Testing the independence number of hypergraphs Michael Langberg California Institute of Technology

k-uniform hypergraph I G=(V,E) Each edge is of size k. I  IS: no edges included in I. (G) = size of max IS.

Property testing of (G) Input: G=(V,E). Goal: distinguish between 2 cases G has a large IS. G is far from having a large IS. Design efficient (,)-distinguishing algorithm: Case I: (G)  n. Case II: Must remove at least nk edges from G for it to have an IS of size n (-far). Efficient = few samples of G (constant).

Naïve PT algorithm Let s be constant (will depend on  and ). Sample s of vertices of G randomly = H. Compute (H). If (H)  s declare “case I” o.w. “case II”. Like to prove: (G)  n (H)  s. G -far from having (G)  n (H) < s. Case I: (G)  n Case II: G -far from (G)  n

Naïve PT algorithm (G)  n (H)  s Exp. |I H| = s. G -far from (G)  n (H) < s. (G) < n (H) < s. Must use “-far” property of G. H I =½ =1

Our result Let G be -far from (G)  n. H random subgraph of G. Thm: If |H| > exp(k)2k/3 then w.h.p. (H) < |H|. Repeating Naïve alg: (G)  n Pr[Output = Case I] = large. G -far from (G)  n Pr[Output = Case I] = small.

Previous work Testing (G) in standard graphs (2-uniform): [Goldreich,Goldwasser,Ron]: prove similar theorem for |H| ~ /4. [Feige,L,Schechtman]: improve to |H|~ 4/3. PT of hypergraphs: Chromatic number: Considered by [Czumaj,Sohler]. Max-k-CNF: [Alon,Fernandez de la Vega,Kannan,Karpinski]. [Alon,Shapira],[Frieze,Kannan],[Andersson,Engebretsen] We combine ideas from [FLS] and [AS] to obtain: Thm: G -far, HrG, |H|~2k/3 then w.h.p. (H) < |H|.

Remainder of talk Present [FLS] proof paradigm for testing of (G) in standard graphs. Present our proof.

[FLS] proof Let G be -far from (G)  n. Let H be large random sample of G. Thm: W.h.p. (H) < |H|. Let R be random subgraph of G. Analyze Pr[R is IS]. Can be used to prove Thm. Use union bound on all large R in H. Pr[(H)  |H|] ≤ #(RH, |R|=|H|)Pr[R is IS]. Pr[R is IS] = ? R H G=(V,E)

[FLS] proof cont. ( ) R G may have an IS of size ~ n. Let G be -far from (G)  n. Let R be random subset of G. Pr[R is IS] = ? G may have an IS of size ~ n. Thus Pr[R is IS] > |R|. [FLS] show that this is “tight”. Union bound for |H|=s, |R|=s: Pr[(H)  s] ≤ [FLS] fix this by considering Pr[R is a maximum IS in H]. R IS G=(V,E) |H| |R| ( ) |R| = s s s = large !!

What about hypergraphs? Let G be -far from (G)  n. Let R be random subset of G. Pr[R is IS] = ? We show that Pr[R is IS] ~ |R|. Use ideas of [AS]: formalize “set of neighbors” (R) = set system of all subsets “adjacent” to R. A adjacent to R:  edge that consists of A portion of R. (R) = { { },{ },{ } }.

Proof: Pr[R is IS]~|R| Consider choosing the set R one by one. R is IS iff each inter. subset is IS. If R is IS, then most steps: vertices of R must be chosen from subset of size n. Initially, R0 = and (R0) = . Consider new random vertex v. Lemma: At each step, v must be chosen from a set of size n, otherwise “size” of (Riv)) >> “size” of (Ri). Corollary: The size of (R) is bounded. For “large” R, most vertices v must be chosen from a set of size n. (R) R v

Proof of lemma Ri v Consider step i: Ri = {r1,…,ri}. Define “degree” of vertex v as: dv = size of (Riv) – size of (Ri). Claim: only n vertices have low degree. Prf: Look at n vertices of lowest degree, the induced subgraph has many edges  vertex of high degree. Lemma: At each step, v must be chosen from a set of size n otherwise size of (Riv) >> size of (Ri). v High deg. dense subgraph n of low deg.

Concluding remarks PT algorithm with sample size ~ 2k/3. Lower bound of ~ 1/2. Similar gap between 1/2 (upper) and 1/ (lower) exists in the case of testing chromatic number. Thanks!