# 99 questions/54A to 60

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This is part of [[H-99:_Ninety-Nine_Haskell_Problems|Ninety-Nine Haskell Problems]], based on [https://prof.ti.bfh.ch/hew1/informatik3/prolog/p-99/ Ninety-Nine Prolog Problems]. |
This is part of [[H-99:_Ninety-Nine_Haskell_Problems|Ninety-Nine Haskell Problems]], based on [https://prof.ti.bfh.ch/hew1/informatik3/prolog/p-99/ Ninety-Nine Prolog Problems]. |
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− | If you want to work on one of these, put your name in the block so we know someone's working on it. Then, change n in your block to the appropriate problem number, and fill in the <Problem description>,<example in lisp>,<example in Haskell>,<solution in haskell> and <description of implementation> fields. |
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== Binary trees == |
== Binary trees == |
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A binary tree is either empty or it is composed of a root element and two successors, which are binary trees themselves. |
A binary tree is either empty or it is composed of a root element and two successors, which are binary trees themselves. |
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− | http://www.hta-bi.bfh.ch/~hew/informatik3/prolog/p-99/p67.gif |
+ | https://prof.ti.bfh.ch/hew1/informatik3/prolog/p-99/p67.gif |

+ | |||

+ | In Haskell, we can characterize binary trees with a datatype definition: |
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+ | |||

+ | <haskell> |
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+ | data Tree a = Empty | Branch a (Tree a) (Tree a) |
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+ | deriving (Show, Eq) |
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+ | </haskell> |
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+ | |||

+ | This says that a <tt>Tree</tt> of type <tt>a</tt> consists of either an <tt>Empty</tt> node, or a <tt>Branch</tt> containing one value of type <tt>a</tt> with exactly two subtrees of type <tt>a</tt>. |
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+ | |||

+ | Given this definition, the tree in the diagram above would be represented as: |
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+ | |||

+ | <haskell> |
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+ | tree1 = Branch 'a' (Branch 'b' (Branch 'd' Empty Empty) |
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+ | (Branch 'e' Empty Empty)) |
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+ | (Branch 'c' Empty |
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+ | (Branch 'f' (Branch 'g' Empty Empty) |
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+ | Empty)) |
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+ | </haskell> |
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+ | |||

+ | Since a "leaf" node is a branch with two empty subtrees, it can be useful to define a shorthand function: |
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+ | |||

+ | <haskell> |
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+ | leaf x = Branch x Empty Empty |
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+ | </haskell> |
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+ | |||

+ | Then the tree diagram above could be expressed more simply as: |
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+ | |||

+ | <haskell> |
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+ | tree1' = Branch 'a' (Branch 'b' (leaf 'd') |
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+ | (leaf 'e')) |
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+ | (Branch 'c' Empty |
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+ | (Branch 'f' (leaf 'g') |
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+ | Empty))) |
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+ | </haskell> |
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+ | |||

+ | Other examples of binary trees: |
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+ | |||

+ | <haskell> |
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+ | -- A binary tree consisting of a root node only |
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+ | tree2 = Branch 'a' Empty Empty |
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+ | |||

+ | -- An empty binary tree |
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+ | tree3 = Empty |
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+ | |||

+ | -- A tree of integers |
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+ | tree4 = Branch 1 (Branch 2 Empty (Branch 4 Empty Empty)) |
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+ | (Branch 2 Empty Empty) |
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+ | </haskell> |
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== Problem 54A == |
== Problem 54A == |
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Line 18: | Line 16: | ||

Example in Lisp: |
Example in Lisp: |
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+ | |||

<pre> |
<pre> |
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* (istree (a (b nil nil) nil)) |
* (istree (a (b nil nil) nil)) |
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Line 25: | Line 24: | ||

</pre> |
</pre> |
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− | In Haskell, we characterize binary trees with a datatype definition: |
+ | Non-solution: |

