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Metric Embeddings and Expanders

Based on Chapter 13 of a survey "Expander graphs and their applications" by Hoory, Linial and Wigderson.

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Metric Embeddings and Expanders

  1. 1. Metric Embeddings and Expanders Grigory Yaroslavtsev (based on Chapter 13 of a survey “Expander graphs and their applications” by Hoory, Linial and Wigderson) Pennsylvania State University December 8, 2011Grigory Yaroslavtsev (PSU) December 8, 2011 1 / 11
  2. 2. Metric embeddings A finite metric space is a pair (X , d), where X is a set of n points and d : X × X → R+ is a distance function (three axioms). Let f : X → Rn be an embedding of (X , d) into (Rn , 2 ) n 2 (d 2 (x, y ) = ||x − y || = i=1 (xi − yi ) ). expansion(f ) = max ||f (x1 ) − f (x2 )||/d(x1 , x2 ) x1 ,x2 ∈X contraction(f ) = max d(x1 , x2 )/||f (x1 ) − f (x2 )|| x1 ,x2 ∈X distortion(f ) = expansion(f ) · contraction(f ) Example, that requires distortion (shortest-path metric for unit-length edges): Grigory Yaroslavtsev (PSU) December 8, 2011 2 / 11
  3. 3. Background on 2 -embeddings If (X , d) is 2 -embeddable ⇒ it is p -embeddable for 1 ≤ p ≤ ∞. Let c2 (X , d) denote the least possible distortion of an embedding of (X , d) into (Rn , 2 ) (dimension n is sufficient). For any n-point metric space c2 (X , d) = O(log n) [Bourgain’85]. We will see how to compute such an embedding later (via SDP), together with a Ω(log n) lower bound for expanders (via dual SDP).Theorem (Johnson-Lindenstrauss ’84) log nAny n-point 2 -metric can be embedded into an O 2 -dimensionalEuclidean space with distortion 1 + . The bound on dimension was shown to be optimal by Jayram and Woodruff (SODA’11), previous Ω( 2 log n ) was by Alon ’03. log 1/ Such dimension reduction is impossible for 1 (Brinkman, Charikar ’03, Lee, Naor ’04, . . . ?). Grigory Yaroslavtsev (PSU) December 8, 2011 3 / 11
  4. 4. Computing minimal distortionTheorem (Linial, London, Rabinovich ’95)Given a metric space (X , d), the minimal 2 -distortion c2 (X , d) can becomputed in polynomial time.Proof. Scale f : X → Rn , so that contraction(f ) = 1, so distortion(f ) ≤ γ iff: d(xi , xj )2 ≤ ||f (xi ) − f (xj )||2 ≤ γ 2 d(xi , xj )2 ∀i, j A symmetric matrix Z ∈ Rn×n is positive semidefinite (PSD), if (all four are equivalent): 1 v T Zv ≥ 0 for all v ∈ Rn . 2 All eigenvalues λi ≥ 0. 3 Z = WW T for some matrix W . n 4 Z= λi wi wiT for λi ≥ 0 and orthonormal vectors wi ∈ Rn . i=1 Grigory Yaroslavtsev (PSU) December 8, 2011 4 / 11
  5. 5. Computing minimal distortion (continued)Proof. Any embedding f : X → Rn can be represented as a matrix U ∈ Rn×n , where row ui = f (xi ). Let Z = UU T , so we need to find a PSD Z , such that: d(xi , xj )2 ≤ zii + zjj − 2zij ≤ γ 2 d(xi , xj )2 , ∀i, j, since ||ui − uj ||2 = zii + zjj − 2zij . Linear optimization problem with an additional constraint that a matrix of variables is PSD ⇒ solvable by ellipsoid in polynomial time. Grigory Yaroslavtsev (PSU) December 8, 2011 5 / 11
