Small Set Expansion in The Johnson Graph

Transcription

1 lectronic Colloquium on Computational Complexity, Report No 78 (018) Small Set xpansion in The Johnson Graph Subhash Khot Dor Minzer Dana Moshovitz Muli Safra Abstract This paper studies expansion properties of the (generalized) Johnson Graph For natural numbers t < l <, the nodes of the graph are sets of size l in a universe of size Two sets are connected if their intersection is of size t The Johnson graph arises often in combinatorics and theoretical computer science: it represents a slice of the noisy hypercube, and it is the graph that underlies direct product tests as well as a candidate hard unique game We prove that any small set of vertices in the graph either has near perfect edge expansion or is not pseudorandom Here not pseudorandom means that the set becomes denser when conditioning on containing a small set of elements In other words, we show that slices of the noisy hypercube while not small set expanders lie the noisy hypercube only have non-expanding small sets of a certain simple structure This paper is related to a recent line of wor establishing the -to- Theorem in PCP The result was motivated, in part, by 7] which hypothesized and made partial progress on similar result for the Grassmann graph In turn, our result for the Johnson graphs served as a crucial step towards the full result for the Grassmann graphs completed subsequently in 0] 1 Introduction xpanding graphs are ubiquitous in mathematics, theoretical computer science, information theory and other areas of study (See, eg, 17]) They are well-connected graphs that are robust to deletion of edges, and random wals on them mix rapidly In an expander, for every set S of at most half of the nodes, many of S s neighbors are outside S Formally, we define edge expansion as follows: Definition 11 (dge expansion) Let G = (U, ) be a d-regular graph, and let S U be a set of vertices The expansion of S is defined as (S, S) Φ(S) =, d S where (S, S) is the set of edges from S outside it In recent years there has been considerable interest in small set expanders, which are graphs where only sets containing a small fraction of the nodes are required to expand The interest in small set expansion was motivated by its connection to the Unique Games Conjecture 3, 6] and small set expanders were used in SDP integrality gaps 3], lower bounds on metric embeddings 3] and algorithms 1] Department of Computer Science, Courant Institute of Mathematical Sciences, New Yor University Research supported by NSF CCF-14159, Simons Collaboration on Algorithms and Geometry, and Simons Investigator Award School of Computer Science, Tel Aviv University School of Computer Science, University of Texas at Austin This material is based upon wor supported by the National Science Foundation under grants number and School of Computer Science, Tel Aviv University ISSN

2 An important example of a small set expander (that is not an expander) is the noisy Boolean hypercube Its nodes are the vectors in {0, 1}, or alternatively sets in a universe of size Two sets are connected, roughly, if their intersection is of size t, typically close to the size of the sets 1 In the noisy hypercube small sets have near perfect expansion, and this property is ey to many of the aforementioned results 3] The reader is referred to 5] for information about the hypercube, noise and the rich theory underlying them 11 Our Contribution In this wor we focus on slices of the hypercube, also nown as Johnson graphs For parameters t < l < we define the Johnson graph J(, l, t) to be the graph whose nodes are sets of size l in a universe of size There is an edge between two sets if their intersection is of size t Typically l is much smaller than and t is close to l Johnson graphs are the subject of considerable interest recently (See, eg, 10, 1, 13]) In combinatorics they are ey to the study of sharp thresholds of graph properties 14] In theoretical computer science they underlie direct product tests 15, 5, 19, 9], and a recent candidate hard unique game ] Johnson graphs are not small set expanders as we demonstrate next Consider the family S x of all sets containing a certain element x in the size- universe ( zoom in ) This is a very small family (The fraction of S x is roughly l/, which is small for large and much smaller l), yet many of the edges with one endpoint in S x have their other endpoint in S x as well (roughly t/l fraction, which is large for t close to l) In this paper we show that essentially only families lie S x are small and non-expanding in Johnson graphs Other families have near-perfect expansion, that is, almost all of their neighbors are outside them We thin of a set being lie an S x if it becomes substantially denser when conditioning on containment of one, or a few, elements x We define a pseudo-random set as one that is unlie any S x : Definition 1 (Pseudo-randomness on the Generalized Johnson Graph) Let J(, l, t) be a Johnson Graph, and let r N, ε > 0 where r l We say that a set of vertices S is (r, ε) pseudo-random if for every R ] of size at most r, Pr A R A S] Pr A A S] ε Our main theorem is that small sets of nodes in the Johnson graph that are pseudorandom have near perfect expansion Theorem 13 (Main Theorem) For every 0 < α < 1 and 0 < η < 1 there exist r N, ε, δ > 0, such that for large enough > l the following holds If S is a set of vertices of J(, l, αl) of density at most δ such that Φ(S) 1 η, then S is not (r, ε) pseudo-random Theorem 16 in the body of the paper gives a concrete dependence of the parameters on α and η Theorem 17 gives an improved dependence of ε at the expense of a worse dependence of the other parameters Theorem 13 is established via spectral analysis on the adjacency matrix of the Johnson graph We show a relation between pseudo-randomness of a set and its correlation with eigenspaces whose eigenvalue is non-negligible (formally stated in Theorem 15) Namely, if a function is pseudo-random, then its weight on the low-level eigenspaces is small We remar that the conclusion of Theorem 13 is in fact slightly stronger Namely, we may conclude that there is non-negligible fraction of zoom-ins that increases the density by at least ε This stronger conclusion 1 One typically considers the uniform distribution over the nodes, so mostly nodes are sets of size roughly l = / One considers the following probability distribution over neighbors y {0, 1} of a vector x {0, 1} : for each 1 i re-sample x i with probability δ to get y i, and set y i = x i with probability 1 δ Hence, edges mostly correspond to intersection of size roughly t = (1 δ)l

