# 99 questions/54A to 60

### From HaskellWiki

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The typical solution in Haskell is to introduce an algebraic data type: | The typical solution in Haskell is to introduce an algebraic data type: | ||

<haskell> | <haskell> | ||

− | data Tree a = | + | data Tree a = Empty | Branch a (Tree a) (Tree a) |

+ | deriving (Show, Eq) | ||

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

Line 37: | Line 38: | ||

<haskell> | <haskell> | ||

− | *M> istree | + | *M> istree Empty |

True | True | ||

− | *M> istree (Branch 1 | + | *M> istree (Branch 1 Empty Empty) |

True | True | ||

− | *M> istree (Branch 1 | + | *M> istree (Branch 1 Empty (Branch 2 Empty Empty)) |

True | True | ||

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

Line 66: | Line 67: | ||

<pre> | <pre> | ||

*Main> cbalTree 4 | *Main> cbalTree 4 | ||

− | [Branch 'x' (Branch 'x' | + | [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 (Branch 'x' Empty Empty)) (Branch 'x' Empty Empty),Branch 'x' (Branch 'x' (Branch 'x' Empty Empty) Empty) (Branch 'x' Empty Empty)] |

</pre> | </pre> | ||

Solution: | Solution: | ||

<haskell> | <haskell> | ||

− | cbalTree 0 = [ | + | cbalTree 0 = [Empty] |

cbalTree n = [Branch 'x' l r | i <- [q .. q + r], l <- cbalTree i, r <- cbalTree (n - i - 1)] | cbalTree n = [Branch 'x' l r | i <- [q .. q + r], l <- cbalTree i, r <- cbalTree (n - i - 1)] | ||

where (q, r) = quotRem (n-1) 2 | where (q, r) = quotRem (n-1) 2 | ||

Line 86: | Line 87: | ||

Example in Haskell: | Example in Haskell: | ||

<pre> | <pre> | ||

− | *Main> symmetric (Branch 'x' (Branch 'x' | + | *Main> symmetric (Branch 'x' (Branch 'x' Empty Empty) Empty) |

False | False | ||

− | *Main> symmetric (Branch 'x' (Branch 'x' | + | *Main> symmetric (Branch 'x' (Branch 'x' Empty Empty) (Branch 'x' Empty Empty)) |

True | True | ||

</pre> | </pre> | ||

Line 94: | Line 95: | ||

Solution: | Solution: | ||

<haskell> | <haskell> | ||

− | mirror | + | mirror Empty Empty = True |

mirror (Branch _ a b) (Branch _ x y) = mirror a y && mirror b x | mirror (Branch _ a b) (Branch _ x y) = mirror a y && mirror b x | ||

mirror _ _ = False | mirror _ _ = False | ||

− | symmetric | + | symmetric Empty = True |

symmetric (Branch _ l r) = mirror l r | symmetric (Branch _ l r) = mirror l r | ||

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

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<pre> | <pre> | ||

*Main> construct [3, 2, 5, 7, 1] | *Main> construct [3, 2, 5, 7, 1] | ||

− | Branch 3 (Branch 2 (Branch 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] | *Main> symmetric . construct $ [5, 3, 18, 1, 4, 12, 21] | ||

True | True | ||

− | *Main> symmetric . construct $ [3, 2, 5, 7, 1] | + | *Main> symmetric . construct $ [3, 2, 5, 7, 1] |

True | True | ||

</pre> | </pre> | ||

Line 137: | Line 138: | ||

<haskell> | <haskell> | ||

add :: Ord a => a -> Tree a -> Tree a | add :: Ord a => a -> Tree a -> Tree a | ||

− | add x | + | add x Empty = Branch x Empty Empty |

add x t@(Branch y l r) = case compare x y of | add x t@(Branch y l r) = case compare x y of | ||

LT -> Branch y (add x l) r | LT -> Branch y (add x l) r | ||

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EQ -> t | EQ -> t | ||

− | construct xs = foldl (flip add) | + | construct xs = foldl (flip add) Empty xs |

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

Line 163: | Line 164: | ||

<pre> | <pre> | ||

*Main> symCbalTrees 5 | *Main> symCbalTrees 5 | ||

− | [Branch 'x' (Branch 'x' | + | [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))] |

</pre> | </pre> | ||

## Revision as of 02:21, 14 December 2006

These are Haskell translations of Ninety Nine Lisp Problems.

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.

## 1 Problem 54A

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

Example:

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

The typical solution in Haskell is to introduce an algebraic data type:

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

Tree a

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 -- the istree predicate is trivially true, for

anything of typeistree :: Tree a -> Bool istree _ = True

Running this:

*M> istree Empty True *M> istree (Branch 1 Empty Empty) True *M> istree (Branch 1 Empty (Branch 2 Empty Empty)) True

## 2 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:

*Main> cbalTree 4 [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 (Branch 'x' Empty Empty)) (Branch 'x' Empty Empty),Branch 'x' (Branch 'x' (Branch 'x' Empty Empty) Empty) (Branch 'x' Empty Empty)]

Solution:

cbalTree 0 = [Empty] cbalTree n = [Branch 'x' l r | i <- [q .. q + r], l <- cbalTree i, r <- cbalTree (n - i - 1)] where (q, r) = quotRem (n-1) 2

Here we use the list monad to enumerate all the trees, in a style that is more natural than standard backtracking.

## 3 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

Solution:

mirror Empty Empty = True mirror (Branch _ a b) (Branch _ x y) = mirror a y && mirror b x mirror _ _ = False symmetric Empty = True symmetric (Branch _ l r) = mirror l r

## 4 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([3,2,5,7,1]). 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

Solution:

add :: Ord a => a -> Tree a -> Tree a add x Empty = Branch x Empty Empty add x t@(Branch y l r) = case compare x y of LT -> Branch y (add x l) r GT -> Branch y l (add x r) EQ -> t construct xs = foldl (flip add) Empty xs

Here, the definition of construct is trivial, because the pattern of accumulating from the left is captured by the standard function foldl.

## 5 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))]

Solution:

symCbalTrees = filter symmetric . cbalTree

## 6 Problem 59

<Problem description>

Example: <example in lisp> Example in Haskell: <example in Haskell>

Solution:

<solution in haskell>

<description of implementation>

## 7 Problem 60

<Problem description>

Example: <example in lisp> Example in Haskell: <example in Haskell>

Solution:

<solution in haskell>

<description of implementation>