1. Xiaoming Sun Tsinghua University David Woodruff MIT
The Communication and Streaming Complexity of Computing the Longest Common and Increasing Subsequences
Xiaoming Sun
Tsinghua University
David Woodruff
MIT

2. The Problem 4 3 7 3 1 1 Stream of elements a1, …, an 2 
Stream of elements a1, …, an 2 
Algorithm given one pass over stream
Problem: Compute the longest increasing subsequence (LIS) – in this case answer is (3,7)

3. Previous Work Let k be the length of the LIS of the stream
There exists an algorithm which computes the LIS with O(k2 log ||) space [LNVZ05]
Trivial (k) lower bound
Our first result: Improve both bounds to a tight (k2 log ||/k)

4. Our Lower Bound Reduction from indexing function: What is xi?
Alice
Bob
What is xi?
x 2 {0,1}n
i 2 [n] = {1, 2, …, n}
Randomized 1-way communication is (n)

5. What is xi? x 2 {0,1}n i 2 [n] = {1, 2, …, n} Alice Bob
Construct a stream A
Construct a stream B
From LIS(A, B), Bob can get xi
2. |LIS(A, B)| = k, where k is input parameter

6. Alice
Ak-1
Value …
x 2 {0,1}n A: A2 A1
Position in stream
Alice uses x to create k-1 increasing sequences A1, …, Ak-1 For each j, Aj has length j. Each bit of x is encoded in some sequence Aj
Every element in Ak-1 is larger than every element in Ak-2, every element in Ak-2 larger than every element in Ak-3, etc.
Set A = Ak-1 ,…, A2 , A1

7. Aj-1
Aj+1
Value
Position in stream Aj B: B
Bob
i 2 [n]
Bob uses i to recover Aj, the sequence encoding xi
Bob creates an increasing subsequence B of length k-j,
Every element in B is greater than Ar if r < j, and every element in B is less than Ar if r > j

8. What is xi? x 2 {0,1}n i 2 [n] Alice Bob Aj+1 B Aj Aj-1 B
Value B Aj
Aj-1 B
A = Ak-1, …, A2, A1
Position in stream
LIS(A, B) = Aj, B, and |LIS(A, B)| = k
But xi encoded in Aj, so Bob recovers xi

9. Thus, any streaming algorithm must use (n) space.
But what is n? We need to construct k increasing sequences that are different for different x in {0,1}n
Assume || large. Divide  into k-1 blocks of size ||/(k-1)
Let Aj be a random increasing sequence of length j in block j.
The space to represent Aj is (k log ||/k) for j > k/2
Set n = (k2 log ||/k).

10. Our Upper Bound
When processing the stream, keep lists A[1], A[2], …, A[k].
A[j] is an LIS of length j in the stream with minimal last element.
Let L[1], L[2], …, L[k] be last elements of A[1], A[2], …, A[k]
To process item x, find i for which L[i] < x < L[i+1], and replace A[i+1] with A[i], x

11. So we have k arrays A[1], …, A[k], each of length at most k.
Naively, this takes O(k2 log ||) space.
But the Ai are increasing, so can compress the list by storing differences.
Total space is O(k2 log ||/k).

12. This talk First result: a tight space bound for the LIS problem
Second result: tight bounds for longest common subsequence (LCS)

13. LCS Bounds
Problem: Alice has a permutation  of [N], Bob has a permutation  of [N]. Decide if |LCS(, )| ¸ k.
Previous space bound: (k) [LNVZ05]
Our space bound: (N) for 3 · k · N/2
(holds for randomized O(1)-pass algorithms)

14. LCS Bounds Why can we only prove (N) for 3 · k · N/2?
If k = 2, reduces to equality test.
If k large, there are at most O(N2(N-k)) permutations  with |LCS(, )| > k, so just use an equality test with error O(1/N2(N-k))

15. Our Lower Bound
Padding lemma: if for k = 3 the randomized communication complexity is (N), then it’s (N) for all k · N/2
Proof: just pad each of the inputs by some common subsequence of length k-3

16. Remains to show high complexity for k =3. We reduce from disjointness
Is there an
i such that
xi = yi = 1?
Alice
Bob
x 2 {0,1}n
y 2 {0,1}n
Randomized multi-way communication is (n)

17. Is there an i such that xi = yi = 1? x 2 {0,1}N/3 y 2 {0,1}N/3 Alice
Bob
x 2 {0,1}N/3
y 2 {0,1}N/3
Construct 
Construct 
Want |LCS(, )| ¸ 3 iff x and y are disjoint

18. Divide 1, …, N into N/3 groups
Alice
 = 1, 2, …, N/3
x 2 {0,1}N/3
Divide 1, …, N into N/3 groups
G1 = (1, 2, 3), G2 = (4, 5, 6), …, GN/3 = (N-2, N-1, N).
Use x to choose 1, …, N/3
i acts on Gi
If xi = 0, i (m+1, m+2, m+3) = (m+1, m+2, m+3).
If xi = 1, i (m+1, m+2, m+3) = (m+1, m+3, m+2).

19. Divide 1, …, N into N/3 groups
Bob
y 2 {0,1}N/3
 = N/3 , …, 1
Divide 1, …, N into N/3 groups
G1 = (1, 2, 3), G2 = (4, 5, 6), …, GN/3 = (N-2, N-1, N).
Use y to choose 1, …, N/3
i acts on Gi
If yi = 0, i (m+1, m+2, m+3) = (m+3, m+2, m+1).
If yi = 1, I (m+1, m+2, m+3) = (m+1, m+3, m+2).

20. N/3(GN/3)
N/3(GN/3) … …
3(G3)
3(G3)
2(G2)
2(G2)
1(G1)
1(G1)
Claim: |LCS(, )| · 3.
Proof: Use the fact that LCS(, ) intersects at most one Gi
Claim: |LCS(, )| = 3 iff there is some i with xi = yi = 1
Proof: Use the way we defined i and i
Thus, can decide disjointness, so (N) communication.

21. Other results Tight space bounds for computing the LIS length.
Generalization to approximate LIS and LCS. Still many gaps here.
Example: approximate LIS length, we have (1/) and O(k log ||). Recent work [GJKK07] has shown O(sqrt(N/) log ||), but still large gap.

22. Conclusion First result: a tight bound for the LIS
Second result: an (N) space bound for the LCS k-decision problem for 3 · k · N/2
Other results for approximation problems
Another open question: extend our lower bound for LIS to randomized multi-round