The longest common subsequence algorithm implemented by Ruby

This article mainly introduces the longest common subsequence algorithm implemented by Ruby, this article gives the implementation code directly, and the friends who need it can refer to the following

The longest common subsequence, LCS, dynamic programming implementation.

#encoding: utf-8
#author: xu jin, 4100213
#date: Nov 01, 2012
#Longest-Commom-Subsequence
#to find a longest commom subsequence of two given character arrays by using LCS algorithm
#example output:
#The random character arrays are: ["b", "a", "c", "a", "a", "b", "d"] and ["a", "c", "a", "c", "a", "a", "b"]
#The Longest-Commom-Subsequence is: a c a a b
 
chars = ("a".."e").to_a
x, y = [], []
1.upto(rand(5) + 5) { |i| x << chars[rand(chars.size-1)] }
1.upto(rand(5) + 5) { |i| y << chars[rand(chars.size-1)] }
printf("The random character arrays are: %s and %s\n", x, y)
c = Array.new(x.size + 1){Array.new(y.size + 1)}
b = Array.new(x.size + 1){Array.new(y.size + 1)}
 
def LCS_length(x, y ,c ,b) 
   m, n = x.size, y.size
   (0..m).each{|i| c[i][0] = 0}
   (0..n).each{|j| c[0][j] = 0}
   for i in (1..m) do
    for j in(1..n) do
    if(x[i - 1] == y [j - 1])
     c[i][j] = c[i - 1][j - 1] + 1;
     b[i][j] = 0
    else
     if(c[i - 1][j] >= c[i][j - 1])
      c[i][j] = c[i - 1][j]
      b[i][j] = 1
     else
      c[i][j] = c[i][j - 1]
      b[i][j] = 2
     end
    end
   end
   end
end
 
def Print_LCS(x, b, i, j)
  return if(i == 0 || j == 0)
  if(b[i][j] == 0)
    Print_LCS(x, b, i-1, j-1)
    printf("%c ", x[i - 1])
  elsif(b[i][j] == 1)
    Print_LCS(x, b, i-1, j)
  else
    Print_LCS(x, b, i, j-1)
  end
end
 
LCS_length(x, y, c ,b) 
print "The Longest-Commom-Subsequence is: "
Print_LCS(x, b, x.size, y.size)