Algorithmic Graph Theory (SS2016) Chapter 8 Gossiping Walter Unger

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(1)Algorithmic Graph Theory (SS2016) Chapter 8 Gossiping. Walter Unger Lehrstuhl für Informatik 1. 14:01 , December 21, 2018.

(2) Introduction 8. Simple Graphs. Networks. Complexity. Telephone-Mode. Inhaltsverzeichnis. Contents I. 1. Introduction Recall and Motivation First Results. 2. Simple Graphs Lines Trees Graphs with Bridges. 3. Networks Cycles HQ Hypercube CCC and BF. Telegraph-Mode. Walter Unger 21.12.2018 14:01. 4. Complexity Results. 5. Telephone-Mode Even Number of Nodes Odd Number of Nodes. 6. Telegraph-Mode Upper Bound Lower Bound. 7. Summary Telefon-Mode Telegraph-Mode. Z. Sum.. SS2016. x.

(3) Introduction 8:1. Simple Graphs. Recall and Motivation. Networks. Complexity. Telephone-Mode. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Recall Definition (Gossip): Given is G = (V , E ). Each node w ∈ V has some information I (w ) and no node of V \ {w } knows I (w ). Construct algorithm, where each node v ∈ V collects information ∪w ∈V I (w ). By comm(A) we denote the complexity (number of rounds) of a communication-algorithm. r (G ) = min{comm(A) | A is a one-way algorithm for the gossip-problem on G } r2 (G ) = min{comm(A) | A is a two-way algorithm for the gossip-problem on G }. Z. Sum.. SS2016. g.

(4) Introduction 8:2. Simple Graphs. Networks. Complexity. Telephone-Mode. Recall and Motivation. Motivation Broadcast is a part of gossip. Many broadcasts have to “cooperate”. This makes the problem interesting. More important for algorithms on networks. Example: Distribute lower bounds for “Branch and Bound”. For gossip we get a difference between telegraph- and telephone-mode. We start with gossiping in the telephone-mode.. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Z. Sum.. SS2016. g.

(5) Introduction 8:3. Simple Graphs. Networks. Complexity. Telephone-Mode. First Results. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Lower Bound Lemma: Let G = (V , E ) agraph with n nodes. Then we have: dlog2 ne n even, r (G ) > r2 (G ) > dlog2 ne + 1 n odd. Proof: Only the case, where n is odd, has to be proven. Show: r2 (G ) > dlog2 ne + 1. Let A be a communication-algorithm for the gossip-problem. A has communication rounds (matchings) E1 , E2 , · · · , Ek . Show by induction: After i rounds has each node at most 2i pieces of information. i = 0: Each node has 20 = 1 pieces of information. i − 1 → i: at most 2i−1 + 2i−1 = 2i pieces of information may be collected by any node.. In round k is at least one node v inactive. v has after k rounds at most 2k−1 pieces of information.. Z. Sum.. SS2016. g.

(6) Introduction 8:4. Simple Graphs. Networks. Complexity. Telephone-Mode. First Results. Simple Algorithm Lemma: For any graph G = (V , E ) with |V | = n we have: r (G ) 6 2n − 2, and r2 (G ) 6 2n − 3. Proof: Follows from the following known statements: minb(G ) 6 n − 1 for any graph G = (V , E ) with |V | = n. r (G ) 6 2 · minb(G ) r2 (G ) 6 2 · minb(G ) − 1. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Z. Sum.. SS2016. g.

(7) Introduction 8:5. Simple Graphs. Networks. First Results. Complexity. Telephone-Mode. Telegraph-Mode SS2016. Simple Algorithm (Continuation) r (G ) r2 (G ). Lemma: We have: r (Tk (1)) = 2k r2 (Tk (1)) = 2k − 1 Proof: Show: r (Tk (1)) > 2k. r (Tk (1)) has one root and k leaves. The maximal matching is 1. In each round is only one leaf active. Each leaf has to send at least once. Each leaf has to receive at least once. Thus in total 2k rounds necessary. r2 (Tk (1)) > 2k − 1, is a simple exercise.. Z. Sum.. Walter Unger 21.12.2018 14:01. 6 6. g. 2n − 2 2n − 3.

(8) Introduction 8:6. Simple Graphs. Networks. Complexity. Lines. Telephone-Mode. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Gossip on Lines Theorem: We have: r2 (L(n)) = n − 1 for any even number n > 2, r2 (L(n)) = n for any odd number n > 3, r (L(n)) = n for any even number n > 2 and r (L(n)) = n + 1 for any odd number n > 3. Proof: Show: r2 (L(n)) > n − 1. Note: r2 (L(n)) > b(L(n)) > diam(L(n)) = n − 1. Z. Sum.. SS2016. n.

(9) Introduction 8:7. Simple Graphs. Networks. Complexity. Telephone-Mode. Lines. Gossip on Lines (Proof I). r2 (L(n)) r2 (L(n)) r (L(n)) r (L(n)). Show: r2 (L(n)) 6 n − 1 for n even. Consider algorithm A, given by the following matchings: 1 2 3 4 5 6 7 8 9. {{0, 1}, {n − 1, n − 2}}, {{1, 2}, {n − 2, n − 3}}, {{2, 3}, {n − 3, n − 4}}, ··· {{n/2 − 1, n/2}} ··· {{2, 3}, {n − 3, n − 4}}, {{1, 2}, {n − 2, n − 3}}, {{0, 1}, {n − 1, n − 2}} v0. v1. v2. v3. v4. v5. v6. v7. v8. Z. Telegraph-Mode. Walter Unger 21.12.2018 14:01. v9. = = = =. n−1 n n n+1. Sum.. SS2016. (n (n (n (n. ≡ ≡ ≡ ≡. 0 1 0 1. (mod (mod (mod (mod. n. 2)) 2)) 2)) 2)).