− | <haskell> |
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− | data Tree a = Empty | Branch a (Tree a) (Tree a) |
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− | deriving (Show, Eq) |
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− | </haskell> |
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− | The above tree is represented as: |
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− | <haskell> |
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− | tree1 = Branch 'a' (Branch 'b' (Branch 'd' Empty Empty) |
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− | (Branch 'e' Empty Empty)) |
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− | (Branch 'c' Empty |
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− | (Branch 'f' (Branch 'g' Empty Empty) |
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− | Empty))) |
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− | </haskell> |
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− | Other examples of binary trees: |
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− | <haskell> |
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− | tree2 = Branch 'a' Empty Empty -- a binary tree consisting of a root node only |
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− | tree3 = nil -- an empty binary tree |
+ | Haskell's type system ensures that all terms of type <hask>Tree a</hask> are binary trees: it is just not possible to construct an invalid tree with this type. Hence, it is redundant to introduce a predicate to check this property: it would always return <hask>True</hask>. |

− | tree4 = Branch 1 (Branch 2 Empty (Branch 4 Empty Empty)) |
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− | (Branch 2 Empty Empty) |
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− | </haskell> |
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− | The type system ensures that all terms of type <hask>Tree a</hask> |
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− | are binary trees: it is just not possible to construct an invalid tree |
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− | with this type. Hence, it is redundant to introduce a predicate to |
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− | check this property: it would always return <hask>True</hask>. |
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== Problem 55 == |
== Problem 55 == |
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Line 46: | Line 38: | ||

Example: |
Example: |
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+ | |||

<pre> |
<pre> |
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* cbal-tree(4,T). |
* cbal-tree(4,T). |
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Line 54: | Line 47: | ||

Example in Haskell, whitespace and "comment diagrams" added for clarity and exposition: |
Example in Haskell, whitespace and "comment diagrams" added for clarity and exposition: |
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− | <pre> |
+ | |

+ | <haskell> |
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*Main> cbalTree 4 |
*Main> cbalTree 4 |
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[ |
[ |
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Line 97: | Line 90: | ||

(Branch 'x' Empty Empty) |
(Branch 'x' Empty Empty) |
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] |
] |
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− | </pre> |
+ | </haskell> |

− | Solution: |
+ | [[99 questions/Solutions/55 | Solutions]] |

− | <haskell> |
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− | cbalTree 0 = [Empty] |
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− | cbalTree n = [Branch 'x' l r | i <- [q .. q + r], l <- cbalTree i, r <- cbalTree (n - i - 1)] |
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− | where (q, r) = quotRem (n-1) 2 |
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− | </haskell> |
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− | Here we use the list monad to enumerate all the trees, in a style that is more natural than standard backtracking. |
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== Problem 56 == |
== Problem 56 == |
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Line 110: | Line 102: | ||

Example in Haskell: |
Example in Haskell: |
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− | <pre> |
+ | |

+ | <haskell> |
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*Main> symmetric (Branch 'x' (Branch 'x' Empty Empty) Empty) |
*Main> symmetric (Branch 'x' (Branch 'x' Empty Empty) Empty) |
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False |
False |
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*Main> symmetric (Branch 'x' (Branch 'x' Empty Empty) (Branch 'x' Empty Empty)) |
*Main> symmetric (Branch 'x' (Branch 'x' Empty Empty) (Branch 'x' Empty Empty)) |
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True |
True |
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− | </pre> |
+ | </haskell> |

− | Solution: |
+ | [[99 questions/Solutions/56 | Solutions]] |

− | <haskell> |
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− | mirror Empty Empty = True |
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− | mirror (Branch _ a b) (Branch _ x y) = mirror a y && mirror b x |
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− | mirror _ _ = False |
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− | symmetric Empty = True |
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− | symmetric (Branch _ l r) = mirror l r |
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− | </haskell> |
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== Problem 57 == |
== Problem 57 == |
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Line 130: | Line 119: | ||

Example: |
Example: |
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+ | |||

<pre> |
<pre> |
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* construct([3,2,5,7,1],T). |
* construct([3,2,5,7,1],T). |
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Line 138: | Line 128: | ||

