My boss left a note on my last article noting how similar my solution to pattern matching was to file diff algorithms he had been reading about. So naturally I had to take it as a great chance to explore how to do a basic diff between two strings. The solution I've created would be pretty easy to upgrade to support whole lines of text rather than individual letters but I'll leave that as an exercise to you. I'll probably do it if/when I get a spare moment I just doubt I'll blog about it. :)
What's the Real Problem?
We need to figure out what the minimum amount of editing is that's needing to convert string #1 (the "old string") into string #2 (the "new string"). If we didn't need to minimize the editing we could just say delete all of the old string and insert each character of the new string. That's pretty useless though. That's would be the equivalent of your favorite SCM software telling you that the difference between file A and file B is the whole file. Like I said, useless.
So what do I mean when I say editing distance? Imagine you only have three ways to modify a string: insert, delete, and flat out changing letters in your new string to match letters in your old string. So for example to change "ba" into "fg" you would need to swap the b for an f and the a for a g. That would mean that your minimum editing distance would be two. Another example would be to change "adbed" to "abcde". This would have a minimum editing distance of 3 since you would delete the first "d", change the "e" to a "c", and then add an "e" to the end of the string.
Getting Technical
Remember the technique I used last time for string pattern matching? No? Read it. I'm using the same basic idea here with using a matrix, placing scores for the different combinations of two characters and figuring out how to traverse this matrix in an optimal fashion.
The trick in this instance is use the old string as your column headings and the new string as your row headings. The values in the cells that make up the table should be comprised of how many editing steps it would take for you to change the given old character (in its current position!) into the new character. Since, for us, deleting, inserting, or swapping a character all have the same cost (the value 1) and since the edit distance between two identical characters is the value 0 initially it would seem as though we'll just be traversing a matrix of 0's and 1's. However in this case since we are looking for the minimum edit distance we need to take into account the minimum edit distance from the previous step.
Remember that we will be traversing this list from one corner of the matrix to the other. This is important because it limits the number of cells we need to take into account to figure out the minimum edit distance up to our cell.
In an attempt to quickly clarify what I mean let's first look at my code for calculating the edit distance of each pair of characters (which represents a single cell in our matrix at position oldCharacterIndex, newCharacterIndex):
private static int _CalculateMinimumCellEditDistance(List<List<int>> matrix, string oldString, int oldCharacterIndex, string newString, int newCharacterIndex)
{
var oldChar = oldString[oldCharacterIndex];
var newChar = newString[newCharacterIndex];
var localEditDistance = oldChar == newChar ? 0 : 1;
var leftArrayValue = localEditDistance + _GetCellValue(matrix, oldCharacterIndex - 1, newCharacterIndex);
var topLeftArrayValue = localEditDistance + _GetCellValue(matrix, oldCharacterIndex - 1, newCharacterIndex - 1);
var topArrayValue = localEditDistance + _GetCellValue(matrix, oldCharacterIndex, newCharacterIndex - 1);
var minEditDistance = Math.Min(Math.Min(leftArrayValue, topLeftArrayValue), topArrayValue);
return minEditDistance;
}
That makes perfect sense if you don't think about the left and top edges. Here we have to introduce a lil finagling to calculate those cells appropriately. Basically the following code means that if you're in the top row looking up the edit distance in column number oldCharacterIndex, the only way to have gotten there is to have made (oldCharacterIndex + 1) count deletions. By the same token, if we're in the left-most column looking for the value in row number newCharacterIndex, the only way to have gotten there is to have made (newCharacterIndex + 1) count insertions.
private static int _GetCellValue(List<List<int>> matrix, int oldCharacterIndex, int newCharacterIndex)
{
if (oldCharacterIndex < 0 && newCharacterIndex < 0)
return 0;
if (oldCharacterIndex < 0)
return newCharacterIndex + 1;
if (newCharacterIndex < 0)
return oldCharacterIndex + 1;
return matrix[oldCharacterIndex][newCharacterIndex];
}
Now I'd like to point out, I'm 100% certain that this is **A** "right" way but my code might have bugs so take it with a grain of salt. The main purpose of this post is to show another way of tackling a different dynamic programming problem and the thought process I went through.
Once I've scored each pair of characters I then set out to traverse the given matrix by going to the smallest edits possible. I also created a class to handle tracking the deletes and edits that would be needed along the way. This class can be used to play back my changes to tell someone how they could change the old string into the newer one. Also, since dynamic programming calls for us to work through the problem backwards, this class reorders the steps to read correctly.
Now, once again we create a matrix of how well each cell performs according to the above method. In our context a cell's value should essentially be thought of as the number of steps required to convert the character at the cell's column index into the character at cell's row index. In my calculation I'm also taking into account shifting the character into the appropriate position. That means that every cell has a been scored in such a way that they all solved their own version of the problem as though they were the only portion of it. What's beautiful about this is that by bringing these results together we can find the optimal string editing just by following the smallest values backwards from the lower right cell of our matrix. Basically you start at what has to be your end point and assume that you took the optimal path there from the previous three optional cells (the cell immediately to the left, to the top-left, and to the top). If you have a tie between top or left just choose whichever you like. In the case of a tie between the top-left and one of the others though I default to the top-left since that's roughly half the number of cells (and thus string edit steps) we'll need to process.
Quick Example
To help pull together all of this here's an edit distance matrix from my program for diffing "bcdeffghi" and "abcdefghij". This string is useful because it exercises all 3 diff operations (insert, swap, and delete):
b c d e f f g h i
a 1 2 3 4 5 6 7 8 9
b 1 2 3 4 5 6 7 8 9
c 2 1 2 3 4 5 6 7 8
d 3 2 1 2 3 4 5 6 7
e 4 3 2 1 2 3 4 5 6
f 5 4 3 2 1 2 3 4 5
g 6 5 4 3 2 2 2 3 4
h 7 6 5 4 3 3 3 2 3
i 8 7 6 5 4 4 4 3 2
j 9 8 7 6 5 5 5 4 3
The lower right hand corner (3 in this case) is our optimal edit distance. This means the string on the top of the table can be transformed into the string on the left of the table with exactly three separate steps.
As always here's the link to my code (It's in the StringMatching folder along with the pattern matcher):