A method for calculating the shortest editing distance in Ruby _ruby special topics

The calculation of the shortest editing distance is realized by using dynamic programming algorithm.

Copy Code code as follows:

#encoding: Utf-8
#author: Xu Jin
#date: Nov 12, 2012
#EditDistance
#to find the minimum cost by using editdistance algorithm
#example output:
# "Please input a string:"
# exponential
# "Please input to the other string:"
# polynomial
# "The expected cost is 6"
# The result is:
# ["E", "X", "P", "O", "n", "E", "N", "-", "T", "I", "a", "L"]
# ["-", "-", "P", "O", "L", "Y", "n", "O", "M", "I", "a", "L"]

P "Please input a string:"
x = gets.chop.chars.map{|c| C}
P "Please input the other string:"
y = gets.chop.chars.map{|c| C}
X.unshift ("")
Y.unshift ("")
E = Array.new (x.size) {array.new (y.size)}
Flag = Array.new (x.size) {array.new (y.size)}
DEL, INS, CHA, fit = (1..4). To_a #deleat, inserts, change, and fit

def edit_distance (x, Y, E, flag)
(0..x.length-1). each{|i| e[i][0] = i}
(0..y.length-1). each{|j| e[0][j] = j}
diff = array.new (x.size) {array.new (y.size)}
For I in (1..x.length-1) do
For J. (1..y.length-1) do
DIFF[I][J] = (x[i] = = Y[j])? 0:1
E[I][J] = [E[i-1][j] + 1, e[i][j-1] + 1, e[i-1][j-1] + diff[i][j]].min
If e[i][j] = = E[i-1][j] + 1
FLAG[I][J] = DEL
elsif E[i][j] = = E[i-1][j-1] + 1
FLAG[I][J] = CHA
elsif E[i][j] = = E[i][j-1] + 1
FLAG[I][J] = INS
else flag[i][j] = fit
End
End
End
End

out_x, out_y = [], []

def solution_structure (x, y, Flag, I, J, out_x, Out_y)
Case Flag[i][j]
When fit
Out_x.unshift (X[i])
Out_y.unshift (Y[j])
Solution_structure (x, Y, Flag, i-1, J-1, out_x, out_y)
When DEL
Out_x.unshift (X[i])
Out_y.unshift ('-')
Solution_structure (x, Y, Flag, I-1, J, out_x, Out_y)
When INS
Out_x.unshift ('-')
Out_y.unshift (Y[j])
Solution_structure (x, y, flag, I, J-1, out_x, out_y)
When CHA
Out_x.unshift (X[i])
Out_y.unshift (Y[j])
Solution_structure (x, Y, Flag, i-1, J-1, out_x, out_y)
End
#if flag[i][j] = = Nil, go
return if i = = 0 && J = = 0
If j = = 0
Out_y.unshift ('-')
Out_x.unshift (X[i])
Solution_structure (x, Y, Flag, I-1, J, out_x, Out_y)
elsif i = = 0
Out_x.unshift ('-')
Out_y.unshift (Y[j])
Solution_structure (x, y, flag, I, J-1, out_x, out_y)
End
End

Edit_distance (x, Y, E, flag)
P "The expected edit distance is #{e[x.length-1][y.length-1]}"
Solution_structure (x, Y, Flag, x.length-1, Y.length-1, out_x, out_y)
Puts "The result is: \ n #{out_x}\n #{out_y}"