Lecture 7: Arora Rao Vazirani Lecture Outline Part I: Semidefinite - - PowerPoint PPT Presentation

Lecture 7: Arora Rao Vazirani

Lecture Outline

• Part I: Semidefinite Programming Relaxation for

Sparsest Cut

• Part II: Combining Approaches
• Part III: Arora-Rao-Vazirani Analysis Overview
• Part IV: Analyzing Matchings of Close Points
• Part V: Reduction to the Well-Separated Case
• Part VI: Open Problems

Part I: Semidefinite Programming Relaxation for Sparsest Cut

• Reformulation: Want to minimize

σ𝑗,𝑘:𝑗<𝑘, 𝑗,𝑘 ∈𝐹(𝐻) 𝑦𝑘 − 𝑦𝑗

2 over all cut

pseudo-metrics normalized so that σ𝑗,𝑘:𝑗<𝑘 𝑦𝑘 − 𝑦𝑗

2 = 1

• More precisely, take 𝑒2 𝑗, 𝑘 = 𝑦𝑘 − 𝑦𝑗

2 and

minimize σ𝑗,𝑘:𝑗<𝑘, 𝑗,𝑘 ∈𝐹(𝐻) 𝑒2(𝑗, 𝑘) subject to:

1. ∃𝑑: ∀𝑗, xi ∈ {−𝑑, +𝑑} 2. σ𝑗,𝑘:𝑗<𝑘 𝑒2(𝑗, 𝑘) = 1

Problem Reformulation

• Reformulation: Minimize

σ𝑗,𝑘:𝑗<𝑘, 𝑗,𝑘 ∈𝐹(𝐻)(𝑦𝑗

2 − 2𝑦𝑗𝑦𝑘 + 𝑦𝑘 2) subject to:

1. ∃𝑑: ∀𝑗, xi ∈ {−𝑑, +𝑑} 2. σ𝑗,𝑘:𝑗<𝑘 (𝑦𝑗

2 − 2𝑦𝑗𝑦𝑘 + 𝑦𝑘 2) = 1

• Relaxation: Minimize

σ𝑗,𝑘:𝑗<𝑘, 𝑗,𝑘 ∈𝐹(𝐻)(𝑁𝑗𝑗−2𝑁𝑗𝑘 + 𝑁

𝑘𝑘) subject to:

1. ∀𝑗, 𝑘, 𝑁𝑗𝑗 = 𝑁

𝑘𝑘

2. σ𝑗,𝑘:𝑗<𝑘 (𝑁𝑗𝑗−2𝑁𝑗𝑘 + 𝑁

𝑘𝑘) = 1

3. 𝑁 ≽ 0

Problem Relaxation

• Consider the cycle of length 𝑜. The semidefinite

program can place the cycle on the unit circle and assign each 𝑦𝑗 the corresponding vector 𝑤𝑗.

Bad Example: The Cycle

4 3 1 5 2 𝑤1 𝑤2 𝑤3 𝑤4 𝑤5

G

• σ𝑗,𝑘:𝑗<𝑘(𝑒2(𝑗, 𝑘)) = Θ(𝑜2)
• σ𝑗,𝑘:𝑗<𝑘, 𝑗,𝑘 ∈𝐹(𝐻)(𝑒2(𝑗, 𝑘)) = Θ(𝑜 ⋅ 1/𝑜2)
• Gives sparsity Θ(1/𝑜3), true value is Θ(1/n2)
• Gap is Ω(𝑜), which is horrible!

Bad Example: The Cycle

4 3 1 5 2 𝑤1 𝑤2 𝑤3 𝑤4 𝑤5

G

Part II: Combining Approaches

• Why did the semidefinite program do so much

worse than the linear program?

• Missing: Triangle inequalities

𝑒2 𝑗, 𝑙 ≤ 𝑒2(𝑗, 𝑘) + 𝑒2(𝑘, 𝑙)

• What happens if we add the triangle inequalities

to the semidefinite program?

