog example given is incorrect.)

Example in Haskell:

Tree> display lisp tree1 "a" Tree> display lisp tree2 "(a b)" Tree> display lisp tree3 "(a (b c))" Tree> display lisp tree4 "(b d e)" Tree> display lisp tree5 "(a (f g) c (b d e))"

As a second, even more interesting exercise try to rewrite tree_ltl/2 in a way that the inverse conversion is also possible.