Example: |
Example: |
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+ | |||

<pre> |
<pre> |
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* test-symmetric([5,3,18,1,4,12,21]). |
* test-symmetric([5,3,18,1,4,12,21]). |
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Yes |
Yes |
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− | * test-symmetric([3,2,5,7,1]). |
+ | * test-symmetric([3,2,5,7,4]). |

No |
No |
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</pre> |
</pre> |
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Example in Haskell: |
Example in Haskell: |
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− | <pre> |
+ | |

+ | <haskell> |
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*Main> construct [3, 2, 5, 7, 1] |
*Main> construct [3, 2, 5, 7, 1] |
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Branch 3 (Branch 2 (Branch 1 Empty Empty) Empty) (Branch 5 Empty (Branch 7 Empty Empty)) |
Branch 3 (Branch 2 (Branch 1 Empty Empty) Empty) (Branch 5 Empty (Branch 7 Empty Empty)) |
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Line 153: | Line 144: | ||

*Main> symmetric . construct $ [3, 2, 5, 7, 1] |
*Main> symmetric . construct $ [3, 2, 5, 7, 1] |
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True |
True |
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− | </pre> |
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− | |||

− | Solution: |
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− | <haskell> |
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− | add :: Ord a => a -> Tree a -> Tree a |
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− | add x Empty = Branch x Empty Empty |
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− | add x t@(Branch y l r) = case compare x y of |
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− | LT -> Branch y (add x l) r |
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− | GT -> Branch y l (add x r) |
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− | EQ -> t |
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− | |||

− | construct xs = foldl (flip add) Empty xs |
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</haskell> |
</haskell> |
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− | Here, the definition of construct is trivial, because the pattern of accumulating from the left is captured by the standard function foldl. |
+ | [[99 questions/Solutions/57 | Solutions]] |

+ | |||

== Problem 58 == |
== Problem 58 == |
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Line 176: | Line 155: | ||

Example: |
Example: |
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+ | |||

<pre> |
<pre> |
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* sym-cbal-trees(5,Ts). |
* sym-cbal-trees(5,Ts). |
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Example in Haskell: |
Example in Haskell: |
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− | <pre> |
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− | *Main> symCbalTrees 5 |
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− | [Branch 'x' (Branch 'x' Empty (Branch 'x' Empty Empty)) (Branch 'x' (Branch 'x' Empty Empty) Empty),Branch 'x' (Branch 'x' (Branch 'x' Empty Empty) Empty) (Branch 'x' Empty (Branch 'x' Empty Empty))] |
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− | </pre> |
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− | Solution: |
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<haskell> |
<haskell> |
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− | symCbalTrees = filter symmetric . cbalTree |
+ | *Main> symCbalTrees 5 |

+ | [Branch 'x' (Branch 'x' Empty (Branch 'x' Empty Empty)) (Branch 'x' (Branch 'x' Empty Empty) Empty),Branch 'x' (Branch 'x' (Branch 'x' Empty Empty) Empty) (Branch 'x' Empty (Branch 'x' Empty Empty))] |
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</haskell> |
</haskell> |
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+ | |||

+ | [[99 questions/Solutions/58 | Solutions]] |
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+ | |||

== Problem 59 == |
== Problem 59 == |
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Line 199: | Line 177: | ||

Example: |
Example: |
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+ | |||

<pre> |
<pre> |
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?- hbal_tree(3,T). |
?- hbal_tree(3,T). |
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Example in Haskell: |
Example in Haskell: |
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− | <pre> |
+ | |

+ | <haskell> |
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*Main> take 4 $ hbalTree 'x' 3 |
*Main> take 4 $ hbalTree 'x' 3 |
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[Branch 'x' (Branch 'x' Empty Empty) (Branch 'x' Empty (Branch 'x' Empty Empty)), |
[Branch 'x' (Branch 'x' Empty Empty) (Branch 'x' Empty (Branch 'x' Empty Empty)), |
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Line 213: | Line 192: | ||

