# Difference between revisions of "99 questions/54A to 60"

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.

## 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 = Leaf | Branch a (Tree a) (Tree a) deriving Show
```
The type system ensures that all terms of type
```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 type `Tree a`.

```istree :: Tree a -> Bool
istree _ = True
```

Running this:

```*M> istree Leaf
True

*M> istree (Branch 1 Leaf Leaf)
True

*M> istree (Branch 1 Leaf (Branch 2 Leaf Leaf))
True
```

## 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
```

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

Solution:

```cbalTree 0 = [Leaf]
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.

## 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.

```*Main> symmetric (Branch 'x' (Branch 'x' Leaf Leaf) Leaf)
False
*Main> symmetric (Branch 'x' (Branch 'x' Leaf Leaf) (Branch 'x' Leaf Leaf))
True
```

Solution:

```mirror Leaf           Leaf           = True
mirror (Branch _ a b) (Branch _ x y) = mirror a y && mirror b x
mirror _              _              = False

symmetric Leaf           = True
symmetric (Branch _ l r) = mirror l r
```

## 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
```

```*Main> construct [3, 2, 5, 7, 1]
Branch 3 (Branch 2 (Branch 1 Leaf Leaf) (Branch 5 Leaf Leaf)) (Branch 7 Leaf Leaf)
*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 Leaf             = Branch x Leaf Leaf
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) Leaf xs
```

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

## Problem 58

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)))]
```

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

Solution:

```symCbalTrees = filter symmetric . cbalTree
```

## Problem 59

<Problem description>

```Example:
<example in lisp>

```

Solution:

```<solution in haskell>
```

<description of implementation>

## Problem 60

<Problem description>

```Example:
<example in lisp>

```<solution in haskell>