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Independence in direct-product

graphs

Pranava K Jha

Dept of Computer Engineering

Delhi Institute of Technology: Delhi

Kashmere Gate, Delhi 110 006, INDIA

e-mail: pkj@dit.ernet.in

Sandi Klavˇ zar∗

Department of Mathematics, PEF

University of Maribor

Koroˇ ska cesta 160, 62000 Maribor, SLOVENIA

e-mail: sandi.klavzar@uni-lj.si

Abstract

Let α(G) denote the independence number of a graph G and let

G × H be the direct product of graphs G and H. Set α(G × H) =

max{α(G) · |H|, α(H) · |G|}. If G is a path or a cycle and H is a

path or a cycle then α(G×H) = α(G×H). Moreover, this equality

holds also in the case when G is a bipartite graph with a perfect

matching and H is a traceable graph. However, for any graph G

with at least one edge and for any i ∈ IN there is a graph H such

that α(G × H) > α(G × H) + i.

1Introduction

Problems of determining the independence number and the matching num-

ber of a graph are also important because of applications of these invariants

in many areas, notably, (i) selection by PRAM, (ii) VLSI layout, (iii) wire

coloring, and (iv) processor scheduling. While independence number prob-

lem is NP-hard [5], matching is solvable in polynomial time [17]. Recently

it was shown that independence number is not even approximable within a

∗This work was supported in part by the Ministry of Science and Technology of

Slovenia under the grant J1-7036.

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factor of ncfor any c > 0 unless P=NP [2, 3]. For a product graph, solving

these problems via factor graphs is economical, since problem size is much

smaller in the factors than in the product. This natural view forms the

basis of several studies, for example, [7, 13, 19].

The present paper addresses the twin problems with respect to the

direct product. The direct product G×H of graphs G and H is a graph with

V (G×H) = V (G)×V (H) and E(G×H) = {{(u,x),(v,y)}|{u,v} ∈ E(G)

and {x,y} ∈ E(H)}. This product (which is also known as Kronecker

product, tensor product, categorical product and graph conjunction) is the

most natural graph product. It is commutative and associative in a natural

way. However, dealing with this product is also most difficult in many

respects among standard products. For instance, a Cartesian product or a

strong product of two graphs is connected if and only if both factors are

connected, and this fact is easily provable. On the other hand, it is not

completely straightforward to see that G × H is connected if and only if

both G and H are connected and at least one of them is non-bipartite,

cf. [20]. Furthermore, if both G and H are connected and bipartite, then

G × H consists of two connected components.

The direct product has several applications, for instance it may be used

as a model for concurrency in multiprocessor systems [15]. Some other

applications are listed in [12].

By a graph is meant a finite, simple, undirected graph. Unless indicated

otherwise, graphs are also connected and have at least two vertices. Let

|G| stand for |V (G)|. For X ⊆ V (G), ?X? denotes the subgraph induced

by X. By χ(G), α(G) and τ(G) we will denote the chromatic number,

the independence number and the matching number of G, respectively. A

graph has a perfect matching if τ(G) = |G|/2. If G is a bipartite graph

with V (G) = V0+ V1and |V0| ≤ |V1| then a complete matching from V0to

V1is a matching which includes every vertex of V0.

The main open problem concerning the direct product is the Hedet-

niemi’s conjecture. Let χ(G × H) = min{χ(G), χ(H)}. It is easily seen

that χ(G×H) ≤ χ(G×H) holds for any graphs G and H. In 1966, Hedet-

niemi [9] conjectured that for all graphs G and H, χ(G×H) = χ(G×H).

For surveys on the conjecture we refer to [4, 14]. The chromatic number of a

graph G and its independence number are closely related via the inequality

χ(G) ≥ |G|/α(G). It is easy to see (and well-known [13, 18]) that

α(G × H) ≥ max{α(G) · |H|, α(H) · |G|} =: α(G × H).