3 follows from the current one: we find a zoom-in set that increases the density by more than ε iteratively As long as the fraction of vertices we have covered with our zoom-in sets is o(δ), we may randomized the set S on the zoom-in sets so far After this randomization, the density of each zoom-in set we have found so far is δ + o(δ) and the expansion of S is bounded away from 1, and in particular we may find an additional zoom-in set 1 Related Wor This paper is related to a recent line of wor establishing the -to- Theorem in PCP 0, 6, 7, 1] The heart of this line of wor is a theorem similar to Theorem 13 for the Grassman graph rather than the Johnson graph, which was subsequently established in 1] The nodes of the Grassman graphs are l-dimensional linear subspaces rather than size-l sets dges correspond to largely intersecting linear subspaces The structure of the small non-expanding sets in the Grassman graph is more complicated than that of the Johnson graph, and our Theorem 13 was a crucial step towards the analogous theorem about the Grassman graph 1] Our result was motivated, in part, by 7] which hypothesized and made partial progress on the expansion of Grassmann graphs The current wor builds on their technique and further develops it To prove Theorem 13, one has to analyze expectations of the form f(x 1,, x r )f(y 1,, y r )f(z 1,, z r )f(w 1,, w r )], (1) for a certain function of interest f emerging from the spectral decomposition on the Johnson Graph, where the expectation is taen uniformly over x, y, z, w ] r that satisfy a predetermined set of equalities of the form a = b for a, b {x i, y i, z i, w i i = 1,, r} We call such expectations four-wise correlations, since it is an expectation of the product of four values of f on correlated inputs For the analog in the Grassmann Graph, one has to study the four-wise correlations f(x 1,, x r )f(y 1,, y r )f(z 1,, z r )f(w 1,, w r )], () where now the expectation is uniform over x i, y i, z i, w i F, that satisfy a predetermined set of linear equations in x i, y i, w i, z i In particular, this set of equations could contain equalities as before, say x 1 = y In 7], the authors could analyze the expectations of the form () for r = 1, For r =, they use brute force analysis to enumerate over all possible linear equation systems that determine the constraints among the variables, and upper bound each such configuration separately The analysis uses a combination of fourier analysis on f and the Cauchy-Schwarz inequality, along with the (additional) notion of zoom-outs As r increases, the number of configurations grows quicly and a case by case analysis becomes infeasible It was very unclear if a more systematic approach is possible that is able to deal with larger r In this paper, we present a systematic approach for the Johnson Graph (wherein fourier analysis on f and the notion of zoom-outs is not needed) It turns out that (as far as this analysis is concerned), the Johnson analysis is a special but crucial case of the Grassmann analysis Indeed, in the Grassmann analysis, if the only linear dependencies are equalities, then the analysis, as far as its high level structure/strategy is concerned, reduces to the Johnson analysis This turned out to be an insightful step in completing the Grassmann analysis in 1] We also show how to analyze higher than four-wise correlations in the Johnson Graph This leads to improved quantitative results in Theorem 13 and in the main technical result used to achieve it, Theorem 15 and its improved form Theorem 17 3

4 Preliminaries In this section we present the necessary bacground on the Johnson Graph We shall use properties of it that are analogous to properties from 7] about the Grassmann Graph We often omit proofs if they are nearly identical Definition 1 Let t < l < be integers The Generalized Johnson Graph J(, l, t) is defined as follows: the vertex set is ( ]) l, the set of all subsets of ] of size l Two vertices A, B ] are connected by an edge if and only if A B = t Abusing notation, we denote the normalized adjacency operator of the generalized Johnson Graph by J(, l, t) We thin of it as operating on real valued functions F : ( ]) l R 1 Fourier Analysis on the generalized Johnson Graph Any graph, and in particular J(, l, t) induces a spectral decomposition of real-valued functions on it In this section we introduce some properties of this decomposition It has a similar structure to the spectral decomposition on the Grassmann Graph 7] We endow the space of real-valued functions on J(, l, t) with the inner product for any F, G: ( ]) l R Level Functions F, G = F A]GA]] A ( ] l ) Definition Let t < l < be integers For any i = 0,, l we define the space spanned by the first i levels J i as follows F J i if and only if there exists f : ( ) ] i R such that F A] = f(i) I A, I =i One can easily verify that each J i is a linear subspace, and that J l contains all real-valued functions on J(, l, t) Furthermore, we have that J i J i+1 Definition 3 We define the space of level i functions by J =i = J i Ji 1 In words, it is the space of all functions from J i perpendicular to J i 1 It follows by the definition that the space of real-valued functions on J(, l, t) can be written as J =0 J =1 J =l Definition 4 Let t < l < be integers For i = 0,, t define ( t i) λ i (, l, t) = ( l i) For i > t, define λ i (, l, t) = 0 4

5 Definition 5 Let j < i < be integers and f : ( ) ] i R For J ], J = j we denote µ J (f) = I J f(i)] Lemma 6 Suppose l 8l, l, and let F J i Then F J =i iff for every R ], R < i we have that µ R (f) = 0 Proof The proof is similar to the proof of 7, Lemma 19] Theorem 7 Let t < l < be integers For any i = 0,, l, and F J =i given by F A] = f(i) we have I A, I =i J(, l, t)f λ i (, l, t)f 4l f In words, F is a near eigenvector of J(, l, t) of eigenvalue λ i (, l, t) Proof Let F be such function and fix a vertex A in J(, l, t) Then J(, l, t)f A] = B: B A =t F B]] = B: B A =t f(i) = I B, I =i B: B A =t min(t,i) r=max(0,t+i l) I B, A I =r We consider different values of i separately, starting with i t In this case, we separate the summand for r = i to get J(, l, t)f A] = i 1 f(i) + f(i) (3) The first sum in (3) equals B: B A =t B: B A =t (t ) i I A B The second sum in (3) is i 1 ( )( ) t l t r i r B: B A =t r=max(0,t+i l) f(i)] I A B, I =i I B, A I =r B: B A =t ] = f(i)] = We show that the last expectation is negligible Fix r i 1, then f(i)] = I: I A =r R A, R =r r=max(0,t+i l) I B, A I =r ( ) t f(i)] = i I A i 1 r=max(0,t+i l) f(i)] I:I A=R, I =i ( t i) ( l i)f A] ( )( t l t r i r ] ) f(i)] I: I A =r f(i) 5