(10) Introduction 8:8. Simple Graphs. Networks. Complexity. Telephone-Mode. Lines. Gossip on Lines (Proof II). r2 (L(n)) r2 (L(n)) r (L(n)) r (L(n)). Show: r2 (L(n)) 6 n for n odd. Consider algorithm A, given by the following matchings: 1 2 3 4 5 6 7 8 9. {{0, 1}, {{1, 2}, {n − 1, n − 2}}, {{2, 3}, {n − 2, n − 3}}, ··· {{bn/2c, dn/2e}} ··· {{2, 3}, {n − 2, n − 3}}, {{1, 2}, {n − 1, n − 2}}, {{0, 1}} v0. v1. v2. v3. v4. v5. v6. v7. Z. Telegraph-Mode. Walter Unger 21.12.2018 14:01. v8. = = = =. n−1 n n n+1. Sum.. SS2016. (n (n (n (n. ≡ ≡ ≡ ≡. 0 1 0 1. (mod (mod (mod (mod. n. 2)) 2)) 2)) 2)).

(11) Introduction 8:9. Simple Graphs. Networks. Complexity. Telephone-Mode. Lines. Gossip on Lines (Proof II). r2 (L(n)) r2 (L(n)) r (L(n)) r (L(n)). Show: r2 (L(n)) > n for n odd. Consider the flow of messages from the left to the right node. These could not be forwarded without delay. Because we would get a time-conflict in the center. Thus at least one messages has to be delayed. This provides the lower bound. v0. v1. v2. v3. v4. v5. v6. v7. Z. Telegraph-Mode. Walter Unger 21.12.2018 14:01. v8. = = = =. n−1 n n n+1. Sum.. SS2016. (n (n (n (n. ≡ ≡ ≡ ≡. 0 1 0 1. (mod (mod (mod (mod. n. 2)) 2)) 2)) 2)).

(12) Introduction 8:10. Simple Graphs. Networks. Complexity. Telephone-Mode. Lines. Gossip on Lines (Proof III). r2 (L(n)) r2 (L(n)) r (L(n)) r (L(n)). Show: r (L(n)) 6 n for n even. Consider algorithm A, given by the following matchings: 1 2 3 4 5 6 7 8 9 10. {(0, 1), (n − 1, n − 2)}, {(1, 2), (n − 2, n − 3)}, {(2, 3), (n − 3, n − 4)}, ··· {(n/2 − 1, n/2)} {(n/2, n/2 − 1)} ··· {(3, 2), (n − 4, n − 3)}, {(2, 1), (n − 3, n − 2)}, {(1, 0), (n − 2, n − 1)} v0. v1. v2. v3. v4. v5. v6. v7. v8. Z. Telegraph-Mode. Walter Unger 21.12.2018 14:01. v9. = = = =. n−1 n n n+1. Sum.. SS2016. (n (n (n (n. ≡ ≡ ≡ ≡. 0 1 0 1. (mod (mod (mod (mod. n. 2)) 2)) 2)) 2)).

(13) Introduction 8:11. Simple Graphs. Networks. Complexity. Telephone-Mode. Lines. Gossip on Lines (Proof IV). r2 (L(n)) r2 (L(n)) r (L(n)) r (L(n)). Show: r (L(n)) > n for n even. The proof is similar to the above one: Consider the flow of messages from the left to the right node. These could not be forwarded without delay. Because we would get a time-conflict in the center. Thus at least one messages has to be delayed. This provides the lower bound. v0. v1. v2. v3. v4. v5. v6. v7. v8. Z. Telegraph-Mode. Walter Unger 21.12.2018 14:01. v9. = = = =. n−1 n n n+1. Sum.. SS2016. (n (n (n (n. ≡ ≡ ≡ ≡. 0 1 0 1. (mod (mod (mod (mod. n. 2)) 2)) 2)) 2)).

(14) Introduction 8:12. Simple Graphs. Networks. Complexity. Telephone-Mode. Lines. Gossip on Lines (Proof V). r2 (L(n)) r2 (L(n)) r (L(n)) r (L(n)). Show: r (L(n)) 6 n + 1 for n odd. Consider algorithm A, given by the following matchings: 1 2 3 4 5 6 7 8 9 10. {(0, 1)}, {(1, 2), (n − 1, n − 2)}, {(2, 3), (n − 2, n − 3)}, ··· {(bn/2c, dn/2e)} {(dn/2e, bn/2c)} ··· {(3, 2), (n − 3, n − 2)}, {(2, 1), (n − 2, n − 1)}, {(1, 0)} v0. v1. v2. v3. v4. v5. v6. v7. Z. Telegraph-Mode. Walter Unger 21.12.2018 14:01. v8. = = = =. n−1 n n n+1. Sum.. SS2016. (n (n (n (n. ≡ ≡ ≡ ≡. 0 1 0 1. (mod (mod (mod (mod. n. 2)) 2)) 2)) 2)).

(15) Introduction 8:13. Simple Graphs. Networks. Complexity. Telephone-Mode. Lines. Gossip on Lines (Proof VI). r2 (L(n)) r2 (L(n)) r (L(n)) r (L(n)). Show: r (L(n)) > n + 1 for n odd.. = = = =. n−1 n n n+1. Sum.. SS2016. (n (n (n (n. The proof is similar to the above one: Consider the flow of messages from the left to the right node. These could not be forwarded without delay. Because we would get a time-conflict in the center. Thus at least one messages (w.l.o.g. the right) has to be delayed. Now the right message has to move, because otherwise we would have already a delay of two. But now we still do get a further delay. Thus we have proven the lower bound. v0. v1. v2. v3. v4. v5. v6. v7. Z. Telegraph-Mode. Walter Unger 21.12.2018 14:01. v8. ≡ ≡ ≡ ≡. 0 1 0 1. (mod (mod (mod (mod. n. 2)) 2)) 2)) 2)).