Branch 'x' (Branch 'x' Empty Empty) (Branch 'x' (Branch 'x' Empty Empty) (Branch 'x' Empty Empty)), |
Branch 'x' (Branch 'x' Empty Empty) (Branch 'x' (Branch 'x' Empty Empty) (Branch 'x' Empty Empty)), |
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Branch 'x' (Branch 'x' Empty (Branch 'x' Empty Empty)) (Branch 'x' Empty Empty)] |
Branch 'x' (Branch 'x' Empty (Branch 'x' Empty Empty)) (Branch 'x' Empty Empty)] |
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− | </pre> |
+ | </haskell> |

+ | |||

+ | [[99 questions/Solutions/59 | Solutions]] |
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− | Solution: |
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− | <haskell> |
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− | hbalTree x = map fst . hbalTree' |
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− | where hbalTree' 0 = [(Empty, 0)] |
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− | hbalTree' 1 = [(Branch x Empty Empty, 1)] |
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− | hbalTree' n = |
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− | let t = hbalTree' (n-2) ++ hbalTree' (n-1) |
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− | in [(Branch x lb rb, h) | (lb,lh) <- t, (rb,rh) <- t |
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− | , let h = 1 + max lh rh, h == n] |
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− | </haskell> |
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− | Alternative solution: |
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− | <haskell> |
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− | hbaltree :: a -> Int -> [Tree a] |
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− | hbaltree x 0 = [Empty] |
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− | hbaltree x 1 = [Branch x Empty Empty] |
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− | hbaltree x h = [Branch x l r | |
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− | (hl, hr) <- [(h-2, h-1), (h-1, h-1), (h-1, h-2)], |
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− | l <- hbaltree x hl, r <- hbaltree x hr] |
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− | </haskell> |
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− | If we want to avoid recomputing lists of trees (at the cost of extra space), we can use a similar structure to the common method for computation of all the Fibonacci numbers: |
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− | <haskell> |
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− | hbaltree :: a -> Int -> [Tree a] |
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− | hbaltree x h = trees !! h |
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− | where trees = [Empty] : [Branch x Empty Empty] : |
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− | zipWith combine (tail trees) trees |
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− | combine ts shortts = [Branch x l r | |
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− | (ls, rs) <- [(shortts, ts), (ts, ts), (ts, shortts)], |
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− | l <- ls, r <- rs] |
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− | </haskell> |
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== Problem 60 == |
== Problem 60 == |
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(**) Construct height-balanced binary trees with a given number of nodes |
(**) Construct height-balanced binary trees with a given number of nodes |
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− | Consider a height-balanced binary tree of height H. |
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− | What is the maximum number of nodes it can contain? |
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− | Clearly, MaxN = 2**H - 1. |
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− | However, what is the minimum number MinN? |
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− | This question is more difficult. |
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− | Try to find a recursive statement and turn it into a function <hask>minNodes</hask> that returns the minimum number of nodes in a height-balanced binary tree of height H. |
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− | On the other hand, we might ask: what is the maximum height H a height-balanced binary tree with N nodes can have? |
+ | Consider a height-balanced binary tree of height H. What is the maximum number of nodes it can contain? |

− | Write a function <hask>maxHeight</hask> that computes this. |
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− | Now, we can attack the main problem: construct all the height-balanced binary trees with a given nuber of nodes. |
+ | Clearly, MaxN = 2**H - 1. However, what is the minimum number MinN? This question is more difficult. Try to find a recursive statement and turn it into a function <hask>minNodes</hask> that returns the minimum number of nodes in a height-balanced binary tree of height H. |

− | Find out how many height-balanced trees exist for N = 15. |
+ | On the other hand, we might ask: what is the maximum height H a height-balanced binary tree with N nodes can have? Write a function <hask>maxHeight</hask> that computes this. |

+ | |||

+ | Now, we can attack the main problem: construct all the height-balanced binary trees with a given number of nodes. Find out how many height-balanced trees exist for N = 15. |
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Example in Prolog: |
Example in Prolog: |
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+ | |||