Adding the Triangle Inequalities

• Let Θ be the angle between 𝑤𝑗 − 𝑤𝑘 and 𝑤𝑙 − 𝑤𝑘
• 𝑤𝑙 − 𝑤𝑗

2 =

𝑤𝑘 − 𝑤𝑗

2 + 𝑤𝑙 − 𝑤𝑘 2 if Θ = 𝜌 2

• 𝑤𝑙 − 𝑤𝑗

2 >

𝑤𝑘 − 𝑤𝑗

2 + 𝑤𝑙 − 𝑤𝑘 2 if Θ > 𝜌 2

• 𝑤𝑙 − 𝑤𝑗

2 <

𝑤𝑘 − 𝑤𝑗

2 + 𝑤𝑙 − 𝑤𝑘 2 if Θ < 𝜌 2

• Triangle inequalities ⬄ no obtuse angles

Geometric Picture

𝑤𝑗 𝑤𝑘 𝑤𝑙 Θ

• Putting 𝑜 > 4 vectors in a circle violates triangle

inequality, so the semidefinite program no longer behaves badly on the cycle. In fact, it gets very close to the right answer.

Fixing Cycle Example

𝑤𝑗 𝑤𝑘 𝑤𝑙

Θ

Goemans-Linial Relaxation

• Semidefinite program (proposed by Goemans

and Lineal): Minimize σ𝑗,𝑘:𝑗<𝑘: 𝑗,𝑘 ∈𝐹(𝐻)(𝑁𝑗𝑗 − 2𝑁𝑗𝑘 + 𝑁

𝑘𝑘) subject to:

1.

∀𝑗, 𝑘, M𝑗𝑗 = 𝑁

𝑘𝑘

2. ∀𝑗, 𝑘, 𝑙, 𝑒2(𝑗, 𝑙) ≤ 𝑒2(𝑗, 𝑘) + 𝑒2(𝑘, 𝑙) where 𝑒2(𝑗, 𝑘) = 𝑁𝑗𝑗 − 2𝑁𝑗𝑘 + 𝑁

𝑘𝑘

3. σ𝑗,𝑘:𝑗<𝑘 𝑁𝑗𝑗 − 2𝑁𝑗𝑘 + 𝑁

𝑘𝑘 = 1

4. 𝑁 ≽ 0

Arora-Rao-Vazirani Theorem

• Theorem [ARV]: The Goemans-Linial

relaxation for sparsest cut gives an 𝑃 𝑚𝑝𝑕𝑜 -approximation and has a polynomial time rounding algorithm.

• Also called metrics of negative type
• Definition: A metric is an 𝑀2

2 metric if it is possible

to assign a vector 𝑤𝑦 to every point 𝑦 such that 𝑒 𝑦, 𝑧 = 𝑤𝑧 − 𝑤𝑦

2.

• Last time: General metrics can be embedded into

𝑀1 with 𝑃 log 𝑜 distortion.

• Theorem [ALN08]: Any 𝑀2

2 metric embeds into 𝑀1

with 𝑃 𝑚𝑝𝑕𝑜(𝑚𝑝𝑕𝑚𝑝𝑕𝑜) distortion.

• [ARV] analyzes the algorithm more directly

𝑀2

2 Metric Spaces

• Degree 4 SOS captures the triangle inequality: if

𝑦𝑗

2 = 𝑦𝑘 2 = 𝑦𝑙 2 then

𝑦𝑗

2 𝑦𝑙 − 𝑦𝑗 2 ≤ 𝑦𝑗 2 𝑦𝑘 − 𝑦𝑗 2 + 𝑦𝑗 2 𝑦𝑙 − 𝑦𝑘 2

⬄2𝑦𝑗

2 𝑦𝑗 2 − 𝑦𝑗𝑦𝑙 ≤ 2𝑦𝑗 2(2𝑦𝑗 2 − 𝑦𝑗𝑦𝑘 − 𝑦𝑗𝑦𝑘)

• Proof:

𝑦𝑗 − 𝑦𝑘

2 𝑦𝑘 − 𝑦𝑙 2 = 4(𝑦𝑗 2 − 𝑦𝑗𝑦𝑘)(𝑦𝑗 2 − 𝑦𝑘𝑦𝑙)

= 4𝑦𝑗

2 𝑦𝑗 2 − 𝑦𝑗𝑦𝑘 − 𝑦𝑘𝑦𝑙 + 𝑦𝑗𝑦𝑙 ≥ 0

• Thus, degree 4 SOS captures the Goemans-Linial

relaxation

Goemans-Linial Relaxation and SOS

Part III: Arora-Rao-Vazirani Analysis Overview

• Semidefinite program gives us one vector 𝑤𝑗 for

each vertex 𝑗.

