1 On the Benefits of Adaptivity in Property Testing of Dense Graphs Joint work with Mira Gonen Dana Ron Tel-Aviv University

2 Let G = (V,E) be a graph.  If G has property P: accept w.h.p.  If G is “  -far” from P: reject w.h.p. “  -far”: “  -fraction of the graph” should be modified to obtain P. A Testing Algorithm for graph property P can query the graph G on the neighborhood relations of vertices in G. Graph Property Testing

3 Models used for Testing Graph Properties 1 2 … d 1 n Bounded-Degree Graphs Model [GR] : (graph is represented by n incidence lists of size d)  queries: who is i’th neighbor of v?   -far:  d  n edges should be modified.  suitable: (almost)-regular sparse graphs (in particular, constant-degree graphs) Dense Graphs Model [GGR] : (graph is represented by n x n adjacency matrix)  queries: is (u,v)  E ?   -far:  n 2 edges should be modified.  suitable: dense graphs v 1 u

4 Adaptive vs. non-adaptive testers A tester is adaptive if its queries depend on answers to previously asked queries. In the bounded degree model testers are “adaptive by nature”- [RS]. Can adaptivity be beneficial in the dense graph model ? [AFKS,GT] showed that there is at most a quadratic gap in the query complexity between adaptive and non- adaptive testers in the dense graphs model. Is there an actual a gap in the dense graphs model? We reveal a gap by considering the natural problem of testing bipartiteness.

5 Testing Bipartiteness in Dense Graphs Bipartiteness Algorithm of [GGR]: –Uniformly and independently select  (log(1/  )/  2 ) vertices in graph. –If subgraph induced by selected vertices is bipartite, then accept, otherwise, reject. Query complexity and running time of algorithm: Õ(1/  4 ). Slight variant yields Õ(1/  3 ) Improved analysis of [AK]: –Sufficient to randomly select only Õ(1/  ) vertices. The query complexity and running time : Õ( 1/  2 ) Are Õ( 1/  2 ) queries on pairs of vertices necessary?

6 Lower Bounds [BT]  (1/  3/2 ) for adaptive algorithms  (1/  2 ) for non-adaptive algorithms The lower bounds hold for graphs of small degree, that is, the degree of every vertex is  (  n 

7 Our Results Main Result: We describe an adaptive bipartiteness tester for graphs with maximum degree O(  n ) that performs Õ( 1/   ) queries. A variant of the algorithm tests the combined property of having degree O(  n ) and being bipartite. A Few notes The  (1/  2 ) lower bound of [BT] for non-adaptive algorithms holds in this low-degree case. Our tester matches the  (1/  3/2 ) lower bound of [BT] up to polylogarithmic factors in 1/  We also show that Õ( 1/   ) queries suffice when (almost) all vertices have degree  (  1/2 n).

8 Main Idea behind Algorithm: Apply techniques from the Bounded-Degree model Algorithm selects, uniformly at random a subset of vertices T, where |T|=  (log(1/  Let G T denote the subgraph induced by T. Emulate the bipartiteness tester of [GR] for bounded-degree graphs on G T. The [GR] algorithm performs O((t) 1/2 poly(log(t),1/  ) neighbor queries when run on graphs with t vertices and with dist par  (i.e., is sublinear in size of graph) Will need to show: (1) G T is  -far from bipartite for large  (distance measured w.r.t. bounded-deg model). (2) Can emulate algorithm efficiently using vertex-pair queries

9 Main Lemma: If a graph G is  -far  from being bipartite and has maximum degree O(  n ), then w.h.p over a random choice of Õ (1/  ) vertices in G, the subgraph G’ induced by the selected vertices is  ’-far from being bipartite for  ’ =  (  /log(1/  )). Analysis G  ’  far T1T1 T2T2  far for every partition (T 1,T 2 ) of T have  ’|T| 2 violating edges

10 Analysis Proving the main result from the main Lemma: 1. Assume main Lemma holds. 2. w.h.p the degree of vertices in G T  d=polylog(1 /   The number of edges that should be removed from G T is at least  d|T|, where   1/polylog (1/   G T is  –far from being bipartite in the bounded degree model.  Applying techniques from the bounded-degree graphs model we get our main result.  ’|T| 2 edges need to be removed from G T |T|= Õ (1/  ).  ’=  (  /log(1/  )). Run the [GR] algorithm on G T

11 Proof Sketch of the main Lemma  View the sample T as consisting of two subsets S and R, each of size  (log(1/  )/  ).  Def: for a partition (S 1,S 2 ) of S, an edge (u,v)  E is conflicting with (S 1,S 2 ) if u and v both have a neighbor in S 1 or both have a neighbor in S 2. S1S1 S2S2 u v

12 Proof Sketch of the main Lemma  Property 1: for every partition (S 1,S 2 ) of S, the subset R spans at least (  /16)|R| 2 edges that conflict with (S 1,S 2 ). S1S1 S2S2 (  /16)|R| 2 R In order to prove the lemma will prove that w.h.p the sample T = S  R has several properties.

13  Property 2: the maximum degree in G T is O(  |T|)=O(log(1/  )).  Claim: let G be a graph that is  –far from being bipartite and has max degree O(  n). Then w.h.p S and R have Property 1, and G T has Property 2. Proof Sketch of the main Lemma S R GTGT O(  |T|) R S1S1 S2S2 (  /16)|R| 2 GTGT O(  |T|) By one of Janson’s inequalities

14 Proof sketch of the main Lemma  Assume S,R has Property 1 and G T has Property 2.  Consider any fixed partition (S 1  R 1,S 2  R 2 ) of T.  Property 1  R spans at least (  /16)|R| 2 conflicting edges. #conflicting edges mapped to each violating edge  c’’log(1/  ) (using Property 2)  there are at least  ’’|R| 2 violating edges with respect to (S 1  R 1,S 2  R 2 ),  ’’   /( c log (1/  ))  there are at least  ’|T| 2 violating edges with respect to (S 1  R 1,S 2  R 2 ),  ’=  ’’/4. Conflicting edges mapping Violating edges Lemma: w.h.p. G T is  ’-far from being bipartite for  ’ =  (  /log(1/  )).

