99 questions/61 to 69
This is part of Ninety-Nine Haskell Problems, based on Ninety-Nine Prolog Problems.
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Binary trees
As defined in problem 54A.
An example tree:
tree4 = 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 tree4 2
Solution:
count_leaves Empty = 0
count_leaves (Branch a Empty Empty) = 1
count_leaves (Branch a left right) = count_leaves left + count_leaves rightProblem 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 tree4 [4, 2]
Solution:
leaves :: Tree a -> [a]
leaves Empty = []
leaves (Branch a Empty Empty) = [a]
leaves (Branch a left right) = leaves left ++ leaves rightProblem 62
Collect the internal nodes of a binary tree in a list An internal node of a binary tree has either one or two non-empty successors. Write a predicate internals/2 to collect them in a list.
Example: % internals(T,S) :- S is the list of internal nodes of the binary tree T. Example in Haskell: Prelude>internals tree4 Prelude>[1,2]
Solution:
internals :: Tree a -> [a]
internals Empty = []
internals (Branch a Empty Empty) = []
internals (Branch a left right) = [a] ++ (internals left) ++ (internals right)Problem 62B
Collect the nodes at a given level in a list A node of a binary tree is at level N if the path from the root to the node has length N-1. The root node is at level 1. Write a predicate atlevel/3 to collect all nodes at a given level in a list.
Example: % atlevel(T,L,S) :- S is the list of nodes of the binary tree T at level L Example in Haskell: Prelude>atlevel tree4 2 Prelude>[2,2]
Solution:
atlevel :: Tree a -> Int -> [a]
atlevel t level = loop t 1
where
loop Empty _ = []
loop (Branch a l r) n
| n == level = [a]
| otherwise = loop l (n+1) ++ loop r (n+1)Another possibility is to decompose the problem:
levels :: Tree a -> [[a]]
levels Empty = repeat []
levels (Branch a l r) = [a] : zipWith (++) (levels l) (levels r)
atlevel :: Tree a -> Int -> [a]
atlevel t n = levels t !! (n-1)Problem 63
Construct a complete binary tree
A complete binary tree with height H is defined as follows:
- The levels 1,2,3,...,H-1 contain the maximum number of nodes (i.e 2**(i-1) at the level i)
- In level H, which may contain less than the maximum possible number of nodes, all the nodes are "left-adjusted". This means that in a levelorder tree traversal all internal nodes come first, the leaves come second, and empty successors (the nil's which are not really nodes!) come last.
Particularly, complete binary trees are used as data structures (or addressing schemes) for heaps.
We can assign an address number to each node in a complete binary tree by enumerating the nodes in level-order, starting at the root with number 1. For every node X with address A the following property holds: The address of X's left and right successors are 2*A and 2*A+1, respectively, if they exist. This fact can be used to elegantly construct a complete binary tree structure.
Write a predicate complete_binary_tree/2.
Example: % complete_binary_tree(N,T) :- T is a complete binary tree with N nodes. Example in Haskell: Main> complete_binary_tree 4 Branch 'x' (Branch 'x' (Branch 'x' Empty Empty) Empty) (Branch 'x' Empty Empty) Main> is_complete_binary_tree $ Branch 'x' (Branch 'x' Empty Empty) (Branch 'x' Empty Empty) True
Solution:
import Data.List
data Tree a = Empty | Branch a (Tree a) (Tree a)
deriving (Show, Eq)
filled :: Tree a -> [[Bool]]
filled Empty = repeat [False]
filled (Branch _ l r) = [True] : zipWith (++) (filled l) (filled r)
complete_binary_tree :: Int -> Tree Char
complete_binary_tree n = generate_tree 1
where generate_tree x
| x > n = Empty
| otherwise = Branch 'x' (generate_tree (2*x) )
(generate_tree (2*x+1))
is_complete_binary_tree :: Tree a -> Bool
is_complete_binary_tree Empty = True
is_complete_binary_tree t = and $ last_proper : zipWith (==) lengths powers
where levels = takeWhile or $ filled t
-- The upper levels of the tree should be filled.
