Lecture 16: Subexponential Time Algorithm for Small Set Expansion - PowerPoint PPT Presentation

Lecture 16: Subexponential Time Algorithm for Small Set Expansion and Unique Games

Lecture Outline • Part I: Unique Games and Small Set Expansion • Part II: Cheeger’s Inequality and Threshold Rank • Part III: Low Threshold Rank Case • Part IV: High Threshold Rank Case • Part V: Sketch of the Extension to Unique Games • Part VI: Open Problems

Part I: Unique Games and Small Set Expansion

Review: Unique Games Problem • Unique Games: Have a graph 𝐻 where we wish to assign each vertex 𝑤 of 𝐻 a label l 𝑤 ∈ [1, 𝑙] where 𝑙 is a large constant. • For each edge (𝑤, 𝑥) in 𝐻 , we have a constraint specifying that 𝑚 𝑥 = 𝜏(𝑚 𝑤 ) where 𝜏 is a permutation of [1, 𝑙] . • Goal: Maximize the number of satisfied constraints.

Unique Games Picture 𝑤 2 In this example, we can satisfy two of the three costraints. 𝑤 3 𝑤 1

Review: Unique Games Conjecture • Unique Games Conjecture (UGC): For all 𝜗 > 0 , there exists a constant 𝑙 such that it is NP-hard to distinguish between the case when at most 𝜗 of the constraints can be satisfied and the case when at least (1 − 𝜗) of the constraints can be satsified • UGC is a central open problem in theoretical computer science • If true, implies optimal inapproximability results for MAX CUT and other problems

Expansion of a Graph • Definition: If 𝐻 is a 𝑒 -regular graph on a set of 𝑜 vertices 𝑊 and 𝑇 ⊆ 𝑊 is a subset of size at 𝑜 2 , the expansion 𝛸 𝐻 (𝑇) of 𝑇 is 𝛸 𝐻 𝑇 = most |𝐹 𝑇,𝑊∖𝑇 | where 𝐹(𝑇, 𝑊 ∖ 𝑇) is the set of edges 𝑒|𝑇| between 𝑇 and 𝑊 ∖ 𝑇 . • Definition: the expansion of a graph 𝐻 is 𝛸 𝐻 = min 𝛸 𝐻 (𝑇) 𝑇:0< 𝑇 ≤ 𝑜 2

Small Set Expansion Problem • What if we want to restrict ourselves to subsets of a certain small density 𝜀 ? • Definition: Define 𝛸 𝐻 (𝜀) = min 𝛸 𝐻 (𝑇) 𝑇: |𝑇| 𝑜 =𝜀 • Gap small set expansion problem (SSE): Given small constants 𝜃, 𝜀 > 0 and a graph 𝐻 on 𝑜 vertices, distinguish whether 𝛸 𝐻 𝜀 ≥ 1 − 𝜃 or 𝛸 𝐻 𝜀 ≤ 𝜃 .

Relation between UG and SSE • One direction: Given a unique games instance 𝐻 where each vertex is involved in the same number of constraints, we can form the graph ෡ G whose vertices correspond to pairs (𝑤, 𝑗) where 𝑤 ∈ 𝑊(𝐻) and 𝑗 ∈ [1, 𝑙] and whose edges correspond to satisfied constraints. • Call ෡ G the label extended graph of 𝐻 • A solution to the unique games instance satisfying almost all constraints gives a subset of vertices of density 𝜀 = 1 𝑙 with small expansion.

Label Extended Graph Picture 𝑤 2 𝑤 3 𝑤 1

Relation between UG and SSE • Unfortunately, there could be other sets of the same size which have small expansion. • For example, we could take a subset of 𝑜/𝑙 vertices {v j } and then take all of the pairs (𝑤 𝑘 , 𝑗) . • Still, this suggests that UG and SSE are closely related.