The main topic of this paper is the study of relation between α(G × H)

and α(G × H). For instance, is it the case (analogous to the Hedetniemi’s

conjecture) that α(G × H) = α(G × H) for any two graphs G and H ? In

particular, does α(G × Cn) = α(G × Cn) hold for any graph G?

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Independence numbers of direct products have also been studied earlier.

Greenwell and Lov´ asz [6] proved that the independence number of the direct

product of k copies of Knis equal to nk−1. For some other results we refer

to [13].

In the next section we prove that for any graph G with at least one edge

and for any i ∈ IN there is a graph H such that α(G×Hi) > α(G×Hi)+i.

On the other hand we show that α(G×H) = α(G×H) when G is a bipartite

graph with a perfect matching and H is a Hamiltonian graph. In Section 3

we compute the independence numbers of the direct products of paths and

cycles. As a by-product we also give the matching numbers of these graphs.

We conclude the paper with some remarks concerning dense graphs with

high independence number related to the direct graph construction.

2α(G × H) versus α(G × H)

As we will show later, for many graphs G and H, α(G × H) indeed equals

α(G × H). However, we can answer both questions from the introduction

negatively with the next theorem.

Theorem 2.1 For any graph G with at least one edge and for any i ∈ IN,

there is a graph Hisuch that

α(G × Hi) > α(G × Hi) + i.

Proof. Let G be an arbitrary graph with at least one edge. Let |G| = n

and α(G) = k. Clearly n ≥ 2 and k ≤ n − 1.

Define a graph H(p,q) as follows. Let V (H(p,q)) = P ∪ Q, where P

induces a complete graph on p ≥ 2 vertices and Q induces an independent

set on q ≥ 1 vertices. In addition, every vertex of Q is adjacent to a fixed

vertex, say w, of P. Clearly, |H(p,q)| = p + q and α(H(p,q)) = q + 1.

Hence

α(G × H(p,q)) = max{k(p + q), (q + 1)n}.

Let u be an arbitrary vertex of G and let X be the subset of V (G×H(p,q))

defined by

X = (V (G) × Q) ∪ ({u} × (P \ {w})).

Clearly, |X| = nq + (p − 1). Furthermore, it is straightforward to see that

X is an independent set of G×H(p,q). Thus α(G×H(p,q)) ≥ nq+(p−1).

For i ∈ IN we now define

Hi= H(pi,qi) = H(n + i + 2, n(n + i)).

Then we have

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nqi+ (pi− 1)=

>

nqi+ n + i + 1

(qi+ 1)n + i

and also

nqi+ (pi− 1)=

=

=

≥

>

nqi− qi+ n(n + i) + pi− 1 − (2i + 3) + (2i + 3)

qi(n − 1) + (n + i + 2)(n − 1) + (2i + 3)

(n − 1)(pi+ qi) + (2i + 3)

k(pi+ qi) + (2i + 3)

k(pi+ qi) + i.

It follows that

α(G × Hi) > α(G × Hi) + i

and the theorem is proved.

2

Theorem 2.1 can be reinterpreted by saying that there is no graph G

which is universal in the sense that α(G × H) = α(G × H) holds for any

graph H.

A graph is called traceable if it contains a Hamiltonian path. Clearly,

paths and cycles are traceable. We are going to show that α is equal to

α for the direct product of a bipartite graph with a perfect matching and

a traceable graph. But first we need several lemmas.

Lemma 2.2 (i) For any graphs G and H, τ(G × H) ≥ 2 · τ(G) · τ(H).

(ii) α(K2× Cn) = n.

Proof. (i) If M is a matching of G and M?is a matching of H then M×M?

is a matching of G × H.

(ii) C2i× K2consists of two disjoint copies of C2iwhile C2i+1× K2is

isomorphic to C4i+2.

The bound from Lemma 2.2 (i) can be arbitrarily smaller than the

matching number of G×H. Consider, for instance, the graph K1,m×K1,n.

It consists of two connected components K1,mnand Km,n. Thus τ(K1,m×

K1,n) = 1 + min{m,n} whereas the bound from Lemma 2.2 (i) gives only

2.