6 For a fixed R, denote by D 1 the uniform distribution over I where I A = R, and by D uniform distribution over I where I R Note that these distributions are il close Hence f(i)] µ R (f) il f, I:I A=R, I =i and since µ R (f) = 0 (Lemma 6) we conclude that the second sum in (3) is upper bounded by i 1 ( )( ) t l t f(i)] r i r it l il f 4l f r=max(0,t+i l) I: I A =r Next, we prove for i > t In this case we have J(, l, t)f A] = B: B A =t t r=max(0,t+i l) I B, A I =r f(i) The argument is identical to the analysis of the error term in the previous case, and is hence omitted Remar 8 With a more careful analysis, one can prove that J =i is actually an eigenspace of J(, l, t) with eigenvalue nearly λ i 3 Decomposition Recall that we have seen that the space of all real-valued functions on ( ]) l can be decomposed as J=0 J =1 J =l For any function F : ( ]) l R, we denote this decomposition by F = F=0 +F =1 ++F =l where F =i J =i We also define f =i : ( ) ] i R to be a function that satisfies F =i A] = f =i (I) I A, I =i It is easy to verify that F =0 µ(f ) For large i, the functions F =i are harder to compute and we approximate them instead Definition 9 Let F : ( ]) l R be a function, and let i {0, 1,, l} We define the weight of F on level i to be W =i F ] def = F =i, F =i 4 Approximate Decomposition As is the case in the Grassmann Graph, computing the exact decomposition is not easy, so we wor with an approximate decomposition instead Given a function F : ( ]) l R, define f 0 µ(f ) Inductively once f j have been defined for all j < i, we define f : ( ) ] i R by f (I) def = µ I (F ) J I f J (J) for all I ( ) ] i Define F A] def = I A, I =i f (I) 6

7 Basic Facts The following theorem is an analog of a theorem from the Grassmann Graph The proof is an adaptation of the proof therein and is hence omitted Theorem 10 Analog of 7, Theorem 5]] Let, l be integers such that 00l3, and let F : ( ]) l R be a function Then for all i = 0,, l, F =i F 100l3 F The following fact will also be used Its an analog of a fact from the Grassmann Graph, and the proofs are similar Fact 11 Analog of 7, Lemma A]] Let l be integers, and let F : ( ]) l R be a function Then for all i = 1,, l, 0 j < i and J ] of size j, 10i f (I)] F I J, I =i Fact 1 Analog of 7, Claim A1]] Let l be integers, and let F : ( ]) l R be a function Then for all i = 0,, l, f i F Note that f depends of course on the underlying function F Often times the function F will be clear from the context, however sometimes we will be dealing with more than one function simultaneously In this case we shall denote the function f by f,f 5 Restrictions Given a function F : ( ]) l R and X ] of size at most l 1, we define the restricted function F X : ( ]\X l X ) R by F X A] = F X A] The following lemma expresses level i + 1 components of a function F as a function of the level i components of F and its restrictions Lemma 13 Let F : ( ]) l R be function, 0 i l 1 and x ] Then for any I ] \ {x} of size i, f +1,F (I {x}) = f,f {x} (I) f,f (I) Proof The proof is by induction on i For i = 0 the claim is obvious, since we have I = and both sides are easily seen to be equal to µ {x} (F ) µ(f ) Let i > 0, assume the claim for every j < i and prove for i Fix I, then by definition f +1,F (I {x}) = µ I {x} (F ) J I f J,F (J) JI f J +1,F (J {x}) (4) Consider the second sum Since we sum only over J strictly contained in I, we may apply the induction hypothesis to get that f J +1,F (J {x}) = f J,F {x} (J) f J,F (J) 7

8 Plug it into (4) to get f +1,F (I {x}) = µ I {x} (F ) f J,F (J) (f J,F {x} (J) f J,F (J)) J I JI = µ I (F {x} ) JI f J,F {x} (J) f,f (I) = f,f {x} (I) f,f (I), in the last equality we used the definition of f,f {x} Corollary 14 Let F : ( ]) l R be function, 0 i l 1 and X ], X < i Then for any I ] \ X of size j = i X, f,f (I X) = ( 1) Y f j,f X\Y (I) Y X Proof By induction on X For X = 0 this is obvious Assuming the statement for X s of size at most j, let us prove for the statement for X = j + 1 Write X = {x} X for X = j Then by Lemma 13 f,f (I X) = f 1,F {x} (I X ) f 1,F (I X ) (5) Applying the induction hypothesis on each term, we get that f 1,F {x} (I X ) = Y X ( 1) Y f j,f {x} X \Y (I) = and f 1,F (I X ) = Y X ( 1) Y f j,f X \Y (I) = Plugging the two into (5) completes the proof 6 Main Results Y X, Y x Y X, Y x ( 1) Y f j,f X\Y (I) ( 1) Y +1 f j,f X\Y (I) In this section, we state our main results Below, we are given a (r, ε) pseudo-random set, whose density is δ We suggest to thin of r as a constant, and on ε as cδ for a large, constant c Theorem 15 Suppose 00l3 δ 4 If S is (r, ε) pseudo-random, then for every i = 0, r, We prove this theorem in Section 3 W =i S] exp(i)δε 1/4 + exp(l 3 1 ) 1/16 Theorem 16 Let α (0, 1), and S be a subset of vertices in J(, l, αl) of density δ Suppose 00l3 δ 4, and let ε > 0, r N If S is (r, ε) pseudo-random, then Φ(S) 1 α r+1 exp(r)ε 1/4 poly(l3, 1/δ) 1/16 8