<pre> |
<pre> |
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?- count_hbal_trees(15,C). |
?- count_hbal_trees(15,C). |
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Line 268: | Line 213: | ||

Example in Haskell: |
Example in Haskell: |
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− | <pre> |
+ | |

+ | <haskell> |
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*Main> length $ hbalTreeNodes 'x' 15 |
*Main> length $ hbalTreeNodes 'x' 15 |
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1553 |
1553 |
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− | *Main> map (hbalTreeNodes 'x') [0,1,2,3] |
+ | *Main> map (hbalTreeNodes 'x') [0..3] |

[[Empty], |
[[Empty], |
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[Branch 'x' Empty Empty], |
[Branch 'x' Empty Empty], |
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[Branch 'x' Empty (Branch 'x' Empty Empty),Branch 'x' (Branch 'x' Empty Empty) Empty], |
[Branch 'x' Empty (Branch 'x' Empty Empty),Branch 'x' (Branch 'x' Empty Empty) Empty], |
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[Branch 'x' (Branch 'x' Empty Empty) (Branch 'x' Empty Empty)]] |
[Branch 'x' (Branch 'x' Empty Empty) (Branch 'x' Empty Empty)]] |
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− | </pre> |
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− | |||

− | Solution: |
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− | <haskell> |
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− | hbalTreeNodes _ 0 = [Empty] |
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− | hbalTreeNodes x n = concatMap toFilteredTrees [minHeight .. maxHeight] |
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− | where toFilteredTrees = filter ((n ==) . countNodes) . hbalTree x |
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− | |||

− | -- Similar to the Fibonacci sequence but adds 1 in each step. |
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− | minNodesSeq = 0:1:zipWith ((+).(1+)) minNodesSeq (tail minNodesSeq) |
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− | minNodes = (minNodesSeq !!) |
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− | |||

− | minHeight = ceiling $ logBase 2 $ fromIntegral (n+1) |
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− | maxHeight = (fromJust $ findIndex (>n) minNodesSeq) - 1 |
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− | |||

− | countNodes Empty = 0 |
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− | countNodes (Branch _ l r) = countNodes l + countNodes r + 1 |
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− | |||

</haskell> |
</haskell> |
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− | Another solution generates only the trees we want: |
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− | <haskell> |
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− | -- maximum number of nodes in a weight-balanced tree of height h |
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− | maxNodes :: Int -> Int |
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− | maxNodes h = 2^h - 1 |
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− | -- minimum height of a weight-balanced tree of n nodes |
+ | [[99 questions/Solutions/60 | Solutions]] |

− | minHeight :: Int -> Int |
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− | minHeight n = ceiling $ logBase 2 $ fromIntegral (n+1) |
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− | -- minimum number of nodes in a weight-balanced tree of height h |
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− | minNodes :: Int -> Int |
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− | minNodes h = fibs !! (h+2) - 1 |
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− | |||

− | -- maximum height of a weight-balanced tree of n nodes |
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− | maxHeight :: Int -> Int |
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− | maxHeight n = length (takeWhile (<= n+1) fibs) - 3 |
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− | |||

− | -- Fibonacci numbers |
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− | fibs :: [Int] |
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− | fibs = 0 : 1 : zipWith (+) fibs (tail fibs) |
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− | |||

− | hbalTreeNodes :: a -> Int -> [Tree a] |
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− | hbalTreeNodes x n = [t | h <- [minHeight n .. maxHeight n], t <- baltree h n] |
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− | where |
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− | -- baltree h n = weight-balanced trees of height h with n nodes |
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− | -- assuming minNodes h <= n <= maxNodes h |
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− | baltree 0 n = [Empty] |
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− | baltree 1 n = [Branch x Empty Empty] |
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− | baltree h n = [Branch x l r | |
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− | (hl,hr) <- [(h-2,h-1), (h-1,h-1), (h-1,h-2)], |
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− | let min_nl = max (minNodes hl) (n - 1 - maxNodes hr), |
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− | let max_nl = min (maxNodes hl) (n - 1 - minNodes hr), |
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− | nl <- [min_nl .. max_nl], |
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− | let nr = n - 1 - nl, |
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− | l <- baltree hl nl, |
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− | r <- baltree hr nr] |
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− | </haskell> |
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[[Category:Tutorials]] |
[[Category:Tutorials]] |

## Revision as of 14:09, 21 January 2012

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

## 1 Binary trees

A binary tree is either empty or it is composed of a root element and two successors, which are binary trees themselves.