(16) Introduction 8:14. Simple Graphs. Networks. Complexity. Trees. Telephone-Mode. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Gossip on arbitrary Trees Lemma: For any tree T we have: r (T ) = 2 · minb(T ) r2 (T ) = 2 · minb(T ) − 1 Idea of the proof: We have already for any graph G : r (G ) 6 2 · minb(G ). We have to show: r (G ) > 2 · minb(G ). Let W = ∪w ∈V I (v ) be the total information. Let A be any communication algorithm on T . Let t be the point in time, when some node knows W . Let v one node, which after t steps know W . Show: at time t only node v knows W .. Z. Sum.. SS2016. n.

(17) Introduction 8:15. Simple Graphs. Networks. Complexity. Telephone-Mode. Trees. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Gossip on arbitrary Trees (Proof I) Let u 6= v be an other node which knows W after t steps. Let (u, y1 , y2 , · · · , yk , v ) be the unique path connecting u and v . If v sends to yk at time t, then v did know W at time t − 1. So we have to consider the case: yk sends to v at time t: In this case yk sends v some missing information. yk knows at time t − 1 the full information, which has to be send from yk to v . The information, which has to be send from v to yk , is already send. Then the node yk know W at time t − 1.. Contradiction, the node u does not exist. Thus we have: t > minb(T ) = b(v , T ). u. y1. y2. y3. yk. v. Z. Sum.. SS2016. n.

(18) Introduction 8:16. Simple Graphs. Networks. Complexity. Telephone-Mode. Trees. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Gossip on arbitrary Trees (Proof II) Consider the situation at node v after round t. Let w.l.o.g. v be the root of T . Let v1 , v2 , · · · , vk be the successors of v . Let T1 , T2 , · · · , Tk be the subtrees with roots v1 , v2 , · · · , vk . In each subtree Ti is some information wi missing. Only the node v knows ∪kj=1 wj . Thus there are b(v , T ) steps to be done. We finally have r (T ) > minb(T ) + b(v , T ) > 2 · minb(T ) v. v1. v2. v3. vk. T1. T2. T3. Tk. Z. Sum.. SS2016. n.

(19) Introduction 8:17. Simple Graphs. Networks. Complexity. Telephone-Mode. Trees. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Gossip on arbitrary Trees (Proof III) Consider the two-way mode: by a similar way we may prove: At time t only two neighbours nodes u and v know the total information. We get in the similar way the second statement. u. y1. y2. y3. yk. w. v1. v2. v3. vk. T1. T2. T3. Tk. v. Z. Sum.. SS2016. n.

(20) Introduction 8:18. Simple Graphs. Networks. Complexity. Trees. Telephone-Mode. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Implication Lemma: For all m > 1 and k > 2 we have: r (Tk (m)) = 2 minb(Tk (m)) = 2 · k · m. r2 (Tk (m)) = 2 minb(Tk (m)) − 1 = 2 · k · m − 1.. Z. Sum.. SS2016. n.

(21) Introduction 8:19. Simple Graphs. Networks. Complexity. Telephone-Mode. Graphs with Bridges. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Graphs with Bridges. Z. Sum.. SS2016. Lemma: Let G = (V , E ) be a graph with bridge e ∈ E , which is separated by e in components G1 and G2 , then we have r (G ) > minb(G ) + 1 + min{minb(G1 ), minb(G2 )} Proof: Let W = ∪v ∈V I (v ) be the total information. Let t > minb(G ) the time, when a node w knows W . If w ∈ G1 hold, then do no node from G2 know W . Then there are still 1 + minb(G2 ) steps to do. If w ∈ G2 hold, then do no node from G1 know W . Then there are still 1 + minb(G1 ) steps to do. Thus we have: r (G ) > minb(G ) + 1 + min{minb(G1 ), minb(G2 )}. G1. v1. v2. G2. n.

(22) Introduction 8:20. Simple Graphs. Networks. Complexity. Telephone-Mode. Graphs with Bridges. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Z. Sum.. SS2016. Graphs with Bridges Lemma:. Let G = (V , E ) be a graph with bridge e ∈ E , which is separated by e in components G1 and G2 , then we have: r2 (G ) > minb(G ) + min{minb(G1 ), minb(G2 )} Proof: Let t > minb(G ) be the time, when node w knows W the first time. As before we may prove: Let i ∈ {1, 2}. If w ∈ Gi and v3−i does not know W , then no node from G3−i knows W . There are still 1 + minb(G3−i ) steps to do. If v1 and v2 know W at time t, then no other node knows W . There are still min{minb(G1 ), minb(G2 )} Steps to do. Thus we have: r2 (G ) > minb(G ) + min{minb(G1 ), minb(G2 )}. G1. v1. v2. G2. n.

(23) Introduction 8:21. Simple Graphs. Networks. Complexity. Telephone-Mode. Cycles. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Gossip on Cycles Theorem: We have: r2 (C (k)) = k/2 for even k. r2 (C (k)) = dk/2e + 1 for odd k. Idea of the proof (k even): [k odd: an easy exercise] Let k be even. r2 (C (k)) > k/2 results by the diameter. r2 (C (k)) 6 k/2 is true by the following algorithm: 1 2 3 4 5. {{0, 1}, {2, 3}, {4, 5}, · · · {{1, 2}, {3, 4}, {5, 6}, · · · {{0, 1}, {2, 3}, {4, 5}, · · · {{1, 2}, {3, 4}, {5, 6}, · · · ···. , {2i, 2i + 1}, · · · , {2i − 1, 2i}, · · · , {2i, 2i + 1}, · · · , {2i − 1, 2i}, · · ·. , {n − 2, n − 1} , {n − 1, 0} , {n − 2, n − 1} , {n − 1, 0}. Note: After i rounds knows each node 2 · i Informationen.. Z. Sum.. SS2016. i.

