編輯距離的Ruby實現

利用動態規划算法，實現最短編輯距離的計算。

#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  endout_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)  endendedit_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}"