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* Retranslate the complete English edition from Simplified Chinese * Polish three English explanations after reader-focused review
116 lines
4.0 KiB
Ruby
116 lines
4.0 KiB
Ruby
=begin
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File: edit_distance.rb
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Created Time: 2024-05-29
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Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
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=end
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### Edit distance: brute-force search ###
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def edit_distance_dfs(s, t, i, j)
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# Return 0 if both prefixes s[0...i] and t[0...j] are empty
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return 0 if i == 0 && j == 0
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# If the prefix of s is empty, return the length j of the prefix of t
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return j if i == 0
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# If the prefix of t is empty, return the length i of the prefix of s
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return i if j == 0
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# If the two characters are equal, skip both of them
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return edit_distance_dfs(s, t, i - 1, j - 1) if s[i - 1] == t[j - 1]
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# Minimum edits = the minimum among the insertion, deletion, and replacement subproblems + 1
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insert = edit_distance_dfs(s, t, i, j - 1)
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delete = edit_distance_dfs(s, t, i - 1, j)
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replace = edit_distance_dfs(s, t, i - 1, j - 1)
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# Return the minimum number of edits
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[insert, delete, replace].min + 1
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end
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def edit_distance_dfs_mem(s, t, mem, i, j)
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# Return 0 if both prefixes s[0...i] and t[0...j] are empty
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return 0 if i == 0 && j == 0
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# If the prefix of s is empty, return the length j of the prefix of t
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return j if i == 0
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# If the prefix of t is empty, return the length i of the prefix of s
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return i if j == 0
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# If a result has already been recorded, return it directly
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return mem[i][j] if mem[i][j] != -1
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# If the two characters are equal, skip both of them
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return edit_distance_dfs_mem(s, t, mem, i - 1, j - 1) if s[i - 1] == t[j - 1]
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# Minimum edits = the minimum among the insertion, deletion, and replacement subproblems + 1
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insert = edit_distance_dfs_mem(s, t, mem, i, j - 1)
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delete = edit_distance_dfs_mem(s, t, mem, i - 1, j)
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replace = edit_distance_dfs_mem(s, t, mem, i - 1, j - 1)
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# Record and return the minimum number of edits
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mem[i][j] = [insert, delete, replace].min + 1
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end
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### Edit distance: dynamic programming ###
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def edit_distance_dp(s, t)
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n, m = s.length, t.length
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dp = Array.new(n + 1) { Array.new(m + 1, 0) }
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# State transitions: first row and first column
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(1...(n + 1)).each { |i| dp[i][0] = i }
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(1...(m + 1)).each { |j| dp[0][j] = j }
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# State transitions: remaining rows and columns
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for i in 1...(n + 1)
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for j in 1...(m +1)
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if s[i - 1] == t[j - 1]
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# If the two characters are equal, skip both of them
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dp[i][j] = dp[i - 1][j - 1]
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else
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# Minimum edits = the minimum among the insertion, deletion, and replacement subproblems + 1
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dp[i][j] = [dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1]].min + 1
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end
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end
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end
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dp[n][m]
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end
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### Edit distance: space-optimized dynamic programming ###
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def edit_distance_dp_comp(s, t)
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n, m = s.length, t.length
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dp = Array.new(m + 1, 0)
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# State transitions: first row
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(1...(m + 1)).each { |j| dp[j] = j }
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# State transitions: remaining rows
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for i in 1...(n + 1)
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# State transitions: first column
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leftup = dp.first # Temporarily store dp[i-1, j-1]
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dp[0] += 1
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# State transitions: remaining columns
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for j in 1...(m + 1)
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temp = dp[j]
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if s[i - 1] == t[j - 1]
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# If the two characters are equal, skip both of them
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dp[j] = leftup
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else
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# Minimum edits = the minimum among the insertion, deletion, and replacement subproblems + 1
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dp[j] = [dp[j - 1], dp[j], leftup].min + 1
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end
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leftup = temp # Update to dp[i-1, j-1] for the next iteration
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end
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end
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dp[m]
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end
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### Driver code ###
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if __FILE__ == $0
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s = 'bag'
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t = 'pack'
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n, m = s.length, t.length
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# Brute-force search
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res = edit_distance_dfs(s, t, n, m)
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puts "Minimum edits needed to change #{s} into #{t}: #{res}"
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# Memoized search
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mem = Array.new(n + 1) { Array.new(m + 1, -1) }
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res = edit_distance_dfs_mem(s, t, mem, n, m)
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puts "Minimum edits needed to change #{s} into #{t}: #{res}"
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# Dynamic programming
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res = edit_distance_dp(s, t)
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puts "Minimum edits needed to change #{s} into #{t}: #{res}"
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# Space-optimized dynamic programming
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res = edit_distance_dp_comp(s, t)
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puts "Minimum edits needed to change #{s} into #{t}: #{res}"
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end
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