Calculation method of the shortest editing distance implemented by ruby

Time:2022-4-25

Dynamic programming algorithm is used to calculate the shortest editing distance.

 

Copy codeThe code is 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 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, insert, 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 in(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 here
  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}”

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