9 The proof is given in Section 4 We remar that this theorem easily implies Theorem 13 from the introduction Finally, we prove the following quantitative improvement of Theorem 15 in Section 5 Theorem 17 There exists c > 0, such that the following holds Let r, m N, ε δ > 0 Suppose 000m (m+1)l 3 δ 4 If S is (r, ε) pseudo-random of density δ, then for every i = 0, r, W =i S] (c m) i δε 1 1 m + exp(l 3 ) 1/(8m (m+1)) In words, the above theorem asserts that for constant r, a (r, ε) pseudo-random set S can have at most δε 1 o(1) of its weight on the first r levels This statement should be compared to the r-level inequalities on the Boolean hypercube, stating that if f : { 1, 1} n {0, 1} has density δ, then its fourier weight on level r is at most δ log r (1/δ) 3 Proof of Theorem 15 Notation In this secton we sometimes use the notation x = y ± ε to say that x y ε Let S be a (r, ε) pseudo-random set of density δ Let F : J(, l, t) be the indicator function of S, and let i {0,, r} For i = 0 the claim is obvious, so consider i > 0 Write the function F according to its decomposition F = F =0 + F =1 + + F =l, and denote η = W =i F ] Assume η δ since otherwise we are done Then we have A F=i A] ] = η, and by orthogonality A (F F=i )A] ] = δ η 31 Information about the second moments Claim 31 A F A] ] = η ± 3 50l3 Proof By the triangle inequality F =i F F =i F 50l3 and therefore F = F =i ± 3 50l3 = η ± 3 50l3 Claim 3 I f (I) ] = W =i F ] ( l i) ± 4 50l3 = η ( l i) ± 4 50l 3 Proof xpand out A F A] ] On the one hand it is equal η ± 3 50l3 by the previous claim On the other 9

10 hand, it is equal f (I) = A A I A, I =i = = I A, I =i f (I) + A ( ) l f (I) ] + i I A ( ) l f (I) ] + i I i 1 i 1 d=0 I I A, I = I =i d=0 I,I A I = I =i, I I =d ( l d )( )( ) l d l i i d i d f (I)f (I ) f (I)f (I ) I,I, I = I =i I I =d f (I)f (I ) ] We shall show that the second sum is small Fix d {0,, i 1} and consider the last expectation Then it is equal f (I) f (I ) ] I, I =i I, I =i I I =d Fix I in the outside expectation and consider the inner expectation is f (I ) ] = D I I, I =i I I =D D I µ D (f ) ± i f ], where we used the fact that the uniform distribution of I over I that contain D, and over I such that I I = D, is i / close Applying Facts 11 and 1 we get that the overall expectation in absolute value is at most 1i Therefore for each d the corresponding expectation is at most ] 1i f 13i, I, I =i we used Fact 1 Plugging it into the first equation in the proof, we see that and therefore A F A] ] ( ) l f (I) ] i 1 i I d=0 ( l d f (I) ] = 1 ( I l ) F A] ] ± ( 1 A l i i )( )( ) l d l i 13i i d i d ) 17l = ( η l ) ± 4 50l3 i 17l, Corollary 33 Let A ], A a Then f a,f A (B) ] ii a ε B =i a l i a l3 10

11 Proof By Claim 3, B =i a f a,f A (B) ] W =i a F A ] ) ± 4 50l3 Note that since F is (r, ε) pseudo-random, W =i a F A ] µ(f A ) µ(f ) + ε ε Additionally, ( ) ( ) l a l a i a ( ) l i a = l i a i (i a) i a i a i Combining these two yields the result Corollary 34 Let A ], A = a Then ( l a i a f,f (A B) ] i+1 i i a ε B ]\A l i a + 4 5l3 B =i a Proof By Corollary 14 f,f (A B) = ( 1) Y f a,f A\Y (B) Y A A Y A f a,f A\Y (B), the last inequality is by Cauchy-Schwarz Therefore by linearity of expectation and Corollary 33 f,f (A B) ] A f a,f A\Y (B) ] B ]\A B ]\A Y A B =i a A Y A i+1 i i a ε l i a B =i a i i a ε l i a l l3 3 A lower bound on the fourth moment Claim 35 Pr A F=i A] η ] 4δ η Proof By Marov s inequality Pr A F A] F =i A] 1 η δ ] δ η 1 η δ δ η = δ δ η δ η Since the probability F A] = 1 is δ, we conclude probability that F A] F =i A] 1 η δ and F A] = 1 is at least η For any such A we have 1 F =i A] 1 η δ, 11

12 implying F =i A] η 4δ Claim 36 Pr A F A] η 8δ ] η 4 Proof Let 1 = { A F =i A] η 4δ }, and = { A F =i A] F A] η 8δ } By Theorem 10 we have F =i F Pr ] = Pr F =i A] F A] η ] 100l3 A 8δ 100l3, and hence by Marov s inequality ( ) 8δ l3 η δ η 4, where we used η δ and the condition on Noting that whenever 1, occur we have F A] η 8δ, we conclude that ] Pr A we used Claim 35 in the last inequality F A] η 8δ The above claim immediately implies the following Claim 37 A F 4 A] ] η5 14 δ 4 33 An upper bound on the fourth moment Pr 1 ] Pr ] η 4, The following lemma is the main technical lemma of this section Lemma 38 Suppose 100l3, l If F is (i, ε) pseudo-random, then Proof By opening the bracets we have Denote F 4 A] ] = A A A F 4 A] ] exp(i)εη + 7l3 1/4 I 1,I,I 3,I 4 A f (I 1 )f (I )f (I 3 )f (I 4 ) D(A, d) = {(I 1, I, I 3, I 4 ) I 1,, I 4 A, I 1 I I 3 I 4 = d}, and note that D(A, d) depends only on i, d denote it by β i,d Then F 4 A] ] ] 4i = β i,d f (I 1 )f (I )f (I 3 )f (I 4 )] A A (I 1,I,I 3,I 4 ) D(A,d) d=i An intersection pattern σ d of a tuple (I 1, I, I 3, I 4 ) (whose direct sum is d dimensional) is a vector indicating the size of the intersection of any pair, triple and quadruple of sets from the tuple Note that when sampling the quadruple (I 1,, I 4 ), the probability of getting each quadruple depends only on its intersection 1

