Distributed Verification and Hardness of Distributed Approximation Atish Das Sarma Stephan Holzer...

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Transcript of Distributed Verification and Hardness of Distributed Approximation Atish Das Sarma Stephan Holzer...

  • Slide 1
  • Distributed Verification and Hardness of Distributed Approximation Atish Das Sarma Stephan Holzer Danupon Nanongkai Gopal Pandurangan David Peleg 1 Weizmann Google Research Liah Kor Roger Wattenhofer ETH Zurich U. of Vienna & Georgia Tech Nanyang Technological University & Brown University ETH ZurichWeizmann Amos Korman U. Paris 7
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  • PLAN 2 Result summary Techniques Overview From communication complexity to distributed algo. lower bound
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  • Distributed network 3
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  • Distributed network Distributed network A graph G of n nodes, diameter D 4 n= 4, D=2 1 1 2 2 3 3 4 4
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  • Main issue: LOCALITY and BANDWIDTH 5 ? 1 1 2 2 3 3 4 4 4 4 2 2 3 3
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  • Time complexity = number of rounds 6 1 1 2 2 3 3 4 4 log n
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  • Example: Spanning tree in O(D) time 7 1 1 2 2 3 3 4 4
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  • Weighted distributed network 8 ? 10 2 1 9 5 9 5 1 1 2 2 3 3 4 4 4 4 2 2 3 3
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  • Fundamental problems Spanning Tree Spanning Tree Broadcasting, Aggregation, etc Minimum Spanning Tree Minimum Spanning Tree Efficient broadcasting, leader election, etc. Shortest path Shortest path Routing, etc. Steiner tree Steiner tree Multicasting, etc. Many other graph problems. 9
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  • How fast can we compute distributively? 10
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  • Three points of this work 1. A new generic technique to prove lower bounds of distributed algorithms. Works for approximation algorithms. Connection to communication complexity 2. New bounds for many problems. Tight in some cases. 3. A systematic study of distributed verification. 11
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  • 12 Distributed algorithms for the above problems require (n 1/2 +D) time Distributed algorithms for the above problems require (n 1/2 +D) time
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  • Two main ingredients 1.Verification Approximation 2.Connection to communication complexity. 13
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  • 14 Showcase Minimum Spanning Tree
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  • Time of Distributed Algorithms ProblemsUpper boundLower bound Spanning tree (ST) O(D) (D) MSTO(D + n ) (D + n 1/2 ) -approx. MST (D + (n / ) 1/2 ) MST VerificationO(D + n ) (D + n 1/2 ) 15 [trivial] [Garay, Kutten, Peleg FOCS93][Peleg, Rubinovich FOCS99] [Elkin STOC04] [Kor, Korman, Peleg STACS11]
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  • Time of Distributed Algorithms ProblemsUpper boundLower bound Spanning tree (ST) O(D) (D) MSTO(D + n ) (D + n 1/2 ) -approx. MST (D + (n / ) 1/2 ) MST VerificationO(D + n ) (D + n 1/2 ) ST VerificationO(D + n ) 16 [trivial] [Garay, Kutten, Peleg FOCS93][Peleg, Rubinovich FOCS99] [Elkin STOC04] [Kor, Korman, Peleg STACS11]
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  • 17 Implication of our results
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  • Time of Distributed Algorithms ProblemsUpper boundLower bound Spanning tree (ST) O(D) (D) MSTO(D + n ) (D + n 1/2 ) -approx. MST (D + (n / ) 1/2 ) MST VerificationO(D + n ) (D + n 1/2 ) ST VerificationO(D + n ) 18 [trivial] [Garay, Kutten, Peleg FOCS93][Peleg, Rubinovich FOCS99] [Elkin STOC04] [Kor, Korman, Peleg STACS11] (D + n 1/2 )
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  • Previous lower bound proofs Deterministic : Count the number of states. Argue that the number is not enough. Randomized: Come up with a good input distributions. 19 Our proof Simple reduction from communication complexity. Avoid complication in proving randomized lower bounds.
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  • PLAN 20 Result summary Techniques Overview From communication complexity to distributed algo. lower bound
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  • 21 Approx MST lower bound (n 1/2 ) Distributed equality verification lower bound (n 1/2 ) Distributed equality verification lower bound (n 1/2 ) ST verification lower bound (n 1/2 ) Distributed equality verification lower bound (n 1/2 ) Distributed equality verification lower bound (n 1/2 ) Direct equality verification lower bound (n 1/2 ) Direct equality verification lower bound (n 1/2 ) Well-known result in communication complexity Similar to hardness of TSP Similar to lower bounds of graph streaming algorithms Three steps of reduction Distributed Algorithms Communication Complexity simulation theorem simulation theorem
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  • PLAN 22 Result summary Techniques Overview From communication complexity to distributed algo. lower bound
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  • Communication complexity of EQUALITY 23
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  • How many bits do they have to exchange? 24 Alice Bob x {0, 1} 100 y {0, 1} 100 x=y? Yes, x=y
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  • 25 One solution: Alice sends everything... time=100 Alice Bob x {0, 1} 100 y {0, 1} 100 x=y?
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  • 26 Theorem: Any algorithm needs 100 bits Alice Bob x {0, 1} 100 y {0, 1} 100 x=y?
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  • Distributed time complexity of EQUALITY 27
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  • 28 Alice x {0, 1} 100 Bob y {0, 1} 100 100 green nodes Alice and Bob are connected by many paths of length 100
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  • 29 Alice x {0, 1} 100 Bob y {0, 1} 100 100 green nodes In each step, one edge can carry one bit on each direction
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  • How many steps do they need to check whether x=y? 30
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  • 31 Alice Bob 100 green nodes A: 100 steps because the network diameter is 100
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  • Lets make the diameter smaller 32
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  • 33 Alice Bob 100 green nodes 10 green nodes Now the diameter is 30 How many steps do we need?
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  • Claim: Need > 50 steps. 34
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  • Proof: Assume there is a distributed algorithm A that uses 50 steps 35
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  • 36 AliceBob x {0, 1} 100 y {0, 1} 100 50 bits x=yx=yx=yx=y Contradiction
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  • Proof: Assume there is a distributed algorithm A that uses 50 steps 37 Goal: Show that Alice & Bob can use A to compute EQUALITY using 50 bits
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  • 38 Alice x {0, 1} 100 Bob y {0, 1} 100 x=yx=y x=yx=y ? ? ? ?
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  • 39 AliceBob x {0, 1} 100 y {0, 1} 100 Alices network Bobs network Run A
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  • 40 AliceBob x {0, 1} 100 y {0, 1} 100 x y ? ? ? ? Alices network Bobs network 0 Step Run A
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  • 41 In step 0, Alice can run A on all machines except Bobs
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  • 42 AliceBob x y ? ? ? ? 1 Step
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  • 43 AliceBob x y ? ? ? ? 1 Step
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  • 44 AliceBob x y ? ? ? ? 1 Step ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? b1b1b1b1 a1a1a1a1 b 1 = b 1 = bit sent by A run on Bobs machine
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  • 45 AliceBob x y ? ? ? ? 1 Step ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? b1b1b1b1 a1a1a1a1 a1a1a1a1 b1b1b1b1 b 1 = b 1 = bit sent by A from Bobs machine keep this
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  • 46 AliceBob x y ? ? ? ? 2 Step ? ? ? ? ? ? ? ? ? ? ? ? b2b2b2b2 a2a2a2a2 a2a2a2a2 b2b2b2b2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? b 2 = b 2 = bit sent by A from Bobs machine
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  • 47 AliceBob x y ? ? ? ? 3 Step ? ? ? ? ? ? ? ? ? ? ? ? b3b3b3b3 a3a3a3a3 a3a3a3a3 b3b3b3b3 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? b 3 = b 3 = bit sent by A from Bobs machine
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  • 48 AliceBob x y ? ? ? ? 4 Step ? ? ? ? ? ? ? ? ? ? ? ? b4b4b4b4 a4a4a4a4 a4a4a4a4 b4b4b4b4 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
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  • 49 AliceBob x y ? ? ? ? 5 Step ? ? ? ? ? ? ? ? ? ? ? ? b5b5b5b5 a5a5a5a5 a5a5a5a5 b5b5b5b5 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
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  • 50 AliceBob x y ? ? ? ? Step ? ? ? ? ? ? ? ? ? ? ? ? b 50 a 50 b 50 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? A finishes x=y
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  • 51 AliceBob x {0, 1} 100 y {0, 1} 100 50 bits x=yx=yx=yx=y Contradiction
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  • Remarks 52
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  • 53 1. By replacing 100 by n 1/2, we can reduce distributed EQUALITY to ST verification x=y?Do red edges form a spanning tree?
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  • 2. Reduce diameter... 54
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  • 55 Alice Bob n 1/2 n 1/2 paths n 1/2 n 1/2 green nodes n 1/4 n 1/4 orange nodes n 1/4 n 1/4 green nodes Diameter = n 1/4
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  • 56 Alice Bob Diameter = log n n 1/2 n 1/2 paths n 1/2 n 1/2 green nodes
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  • 3. Getting randomized lower bound EQAULITY does not give randomized lower bound. Simulation theorem holds for all functions. Reduce from communication complexity of HAMILTONIAN CYCLE [Spieker, Raz FOCS93] 57
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  • Recap 1. A new generic technique to prove lower bounds of distributed algorithms. Works for approximation algorithms. 2. New bounds for many problems. Tight in some cases. 3. A systematic study