Lecture 6: Linear Programming for Sparsest Cut
Sparsest Cut and SOS • The SOS hierarchy captures the algorithms for sparsest cut, but they were discovered directly without thinking about SOS (and this is how we’ll present them) • Why we are covering sparsest cut in detail: 1. Quite interesting in its own right 2. Illustrates the kinds of things SOS can capture 3. Determining if SOS can do better is a major open problem on SOS.
Lecture Outline • Part I: Sparsest cut • Part II: Linear programming relaxation and analysis via metric embeddings • Part III: Bourgain’s Theorem • Part IV: Tight example: expanders
Part I: Sparsest Cut
Flaw of Minimum Cut • We’ve seen that MIN-CUT can be solved efficiently • However, MIN-CUT may not be the best way to decompose a graph • Example:
Flaw of Minimum Cut • MIN-CUT: • Desired Cut:
ҧ Sparsest Cut Problem • Idea: Divide # of cut edges by # of possible which could have been cut • Definition: Given a cut 𝐷 = (𝑇, ҧ 𝑇) , define 𝜚 𝐷 = # 𝑝𝑔 𝑓𝑒𝑓𝑡 𝑑𝑣𝑢 𝑇 ⋅ 𝑇 • Sparsest cut problem: Minimize 𝜚(𝐷) • Can also have a weighted version: σ 𝑗,𝑘:𝑗∈𝑇,𝑘∈ ҧ 𝑇, 𝑗,𝑘 ∈𝐹(𝐻) 𝑥(𝑗, 𝑘) 𝜚 𝐷 = σ 𝑗,𝑘:𝑗∈𝑇,𝑘∈ ҧ 𝑇 𝑥(𝑗, 𝑘)
Linear Programming for Sparsest Cut • Theorem [LR99]: There is a linear programming relaxation for sparsest cut which gives an 𝑃(log 𝑜) approximation.
Part II: Linear Programming Relaxation and Analysis via Metric Embeddings
Metric and Pseudo-metric Spaces • Definition: A metric space (𝑌, 𝑒) is a set of points 𝑌 and a distance function 𝑒: 𝑌 × 𝑌 → ℝ ≥0 where 1. ∀𝑦 1 , 𝑦 2 ∈ 𝑌, 𝑒 𝑦 1 , 𝑦 2 = 𝑒(𝑦 1 , 𝑦 2 ) 2. ∀𝑦 1 , 𝑦 2 ∈ 𝑌, 𝑒 𝑦 1 , 𝑦 2 = 0 ⬄ 𝑦 1 = 𝑦 2 3. ∀𝑦 1 , 𝑦 2 , 𝑦 3 ∈ 𝑌, d x 1 , x 3 ≤ 𝑒 𝑦 1 , 𝑦 2 + 𝑒(𝑦 2 , 𝑦 3 ) • Example 1: Euclidean Space: 𝑒 𝑦, 𝑧 = 𝑧 − 𝑦 • Example 2: 𝑀 1 distance: 𝑒 𝑦, 𝑧 = σ 𝑗 |𝑧 𝑗 − 𝑦 𝑗 | • Without the second condition, this is called a pseudo-metric space
ҧ Cut Spaces • A cut 𝐷 = (𝑇, ҧ 𝑇) induces a pseudo-metric space on a graph 𝐻 : Take 𝑒(𝑣, 𝑤) = 0 if 𝑣, 𝑤 ∈ 𝑇 or 𝑣, 𝑤 ∈ 𝑇 and otherwise take 𝑒 𝑣, 𝑤 = 𝑑 for some 𝑑 > 0 . • We call this a cut space.
Problem Reformulation σ 𝑗,𝑘:𝑗<𝑘, 𝑗,𝑘 ∈𝐹(𝐻) 𝑒(𝑗,𝑘) • Reformulation: Minimize over σ 𝑗,𝑘:𝑗<𝑘 𝑒(𝑗,𝑘) all cut spaces • First issue: Objective function is nonlinear • Fix: Set denominator equal to 1 . • Modified Reformulation: Minimize σ 𝑗,𝑘:𝑗<𝑘, 𝑗,𝑘 ∈𝐹(𝐻) 𝑒(𝑗, 𝑘) over all cut spaces normalized so that σ 𝑗,𝑘:𝑗<𝑘 𝑒(𝑗, 𝑘) = 1
Problem Relaxation • Want to minimize σ 𝑗,𝑘:𝑗<𝑘, 𝑗,𝑘 ∈𝐹(𝐻) 𝑒(𝑗, 𝑘) over all cut spaces normalized so that σ 𝑗,𝑘:𝑗<𝑘 𝑒(𝑗, 𝑘) = 1 • Relaxation: Minimize σ 𝑗,𝑘:𝑗<𝑘, 𝑗,𝑘 ∈𝐹(𝐻) 𝑒(𝑗, 𝑘) over all pseudo-metrics normalized so that σ 𝑗,𝑘:𝑗<𝑘 𝑒(𝑗, 𝑘) = 1 . Linear program constraints: ∀𝑗, 𝑘 , 𝑒 𝑗, 𝑘 = 𝑒(𝑘, 𝑗) ≥ 0 1. ∀𝑗, 𝑘, 𝑙, 𝑒(𝑗, 𝑙) ≤ 𝑒(𝑗, 𝑘) + 𝑒(𝑘, 𝑙) 2. σ 𝑗,𝑘:𝑗<𝑘 𝑒(𝑗, 𝑘) = 1 3.
