Longest Increasing Subsequences in Windows Based on Canonical Antichain Partition

1. Longest Increasing Subsequences in WindowsBased on Canonical Antichain Partition Erdong Chen (Joint work with Linji Yang & Hao Yuan) Shanghai Jiao Tong Univ.

2. Outline • Problem Definition • Canonical Antichain Partition • Sweep Algorithm • Complexity Analysis • Conclusion

3. Longest Increasing Subsequence (LIS) Input sequence: All Longest Increasing Subsequences:

4. LIS in a Window Sequence a Window The length of LIS within the window is 2

5. 6 6 9 9 9 8 8 2 2 3 3 5 5 3 3 5 5 1 4 6 9 8 2 2 3 3 5 5 1 4 7 7 Longest Increasing Sequence in A Set of Variable-size Windows (LISSET) Length of LIS = 2 + Length of LIS = 3 + Length of LIS = 2 + Length of LIS = 4 = OUTPUT = 11

6. 6 9 2 3 5 3 5 2 3 5 7 Longest Increasing Sequence in A Set of Variable-size Windows (LISSET) Length of LIS = 2 + Length of LIS = 3 + Length of LIS = 2 + Length of LIS = 4 = OUTPUT = 11

7. Related Works • Knuth proposed an O(n log n) algorithm for LIS problem • Fredman proved an Ω(n log n) lower bound under the decision tree model • An O(n log log n) algorithm is possible by using van Emde Boas tree on a permutation.

8. Related Works (Cont.) • Longest Increasing Subsequences in Sliding Windows (LISW problem) (by Michael H. Albert et al), Time Complexity O(OUTPUT + n log log n) • We called it Longest Increasing Subsequences in Fixed-size windows

9. OUTPUT = 14 = 6 Length of LIS = 1 + 6 9 Length of LIS = 2 9 2 6 8 2 3 8 9 8 3 2 5 + 6 9 Length of LIS = 2 + 9 Length of LIS = 1 + 2 3 Length of LIS = 2 + 2 3 5 Length of LIS = 3 + n+w-1 = 8 windows 3 5 Length of LIS = 2 + Length of LIS = 1 5 LISW Problem n = 6 w = 3

10. Our Contribution • A algorithm for the generalized problem LISSET • To solve the sub case LISW problem, our algorithm runs in O(OUTPUT) time. The best result among previous attempts on LISW is O(OUTPUT + n log log n)

11. p2(2,9) p3(3,8) p9(9,7) p1(1,6) p6(6,5) p8(8,4) p5(5,3) p4(4,2) p7(7,1) Canonical Antichain Partition The sequence: 6, 9, 8, 2, 3, 5, 1, 4, 7

12. p2(2,9) p3(3,8) p9(9,7) p1(1,6) p6(6,5) p8(8,4) p5(5,3) p4(4,2) p7(7,1) Points in this region dominates p1 Points in this region dominates p4 Dominance Order a<b iff xa<xb & ya<yb

13. p2 p3 p9 p1 p6 p8 p5 p4 p7 Height of points by Dominance Order Height = 1

14. p2 p3 p1 p6 p8 p5 p4 p7 Height of points by Dominance Order Height = 2 Height = 1

15. p2 p3 p1 p6 p8 p5 p4 p7 Height of points by Dominance Order Height = 4 Height = 2 Height = 3 Height = 1

16. L(1) L(3) L(2) L(4) p2=HEAD(2) p3 p9=HEAD(4) =TAIL(4) p1=HEAD(1) p6=HEAD(3) p8=TAIL(3) p5=TAIL(2) p4 p7=TAIL(1) Antichain: xi<xi+1 & yi>=yi+1 Antichain and Chain

17. L(1) L(3) L(2) L(4) p2 p3 p9 p1 p6 p8 p5 p4 p7 Chain: xi<xi+1 & yi<yi+1 Max Element Min Element Antichain and Chain The sequence: 6, 9, 8, 2, 3, 5, 1, 4, 7 The longest chain <p4,p5,p6,p9> corresponds the LIS <2,3,5,7>

18. 6 9 8 9 8 2 Sweep Algorithm • Sweep from left to right • Operations • DELETE operation (Delete at left) e.g.: delete 6 • INSERT operation (Insert at right) e.g.: insert 2 • QUERY operation • Algorithm Flow

19. L(1) L(3) L(2) p4 p5 p2 p6 p1=pdel p7 p3 DELETE operation D(1) = {p1} D(2) = {p2} D(3) = {p4,p5}

20. p4 p5 p2 p6 p1=pdel p7 p3 D(2) D(3) L(3)/D(3) L(2)/D(2) L(1)/D(1) DELETE operation (Cont.)

21. L’(3) L’(2) L’(1) p4 p5 p2 p6 p1=pdel p7 p3 After the Delete operation

22. Analysis of Delete operation • Theorem 2. The cost of one DELETE operation equals the total number of points whose height decreases, i.e., O(|D|).

23. INSERT & QUERY operations • Theorem 3. The cost of INSERT operation equals the length of the LIS with the pINS as the maximum element. • Theorem 4. The cost of outputting a longest chain equals to the length of the output subsequence.

24. Algorithm Flow • Step 1: Sort the windows Wi by their left endpoints (if two windows share the same left endpoint, the longer window comes first) O(n+m) • Step 2: initialize current window to ∅ • Step 3: slide the window from Wj to Wj+1 (j=1,2,…m-1)

25. Details of Step 3 • Move from Wj to Wj+1, Wj=(a1,b1), Wj+1=(a2,b2) • Disjoint • Overlap • Contain • Same left endpoint • Different left endpoints • QUERY(r2) to output a LIS within Wj+1

26. Amotized Complexity Analysis • Given a sequence π = π1π2 . . . πn • depthiis defined to be the largest height that πiachieved in the m windows. In other words, among all increasing subsequences in m windows, depthiis the length of the longest one with πias the maximal element.

27. Complexity of each operation by Amortize Analysis • QUERY operation: O(OUTPUT) • INSERT operation: • DELETE operations: (A point pi can decrease at most depthi times)

28. Complexity Analysis of LISSET • Theorem 5 (LISSET Problem). The algorithm described above computes the m longest increasing subsequences, one for each window, in total time:

29. Complexity Analysis of LISW • depthi equals the length of the output in window πi-w+1, πi-w+2, …, πi • So, • And, • Thus,

30. Complexity Analysis of LISW • Theorem 6 (LISW Problem). Our algorithm finds the longest increasing subsequence in a sliding window over a sequence of n elements in O(OUTPUT) time.

31. Future Works…

32. Questions?