The daily walkthrough of the classic Algorithm question--the fifth question string similarity

The original: A daily walkthrough of the classic Algorithm--the fifth question string similarity

Let's look at another version of the longest common subsequence, find the string similarity (edit distance), I also said, this is a very practical algorithm, in the DNA contrast, the net

Page clustering and other aspects are useful.

One: Concept

For two strings A and B, change the string A to B through basic additions or deletions, or change B to a, the least steps we use in the process of change are called "editing distances".

For example, the following string: We through a variety of operations, after the seizure of the editing distance of 3, do not know you see it?

Second: Analysis

It may be a bit complicated and difficult to understand, we try to split the big question, "String vs string", break it down to "character vs string", then decompose

into "character vs. character".

<1> "character" vs "character"

This is the simplest case, such as the editing distance between "A" and "B" is obviously 1.

<2> "character" vs "string"

"A" to "AB" editing distance of 1, "a" and "ABA" editing distance of 2.

<3> "string" vs "string"

The editorial distance of "ABA" and "BBA" is 1, and we can draw a conclusion that "ABA" is a set of editing distances from 23 subsequence and "BBA" strings.

The minimum editing distance taken out in this case, which means that we have the problem of repeated computations, and I am seeking the editing distance of the subsequence "AB" and "BBA"

A minimum value is chosen for the editing distance between the subsequence "a" and "BBA" and "B" and "BBA", but I have already calculated the sequence A and sequence B earlier, and this repeated calculation

The problem is a bit like "Fibonacci", just to meet the "dynamic planning" in the optimal sub-structure and overlap sub-problem, so we decided to use dynamic programming to solve.

Three: Formula

As with the longest common subsequence, we use a two-dimensional array to hold the minimum editing distance for the current position of the string x and Y.

Existing two sequence X={x1,x2,x3,...xi},y={y1,y2,y3,....,yi},

Set a C[I,J]: Saves the current smallest LD of Xi and YJ.

①: When Xi = Yi, then c[i,j]=c[i-1,j-1];

②: when Xi! = Yi, then c[i,j]=min{c[i-1,j-1],c[i-1,j],c[i,j-1]};

Eventually our C[i,j] kept the smallest ld.

Four: Code

1 usingSystem;2 3 namespaceConsoleApplication24 {5      Public class Program6     {7         Static int[,] Martix;8 9         Static stringSTR1 =string. Empty;Ten  One         Static stringSTR2 =string. Empty; A  -         Static voidMain (string[] args) -         { the              while(true) -             { -STR1 =console.readline (); -  +STR2 =console.readline (); -  +Martix =New int[STR1. Length +1, str2. Length +1]; A  atConsole.WriteLine ("the editing distance for the string {0} and {1} is: {2}\n", str1, str2, LD ()); -             } -         } -  -         /// <summary> -         ///calculating the editing distance of a string in         /// </summary> -         /// <returns></returns> to          Public Static intLD () +         { -             //Initialize boundary values (ignoring boundary conditions at calculation) the              for(inti =0; I <= str1. Length; i++) *             { $Martix[i,0] =i;Panax Notoginseng             } -  the              for(intj =0; J <= str2. Length; J + +) +             { Amartix[0, j] =J; the             } +  -             //the X-coordinate of the matrix $              for(inti =1; I <= str1. Length; i++) $             { -                 //the Y-coordinate of the matrix -                  for(intj =1; J <= str2. Length; J + +) the                 { -                     //Equality CasesWuyi                     if(Str1[i-1] = = Str2[j-1]) the                     { -Martix[i, j] = Martix[i-1, J-1]; Wu                     } -                     Else About                     { $                         //take "left Front", "Top", and "left" minimum value -                         varTemp1 = Math.min (Martix[i-1, j], Martix[i, J-1]); -  -                         //Get Minimum value A                         varmin = Math.min (Temp1, Martix[i-1, J-1]); +  theMartix[i, J] = min +1; -                     } $                 } the             } the  the             //returns the editing distance of a string the             returnmartix[str1. Length, str2. Length]; -         } in     } the}

The daily walkthrough of the classic Algorithm question--the fifth question string similarity