2

Lemma 2.3 If G is a bipartite graph with a perfect matching and H is a

Hamiltonian graph, then

α(G × H) = α(G × H) = |G| · |H|/2.

Proof. Let S be an independent set of G × H and let e be an edge of G.

Since H contains a Hamiltonian cycle, Lemma 2.2 (ii) implies

|S ∩ (e × H)| ≤ |H|.

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As G has a perfect matching it follows that α(G × H) ≤ |G| · |H|/2.

Conversely, since G is bipartite and has a perfect matching, α(G) =

|G|/2. But then α(G × H) ≥ |G| · |H|/2.

In fact, Lemma 2.3 can be slightly generalized by assuming that G is a

graph with a perfect matching and with α(G) = |G|/2.

Lemma 2.4 If G is a bipartite graph with a perfect matching and H is a

traceable graph, then

?|H|

Proof. Since G has a perfect matching, τ(G) = |G|/2 and as H is traceable,

τ(H) = ?|H|/2?. Thus by Lemma 2.2 (i),

τ(G × H) ≥ 2 ·|G|

22

Because G is bipartite, G×H is bipartite as well. Since for bipartite graphs

τ + α equals the number of vertices (cf. [8]) we have

2

α(G × H) ≤ |G| ·

2

?

.

·

?|H|

?

= |G| ·

?|H|

2

?

.

α(G × H) = |G| · |H| − τ(G × H).

By the above we obtain

α(G × H)

≤

=

|G| · |H| − |G| ·

|G| · (|H| −

|G| ·

?

|H|

2

?

?

?

|H|

2

)

=

?

|H|

2

?

.

2

Theorem 2.5 Let G be a bipartite graph with a perfect matching and let

H be a traceable graph. Then

α(G × H) = α(G × H) = |G| · |H|/2.

Proof. If |H| is even, then the bound of Lemma 2.4 coincides with the α.

Let |H| = 2i + 1. If α(H) = i + 1, then |G| · α(H) equals the bound of

Lemma 2.4 and the proof is done also for this case.

The last case to consider is when |H| = 2i + 1 and α(H) ≤ i. Let

v1v2...v2i+1be a Hamiltonian path of H. Then, at least two vertices with

odd indices must be adjacent, for otherwise these vertices would form an

independent set of size i+1. Let v2j+1and v2k+1be adjacent vertices, j < k.

Consider now the subgraphs H1, H2 and H3 of H which are induced by

the vertices v1,...,v2j; v2j+1,...,v2k+1and v2k+2,...,v2i+1, respectively.

Note that H1and H3contain paths on 2j and 2i−2k vertices, respectively,

and that H2contains a (Hamiltonian) cycle on 2k−2j+1 vertices. Clearly,

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α(G × H) ≤ α(G × H1) + α(G × H2) + α(G × H3),

which is in turn by Lemmas 2.3 and 2.4 at most

|G|·j+|G|·(2k−2j+1)/2+|G|·(i−k) = |G|·(i+1/2) = |G|·|H|/2

and the theorem is proved.

2

We will apply Theorem 2.5 in the next section for the case of paths and

cycles. But before we close this section let us give the following connection

of our study with the Hedetniemi’s conjecture.

Proposition 2.6 Let G and H be graphs such that χ(G)α(G) = |G|,

χ(H)α(H) = |H| and α(G×H) = α(G×H). Then χ(G×H) = χ(G×H).

Proof. It suffices to show that χ(G × H) ≥ χ(G × H). We may without

loss of generality assume

χ(G) =

|G|

α(G)≥

|H|

α(H)= χ(H).

Then we have

χ(G × H)

≥

=

|G × H|

α(G × H)=

|H|

α(H)= χ(H)

min{χ(G), χ(H)} = χ(G × H).

|G||H|

|G|α(H)

=

We now give an example which illustrates Proposition 2.6. For m ≥ 1

let Gmdenote the direct product of n copies of the complete graph Kn.

