On the Sizes of Maximal Independent Sets of Cylindrical Grid Graphs

ABSTRACT.

If a graph G has exactly t different sizes of maximal independent sets, G belongs to a collection called ℳ_t . For the Cartesian product of the graph P_n , the path of length n, and C_m , the cycle of length m, called cylindrical grid, we present a method to find maximal independent sets having different sizes and a lower bound on t, such that these graphs belong to ℳ_t .

Keywords:
Well-covered graph; maximal independent set; Cartesian product

RESUMO.

Se um grafo G tem exatamente t tamanhos diferentes de conjuntos independentes maximais, G pertence a uma coleção chamada ℳ_t . Para o produto Cartesiano do grafo P_n , o caminho de tamanho n, e C_m , o ciclo de tamanho m, chamado grade cilíndrica, apresentamos um método para encontrar conjuntos independentes maximais com diferentes tamanhos e um limite inferior para t, tal que estes grafos pertençam à ℳ_t .

Palavras-chave:
Grafo bem-coberto; conjunto independente maximal; produto Cartesiano

1 INTRODUCTION

In ⁹ Plummer defines a graph to be well-covered if all its maximal independent sets have the same size. Generalizing this concept, Finbow, Hartnell, and Whitehead ⁴ define, for every t ∈ ℕ, the set ℳ_t as the set of graphs that have maximal independent sets of exactly t different sizes. With this notation, ℳ₁ is precisely the set of well-covered graphs.

Well-covered graphs have been investigated from several different parameters. See the survey ¹⁰, for more details. Topp and Volkmann ¹¹ proposed the following question about Cartesian products and well-covered graphs: "Do there exist non well-covered graphs whose Cartesian product is well-covered? ". This question was partially answered by Fradkin ⁵ for some classes of triangle-free graphs and recently a negative answer was given by Hartnell and Rall ⁷, for arbitrary graph. For G ∈ ℳ_t and v ∈ V(G), Barbosa and Hartnell ² determine the extreme values that r can assume where G\v belongs to ℳ_r . Additional properties of graphs in this class are given in ¹. Results on ℳ_t , for t ≥ 2, related to graphs without small cycles are also given in ^{3), (4), (6}.

The Cartesian product P_n □ C_m is a cylindrical grid graph. Every maximal independent set in a graph is a dominating set, although the converse is not always true. In ⁸, methods to find the domination number of cylindrical grid graphs P_n □ C_m with m ≥ 3 and n = 2,3, and 4 were proposed. Moreover, bounds on the domination numbers were found when n = 5 and m ≥ 3. We present a method to find maximal independent sets having different sizes and a lower bound for t, such that P_n □ C_m belongs to ℳ_r .

Before we present our results and proofs, we summarize our notation.

We consider finite, simple, and undirected graphs. For a graph G, the vertex set and the edge set are denoted V(G) and E(G), respectively. For a vertex u of G, its neighbourhood is denoted N_G (u) and its closed neighbourhood denoted N_G [u] is the set N_G (u) ∪ {u}. For a set U of vertices of G, let

A set D ⊆ V(G) of a graph G is a dominating set if every vertex v ∈ V(G)\D is adjacent to some vertex u ∈ D. The domination number of G is the cardinality of a smallest dominating set of the graph G and is usually denoted by γ(G).

A set I of vertices of a graph G is independent if no two vertices in I are adjacent. An independent set I of G is maximal if every vertex u in V(G)∖I has a neighbour in I. An independent set I of G is maximum if G has no independent set J with |J| > |I|. We denote i(G) the cardinality of a smallest maximal independent set in G. We denote α(G) the cardinality of a maximum independent set in G and ms(G) the set of all sizes of maximal independent sets in G. Hence, if G is well-covered, then i(G) = α(G) and |ms(G)| = 1.

