Ruby实现的最优二叉查找树算法

算法导论上的伪码改写而成，加上导论的课后练习第一题的解的构造函数。

#encoding: utf-8

=begin

author: xu jin

date: Nov 11, 2012

Optimal Binary Search Tree

to find by using EditDistance algorithm

refer to <<introduction to algorithms>>

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]

          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< i

    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]}."

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