Testing the independence number of hypergraphs

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

2. 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.

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

4. 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

5. 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

6. 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.

7. 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|.

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

9. [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] = ? H R G=(V,E)

10. s s |H| |R| ( ) ( ) s |R| = [FLS] proof cont. R • 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]. IS G=(V,E) = large !!

11. What about hypergraphs? R • 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. • Aadjacent to R:  edge that consists of A portion of R. • (R) = { { },{ },{ } }.

12. Proof: Pr[R is IS]~|R| (R) 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 chosenfrom 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. v

13. Proof of lemma Ri • 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 chosenfrom a set of size n otherwise size of (Riv) >> size of (Ri). v High deg. dense subgraph n of low deg.

14. 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!