## Tuesday, December 6, 2011

### No. 25 - Edit Distance

Problem: Implement a function which gets the edit distance of two input strings. There are three types of edit operations: insertion, deletion and substitution. Edit distance is the minimal number of edit operations to modify a string from one to the other.

For example, the edit distance between “Saturday” and “Sunday” is 3, since the following three edit operations are required to modify one into the other:
1.       Saturday → Sturday (deletion of ‘a’)
2.       Sturday→ Surday (deletion of ‘t’)
3.       Surday → Sunday (substitution of ‘n’ for ‘r’)

There is no way to achieve it with fewer than three operations.

Analysis: If a function f(i, j) is defined to indicate the edit difference between the substring of the  first string ending with the jth character and the substring of the second string ending with the ith character. It is obvious that f(i, 0) = i, because when we delete i characters from the substring of the first string ending with the ith character, we get an empty string (it is also the substring of the second string ending with the 0-th string). Similarly, f(0, j) = j.

Let discuss the cases when both i and j are greater than 1. If the ith character of the second string is same as the jth character of the first string, no edit operations are necessary. Therefore, f(i, j) = f(i-1, j-1).

When the jth character of the first string is different from the ith character of the second string, there are three options available: (1) Insert the ith character of second string into the first string. In this case, f(i, j) = f(i-1, j) + 1. (2) Delete the jth character of the first string. In this case, f(i, j) = f(i, j-1) + 1. (3) Replace the jth character of the first string with the ith character of the second string. In this case, f(i, j) = f(i-1, j-1) + 1. What is the final value for f(i, j)? It should be the minimal value of the three cases.

If we draw a table to show the edit distance values f(i, j) between “Saturday” and “Sunday”, it looks like the Table 1.

 S a t u r d a y 0 1 2 3 4 5 6 7 8 S 1 0 1 2 3 4 5 6 7 u 2 1 1 2 2 3 4 5 6 n 3 2 2 2 3 3 4 5 6 d 4 3 3 3 3 4 3 4 5 a 5 4 3 4 4 4 4 3 4 y 6 5 4 4 5 5 5 4 3
Table 1: Edit distance value f(i, j) between “Saturday” and “Sunday”.

The edit distance of two strings is at the right-bottom corner of the table for edit distance values.
A formal equation can be defined for this problem:

It is not difficult to develop code based on the equation above. An edit distance value table can be implemented as a 2-D array. Some sample code is shown as below:

int getEditDistance(char* str1, char* str2)
{
if(str1 == NULL || str2 == NULL)
return -1;

int len1 = strlen(str1);
int len2 = strlen(str2);

int** distances = (int**)new int[len2 + 1];
for(int i = 0; i < len2 + 1; ++ i)
distances[i] = new int[len1 + 1];

int editDistance = getEditDistance(str1, str2, distances, len1, len2);

for(int i = 0; i < len2 + 1; ++ i)
delete[] distances[i];
delete[] distances;

return editDistance;
}

int getEditDistance(char* str1, char* str2, int** distances, int len1, int len2)
{
for(int i = 0; i < len2 + 1; ++ i)
distances[i] = i;
for(int j = 0; j < len1 + 1; ++ j)
distances[j] = j;

for(int i = 1; i < len2 + 1; ++ i)
{
for(int j = 1; j < len1 + 1; ++ j)
{
if(str1[j - 1] == str2[i - 1])
distances[i][j] = distances[i - 1][j - 1];
else
{
int deletion = distances[i][j - 1] + 1;
int insertion = distances[i - 1][j] + 1;
int substitution = distances[i - 1][j - 1] + 1;
distances[i][j] = min(deletion, insertion, substitution);
}
}
}

return distances[len2][len1];
}

int min(int num1, int num2, int num3)
{
int less = (num1 < num2) ? num1 : num2;
return (less < num3) ? less : num3;
}

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1. When the jth character of the first string is different from the ith character of the second string, there are three options available: (1) Insert the ith character of second string into the first string. In this case, f(i, j) = f(i-1, j) + 1. (2) Delete the jth character of the first string. In this case, f(i, j) = f(i, j-1) + 1. (3) Replace the jth character of the first string with the ith character of the second string. In this case, f(i, j) = f(i-1, j-1) + 1. What is the final value for f(i, j)? It should be the minimal value of the three cases.

Why only three options available options?
Case (4) Insert the jth character of first string into the second string. In this case, f(i, j) = f(i-1, j) + 1. Case (5) Delete the ith character of the second string. In this case, f(i, j) = f(i, j-1) + 1.

1. there should be an assumption that you can only modify the first.

2. 3. This comment has been removed by the author.

4. made csharp site on csharp code examples

5. NayHoh,

This is indeed great! But I think perhaps you are generally referring No. 25 - Edit Distance which is getting unsustainable.

We recently received our first CloudWatch billing alarm and I had a really difficult time figuring out what was going on with our above-average S3 AWS Training USA
charges. My research led me to the CloudWatch S3 metrics page where I found that in the last few weeks our consumption has suddenly spiked and the number of objects grew out of control. More and more searching led me to discover that aborted multipart uploads leave behind invisible S3 data. We've never had this problem before, so I was very surprised to see that our PowerShell automation, running flawlessly for over a year, is suddenly leaving files behind.

Great effort, I wish I saw it earlier. Would have saved my day :)

Many Thanks,
Ajeeth

6. Selamat Petang,

Great post. Well though out. This piece reminds me when I was starting out No. 25 - Edit Distance after graduating from college.

Route 53 is not refreshing in less than 30 minutes. I create and assign a elastic ip adress. i point to A record from route 53 to EC2 instance to North California and seems not working. AWS Training

By the way do you have any YouTube videos, would love to watch it. I would like to connect you on LinkedIn, great to have experts like you in my connection (In case, if you don’t have any issues).

MuchasGracias,
Ajeeth