1 Maximal Independent Set

2 Independent Set (IS): In a graph G=(V,E), |V|=n, |E|=m, any set of nodes that are not adjacent

3 Maximal Independent Set (MIS): An independent set that is no subset of any other independent set

4 Maximum Independent Set: A MIS of maximum size A graph G……a MIS of G… …a MIS of max size Remark: finding a maximum IS is an NP- hard problem, while a MIS can be found in polynomial time!

5 Applications in DS: network topology In a network graph consisting of nodes representing processors, a MIS defines a set of processors which can operate in parallel without interference For instance, in wireless ad hoc networks, to avoid interferences, a conflict graph is built, and a MIS on that defines a clustering of the nodes enabling efficient routing

6 Applications in DS: network monitoring A MIS is always a Dominating Set (DS) of the graph (the converse in not true, unless the DS is independent), namely every node in G is at distance at most 1 from at least one node in the MIS (otherwise the MIS could be enlarged, against the assumption of maximality!)  In a network graph G consisting of nodes representing processors, a MIS defines a set of processors which can monitor the correct functioning of all the nodes in G (in such an application, one should find a minimum MIS)

7 A Sequential Greedy algorithm Suppose that will hold the final MIS Initially

8 Pick a node and add it to Phase 1:

9 Remove and neighbors

10 Remove and neighbors

11 Pick a node and add it to Phase 2:

12 Remove and neighbors

13 Remove and neighbors

14 Repeat until all nodes are removed Phases 3,4,5,…:

15 Repeat until all nodes are removed No remaining nodes Phases 3,4,5,…,x:

16 At the end, set will be a MIS of

17 Worst case graph (for number of phases): n nodes, n-1 phases Running time of the algorithm: Θ(m) Number of phases of the algorithm: O(n)

18 Homework Can you see a distributed version of the algorithm just given?

19 A General Algorithm For Computing MIS Same as the sequential greedy algorithm, but at each phase, instead of a single node, we now select any independent set (this selection should be seen as a black box at this stage)  The underlying idea is that this approach will be useful for a distributed algorithm

20 Suppose that will hold the final MIS Initially Example:

21 Find any independent set Phase 1: And insert to :

22 remove and neighbors

23 remove and neighbors

24 remove and neighbors

25 Phase 2: Find any independent set And insert to : On new graph

26 remove and neighbors

27 remove and neighbors

28 Phase 3: Find any independent set And insert to : On new graph

29 remove and neighbors

30 remove and neighbors No nodes are left

31 Final MIS

32 1.The algorithm is correct, since independence and maximality follow by construction 2.The number of phases depends on the choice of the independent set in each phase: The larger the subgraph removed at the end of a phase, the smaller the residual graph, and then the faster the algorithm. Then, how do we choose such a set, so that independence is guaranteed and the convergence is fast? Analysis

33 Example: If is MIS, 1 phase is needed Example: If each contains one node, phases may be needed (sequential greedy algorithm)

34 A Randomized Sync. Distributed Algorithm Follows the general MIS algorithm paradigm, by choosing randomly at each phase the independent set, in such a way that it is expected to remove many nodes from the current residual graph Works with synchronous, anonymous, uniform models Remark: It is randomized in a Las Vegas sense, i.e., it uses randomization only to reduce the expected running time, but always terminates with a correct result

35 Let be the maximum node degree in the whole graph 1 2 Suppose that d is known to all the nodes (this may require a pre-processing)

36 Elected nodes are candidates for independent set Each node elects itself with probability At each phase : 1 2

37 However, it is possible that neighbor nodes are elected simultaneously (nodes can check it out by testing their neighborhood) Problematic nodes

38 All the problematic nodes step back to the unelected status, and proceed to the next phase. The remaining elected nodes form independent set

39 Success for a node in phase : disappears at end of phase (enters or ) Analysis: 1 2 No neighbor elects itself A good scenario that guarantees success elects itself

40 Basics of Probability Let A and B denote two events in a probability space; let 1.  A (i.e., not A) be the event that A does not occur; 2.A Ո B be the event that both A and B occur; 3.A U B be the event that A or (non-exclusive) B occurs. Then, we have that: 1.P(  A)=1-P(A); 2.P(A Ո B)=P(A)P(B) (if A and B are independent) 3.P(A U B)=P(A)+P(B)-P(A Ո B) (if A and B are mutually exclusive, then P(A Ո B)=0, and P(A U B)=P(A)+P(B)).

41 Fundamental inequalities

42 Probability for a node z of success in a phase: P(success z) = P((z enters I k ) OR (z enters N(I k )) i.e., is at least the probability that it enters I k, i.e., it elects itself AND no neighbor elects itself, i.e. (notice that events are independent): 1 2 No neighbor should elect itself elects itself P(success z) ≥ p(1-p) y (recall that p=1/d)

43 Probability of success for a node in a phase: At least For First (left) ineq. with t =-1

44 Therefore, node disappears at the end of phase with probability at least 1 2  Node z does not disappear at the end of phase k with probability at most

45 after phases Definition: Bad event for node : node did not disappear This happens with probability P(AND k=1,..,4ed lnn (z does not disappear at the end of phase k)) i.e., at most: (first (right) ineq. with t =-1 and n =2ed) Independent events

46 after phases Bad event for G: at least one node did not disappear This happens with probability (notice that events are not mutually exclusive): P(OR x  G (bad event for x)) ≤

47 within phases Good event for G: all nodes disappear This happens with probability: (i.e., with high probability)

48 Total number of phases: # rounds for each phase: 3 1.In round 1, each node tries to elect itself and notifies neighbors; 2.In round 2, each node receives notifications from neighbors, decide whether it is in I k, and notifies neighbors; 3.In round 3, each node receiving notifications from elected neighbors, realizes to be in N(I k ), notifies its neighbors about that, and stops.  total # of rounds: (w.h.p.) Time complexity

49 Homework Can you provide a good bound on the number of messages?