Friday, April 11, 2008

* Maximal Independent Set

MAXIMAL INDEPENDENT SET 

A maximal independent set is a maximal set of nodes in a graph, no two
of which are adjacent.  Note that there are several maximal
independent sets, and a maximal set may be much smaller than a maximum
set.  E.g., consider the star graph.

Given a coloring with C colors computed in T time, an MIS can be
obtained in an additional C time.  In step i, a node with color i adds
itself to the MIS if it has no adjacent node already in the MIS.  So
total time = T + C.  Thus, for bounded-degree graphs and trees, we
have an O(log^* n) algorithm for computing the MIS.

Luby's algorithm for general graphs:

Repeat:

1. Node v marks itself with probability 1/(2d(v)), where d(v) is
current degree of node.

2. If no higher degree neighbor of v is also marked (breaking ties by
ID), node v joins the MIS.  Otherwise, node v unmarks itself.

3. If a node has a neighbor in MIS, it removes itself for future
rounds.

We will prove that the above algorithm computes an MIS in O(log n)
rounds with high probability.

Lemma: A node v joins the MIS with probability at least 1/(4d(v)).

Proof: Probability that node is marked in step 1 is 1/(2d(v)).  The
probability that a neighbor w is marked is at most 1/(2(d(w))).  We
have to only worry about marked nodes with degree at least d(v) since
other nodes will be eliminated.  So the probability that any neighbor
with higher degree will be marked is at most d(v)*(1/(2d(v))) = 1/2,
giving the probability of v joining to be at least 1/(4d(v)).

End Proof

Call a node good if Sum_{w neighbor of v} 1/(2d(w)) >= 1/6.  Call an
edge (u,v) good if either u or v is good.

Lemma: A good node will be removed with probability at least 1/36.

Proof: Fix a good node v.  If v has a neighbor of degree 2, then it is
easy.  So assume for the remainder that all neighbors of v have degree
at least 3.  There exists a subset X of neighbors w such that sum of
1/(2d(w)) is at least 1/6 and at most 1/3.  Let P be the probability
that v will be removed.

P >= Pr[there exists u in X such that u is in MIS]

>= Sum_{u in X} Pr[u in MIS] - Sum_{u!=w, both in X} Pr[u in MIS and w
in MIS]

>= Sum_{u in X} (1/4d(u)) - Sum_{u!=w, both in X} Pr[u marked and w marked]

>= Sum_{u in X} (1/4d(u)) - Sum_{u,w in X} Pr[u marked]*Pr[w marked]

>= Sum_{u in X} (1/4d(u)) - (Sum_{u in X} 1/(2d(u)))*(Sum_{u in X} 1/(2d(u))) 

>= Sum_{u in X} (1/2(d(u))) [1/2 - Sum_{u in X} 1/(2d(u))]

>= Sum_{u in X} 1/6 * [1/2 - 1/3] = 1/36.

End Proof

Lemma: Any time, at least half the edges are good.

Proof: Direct edges toward higher degree nodes.  We show that bad
nodes have outdegree at least twice their indegree.  Otherwise, at
least 1/3 of its neighbors have degree at most d(v), but then 

Sum_{w neighbor of v} 1/(2d(w)) >= (d(v)/3)*(1/(2d(v))) = 1/6, 

a contradiction.

Number of edges directed into bad nodes is at most half the number of
edges directed out of bad nodes, thus at most half the total number of
edges.  Thus, the number of edges directed into good nodes is at least
half the number of edges.  These edges are all good.

End Proof