(24) Introduction 8:22. Simple Graphs. Networks. Complexity. Cycles. Telephone-Mode. Telegraph-Mode. Walter Unger 21.12.2018 14:01. 1-Way Gossip on Cycles (Idea) Messages should traverse in both directions. Activate each f (n)-th node on the cycle. This will result in an additional Θ(f (n)) steps. During the distribution we get Θ( 2·fn(n) ) delays. √ Thus we will choose f (n) = Θ( n). By this idea we may get a lower and upper bound.. Z. Sum.. SS2016. s.

(25) Introduction 8:23. Simple Graphs. Networks. Complexity. Cycles. Telephone-Mode. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Gossip on Cycles (Idea). Conflict. Z. Sum.. SS2016. s.

(26) Introduction 8:24. Simple Graphs. Networks. Complexity. Telephone-Mode. Cycles. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Gossip on Cycles (Idea of the algorithm) √ Split the cycle in Θ( n) blocks Bi .. √ Within block Bi (i ∈ {1, 2, 3, · · · , k} with k ∈ Θ( n)) do the following: Phase 1: The nodes vi [ui ] start a “wave” to the left√[right]. The messages of pvi and ui are delayed Θ( n) times by the other messages. After n/2 + Θ( n)) round know nodes zi the total information.. Phase 2:. √ Each node zi distribute the total information to Θ( n) nodes.. Note: If n is even, we have always a delay of one and the synchronization is easy. f6. f5. f4. f3. f2. f1. e6. e5. e4. e3. e2. e1. d6. d5. d4. d3. d2. d1. a1. a2. a3. a4. a5. a6. b1. b2. b3. b4. b5. b6. c1. c2. c3. c4. c5. c6. Z. Sum.. SS2016. s.

(27) Introduction 8:25. Simple Graphs. Networks. Complexity. Cycles. Telephone-Mode. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Gossip on Cycles (Idea) Theorem: We have:. √ 2n − 1 for even n. p r (C (n)) 6 dn/2e + d2 · dn/2ee − 1 for odd n. √ r (C (n)) > n/2 + 2n − 1 for even n. √ r (C (n)) > dn/2e + d 2n − 1/2e − 1 for odd n. r (C (n)) 6 n/2 +. Proof: See literature.. Z. Sum.. SS2016. s.

(28) Introduction 8:26. Simple Graphs. Networks. Complexity. Hypercube. Gossip on the Hypercube Theorem: For all m ∈ N we have: r2 (HQ(m)) = m Proof: The lower bound is the diameter. Upper bound by the following algorithm: for i = 1 to m do for all a1 , a2 , · · · , am−1 ∈ {0, 1} do in parallel a1 a2 · · · ai−1 0ai ai+1 · · · am−1 sends to a1 a2 · · · ai−1 1ai ai+1 · · · am−1 Corollary: For all m ∈ N we have: r2 (K (2m )) = m. Telephone-Mode. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Z. Sum.. SS2016. s.

(29) Introduction 8:27. Simple Graphs. Networks. Complexity. CCC and BF. Telephone-Mode. Telegraph-Mode. Walter Unger 21.12.2018 14:01. CCC and BF (Idea) Consider one-way mode: Start with the√ first phase of the gossip-algorithm for cycles on all cycles. Then √ each Θ( n)-th node on each cycle knows the total information of its cycles. In Θ( n) waves distribute this information down and between the cycles. √ After Θ(n) steps knows each Θ( n)-th node of each cycle the total information. The final part is the second phase of the gossip-algorithm of cycles on all cycles. All nodes know now the total information.. CCC,BF. Z. Sum.. SS2016. s.

(30) Introduction 8:28. Simple Graphs. Networks. Complexity. CCC and BF. Telephone-Mode. Telegraph-Mode. Walter Unger 21.12.2018 14:01. CCC and BF (Idea) Consider two-way mode: Start with the gossip algorithm for cycles on all cycles. Each node of the cycle knows now the total information of its cycle. In Θ(n/2) waves distribute this information down and between the cycles. After Θ(n) steps knows each node the total information.. CCC,BF. Z. Sum.. SS2016. s.

(31) Introduction 8:29. Simple Graphs. Networks. Complexity. CCC and BF. Telegraph-Mode. Walter Unger 21.12.2018 14:01. CCC and BF Theorem: Let k > 3, then we have: q k r (CCC (k)) 6 r (C (k)) + 3k − 1 6 d 7k e + 2 d e − 2. 2 2 q r (BF (k)) 6 r (C (k)) + 2k 6 d 5k e + 2 d k2 e − 1. 2 r2 (CCC (k) 6. k 2. + 2k = 5 · d k2 e for even k.. r2 (CCC (k) 6 d k2 e + 2k + 2 = 5 · d k2 e for odd k. r2 (BF (k) 6. Telephone-Mode. k 2. + 2k = 5 · d k2 e for even k.. r2 (BF (k) 6 d k2 e + 2k + 2 = 5 · d k2 e for odd k.. Z. Sum.. SS2016. s.

(32) Introduction 8:30. Simple Graphs. Networks. Results. Complexity. Telephone-Mode. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Complexity Definition: The two-way gossip-problem is: Given: G = (V , E ) and k ∈ N. Question: Does r2 (G ) 6 k hold. Definition: The one-way gossip-problem is: Given: G = (V , E ) and k ∈ N. Question: Does r (G ) 6 k hold.. Z. Sum.. SS2016. i.

