Download presentation

Published byAudrey King Modified over 4 years ago

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

Similar presentations

OK

1+eps-Approximate Sparse Recovery Eric Price MIT David Woodruff IBM Almaden.

1+eps-Approximate Sparse Recovery Eric Price MIT David Woodruff IBM Almaden.

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google