𝑀 1 Spaces • Definition: We say that a pseudo-metric (𝑌, 𝑒) is an 𝑀 1 space if there is a mapping f: 𝑌 → ℝ n such that ∀𝑦, 𝑧 ∈ 𝑌 , 𝑒 𝑦, 𝑧 = σ 𝑗 |𝑔 𝑧 𝑗 − 𝑔 𝑦 𝑗 | • In this case, we may as well pretend we are already in ℝ 𝑜 with the 𝑀 1 distance function • Lemma: For the sparsest cut relaxation, there is no gap between 𝑀 1 spacs and cut spaces!
𝑀 1 Space Example • If 𝑦 1 = 1,2 , 𝑦 2 = (0,3) , and 𝑦 3 = (4,4) , then in the 𝑀 1 metric, 𝑒 𝑦 1 , 𝑦 2 = 2 , 𝑒 𝑦 1 , 𝑦 3 = 5 , and 𝑒 𝑦 2 , 𝑦 3 = 5 6 5 𝑦 3 4 𝑦 2 3 𝑦 1 2 1 0 0 1 2 3 4 5 6
Decomposing 𝑀 1 Pseudo-metrics • Lemma: Any finite 𝑀 1 space can be decomposed as a linear combination of cut spaces. • Proof sketch: We can work coordinate by coordinate. For a single coordinate, here is the picture: + 1 × -2 -1 0 1 2 = 2 × + -2 -1 0 1 2 -2 -1 0 1 2 1 × -2 -1 0 1 2
Useful Lemma • Lemma: If 𝑏, 𝑐 ≥ 0 and 𝑑, 𝑒 > 0 then min 𝑏 𝑑 , 𝑐 ≤ 𝑏 + 𝑐 𝑑 + 𝑒 ≤ max 𝑏 𝑑 , 𝑐 𝑒 𝑒 • Proof: Without loss of generality, assume that 𝑏 𝑑 ≤ 𝑐 𝑒 . Take 𝑏 ′ = 𝑐𝑑 𝑒 ≥ 𝑏 and take 𝑐 ′ = 𝑒𝑏 𝑑 ≤ 𝑐 . 𝑑 = 𝑏+𝑐 ′ 𝑑+𝑒 ≤ 𝑏 ′ +𝑐 Now 𝑏 𝑑+𝑒 ≤ 𝑏+𝑐 𝑑+𝑒 = 𝑐 𝑒 • Together with the previous decomposition, this shows that for any 𝑀 1 space, there’s always a cut spacec which is as good or better.
Metric Embeddings and Distortion • Often want to embed a more complicated metric space into a simpler one. This embedding won’t be perfect, but may still be useful • Given metric spaces 𝑌, 𝑒 , (𝑍, 𝑒 ′ ) and a map 𝑔: 𝑌 → 𝑍 : 𝑒 ′ (𝑔 𝑣 ,𝑔(𝑤)) 1. Define the expansion of 𝑔 to be m𝑏𝑦 𝑒(𝑣,𝑤) 𝑣,𝑤∈𝑌 𝑒(𝑣,𝑤) 2. Define the contraction of 𝑔 to be m𝑏𝑦 𝑒 ′ (𝑔 𝑣 ,𝑔(𝑤)) 𝑣,𝑤∈𝑌 3. Define the distortion of 𝑔 to be the product of the expansion and the contraction of 𝑔
Metric Embeddings into 𝑀 1 • If the pseudo-metric given by our linear program can be embedded into 𝑀 1 with distortion 𝛽 , this gives an 𝛽 -approximation for the value of the sparsest cut. • Question: How well can general finite pseudo- metric spaces be embedded into 𝑀 1 ?