(33) Introduction 8:31. Simple Graphs. Networks. Complexity. Results. Complexity Theorem: The two-way and one-way gossip-problem on trees is in P Proof: simple exercise. Theorem: The two-way and one-way gossip-problem is in N PC Proof: Same way as the for the broadcast-problem.. Telephone-Mode. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Z. Sum.. SS2016. n.

(34) Introduction 8:32. Simple Graphs. Networks. Complexity. Telephone-Mode. Even Number of Nodes. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Gossip on Graphs with 2 · m Nodes (0. Idea). 0000. 0001. 0010. 0011. 0100. 0101. 0110. 0111. 1000. 1001. Z. Sum.. SS2016. s.

(35) Introduction 8:33. Simple Graphs. Networks. Complexity. Telephone-Mode. Even Number of Nodes. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Gossip on Graphs with 2 · m Nodes (1. Idea). 0000. 0001. 0010. 0011. 0100. Implication: For all m ∈ N we have: r2 (K (2m )) = m For all m ∈ N we have: r2 (K (m)) 6 dlog me + 1. 0101. 0110. 0111. 1000. 1001. Z. Sum.. SS2016. s.

(36) Introduction 8:34. Simple Graphs. Networks. Complexity. Telephone-Mode. Even Number of Nodes. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Gossip on Graphs with 2 · m Nodes (2. Idea) Too many nodes where inactive for too long time. These nodes could not double their information. Idea: Try to double the information of any node. Detailed idea: In each step each node has an “interval” of information. To make the doubling easy split the nodes into two groups. Both groups should be the same size. In the first step pairs of node from each group share their information.. Z. Sum.. SS2016. s.

(37) Introduction 8:35. Simple Graphs. Networks. Complexity. Telephone-Mode. Even Number of Nodes. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Gossip on Graphs with 2 · m Nodes (2. Idea) 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 w1. w2. w3. w4. w5. w6. w7. v1. v2. v3. v4. v5. v6. v7. 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7. Z. Sum.. SS2016. s.

(38) Introduction 8:36. Simple Graphs. Networks. Complexity. Telephone-Mode. Even Number of Nodes. Gossip on Graphs with 2 · m Nodes Theorem: For all m ∈ N we have: r2 (K (2m)) = dlog 2me Proof: Split the nodes in groups Q[i] and R[i] (0 6 i 6 m − 1}). algorithm: for all i ∈ {0, · · · , m − 1} do in parallel Exchange the information between Q[i] and R[i] for t = 1 to dlog2 me do for all i ∈ {0, . . . , m − 1} do in parallel Exchange the information between Q[i] and R[(i + 2t−1 ) mod m] Invariant: Let α[i] be the information of Q[i] and R[i] after their initial exchange. After round t know nodes Q[i] and R[(i + 2t−1 ) mod m]: ∪06j62t −1 α[(i + j) mod m]. The invariant is easy to be shown.. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Z. Sum.. SS2016. s.

(39) Introduction 8:37. Simple Graphs. Networks. Complexity. Telephone-Mode. Odd Number of Nodes. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Gossip on Graphs with 2 · m + 1 Nodes (a try) 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 0 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 7 1 2 3 4 5 6 0 1 2 3 4 5 6 7 1 2 3 4 5 0 0 1 2 3 4 5 6 7 1 2 3 4 5 6 7 w1. w2. w3. w4. w5. w6. v1. v2. v3. v4. v5. v6. v7. 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 0 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 0 0 0 1 2 3 4 5 6 7 1 2 3 4 5 6 0 1 2 3 4 5 6 7 1 2 3 4 5 0 0 1 2 3 4 5 6 7 1 2 3 4 5 6 0 1 2 3 0 0 0 7. We need an extra round. A nice proof with this idea will become complicated. We will try to put some structure into the proof.. Z. Sum.. SS2016. s.

(40) Introduction 8:38. Simple Graphs. Networks. Complexity. Telephone-Mode. Odd Number of Nodes. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Gossip on Graphs with 2 · m + 1 Nodes (Idea) w1. w2. w3. w4. w5. w6. v1. v2. v3. v4. v5. v6. v7. How could this be an idea? We only have the edges of the first step. Idea: We could now choose a small even number of Nodes, which together have the total information. These nodes may perform the above gossip algorithm. In the last step we repeat the first round.. Z. Sum.. SS2016. s.

(41) Introduction 8:39. Simple Graphs. Networks. Complexity. Telephone-Mode. Odd Number of Nodes. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Gossip on Graphs with 2 · m + 1 Nodes Let n = 2 · m + 1. Let v0 , v1 , v2 , · · · , vn−1 be all nodes. For all i ∈ {0, 1, · · · , m − 1} the node vm+2+i sends to vi . The node {v0 , v1 , v2 , · · · , vm } have now the total information. If m + 1 is even, perform a gossip on the nodes {v0 , v1 , v2 , · · · , vm }. If m + 1 is odd, perform a gossip on the nodes {v0 , v1 , v2 , · · · , vm+1 }. For all i ∈ {0, 1, · · · , m − 1} the nodes vi send to vm+2+i . Correctness follows direct by the construction. Running time for m + 1 even: r2 (K (m + 1)) + 2 = dlog2 (m + 1)e + 2 = dlog2 (n + 1)e + 1. = =. . Running time for m + 1 odd: r2 (K (m + 2)) + 2 = dlog2 (m + 2)e + 2 = dlog2 (n + 3)e + 1. = =. . log2 n+1 +2 2 dlog2 ne + 1 log2 n+3 +2 2 dlog2 ne + 1. Z. Sum.. SS2016. s.

