0) Store the current position, i, in both strings as an “active solution” of length 0 (an active solution is a datastructure containing an offset in string1, an offset in string2 and a current length).

1) check to see if string1[i] eq string2[x] where x is set to the last tested string2 position in each active solution plus one, if so increment its length by 1.

2) Same as above, but testing string2[i] against active solutions

3) At this point, all active solutions not matched in steps 1 or 2, above are removed from the list of active solutions and added to known solutions.

4) For all positions in string2 which are saved in the lookup table (see below) under string1[i], store a new, active solution of length 1.

5) Same as #4, but reversing string1/string2.

6) Append the index i to the list at lookup1[string1[i]]

7) as #6, but save to lookup2[string2[i]]

Once done with the outer loop, return the longest known solution.

Practical concerns include the fact that this approach could be memory intensive for very large strings.