13 pattern Therefore there is a distribution γ( σ d ) over intersection patterns such that the previous expression equals F 4 A] ] 4i = γ( σ d )β i,d A A f (I 1 )f (I )f (I 3 )f (I 4 )] (6) d=i σ d (I 1,I,I 3,I 4 ) γ( σ d ) I 1,,I 4 A Note that now the distribution over (I 1,, I 4 ) is uniform over all quadruples whose direct sum is of dimension d and is intersection pattern is σ d, and thus F 4 A] ] 4i = γ( σ)β i,d f (I 1 )f (I )f (I 3 )f (I 4 )] A (I d=i σ 1,I,I 3,I 4 ) γ( σ d ) d The following lemma is the ey in the previous lemma Lemma 39 Suppose 100l3 Let i, d be integers such that i d 4i and let σ d be any intersection pattern Then f (I 1 )f (I )f (I 3 )f (I 4 )] i+ i d εη l d l3 1/4 (I 1,I,I 3,I 4 ) γ( σ d ) We defer the proof of Lemma 39 to the next section and show how to complete the proof of Lemma 38 based on it Using Lemma 39 we see that ( ) 4i i+ i d εη (6) γ( σ)β i,d l d l3 1/4 Note that β i,d and so by the previous inequality d=i 4i (6) σ ( )( ) l d 4 d i d=i ( σ ( le d ) d ( ) d e 4i l d d d (4e ) 4i, γ( σ)l d d d (4e ) 4i ( i+ i d εη exp(i)εη l3 1/4 exp(i)εη + 7l3 1/4 ) 4i i l d (i/d) d d=i l3 1/4 ) 13

14 34 Wrapping things up Proof of Theorem 15 Combining Claim 37 and Lemma 38 we see that Rearranging we see that the last inequality is by η δ 35 Proof of Lemma 39 η 5 14 δ 4 A F 4 A] ] exp(i)εη + 7l3 1/4 η exp(i)δε 1/ l3 1/16 δ η exp(i) δε 1/ l3 1/16, Proof Let x 1,, x d be chosen uniformly from ], and let P 1,, P 4 be sets depending on x 1,, x d corresponding to the intersection pattern σ d Note that d f (P 1 )f (P )f (P 3 )f (P 4 )] f (I 1 )f (I )f (I 3 )f (I 4 )] x 1,,x d f 4, since the distributions of the I s and the P s are d upper bound of the expectation involving the P s (I 1,I,I 3,I 4 ) γ( σ d ) Proposition 310 If there is x j that appears in only one of the P s, then f (P 1 )f (P )f (P 3 )f (P 4 )] x 1,,x d close to each other Therefore, it suffices to show an 14id Proof Assume without loss of generality j = 1 and that P 1 contains x 1 Then f (P 1 )f (P )f (P 3 )f (P 4 )] = x 1,,x d f (P )f (P 3 )f (P 4 ) x1 f (P 1 ) x,, x d ] x,,x d (7) Fix x,, x d and denote J = {x,, x d } P 1 Since x 1 P 1, we have that J P 1 1 = i 1 Additionally x1 f (P 1 ) x,, x d ] f (I)] + d f The latter expectation is upper bounded by 10i from Fact 11 Combining this with Fact 1 we conclude that x1 f (P 1 ) x,, x d ] 11i d Using this, the triangle inequality and Fact 1 on (7) completes the proof I J ] 14

15 Thus we may assume that every x j appears at least in two sets Next, denote by H 4 = P 1 P P 3 P 4 the set of x j s appearing in all sets, h 4 = H 4, H 3 the set of x j s appearing in three of the sets, h 3 = H 3, H the set of x j s appearing in precisely two sets, h = H Claim 311 f (P 1 )f (P )f (P 3 )f (P 4 )] H 4,H 3,H H 4,H 3 f (P 1 ) ] f (P ) ] f (P 3 ) ] f (P 4 ) ] ] H H H H Proof Let x H, and suppose P j1, P j contain it but not P j3, P j4 Then the left hand side is at most f (P j3 ) f (P j4 ) f (P j1 ) f (P j ) ]] x H 4,H 3,H \{x} Applying Cauchy-Schwarz inequality on the inner expectation, the above expression is upper-bounded by f (P j3 ) f (P j4 ) f (P j1 ) ] f (P j ) ]] x x H 4,H 3,H \{x} Continuing in this manner - namely picing each time a new variable from H, isolating the two terms that depends on it and applying Cauchy-Schwarz on that expectation, yields the desired bound We next upper bound H 4,H 3 H f (P 1 ) ] H f (P ) ] H f (P 3 ) ] H f (P 4 ) ] ] (8) Using Cauchy-Schwarz inequality, (8) H 4,H 3 max H 3 P,H 4 f (P 1 ) ] f (P ) ] H H ] H 4,H 3 max H 3 P 3,H 4 f (P 3 ) ] f (P 4 ) ] ] H H f (P 1 ) ] f (P 4 ) ] H 3 \P,H H 3 \P 3,H f (P ) ] f (P 3 ) ] (9) H 4,H 3,H H 4,H 3,H The product of the third and the fourth term is equal to f (P 3 ) ] η ii P 3 l i l3 Where the last inequality is by Claim 3 and the estimate ( ( l i) l ) i i Next, we estimate the maximums For j = 1,, 3, 4 define H 3,j = H 3 P j Apply Corollary 34 on the first maximum with A = (H 4, H 3,1 P ) and B = (H P 1, H 3,1 \ P ), we get it is upper bounded by max f (A B) ] i+1 i B ε A B l B + 4 5l3 i+1 i λ 1/ ε l λ + 5l3 1/ 1/4, 15

