An optimal binary lookup tree algorithm implemented by Ruby

This article mainly introduces the best binary lookup tree algorithm in Ruby, this paper gives the implementation code directly, the need for friends can refer to the following

The pseudo code in the introduction of the algorithm is rewritten, coupled with the constructor of the solution of the first question in the introductory lesson practice.

The code is as follows:

#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 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]}."