15  w.h.p over the choice of sample T, all vertices in G T have degree at most d=O( log(1/  ) ), and it is necessary to remove more than  ’ |T| 2 edges in order to make it bipartite, for  ’ =  (  /(log(1/  )))  G T is  -far from being bipartite in the bounded degree model, for  ’  |T| 2 /d|T| =  (1/(log(1/  ))) Emulating the Algorithm for Bounded-Degree Graphs Main Lemma + Claim

16 Emulating the Algorithm for Bounded-Degree Graphs In order to run the [GR] algorithm we have to emulate random walks by using vertex-pair queries only: To perform a random step from a vertex v: perform all queries (v,u) for u in T, and take a random neighbor.  each neighbor query takes |T| vertex-pair queries. # neighbor queries in the alg = |T| 1/2 poly(log |T|,1/    polylog (1/      total cost = polylog (1/    u1u1 v u2u2 u3u3 u d(v) T we started from polylog(1/  ) vertices in G T, performed polylog (1/  )/  1/2 random walks for each such vertex, each of length polylog(1/  )

17 An Adaptive Testing Algorithm for graphs with degree O(  n) Uniformly at random select a subset of vertices T, where |T|=  (log(1/     and let G T denote the subgraph induced by T. Uniformly and independently at random select  (log(1/   vertices from T. Let the set of vertices selected be denoted by W. For each vertex v  W, perform polylog (1/     random walks in G T, each of length polylog (1/  . If an odd-length cycle is detected in the subgraph induced by all random walks then reject, otherwise accept. Emulate the [GR] algorithm for testing bipartiteness of bounded- degree graphs (with particular setting of parameters)

18 An Adaptive Testing Algorithm for graphs with degree O(  n)  Main Theorem: The algorithm described is a testing algorithm for graphs with maximum degree O(  n). Its query complexity and running time are Õ(1/   /2 ).

19 Conclusions  Adaptive testers are stronger than non-adaptive testers in the dense graphs model. Give an adaptive bipartiteness tester for graphs for which all vertices have maximum degree O(  n ) that performs Õ (1/  3/2 ) queries.  The    lower bound of [BT] for non-adaptive algorithms holds for graphs for which the degree of every vertex is  (  n).  A variant of the algorithm tests combined property of being bipartite and having degree O (  n )  Our tester matches the  3/2 ) lower bound of [BT] up to polylogarithmic factors in 1/ . Proved that Õ ( 1/  3/2 ) queries suffice when (almost) all vertices have degree   /2 n ).

20 Thanks

21 Conclusions and Open problems Further Research:  Is there an algorithm whose complexity is Õ ((1/  3/2 ) for graphs with degree  (  n) for some  <  <  1/2 ? for all graphs? In the  (  n) case we can prove: Lemma: Let G be a graph that is  -far from being bipartite, and which all vertices have degree  (  n) for some  > . Then w.h.p over the uniform random selection of a vertex subsets S and R, each of size Õ (1/  ), the subgraph G R  S is  ’-far from being bipartite for  ’ =  (  /log(1/  )). Study other properties with a gap between adaptive and non- adaptive testers.  1/2 n nn ?

22 Proof sketch of the main Lemma  Assume R has property 1 and G T has property 2.  Consider any fixed partition (S 1  R 1,S 2  R 2 ) of T.  Property 1  R spans at least (  /16)|R| 2 conflicting edges. #conflicting edges mapped to each violating edge  c’’log(1/  )  there are at least  ’’|R| 2 violating edges with respect to (S 1  R 1,S 2  R 2 ),  ’’   /( c log (1/  ))  there are at least  ’|T| 2 violating edges with respect to (S 1  R 1,S 2  R 2 ),  ’=  ’’/4. for every partition (S 1,S 2 ) of S, the subset R spans at least (  /16)|R| 2 edges that conflict with (S 1,S 2 ). the maximum degree in G T is at most O(  |T|). Conflicting edges mapping Violating edges emulate

23 w Let (u,v) be a conflicting edge. If it is violating: (u,v)  (u,v) Otherwise: w.l.g. u in R 1, v in R 2. w.l.o.g. u has a neighbor w in S 1, v has a neighbor z in S 1 (u,v)  (u,w) Worst case: for all u all conflicting edges (u,v) are mapped to the same (u,w).  Max num of conflicting edges that are mapped to each violating edge  c’log(1/  ). Proof sketch of Main Lemma - cont Conflicts with (S 1,S 2 ) and spanned by R u,v belong to different sides of the partition (S 1  R 1, S 2  R 2 ) By property 2 u has at most c’log(1/  ) neighbors in R u,v in R 1, or u,v in R 2 (u,v) is conflicting R1R1 R2R2 S1S1 S2S2 u v conclusions

24 Conclusions and Open problems Further Research:  Is there an algorithm whose complexity is Õ((1/  3/2 ) for graphs with degree  (  n) for some  <  <  1/2 ? For all graphs? Study other properties with a gap between adaptive and non-adaptive testers and in particular remove the promise / dependence on   1/2 n nn ?