390 likes | 440 Views
Luby’s MIS Distributed Algorithm. Runs in time. with high probability. This algorithm is asymptotically better than the previous one. Let be the degree of node. 2. 1. At each phase :. Each node elects itself with probability. degree of in. 2. 1.
E N D
Luby’s MIS Distributed Algorithm Runs in time with high probability This algorithm is asymptotically better than the previous one.
At each phase : Each node elects itself with probability degree of in 2 1 Elected nodes are candidates for the independent set
If two neighbors are elected simultaneously, then the higher degree node wins Example: if
If both have the same degree, ties are broken arbitrarily Example: if
Using previous rules, problematic nodes are removed Problematic nodes
The remaining elected nodes form independent set
Analysis Consider phase A good event for node at least one neighbor enters 2 1
If is true, then and will disappear at the end of current phase At the end of phase
At least one neighbor of elects itself with probability at least LEMMA 1: maximum neighbor degree 2 1
PROOF: No neighbor of elects itself with probability (the elections are independent) 2 1
Therefore, at least one neighbor of Elects itself with probability at least 2 1 END OF PROOF
LEMMA 2: If a node elects itself, then it enters with probability at least 2 2 1 1
PROOF: Node enters if no neighbor of same or higher degree elect itself 2 1
Probability that some neighbor of with same or higher degree elects itself 2 1 neighbors of same or higher degree
Probability that that no neighbor of with same or higher degree elects itself 2 1 neighbors of same or higher degree
2 2 1 1 Thus, if elects itself, it enters with probability at least END OF PROOF
LEMMA 3: at least one neighbor of enters 2 1
PROOF: New event neighbor is in and no node is elected 2 1
The events are mutually exclusive
It holds: Therefore:
elects itself and no node elects itself after elects itself, it enters 2 1
after elects itself, it enters (from Lemma 2) 2 2 1 1
elects itself and no node elects itself The events are mutually exclusive
We showed earlier (Lemma 1) that: Therefore:
Therefore node disappears in phase with probability at least END OF PROOF
Let be the maximum node degree in the graph Suppose that in : Then, constant
Thus, in phase a node with degree disappears with probability at least (thus, nodes with high degree will disappear fast)
Consider a node which in initial graph has degree Suppose that the degree of remains at least for the next phases Node does not disappear within phases with probability at most
Take Node does not disappear within phases with probability at most
Thus, within phases either disappears or its degree gets less than with probability at least
Therefore, by the end of phases there is no node of degree higher than with probability at least (ineq. 2)
In every phases, the maximum degree of the graph reduces by at least half, with probability at least
Maximum number of phases until degree drops to 0 (MIS has formed) with probability at least (ineq. 2)
Total number of phases: with high probability Time duration of each phase: Total time: