Covers, Dominations, Independent Sets and

Matchings

AmirHossein Bayegan

Amirkabir University of Technology

Basic Concepts

Vertex Cover◦ If G(V,E) is a graph, then a vertex of G is said to

cover any edge of G that is incident with◦ A set of vertices in G that covers every edge of

G is called a vertex cover

Edge Cover◦ An edge covers its endpoints◦ A set of edges in G that covers every vertex of G

is called an edge cover

Vertex Dominating Set◦ A vertex is said to dominate both itself and any

adjacent vertex◦ A vertex dominating set is a set of vertices S

such that every vertex in G is dominated by some vertex in S

Edge Dominating Set◦ An edge is said to dominate the edges incident

to its endpoints as well as itself◦ An edge dominating set is a set of vertices X

such that every edge in G is dominated by some edge in X

Independent Set of Vertices◦ A set of vertices no two of which are adjacent

Independent Set of Edges or Matching◦ A set of edges no two of which are adjacent

Minimum cardinality of a vertex cover of G: Min-VCov(G)

Minimum cardinality of an edge cover of G: Min-ECov(G)

Minimum cardinality of a vertex dominating set: Min-VDom(G)

Minimum cardinality of a edge dominating set: Min-EDom(G)

Maximum cardinality of a matching on G: Max-Match(G)

Maximum cardinality of a vertex independent set: Max-Indep(G)

Theorem: if G(V,E) is a graph with no isolated

vertices then1. Min-EDom(G) Max-Match(G) Min-VCov(G),

and

2. Min-VDom(G) Max-Indep(G) Min-ECov(G)

Theorem (Gallai’ s Complementarity Relations):

if G(V,E) is a graph with no isolated vertices

then1. Max-Match(G) + Min-ECov(G) = |V(G)|

2. Max-Indep(G) + Min-VCov(G) = |V(G)|

We present an O(n log n) time approximation algorithm with absolute worst-case ratio , due to Chiba, Nishizeki and Saito

The algorithm is based on the following ideas: (i) we can reduce the problem on a general

planar graph to one on a planar graph having no vertices of degree 4 or less;

(ii) we can design an O(n log n) time “on-line” algorithm to execute any sequence of vertex-identifications and edge deletions of a given planar graph.

Approximation algorithm for maximum independent set in planar graphs

Lemma 7.1. Let a planar graph G = (V, E) contain a vertex v of degree 4 with N(v) = {vl , v2 , v3 , v4} , and let H be a subgraph of G induced by {v} U N(v). Then, renaming vertices in N(v) if necessary, we can assume that either

(i) H is isomorphic with H4; or (ii) H contains H’4 or H”4 as a subgraph, and

moreover (v1,v3)E

Recall:

Corollary 1.4. Every planar graph contains a vertex of degree at most 5

Theorem 7.2. For any planar graph G = (V, E) with n vertices, the algorithm finds an independent set of at least vertices.

Proof : We proceed by induction on the number n of vertices of G◦ for any planar graph having n < 4 vertices, our

algorithm always finds an independent vertex set of at least one vertex

1. Inductive hypothesis

2. algorithm always adds at least one vertex to I(G’) to form an independent set I(G)

3. the degree of v is 5 or less

1+2+3

5

n

5/|||)(| VGI

1|)(||)(| GIGI

5|||| VV

5/||)( VGI

Theorem: If G is a planar connected graph with minimum degree 3, 4, or 5, a maximum independent set of G contains less than 2n/3, 2n/4, or 2n/5 vertices, respectively.

Proof. Let I* V be a maximum independent set of G. Let G’ be the spanning subgraph obtained from G by deleting all the edges having both ends in V - I*, and assume that G’ has m’ edges. Since G‘ is a planar bipartite graph with n vertices,

d(G, v) = d(G’, v) for every vertex v in so according to the minimum degree: , or

So , or

42' nm

|*|3' Im |*|4' Im |*|5' Im *I

3/2|*| nI 4/2|*| nI 5/2|*| nI

Theorem 7.1. For any planar graph G = (V, E), the algorithm finds an independent set of G with worst-case ratio

Proof.

