U.C. Berkeley CS94: Beyond Worst-Case Analyss Handout 6 Luca Trevsan September, 07 Scrbed by Theo McKenze Lecture 6 In whch we study the spectrum of random graphs. Overvew When attemptng to fnd n polynomal tme an upper bound certfcate on the max cut and maxmum ndependent set of a graph, we have used the followng property. Proposton If G G n,, then wth hgh probablty A E(A) O( n), where s the spectral norm. Generally, f G G n,p and p > log n n then w.h.p. A E(A) O( np). Today we wll prove how to obtan the bound n Proposton wth an extra term of log n, as well as show an outlne of the method of fndng the bound n Proposton. We wll also show how when p s small ths bound breaks down, namely how when p = Θ( n ), ( ) log n A E(A) Ω. Introducng the Trace Henceforth M k j sgnfes (M k ) j. Take M symmetrc and real. All egenvalues of ths matrx are real, and we can enumerate them λ, λ,..., λ n such that λ λ... λ n. Defnton The trace Tr(A) s defned to be Tr(A) = n = A where A s an n n matrx. Moreover we know that Theorem Tr(A) = n = λ.
If we take k large and even, the egenvalues of M k are λ k λ k... λ n k. Therefore we have M k = Tr(M k ) = λ k λ k = M k. Moreover we have Tr(M k ) k = ( λ k ) k n k λ = n k M. Ths gves us an estmaton of the norm, M ( M k) k gves a constant factor approxmaton of M. n k M, whch for k > log n 3 Usng the Trace to Bound the Spectral Norm Assume that G G n, Theorem 3 and A s the adjacency matrx of G. We wll prove the followng. E (Tr((A E(A)) k )) s bounded above by O(k) n +k/ k k/. If k > log n, G G n, by takng the kth root we acheve a bound of O( n log n) on A E(A). 3. Expected Value of Matrx Entres Frst, we examne the matrx M = A E(A). We have M = 0 and M j {±} wth equal probablty of each when j. Moreover M j = M j. If j, E(Mj k ) = 0 f k s odd and E(Mj k ) = for k even. E( M k) = n E(M k ) by the lnearty of expectaton and symmetry between the entres. We evalute M k. M k = M M M k, {,..., k } where,... k represents the ntermedate steps on a path between vertces that starts at and returns to. For example, M = M M. Note that we can repeat edges n these paths. By the lnearty of expectaton E(M) k = E(M M M k, ). {,..., k } If any par {, j} occurs l tmes n the sequence of pars {, }, {, },..., { k, }, where l s odd, then as the value of ths term s ndependent from all other terms and E M l j = 0 for odd l, then E(M M M k ) = 0. If all pars occur an even number of tmes, ther product s expectaton s. Therefore E(M k ) s the number of sequences,..., k V k such that, n the sequence of pars {, }, {, },..., { k, }, each par occurs an even number of tmes.
3. Encodng argument In order to gve an upper bound on the number of such sequences, we wll show how to encode a sequence where there are m dstnct edges. In the sequence,... k, the element j s represented ether as (0, j ), whch takes + log n bts, f j appears for the frst tme n the sequence at locaton j, and as (0, l) otherwse, where l < j s such that l = j, whch requres + log k bts. Notce that, f j occurs for the frst tme at locaton j, then the par { j, j } also occurs for the frst tme at the locaton j and j. Thus the number of tmes that we encounter a vertex for the frst tme s at most the number of dstnct edges. If we have t dstnct vertces (other than vertex ), then we are usng k + t log n + (k t) log k; for k < n, ths value ncreases wth t, but we have t m k/ (because every edge has to appear an even number of tmes and so there can be at most k/ dstnct edges. Ths means that we use at most k + k log n + k log k bts n the encodng. The number of strngs that can be encoded usng at most L bts s L+. If we assume k < n, we have the bound E(M k ) k k n k k+, meanng Tr(M) = n E(M k ) n + k k+ k k. Therefore usng sutable k and t we acheve our bound on M. For example, choose k = log n and t = 0 n log n. We use Markov s nequalty to obtan P( M > t) = P( M k > t k ) E M k t k ( n ) k k n k e Ω(log n) 0. t 4 Tghtenng the Bound To obtan the sharper bound of O( n), we need to count the number of pars more sharply and remove the k k term, namely mprove the way we talk about repettons. Here we gve an outlne for how to fnd a tghter bound. The worst case n the above analyss s when the number of dstnct vertces (not countng vertex ) s maxmal, whch s k/. In that case, the number of dstnct edges { j, j+ } s k/, and they must form a connected graph over + k/ vertces, that s, they have to form a tree. Furthermore, each edges s repeated exactly twce n the closed walk, otherwse we would not have enough dstnct edges to connect + k/ dstnct vertces. If the pars form a tree, then the only way we can have closed walk n whch every edge s repeated twce s that the closed walk s a depth-frst vst of the tree. In ths case, we can mprove our encodng n the followng way. In a depth-frst vst of a tree only two events are possble at each step: ether we dscover a new vertex, or we backtrack on the edge between the current node and the parent node. Thus we only need to pay + log n bts to encode a new node n the sequence and bt to encode an already seen node, and we obtan a bound of k + k log n+ k = k n k. By takng the kth root we obtan a bound on M of O( n). 3
5 Generalzng to any p Now assume G G n,p and A s the adjacency matrx of G. We also assume p <. We defne M = A E(A). In ths matrx M = 0 and f j, M,j = p wth probablty p and p wth probablty p. Therefore E(M j ) = 0, E(Mj ) = p p p. In fact, E(Mj k ) p for all k. From ths we see we need to sum over sequences such that the multset has each par occurng at least two tmes, as f any par occurs once, the expectaton s 0. Therefore the bound s E(M k ),... k p l where l s the number of dstnct pars and the sum s taken over multsets where each par occurs at least twce. For large l, the number of sequences where each par occurs at least twce wth l dstnct pars s approxmately O(l) n l. Ths would gve us p l = p l O(l) n l O(p k O(k) n k ),... k l so the bound on M s O( np). However, the bound on the number of sequences wth l dstct pars breaks down when l s much smaller than k. In a full proof much more complcated calculatons must be done. 6 Problems wth sparse graphs ( ) Theorem 4 If p = Θ( n ), then A E(A) Ω log n w.h.p. Ths breaks down the nce bound we obtaned n secton 5. Ths follows from the rregularty of sparse graphs. There wll be solated vertces and vertces wth degree much hgher than average. ( ) Lemma If p = Θ( n ) then w.h.p. the hghest degree vertex of G s of order Θ log n. Proposton 5 If G has a node of degree d, then, for every p < 4 d, λ max(a pj) Ω( d). Ths mples that 0 < p <. 4 d, A pj Ω( d). Proof: We have λ max (A pj) = max x x T (A pj)x x where the maxmum s taken over all nonzero vectors x. Call v a node of degree d and call d of ts neghbors u,..., u d. 4
Consder the vector x such that x v =, x u = d and x w = 0 for other vertces w. We have x T Ax d x T pjx = p ( x ) = p ( + d) 4pd x = Therefore f p 4, d x T (A pj)x x d d = Ω( d) yeldng the desred bound. Theorem 4 proceeds mmedately from Proposton 5 and Lemma. 5