Lecture 4: Cheeger s Inequality


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1 Spectral Graph Theory ad Applicatios WS 0/0 Lecture 4: Cheeger s Iequality Lecturer: Thomas Sauerwald & He Su Statemet of Cheeger s Iequality I this lecture we assume for simplicity that G is a dregular graph. We shall work with the ormalized adjacecy matrix M = da. The goal of this class is to prove Cheeger s iequality which establishes a iterestig coectio betwee λ ad the (ormalized edge expasio. Defiitio 4. ((Normalized Edge Expasio of a Regular Graph. The ormalized edge expasio of a dregular graph G is defied as: h(g = S : S V / E(S, V \ S. d S Theorem 4. ([Alo86, SJ89]. For ay dregular graph G = (V, E, or equivaletly, Preparatios h λ h, ( λ / h ( λ. We start with a useful equality that will be applied later o. Lemma 4.3. Proof. Let us compute M u,v (x u x v = M u,v (x u x v = x T x. For the first term: M u,v (x u + x v = M u,v x u = u V Further, Therefore, u V v V M u,v x u x v = u V x u M u,v (x u + x v M u,v x u x v. v V M u,v x v = u V M u,v = x T x. x u v V M u,v (x u x v = x T x. x u (Mx u =.
2 Lecture 4: Cheeger s Iequality Lemma 4.4 (CouratFischer Formula for λ ad λ. Let λ = λ λ be the eigevalues of the ormalized adjacecy matrix M = d A. The, λ = max x T x, λ = max x Moreover, λ x M u,v (x u x v x T x x {u,v} E (x u x v dx T x Proof. We begi with the first statemet. Let v,..., v be a orthoormal basis of eigevectors. Let x be ay ozero vector ad express x i terms of this basis of eigevectors: x = α i v i, i= where α i = v T i x. It follows that ( T ( = α i v i M α i v i = = i= ( ( α i v it i= i= αi λ i. i= i= α i λ i v i Similarly, ( ( x T x = α i v it α i v i = i= i= αi. i= Therefore, x T Mx x T x = i= α i λ i i= α i λ i= α i i= α i = λ, ( so, λ max Moreover, if x is a eigevector with eigevalue oe, the = x T x ad equatio ( becomes a equality. Thus λ = max Cosider ow ay vector x 0 with x. The, α = 0 ad therefore, x T x = i= α i λ i i= α i λ i= α i i= α i = λ,
3 Lecture 4: Cheeger s Iequality 3 ad we have equality agai if x is the eigevector of λ, sice the all α i = 0 except for i =. Therefore, λ = max x For the secod statemet, we kow by Lemma 4.3 that x T x λ = max M u,v (x u x v x T x x v M u,v (x u x v = x v which gives the secod statemet of the lemma. 3 Proof of Cheeger s Iequality x T x For the proof of Cheeger s iequality, we itroduce aother related expasio parameter, the socalled coductace: Φ(G := S V E(S, V \ S. d S Ituitively, the coductace measures how close a graph is to a radom dregular graph. The reaso is that for ay set S, we have d S icidet edges ad the proportio of edges goig to V \ S is V \ S / V. Our ext claim is that: h(g Φ(G h(g. The left iequality follows directly from the defiitio. For the right iequality, we first ote that E(S, V \ S h(g S V d { S, V \ S }. Further, V \S V S V \S V { S, V \ S } ad hece h(g Φ(G. Theorem 4.5. For ay dregular graph G = (V, E,, or equivaletly, h λ h, ( λ / h ( λ. Proof. Proof of λ h. We will relate Φ(G to λ. We first formulate Φ(G as a iteger imizatio problem to fid out that λ is essetially the relaxatio of that problem. Φ(G = x {0,}, x {0, } x {0,}, x {0, } d M u,v x u x v d( u V x u( u V x u M u,v (x u x v ( u V x u( u V x u
4 Lecture 4: Cheeger s Iequality 4 We ow rewrite the deoator usig the followig equality: ( ( x u x u = x u x u x v = x u x u x v = (x u x v. u V u V u V u V so that Φ(G M u,v (x u x v x {0,}, x {0, } (x u x v M u,v (x u x v x R, x {0, } (x u x v M u,v (x u x v x R, x {0, },x (x (x x + α u x v M u,v (x u x v x R, x {0, },x x R, x {0, },x = λ. x T x x ux v M u,v (x u x v x T x ( x u x v = u,v u x u x v = 0 Proof of the other directio. Let x be the correspodig eigevector to λ. Assume without loss of geerality that at most / etries of x are positive (otherwise we work with x. Defie a vector y R as y u = max{x u, 0}. We ow prove the followig iequalities: Claim λ P Mu,v (yu yv y T y.. ( M u,v (y u y v 4y T y M u,v y u yv 3. M u,v y u yv hy T y. Assug that the three iequalities i the above claim hold, we ca fiish the proof as follows: Claim λ Claim Claim 3 h /. M u,v (y u y v y T y ( M u,v y u yv 8(y T y The proof of the first ad secod iequality of the claim are relatively straightforward ad ca be foud i the lecture otes. We oly give the proof of the third oe, which is the most iterestig oe, as the edge expasio comes ito play. v
5 Lecture 4: Cheeger s Iequality 5 The third iequality is the most techical oe. Let V = {,..., } ad order the vertices so that y y y. Let t {,..., } be the largest idex such that y k > 0. The, M u,v y u yv = = t i= j=i+ where the last equality holds sice for every i < j, j k=i i yj t k k+ y, k= i k j>k k k+ y = Mi,j (y i yj Now defie for ay k N, S k := {,..., k}. The, t k t ( y k+ = y k yk+ d E(S k, V \ S k. k= k= i k j>k Moreover, by defiitio of the expasio, we have t ( y k yk+ t d E(S k, V \ S k hk (y k y k+ k= k= = h = h t k (yk y k+ k= t k= = hy T y. y k (sice t / The proof of the iequality λ h / suggests the followig algorithm for fidig a small cut.. Compute the eigevector x correspodig to the largest eigevalue λ. Order the vertices so that x x x 3. Try all cuts of the form ({,,..., k}, {k +,..., } ad retur the smallest oe By the proof of Cheeger s iequality (Theorem 4., it follows that the cut retured by above algorithm is at most λ. 4 Example We apply our algorithm to the followig graph: A =
6 Lecture 4: Cheeger s Iequality 6 Clearly, the correspodig graph G is 3regular Figure : Illustratio of the graph G. The labels of the odes o the right had side display the correspodig etries of the eigevector of the secod largest eigevalue (which is 5/3. The dashed lie describes the best possible cut, which is foud by our algorithm. 5 Further Examples Let us ow apply Cheeger s iequality to three differet graphs: Let G be a complete graph with vertices. The, λ = 0 ad h(g S V \ S V \ S S : S V / d S S : S V / = / /. Let G be a cycle with vertices. Ay set S V with S / has at least two edges from S to V \ S. Hece, h(s d S / =, where the iequality is tight if S is a set of cosecutive vertices of size /. Further, λ = cos(π/ Θ(/, so the spectral ad geometric expasio differ by a square. For the hypercube G with = d vertices, you are asked i a exercise to prove that λ = d. Hece from Cheeger s iequality, h λ = d. Choosig S to be all vertices whose first bit equals zero yields Therefore, h /d. h S d S = d. Refereces [Alo86] N. Alo. Eigevalues ad expaders. Combiatorica, 6(:83 96, 986. [SJ89] Alistair Siclair ad Mark Jerrum. Approximate coutig, uiform geeratio ad rapidly mixig markov chais. If. Comput., 8(:93 33, 989.
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