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On the Benefits of Adaptivity in Property Testing of Dense Graphs

Joint work with Mira Gonen

Dana Ron

Tel-Aviv University

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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 Pcan query the graph G on the neighborhood relations of vertices in G.

Graph Property Testing

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Models used for Testing Graph Properties

1 2 … d

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: dn 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: n2 edges should be modified. suitable: dense graphs

v

1u

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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.

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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?

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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

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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).

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Main Idea behind Algorithm: Apply techniques from the Bounded-

Degree modelAlgorithm selects, uniformly at random a subset of vertices T, where |T|= (log(1/Let GT denote the subgraph induced by T.

Emulate the bipartiteness tester of [GR] for bounded-degree graphs on GT.

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) GT is -far from bipartite for large (distance measured w.r.t. bounded-deg model).(2) Can emulate algorithm efficiently using vertex-pair queries

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Main Lemma:

If a graph G is -farfrom 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

T1

T2

far

for every partition (T1,T2) of T have ’|T|2 violating edges

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Analysis

Proving the main result from the main Lemma:

1. Assume main Lemma holds. 2. w.h.p the degree of vertices in GT d=polylog(1/ The number of edges that should be removed from

GT is at least d|T|, where 1/polylog(1/ GT 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 GT

|T|= Õ(1/).

’= (/log(1/)).

Run the [GR] algorithm on GT

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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 (S1,S2) of S, an edge (u,v)E is conflicting with (S1,S2) if u and v both have a neighbor in S1 or both have a neighbor in S2 .

S1

S2

uv

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Proof Sketch of the main Lemma

Property 1: for every partition (S1,S2) of S, the subset R spans at least (/16)|R|2 edges that conflict with (S1,S2).

S1

S2

)/16|(R|2

R

In order to prove the lemma will prove that w.h.p the sample T = S R has several properties.

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Property 2: the maximum degree in GT 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 GT has Property 2.

Proof Sketch of the main Lemma

SR

GT O(|T|)

R

S1

S2

)/16|(R|2

GT

O(|T|)

By one of Janson’s inequalities

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Proof sketch of the main Lemma

Assume S,R has Property 1 and GT has Property 2.

Consider any fixed partition (S1R1,S2R2) 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 (S1R1,S2R2), ’’ /(c log (1/))

there are at least ’|T|2 violating edges with respect to (S1R1,S2R2), ’= ’’/4.

Conflicting edges mapping Violating edges

Lemma: w.h.p. GT is ’-far from being bipartite for ’ = (/log(1/)).

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w.h.p over the choice of sample T, all vertices in GT 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/)))

GT 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

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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|1/2poly(log |T|,1/ polylog (1/

total cost = polylog (1/

u1

v

u2

u3

ud(v)

T

we started from polylog(1/) vertices in GT, performed polylog (1/)/1/2 random walks for each such vertex, each of length polylog(1/)

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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 GT 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 GT , 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)

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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).

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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).

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Thanks

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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 GRS is ’-far from being bipartite for ’=(/log(1/)).

Study other properties with a gap between adaptive and non-adaptive testers.

1/2nn?

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Proof sketch of the main Lemma

Assume R has property 1 and GT has property 2.

Consider any fixed partition (S1R1,S2R2) 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 (S1R1,S2R2), ’’ /(c log(1/))

there are at least ’|T|2 violating edges with respect to (S1R1,S2R2), ’= ’’/4.

for every partition (S1,S2) of S, the

subset R spans at least (/16)|R|2

edges that conflict with (S1,S2).

the maximum degree in GT is at most O(|T|).

Conflicting edges mapping Violating edges

emulate

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w

Let (u,v) be a conflicting edge.

If it is violating: (u,v) (u,v)

Otherwise: w.l.g. u in R1, v in R2 .

w.l.o.g. u has a neighbor w in S1, v has a neighbor z in S1

)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 (S1,S2) and

spanned by R

u,v belong to different sides

of the partition

(S1R1, S2R2)

By property 2 u has at

most c’log(1/) neighbors

in R

u,v in R1 ,or u,v in

R2

)u,v (is conflicting

R1R2

S1

S2

u

v

conclusions

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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

1/2nn?