On the Sizes of Maximal Independent Sets of Cylindrical Grid Graphs
R.M. BARBOSA and M.R. CAPPELLE*
Received on September 11, 2014 / Accepted on November 28, 2016
ABSTRACT.If a graph Ghas exactly t different sizes of maximal independent sets, G belongs to a collection calledMt. For the Cartesian product of the graphPn, the path of lengthn, andCm, the cycle of lengthm, called cylindrical grid, we present a method to find maximal independent sets having different sizes and a lower bound ont, such that these graphs belong toMt.
Keywords:Well-covered graph, maximal independent set, Cartesian product.
1 INTRODUCTION
In [9] Plummer defines a graph to bewell-coveredif all its maximal independent sets have the same size. Generalizing this concept, Finbow, Hartnell, and Whitehead [4] define, for every
t ∈ N, the setMt as the set of graphs that have maximal independent sets of exactlyt different sizes. With this notation,M1is 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. ForG ∈ Mt andv ∈ V(G), Barbosa and Hartnell [2] determine the extreme values thatrcan assume whereG\vbelongs toMr. Additional properties of graphs in this class are given in [1]. Results onMt, fort ≥ 2, related to graphs without small cycles are also given in [3, 4, 6].
The Cartesian productPn Cm is acylindrical 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 graphsPn Cm withm ≥3 andn =2,3,and 4 were proposed. Moreover, bounds on the domination numbers were found whenn = 5 andm ≥ 3. We present a method to find maximal independent sets having different sizes and a lower bound fort, such thatPnCmbelongs toMt.
*Corresponding author: M´arcia Rodrigues Cappelle.
Before we present our results and proofs, we summarize our notation.
We consider finite, simple, and undirected graphs. For a graphG, the vertex set and the edge set are denotedV(G)andE(G), respectively. For a vertexuofG, its neighbourhood is denoted
NG(u)and its closed neighbourhooddenoted NG[u]is the set NG(u)∪ {u}. For a setU of vertices ofG, let
NG(U)=
u∈U
NG(u)\U and NG[U] =NG(U)∪U.
A set D ⊆ V(G) of a graphG is adominating setif every vertexv ∈ V(G)\Dis adjacent to some vertexu ∈D. Thedomination numberofGis the cardinality of a smallest dominating set of the graphGand is usually denoted byγ (G).
A set Iof vertices of a graphGisindependentif no two vertices inI are adjacent. An indepen-dent setI ofGismaximalif every vertexuinV(G)\I has a neighbour inI. An independent set
I ofGismaximumifGhas no independent setJ with|J|>|I|. We denotei(G)the cardinal-ity of a smallest maximal independent set inG. We denoteα(G)the cardinality of a maximum independent set inGandms(G)the set of all sizes of maximal independent sets inG. Hence, if
Gis well-covered, theni(G)=α(G)and|ms(G)| =1.
For any two graphsGandH, theCartesian product GHis the graph with vertex setV(G)× V(H), such that two vertices(u1, v1)and(u2, v2)ofGHare adjacent wheneverv1=v2and u1u2∈E(G)oru1=u2andv1v2∈ E(H). Here, the vertices of the pathPnor the cycleCnare always denoted 0,1, . . . ,n−1. For the graph Pn Cm, we denote(Cm)i the graphCm {i}, withi ∈V(Pn).
For naturals a,b, andc, withc > 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 (P2Cm)=
m
2, if m≡0(mod 4),
m+1 2
, otherwise;
• γ (P3Cm)=
3m 4
;
• γ (P4Cm)=
m+1, for m=3,5,9,
m, otherwise;
• for m≥6, m+m
5
One of our main results is the following. It will be proved in Section 2.2. Theorem 2. Let n≥3and m≥4. If G=PnCm, then G∈Mtfor some
t ≥
nm4 , if m≡0,2,3(mod 4), nm4 − ⌊n2⌋, if m≡1(mod 4).
We begin showing the size of a maximum independent set in a graphPnCm. Proposition 3.For n≥2and m≥3,α(PnCm)=n
m 2 .
Proof. Since α(Cm) = m2 and we have n cyclesCm, we conclude thatα(Pn Cm) ≤
nm
2 . Now, we show thatα(Pn Cm) ≥ n m
2 presenting a maximal independent set with this cardinality. Let I be the set of vertices (i,j)of Pn Cm such thati = 0, . . . ,n −1,
j =0, . . . ,2m
2 −1, andi+ j is an odd integer. The set I has cardinalityn m
2 and it is a
maximal independent set inPnCm.