• We first consider the case when these vectors

are spread out.

• Definition: We say that a set of 𝑜 vectors {𝑤𝑗} is

well-spread if it can be scaled so that:

1. ∀𝑗, 𝑤𝑗 ≤ 1 2.

1 𝑜2 σ𝑗<𝑘 𝑒𝑗𝑘 2 is Ω(1) (the average squared distance

between vectors is constant)

• We will assume we are using this scaling.

Well-Spread Case

• Theorem: Given a set of 𝑜 vectors {𝑤𝑗} which

are well-spread and obey the triangle inequality, there exist well-separated subsets 𝑌 and 𝑍 of these vectors of linear size. In other words, there exist 𝑌, 𝑍 such that:

1. 𝑌 and 𝑍 are Δ far apart (i.e. ∀𝑤𝑗 ∈ 𝑌, 𝑤𝑘 ∈ 𝑍, 𝑒𝑗𝑘

2 ≥ Δ) where Δ is Ω 1 𝑚𝑝𝑕𝑜

2. |𝑌| and |𝑍| are both Ω(𝑜)

Structure Theorem

• Idea: If we have well-separated subsets 𝑌, 𝑍,

take a random cut of the form (𝑇𝑠, ҧ 𝑇𝑠) where 𝑇𝑠 = {𝑗: 𝑒2 𝑤𝑗, 𝑌 = min

𝑘:𝑤𝑘∈𝑍 𝑒𝑗𝑘 2 ≤ 𝑠} and 𝑠 ∈ 0, Δ

• All 𝑗, 𝑘 ∈ 𝐹(𝐻) contribute at most

𝑒𝑗𝑘

2

Δ to the

expected number of edges cut and 𝑒𝑗𝑘

2 to

σ𝑗,𝑘:𝑗<𝑘, 𝑗,𝑘 ∈𝐹(𝐻) 𝑒𝑗𝑘

2 (the number of edges the

SDP “thinks” are cut)

Finding a Sparse Cut

• Since 𝑌, 𝑍 have size Ω(𝑜) and are always on
• pposite sides of the cut, we always have that

𝑇𝑠 ⋅ | ҧ 𝑇𝑠| is Θ 𝑜2 . This matches σ𝑗,𝑘:𝑗<𝑘 𝑒𝑗𝑘

2 up

to a constant factor. (this is why we need 𝑌 and 𝑍 to have linear size!)

• Thus, the expected ratio of the sparsity to the

SDP value is at most

1 Δ = O

𝑚𝑝𝑕𝑜 , as needed.

Finding a Sparse Cut Continued

• Take the hypercube −

1 log2 𝑜 , 1 log2 𝑜 log2 𝑜

• X = 𝑦: σ𝑗 𝑦𝑗 ≤ −1 and Y = 𝑧: σ𝑗 𝑦𝑗 ≥ 1

have the following properties:

1. 𝑌 and 𝑍 have linear size 2. ∀𝑦 ∈ 𝑌, 𝑧 ∈ 𝑍, 𝑦, 𝑧 differ in ≥ 2 log2 𝑜

• coordinates. Thus, 𝑒2 𝑦, 𝑧 ≥

2 log2 𝑜 log2 𝑜 = 2 log2 𝑜

Tight Example: Hypercube

• Let 𝑒 be the dimension such that ∀𝑗, 𝑤𝑗 ∈ ℝ𝑒.
• Algorithm (Parameters 𝜏 > 0, Δ, 𝑒)
• 1. Choose a random 𝑣 ∈ ℝ𝑒.
• 2. Find a value 𝑏 such that there are Ω(𝑜) vectors 𝑤𝑗

with 𝑤𝑗 ⋅ 𝑣 ≤ 𝑏 and Ω(𝑜) vectors 𝑤𝑘 with 𝑤𝑘 ⋅ 𝑣 ≥ 𝑏 +

𝜏 𝑒. Let 𝑌′ and 𝑍′ be these two sets of vectors

• 3. As long as there is a pair 𝑦 ∈ 𝑌′, 𝑧 ∈ 𝑍′ such that

𝑒 𝑦, 𝑧 < Δ, delete 𝑦 from 𝑌′ and 𝑧 from 𝑍′. The resulting sets will be the desired 𝑌, 𝑍.