Part III: Bourgain’s Theorem
Bourgain’s Theorem • Theorem [Bou85]: Every metric on 𝑜 points can be embedded into an 𝑀 1 metric with distortion 𝑃(log 𝑜) . Moreover, 𝑃( 𝑚𝑝𝑜 2 ) coordinates are sufficient • Note: the bound on the number of coordinates is due to Linial, London, and Rabinovich [LLR95]
Fréchet Embeddings • Def: Given a set of points 𝑇 , define 𝑒 𝑦, 𝑇 = min 𝑡∈𝑇 𝑒 𝑦, 𝑡 • Fréchet embedding: Gives a value to each point based on its distance from some subset 𝑇 of points and takes the distance between. In other words, 𝑒 𝑇 𝑦, 𝑧 = |𝑒 𝑧, 𝑇 − 𝑒(𝑦, 𝑇)| • Proposition: For any 𝑇 , 𝑒 𝑇 𝑦, 𝑧 ≤ 𝑒(𝑦, 𝑧)
Fréchet Embedding Example • Start with the distance metric 𝑒 𝑣, 𝑤 = length of the shortest path from 𝑣 to 𝑤 on the graph shown. If we take 𝑇 to be the set of red vertices, we get the values shown for 𝑒(𝑤, 𝑇) . 1 1 0 1 2 0 1 2 3
Fréchet Embeddings Bound • 𝑒 𝑦, 𝑇 = min 𝑡∈𝑇 𝑒 𝑦, 𝑡 • 𝑒 𝑇 𝑦, 𝑧 = |𝑒 𝑧, 𝑇 − 𝑒(𝑦, 𝑇)| • Proposition: For any 𝑇 , 𝑒 𝑇 𝑦, 𝑧 ≤ 𝑒(𝑦, 𝑧) • Proof: Let 𝑡 be the point in 𝑇 of minimal distance from 𝑦 . 𝑒 𝑧, 𝑇 ≤ 𝑒 𝑧, 𝑡 ≤ 𝑒 𝑦, 𝑡 + 𝑒 𝑦, 𝑧 = 𝑒 𝑦, 𝑧 + 𝑒(𝑦, 𝑇) • By symmetry, d 𝑦, 𝑇 ≤ 𝑒 𝑦, 𝑧 + 𝑒(𝑧, 𝑇) so d S x, y = 𝑒 𝑧, 𝑇 − 𝑒 𝑦, 𝑇 ≤ 𝑒(𝑦, 𝑧) , as needed.
Bourgain’s Theorem Proof Idea • Proof idea: Choose many Fréchet embeddings, have a coordinate for each one. • Resulting expansion is at most the sum of the weights on the embeddings (this will be 𝑃(𝑚𝑝𝑜) for us) • Challenge: Ensure that the contraction is 𝑃(1) . In other words, ensure that some of the Fréchet embeddings preserve some of the distance between each pair of points 𝑦 and 𝑧 .
Bad Case #1 • Issue: Could have that 𝑔 𝑇 𝑦, 𝑧 ≪ 𝑒(𝑦, 𝑧) . In fact, 𝑔 𝑇 (𝑦, 𝑧) can easily be zero! • Case 1: All points in 𝑇 are far from 𝑦 and 𝑧 and 𝑒 𝑦, 𝑇 = 𝑒(𝑧, 𝑇) . • Example: y x Nearest point in 𝑇
Bad Case #2 • Case 2: There two points 𝑡 𝑦 and 𝑡 𝑧 in 𝑇 where 𝑡 𝑦 is very close to 𝑦 and 𝑡 𝑧 is very close to 𝑧 . If so, can have that d x, S = 𝑒 𝑦, 𝑡 𝑦 = 𝑒 𝑧, 𝑡 𝑧 = 𝑒(𝑧, 𝑇) • Example: 𝑡 𝑧 𝑡 𝑦 x y
Attempt #1 • Want 𝑇 to contain exactly one point 𝑞 which is very close to 𝑦 or 𝑧 . • Let 𝑒 = 𝑒(𝑦, 𝑧) . Pick 𝑇 so that 𝑇 has precisely one point 𝑞 which is within distance 𝑒 3 of either 𝑦 or 𝑧 . • Can be accomplished with constant probability by taking a random S of the appropriate size. x y
Attempt #1 • Attempt #1: Pick 𝑇 so that 𝑇 has precisely one point 𝑞 which is within distance 𝑒 3 of either 𝑦 or 𝑧 . • Danger: 𝑇 also contains point(s) of distance slightly more than 𝑒 3 from the other point. x y
Attempt #1 • Possible fix: Require that 𝑇 contains exactly one point within distance 𝑒 3 of 𝑦 or 𝑧 and no other 𝑒 2 of 𝑦 or 𝑧 points within distance 𝑒 • This implies 𝑒 𝑇 𝑦, 𝑧 ≥ 6 • However, may be too much to ask for… x y
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