Sublinear Algorithms Lecture 3 Sofya Raskhodnikova Penn State - PowerPoint PPT Presentation

Sublinear Algorithms Lecture 3 Sofya Raskhodnikova Penn State University 1

Graph Properties

Testing if a Graph is Connected [Goldreich Ron] Input: a graph 𝐻 = (𝑊, 𝐹) on 𝑜 vertices • in adjacency lists representation (a list of neighbors for each vertex) • maximum degree d , i.e., adjacency lists of length d with some empty entries Query (𝑤, 𝑗) , where 𝑤 ∈ 𝑊 and 𝑗 ∈ [𝑒] : entry 𝑗 of adjacency list of vertex 𝑤 Exact Answer: W (dn) time • Approximate version: Is the graph connected or ² -far from connected? # 𝑝𝑔 𝑓𝑜𝑢𝑗𝑠𝑓𝑡 𝑗𝑜 𝑏𝑒𝑘𝑏𝑑𝑓𝑜𝑑𝑧 𝑚𝑗𝑡𝑢𝑡 𝑝𝑜 𝑥ℎ𝑗𝑑ℎ 𝐻 1 𝑏𝑜𝑒 𝐻 2 𝑒𝑗𝑔𝑔𝑓𝑠 dist 𝐻 1 , 𝐻 2 = 𝑒𝑜 1 Time: 𝑃 𝜁 2 𝑒 today No dependence on n! + improvement on HW 3

Testing Connectedness: Algorithm Connectedness Tester(G, d, ε ) Repeat s=16/ e d times: 1. pick a random vertex 𝑣 2. determine if connected component of 𝑣 is small: 3. perform BFS from 𝑣 , stopping after at most 8/ e d new nodes Reject if a small connected component was found, otherwise accept. 4. Run time: O( 𝑒 / e 2 𝑒 2 )=O(1/ e 2 𝑒 ) Analysis: • Connected graphs are always accepted. • Remains to show: 2 If a graph is ² -far from connected, it is rejected with probability ≥ 3 4

Testing Connectedness: Analysis Claim 1 If G is e -far from connected, it has ≥ e 𝑒𝑜 4 connected components. Claim 2 If G is e -far from connected, it has ≥ e 𝑒𝑜 8 connected components of size at most 8/ e d . If Claim 2 holds, at least e 𝑒𝑜 8 nodes are in small connected components. • 2⋅8 16 e 𝑒𝑜/𝑜 = e 𝑒 nodes to detect one • By Witness lemma, it suffices to sample from a small connected component. 5

Testing Connectedness: Proof of Claim 1 Claim 1 If G is e -far from connected, it has ≥ e 𝑒𝑜 4 connected components. We prove the contrapositive: If G has < e 𝑒𝑜 4 connected components, one can make G connected by modifying < e fraction of its representation, i.e., < e 𝑒𝑜 entries. • If there are no degree restrictions, k components can be connected by adding k-1 edges, each affecting 2 nodes. Here, k < e 𝑒𝑜 4 , so 2k-2 < e 𝑒𝑜 . • What if adjacency lists of all vertices in a component are full, i.e., all vertex degrees are d? 6

Freeing up an Adjacency List Entry Claim 1 If G is e -far from connected, it has ≥ e 𝑒𝑜 4 connected components. What if adjacency lists of all vertices in a component are full, i.e., all vertex degrees are d? 𝑤 • Consider an MST of this component. Let 𝑤 be a leaf of the MST. • Disconnect 𝑤 from a node other than its parent in the MST. • • Two entries are changed while keeping the same number of components. • Thus, k components can be connected by adding 2k-1 edges, each affecting 2 nodes. Here, k < e 𝑒𝑜 4 , so 4k-2 < e 𝑒𝑜 . 7

Testing Connectedness: Proof of Claim 2 Claim 1 If G is e -far from connected, it has ≥ e 𝑒𝑜 4 connected components. Claim 2 If G is e -far from connected, it has ≥ e 𝑒𝑜 8 connected components of size at most 8/ e d . If Claim 1 holds, there are at least e 𝑒𝑜 4 connected components. • 𝑜 4 Their average size ≤ e 𝑒𝑜/4 = • e 𝑜 . • By an averaging argument (or Markov inequality), at least half of the components are of size at most twice the average. 8

