Computing Inverse ST in Linear Complexity

1. 1 Computing Inverse ST in Linear Complexity Ge Nong, Sun Yat-Sen University, PRC Sen Zhang, SUNY College at Oneonta, USA Wai Hong Chan, Hong Kong Baptist University, HK

2. 2 Outline Problem introduction Existing solutions The linear solution

3. 3 Introduction Burrows and Wheeler Transfrom (BWT)

4. 4 Introduction Sort Transform (ST) - Introduced in 1996 by Schindler A generalization of the BWT for limited-order contexts

5. 5 IBWT vs. IST

6. 6 Difficulties in Inverting ST Problems to be solved All the k-order contexts must be restored in their correct order To detect if any adjacent pair of k-order contexts are equal or not

7. 7 Existing Results on BWT and ST Results on BWT Suffix array construction algorithms can be directly used to compute BWT, linear algorithms in O(n) time/space are available Computing the inverse BWT is trivial, and can be done in O(n) time/space Results on ST ST can be computed using the algorithm for BWT, with slight modifications The best algorithm for computing the inverse ST requires O(n) space and O(nlogk) time

8. 8 Existing Solutions Question: Is the inverse ST linear computable?

9. 9 Objective To design an algorithm for computing inverse ST in linear time/space, i.e., O(n)

10. 10 The Linear Solution Problem formalization S: a size-n string, ending with a sentinel $ Fk/Lk: the transposed first/last columns of Mk, respectively Pk: mapping Lk to Fk Qk: mapping Fk to Lk, reciprocal to Pk

11. 11 Restoring all the k-order contexts We introduce the concept of “cycle” to solve this problem The characters of Lk are classified into cycles, where each cycle a(i) is defined as

12. 12 Algorithm for finding all the cycles Algorithm framework Initially, mark all the items of Lk as unvisited. Traverse Lk once from left to right. For each unvisited item Lk[i], retrieve the cycle a(i) using Qk in O(ni) time, where ni=|a(i)| is the length of cycle a(i) The algorithm can find all cycles in O(n) time/space

13. 13 Examples of Cycles

14. 14 Computing heights for cycles Height(i) = LCP (i -1,i) Height(i) < k if and only if D(i)=1 How to compute the heights of the characters in all cycles in O(n) time/space?

15. 15 Computing the Heights in O(n) time Classify the cycles into two kinds, according to their lengths For a cycle with ni = k/2, we need O(ni +k)=O(ni) to compute the heights for all the characters in the cycle, see Lemma 2. For a cycle with ni < k/2, we need O(ni) to compute the heights for all the characters in the cycle, see Theorem 4.

16. 16 The main results We have explored a number of properties to see that For a cycle with length ni = k/2, the time for finding the heights of all the characters in this cycle is, in the worst case, k + 2ni - 2 ? O(ni)

17. 17

18. 18 Theorem 5 Given Lk, we can restore the original text S in O(n) time/space, for k in [1, n].

19. 19 Making the overall solution Consisting of these linear steps: Find all the cycles, and classify them into two classes according to whether their length is shorter than k/2 or not Compute the heights for all the characters in the cycles not shorter than k/2, and their lower adjacent cycles shorter than k/2 Compute the heights for all the characters in the cycles shorter than k/2 and not included in the above step From the height vector H, compute the difference vector D Restore S from Lk, Pk and D

20. 20 Complexity

21. 21 Undergoing works Symmetric encryption solutions based on ST/IST Efficient implementation of the proposed linear algorithm Efficient algorithms for direct computing the ST

22. 22 Acknowledgment Thanks to all the reviewers of this paper Thanks to CPM 2008