For any two graphs G and H, the Cartesian product G □ H is the graph with vertex set V(G) × V(H), such that two vertices (u ₁, v ₁) and (u ₂, v ₂) of G □ H are adjacent whenever v ₁ = v ₂ and u ₁ = u ₂ ∈ E(G) or u ₁ = u ₂ and v ₁ = v ₂ ∈ E(H). Here, the vertices of the path P_n or the cycle C_n are always denoted 0, 1, ..., n - 1. For the graph P_n □ C_m , we denote (C_m )_i the graph C_m □ {i}, with i ∈ V(P_n ).

For naturals a, b, and c, with c > b + 1, we shall denote the set {i: a ≤ i ≤ b} ∪ {c} by {a, ..., b, c}.

2 RESULTS

In ⁸ Nandi, Parui, and Adhikari establish the following result regarding the domination number of Cartesian product of paths and cycles.

Theorem 1. ⁽⁸⁾ For all m ≥ 3,

γ(P ₂ □ C_m ) =

γ(P₃ □ Cm) = ;

γ(P₄ □ Cm) =

for m ≥ 6, m + P ₅ □ C_m ) ≤ m + ≤ γ(.

One of our main results is the following. It will be proved in Section 2.2.

Theorem 2. Let n ≥ 3 and m ≥ 4. IfG = Pn □ Cm, then G ∈ ℳt for some

We begin showing the size of a maximum independent set in a graph Pn □ Cm.

Proposition 3. For n ≥ 2 and m ≥ 3, α(Pn □ Cm) = n.

Proof. Since α(Cm) = n cycles C_m , we conclude that α(P_n □ C_m ) ≤ nP_n □ C_m ) ≥ nI be the set of vertices (i, j) of P_n □ C_m such that i = 0, ..., n - 1, j = 0, ..., 2i + j is an odd integer. The set I has cardinality nP_n □ C_m . □ presenting a maximal independent set with this cardinality. Let and it is a maximal independent set in and we have - 1, and . Now, we show that α(

In Theorem 7, we determine t such that P ₂ □ C_m belongs to ℳ_t . Before, we prove some preliminary results.

Lemma 4. Let m ≥ 3 and G = P2 □ Cm. If I is a maximal independent set of G, then I has even cardinality.

Proof. Let I be a maximal independent set in G, X_i = I ∩ (C_m )_i and x_i = |X_i |, for i = 0, 1. We show that x ₀ = x ₁. By symmetry, consider the subgraph induced by (C_m )₀\X ₀, denoted by H. The graph H has exactly x ₀ disjoint connected paths. We denote them by P(k), k = 1, ..., x ₀. If |V(P(k))| > 3, for some k ∈ {1, ..., x ₀}, (C_m )₀\N_G [X ₀] has some path with at least two vertices, which implies at least two consecutive vertices in I ∩ (C_m )₁. Thus we may assume |V(P(k))| ≤ 3, k = 1, ..., x ₀.

Fix some k ∈ {1, ..., x ₀}. Let (v ₁, v ₂, ..., v_q ), with q = |V(P(k))|, be one of the two orderings of the vertices of P(k) such that adjacent vertices are consecutive. For each j ∈ {1, 2, ..., |V(P(k))|}, let v'_j be the neighbour of v_j in (C_m )₁. Note that exactly x ₀ vertices in (C_m )₁ are in N_G (X ₀) and they are separated by at most three vertices. We have three cases for each P(k). If |V(P(k))| = 1, v'₁ must belong to I since its neighbours in the same cycle are dominated by I. If |V(P(k))| = 2, v ₁ and v ₂ belong to N_G (I) and exactly one of v'₁ and v'₂ must belong to I. If |V(P(k))| = 3, the vertices, v ₂, v'₁, v'₂, and v'₃ are not in N_G (X ₀). Hence v'₂ must be in I. Therefore, in all cases, for k = 1, ..., x ₀, for every path P(k), there is exactly one vertex in X ₁, resulting in |I ∩ (C_m )₀| = |I ∩ (C_m )₁|. □

Lemma 5. Let n ≥ 2, m ≥ 3 and G = P_n □ C_m. If I is a maximal independent set in G, then

for i {0, n - 1}.