In Haskell, we can characterize binary trees with a datatype definition:

data Tree a = Empty | Branch a (Tree a) (Tree a) deriving (Show, Eq)

This says that a `Tree` of type `a` consists of either an `Empty` node, or a `Branch` containing one value of type `a` with exactly two subtrees of type `a`.

Given this definition, the tree in the diagram above would be represented as:

tree1 = Branch 'a' (Branch 'b' (Branch 'd' Empty Empty) (Branch 'e' Empty Empty)) (Branch 'c' Empty (Branch 'f' (Branch 'g' Empty Empty) Empty))

Since a "leaf" node is a branch with two empty subtrees, it can be useful to define a shorthand function:

`leaf x = Branch x Empty Empty`

Then the tree diagram above could be expressed more simply as:

tree1' = Branch 'a' (Branch 'b' (leaf 'd') (leaf 'e')) (Branch 'c' Empty (Branch 'f' (leaf 'g') Empty)))

Other examples of binary trees:

-- A binary tree consisting of a root node only tree2 = Branch 'a' Empty Empty -- An empty binary tree tree3 = Empty -- A tree of integers tree4 = Branch 1 (Branch 2 Empty (Branch 4 Empty Empty)) (Branch 2 Empty Empty)

## 2 Problem 54A

(*) Check whether a given term represents a binary tree

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

Example in Lisp:

* (istree (a (b nil nil) nil)) T * (istree (a (b nil nil))) NIL

Non-solution:

Haskell's type system ensures that all terms of type

## 3 Problem 55

(**) Construct completely balanced binary trees

In a completely balanced binary tree, the following property holds for every node: The number of nodes in its left subtree and the number of nodes in its right subtree are almost equal, which means their difference is not greater than one.

Write a function cbal-tree to construct completely balanced binary trees for a given number of nodes. The predicate should generate all solutions via backtracking. Put the letter 'x' as information into all nodes of the tree.

Example:

* cbal-tree(4,T). T = t(x, t(x, nil, nil), t(x, nil, t(x, nil, nil))) ; T = t(x, t(x, nil, nil), t(x, t(x, nil, nil), nil)) ; etc......No

Example in Haskell, whitespace and "comment diagrams" added for clarity and exposition:

*Main> cbalTree 4 [ -- permutation 1 -- x -- / \ -- x x -- \ -- x Branch 'x' (Branch 'x' Empty Empty) (Branch 'x' Empty (Branch 'x' Empty Empty)), -- permutation 2 -- x -- / \ -- x x -- / -- x Branch 'x' (Branch 'x' Empty Empty) (Branch 'x' (Branch 'x' Empty Empty) Empty), -- permutation 3 -- x -- / \ -- x x -- \ -- x Branch 'x' (Branch 'x' Empty (Branch 'x' Empty Empty)) (Branch 'x' Empty Empty), -- permutation 4 -- x -- / \ -- x x -- / -- x Branch 'x' (Branch 'x' (Branch 'x' Empty Empty) Empty) (Branch 'x' Empty Empty) ]

## 4 Problem 56

(**) Symmetric binary trees

Let us call a binary tree symmetric if you can draw a vertical line through the root node and then the right subtree is the mirror image of the left subtree. Write a predicate symmetric/1 to check whether a given binary tree is symmetric. Hint: Write a predicate mirror/2 first to check whether one tree is the mirror image of another. We are only interested in the structure, not in the contents of the nodes.