• Need to show: P[𝑌, 𝑍 have size Ω(𝑜)] is Ω(1)

Finding Well-Separated Sets

• Will first explain why step 1,2 succeed with

probability 2𝜀 > 0.

• Will then show that the probability step 3

deletes a linear number of points is ≤ 𝜀

• Together, this implies that the entire algorithm

succeeds with probability at least 𝜀 > 0.

Finding Well-Separated Sets

• What happens if we project a vector 𝑤 of length

𝑚 in a random direction in ℝ𝑒?

• Without loss of generality, assume 𝑤 = 𝑓1
• To pick a random unit vector in ℝ𝑒, choose

each coordinate according to 𝑂 0,

1 𝑒 (the

normal distribution with mean 0 and standard deviation

1 𝑒), then rescale.

• If 𝑒 is not too small, w.h.p. very little rescaling

will be needed.

Behavior of Gaussian Projections

• What happens if we project a vector of length 𝑚

in a random direction in ℝ𝑒?

• Resulting value has a distribution which is ≈

normal distribution of mean 0, standard deviation

1 𝑒 (difference comes from the

rescaling step)

Behavior of Gaussian Projections

• If we take a random 𝑣 ∈ ℝ𝑒, with probability

Ω(1), σ𝑗<𝑘 (𝑤𝑘 − 𝑤𝑗) ⋅ 𝑣 is Ω

𝑜2 𝑒

• Note: this can fail with non-negligible

probability, consider the case when ∀𝑗, 𝑤𝑗 = ±𝑤. If 𝑣 is orthogonal to 𝑤 then everything is projected to 0.

• For arbitrarily small 𝜗 > 0, with very high

probability, |𝑤𝑗 ⋅ 𝑣| is 𝑃

1 𝑒 for 1 − 𝜗 𝑜 of the

𝑗 ∈ [1, 𝑜]

Success of Steps 1,2

• Together, these facts imply that if we choose a

random unit vector 𝑣, with probability Ω(1), there exist 𝑌′, 𝑍′, 𝑏1, 𝑏2 such that

1. 𝑌′, 𝑍′ have size Ω(𝑜) 2. ∀𝑦 ∈ 𝑌′, 𝑣 ⋅ 𝑦 ≤ 𝑏1 3. ∀𝑧 ∈ 𝑍′, 𝑣 ⋅ 𝑧 ≥ 𝑏2 4. 𝑏2 − 𝑏1 is Ω(1)

Success of Steps 1,2

• We need to show that the probability step 3

eliminates

𝑛𝑗𝑜{ 𝑌 ,|𝑍|} 2

pairs of points is at most 𝜀

• We also need to show how the general case can

be reduced to the well-spread case.

Remaining Steps

Part IV: Analyzing Matchings of Close Points

Matching Covers

• If part 3 of the algorithm causes it to fail with

probability 𝜀, then for 𝜀 fraction of the directions 𝑣 there is a matching 𝑁𝑣 of points of size 𝑑′𝑜 such that for each pair (𝑤𝑗, 𝑤𝑘) in the matching:

1. d2 𝑤𝑗, 𝑤𝑘 ≤ Δ 2. 𝑤𝑘 − 𝑤𝑗 ⋅ 𝑣 ≥

2𝜏 𝑒

where 𝜀, 𝑑′, 𝜏 > 0 are constants

• Note: Corresponds to Definition 4 in [ARV]
• Define the matching graph 𝑁 to be 𝑁 =∪𝑣 𝑁𝑣

• Assume that 𝑒 𝑤𝑗, 𝑤𝑘 ≤

Δ for some 𝑤𝑗, 𝑤𝑘

• P

𝑤𝑘 − 𝑤𝑗 ⋅ 𝑣 ≥ 2𝜏

𝑒 ∼ 𝑓 −

4𝜏2 𝑒2(𝑤𝑗,𝑤𝑘) ≤ 𝑓−4𝜏2 Δ

• If Δ is a sufficiently small constant times

1 𝑚𝑝𝑕𝑜 ,

with high probability there are no pairs of close points at all between 𝑌′ and 𝑍′!

