Edge Isoperimetric Inequalities


 Bernard Homer Floyd
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1 November 7, 2005 Ross M. Rchardson Edge Isopermetrc Inequaltes 1 Four Questons Recall that n the last lecture we looked at the problem of sopermetrc nequaltes n the hypercube, Q n. Our noton of boundary was that of vertex boundary, defned by δ(g) = {v S v u, u S}. We found that when tryng to mze the vertex boundary whle holdng the sze of our set S fxed, the (near) Hamg balls characterze those S wth mal vertex boundary. In ths lecture, we consder a dfferent noton of boundary based on edges. We defne the edge boundary of S to be (S) = {{u, v} u S, v S}. Ths s exactly the set of edges requred to dsconnect S from any vertex not n S. 1.1 Four Problems We shall look at four related sopermetrc problems whch all utlze the concept of edge boundary Queston 1 We begn wth the smplest queston: Gven a fxed postve nteger m, what s the smallest edge boundary for a set of m vertces? We may formalze ths queston by defnng g(m) = (S). We may then ask: Can we place bounds on g(m)? Can we characterze those subsets S whch acheve the value of g(m)? Whereas n the case of vertex boundary, the optmal subsets S were found to be (near) Hamg balls, we shall see that for the edge sopermetrc problem these optmal sets are (near) subcubes. Because the problem of fndng a set S wth mum edge boundary s dual to the problem of maxmzng the number of edges between vertces n S, ths result makes ntutve sense. 1
2 1.1.2 Queston 2 For a subset S V (Q n ) of sze m, let E be all those edges n E(Q n ) whose vertces dffer n the th coordnate. Set (S) = (S) E. Ths provdes a partton of (S), wth (S) = (S). We may then defne our functon of nterest as g 1 (m) = max (S). Thus we have that for every subset S of sze m, (S) g 1 (m) for some. Trvally we obtan the relaton g 1 (m) g(m)/n. Can we further characterze g 1 (m)? Ths queston was frst proposed by BenOr and Lnal [1] Queston 3 An alternate drecton s to look at subgraphs nduced by a subset S of vertces. For S V (Q n ), we denote by Q n [S] the subgraph of Q n nduced by S, that s, the graph on vertex set S and contanng all edges of E(Q n ) for whch both endponts are n S. For postve nteger m, defne g 2 (m) = E (Q n [S]). Note that f m 2 n 1 we have g 2 (m) = 0, snce the hypercube s bpartte. However, when m = 2 n 1 + 1, we obtan a sharp jump n the number of edges of Q n [S] Queston 4 Let (G) denote the maxmum degree of a graph G. We may agan look at nduced subgraphs of the hypercube, but measure ther maxmum degree. Formally we defne g 3 (m) = (Q n [S]). Once agan, we must restrct our attenton to m > 2 n 1. Can we characterze g 3 (m)? Currently, both g 2 (m) and g 3 (m) are not fully understood even for m = 2 n Theorems We shall explore the frst two questons. 2
3 2.0.5 On the Frst Queston Our frst queston asks smply what subsets have mal edge boundares. If we let m be a power of 2 then our answer s smple: a subcube of sze m. If m s not an nteger the answer s not too dfferent, but unfortunately t becomes a bt less beautful to both state and prove. Nonetheless, we have the followng Theorem: Theorem 1. Fx a postve nteger m. For 0 γ m, we may express γ by ts bnary expanson γ = n 1 =0 γ 2. Let S = {(γ 0,..., γ n 1 ) 0 γ m} be the set of all bnary expansons of postve ntegers γ m. Then S acheves the mum edge boundary of all subsets wth m vertces. Proof. By Inducton. Ths s left as Exercse 1. If we sacrfce an exact descrpton, we obtan a much cleaner bound. Theorem 2. For S V (Q n ) wth S = m, (S) m(n log 2 m). In order to prove Theorem 2, we shall requre an addtonal result. Recall that the average degree d of a graph G s defned as d = v V (G) d G(v) G. Theorem 3. Let G be a subgraph of Q n [S] wth average degree d. Then V (G) 2 d. Let us frst show how Theorem 3 mples Theorem 2. Note that (S) = m(n d), by defnton. By Theorem 3, we have 2 d m or d log 2 m. Combnng these two observatons yelds (S) m(n log 2 m). We thus need only prove Theorem 3. Proof (Theorem 3). We prove ths by nducton. We vew Q n as composed of two (n 1)cubes, whch we label Q (1) and Q (2). We let G 1 and G 2 be the ntersecton of G wth these two subcubes, and set m = V (G ). Wthout loss of generalty, assume 0 m 1 m 2. Fnally assume there are s edges of G between G 1 and G 2. Note that for each vertex n G 1 there can be at most one edge adjacent to ths vertex crossng to G 2, so s m 1. We may conclude by nducton that m log 2 m v V (G ) d G (v) = v V (G ) d G (v) s, = 1, 2. 3
