Stanford Unversty CS359G: Graph Parttonng and Expanders Handout 4 Luca Trevsan January 3, 0 Lecture 4 In whch we prove the dffcult drecton of Cheeger s nequalty. As n the past lectures, consder an undrected d-regular graph G = (V, E), call A ts adjacency matrx, and M := d A ts scaled adjacency matrx. Let λ λ n be the egenvalues of M, wth multplctes, n non-ncreasng order. We have been studyng the edge expanson of a graph, whch s the mnmum of h(s) over all nontrval cuts (S, V S) of the vertex set (a cut s trval f S = or S = V ), where the expanson h(s) of a cut s h(s) := Edges(S, V S) d mn{ S, V S } We have also been studyng the (unform) sparsest cut problem, whch s the problem of fndng the non-trval cut that mnmzes φ(s), where the sparssty φ(s) of a cut s φ(s) := We are provng Cheeger s nequaltes: Edges(S, V S) d S V S n λ h(g) ( λ ) () and we establshed the left-hand sde nequalty n the prevous lecture, showng that the quantty λ can be seen as the optmum of a contnuous relaxaton of φ(g), so that λ φ(g), and φ(g) h(g) follows by the defnton. Today we prove the more dffcult, and nterestng, drecton. The proof wll be constructve and algorthmc. The proof can be seen as an analyss of the followng algorthm.
Algorthm: SpectralParttonng Input: graph G = (V, E) and vector x R V Sort the vertces of V n non-decreasng order of values of entres n x, that s let V = {v,..., v n } where x v x v... x vn Let {,..., n } be such that h({v,..., v }) s mnmal Output S = {v,..., v } We note that the algorthm can be mplemented to run n tme O( V + E ), assumng arthmetc operatons and comparsons take constant tme, because once we have computed h({v,..., v }) t only takes tme O(degree(v + )) to compute h({v,..., v + }). We have the followng analyss of the qualty of the soluton: Lemma (Analyss of Spectral Parttonng) Let G = (V, E) be a d-regular graph, x R V be a vector such that x, let M be the normalzed adjacency matrx of G, defne δ :=,j M,j x x j n,j x x j and let S be the output of algorthm SpectralParttonng on nput G and x. Then h(s) δ Remark If we apply the lemma to the case n whch x s an egenvector of λ, then δ = λ, and so we have h(g) h(s) ( λ ) whch s the dffcult drecton of Cheeger s nequaltes. Remark 3 If we run the SpectralParttonng algorthm wth the egenvector x of the second egenvalue λ, we fnd a set S whose expanson s h(s) ( λ ) h(g) Even though ths doesn t gve a constant-factor approxmaton to the edge expanson, t gves a very effcent, and non-trval, approxmaton. As we wll see n a later lecture, there s a nearly lnear tme algorthm that fnds a vector x for whch the expresson δ n the lemma s very close to λ, so, overall, for any graph G we can fnd a cut of expanson O( h(g)) n nearly lnear tme.
Proof of Lemma In the past lecture, we saw that λ can be seen as the optmum of a contnuous relaxaton of sparsest cut. Lemma provdes a roundng algorthm for the real vectors whch are solutons of the relaxaton. In ths secton we wll thnk of t as a form of randomzed roundng. Later, when we talk about the Leghton-Rao sparsest cut algorthm, we wll revst ths proof and thnk of t n terms of metrc embeddngs. To smplfy notaton, we wll assume that V = {,..., n} and that x x x n. Thus our goal s to prove that there s an such that h({,..., }) δ We wll derve Lemma by showng that there s a dstrbuton D over sets S of the form {,..., } such that ES D Edges(S, V S) d ES D mn{ S, V S } δ () We need to be a bt careful n dervng the Lemma from (). In general, t s not true that a rato of averages s equal to the average of the ratos, so () does not mply that E h(s) δ. We can, however, apply lnearty of expectaton and derve from () the nequalty E S D d Edges(S, V S) δ mn{ S, V S } 0 So there must exst a set S n the sample space such that d Edges(S, V S) δ mn{ S, V S } 0 meanng that, for that set S, we have h(s) δ. (Bascally we are usng the fact that, for random varables X, Y over the same sample space, although t mght not be true that E X = E Y E X, we always have Y P[ X E X ] > 0, provded that Y > 0 over Y E Y the entre sample space.) From now on, we wll assume that. x n/ = 0, that s, the medan of the entres of x s zero. x + x n = whch can be done wthout loss of generalty because addng a fxed constant c to all entres of x, or multplyng all the entres by a fxed constant does not change the 3
value of δ nor does t change the property that x x n. The reason for these choces s that they allow us to defne a dstrbuton D over sets such that E mn{ S, V S } = S D x (3) We defne the dstrbuton D over sets of the form {,..., }, n, as the outcome of the followng probablstc process: We pck a real value t n the range [x, x n ] wth probably densty functon f(t) = t. That s, for x a b x n, P[a t b] = b a t dt. Dong the calculaton, ths means that P[a t b] = a b f a, b have the same sgn, and P[a t b] = a + b f they have dfferent sgns. We let S := { : x t} Accordng to ths defnton, the probablty that an element n/ belongs to the smallest of the sets S, V S s the same as the probablty that t belongs to S, whch s the probablty that the threshold t s n the range [x, 0], and that probablty s x. Smlarly, the probablty that an element > n/ belongs to the smallest of S, V S s the same as the probablty that t belongs to V S, whch s the probablty that t s n the range [0, x ], whch s agan x. So we have establshed (3). We wll now estmate the expected number of edges between S and V S. E d Edges(S, V S) = M,j P[(, j) s cut by (S, V S)],j The event that the edge (, j) s cut by the partton (S, V S) happens when the value t falls n the range between x and x j. Ths means that If x, x j have the same sgn, If x, x j have dfferent sgn, P[(, j) s cut by (S, V S)] = x x j P[(, j) s cut by (S, V S)] = x + x j Some attempts, show that a good expresson to upper bound both cases s P[(, j) s cut by (S, V S)] x x j ( x + x j ) 4
Pluggng nto our expresson for the expected number of cut edges, and applyng Cauchy-Schwarz E d Edges(S, V S) M,j x x j ( x + x j ),j M (x x j ) M ( x + x j ) The assumpton of the Lemma tell us that And we can rewrte M (x x j ) = δ (x x j ) n (x x j ) = n x x x j = n ( ) x x n x whch gves us M (x x j ) δ x Fnally, t remans to study the expresson M ( x + x j ). By applyng the nequalty (a + b) a + b (whch follows by notng that a + b (a + b) = (a b) 0), we derve M ( x + x j ) Puttng all the peces together we have M (x + x j) = 4 E d Edges(S, V S) δ x (4) whch, together wth (3) gves (), whch, as we already dscussed, mples the Man Lemma. x 5