Analyzing Δ = Ω

1 𝑚𝑝𝑕𝑜

• When the algorithm fails in step 3, this gives us

pairs of points (𝑤𝑗, 𝑤𝑘) which are edges of the matching graph 𝑁, implying that 𝑒2 𝑤𝑗, 𝑤𝑘 ≤ Δ and 𝑤𝑘 − 𝑤𝑗 ⋅ 𝑣 ≥

2𝜏 𝑒

• We will use this to find pairs of points (𝑤𝑗, 𝑤𝑘)

which are 𝑙 steps apart in the matching graph where 𝑤𝑘 − 𝑤𝑗 ⋅ 𝑣 ≥

𝑙𝜏 𝑒

Key Idea for Larger Δ

• We will find pairs of points (𝑤𝑗, 𝑤𝑘) which are 𝑙

steps apart in the matching graph where 𝑤𝑘 − 𝑤𝑗 ⋅ 𝑣 ≥ 𝑙𝜏

𝑒

• Using triangle inequality, 𝑒2 𝑤𝑗, 𝑤𝑘 ≤ 𝑙Δ
• P

𝑤𝑘 − 𝑤𝑗 ⋅ 𝑣 ≥ 𝑙𝜏

𝑒 ∼ 𝑓 −

𝑙2𝜏2 𝑒2(𝑤𝑗,𝑤𝑘) ≤ 𝑓−𝑙𝜏2 Δ

• For Δ = Ω

1 log 𝑜 , if we can apply this with 𝑙 =

Ω log 𝑜 , we again obtain a contradiction.

Key Idea for Larger Δ Continued

• Lemma: If a graph 𝐻 has average degree 𝑒, we

can find a non-empty subgraph of 𝐻 which has minimal degree

𝑒 4.

• Proof: Iteratively delete vertices which have

degree ≤ 𝑒

• 4. The total number of edges deleted

is at most 𝑜𝑒

4 . However, 2|𝐹 𝐻 | ≥ 𝑜𝑒, so there

must be ≥ 𝑜𝑒

4 edges remaining.

Average Degree to Minimal Degree

• Average probability that a vertex is matched is

at least 𝑑′𝜀

• Can apply a similar idea and delete any vertex

which is matched with probability ≤ 𝑑′𝜀

4

• By similar logic, at least half the edges are

preserved.

• This implies that there are at least 𝑑′𝑜 vertices

remaining (otherwise more than half of every matching of ≥ 𝑑′𝑜 edges is deleted)

• Note: Corresponds to Lemma 4 of [ARV09]

Minimal Probability Guarantee

• Corollary: There is a set of vertices 𝑌 of size ≥

𝑑′𝑜 such that ∀𝑦 ∈ 𝑌, 𝑄 𝑦 is matched with an 𝑦′ ∈ 𝑌 ≥ 𝜀′ where 𝜀′ = 𝑑′𝜀

4

Minimal Probability Guarantee

• How can we find pairs of points whose

projected distance is larger and larger by taking steps in the matching graph?

• Let’s assume we have a very convenient

inductive setup.

Building Up Projection Distances

• Have a set of points 𝑌 of size ≥ 𝑑′𝑜

∀𝑦 ∈ 𝑌, 𝑄 𝑦 is matched with an 𝑦′ ∈ 𝑌 ≥ 𝜀′

• Inductive setup: Assume we also have a subset

𝑎 ⊆ 𝑌 of points of size 𝜐|𝑌| such that

∀𝑨 ∈ 𝑎, 𝑄 ∃𝑨′ ∈ 𝑌: 𝑒𝑁 𝑨, 𝑨′ ≤ 𝑙, 𝑨 − 𝑨′ ⋅ 𝑣 ≥ 𝑙𝜏 𝑒 ≥ 1 − 𝜀′ 4

where 𝑒𝑁(𝑨, 𝑨′) is the number of steps required to reach 𝑨′ from 𝑨 in the matching graph

• Note: This corresponds to Definitions 6,8 of

[ARV]

Setup

• 𝑌 is a set of points where every 𝑦 ∈ 𝑌 is

matched to another 𝑦′ ∈ 𝑌 for ≥ 𝜀′ fraction of the directions

• Have a subset 𝑎 ⊆ 𝑌 of size ≥ 𝜐|𝑌| where each

𝑨 ∈ 𝑎 is “covered” in ≥ 1 −

𝜀′ 4 fraction of the

directions by points which are ≤ 𝑙 steps away in the matching graph whose projected distance is ≥

𝑙𝜏 𝑒

Setup Rephrased

Composition Step

𝑣 𝑨 𝑨′ 𝑦′

• r

𝑦′ 𝑨 𝑨′

• Given a direction 𝑣, for each point 𝑨 ∈ 𝑎:
• 1. Check if 𝑨 is matched in 𝑁𝑣 = 𝑁−𝑣
• 2. If so, let 𝑦′ be the point 𝑨 is matched with.

