Presentation For VLSI Application of Suffix Trees

1. Presentation For VLSIApplication of Suffix Trees Chunkai Yin Apr 2002

2. Topics • Approximate palindromes and repeats • Multiple common substring problem

3. Approximate palindromes and repeats • Definitions • Find all tandem repeats (naive method) • Find all tandem repeats (Landau-Schmidt) • Find all k-mismatch tandem repeats (L-S)

4. Definitions • Tandem repeat string written as AA, where A is a substring abcabc Specified by starting position and length of A • K-mismatch palindrome/tandem a substring that becomes a palindrome/tandem after k or fewer characters are changed zzcbbcca 2-mismatch palindrome pzcb + bcca axabaybb 2-mismatch tamdem axab + aybb

5. Find all tandem repeats(naive method) • Guess a start position i , middle position j for the tandem • Do a longest common extension query from i and j. If the extension from i reaches j, the tandem exists starting at position i. • Time complexity O(n2) • EX1 fababd

6. Problem: Find all tandem repeats The Landau-Schmidt method divide the problem into four subproblems (Recursive, divide-and-conquer) • Find all tandem repeats contained entirely in the first half of S (up to h = n/2) • Find all tandem repeats contained entirely in the second half of S (after h ) • Find all tandem repeats where the first copy contains position h of S. • Find all tandem repeats where the second copy contains position h of S.

7. How to handle this ? • Subproblem A, B will be solved by recursively applying the method • Subproblem C and D are symmetric, just consider subproblem C. Therefore, the core problem is subproblem C.

8. Solve the subproblem 3 • Initialization: h= n/2; q=h+L; L is a fixed number each time • Compute the longest common extension (from left to right) from h and q. L1 is the length of extension. • Compute the longest common extension (from right to left) from h-1 and q-1. L2

9. Solve the subproblem 3 • If and only if L1+L2>=L, a tandem repeat exists whose first copy contains position h with the length 2L, and it can begin at any position between h-L2 and h+L1-L inclusive. The second copy begins L places to the right. Output the starting point. • EX2

10. L1, L2 and range for starting point L q A B h Δ1=L-L2, L= Δ1 +L2 A=h-L2 Δ2=L-L1, L= Δ2 +L1 B=h- Δ2=h+L1-L Δ2 L1 Δ2 L1 L2 Δ1 L2 Δ1

11. An Example ….b d c a b b d c a b b t t t…. h q L=5 L1=3 L2=3 A:1 B:2 Two forms available: bdcab + bdcab dcabb + dcabb

12. Time Complexity • Time used for output O(Z) Z is the total number of tandem repeats in S • Time used for the extension queries which is proportional to the number of extension queries ext. T(n)=T(n/2) + T(n/2)+ n + n n=n/2 (->) +n/2(<-) A B C D T(n)= O(nlogn), total=O(nlogn)+Z

13. Find all k-mismatch tandem repeats (L-S) • Immediate extension of the O(nlogn+Z) method for finding exact tandem repeats. • Run k successive longest common extension queries forward from h and q and run k successive longest common extension queries backward from h-1 and q-1. • Then try to find t, a midpoint of the tandem repeat. In the interval between h and q, the number of mismatches from h to t (found during the forward extension) plus the the number of mismatches from t+1 to q-1(found during the backward ext) is <=k.

14. Time complexity O(knlogn +Z)

15. Multiple common substring prob. • Problem: What substring are common to a large number of distinct strings? • Definitions T: a generalized suffix tree for K strings **The identical suffixes appearing in more than one string end at distinct leaves. Each leaf has one of K unique string identifier.

16. C(v): the number of distinct leaf string identifiers in the subtree of v • S(v): the total number of leaves in the subtree of v • U(v): how many duplicate suffixes from the same string occur in v’s subtree. • C(v)=S(v)-U(v) • L(k): the length of the longest substring common to at least k of the strings. • U(v)=i:ni(v)>0(ni(v)-1) • ni(v): the number of leaves with identifier i in the subtree rooted at v

17. The key work we need do • Our object: get C(v) • What is the algorithm? • Define Li: the list of leaves with identifier i, in increasing order of their DFS numbers

18. Algorithm for the problem • Build a generalized suffix tree T for the K strings. • Number the leaves of T as they are encountered in a depth-first traversal of T • For each string identifier i, extract the ordered list Li, of leaves with identifier i. • For each identifier i, compute the lca of each consecutive pair of leaves in Li, and increment by one each time that w is the computed lca.

19. Algorithm Cont. Lemma: If we compute the lca for each consecutive pair of leaves in Li, then for any node v, exactly ni(v)-1 of the computed lcas will lie in the substring of v. • With a bottom-up traversal of T, compute, for each node v, S(v), and U(v)=i:ni(v)>0(ni(v)-1)= [h(w):w is in the subtree of v] • Set C(v)=S(v)-U(v) for each v • Accumulate the table of L(k) values. • Time complexity: O(n)

20. End of the presentation Thank you Thank you Thank you