Testing if a Graph is Connected [Goldreich Ron] Input: a graph 𝐻 = (𝑊, 𝐹) on 𝑜 vertices • in adjacency lists representation (a list of neighbors for each vertex) • maximum degree d Connected or 𝜁 -far from connected? 1 𝑃 𝜁 2 𝑒 time (no dependence on 𝑜 ) 9

Randomized Approximation in sublinear time Simple Examples

Randomized Approximation: a Toy Example Input: a string 𝑥 ∈ 0,1 𝑜 0 0 0 1 … 0 1 0 0 Goal: Estimate the fraction of 1’s in 𝑥 (like in polls) It suffices to sample 𝑡 = 1 ⁄ 𝜁 2 positions and output the average to get the fraction of 1’s ±𝜁 (i.e., additive error 𝜁 ) with probability ¸ 2/3 Hoeffding Bound Let Y 1 , … , Y s be independently distributed random variables in [0,1] and 𝑡 ≥ δ ≤ 2e −2𝜀 2 /𝑡 . let Y = ∑ Y i (sample sum). Then Pr Y − E Y 𝑗=1 𝑡 Y i = value of sample 𝑗 . Then E[Y] = ∑ E[Y i ] = 𝑡 ⋅ (fraction of 1’s in 𝑥 ) 𝑗=1 Pr (sample average) − fraction of 1′s in 𝑥 ≥ 𝜁 = Pr Y − E Y ≥ 𝜁𝑡 ≤ 2e −2𝜀 2 /𝑡 = 2𝑓 −2 < 1/3 substitute 𝑡 = 1 ⁄ 𝜁 2 Apply Hoeffding Bound with 𝜀 = 𝜁𝑡 11

Approximating # of Connected Components [Chazelle Rubinfeld Trevisan] Input: a graph 𝐻 = (𝑊, 𝐹) on n vertices • in adjacency lists representation (a list of neighbors for each vertex) • maximum degree d Exact Answer: W (dn) time Additive approximation: # of CC ± ε n with probability ¸ 2/3 Time: 𝑒 1 𝑒 • Known: 𝑃 𝜁 2 log 𝜁 , W 𝜁 2 𝑒 • Today: 𝑃 𝜁 3 . No dependence on n! Partially based on slides by Ronitt Rubinfeld: 12 http://stellar.mit.edu/S/course/6/fa10/6.896/courseMaterial/topics/topic3/lectureNotes/lecst11/lecst11.pdf

Approximating # of CCs: Main Idea Let 𝐷 = number of components • For every vertex 𝑣 , define • 𝑜 𝑣 = number of nodes in u’s component Breaks C up into 1 for each component A : ∑ 𝑜 𝑣 = 1 – 𝑣∈𝐵 contributions 1 of different nodes ∑ = 𝐷 𝑜 𝑣 𝑣∈𝑊 • Estimate this sum by estimating 𝑜 𝑣 ’s for a few random nodes – If 𝑣 ’s component is small, its size can be computed by BFS. – If 𝑣 ’s component is big, then 1/𝑜 𝑣 is small, so it does not contribute much to the sum – Can stop BFS after a few steps Similar to property tester for connectedness [Goldreich Ron] 13

Approximating # of CCs: Algorithm Estimating 𝑜 𝑣 = the number of nodes in 𝑣 ’s component : 2 Let estimate 𝑜 𝑣 = min 𝑜 𝑣 , • 𝜁 𝑏 – When 𝑣 ’s component has · 2/ e nodes , 𝑜 𝑣 = 𝑜 𝑣 𝑑 1 − 1 ≤ 𝜁 𝑐 𝑣 = 2/ e , and so 0 < 1 𝑣 − 1 𝑜 𝑣 < 1 𝑣 = 𝜁 2 – Else 𝑜 2 𝑜 𝑣 𝑜 𝑣 𝑜 𝑜 1 = ∑ Corresponding estimate for C is 𝐷 • . It is a good estimate: 𝑣∈𝑊 𝑣 𝑜 1 1 1 1 𝜁𝑜 − 𝐷 = ∑ − ∑ ≤ ∑ 𝐷 𝑣 − 𝑜 𝑣 ≤ 𝑣∈𝑊 𝑣∈𝑊 𝑣∈𝑊 𝑜 𝑣 𝑜 𝑣 𝑜 2 APPROX_#_CCs (G, d, ε ) Repeat s= Θ (1/ e 2 ) times: 1. pick a random vertex 𝑣 2. 𝑣 via BFS from 𝑣 , stopping after at most 2/ e new nodes compute 𝑜 3. = (average of the values 1/𝑜 4. Return 𝐷 𝑣 ) ∙ 𝑜 Run time: O(d / e 3 ) 14

