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The Quasi-Randomness of Hypergraph Cut Properties. Asaf Shapira & Raphael Yuster. Background. Abstract Question: When can we say that a single graph behaves like a random graph ?. “Concrete” problem: Which graph properties “ force” a graph to behave like a “ truly” random one.
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The Quasi-Randomness of Hypergraph Cut Properties Asaf Shapira & Raphael Yuster
Background Abstract Question: When can we say that a single graph behaves like a random graph? “Concrete” problem: Which graph properties “force” a graph to behave like a “truly” random one. Chung, Graham, and Wilson ’89, Thomason ‘87: 1. Defined the notion of a p-quasi-random graph= A graph that “behaves” like a typical graph generated by G(n,p). 2. Proved that several “natural” properties that hold in G(n,p) whp “force” a graph to be p-quasi-random.
The CGW Theorem • Theorem [CGW ‘89]: Fix any 0<p<1, and let G=(V,E)be a • graph on n vertices. The following are equivalent: • Any set U V spans ½p|U|2 edges • Any set U V of size ½n spans ½p|U|2 edges • G contains ½pn2edges and p4n4 copies of C4 • Most pairs u,v have co-degree p2n Note: All the above hold whp in G(n,p). Definition: A graph that satisfies any (and therefore all) the above properties is p-quasi-random, or just quasi-random.
Quasi-Random Properties Definition: Say that a graph property is quasi-random if it is equivalent to the properties in the CGW theorem. “The” Question: Which graph properties are quasi-random? No! Any (reasonable) property that holds in G(n,p)whp? Example 1:Having ½pn2 edges and p3n3 copies of K3 is not a quasi-random property. Recall that if we replace K3 with C4 we do get a quasi-random property. Example 2:Having degrees pn is not a quasi-random prop. …but having co-degreesp2nis a quasi-random property.
The Chung-Graham Theorem Theorem [Chung-Graham ’89]: 1. Having ¼ pn2 edges crossing all cuts of size (½n,½n)is not a quasi-random property. 2. For any 0<<½, having (1-)pn2 edges crossing all cuts of size (n,(1-)n)is a quasi-random property. To get (1), take an Independent set on n/2 vertices, a clique on the rest, and connect them with a random graph. [CG ‘89] Gave two proofs of (2). One using a counting argument, and another algebraic proof based on the rank of certain intersection matrices. [Janson ‘09] Gave another proofs of (2), using graph limits.
Quasi-Random Hypergraphs What is a quasi-random hypergraph? Answer 1: The “obvious” generalization of quasi-random graphs. Every set of vertices has the “correct” edge density. Definition: This is called “weak” quasi-randomness. Why? Because it does not imply certain things that are implied by quasi-randomness in graphs. Answer 2: “Strong” quasi-randomness. Fact: Strong Quasi-Randomness Weak Quasi-Randomness Notation: “P is Quasi-Random” means P Weak Quasi-Rand
Our Main Results Definition:Let =(1,…,k) satisfy 0<i<1 and I=1. Let P be the following property of k-uniform hypergraphs: Any (1n,…,kn)-cut has the correct number of edges crossing it. Theorem 1 [S-Yuster ’09]: 1. If = (1/k,…,1/k) then P is not quasi-random. 2. If (1/k,…,1/k) then P is quasi-random. Theorem 2 [S-Yuster ’09]: 1. When = (½,½) the only way a non-quasi random graph can satisfy P is the “trivial” one. 2. Same result conditionally holds in hypergraphs.
Proof Overview p 2p 0 Definition:Let =(1,…,k) satisfy 0<i<1 and I=1. We let P be the following property of k-uniform hypergraphs: Any (1n,…,kn)-cut has the correct number of edges crossing it. Theorem:If = (1/k,…,1/k) then P is not quasi-random. [CG‘89] For k=2: i vertices k-i vertices 2ip/k For arbitrary k2:
Proof Overview p2 p1 p3 V-A |V-A|=(1-)n A |A|=n Theorem:If (1/k,…,1/k) then P is not quasi-random. Proof (of k=2): It is enough to show that every set of vertices of size n has the correct edge density. Let A be such a set. Let 0c, and “re-shuffle” the partition by randomly picking cn vertices from A and (-c)n vertices from V-A. 1. We know the expected number of edges in the new cut. 2. This expectation is a linear function in p1 , p2 , p3. 3. Using c{0,,/2} we get 3 linear equations, which have a unique solution p1=p2=p3=p when 1/2.