Reduction from SSE to UG • Theorem [RS10]: There is a reduction from SSE to UG. • Idea: Consider the following game with a verifier and two provers. Given a 𝑒 -regular graph 𝐻 : 1 1. the verifier chooses k = 𝜀 edges 𝑣 1 , 𝑤 1 , … , (𝑣 𝑙 , 𝑤 𝑙 ) at random, sends the permuted set (𝑣 1 , … , 𝑣 𝑙 ) to one prover, and sends the permuted set (𝑤 1 , … , 𝑤 𝑙 ) to the other prover. 2. Each prover chooses one vertex from their set 3. The provers win if they selected some edge (𝑣 𝑗 , 𝑤 𝑗 )

Unique Games Instance • This corresponds to a unique games instance: • The vertices are possible subsets of 𝑙 vertices sent to a prover. • Each randomly chosen set of 𝑙 edges 𝑣 1 , 𝑤 1 , … , (𝑣 𝑙 , 𝑤 𝑙 ) gives a constraint between the vertices (𝑣 1 , … , 𝑣 𝑙 ) and (𝑤 1 , … , 𝑤 𝑙 )

Unique Games Partial Strategy • Key idea: If there is a set 𝑇 of size 𝜀𝑜 which has small expansion, the provers can use the following partial strategy: • If they are given a set which contains precisely one vertex in 𝑇 , take that vertex. Otherwise, do not answer. • Because of the small expansion of 𝑇 , when one prover answers, with high probability the other prover answers as well and they will be correct.

From Partial to Full Strategies • In a unique game, we must select a choice for every vertex. • Idea: Play the game multiple times independently and allow the provers to choose one game which they will play. To win, the provers must choose the same game and win it. • Repeating the game a constant number of times, with high probability the provers’ partial strategy will work for at least one game (and they can choose the first such game)

Soundness • Also need to show that this unique game is sound, i.e. if there is no set of size 𝜀𝑜 with small expansion then the provers have no strategy to succeed. • We won’t discuss this here, see [RS10] for details.

Subexponential Time Algorithm • Theorem [ABS10]: There is an absolute constant 𝑑 such that 1. There is a 2 𝑃(𝑙𝑜 𝜗 ) time algorithm that takes a unique games instance with alphabet size 𝑙 which has a solution satisfying 1 − 𝜗 𝑑 of its constraints and outputs a solution satisfying 1 − 𝜗 of its constraints. 𝑃 𝑜𝜗 2. There is a 2 𝜀 time algorithm that takes a d- regular graph 𝐻 such that 𝛸 𝐻 𝜀 ≤ 𝜗 𝑑 and outputs a set of vertices 𝑇′ such that 𝑇 ′ ≤ 𝜀𝑜 and 𝛸 𝐻 𝑇′ ≤ 𝜗

Subexponential Time Algorithm • For most of the remainder of this lecture, we will focus on the subexponential time algorithm for SSE • The subexponential time algorithm for UG is an extension of this algorithm.

Part II: Cheeger’s Inequality and Threshold Rank

Review: Cheeger’s Inequality • Cheeger’s inequality : Let 𝐻 be a d-regular graph, let 𝐵 be its adjacency matrix, and let 1 = 𝜇 1 ≥ 𝜇 2 ≥ ⋯ ≥ 𝜇 𝑜 be the eigenvalues of 𝐵 𝑒 . Then 1−𝜇 2 ≤ 𝛸 𝐻 ≤ 2(1 − 𝜇 2 ) 2 • The subexponential time algorithm for SSE can be thought of as an analogue of Cheeger’s inequality which looks at many top eigenvalues, not just the second.

Easy Direction of Cheeger’s Inequality 1−𝜇 2 • Proof that 𝛸 𝐻 ≥ 2 : 𝑜 • Let 𝑇 be the subset of size ≤ 2 such that |𝐹(𝑇,𝑊∖𝑇)| 𝛸 𝐻 𝑇 = = 𝛸 𝐻 and take 𝑤 to be the 𝑒|𝑇| vector 𝑤 𝑗 = 𝑜 − |𝑇| if 𝑗 ∈ 𝑇 and 𝑤 𝑗 = −|𝑇| if 𝑗 ∉ 𝑇 . Note that 𝑤 2 = 𝑜 𝑜 − 𝑇 |𝑇| 𝑤 𝑈 𝐵 𝑒 𝑤 • 𝑤 ⊥ 1 , so 𝜇 2 ≥ 𝑤 2