Let k,s ≥ 1. Then by the result of Greenwell and Lov´ asz from [6] we have

α(Gk) = nk−1. Since χ(Gk) = n we have χ(Gk)α(Gk) = nk= |Gk|. The

same holds for Gsas well. Furthermore, using the result of Greenwell and

Lov´ asz again we obtain

α(Gk× Gs) = nk+s−1= α(Gk)|Gs| = α(Gk× Gs).

Proposition 2.6 now implies that χ(Gk+s) = χ(Gk×Gs) = χ(Gk×Gs) = n.

3Products of paths and cycles

In this section we will obtain the independence numbers for the direct

product of paths and cycles. As a byproduct we will also list the matching

numbers of these graphs.

For the path Pmand the cycle Cn, let V (Pk) = V (Ck) = {0,···,k−1}.

As we already mentioned, the graph Pm× Pn consists of two connected

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components. In addition, vertices (p,q) and (r,s) of Pm× Pn belong to

the same component if and only if p + q and r + s are of the same parity.

A component of Pm× Pn will be called an even component (resp. odd

component) if vertices (p,q) of that component are such that p + q is even

(resp. odd). Note that the even component has ?mn/2? vertices while

the odd component has ?mn/2? vertices. Further, each component has

(m − 1) · (n − 1) edges. The graph P9× P5appears in Fig. 1.

00

11

02

13

04

20

31

22

33

24

40

51

42

53

44

60

71

62

73

64

8082 84

10

01

12

03

14

30

21

32

23

34

50

41

52

43

54

70

61

72

63

74

8183

Even component

Odd component

@@

@@

@@

@@

@@

@@

@@

@@

@@

@@

@@

@@

@@

@@

@@

@@

??

??

??

??

??

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??

Figure 1: Graph P9× P5

First we present two results which might be of independent interest.

Proposition 3.1 Each component of P2i+1×P2j+1has a complete match-

ing from the smaller partite set to the bigger one.

Proof. Let i ≥ j ≥ 1. Let E1be a (maximum) matching of P2i+1with i

edges and let E2be a (maximum) matching of P2j+1with j edges. Then it

is easy to see that the matching E1×E2(which is of size 2ij) of the graph

P2i+1× P2j+1 is evenly divided between the two components. Thus the

matching number of each component of P2i+1× P2j+1is at least ij, which

coincides with the size of the smaller partite set of the even component,

and hence the result follows for this component.

For the odd component, the partite sets are {0,2,...,2i}×{1,3,...,2j−

1} and {1,3,...,2i−1}×{0,2,...,2j}, which are of cardinalities (i+1)·j

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and i·(j +1), respectively. Partition the vertex set of this component into

the following subsets: V1,V3,...,V2(i+j)−1, where V2k−1= {(p,q) | p+q =

2k−1}, 1 ≤ k ≤ i+j. Clearly, this is a well-defined partition. Further, the

reader may verify that for 1 ≤ 2k − 1 ≤ 2j − 1, ?V2k−1? and ?V2(i+j−k)+1?

are both isomorphic to P2kand that for 2j + 1 ≤ 2k − 1 ≤ 2i − 1, ?V2k−1?

is isomorphic to P2j+1. It follows that this component contains a matching

of size 2·(1+2+···+j)+(i−j)·j = (j +1)·j +(i−j)·j = (i+1)·j,

which coincides with the size of the smaller partite set of this component.

Hence the result.

2

Proposition 3.2 If m and n be both odd and m ≥ n, then Cm×Cnadmits

of a vertex decomposition into n m-cycles.

Proof. Consider the following vertex subset of Cm× Cn:

{(0,b0), (1,b1), ..., (m − 1,bm−1)},

where bi= i for 0 ≤ i ≤ n − 1, and bi= (i + 1) mod 2 for n ≤ i ≤ m − 1.