In Theorem 7, we determine t such that P2 Cm belongs to Mt. Before, we prove some preliminary results.
Lemma 4. Let m ≥ 3and G = P2Cm. If I is a maximal independent set of G, then I has even cardinality.
Proof. LetI be a maximal independent set inG,Xi =I ∩(Cm)i andxi = |Xi|, fori =0,1. We show thatx0=x1. By symmetry, consider the subgraph induced by(Cm)0\X0, denoted by H. The graphHhas exactlyx0disjoint connected paths. We denote them byP(k),k=1, . . . ,x0. If|V(P(k))| > 3, for somek ∈ {1, . . . ,x0},(Cm)0\NG[X0]has some path with at least two vertices, which implies at least two consecutive vertices in I ∩(Cm)1. Thus we may assume |V(P(k))| ≤3,k=1, . . . ,x0.
Fix some k ∈ {1, . . . ,x0}. Let (v1, v2, . . . , vq), 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))|}, letv′j be the neighbour ofvj in(Cm)1. Note that exactlyx0vertices in
(Cm)1are inNG(X0)and they are separated by at most three vertices. We have three cases for each P(k). If|V(P(k))| = 1,v1′ must belong to I since its neighbours in the same cycle are dominated by I. If|V(P(k))| = 2,v1andv2 belong toNG(I)and exactly one ofv1′ andv2′ must belong toI. If|V(P(k))| =3, the verticesv2, v1′, v2′, andv3′ are not inNG(X0). Hencev2′ must be inI. Therefore, in all cases, fork=1, . . . ,x0, for every pathP(k), there is exactly one vertex inX1, resulting in|I∩(Cm)0| = |I∩(Cm)1|. Lemma 5.Let n ≥2, m ≥3and G =PnCm. If I is a maximal independent set in G, then
|I∩(Cm)i| ≥ m
4
,
Proof. Let I be a maximal independent set in G, Xi = I ∩ (Cm)i and xi = |Xi|, for i ∈ {0,n−1}. Supposexi <m4, for somei∈ {0,n−1}. By symmetry, we may assumei=0. Consider the graph (Cm)0 and the subgraph H induced by (Cm)0\NG[X0]. Since at most 3m
4
−3 vertices of(Cm)0are inNG[X0], at least⌈m4⌉of those vertices are not inNG[X0]. Moreover,Hhas at mostx0disjoint connected paths. Therefore, at least one of these paths has at least two vertices. SinceI is maximal, two adjacent vertices in(Cm)1are inI. This contradicts the independence ofIand completes the proof.
Proposition 6.For m≥3, i(P2Cm)=2m4.
Proof. Let J = {(0,i):i ≡1(mod 4)} ∪ {(1,i): i ≡3(mod 4)}. We consider a set I with
I = J, ifm≡0(mod 4),I =J∪ {(1,0), (0,m−1)}, ifm≡1(mod 4), andI =J∪ {(1,0)}, otherwise.
Note thatIis a maximal independent set of P2Cm, and|I| =2m4. Hence,i(P2Cm)≤ 2m
4
. By Lemma 5,|I∩(Cm)0| + |I∩(Cm)1| ≥2m4and the desired statement follows. Now, we can show the quantity of different sizes of maximal independent sets inP2Cm.
Theorem 7.For m≥3and G=P2Cm, G∈
M⌊m
4⌋, if m≡1(mod 4), M⌊m
4⌋+1, otherwise.
Proof. By Proposition 3, α(G) = 2m
2 , and by Proposition 6,i(G) = 2 m
4
. We present maximal independent sets ofGwith every even cardinality between 2m
4
and 2m
2 , in view of Lemma 4.
First, let l= ⎧ ⎪ ⎨
⎪ ⎩
m, if m≡0,3(mod 4), m−3, if m≡1(mod 4), m−2, if m≡2(mod 4),
and
k=
⌊m4⌋ −1, if m≡1(mod 4), ⌊m4⌋, otherwise.