𝑨 − 𝑦′ ⋅ 𝑣 ≥

2𝜏 𝑒

• 3. If 𝑨 − 𝑦′ ⋅ 𝑣 > 0, check if 𝑨 is covered in direction

𝑣. If 𝑨 − 𝑦′ ⋅ 𝑣 < 0 check if 𝑨 is covered in direction −𝑣. With probability ≥ 1 −

𝜀′ 2 , 𝑨 is

covered in both directions. Let 𝑨′ = covering point.

• 4. Observe that 𝑨′ − 𝑦′ ⋅ 𝑣 ≥

𝑙𝜏+2𝜏 𝑒

and 𝑒𝑁 𝑦′, 𝑨′ ≤ 𝑙 + 1

Composition Step

• Have that the density of the new covering edges

is at least

𝜐𝜀′ 2 .

• Following the same kind of logic we used to go

from average to minimal degree, can find a subset 𝑎′ ⊆ 𝑌 of size ≥

𝜐𝜀′ 8 |𝑌| where every

vertex 𝑨′ ∈ 𝑎′ is covered in ≥ 𝜐𝜀′

8 of the

directions.

• Note: Corresponds to Lemma 11 of [ARV]

Composition Step

• How can we recover the inductive hypothesis?
• Can boost the covering probability to almost 1

with a small loss in the projection length!

• Corollary 12 of [ARV] rephrased: If the covering

vectors have length at most

σ 16 log

16 τδ′ +8 log 8 δ′

then if z is covered with probability

𝜐𝜀′ 8 with projection

length

𝑙𝜏+2𝜏 𝑒 , it is covered with probability 1 −

𝜀′/4 with projection length (𝑙+1)𝜏

𝑒

Boosting Lemma

• If we apply this directly:

– 𝜐 ∼ 𝜀′ −𝑙 – Need covering vectors to have length 𝑃

1 𝑚𝑝𝑕𝜐

= 𝑃

1 𝑙

– Guaranteed to have length ≤ 𝑙Δ – We can take 𝑙 = Ω(Δ−1

2). We want

𝑙 Δ to be a large

constant times log 𝑜 , which means we can take Δ = Ω( 𝑚𝑝𝑕𝑜 −2/3)

Bound on 𝑙 and Δ

• To reach 𝑙 = Ω

𝑚𝑝𝑕𝑜 , a more careful argument is needed, see [ARV].

• Note: We should not expect 𝑙 to be any higher

than O 𝑚𝑝𝑕𝑜 . Recalling that the projection length with 𝑙 steps is

𝑙𝜏 𝑒, if 𝑒 = Θ(𝑚𝑝𝑕𝑜)

(matching the hypercube example) and 𝑙 is 𝜕 𝑚𝑝𝑕𝑜 then this is 𝜕(1), which is too large!

Reaching 𝑙 = Ω 𝑚𝑝𝑕𝑜

Part V: Reduction to the Well- Separated Case

• Take the scaling where σ𝑗,𝑘:𝑗<𝑘 𝑒2(𝑗, 𝑘) =

𝑜 2

(i.e. the average squared distance between pairs

• f points is 1)
• One of the following two cases holds:
• 1. There exists a point 𝑦0 such that

𝑜 10 other points

are within squared distance

1 10 of 𝑦0

• 2. For all points 𝑦, less than

𝑜 10 other points are within

squared distance

1 10 of 𝑦

Two Cases

• Assume there exists a point 𝑦0 such that

𝑜 10 other

points are within squared distance

1 10 of 𝑦0

• Let 𝑌 = {x: d2 x, x0 ≤

1 10}

• Key idea: Take the Fréchet embedding with

respect to 𝑌!