(42) Introduction 8:40. Simple Graphs. Networks. Complexity. Telephone-Mode. Upper Bound. Telegraph-Mode. Walter Unger 21.12.2018 14:01. 1st Idea (Let the Knowledge grow) 1 2 0 0 0 0 0 0 2 3 0 0 0 0 0 0 3 4 0 0 0 0 0 0 4 5 0 0 0 0 0 0 5 6 0 0 0 0 0 0 6 7 1 0 0 0 0 0 7 1 2 3 0 0 0 0 0 2 3 4 0 0 0 0 0 3 4 5 0 0 0 0 0 4 5 6 0 0 0 0 0 5 6 7 1 0 0 0 0 6 7 1 2 0 0 0 0 7 w1. w2. w3. w4. w5. w6. w7. v1. v2. v3. v4. v5. v6. v7. 0 0 0 4 5 0 7 1 0 0 0 5 6 0 0 2 0 0 0 6 7 1 0 3 0 0 0 7 1 2 0 4 0 0 0 0 2 3 0 5 0 0 0 0 3 4 0 6 0 1 0 0 4 5 6 7 1 2 0 0 5 6 7 1 2 3 0 0 6 7 1 2 3 4 0 0 7 1 2 3 4 5 0 0 0 2 3 4 5 6 0 0 0 3 4 5 6 7. We need more rounds. A nice proof with this idea will become complicated. We will try to put some structure into the proof.. Z. Sum.. SS2016. w.

(43) Introduction. Simple Graphs. Networks. Complexity. 8:41. Upper Bound. 2nd. Idea (Let the Knowledge grow in a structured way). Telephone-Mode. Telegraph-Mode. Walter Unger 21.12.2018 14:01. 1 0 0 4 5 6 7 1 2 0 0 5 6 7 1 2 3 0 0 6 7 1 2 3 4 0 0 7 1 2 3 4 5 0 0 0 2 3 4 5 6 0 0 0 3 4 5 6 7 1 0 0 4 5 6 7 1 2 0 0 5 6 7 1 2 3 0 0 6 7 1 2 3 4 0 0 7 1 2 3 4 5 0 0 0 2 3 4 5 6 0 0 0 3 4 5 6 7 w1. w2. w3. w4. w5. w6. w7. v1. v2. v3. v4. v5. v6. v7. 1 0 0 0 0 6 7 1 2 0 0 0 0 7 1 2 3 0 0 0 0 0 2 3 4 0 0 0 0 0 3 4 5 0 0 0 0 0 4 5 6 0 0 0 0 0 5 6 7 1 0 0 0 0 6 7 1 2 0 0 0 0 7 1 2 3 0 0 0 0 0 2 3 4 0 0 0 0 0 3 4 5 0 0 0 0 0 4 5 6 0 0 0 0 0 5 6 7. We need an additional two rounds. vx and wy alternate as sender and receiver. The information grows in blocks (intervals) in the nodes. With this idea we may do the proof. Only the first two rounds are special.. Z. Sum.. SS2016. w.

(44) Introduction. Simple Graphs. Networks. Complexity. 8:42. Upper Bound. 2nd. Idea (Let the Knowledge grow in a structured way). Telephone-Mode. Telegraph-Mode. Walter Unger 21.12.2018 14:01. After the first two rounds some node-pairs share their information. Consider this situation as the start: All vx and wx have one information pair. vi sends to wj and the wx have 2 information pairs. wi sends to vj and the vx have 3 information pairs. vi sends to wj and the wx have 5 information pairs. wi sends to vj and the vx have 8 information pairs. vi sends to wj and the wx have 13 information pairs. wi sends to vj and the vx have 21 information pairs. Thus the grow-rate and the algorithm is clearly visible.. Z. Sum.. SS2016. w.

(45) Introduction 8:43. Simple Graphs. Networks. Complexity. Telephone-Mode. Upper Bound. algorithm. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Z. Sum.. SS2016. w. fib(0) = fib(1) = 1 fib(i) = fib(i − 1) + fib(i − 2). Let n = 2m. Gossip-Algorithm: t := 0; for all i ∈ {0, . . . , m − 1} do in parallel R[i] sends to Q[i]; for all i ∈ {0, . . . , m − 1} do in parallel Q[i] sends to R[i]; while fib(2t + 1) < m do begin t := t + 1; for all i ∈ {0, . . . , m − 1} do in parallel R[(i + fib(2t − 1)) mod m] sends to Q[i]; if fib(2t) < m then for all i ∈ {0, . . . , m − 1} do in parallel Q[(i + fib(2t)) mod m] sends to R[i] end;.

(46) Introduction 8:44. Simple Graphs. Networks. Complexity. Telephone-Mode. Upper Bound. Telegraph-Mode. Walter Unger 21.12.2018 14:01. One-Way-Gossip. Z. Sum.. SS2016. w. fib(0) = fib(1) = 1 fib(i) = fib(i − 1) + fib(i − 2). Theorem: Let n = 2m and k = min{x | fib(x) > m}. Then we have r (K (n)) 6 k + 1. Proof: The algorithm stops, if fib(2t + 1) > m or fib(2t) > m holds. The number of rounds within the loop is 2t or 2(t − 1) + 1. The total number of rounds is (k − 1) + 2. Correctness may be proven by the following invariant: Let a[i] be the information, which share R[i] and Q[i] after two rounds. After t loops we have: Q[i] knows ∪06j6fib(2t+1)−1 α[(i + j) mod m] R[i] knows ∪06j6fib(2t+2)−1 α[(i + j) mod m]. The correctness is a direct result of this..