16 where λ 1 = H P 1 + H 3,1 \ P Similarly, the second maximum is upper bounded by where λ = H P 4 + H 3,4 \ P 3 Combining everything, we see that i+1 i λ / ε l 1 λ + 5l3 1/4, 9 i+ i 1 (i+λ 1+λ ) εη l 1 (i+λ 1+λ ) + 9 5l3 (10) 1/4 By counting occurrences of points in the P, we get that i = h 4 + H P + H 3, and by counting occurrences of points in P 3, i = h 4 + H P 3 + H 3,3 Also, note that H 3,1 \ P = H 3 \ H 3, since any x H 3 \ H 3, is in 3 of the sets but not in P, and hence it is in P 1 Thus, H 3,1 \ P = H 3 \ H 3, = h 3 H 3, and similarly H 3,4 \ P 3 = H 3 \ H 3,3 = h 3 H 3,3 Plugging everything into (11), we see that the exponent of l (and also i) is equal to h 4 + H P + H 3, + h 4 + H P 3 + H 3,3 + H P 1 + H P 4 + h 3 H 3,3 + h 3 H 3, which is (h 4 + h 3 + h ) = d Thus, 9 i+ i d εη l d + 9 5l3 (11) 1/4 4 Pseudorandomness implies xpansion Proof of Theorem 16 Let S be a set as in the Theorem, and let F be its indicator functions in J(, l, αl) Note that δ(1 Φ(S)) = F, J(, l, αl)f (1) Claim 41 F, J(, l, αl)f Proof Writing F = F =0 + + F =l we have F, J(, l, αl)f = F, The first term equals l i=0 l J(, l, αl)f =i = F, i=0 λ i (, l, αl)w =i F ] + 101l3 + F, l λ i (, l, αl)f =i i=0 l J(, l, αl)f =i λ i (, l, αl)f =i i=0 l l F =i, λ i (, l, αl)f =i = i=0 i=0 l λ i (, l, αl)w =i F ] i=0 16

17 by orthogonality The second term can be upper bounded using Cauchy-Schwarz and the triangle inequality by ( l l ) F, J(, l, αl)f =i λ i (, l, αl)f =i F J(, l, αl)f =i λ i (, l, αl)f =i i=0 Using Theorem 10 and F 1, the last expression is at most i=0 l i=0 100l3 + J(, l, αl)f λ i (, l, αl)f l i=0 100l3 + 4l f the last inequality is by Theorem 7 Therefore using Fact 1 and simplification last expression is at most (l + 1)3 100l3 101l3 The following upper bound follows easily by the definition of λ i Fact 4 λ i (, l, αl) α i Therefore we conclude that F, J(, l, αl)f r α i W =i F ] + i=0 αl i=r+1 α i W =i F ] + 101l3 Let i r Observe that since S is (r, ε) pseudo-random it follows that S is (i, ε) pseudo-random Therefore, applying Theorem 15 we see that for i = 0, 1,, r, W =i F ] exp(i)δε 1/ l3 1/16 For the second sum, note that by Parseval the sum of all weights of F is at most δ Hence, combining these two facts we conclude F, J(, l, αl)f r α i exp(i)δε 1/4 + α r+1 i=0 αl i=r+1 W =i F ] + 0 7l3 1/16 Plugging this into (1) and simplifying yields exp(r)δε 1/4 + δα r l3 1/16 Φ(S) 1 α r+1 exp(r)ε 1/4 0 7l3 1/16 δ 17

18 5 Proof of Theorem 17 Let S be a (r, ε) pseudo-random set, and F be its indicator function The proof of Theorem 17 follows the same outline as the proof of Theorem 15, except that we use higher moment Let 0 i r Claim 51 Proof Immediate from Claim 36 A F m A] ] ηm+1 6m+ δ m The intersection pattern of P 1,, P m is a vector σ indicating the sizes of all all intersections of any collection of the sets Lemma 5 Let σ d be an intersection pattern for m sets, and let P 1,, P m be sets that match this intersection pattern, that have formal elements x 1,, x d Then m f (P j ) x 1,,x d 4mi+m id ε m 1 η exp(l 3 ) l d + (13) 1/(4m(m+1)) The rest of this section is devoted for the proof of this lemma Let σ d be an intersection pattern If among x 1,, x d there is a variable that appears only in one of the P s then lemma holds trivially: Proposition 53 If there is x j that appears in only one of the P s, then m (11+m)id f (P j ) x 1,,x d Proof The same proof as in Proposition 310 We thus assume from now on that each x j appears in at least of the P s For s =, 3, m, let H s be the set of x j s that appear in exactly s of the P s For convenience, let us also denote H s = H s H s+1 H m and H s = H H 3 H s For each one of the P -sets, define e 1 (P ) def = f (P ) and for s define e s (P ) def = Hs f (P ) ] For each a =, 3,, m, define an operator ρ a on random variables defined by ρ a (Z) = Ha Z a/(a 1)] For j = 1,,, m 1, define T j,j (P ) def = e j (P ) Inductively for a > j, define T j,a (P ) def = ρ a (T j,a 1 (P )) 18