• In particular, take

𝑒𝑌 𝑧, 𝑨 = |𝑒2 𝑧, 𝑌 − 𝑒2 𝑨, 𝑌 |

Case #1

• We will show that

σ𝑗,𝑘:𝑗<𝑘, 𝑗,𝑘 ∈𝐹(𝐻) 𝑒𝑌 𝑗,𝑘 σ𝑗,𝑘:𝑗<𝑘 𝑒𝑌 𝑗,𝑘

is 𝑃

σ𝑗,𝑘:𝑗<𝑘, 𝑗,𝑘 ∈𝐹 𝐻 𝑒2 𝑗,𝑘 σ𝑗,𝑘:𝑗<𝑘 𝑒2 𝑗,𝑘

• 𝑒𝑌 is an 𝑀1 metric, so this gives an 𝑃(1)-

approximation!

• First note that σ𝑗,𝑘:𝑗<𝑘, 𝑗,𝑘 ∈𝐹(𝐻) 𝑒𝑌 𝑗, 𝑘 is less

than or equal to σ𝑗,𝑘:𝑗<𝑘, 𝑗,𝑘 ∈𝐹(𝐻) 𝑒2 𝑗, 𝑘

• We just need to show that σ𝑗,𝑘:𝑗<𝑘 𝑒𝑌 𝑗, 𝑘 is

Ω(𝑜2)

Case #1 Continued

• Proposition: The average squared distance of

points outside of 𝑌 from 𝑌 is at least

1 5

• Proof: If this were not the case then the average

squared distance between points would be < 1 as for all 𝑧, z, 𝑒2 𝑧, 𝑨 ≤ 𝑒2 𝑧, 𝑌 + 𝑒2 𝑨, 𝑌 + 1 5

• Corollary: σ𝑗,𝑘:𝑗<𝑘 𝑒𝑌 (𝑗, 𝑘) is Θ 𝑜2 . To show

this, it is sufficient to consider the pairs where exactly one of 𝑗, 𝑘 are in 𝑌.

Case #1 Continued

• Assume that for all points 𝑦, there are fewer than

𝑜 10 other points which are within squared

distance 1

10 of 𝑦

• Proposition: There is a point 𝑦0 such that at least

𝑜 2 other points are within distance 2 of 𝑦0

• Proof: If this were not the case then the average

distance between points would be > 1.

• Let 𝑌 be the set of points within distance 2 of 𝑦0.

Case #2

• Key idea: Subtract 𝑦0 from all vectors!
• After this translation:

– All points in 𝑌 have length ≤ 2 – For all points 𝑦 ∈ 𝑌, there are at least

n 2 − n 10 = 2𝑜 5

points in 𝑌 which have squared distance more than

1 10 from 𝑦. Thus, the average squared distance

between points in 𝑌 is Ω(1)

• Restricting to 𝑌 and scaling down by a factor of

2, we are now in the well-spread case

Case #2 Continued

Part VI: Open Problems

• Lower Bounds have been shown for this

semidefinite program

• Khot and Vishnoi [KV05] proved the first super-

constant lower bound.

• For weighted graphs, Naor and Young [NY17]

showed an Ω 𝑚𝑝𝑕𝑜 lower bound (which is tight up to a 𝑚𝑝𝑕𝑚𝑝𝑕𝑜 factor).

• However, these lower bounds don’t apply even

to degree 4 SOS!

Lower Bounds

• Is this also true for unweighted graphs?
• Does degree 4 SOS or higher degree SOS give

further improvements? Can we show a superconstant lower bound for a constant number of rounds of SOS?

Open Questions

References

• [ALN08] S. Arora, J. R. Lee, A. Naor. Euclidean distortion and the sparsest cut. J.
• Amer. Math. Soc. 21 (1), p. 1–21. 2008
• [ARV] S. Arora, S. Rao, U. Vazirani. Expander Flows, Geometric Embeddings and

Graph Partitioning. https://www.cs.princeton.edu/~arora/pubs/arvfull.pdf

• [KV05] S. Khot and N. Vishnoi. The unique games conjecture, integrality gap for cut

problems and embeddability of negative type metrics into 𝑀1. FOCS 2005

• [NY17] A. Naor and R. Young. The integrality gap of the Goemans--Linial SDP

relaxation for Sparsest Cut is at least a constant multiple of 𝑚𝑝𝑕𝑜. STOC 2017