-- Every level has twice the number of nodes as the one above it,
-- so [1,2,4,8,16,...]
lengths = map (length . filter id) $ init levels
powers = iterate (2*) 1
-- The last level should contain a number of filled spots,
-- and (maybe) some empty spots, but no filled spots after that!
last_filled = map head $ group $ last levels
last_proper = head last_filled && (length last_filled) < 3Alternative solution which constructs complete binary trees from a given list (also includes a lookup function as per the Prolog solution):
completeBinaryTree :: Int -> a -> Tree a
completeBinaryTree n = cbtFromList . replicate n
cbtFromList :: [a] -> Tree a
cbtFromList [] = Empty
cbtFromList xs = head $ build 1 xs
where build _ [] = []
build n xs = let (l,rest) = splitAt n xs
as = pairs $ build (2*n) rest ++ repeat Empty
in map (\ (x, (lt,rt)) -> Branch x lt rt) $ zip l as
pairs (lt:rt:rest) = (lt,rt) : pairs rest
lookupIndex :: Tree a -> Integer -> a
lookupIndex t = lookup t . path
where lookup Empty _ = error "index to large"
lookup (Branch x _ _) [] = x
lookup (Branch x l r) (p:ps) = lookup (if even p then l else r) ps
path = reverse . takeWhile (>1) . iterate (`div` 2) . (1+)Another version of cbtFromList using local recursion:
cbtFromList :: [a] -> Tree a
cbtFromList xs = let (t, xss) = cbt (xs:xss) in t
where cbt ((x:xs):xss) =
let (l, xss') = cbt xss
(r, xss'') = cbt xss'
in (Branch x l r, xs:xss'')
cbt _ = (Empty, [])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:
http://www.hta-bi.bfh.ch/~hew/informatik3/prolog/p-99/p64.gif
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:
tree64 = 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
)Example in Haskell:
> layout tree64
Branch ('n',(8,1)) (Branch ('k',(6,2)) (Branch ('c',(2,3)) ...
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) rThe 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:
http://www.hta-bi.bfh.ch/~hew/informatik3/prolog/p-99/p65.gif
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:
tree65 = 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
)Example in Haskell:
> layout tree65
Branch ('n',(15,1)) (Branch ('k',(7,2)) (Branch ('c',(3,3)) ...
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 + 1The 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:
http://www.hta-bi.bfh.ch/~hew/informatik3/prolog/p-99/p66.gif
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?
Example in Haskell:
> layout tree65
Branch ('n',(5,1)) (Branch ('k',(3,2)) (Branch ('c',(2,3)) ...
Solution:
layout :: Tree a -> Tree (a, Pos)
layout t = t'
where (l, t', r) = layoutAux x1 1 t
x1 = maximum l + 1
layoutAux :: Int -> Int -> Tree a -> ([Int], Tree (a, Pos), [Int])
layoutAux x y Empty = ([], Empty, [])
layoutAux x y (Branch a l r) = (ll', Branch (a, (x,y)) l' r', rr')
where (ll, l', lr) = layoutAux (x-sep) (y+1) l
(rl, r', rr) = layoutAux (x+sep) (y+1) r
sep = maximum (0:zipWith (+) lr rl) `div` 2 + 1
ll' = 0 : overlay (map (+sep) ll) (map (subtract sep) rl)
rr' = 0 : overlay (map (+sep) rr) (map (subtract sep) lr)
-- overlay xs ys = xs padded out to at least the length of ys
-- using any extra elements of ys
overlay :: [a] -> [a] -> [a]
overlay [] ys = ys
overlay xs [] = xs
overlay (x:xs) (y:ys) = x : overlay xs ysThe auxiliary function is passed the x- and y-coordinates for the root of the subtree and the subtree itself. It returns
- a list of distances the laid-out tree extends to the left at each level,
- the subtree annotated with positions, and
- a list of distances the laid-out tree extends to the right at each level.
These distances are usually positive, but may be 0 or negative in the case of a skewed tree. To put two subtrees side by side, we must determine the least even separation so that they do not overlap on any level. Having determined the separation, we can compute the extents of the composite tree.
The definitions of layout and its auxiliary function use local recursion to compute the x-coordinates. This works because nothing else depends on these coordinates.
Problem 67
<Problem description>
Example: <example in lisp> Example in Haskell: <example in Haskell>
Solution:
<solution in haskell><description of implementation>
Problem 68
<Problem description>
Example: <example in lisp> Example in Haskell: <example in Haskell>
Solution:
<solution in haskell><description of implementation>
Problem 69
<Problem description>
Example: <example in lisp> Example in Haskell: <example in Haskell>
Solution:
<solution in haskell><description of implementation>