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 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).

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] = ? R H
G=(V,E)

10. [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 !!

11. 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) = { { },{ },{ } }.

12. 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

13. 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.

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!