Proof. Let I be a maximal independent set in G, X_i = I ∩ (C_m )_i and x_i = |X_i |, for i ∈ {0, n - 1}. Suppose x_i < C_m )₀ are in N_G [X ₀], at least N_G [X ₀]. Moreover, H has at most x ₀ disjoint connected paths. Therefore, at least one of these paths has at least two vertices. Since I is maximal, two adjacent vertices in (C_m )₁ are in I. This contradicts the independence of I and completes the proof. □ - 3 vertices of (, for some i ∈ {0, n - 1}. By symmetry, we may assume i = 0. Consider the graph (Cm)0 and the subgraph H induced by (Cm)0\NG[X0]. Since at most 3 of those vertices are not in

Proposition 6. For m ≥ 3, i(P ₂ □ C_m) = 2.

Proof. Let J = {(0, i) : i ≡ 1 (mod 4)} ∪ {(1, i) : i ≡ 3 (mod 4)}. We consider a set I with I = J, if m ≡ 0 (mod 4), I = J ∪ {(1, 0), (0, m - 1)}, if m ≡ 1 (mod 4), and I = J ∪ {(1, 0)}, otherwise.

Note that I is a maximal independent set of P ₂ □ C_m , and |I| = 2i(P ₂ □ C_m ) ≤ 2I ∩ (C_m )₀| + |I ∩ (C_m )₁| ≥ 2. Hence, and the desired statement follows. □. By Lemma 5, |

Now, we can show the quantity of different sizes of maximal independent sets in P ₂ □ C_m .

Theorem 7. For m ≥ 3 and G = P ₂ □ C_m, G ∈

Proof. By Proposition 3, α(G) = 2 and 2. We present maximal independent sets of G with every even cardinality between 2, and by Proposition 6, i(G) = 2, in view of Lemma 4.

First, let l =

and

Let I(0)' = {(1, i), i ≡ 0 (mod 4)}, i ≤ l} ∪ {(0, i), i ≡ 2 (mod 4),I ≤ l} and

Note that the set I(0) is a maximal independent set in G and |I(0)| = 2i(G). For every j ∈ {1, ..., k}, let =

I(j) = I(j - 1)∖{(0, 4j - 2)} ∪ {(0, 4j - 3), (0, 4j - 1), (1, 4j - 2)}.

For every j ∈ {0, ..., k}, I(j) is a maximal independent set in G and |I(j)| = 2j. + 2

Hence G has maximal independent sets of k + 1 different sizes which results in m (mod 4), or if 1 ≡ + 1, otherwise. This completes the proof. □

By Theorem 2, when n ≥ 3 and m ≥ 4, P_n □ C_m is not well-covered. The graph P ₂ □ C ₅ is well-covered and also P_n □ C ₃ for n ≥ 2, as we show in next proposition.

Proposition 8. For n ≥ 2, if G = P_n □ C ₃ , then G is well-covered.

Proof. By Proposition 3, α(G) = n. Let I be a maximal independent set in G. We show that |I| = n. For a contradiction, suppose |I| < n. Hence there is some j ∈ {0, ..., n - 1}, with I ∩ (C ₃)_j = Ø. Let S denote the set N_G (V((C ₃)_j )). Since I is maximal and from the structure of G, |S ∩ I| ≥ 3. But S induces one or two cycles C ₃, which contradicts the independence of I.

Proposition 9. Let n ≥ 3, m ≥ 4 and G = P_n □ C_m. If m is even, then there is no maximal independent set I in G such that |I| = - 1;