Proof Overview p p 2p 0 The random graph The example showing that P is not quasi random Theorem 2: When = (½,½) the only way a non-quasi random graph can satisfy P is the “trivial” one. Instead of thinking about graphs, let’s consider the problem of assigning weights to the edges of the complete graph, s.t. for any (n/2,n/2)-cut, the total weight crossing it is p. What is “trivial”? Two ways a graph can satisfy P There are many such solutions
Proof Overview p p 2p 0 The random graph (1) The example showing that P is not quasi random (2) What is “trivial”? Two ways a graph can satisfy P • Satisfying P is equivalent to satisfying a set of linear equations: • Unknowns are the weights of the edges. • We have one linear equation for any (n/2,n/2)-cut Definition: A trivial solution is any affine combination of solution (1) and (a collection of) solution (2).
Proof Overview p p 2p 0 The random graph (1) The example showing that P is not quasi random (2) Definition: A trivial solution is any affine combination of solution (1) and (a collection of) solution (2). Theorem 2:When = (½,½) the only solutions satisfying P are the trivial ones.
Proof Overview p 2p The example showing that P is not quasi random (2) 0 p The random graph (1) Note:A is an matrix. Step 1: rank(A) Definition: A trivial solution is any affine combination of solution (1) and (2). Theorem 2:When = (½,½) the only solutions satisfying P are the trivial ones. Recall Satisfying P is equivalent to satisfying a set of linear equations: 1. Unknowns are the weights of the edges. 2. We have one equation for any (n/2,n/2)-cut Definition:Let us write this as Ax=p. Proof:Any solution is a solution of the linear system Ax=p. Step 2: span[solutions (2) – (1)] has dimension n-1.
Proof Overview p 2p The example showing that P is not quasi random (2) 0 p The random graph (1) p 0 2p • This collection spans Rn-1 Definition: A trivial solution is any affine combination of solution (1) and (2). Step 2:trivial solutions have dimension n-1. Proof: For every solution (2) consider the vector of pairs (v1,vi). n/2-1 of the entries are 2p, the other are p. After subtracting (1) from these vectors, we get, for every subset S [n-1] of size n/2, a vector vS, satisfying: • vS(i) = 0 if iS. • vS(i) = p if i S .
Proof Overview Step 1:rank(A) Note:A is an matrix 1 -1 -c C (vS,t – vS) vs- t SV-S n/2n/2 n/2n/2-1 We first prove that matrix of (n/2,n/2-1)-cuts has full rank. Proof: Take the vector vS corresponding to some cut. vS,t = vector of cut obtained by moving t from S to V-S. (vS,t – vS) t
Proof Overview Step 1:rank(A) [Gottlieb ‘66]:rank(I(2,h,k)) =. Conclusion:A spans the rows of the matrix I(2,n/2,n-1) 2-element subsets of [n-1] IS,T = 1 iif ST n/2-element subsets of [n-1]
Concluding Remarks Definition:Let =(1,2,3) satisfy 0<i<1 and i=1. Let P be the following graph property: Any (1n,2n,3n)-cut is crossed by the “correct” number of K3. Coro: If (1/3,1/3,1/3) then P is quasi-random Proof: Replace every K3 with a 3-hyper edge. We get a hypergraph satisfying P, which must be quasi-random by Theorem 1. This means that in the graph, any set of vertices has the “correct” number of K3. A theorem of Simonovits-Sos implies that the graph must be quasi-random. Open Problem: What happens when = (1/3,1/3,1/3)?
Background Relation to (theoretical) computer science: 1. Conditions of randomness that are verifiable in polynomial time. For example, using number of C4, or using 2(G). 2. Algorithmic version of Szemeredi’s regularity-lemma: [Alon et al. ’95] Uses equivalence between quasi-randomenss and co-degrees.
Background Relation to Extremal Combinatorics: 1. Central in the stronghypergraph generalizations of Szemeredi’s regularity-lemma [RSSN’04, Gowers’06, Tao’06]. Quasi-Random Groups[Gowers ‘07] Generalized Quasi-Random Graphs [Lovasz-Sos ‘06] Quasi-Random Set Systems [Chung-Graham ‘91]