19 Denote for each s, Q s H s+1 ] def = Hs Note that for s = m, this is the term we wish to bound Proposition 54 For every s m, Q s H s+1 ] m m f (P j ) T 1,s (P j ) We shall use the following fact in the proof, which is a direct corollary of Holder s inequality Fact 55 Let h 1,, h n : ( m) R Then 1/s h 1 h n 1 h 1 n h n n Proof of Proposition 54 The proof is by induction on s The base case s = 1 is trivial Let s 1, assume we have proven for s, and prove for s + 1 Note that Q s+1 H s+ ] = Applying the induction hypothesis, we get that Q s+1 H s+ ] Hs+1 Q s H s+1 ]] Hs+1 m (T 1,s (P j )) 1/s Write H s+1 = { } x j1,, x jq Iteratively, for each y Hs+1 we consider the s + 1 P s it appears in, and then apply Fact 55 on them Thus, for instance suppose we have that y is in P 1,, P s+1, then we would get s+1 Q s+1 H s+ ] T 1,s (P j ) (y (s+1)/s]) 1/(s+1) m T 1,s (P j ) H s+1 \{y} Repeating this process for every y H s+1, one gets Q s+1 H s+ ] m ( T 1,s (P j ) (s+1)/s]) 1/(s+1) = H s+1 P j m j=s+ T 1,s+1 (P j ) 1/(s+1) Thus, we have that LHS(13) m T 1,m (P j ) 1/m (14) 19

20 Proposition 56 For any P -set P and 1 j m 1 T j,m (P ) T j+1,m (P ) max e j (P ) m j(j+1) H j Proof The proof is by induction on j The induction basis j = m 1 follows similar lines of the induction step and is hence omitted Assume the claim for j m 1 and prove for j 1 Clearly, H j T j 1,m (P ) = ρ m ρ j (e j 1 (P )) e j 1 (P ) j/(j 1)] Hj e j 1 (P )] max ( = ρ m ρ j+1 e j 1 (P ) j/(j 1)]) H j H j e j 1 (P ) 1/(j 1) = e j (P ) max H j e j 1 (P ) 1/(j 1) Also, note that each operator ρ a is monotone on non-negative random variables, and for a random variable Z and a constant c 0 we have that ρ a (cz) = ρ a (c)ρ a (Z) Thus, combining the above two we get that T j 1,m (P ) ρ m ρ j+1 ](e j (P )) ρ m ρ j+1 ](max H j e j 1 (P ) 1/(j 1) ) = T j,m (P ) max H j e j 1 (P ) m/j(j 1) Repeated application of the above proposition yields that for any P -set, m 1 T 1,m (P ) T m,m (P ) max e a (P ) m H a+1 a=1 a(a+1) = m a=1 max e a (P ) m g(a) H a+1 where g(a) = 1 1 a(a+1) for a m 1 and g(m) = m We used the fact that T m,m(p ) = e m (P ) Plugging this into (14) yields LHS(13) m m a=1 Using Corollary 34, we see that that for a < m, max e a 1 (P j ) g(a) H a+1 For a = m, by Claim 3 max e a (P ) i+1 i P Ha ε + exp(l3 ) H a+1 l P H a max e a (P ) i P Ha η + exp(l3 ) H a+1 l P H a 0

21 Therefore, LHS(13) m m ( i P j H m η ( l P j H m i P j H m η l P j H m 4mi+m ηε m 1 m ) g(m) + exp(l3 ) ) g(m) ( m m a=1 a=1 m i g(a) P j H a m m 1 a=1 ( i+1 i P j H a ε l P j H a l + exp(l3 ) g(a) P j H a where to compute the power of ε we used the fact that m 1 last product It is equal to m m a a=1 r=1 i g(a) P j H r l g(a) P j H r = = = = a=1 i+1 i P j H a ε l P j H a ) g(a) + ) g(a) + exp(l3 ) exp(l 3 ) 1/(4m(m+1)) 1/(4m(m+1)) (15) g(a) = 1 1 m (telescopic sum) Consider the m a i g(a) m P j H r a=1 r=1 l g(a) m P j H r m a a=1 r=1 m m i g(a) r Hr l g(a) r Hr i g(a) r Hr l g(a) r Hr r=1 a=r m i r Hr m g(a) a=r r=1 l r Hr m g(a) a=r In the second equality, we used the fact that each element in H r is counted r times by the P j s In the third equality we interchanged the order of multiplication Note that m g(a) = 1 r, hence the last product is equal to Plugging this into 15 yields concluding the proof m i Hr r=1 l Hr = i m H r r=1 l m H r r=1 a=r = id l d LHS(13) 4mi+m id ηε m 1 exp(l 3 ) l d + 1/(4m(m+1)), The following corollary is proven using a similar argument to the one in Lemma 38 Since the proof is almost identical, we omit it Corollary 57 Suppose 100l3, l If F is (i, ε) pseudo-random, then F m A] ] (10m) 4mi ηε m 1 exp(l 3 ) + A 1/(4m(m+1)) 1