It is clear that these m vertices induce a cycle of length m, say σ0. For

i ∈ {1,...,n − 1}, consider the vertex subsets

{(0,b0+ i), (1,b1+ i), ..., (m − 1,bm−1+ i)},

where the sum bj+ i is taken modulo n. It is easy to check that these

sets of vertices induce cycles of length m, say σi, in Cm× Cn. Finally, the

cycles σ0,σ1,...,σn−1thus constructed form a partition of the vertex set

of Cm× Cn.

Theorem 3.3 If G is a path or a cycle and H is a path or a cycle, then

α(G × H) = α(G × H).

Proof. By Theorem 2.5, the result is true for all the cases except when

both G and H have odd number of vertices.

Consider first the case P2i+1× P2j+1and assume without loss of gen-

erality that i ≥ j. By Proposition 3.1, the independence number of a

component of P2i+1×P2j+1equals to the size of a larger partite set. Hence

α(P2i+1× P2j+1) = (i + 1) · (j + 1) + i · (j + 1) = (2i + 1) · (j + 1) which is

in turn equal to α(P2i+1× P2j+1).

For the case C2i+1×P2j+1we recall from [10] that this graph admits a

decomposition into two cycles of length 2·(2i+1)·j. Thus, C2i+1×P2j+1

contains a matching of size (2i+1)·j. (We note, however, that considering

subproducts of the type C2i+1×K2, we can also see that directly.) It follows

that α(C2i+1×P2j+1) ≤ (2i+1)·(2j +1)−(2i+1)·j = (2j +1)·(j +1).

This is equal to α(C2i+1× P2j+1).

The last case C2i+1× C2j+1follows immediately from Proposition 3.2.

2

2

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Note that all the products from Theorem 3.3 are bipartite with the

exception of C2i+1× C2j+1. Thus, since for bipartite graphs τ + α equals

the number of vertices, Theorem 3.3 also gives all matching numbers but

one for these graphs. The remaining matching number τ(C2i+1×C2j+1) is

equal to ((2i+1)(2j +1)−1)/2. This follows from the fact that this graph

is Hamiltonian, see [11]. Results of this section are collected in Table 1.

m

odd

odd

odd

even

even

odd

nG

Cm

Cm

Cm

Hα(G × H)

(m − 1) · n/2

mn/2

m · (n + 1)/2

mn/2

m · (n + 1)/2

m · (n + 1)/2

τ(G × H)

(mn − 1)/2

mn/2

m · (n − 1)/2

mn/2

m · (n − 1)/2

m · (n − 1)/2

odd

even

odd

even

odd

odd

Cn, m ≥ n

Cnor Pn

Pn

Cnor Pn

Pn

Pn, m ≥ n

Cmor Pm

Cmor Pm

Pm

Table 1: α and τ in products of paths and cycles

Note that each of C2i× C2j, C2i× P2j and P2i× P2j has the same

independence number. Analogous statement holds for C2i+1× C2j and

C2i+1× P2j(resp. C2i× P2j+1and P2i× P2j+1).

4Concluding remarks

Albertson, Chan and Haas [1] established a relationship between the inde-

pendence number and the odd girth of a graph. Their main result states

that if the odd girth of a graph G is at least 7 and the smallest degree

of G is greater than |G|/4 then α(G)/|G| is at least 3/7. The result is

interesting because a dense graph is suspected to have a low independence

number. Results of this paper indicate that the direct product construction

can be used to obtain examples of graphs in this direction. To make this

more precise we give the following example. Let G = K2i+1×C2i+1, and let

n = |V (G)|. It is easy to see that |E(G)| = n·(√n−1), α(G) ≥ (n−√n)/2,

τ(G) = (n − 1)/2 (since this graph is Hamiltonian), the odd girth of G is

√n, and the even girth of G is 4. Thus we have a graph with (i) density of

order√n, (ii) independence number of order n/2, (iii) high odd girth, and

(iv) low even girth. Importance of graphs having high density and high

girth appears in [16].

Finally, we want to add that Proposition 3.1 is true not only in the case

of two odd paths but also (with natural changes of the statement) for the

graphs Cm× C2j, Cm× Pnand Pm× Pn. This can be seen from Table 1.

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