LetI(0)′= {(1,i),i ≡0(mod 4)},i ≤l} ∪ {(0,i),i ≡2(mod 4),i≤l}and
I(0)= ⎧ ⎪ ⎨
⎪ ⎩
I(0)′, ifm≡0,3(mod 4), I(0)′∪ {(0,m−1), (1,m−2)}, if m≡1(mod 4), I(0)′∪ {(0,m−1)}, if m≡2(mod 4).
Note that the set I(0)is a maximal independent set inGand|I(0)| =2⌈m4⌉ =i(G). For every
j∈ {1, . . . ,k}, let
For every j ∈ {0, . . . ,k},I(j)is a maximal independent set inGand|I(j)| =2⌈m4⌉ +2j. Hence G has maximal independent sets ofk+1 different sizes which results inm
4 if 1 ≡ m(mod 4), orm4 +1, otherwise. This completes the proof. By Theorem 2, whenn ≥ 3 andm ≥4, Pn Cm is not well-covered. The graph P2 C5is well-covered and alsoPnC3forn ≥2, as we show in next proposition.
Proposition 8.For n≥2, if G=PnC3, then G is well-covered.
Proof. By Proposition 3,α(G)=n. Let I be a maximal independent set inG. We show that
|I| = n. For a contradiction, suppose|I| < n. Hence there is some j ∈ {0, . . . ,n −1}, with
I∩(C3)j = ∅. LetSdenote the setNG(V((C3)j)). SinceI is maximal and from the structure ofG, |S∩ I| ≥ 3. But S induces one or two cyclesC3, which contradicts the independence
ofI.
Proposition 9. Let n ≥3, m ≥4and G = Pn Cm. If m is even, then there is no maximal independent set I in G such that|I| = nm2 −1;
Proof. By Proposition 3,α(G) = nm2 . A maximum independent set ofGhas m2 vertices of every cycle(Cm)i, fori =0, . . . ,n−1. Suppose that there exists a maximal independent setI ofGof size nm2 −1. Any independent set ofGhas at most m2 vertices on any cycle(Cm)i, for i =0, . . . ,n−1. Hence there exists somei ∈ {0, . . . ,n−1}such that|(Cm)i ∩I| = m2 −1. At least one between(Cm)i−1and(Cm)i+1exists. Let j∈ {i−1,i+1}such that(Cm)j exists. Set(Cm)j ∩I has size m2. Therefore, either all vertices(j,k)withkeven or all vertices(j,k) withkodd are in(Cm)j ∩I. Adjust notation of the vertex labeling ofCm such that all vertices
(j,k)withkeven are inI. Therefore, all vertices(i, f)of(Cm)i∩I will be such that f is odd. Consideringm ≥ 4, m2 −1 is at least one, thus(Cm)i ∩I has at least one vertex(i, f)with f odd. Therefore, if both(Cm)i−1and(Cm)i+1exist, then on both of them all vertices(i −1,k) and(i +1,k)fork even are in I. Considering there is exactly m2 −1 vertices in(Cm)i ∩I, there is a vertex(i, f)with f odd without a neighbour in(Cm)i ∩I. However,(i, f)is not a neighbour of any vertex inI. Thus,I is not maximal.
2.1 Two useful classes of graphs
Let the graphH1have the vertex setV(1) = {v1(1), v2(1), v3(1)}and the edge setE(1) = {v1(1)v2(1),
v2(1)v3(1)}. Forr≥2, let the graphHr have the vertex set
V(r)=V(r−1)∪ {v1(r)}, forreven, V(r)=V(r−1)∪ {v1(r), v(2r), v3(r)}, forrodd,
and the edge set
E(r) =E(r−1)∪ {v1(r)v(2r−1)}, forr even, E(r) =E(r−1)∪ {v1(r)v(2r), v2(r)v3(r), v2(r)v1(r−1)}, forr odd.
See Figure 1 for an example.
✉ ✉ ✉
✉ ✉ ✉ ✉ ✉
✉ ✉ ✉
v1(4)
v1(5) v2(5) v3(5)
Figure 1: The graphH5.
From the recursive construction, we can determine the sizes of maximal independent sets inHr, as we prove in Proposition 10.
Proposition 10.For r ≥2, Hr ∈Mr. Furthermore,ms(Hr)= {r2, . . . ,
3r 2
−2,3r2}.
