99 questions/61 to 69
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.
Binary trees
The type of binary trees:
data Tree a = Empty | Branch a (Tree a) (Tree a)
deriving (Show, Eq)
An example tree:
tree1 = Branch 1 (Branch 2 Empty (Branch 4 Empty Empty))
(Branch 2 Empty Empty)
Problem 61
Count the leaves of a binary tree
A leaf is a node with no successors. Write a predicate count_leaves/2 to count them.
Example: % count_leaves(T,N) :- the binary tree T has N leaves Example in Haskell: > count_leaves tree1 2
Solution:
count_leaves Empty = 0
count_leaves (Branch a Empty Empty) = 1
count_leaves (Branch a left right) = count_leaves left + count_leaves right
Problem 61A
Collect the leaves of a binary tree in a list
A leaf is a node with no successors. Write a predicate leaves/2 to collect them in a list.
Example: % leaves(T,S) :- S is the list of all leaves of the binary tree T Example in Haskell: > leaves tree1 [4, 2]
Solution:
leaves :: Tree a -> [a]
leaves Empty = []
leaves (Branch a Empty Empty) = [a]
leaves (Branch a left right) = leaves left ++ leaves right
Problem 62
<Problem description>
Example: <example in lisp> Example in Haskell: <example in Haskell>
Solution:
<solution in haskell>
<description of implementation>
Problem 62B
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Example: <example in lisp> Example in Haskell: <example in Haskell>
Solution:
<solution in haskell>
<description of implementation>
Problem 63
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Example: <example in lisp> Example in Haskell: <example in Haskell>
Solution:
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<description of implementation>
Problem 64
Given a binary tree as the usual Prolog term t(X,L,R) (or nil). As a preparation for drawing the tree, a layout algorithm is required to determine the position of each node in a rectangular grid. Several layout methods are conceivable, one of them is shown in the illustration below:
In this layout strategy, the position of a node v is obtained by the following two rules:
- x(v) is equal to the position of the node v in the inorder sequence
- y(v) is equal to the depth of the node v in the tree
Write a function to annotate each node of the tree with a position, where (1,1) in the top left corner or the rectangle bounding the drawn tree.
Here is the example tree from the above illustration:
tree2 = Branch 'n'
(Branch 'k'
(Branch 'c'
(Branch 'a' Empty Empty)
(Branch 'h'
(Branch 'g'
(Branch 'e' Empty Empty)
Empty
)
Empty
)
)
(Branch 'm' Empty Empty)
)
(Branch 'u'
(Branch 'p'
Empty
(Branch 's'
(Branch 'q' Empty Empty)
Empty
)
)
Empty
)
Solution:
type Pos = (Int, Int)
layout :: Tree a -> Tree (a, Pos)
layout t = fst (layoutAux 1 1 t)
where layoutAux x y Empty = (Empty, x)
layoutAux x y (Branch a l r) = (Branch (a, (x',y)) l' r', x'')
where (l', x') = layoutAux x (y+1) l
(r', x'') = layoutAux (x'+1) (y+1) r
The auxiliary function is passed the x-coordinate for the left-most node of the subtree, the y-coordinate for the root of the subtree, and the subtree itself. It returns the subtree annotated with positions, plus the count of Branch nodes in the subtree.
Problem 65
An alternative layout method is depicted in the illustration below:
Find out the rules and write the corresponding function. Hint: On a given level, the horizontal distance between neighboring nodes is constant.
Use the same conventions as in problem P64 and test your function in an appropriate way.
Here is the example tree from the above illustration:
tree3 = Branch 'n'
(Branch 'k'
(Branch 'c'
(Branch 'a' Empty Empty)
(Branch 'e'
(Branch 'd' Empty Empty)
(Branch 'g' Empty Empty)
)
)
(Branch 'm' Empty Empty)
)
(Branch 'u'
(Branch 'p'
Empty
(Branch 'q' Empty Empty)
)
Empty
)
Solution:
layout :: Tree a -> Tree (a, Pos)
layout t = layoutAux x1 1 sep1 t
where d = depth t
ld = leftdepth t
x1 = 2^(d-1) - 2^(d-ld) + 1
sep1 = 2^(d-2)
layoutAux x y sep Empty = Empty
layoutAux x y sep (Branch a l r) =
Branch (a, (x,y))
(layoutAux (x-sep) (y+1) (sep `div` 2) l)
(layoutAux (x+sep) (y+1) (sep `div` 2) r)
depth :: Tree a -> Int
depth Empty = 0
depth (Branch a l r) = max (depth l) (depth r) + 1
leftdepth :: Tree a -> Int
leftdepth Empty = 0
leftdepth (Branch a l r) = leftdepth l + 1
The auxiliary function is passed the x- and y-coordinates for the root of the subtree, the horizontal separation between the root and its child nodes, and the subtree itself. It returns the subtree annotated with positions.
Problem 66
Yet another layout strategy is shown in the illustration below:
The method yields a very compact layout while maintaining a certain symmetry in every node. Find out the rules and write the corresponding Prolog predicate. Hint: Consider the horizontal distance between a node and its successor nodes. How tight can you pack together two subtrees to construct the combined binary tree?
Use the same conventions as in problem P64 and P65 and test your predicate in an appropriate way. Note: This is a difficult problem. Don't give up too early!
Which layout do you like most?
<Problem description>
Example in Haskell:
<example in Haskell>
Solution:
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<description of implementation>
Problem 67
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Example: <example in lisp> Example in Haskell: <example in Haskell>
Solution:
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<description of implementation>
Problem 68
<Problem description>
Example: <example in lisp> Example in Haskell: <example in Haskell>
Solution:
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<description of implementation>
Problem 69
<Problem description>
Example: <example in lisp> Example in Haskell: <example in Haskell>
Solution:
<solution in haskell>
<description of implementation>