Università
della
Svizzera
italiana
Network Algorithms
Prof. F. Kuhn
Spring 2010
Sample Solution 6
Exercise 1 (Dominating Sets in Special Graphs)
a) Because in a ring, no node in the dominating set can cover more than 3 (including itself)
nodes, every dominating set of an n-node ring contains at least dn/3e nodes. Further since
no two neighbors can be in a maximal independent set (MIS), an MIS of a ring contains at
most bn/2c nodes. Because every MIS is also a dominating set, computing an MIS gives a
3/2-approximation for the minimum dominating set problem in rings. We have seen in the
first chapter (and exercises) that an MIS can be computed in O(log∗ n) rounds in a ring.
b) The simplest possible algorithm is to let each node decide to join a set S independently with
probability p. Let us see how large we have to choose p such that the resulting set of nodes S
is a dominating set with probability at least 1 − 1/n. For a particular node u, the probability
that u is not in S and no neighbor of u is in S is (1 − p)d+1 . By a union bound, we therefore
d+1
get that the probability
othat one of the n nodes is not covered is at most n(1 − p) . We
n
ln n
. By using that 1 + x ≤ ex for all x ∈ R, we then get
choose p = min 1, 2d+1
n(1 − p)d+1 ≤ ne−p(d+1) ≤ ne−2 ln n =
1
.
n
Hence, for the given choice of p, S is a dominating set with probability 1−1/n. The expected
size of S is
2n ln n
.
E[|S|] = np ≤
d+1
Because no node can cover more than d + 1 nodes, every dominating set has size at least
n/(d + 1). Therefore, the algorithm computes an 2 ln(n)-approximation.
Remarks: Note that |S| is a binomial random variable. By using standard bounds on the
concentration of binomial random variables, it can also be shown that the approximation
ratio is O(log n) with high probability. The expected approximation ratio can be improved
by choosing a smaller value for p (p = min {1, ln(d + 1)/(d + 1)}) and adding all uncovered
nodes to S in a second step.
Exercise 2 (Vertex Cover Approximation)
a) Let m = nd/2 be the number of edges of G. Because a node with degree δ can cover at most
its δ edges, a vertex cover of a d-regular graph needs to have at least m/d = n/2 nodes. The
set V of all nodes is therefore a 2-approximation in d-regular graphs.
b) Using the same argument as before, no node can cover more than ∆ edges. Because all the
nodes in A have degree at least ∆/2 and because each edge is incident to at most 2 nodes in
A, G has at least |A|∆/4 edges. A dominating set of G therefore needs to have size at least
|A|/4. The set A can therefore be at most 4 times as large as the smallest vertex cover of G.
c) Assume that the largest degree in G is ∆ and for simplicity assume that the nodes know ∆.
We get a distributed vertex cover algorithm as follows. The algorithm consists of O(log ∆)
phases. Initially the vertex cover is empty. In Phase i (the first phase is Phase 1), nodes with
degree at least ∆/2i join the vertex cover. After each phase, we remove all the edges that
have been covered. Note that because in Phase i, all nodes with degree at least ∆/2i join
the vertex cover and all their incident edges are therefore removed at the end of Phase i, the
maximum degree at the beginning of Phase i+1 is less than ∆/2i . Hence, by the observation
in the b), in each phase, the number of nodes added to the vertex cover is at most 4 times
as large as the size of the smallest vertex cover of G. Consequently, the algorithm computes
an O(log ∆)-approximation in O(log ∆) rounds.