# Ruby如何实现最优二叉查找树算法

#encoding: utf-8
=begin
author: xu jin
date: Nov 11, 2012
Optimal Binary Search Tree
to find by using EditDistance algorithm
refer to >
example output:
"k2 is the root of the tree."
"k1 is the left child of k2."
"d0 is the left child of k1."
"d1 is the right child of k1."
"k5 is the right child of k2."
"k4 is the left child of k5."
"k3 is the left child of k4."
"d2 is the left child of k3."
"d3 is the right child of k3."
"d4 is the right child of k4."
"d5 is the right child of k5."

The expected cost is 2.75.
=end

INFINTIY = 1 / 0.0
a = ['', 'k1', 'k2', 'k3', 'k4', 'k5']
p = [0, 0.15, 0.10, 0.05, 0.10, 0.20]
q = [0.05, 0.10, 0.05, 0.05, 0.05 ,0.10]
e = Array.new(a.size + 1){Array.new(a.size + 1)}
root = Array.new(a.size + 1){Array.new(a.size + 1)}

def optimalBST(p, q, n, e, root)
w = Array.new(p.size + 1){Array.new(p.size + 1)}
for i in (1..n + 1)
e[i][i - 1] = q[i - 1]
w[i][i - 1] = q[i - 1]
end
for l in (1..n)
for i in (1..n - l + 1)
j = i + l -1
e[i][j] = 1 / 0.0
w[i][j] = w[i][j - 1] + p[j] + q[j]
for r in (i..j)
t = e[i][r - 1] + e[r + 1][j] + w[i][j]
if t           e[i][j] = t
root[i][j] = r
end
end
end
end
end

def printBST(root, i ,j, signal)
return if i > j
if signal == 0
p "k#{root[i][j]} is the root of the tree."
signal = 1
end
r = root[i][j]
#left child
if r - 1     p "d#{r - 1} is the left child of k#{r}."
else
p "k#{root[i][r - 1]} is the left child of k#{r}."
printBST(root, i, r - 1, 1 )
end
#right child
if r >= j
p "d#{r} is the right child of k#{r}."
else
p "k#{root[r + 1][j]} is the right child of k#{r}."
printBST(root, r + 1, j, 1)
end

end

optimalBST(p, q, p.size - 1, e, root)
printBST(root, 1, a.size-1, 0)
puts "\nThe expected cost is #{e[1][a.size-1]}."