Proof. By Proposition 3, α(G) = C_m )_i , for I = 0, ..., n - 1. Suppose that there exists a maximal independent set I of G of size C_m )_i , for i = 0, ..., n - 1. Hence there exists some i ∈ {0, ..., n - 1} such that |(C_m )_i ∩ I| = C_m )_{i- 1} and (C_m )_{i+ 1} exists. Let j ∈ {i - 1, i + 1} such that (C_m )_j exists. Set (C_m )_j ∩ I has size C_m )_i ∩ I has at least one vertex (i, f) with f odd. Therefore, if both (C_m )_{i- 1} and (C_m )_{i+ 1} exist, then on both of them all vertices (I - 1, k) and (I + 1, k) for k even are in I. Considering there is exactly C_m )_i ∩ I, there is a vertex (i, f) with f odd without a neighbour in (C_m )_i ∩ I. However, (i, f) is not aneighbour of any vertex in I. Thus, I is not maximal. □. A maximum independent set of G has . Therefore, either all vertices (j, k) with k even or all vertices (j, k) with k odd are in (Cm)j ∩ I. Adjust notation of the vertex labeling of Cm such that all vertices (j, k) with k even are in I. Therefore, all vertices (i, f) of (Cm)i ∩ I will be such that f is odd. Considering m ≥ 4, - 1 vertices in ( vertices of every cycle (. Any independent set of G has at most vertices on any cycle( - 1 is at least one, thus ( - 1. At least one between (

2.1 Two useful classes of graphs

Before we prove Theorem 2, we show how to construct recursively two useful classes of graphs, denoted H_r and F_s . Furthermore, we show the different sizes of maximal independent sets in these graphs.

Let the graph H ₁ have the vertex set V ⁽¹⁾ = {v ₁ ⁽¹⁾, v ₂ ⁽¹⁾, v ₃ ⁽¹⁾} and the edge set E ⁽¹⁾ = {v ₁ ⁽¹⁾, v ₂ ⁽¹⁾, v ₃ ⁽¹⁾}. For r ≥ 2, let the graph H_r have the vertex set

and the edge set

See Figure 1 for an example.

Figure 1:
The graph H ₅.

From the recursive construction, we can determine the sizes of maximal independent sets in H_r , as we prove in Proposition 10.

Proposition 10. For r ≥ 2, H_r ∈ ℳ_r. Furthermore, ms(H_r ) = .

Proof. We prove the statement by induction on r. For r = 2, 3, 4, the result is trivial. Now let r ≥ 5. We have to consider two cases. If r is even, by construction of Hr, it was obtained from Hr-1 by the addition of the vertex v1(r) and the edge v1(r) v2(r-1). Since v1(r) is a leaf, every maximal independent set in Hr contains either v1(r) or v2(r-1). Considering the first case, Hr\NG[v1(r)] has a component isomorphic to Hr-2 and two isolated vertices. By induction hypothesis, Hr-2 ∈ ℳr-2 and ms(Hr-2) = . By adding the isolated vertices and v1(r) we obtain maximal independent sets in Hr with sizes in A = . SinceA ∪ B = ms(Hr) = . By adding v2(r-1) we obtain maximal independent sets in Hr with orders in B = , the desired statement follows for r even, with r ≥ 5.. The graph Hr\NG[v2(r-1)] is isomorphic to Hr-3. By induction hypothesis, Hr-3 ∈ ℳr-3 and ms(Hr-3) =

If r is odd, by construction of Hr, it was obtained from Hr-1 by the addition of the vertices v1(r), v2(r) and v3(r), and the edges v1(r)v2(r), v2(r)v3(r) and v2(r)v1(r-1). Since v1(r) and v3(r) have degree one, every maximal independent set in Hn contains either v1(r) and v3(r), or v2(r). Considering the first case, Hr\NG[v1(r), v3(r)] is isomorphic to Hr-1. By induction hypothesis, Hr-1 ∈ ℳr-1 and ms(Hr-1) = . By adding v1(r) and v3(r) we have maximal independent sets in Hn with sizes in C = . Since C ∪ D = ms(Hr) = . By adding the vertex v2(r) we obtain maximal independent sets in Hr with sizes in D = . The graph Hr\NG[v2(r)] is isomorphic to Hr-2. By induction hypothesis, Hr-2 ∈ ℳr-2 and ms(Hr-2) = , the desired statement follows for r odd, with r ≥ 5. This completes the proof. □