Example in Haskell:

*Main> symmetric (Branch 'x' (Branch 'x' Empty Empty) Empty) False *Main> symmetric (Branch 'x' (Branch 'x' Empty Empty) (Branch 'x' Empty Empty)) True

## 5 Problem 57

(**) Binary search trees (dictionaries)

Use the predicate add/3, developed in chapter 4 of the course, to write a predicate to construct a binary search tree from a list of integer numbers.

Example:

* construct([3,2,5,7,1],T). T = t(3, t(2, t(1, nil, nil), nil), t(5, nil, t(7, nil, nil)))

Then use this predicate to test the solution of the problem P56.

Example:

* test-symmetric([5,3,18,1,4,12,21]). Yes * test-symmetric([3,2,5,7,4]). No

Example in Haskell:

*Main> construct [3, 2, 5, 7, 1] Branch 3 (Branch 2 (Branch 1 Empty Empty) Empty) (Branch 5 Empty (Branch 7 Empty Empty)) *Main> symmetric . construct $ [5, 3, 18, 1, 4, 12, 21] True *Main> symmetric . construct $ [3, 2, 5, 7, 1] True

## 6 Problem 58

(**) Generate-and-test paradigm

Apply the generate-and-test paradigm to construct all symmetric, completely balanced binary trees with a given number of nodes.

Example:

* sym-cbal-trees(5,Ts). Ts = [t(x, t(x, nil, t(x, nil, nil)), t(x, t(x, nil, nil), nil)), t(x, t(x, t(x, nil, nil), nil), t(x, nil, t(x, nil, nil)))]

Example in Haskell:

*Main> symCbalTrees 5 [Branch 'x' (Branch 'x' Empty (Branch 'x' Empty Empty)) (Branch 'x' (Branch 'x' Empty Empty) Empty),Branch 'x' (Branch 'x' (Branch 'x' Empty Empty) Empty) (Branch 'x' Empty (Branch 'x' Empty Empty))]

## 7 Problem 59

(**) Construct height-balanced binary trees

In a height-balanced binary tree, the following property holds for every node: The height of its left subtree and the height of its right subtree are almost equal, which means their difference is not greater than one.

Example:

?- hbal_tree(3,T). T = t(x, t(x, t(x, nil, nil), t(x, nil, nil)), t(x, t(x, nil, nil), t(x, nil, nil))) ; T = t(x, t(x, t(x, nil, nil), t(x, nil, nil)), t(x, t(x, nil, nil), nil)) ; etc......No

Example in Haskell:

*Main> take 4 $ hbalTree 'x' 3 [Branch 'x' (Branch 'x' Empty Empty) (Branch 'x' Empty (Branch 'x' Empty Empty)), Branch 'x' (Branch 'x' Empty Empty) (Branch 'x' (Branch 'x' Empty Empty) Empty), Branch 'x' (Branch 'x' Empty Empty) (Branch 'x' (Branch 'x' Empty Empty) (Branch 'x' Empty Empty)), Branch 'x' (Branch 'x' Empty (Branch 'x' Empty Empty)) (Branch 'x' Empty Empty)]

## 8 Problem 60

(**) Construct height-balanced binary trees with a given number of nodes

Consider a height-balanced binary tree of height H. What is the maximum number of nodes it can contain?

Clearly, MaxN = 2**H - 1. However, what is the minimum number MinN? This question is more difficult. Try to find a recursive statement and turn it into a functionNow, we can attack the main problem: construct all the height-balanced binary trees with a given number of nodes. Find out how many height-balanced trees exist for N = 15.

Example in Prolog:

?- count_hbal_trees(15,C). C = 1553

Example in Haskell:

*Main> length $ hbalTreeNodes 'x' 15 1553 *Main> map (hbalTreeNodes 'x') [0..3] [[Empty], [Branch 'x' Empty Empty], [Branch 'x' Empty (Branch 'x' Empty Empty),Branch 'x' (Branch 'x' Empty Empty) Empty], [Branch 'x' (Branch 'x' Empty Empty) (Branch 'x' Empty Empty)]]