Proof. We prove the statement by induction onr. Forr=2,3,4, the result is trivial. Now let
r ≥5. We have to consider two cases. Ifris even, by construction ofHr, it was obtained from Hr−1by the addition of the vertexv1(r)and the edgev
(r)
1 v
(r−1)
2 . Sincev
(r)
1 is a leaf, every maximal independent set inHr contains eitherv1(r)orv2(r−1). Considering the first case, Hr\NG[v1(r)]has a component isomorphic toHr−2and two isolated vertices. By induction hypothesis, Hr−2 ∈ Mr−2 and ms(Hr−2) = {r
2 −1, . . . , 3r
2 −5, 3r
2 −3}.By adding the isolated vertices andv
(r)
1 we obtain maximal independent sets in Hr with sizes in A = {r2 +2, . . . ,3r2 −2,3r2}.The graph Hr \NG[v(2r−1)] is isomorphic to Hr−3. By induction hypothesis, Hr−3 ∈ Mr−3 and ms(Hr−3)= {r2−1, . . . ,3r2 −6,3r2 −4}.By addingv2(r−1)we obtain maximal independent sets inHrwith orders inB= {r2, . . . ,3r2 −5,3r2 −3}.SinceA∪B=ms(Hr)= {r2, . . . ,3r2 −2,3r2}, the desired statement follows forreven, withr≥5.
Ifr is odd, by construction of Hr, it was obtained from Hr−1by the addition of the vertices
and ms(Hr−1) = {r−12 , . . . ,3r−12 −3,3r−12 −1}. By addingv1(r) andv3(r) we have maximal independent sets in Hn with sizes in C = {r+12 +1, . . . ,3r2+1 −2,3r+12 }. The graph Hr \ NG[v2(r)] is isomorphic to Hr−2. By induction hypothesis, Hr−2 ∈ Mr−2 and ms(Hr−2) = {r−12 , . . . ,3r+12 −5,3r2+1−3}.By adding the vertexv2(r)we obtain maximal independent sets in
Hr with sizes inD= {r+12 , . . . ,3r2+1−4,3r2+1−2}.SinceC∪D=ms(Hr)= {r+12 , . . . ,3r+12 − 2,3r+12 }, the desired statement follows forrodd, withr≥5. This completes the proof.
Proposition 11. Let r be a fixed positive integer with r ≥ 3and t ≥1. If G is a graph with t components, each one isomorphic to Hr, then G∈Mrt and,
ms(G)=
tr
2
, . . . ,t
r
2
+r
−2,t
r
2
+r .
Proof. Fort = 1 the result is trivial in view of Proposition 10. We considert ≥ 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)| ∈ {⌈r2⌉, . . . ,⌈2r⌉ +r −2,⌈r2⌉ +r}. Thus, all values on the interval
[⌈r2⌉,⌈r2⌉+r−2]are present in ms(Lj). Therefore, it is possible to obtain a maximal independent setI ofGwith|I| =xfor allxin the interval[t⌈2r⌉,t(⌈r2⌉ +r−2)], through the combination of maximal independent sets of graphs Lj. We are still missing all integers on the interval [t(⌈r2⌉ +r−2)+1,t(⌈r2⌉ +r)−2]andt(⌈r2⌉ +r)for ms(G). For any integerk∈ {0, . . . ,t}, it is possible to combinekmaximal independent sets ofLj′, j′=0, . . . ,k, of size⌈r
2⌉ +r−2 witht−kmaximal independent sets of the remainingLj′′, j′′=k+1, . . . ,t, with size⌈r
2⌉ +r, resulting in a maximal independent set ofGwith sizet(⌈r2⌉ +r)−2k. Thus, all integers on the set{t(⌈r2⌉ +r)−2k : 0 ≤ k ≤ t}are also on the set ms(G). Forr ≥ 3, both(⌈r2⌉ +r−2)
and(⌈r2⌉ +r−3)belong to ms(Lj). So, by taking one maximal independent set of L1 with size⌈r2⌉ +r−3,k−1 maximal independent sets ofLj′, j′ =2, . . . ,k, of size⌈r
2⌉ +r−2, andt −kmaximal independent sets ofLj′′, j′′ =k+1, . . . ,t, with size⌈r
2⌉ +r, we obtain a maximal independent set ofG with sizet(⌈r2⌉ +r)−2k−1, fork =1, . . . ,t. Therefore,
ms(G)= {t⌈r2⌉, . . . ,t(⌈r2⌉ +r)−2,t(⌈r2⌉ +r)}.