Proposition 11. Let r be a fixed positive integer with r ≥ 3 and t ≥ 1. If G is a graph with t components, each one isomorphic to Hr, then G ∈ ℳrt and,

Proof. For t = 1 the result is trivial in view of Proposition 10. We consider t ≥ 2. Let Lj, for j ∈ {1, ..., t}, be the components of G. Let I(j) be a maximal independent set in the subgraph Lj. For every j ∈ {1, ..., t}, ∪tj=1I(j) is a maximal independent set of G. By Proposition 10, |I(j)| ∈ {r - 2, r}. Thus, all values on the interval [r - 2] are present in ms(L_j ). Therefore, it is possible to obtain a maximal independent set I of G with |I| = x for all x in the interval [tt(r - 2)], through the combination of maximal independent sets of graphs L_j . We are still missing all integers on the interval [t(r - 2) + 1, t(r) - 2] and t(r) for ms(G). For any integer k ∈ {0, ..., t}, it is possible to combine k maximal independent sets of L_j' , j' = 0, ..., k, of size r - 2 with t - k maximal independent sets of the remaining L_j" , j'' = k + 1, ..., t, with size r, resulting in a maximal independent set of G with size t(r) - 2k. Thus, all integers on the set {t(r) - 2k: 0 ≤ k ≤ t} are also on the set ms(G). For r ≥ 3, both (r - 2) and (r - 3) belong to ms(L_j ). So, by taking one maximal independent set of L ₁ with size k - 1 maximal independent sets of L_j' , j' = 2, ..., k, of size r - 2, and t - k maximal independent sets of L_j" , j'' = k + 1, ..., t, with size r, we obtain a maximal independent set of G with size t(r) - 2k - 1, for k = 1, ..., t. Therefore, ms(G) = {tt(r) - 2, t(r)}. □ + + + r - 3, + + + + + , , ..., + , ..., + + , + + + + + + + +

Now we show how to construct a graph F_s , recursively.

Let the graph F ₁ have the vertex set V ⁽¹⁾ = {u ₁ ⁽¹⁾, u ₂ ⁽¹⁾} and the edge set E ⁽¹⁾ = {u ₁ ⁽¹⁾, u ₂ ⁽¹⁾}. For s ≥ 2, let the graph F_s have the vertex set V ^{( s )} = V ^{( s -1)} ∪ {u ₁ ^{( s )}, u ₂ ^{( s )}} and the edge set

Proposition 12. For s ≥ 1, the graph F_s ∈ and ms(F_s ) = {s}. + 1, . . . ,

Proof. By construction, F_s has i(F_s ) ≥ i = 1, ..., s, at most one vertex of {u ₁ ^{( i )}, u ₂ ^{( i )}} can be in a maximal independent set of F_s , thus α(F_s ) ≤ s. We conclude the proof showing how to find in F_s maximal independent sets with sizes in the set {s}. Let I(0) be the set containing, for i = 0, ..., u ₂ ^{(2 i +1)}. If s is even, the vertex u ₂ ^{( s )} must be added to I(0). Note that I(0) is a maximal independent set in F_s . For every j ∈ {1, ..., I(j) = (I(j - 1)\u ₂ ^{(2 j -1)}) ∪ {u ₁ ^{(2 j -1)}, u ₁ ^{(2 j )}}. - 1}, let + 1 vertices of degree one. Moreover, no pair of such vertices share a neighbour. Therefore, -1, the vertices + 1. For every + 1, ..,

For j = 0, ..., I(j) is a maximal independent set in F_s and |I(j)| = j + 1, which implies ms(F_s ) = {s}. □ + 1, ..., + - 1,

2.2 Proof of Theorem 2

Now we use the graphs H_n and F_n and their different sizes of maximal independent sets to show a lower bound on the number of possible sizes of maximal independent sets in P_n □ C_m , for n ≥ 3 and m ≥ 4. We restate Theorem 2:

Theorem 2. Let n ≥ 3 and m ≥ 4. IfG = P_n □ C_m, then G ∈ ℳ_t for some

Proof. We consider four cases:

Case 1. m ≡ 0 (mod 4).