4 Observe that m 1 log 2 m 1 + m 2 log 2 m 1 + 2s d G (v). v V (G) Thus we have, notng that m 2 m 1 m log 2 m = (m 1 + m 2 ) log 2 (m 1 + m 2 ) m 1 log 2 m 1 + m 2 log 2 m 2 + 2m 1 d G (v). v V (G) Here, the frst nequalty s a fact whch s proved at the end of these notes (where one can see why the base two logarthm s requred), and the second nequalty follows smply from the calculatons done above and observng that m 1 s. Thus, the proof concludes. 2.1 On the second queston To begn to approach the second queston, we shall ntroduce the noton of Boolean functons. A Boolean functon s smply a functon on 0/1 strngs of length n whch s ether zero or one. It shall be of some utlty to vew each coordnate of the length n strng as a separate 0/1 varables, whch we shall denote x 1,..., x n. Let us now look at three mportant examples. Party: f 1 (x 1,..., x n ) = x (mod 2) Projecton: f 2 (x 1,..., x n ) = x 1 { 1 Majorty: f 3 (x 1,..., x n ) = x n/2, 0 otherwse. To analyze these examples we shall assocate to each Boolean functon a game. Each game has n players assocated to one each of the varables x 1,..., x n. Each player flps a far con to decde the value of x. The value of the game s just the value of f(x 1,..., x n ). But what f one of the players acts ntellgently, choosng hs value deterstcally n order to nfluence the value of the game? We can then ask, how much nfluence can a sngle player have on the value of the game? To formalze ths queston, let us fx some ndex, and form the Boolean functon f (0) on the varables x 1, x 2,..., x 1, x +1,..., x n by settng f (0)(x 1,..., x 1, x +1,..., x n ) = f(x 1, x 2,..., x 1, 0, x +1,..., x n ). We form f (1) smlarly. We may then defne the nfluence I f () of the th player to be I f () = P (f (0) f (1)). Observe that the party functon satsfes I f () = 1, because flppng one bt n a strng changes the party of the entre strng. For the projecton functon, observe that I f () s 1 f = 1 and 0 otherwse,.e. the frst player completely deteres the value of f. The majorty functon s more complcated, and t s a nontrval fact that the nfluence of the th player s 1/ n. 4
5 S f x = 0 x = 1 Fgure 1: An th coordnate boundary edge corresponds to a place of nfluence for player. Say that the conflppng game s far f the probablty that f = 0 s 1/2. We can show that for any far game, there must exst some player wth nfluence at least 1/n. To see ths, we need to relate our game to the hypercube. Observe that a Boolean functon f deteres a partton of Q n. We denote by S f the set of all vertces of the hypercube for whch f = 1. If P (f = 0) = 1/2 then S f conssts of exactly half of the vertces of Q n. We agan note that for a fxed, we can vew the Q n as two copes of Q n 1 (denoted by Q (1) and Q (2) ) joned by edges along the th coordnate. Observe that the number of strngs for whch changng s value changes the value of f s smply the number edges from (S f ) whch cross from Q (1) to Q (2), denoted by (S) (see fgure 1). Thus, the nfluence of player s gven by Sumg over all, we have I f () = (S f ) 2 n 1. I f () = (S f ) 2 n 1. Fnally, we may apply Theorem 2 wth m = S f = 2 n 1 to gve a lower bound on (S f ), whch forces the rght hand sde to be at least one. Thus, I f () 1/n for some. Indeed, more s true. Kahn, Kala, and Lnal [3] have shown that for any far game there exsts some player such that I f () c log n n, where here c s some fxed constant. 3 Appendx Lemma 1. If 0 x y then (x + y) log 2 (x + y) x log 2 x + y log 2 y + 2x. 5
6 Proof. Assume x > 0, otherwse the result s trval. Let γ = y/x. Then (x + y) log 2 (x + y) x log 2 x(1 + γ) + y log 2 y(1 + 1/γ) = x log 2 x + y log 2 y + x log 2 (1 + γ) + y log 2 (1 + 1/γ) = x log 2 x + y log 2 y + x(log 2 (1 + γ) + γ log 2 (1 + 1/γ)) To see that log 2 (1 + γ) + γ log 2 (1 + 1/γ) 2 we observe that log 2 (1 + γ) + γ log 2 (1 + 1/γ) log 2 (1 + γ) + log 2 (1 + 1/γ) log 2 (1 + γ)(1 + 1/γ) log 2 (2 + γ + 1/γ) = 2. Where here we use the basc fact that γ + 1/γ 2. References [1] BenOr, M. and N. Lnal. Collectve Con Flppng, Robust Votng Schemes and Mnma of Banzhaf Values. FOCS 1985: [2] Chung, Fan. Edge Isopermetrc Inequaltes. Draft, [3] Kahn, J., Kala, G. and N. Lnal. The Influence of Varables on Boolean Functons. FOCS 1988:
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