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Polylogarithmic Inapproximability of Radio Broadcast. Guy Kortsarz Joint work with: Michael Elkin. Radio Broadcast. Undirected graph, v V wants to broadcast A vertex receives the message if and only if exactly one its neighbors transmits. The radio broadcast problem. Given : a graph G,v
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PolylogarithmicInapproximability ofRadio Broadcast Guy Kortsarz Joint work with: Michael Elkin
Radio Broadcast • Undirected graph, v V wants to broadcast • A vertex receives the message if and only if exactly one its neighbors transmits
The radio broadcast problem • Given: a graph G,v • Find: a minimum number of rounds schedule. Let opt denote the optimum number of rounds
In the following graph the optimum is 3 Example R2 R1 R3 R3 Figure 1: opt = 3
History • Chlamtac and Weinstein, 87: O(R log2n) upper bound • Bar-Yehuda, Goldreich and Itai, Kowalski and Pelc, 04: O(R log n + log2n)upper bound • Alon, Bar-Noy, Lineal and Peleg, 89: R+(log2n)lower bound • Gaber and Mansour, 95: O(R+log5n) upper bound • Elkin and Kortsarz, 04: R+O(R1/2 log2n)=O(R+log4n) upper bound • Elkin and Kortsarz, 04: R+O(R1/2 log n + log3n)=O(R+log3n) for planar graphs
Approximation status Table1: The summary of previous and our results
A3 A4 A1 A2 a a’ B1 B2 B3 B4 b Figure 2: A MIN-REP instance Min-Rep • Input:G(A,B,E) • Given: A partition A = Ai, B = Bi
A1 A2 A3 A4 a B1 B2 B4 b Goal • Choose overall few (representative) from X A B so that |X| is minimum, and: • All “superedges” are covered B3 Figure 3: An “exact” solution
The MIN-REP hardness result In its full generality, due to Ran Raz • Yes instance is mapped to an “exact cover” • No instance: every choice of complete cover needs average of representatives per Ai, Bi
A B Figure 4: A B induces a collection of stars The star property The hardness result holds even under the assumption of the star property:
Figure 5: SET-COVER Set-Cover • Input: B(V1, V2, H) • S V1 covers V2 if N(S) = V2 • Goal: Minimum size V2 – cover
M(A,B) b’ A a b b’’ The Lund and Yanakakis L.B B Figure 6: SET-COVER
The Lund and Yanakakis L.B • Yes instance: An exact MIN-REP cover gives and exact cover • No instance: In a no instance, every cover is “large” • log( |M(A,B)| ) gap
M(A,B) b’ A a b b’’ s Figure 7: a, b, b’, b’’ are connected to a random half of the complementary half The reduction B
M(A,B) R3 R2 A a B b R1 s Figure 8: First s transmits, then A S transmits, and then B S transmits A 3 rounds schedule for a YES instance
10 Q 9 X Y Z 8 Z Q 7 P Q 6 P 5 Z P 4 Q Y 3 X Y 2 X 1 X Witnesses for NO instance Figure 9: Type 1 and Type 2 witnesses
Choose all deleted vertices as Type 1 witnesses • From every remaining round choose 2 witnesses. Type 2 witnesses • If # of rounds is O (log n), then # of witnesses is O (log n) • If v is notconnected to all Type 1 witnesses, but is connected to all Type 2 witnesses, v doesn’t get the message • Pr = 1/polfor that • Use union bound over all schedules
Open problems • Prove O (R + log2n) upper bound • If R = O (log n) can we do better than log2n approximation? • Prove R + O (log2n)(?) or opt + O (log2n)