Let I be a set {(k, i) : k ≡ 1 (mod 2),i ≡ 0 (mod 4)}. The set I is independent and it has size H_n . See Figure 2 for an illustration.. Let the graph H arise from G by deleting all vertices in NG[I]. The graph H has components, and each one is isomorphic to the graph

Figure 2:
G = P ₅ □ C_m , with m ≡ 0 (mod 4). The independent set I is formed by the circled vertices and G\N_G [I] has m/4 components isomorphic toH ₅.

By Proposition 11, H ∈ and

By adding the vertices of I, we obtain G, which are different sizes of maximal independent sets in .

Case 2. m ≡ 1 (mod 4).

Let I be a set {(k, i) : k ≡ 1 (mod 2), i < m - 1,i ≡ 0 (mod 4)} ∪ {(k, m - 4) : k ≡ 0 (mod 4)} ∪ {(k, m -1) : k ≡ 2 (mod 4)}. The set I is independent and it has size components.. Let G' be the graph arise from G by deleting all vertices in NG[I]. The graph G' has

Denote by H the graph whose components are the G' isomorphic to the graph H_n . The last component is isomorphic to the graph F_n . Denote it by H'. See Figure 3 for an illustration. components of

Figure 3:
G = P ₅ □ C_m , with m ≡ 1 (mod 4). The independent set I is formed by the circled vertices and G\N_G [I] has m - 5/4 components isomorphic to H ₅ and another one isomorphic to F ₅.

By Proposition 11,

By Proposition 12, ms(H') = {I with any maximal independent set of H and any independent set of H' is a maximal independent set of G. + 1, ..., n}. The union of

Considering n ≥ 3, we have that by Proposition 12, ms(H') has at least two elements. Moreover, ms(H) is an interval with one element missing. Then, ms(G) = ms(G)| = . and |

Case 3. M ≡ 2 (mod 4).

Let I be a set {(k, i) : k ≡ 1 (mod 2),i ≡ 0 (mod 4)} ∪ {(k, m - 1):k ≡ 0 (mod 2)}. The set I is independent and it has size H_n . See Figure 4 (a) for an illustration. By Proposition 11,. Let the graph H arise from G by deleting all vertices in NG[I]. The graph H has components, each of which is isomorphic to the graph

Figure 4:
G = P₅ □ C _m, with m ≡ 2 (mod 4) (a) and m ≡3 (mod 4) (b). The independent set I is formed by the circled vertices and G\N_G [I] has ⌊m/4⌋ components isomorphic to H ₅.

By adding the vertices of I, we obtain G, which are . different sizes of maximal independent sets in

Case 4. M ≡ 3 (mod 4).

Let I be a set {(k, i) : k ≡ 1 (mod 2), i ≡ 0 (mod 4)} ∪ {(k, m - 1) : k ≡ 0 (mod 2)}. The set I is independent and it has size H_n . See Figure 4 (b) for an illustration. By Proposition 11, components, each of which is isomorphic to the graph . Let the graph H arise from G by deleting all vertices in NG[I]. The graph H has

By adding the vertices of I, we obtain G, which are . different sizes of maximal independent sets in

3 CONCLUDING REMARKS

Since a vertex in a cylindrical grid graph G can dominate at most 5 vertices, a lower bound for i(G) is G. Furthermore, the bound in Theorem 2 is sharp when m is a multiple of 4 and n is a small value. The technique we have applied may be useful to find maximal independent sets of different sizes in other graph classes or even in the study of other problems in Graph Theory. + 1 for the quantity of different sizes of maximal independent sets in , which gives us an upper bound

ACKNOWLEDGEMENTS

Márcia R. Cappelle was partially suported by the Fundação de Amparo à Pesquisa do Estado de Goiás (FAPEG). We would like to thank the anonymous reviewers for the very useful comments and suggestions.

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Publication Dates

• Publication in this collection
  Sep-Dec 2016

History

• Received
  11 Sept 2014
• Accepted
  28 Nov 2016