  6. 6. Characterization of PSD matricesLemmaA matrix Z is PSD if and only if ij qij zij ≥ 0 for all PSD matrices Q.Proof. ⇐: For v ∈ Rn let Qij = vi · vj . Then Q is PSD and v T Zv = ij (vi · vj )zij = ij qij zij ≥ 0. ⇒: Let Q = k λk wk wk for λi ≥ 0, or equivalently Q = k Ak , T where Ak = λk wk wk . T kz = λ T Because ij Aij ij k ij wki wkj zij = λk wk Zwk ≥ 0, we have k k ij qij zij = ij k Aij zij = k ij Aij zij ≥ 0. Grigory Yaroslavtsev (PSU) December 8, 2011 6 / 11
  7. 7. Lower bound on distortionTheorem (Linial-London-Rabinovich ’95)The least distortion of any finite metric space (X , d) in the Euclideanspace is given by: 2 pij >0 pij d(xi , xj ) c2 (X , d) ≥ max 2 . P∈PSD,P·1=0 − pij <0 pij d(xi , xj )Proof.Primal SDP: qij zij ≥ 0 ∀Q ∈ PSD ij zii + zjj − 2zij ≥ d(xi , xj )2 ∀i, j 2 2 γ d(xi , xj ) ≥ zii + zjj − 2zij ∀i, j Grigory Yaroslavtsev (PSU) December 8, 2011 7 / 11
  8. 8. Lower bound on distortion via a dual SDP solution qij zij ≥ 0 ∀Q ∈ PSD (1) ij zii + zjj − 2zij ≥ d(xi , xj )2 ∀i, j (2) γ 2 d(xi , xj )2 ≥ zii + zjj − 2zij ∀i, j (3) Take Q ∈ PSD, such that j qij = 0. If qij > 0, add corresponding inequality (2) multiplied by qij /2. If qij < 0, add corresponding inequality (3) multiplied by −qij /2. 2 2 qij <0 (zii + zjj − 2zij )qij /2 − γ qij <0 d(xi , xj ) qij /2 ≥ 2 qij >0 d(xi , xj ) qij /2 − qij <0 (zii + zjj − 2zij )qij /2 Because ij qij zij ≥ 0 and i qij = 0, we get a contradiction, if γ 2 qij <0 d(xi , xj )2 qij + qij >0 d(xi , xj )2 qij > 0. Grigory Yaroslavtsev (PSU) December 8, 2011 8 / 11
  9. 9. Example: Hypercube with Hamming metric Let’s denote r -dimensional hypercube with Hamming metric as Qr . √ Identity embedding gives distortion r . 2 pij >0 pij d(xi , xj ) c2 (Qr ) ≥ max 2 . P∈PSD,P·1=0 − pij <0 pij d(xi , xj ) r ×2r Define P ∈ R2 , such that P1 = 0 as:  −1  if d(i, j) = 1   r − 1 if i = j P(x, y ) = 1  if d(i, j) = r   0 otherwise P ∈ PSD: eigenvectors χI (J) = (−1)|I ∩J| for I , J ⊆ {1, . . . , n}. Because pij >0 pij d(xi , xj )2 = 2r · r 2 and √ − pij <0 pij d(xi , xj )2 = 2r · r , we have c2 (Qr ) ≥ r . Grigory Yaroslavtsev (PSU) December 8, 2011 9 / 11
  10. 10. Embedding expanders into 2 √ For the hypercube we’ve got a Ω( log n) lower bound. Now we will get a Ω(log n) lower bound for expanders. Take k-regular expander G with n vertices and λ2 ≤ k − for > 0. √ Embedding vertex i to ei / 2: expansion = 1, contraction = O(log n).Theorem (Linial-London-Rabinovich ’95)For G as above c2 (G ) = Ω(log n), constant depends only on k and . If H = (V , E ) is the graph on the same vertex set, where two vertices are adjacent if their distance in G is at least logk (n) , then H has a perfect matching (by Dirac’s theorem has Hamiltonian cycle). Grigory Yaroslavtsev (PSU) December 8, 2011 10 / 11
  11. 11. Proof of the lower bound for embdedding expanders Let B be the adjacency matrix of a perfect matching in H and P = kI − AG + 2 (B − I ), so P1 = 0. x T (kI − AG )x ≥ (k − λ2 )||x||2 ≥ ||x||2 x T (B − I )x = (2xi xj − xi2 − xj2 ) ≥ −2 (xi2 + xj2 ) = −2||x||2 . (i,j)∈B (i,j)∈B P is PSD, because x T Px = x T (kI − AG )x + x T 2 (B − I )x ≥ 0. − d(i, j)2 pij = kn pij <0 d(i, j)2 pij ≥ · n logk n 2 , 2 pij >0 because distances of edges in B are at least logk n . Thus, c2 (G ) = Ω(log n). Grigory Yaroslavtsev (PSU) December 8, 2011 11 / 11

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