Approximating # of CCs: Analysis 𝜁𝑜 1 − 𝐷 > Want to show: Pr 𝐷 ≤ 3 2 Hoeffding Bound Let Y 1 , … , Y s be independently distributed random variables in [0,1] and 𝑡 ≥ δ ≤ 2e −2𝜀 2 /𝑡 . let Y = ∑ Y i (sample sum). Then Pr Y − E Y 𝑗=1 Let Y i = 1/𝑜 𝑣 for the i th vertex 𝑣 in the sample 𝑡 𝑡 𝑡𝐷 1 1 𝑡𝐷 𝑜 ∑ • Y = ∑ Y i = 𝑜 and E[Y] = ∑ E[Y i ] = 𝑡 ⋅ E[Y 1 ] = 𝑡 ⋅ = 𝑜 𝑣∈𝑊 𝑜 𝑤 𝑗=1 𝑗=1 2 ≤ 2𝑓 − 𝜁2𝑡 𝜁𝑜 𝑜 𝑜 𝜁𝑜 𝜁𝑡 − 𝐷 > Pr 𝐷 = Pr 𝑡 𝑍 − 𝑡 𝐹 𝑍 > = Pr Y − E Y > 2 2 2 1 1 Need 𝑡 = Θ 𝜁 2 samples to get probability ≤ • 3 15

Approximating # of CCs: Analysis 𝜁𝑜 − 𝐷 ≤ So far: 𝐷 2 𝜁𝑜 1 − 𝐷 > Pr 𝐷 ≤ 3 2 2 • With probability ≥ 3 , − 𝐷 ≤ 𝜁𝑜 2 + 𝜁𝑜 − 𝐷 ≤ 𝐷 − 𝐷 + 𝐷 𝐷 2 ≤ 𝜁𝑜 Summary: The number of connected components in 𝑜 -vetex graphs of 𝑒 degree at most 𝑒 can be estimated within ±𝜁𝑜 in time 𝑃 𝜁 3 . 16

Minimum spanning tree (MST) • What is the cheapest way to connect all the dots? Input: a weighted graph 3 with n vertices and m edges 4 2 7 1 5 • Exact computation: – Deterministic 𝑃(𝑛 ∙ inverse-Ackermann (𝑛)) time [Chazelle] – Randomized 𝑃(𝑛) time [Karger Klein Tarjan] Partially based on slides by Ronitt Rubinfeld: 17 http://stellar.mit.edu/S/course/6/fa10/6.896/courseMaterial/topics/topic3/lectureNotes/lecst11/lecst11.pdf

Approximating MST Weight in Sublinear Time [Chazelle Rubinfeld Trevisan] Input: a graph 𝐻 = (𝑊, 𝐹) on n vertices • in adjacency lists representation • maximum degree d and maximum allowed weight w • weights in {1,2,…, w } Output: (1+ ε )-approximation to MST weight, 𝑥 𝑁𝑇𝑈 Time: 𝑒𝑥 𝑒𝑥 𝑒𝑥 No dependence on n! , W • Known: 𝑃 𝜁 3 log 𝜁 2 𝜁 𝑒𝑥 3 log 𝑥 • Today: 𝑃 𝜁 3 18

Idea Behind Algorithm • Characterize MST weight in terms of number of connected components in certain subgraphs of G • Already know that number of connected components can be estimated quickly 19

MST and Connected Components: Warm-up • Recall Kruskal’s algorithm for computing MST exactly. Suppose all weights are 1 or 2. Then MST weight = (# weight-1 edges in MST) + 2 ⋅ (# weight-2 edges in MST) = 𝑜 – 1 + (# of weight-2 edges in MST) MST has 𝑜 − 1 edges = 𝑜 – 1 + (# of CCs induced by weight-1 edges) −1 By Kruskal weight 1 MST connected components weight 2 induced by weight-1 edges