(47) Introduction 8:45. Simple Graphs. Networks. Complexity. Telephone-Mode. Upper Bound. One-Way-Gossip Theorem: Let n = 2m − 1 and k = min{x | fib(x) > m}. Then we have r (K (n)) 6 k + 2. Proof: Using the same idea as for the two-way mode. Theorem: Let n even. Then we have: r (K (n)) > 2 + dlog 1 (1+√5) 2n e. 2. Proof: See literature n 2 Upper Bound 2 Lower Bound 2. Telegraph-Mode. Walter Unger 21.12.2018 14:01. (Idea is 4 6 4 5 4 5. given the following). 8 10 12 14 16 6 6 7 7 7 5 6 6 7 7. 18 8 7. 20 8 7. 22 8 7. Z. Sum.. SS2016. w.

(48) Introduction 8:46. Simple Graphs. Networks. Complexity. Lower Bound. Telephone-Mode. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Idea for the lower Bound Situation: Algorithm with “fibonacci growth”. No idea to enlarge this growth.. Construction of a lower bound: Start with an arbitrary algorithm. Use only the restriction of the algorithm. Abstract.. We will now try to do the abstraction. Try the get the core-problem. The core-problem ist: “Fibonacci growth” could not be improved.. Z. Sum.. SS2016. s.

(49) Introduction 8:47. Simple Graphs. Networks. Complexity. Telephone-Mode. Lower Bound. Telegraph-Mode. Walter Unger 21.12.2018 14:01. 1. Abstraction Definition: The Network Counting Problem: Given a directed graph G = (V , E ). Each node stores a number. Initial just the number 1 is stored. The receiver add the number from the sender to his number after one communication. The objective is: all nodes should store a number larger then |V |. With nc(G ) we denote the minimal rounds to achieve this objective. Lemma: For any graph G we have: r (G ) > nc(G ).. Z. Sum.. SS2016. s.

(50) Introduction 8:48. Simple Graphs. Networks. Complexity. Lower Bound. Telephone-Mode. Telegraph-Mode. Walter Unger 21.12.2018 14:01. 2. Abstraction Let G = ({v1 , v2 , v3 , · · · , vn }, E ) be a directed Graph. Each node vi stores after t rounds the number zit . One situation of the network counting problem could be described by a vector: Initial: (1, 1, 1, · · · , 1)T . After t rounds: (z1t , z2t , z3t , znt )T .. One round of an algorithm for the network counting problem is given by a matrix B: A is a n × n matrix. aij = 1 node j sends to node i. A contains on the diagonal only ones. A has in each row at most two ones. A has in each column at most two ones. If aij = akl = 1 (i 6= j 6= k 6= l), then we have l 6= i 6= k and l 6= j 6= k. Thus we get: A · (z1t , z2t , z3t , znt )T = (z1t+1 , z2t+1 , z3t+1 , znt+1 )T. Z. Sum.. SS2016. s.

(51) Introduction 8:49. Simple Graphs. Networks. Complexity. Telephone-Mode. Lower Bound. Telegraph-Mode. Walter Unger 21.12.2018 14:01. 2. Abstraction (Continuation) We consider now matrices of the above form. These are matrices A, for which there is a transformation T with:  B 0  B   .   .  B TAT −1 =    1   .   . 0 11 and B = . 01. Z. Sum.. SS2016.        .       1. We will estimate the growth, which these matrices provide for the network counting problem.. s.

(52) Introduction 8:50. Simple Graphs. Networks. Complexity. Lower Bound. Telephone-Mode. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Recollection (Norm, 3. Abstraction) Let ||..|| be the vector norm over Rn . Then we have: ||x|| = 0 ⇔ x = 0n , ||α · x|| = |α| · ||x||, ||x + y || 6 ||x|| + ||y || this holds for all α ∈ R, x, y ∈ Rn. The matrix norm for a vector norm ||..|| is defined by ||A|| = supx6=0 ||Ax|| . Then we have: ||x|| ||A|| = 0 ⇔ A = 0 ||A + B|| 6 ||A|| + ||B|| ||αA|| = α · ||A|| ||A · B|| 6 ||A|| · ||B|| ||A · x|| 6 ||A|| · ||x|| 2 this holds for all A, B ∈ Rn , x ∈ Rn , α ∈ R, α > 0.. p n Σi=1 |xi |2 for ein x = (x1 , .., xn ), p Known: ||A|| = Spectral Norm(A) = |λmax (AT · A)| with: λmax is the largest Eigenvalue. Here we use: ||x|| =. Z. Sum.. SS2016. s.

(53) Introduction 8:51. Simple Graphs. Networks. Lower Bound. 2. Abstraction (Continuation) We compute the spectral norm: ||A|| = ||TAT −1 || = ||B||. 10 11 11 T B ·B = = . 11 01 12 ⇒ (2 − λ)(1 − λ) − 1 = 0 ⇒ λ2 − 3λ + 1 = 0 q ⇒ λmax (B T B) = 32 + 54 p √ ||A|| = λmax (AT A) = 21 (1 + 5). Complexity. Telephone-Mode. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Z. Sum.. SS2016. s.

(54) Introduction 8:52. Simple Graphs. Networks. Complexity. Telephone-Mode. Lower Bound. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Theorem: A algorithm, solving the network counting problem needs 2 + dlog 1 (1+√5) 2n e rounds. 2. Proof: Let Aj , 1 6 j 6 r be matrices, which solve the problem in r rounds. α := (α1 , α2 , · · · , αn )T = Ar −2 · ... · A2 · A1 · (1, 1, · · · , 1). rQ −2 √ √ ||α|| 6 ( ||Ai ||) · ||(1, ..., 1)|| 6 ( 12 (1 + 5))r −2 · n i=1. Let inf (i, t) be the number, which have the nodes vi after t rounds. After round t we have: inf (i, t) > n for all i ∈ {1, 2, · · · , n}. After round t − 1 we have: inf (i, t − 1) > n for at least n/2 nodes. There could be some i with: inf (i, t − 2) > n. But if αi < n and inf (i, t − 1) > n, then there exists j with: αi + αj > n.. Z. Sum.. SS2016. s.