Now we show how to construct a graphFs, recursively.
Let the graphF1have the vertex setV(1) = {u(11),u2(1)}and the edge setE(1)= {u1(1)u(21)}. For s≥2, let the graphFs have the vertex setV(s)=V(s−1)∪ {u(1s),u(2s)}and the edge set
E(s)=E(s−1)∪ {u(1s)u2(s−1),u(1s)u2(s)}, forseven, E(s)=E(s−1)∪ {u(2s)u2(s−1),u(1s)u2(s)}, forsodd.
Proposition 12.For s≥1, the graph Fs ∈M⌈s
2⌉andms(Fs)= { s
Proof. By construction, Fs has s2 +1 vertices of degree one. Moreover, no pair of such vertices share a neighbour. Therefore, i(Fs) ≥ 2s +1. For every i = 1, . . . ,s, at most one vertex of{u(1i),u(2i)}can be in a maximal independent set ofFs, thusα(Fs)≤s. We conclude the proof showing how to find inFsmaximal independent sets with sizes in the set{s2 +1, . . . ,s}. Let I(0)be the set containing, fori = 0, . . . ,2s−1, the verticesu(22i+1). Ifs is even, the vertexu(2s)must be added toI(0). Note thatI(0)is a maximal independent set inFs. For every
j∈ {1, . . . ,s2−1}, letI(j)=(I(j−1)\u(22j−1))∪ {u1(2j−1),u(12j)}. For j =0, . . . ,s
2
−1, I(j)is a maximal independent set inFs and|I(j)| = 2s + j +1, which impliesms(Fs)= {s2 +1, . . . ,s}.
2.2 Proof of Theorem 2
Now we use the graphsHnandFnand their different sizes of maximal independent sets to show a lower bound on the number of possible sizes of maximal independent sets in Pn Cm, for n≥3 andm≥4. We restate Theorem 2:
Theorem 2.Let n≥3and m≥4. If G=PnCm, then G∈Mtfor some
t ≥
nm
4 , if m≡0,2,3(mod 4), nm4 − ⌊n2⌋, if m≡1(mod 4).
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 m4 n
2 . Let the graphHarise fromGby deleting all vertices inNG[I]. The graphHhas m 4 components, and each one is isomorphic to the graphHn. See Figure 2 for an illustration. By Proposition 11,H ∈Mn(m
4)and ms(H)=
m
4
n
2
, . . . ,
m
4
n
2
+n
−2,
m
4
n
2
+n
.
By adding the vertices of I, we obtain nm4 different sizes of maximal independent sets in G, which are{nm4 , . . . ,nm2 −2,nm2 }.
Case 2. m≡1(mod 4).
LetIbe a set{(k,i):k≡1(mod2),i<m−1,i ≡0(mod4)}∪{(k,m−4):k≡0(mod4)}∪ {(k,m−1):k≡2(mod4)}. The setIis independent and it has size((m−14 n
2 )+ n
2
). LetG′
t t t t t t t t
t t t t t t t t
t t t t t t t t
t t t t t t t t
t t t t t t t t
❤ ❤ ❤ ❤ ☞ ✌ ☞ ✌ ☞ ✌ ✎ ✍ ✎ ✍ ✎ ✍ ☞ ✌ ☞ ✌ ☞ ✌ ✎ ✍ ✎ ✍ ✎ ✍ . . . . . . . . . . . . . . .
Figure 2:G = P5Cm, withm≡0(mod 4). The independent setI is formed by the circled vertices andG\NG[I]hasm4 components isomorphic toH5.
Denote by H the graph whose components are the m−54 components of G′ isomorphic to the graphHn. The last component is isomorphic to the graphFn. Denote it byH′. See Figure 3 for an illustration.
By Proposition 11,
ms(H)=
m− 5 4 n 2 , . . . , m− 5 4 n 2 +n −2,
m− 5 4 n 2 +n .
By Proposition 12, ms(H′)= {n
2 +1, . . . ,n}. The union ofI with any maximal independent set ofHand any independent set ofH′is a maximal independent set ofG.
Consideringn ≥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)= {nm4 +n2 +1, . . . ,2nm4 }
and|ms(G)| =nm 4 − ⌊
n 2⌋.
✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ☞ ✌ ☞ ✌ ☞ ✌ ✎ ✍ ✎ ✍ ✎ ✍ ☞ ✌ ☞ ✌ ✎ ✍ . . . . . . . . . . . . . . .
Case 3. m≡2(mod 4).
LetI be a set{(k,i):k≡1(mod2),i ≡0(mod 4)} ∪ {(k,m−1):k≡0(mod2)}. The setI
is independent and it has size((m+24 n 2 )+
n 2
). Let the graphHarise fromGby deleting all vertices inNG[I]. The graphHhasm−24 components, each of which is isomorphic to the graph Hn. See Figure 4(a) for an illustration. By Proposition 11,
ms(H)=
m− 2 4 n 2 , . . . , m− 2 4 n 2 +n −2,
m− 2 4 n 2 +n .
By adding the vertices ofI, we obtainnm4 different sizes of maximal independent sets inG, which are{nm
4
, . . . ,nm2 −2,nm2 }.
t t t t t t
t t t t t t
t t t t t t
t t t t t t
t t t t t t
❤ ❤ ❤ ❤ ❤ ❤ ❤ ☞ ✌ ☞ ✌ ☞ ✌ ✎ ✍ ✎ ✍ ✎ ✍ . . . . . . . . . . . . . . . (a)
t t t t t t t
t t t t t t t
t t t t t t t
t t t t t t t
t t t t t t t
❤ ❤ ❤ ❤ ❤ ❤ ❤ ☞ ✌ ☞ ✌ ☞ ✌ ✎ ✍ ✎ ✍ ✎ ✍ . . . . . . . . . . . . . . . (b)
Figure 4:G= P5Cm, withm≡2(mod 4)(a) andm≡3(mod 4)(b). The independent set I is formed by the circled vertices andG\NG[I]has⌊m4⌋components isomorphic toH5.
Case 4. m≡3(mod 4).
LetI be a set{(k,i):k≡1(mod 2),i ≡0(mod 4)} ∪ {(k,m−1):k≡0(mod 2)}. The setI
is independent and it has size((m+14 n2 )+n 2
vertices inNG[I]. The graphH hasm−34 components, each of which is isomorphic to the graph Hn. See Figure 4(b) for an illustration. By Proposition 11,
ms(H)=
m−
3 4
n
2
, . . . ,
m−
3 4
n
2
+n
−2,
m−
3 4
n
2
+n
.
By adding the vertices ofI, we obtainnm4 different sizes of maximal independent sets inG, which are{nm4, . . . ,nm2 −2,nm2 }.
3 CONCLUDING REMARKS
Since a vertex in a cylindrical grid graphGcan dominate at most 5 vertices, a lower bound for
i(G) is⌈nm5 ⌉, which gives us an upper boundn⌊m2⌋ − ⌈nm5 ⌉ +1 for the quantity of different sizes of maximal independent sets inG. Furthermore, the bound in Theorem 2 is sharp whenm
is a multiple of 4 andn 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.
ACKNOWLEDGEMENTS
M´arcia R. Cappelle was partially suported by the Fundac¸˜ao de Amparo `a Pesquisa do Estado de Goi´as (FAPEG). We would like to thank the anonymous reviewers for the very useful comments and suggestions.
RESUMO.Se um grafoGtem exatamentettamanhos diferentes de conjuntos independentes maximais,Gpertence a uma colec¸˜ao chamadaMt. Para o produto Cartesiano do grafoPn, o
caminho de tamanhon, eCm, o ciclo de tamanhom, chamado grade cil´ındrica, apresentamos um m´etodo para encontrar conjuntos independentesmaximaiscom diferentes tamanhos e um limite inferior parat, tal que estes grafospertenc¸am`aMt.
Palavras-chave:Grafo bem-coberto, conjunto independentemaximal, produto Cartesiano.
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![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 .](https://thumb-eu.123doks.com/thumbv2/123dok_br/18983838.458082/9.1063.313.882.964.1249/figure-independent-formed-circled-vertices-components-isomorphic-isomorphic.webp)
![Figure 4: G = P 5 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 .](https://thumb-eu.123doks.com/thumbv2/123dok_br/18983838.458082/10.1063.308.702.504.1088/figure-mod-independent-formed-circled-vertices-components-isomorphic.webp)