Calculation 𝐵 2 • 𝑤 𝑈 𝑒 σ 𝑗,𝑘 ∈𝐹(𝐻) 𝑤 𝑗 𝑤 𝑘 𝑒 𝑤 = • If 𝐹 𝑇, 𝑊\S were 0 , we would have that 𝐵 2 + 𝑜 − 𝑇 𝑇 2 = 𝑤 𝑈 𝑤 2 𝑒 𝑤 = 𝑇 𝑜 − 𝑇 • Each edge between 𝑇 and 𝑇 ∖ 𝑊 reduces the number of edges within 𝑇 and the number of 1 edges within 𝑇 ∖ 𝑊 by 2 , which creates a difference of 2 = − 𝑜 2 1 𝑇 − 𝑇 2 − 𝑜 − 𝑇 d −2 𝑜 − 𝑇 𝑒

Calculation Continued 𝑜 2 𝐹 𝑇,𝑊∖𝑇 𝐵 𝑜 𝑤 2 − 𝑤 2 − 𝑤 𝑈 𝑒 𝑤 = = 𝑜−|𝑇| ⋅ 𝑒 2𝑜 𝑜− 𝑇 𝑇 𝐹 𝑇,𝑊∖𝑇 𝑜 𝑤 2 (1 − = 𝑜−|𝑇| 𝛸 𝐻 (𝑇)) 𝑒 𝑇 𝑜 • 𝑤 ⊥ 1 , so 𝜇 2 ≥ 1 − 𝑜− 𝑇 𝛸 𝐻 𝑇 ≥ 1 − 2𝛸 𝐻 (𝑇)

Hard Direction of Cheeger’s Inequality • Want to show that 𝛸 𝐻 ≤ 2(1 − 𝜇 2 ) • Proof idea: Let 𝑤 be the eigenvector with eigenvalue 𝜇 2 . Show that there exists a cutoff value 𝑑 such that if we take 𝑇 = {𝑗: 𝑤 𝑗 ≤ 𝑑} then 𝛸 𝐻 𝑇 ≤ 2(1 − 𝜇 2 )

Threshold Rank • Definition: Let 𝐻 be a d-regular graph, let 𝐵 be its adjacency matrix, and let 1 = 𝜇 1 ≥ 𝜇 2 ≥ 𝐵 ⋯ ≥ 𝜇 𝑜 be the eigenvalues of 𝑒 . Given 𝜐 ∈ [0,1) , the threshold rank is defined to be 𝑠𝑏𝑜𝑙 𝜐 𝐻 = |{𝑗: 𝜇 𝑗 > 𝜐}| • Example 1: 𝜇 0 is the usual rank of 𝐵 • Example 2: For all 𝜐 > 0 , with high probability 𝑠𝑏𝑜𝑙 𝜐 𝐻 = 1 for a random graph if there are sufficiently many vertices.

Theorem Cases • Given a small 𝜃 > 0 , either 𝑠𝑏𝑜𝑙 1−𝜃 𝐻 ≤ 𝑜 𝜗 or 𝑠𝑏𝑜𝑙 1−𝜃 𝐻 > 𝑜 𝜗 • Case I (analogue of the easy direction of Cheeger’s inequality): For any set 𝑇 with small expansion, there is a corresponding vector 𝑤 which is close to being in the subspace of eigenvectors with eigenvalue > 1 − 𝜃 . Since this subspace has dimension ≤ 𝑜 𝜗 , we can search for an approximation to 𝑤 in subexponential time.

Theorem Cases • Case II (analogue of the hard direction of Cheeger’s inequality): If 𝑠𝑏𝑜𝑙 1−𝜃 𝐻 > 𝑜 𝜗 then we can find a set of vertices 𝑇 of size at most 𝜀𝑜 (but it could be much smaller) which has small expansion.

Part III: Low Threshold Rank Case