1. Approximate Matching of Run-Length Compressed Strings
Algorithmica (2003)
Veli M¨akinen, Gonzalo Navarro, and Esko Ukkonen

2. Run-Length encoding aaabb (a,3),(b,2)
Edit Distance on Run-Length Compressed Strings
Extending to Weighted Edit Distance
Approximate Searching
Improving a Greedy Algorithm for LCS

3. Part1: An O(mn’+m’n) Algorithm for the Levenshtein Distance
String A=a1a2 · · · am compressed length m’
String B=b1b2 · · · bn compressed length n’
Levenshtein distance, DL(A , B)
di, j = min(di-1, j + 1, di, j-1 + 1, di-1, j-1 + if ai = bj then 0 else 1)
DID(A, B)
di, j = min(di-1, j + 1, di, j-1 + 1, di-1, j-1 + if ai = bj then 0 else ∞)
Use Dynamic Programming

4. Relationship between DID and LCS
2 ×|LCS(A, B)| = m+n -DID(A, B)
m + n = 2 ×|LCS(A, B)|+ x + y
DID(A, B) = x + y

5. Notations

6. Known: Top and Left border Goal: Right and Button border
Equal letter box:

7. Different letter box:
Observation: consecutive cells in the (dij) matrix differ at most by one

8. Algorithm:

9. Time Complexity of the Algorithm

10. Part2: Extending to Weighted Edit Distance

11. Which one is correct? or

12. path(d, r ) = Cs min(d, r ) + Cd max(d - r, 0) + Ci max(r - d, 0),
s-t
(q,0)
s-q
(s,t) t
path(d, r ) = Cs min(d, r ) + Cd max(d - r, 0)
+ Ci max(r - d, 0), r r Cs Cs d Cs d Ci Cd
d<r
d=r
d>r

13. How to evaluate min value in constant time
The problem is, path is not a constant any more
+Cs-Ci
(s1,t1)
(s3,t3)
(s2,t2)
+Cs-Cd
(s4,t4)

14. Part3: Approximate Searching
Find all approximate occurrence of A(short pattern) in B(long string)
Let all d0,j=0 and find all dm,j≦k
More efficient approach — evaluate only the first m columns in each long run

15. Time Complexity Short run in B with length r≦m: O(m’r+m)
Long run: O(m’m+m+m)
Total time complexity is O(n’m’m+R), R = number of occurence

16. Part4: Improving a Greedy Algorithm for LCS
Basic idea: Fill the only corner of the boxes
Different letter box:
←x→ +s +t

17. Equal letter box: Recursively tracing an optimal path
Time complexity of tracing a path is O(m’+n’)
The algorithm takes O(m’n’(m’+n’))

18. Analysis of Time Complexity
Observation: each cell in the borders of the boxes can be visited only once
Also achieve O(m’n+n’m) bound
Time complexity is
O(min(m’n’(m’+n’), m’n+n’m))
Space complexity is O(m’n’)