22 Proof of Theorem 17 Combining Claim 51 and Corollary 57, we get that Rearranging (and using η δ ) we get that η m+1 6m+ δ m (10m)4mi ηε m 1 exp(l 3 ) + 1/(4m(m+1)) taing m-root yields finishing the proof η m 6m+ (10m) 4mi δ m ε m 1 + exp(l 3 ) 1/(4m(m+1)), η 16(10m) i δε 1 1 m + exp(l 3 ) 1/(8m (m+1)) 6 Open Problems Derandomizing the hypercube Many results in the PCP literature 16], construction of integrality gaps and non-embeddability results 3] rely on the small-set expansion property of the (noisy) hypercube This property states that for every noise rate ρ > 0, there exists ε > 0 such that a set of density δ in the noisyhypercube has expansion at least 1 O(δ ε ) The quantitative aspect of these results is often determined by the number of vertices in the noisy-hypercube, which is large Improving these results can be achieved by constructing a hypercube lie graphs (eg that have a small-set expansion type property) with significantly smaller number of vertices This was done by Bara et al ] via the Short Code, that in many ways behaves similarly to the noisy hypercube and thus may replace the noisy hypercube in the applications above Can our results be used to get improved results? This tas is not trivial since the Johnson Graph does not have a small-set expansion property We do have however a characterization of sets for which the small-set expansion property fails that is sometimes sufficient For instance, the outer PCP of 0, 7] deviates from classical PCP constructions by incorporating this characterization into the analysis Understanding slices of the hypercube The Johnson Graph represents a slice of the Boolean hypercube, which is a well-studied object from a combinatorial point of view 11, 10, 1, 13] It is related to the study of sharp threshold of Boolean functions and graph properties 14] Can our result lead to new results about slices of the Boolean hypercube the way that small set expansion of the hypercube and hypercontractivity in general led to results about the hypercube? Subsequent to our wor, Lifshitz and the second author have made progress in this direction 4] Direct product testing For the Grassmann Graph, the characterization of small-sets that have near-perfect expansion implies a direct product testing result, ie, an encoding and a (-to-) test on words, such that any word that passes the test with noticeable probability has global structure (inside some subgraph of Grassmann Graph) 3] Our results are related, in the same sense, to the problem of direct product testing on the hypercube 18, 8, 4] Can one obtain new results about direct product testing via our theorem? It is interesting to note that the notion of zoom-ins appears in many of these wors, and often as an intermediate step (if there are more than queries)

23 References 1] Sanjeev Arora, Boaz Bara, and David Steurer Subexponential algorithms for unique games and related problems J ACM, 6(5):4, 015 ] Boaz Bara, Parishit Gopalan, Johan Håstad, Raghu Mea, Prasad Raghavendra, and David Steurer Maing the long code shorter SIAM J Comput, 44(5): , 015 3] Boaz Bara, Pravesh Kothari, and David Steurer Small-set expansion in shortcode graph and the -to-1 conjecture manuscript, 018 4] Roee David, Irit Dinur, lazar Goldenberg, Guy Kindler, and Igor Shinar Direct sum testing SIAM J Comput, 46(4): , 017 5] Irit Dinur and lazar Goldenberg Locally testing direct product in the low error range pages 613 6, 008 6] Irit Dinur, Subhash Khot, Guy Kindler, Dor Minzer, and Muli Safra Towards a proof of the -to-1 games conjecture? lectronic Colloquium on Computational Complexity (CCC), 3:198, 016 7] Irit Dinur, Subhash Khot, Guy Kindler, Dor Minzer, and Muli Safra On non-optimally expanding sets in grassmann graphs lectronic Colloquium on Computational Complexity (CCC), 4:94, 017 8] Irit Dinur and Inbal Livni Navon xponentially small soundness for the direct product z-test In Ryan O Donnell, editor, 3nd Computational Complexity Conference, CCC 017, July 6-9, 017, Riga, Latvia, volume 79 of LIPIcs, pages 9:1 9:50 Schloss Dagstuhl - Leibniz-Zentrum fuer Informati, 017 9] Irit Dinur and David Steurer Direct product testing In Computational Complexity Conference, ] Yuval Filmus Friedgut-Kalai-Naor theorem for slices of the boolean cube Chicago J Theor Comput Sci, 016, ] Yuval Filmus An orthogonal basis for functions over a slice of the boolean hypercube lectr J Comb, 3(1):P13, 016 1] Yuval Filmus, Guy Kindler, lchanan Mossel, and Karl Wimmer Invariance principle on the slice In 31st Conference on Computational Complexity, CCC 016, May 9 to June 1, 016, Toyo, Japan, pages 15:1 15:10, ] Yuval Filmus and lchanan Mossel Harmonicity and invariance on slices of the boolean cube In 31st Conference on Computational Complexity, CCC 016, May 9 to June 1, 016, Toyo, Japan, pages 16:1 16:13, ] hud Friedgut Sharp thresholds of graph properties, and the -sat problem J Amer Math Soc, 1: , ] Oded Goldreich and Shmuel Safra A combinatorial consistency lemma with application to proving the PCP theorem SIAM J Comput, 9(4): , ] Johan Håstad Some optimal inapproximability results J ACM, 48(4): , 001 3

24 17] Shlomo Hoory, Nathan Linial, and Avi Wigderson xpander graphs and their applications Bull Amer Math Soc, 43(04):439 56, ] Russell Impagliazzo, Valentine Kabanets, and Avi Wigderson New direct-product testers and -query pcps SIAM J Comput, 41(6): , 01 19] Russell Impagliazzo, Valentine Kabanets, and Avi Wigderson New direct-product testers and -query PCPs SIAM Journal on Computing, 41(6): , 01 0] Subhash Khot, Dor Minzer, and Muli Safra On independent sets, -to- games and grassmann graphs lectronic Colloquium on Computational Complexity (CCC), 3:14, 016 1] Subhash Khot, Dor Minzer, and Muli Safra Pseudorandom sets in grassmann graph have near-perfect expansion lectronic Colloquium on Computational Complexity (CCC), 5:6, 018 ] Subhash Khot and Dana Moshovitz Candidate hard unique game In Proceedings of the 48th Annual ACM SIGACT Symposium on Theory of Computing, STOC 016, Cambridge, MA, USA, June 18-1, 016, pages 63 76, 016 3] Subhash Khot and Nisheeth K Vishnoi The unique games conjecture, integrality gap for cut problems and embeddability of negative-type metrics into l 1 J ACM, 6(1):8:1 8:39, 015 4] Noam Lifshitz and Dor Minzer, 018 in preparation 5] Ryan O Donnell Analysis of Boolean Functions Cambridge University Press, 014 6] Prasad Raghavendra and David Steurer Integrality gaps for strong SDP relaxations of UNIQU GAMS In 50th Annual I Symposium on Foundations of Computer Science, FOCS 009, October 5-7, 009, Atlanta, Georgia, USA, pages , CCC ISSN