(55) Introduction 8:53. Simple Graphs. Networks. Complexity. Telephone-Mode. Lower Bound. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Continuation. Z. Sum.. SS2016. s. α := (α1 , α2 , · · · , αn )T = Ar −2 · ... · A2 · A1 · (1, 1, · · · , 1). Let c1 be the number of cases with: αi > n, c2 be the number of cases with: αi < n and αj > n, c3 be the number of cases with: αi < n, αj < n and αi + αj > n.. Then we have: c1 > c2 and c1 + c2 + c3 > n/2. Thus we also get: 2c1 + c3 > 2n q p ||α|| = Σni=1 αi2 > c1 n2 + c3 · 2 ·. n2 4. >n·. q. 1 (2c1 2. + c3 ) >. We already have: rQ −2 √ √ ||α|| 6 ( ||Ai ||) · ||(1, ..., 1)|| 6 ( 12 (1 + 5))r −2 · n. i=1. And√we get: √ n · n 6 ||α|| 6 Φr −2 · n, 2 From which we conclude: r > 2 + dlog 1 (1+√5) 2n e 2. n√ n. 2.

(56) Introduction 8:54. Simple Graphs. Networks. Complexity. Telephone-Mode. Lower Bound. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Quality of these Bounds Lemma: Let n = 2m and let: t1 := 1 + k, with k is the smallest number with m 6 F (k) and t2 := 2 + dlog 1 (1+√5) me. 2. Then we have t1 = t2 for infinite manny m and t1 6 t2 + 1 for all m. Proof: Let Φ = 12 (1 +. √ 5).. Then we have: Φ2 = Φ + 1. Furthermore we have Φi−2 6 F (i) 6 Φi−1 for all i > 2. Consider n ∈ N with: n = 2 · F (k) for some k. Then we have: t1 = k + 1 and t2 = 2 + dlogΦ F (k)e = 2 + k − 1 = k + 1. From which we get: t1 = t2 for these n.. Z. Sum.. SS2016. n.

(57) Introduction 8:55. Simple Graphs. Networks. Complexity. Telephone-Mode. Lower Bound. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Quality of these Bounds (Part 2) Lemma: Let n = 2m and let: t1 := 1 + k, with k is the smallest number with m 6 F (k) and t2 := 2 + dlog 1 (1+√5) me. 2. Then we have t1 = t2 for ininite manny m and t1 6 t2 + 1 for all m. Proof: Setze Φ = 21 (1 +. √. 5).. Then we have Φi−2 6 F (i) 6 Φi−1 for all i > 2. Let n = 2 · m arbitrary. Let i be defined by: Φi−1 < m 6 Φi , then we have: t2 = 2 + i. Let k be the smalest number with F (k) > m. Note: Φk−2 6 F (k) 6 Φk−1 . Then we have: i = k − 1 oder i = k − 2. From which we conclude: t1 = k + 1 6 i + 3.. Z. Sum.. SS2016. n.

(58) Introduction 8:56. Simple Graphs. Networks. Complexity. Telefon-Mode. Telephone-Mode. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Summary (Telefon-Mode) Graph Kn Hk Pn Cn CCCk. |V | n 2k n n k · 2k. diam 1 k n−1 b 2n c 5k b 2 c−2. SEk BFk DBk. 2k k · 2k 2k. 2k − 1 b 3k c 2 k. Lower Bound dlog2 ne + odd(n) k n − even(n) n d 2 e + odd(n) d 5k e−2 2 5k d 2 e + 1, k odd 2k − 1 1.9770k 1.5965k. Upper Bound dlog2 ne + odd(n) k n − even(n) n d 2 e + odd(n) d 5k e − 2, k even 2 2k + 5 2.25 · k + o(k) 2k + 5. Z. Sum.. SS2016. n.

(59) Introduction 8:57. Simple Graphs. Networks. Complexity. Telephone-Mode. Telegraph-Mode. Summary (Telegraph-Mode) |V | n 2k n n even. diam 1 k n−1 b 2n c. n odd. b 2n c. Lower Bound 1.44 log2 n 1.44k n +√ odd(n) n + d 2ne − 1 2 √ d 2n e + d 2n − 12 e − 1. CCCk. k · 2k. b 5k c−2 2. d 5k e−2 2. SEk. 2k. 2k − 1. 2k − 1. BFk. k · 2k. b 3k c 2. 1.9770k. 3k +3 q d 5k e + 2 d k2 e − 1 2. DBk. 2k. k. 1.5965k. 3k + 3. Graph Kn Hk Pn Cn. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Upper Bound 1.44 log2 n 1.88k n +√ odd(n) n +d q 2ne − 1 2. d 2n e + d2 d 2n ee − 1 q k d 7k d e + 2 e −2 2 2. Z. Sum.. SS2016. n.

(60) Introduction 8:58. Simple Graphs. Networks. Complexity. Telegraph-Mode. Literatur J. Hromkovič, et al.: Dissemination of Information in Communication Networks : Broadcasting, Gossiping, Leader Election, and Fault-Tolerance. EATCS Series, Springer 2005.. Telephone-Mode. Telegraph-Mode. Walter Unger 21.12.2018 14:01. Z. Sum.. SS2016. n.

(61) Questions 9. Inhaltsverzeichnis. Walter Unger 21.12.2018 14:01. Legend n : Not of relevance g : implicitly used basics i : idea of proof or algorithm s : structure of proof or algorithm w : Full knowledge. SS2016. Z. x.

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