◦ If dm(G) 5 we use last two theorems

◦ If dm(G) 4 it suffice to show that

We consider three cases depending on dm(G):

2/1

2|)'(*||)(*| GIGI

The bound on the worst-case ratio is sharp

2/)1)(*()( GIGI

The data structure of the algorithm:

Lemma 7.2. Let G = (V, E) be a graph, and let s be any sequence of edge deletions on G. There exists an on-line algorithm to execute s in O(m) time.

Lemma 7.3. Let G = (V, E) be a graph, and let s be any sequence of identifications on G. There exists an on-line algorithm to execute s in at most O(min {mlogn, n2}) time.

Proof: The total amount of time T(s) required for s satisfies:

we shall show that T(s) is at most O(m log n):

Consider multigraphs G*i corresponding to Gi , in which the multiple edges and loops produced by identifications are not deleted. We have

The surrogate of ui , in G*i+1 has degree at least twice as large as ui , in G*i . The degree of a vertex is nondecreasing, and the degree of a vertex in G*i+1 is clearly at most 2m.

Therefore the total cost is at most 2m log2m = O(m log n)

Theorem 7.3. The algorithm spends at most O(n logn) time on a planar graph G =(V,E).

we can find a vertex v of minimum degree in a constant time

Identification O(m logn)edge-deletion O(m)deletion of an isolated vertex O(n)

Total O(n logn)

Baker’s algorithm

Approximation algorithm for the maximum independent set problem on planar graphs

Attains the worst-case ratio k/(k + 1) and runs in O(8kkn) time for any positive integer k

K Time worst-case ratio 1 linear ½loglog n O(n(1og n)3 log log n) log log n/(1 + log log n)

1. choose k as an appropriate integer

2. consider a plane embedding of a given graph G and define the level of a vertex as follows: the vertices on the outer boundary have level 1; the vertices on the outer boundary of the subgraph obtained by deleting the level i - 1 vertices have level i;

3. for each i, 0 i k , do the following: construct graph Gi = G - Vi , where Vi is the set of all the vertices of level i(mod k+1) (all the components of G, are “k-outerplanar”, that is, it has an embedding having no vertex of level > k); find a maximum independent set I*(Gi) of G, by taking union of those of the components;

4. output the largest I*(Gi) as the solution of G.

A maximum independent set I*(G) of G is partitioned into k + 1 subsets {I*(G) ∩ Vi | i = 0, 1, , k}

Clearly the smallest one contains at most 1/(k + 1) of I*(G)Therefore the largest I*(Gi) satisfies

|I*(Gi)|≥k|I*(G)|/(k + 1) worst-case ratio k/(k + 1)

Matching and processor schedulingSuppose we have set of tasks each takes a

unit time

The order the tasks can be scheduled is restricted by precedence constraints

We will show how to schedule these tasks in parallel on a pair of processors in such a way as to minimize the earliest completion of the system

We shall bound the completion time in terms of the size of matching

The optimal scheduling algorithm is as follows:

1. Identify tasks Ti1 to Tik having no successors and assign them priorities 1 to k, respectively. Say next to k+1

2. repeat (2a and b) until every task has been assigned a priority

a. For each task T all of whose successor tasks s1,…,sm

have been assigned a priority, let v(T) be an m-vector components as the priority labels of tasks in ascending order

b. Select lexicographically smallest vector-labelled task T, and set priority(T) to next. Then set next to next+1

3. Whenever a processor becomes free assign it to the highest priority task remaining to complete

31

(1,3)

(3) 2

(2,6)

(2,4)(5,8)

(7)

4

5

6

7

89

Theorem(Scheduling Lower Bound). Let a system of tasks constrained by a set of precedence constraints represented by a digraph G(V,E) be defined as shown. Let t min

denote the minimum time in which the task system can be completed. Then

tmin

M() returns a maximum matching C() returns the complement U() returns undirected graph T() returns transitive closure

|))))((((|)(| GTUCMGV

T1 T2 T5

T3 T4

Task precedence graph

T1 T4

T3T2

T5

Associated graph C(U(T(G)))

T1 T4

T3 T2

T5MM

Thanks

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