On the Broadcast Independence Number of Circulant Graphs
aa r X i v : . [ c s . D M ] F e b On the Broadcast Independence Numberof Circulant Graphs
Abdelamin LAOUAR ∗ Isma BOUCHEMAKH ∗ ´Eric SOPENA † February 9, 2021
Abstract
An independent broadcast on a graph G is a function f : V −→ { , . . . , diam( G ) } suchthat ( i ) f ( v ) ≤ e ( v ) for every vertex v ∈ V ( G ), where diam( G ) denotes the diameter of G and e ( v ) the eccentricity of vertex v , and ( ii ) d ( u, v ) > max { f ( u ) , f ( v ) } for every twodistinct vertices u and v with f ( u ) f ( v ) >
0. The broadcast independence number β b ( G ) of G is then the maximum value of P v ∈ V f ( v ), taken over all independent broadcasts on G .We prove that every circulant graph of the form C ( n ; 1 , a ), 3 ≤ a ≤ ⌊ n ⌋ , admits anoptimal 2-bounded independent broadcast, that is, an independent broadcast f satisfying f ( v ) ≤ v , except when n = 2 a + 1, or n = 2 a and a is even. We thendetermine the broadcast independence number of various classes of such circulant graphs,and prove that, for most of these classes, the equality β b ( C ( n ; 1 , a )) = α ( C ( n ; 1 , a )) holds,where α ( C ( n ; 1 , a )) denotes the independence number of C ( n ; 1 , a ). Keywords:
Broadcast; Independent broadcast; Circulant graph.
MSC 2010:
All the graphs considered in this paper are undirected and simple. For such a graph G , wedenote by V ( G ) and E ( G ) its set of vertices and its set of edges, respectively. Let G be anontrivial connected graph, that is, a connected graph with at least one edge. The distance from a vertex u to a vertex v in G , denoted d G ( u, v ), or simply d ( u, v ) when G is clear fromthe context, is the length (number of edges) of a shortest path from u to v . The eccentricity ofa vertex v in G , denoted e G ( v ), is the maximum distance from v to any other vertex of G . Theminimum eccentricity in G is the radius of G , denoted rad( G ), while the maximum eccentricityin G is its diameter , denoted diam( G ). Two vertices u and v with d G ( u, v ) = diam( G ) are saidto be antipodal .A function f : V ( G ) → { , . . . , diam( G ) } is a broadcast on G if f ( v ) ≤ e G ( v ) for everyvertex v ∈ V . For each vertex v , f ( v ) is the f -value of v , or the broadcast value of v if f isclear from the context. Given such a broadcast f , an f -broadcast vertex is a vertex v for which f ( v ) >
0. The set of all f -broadcast vertices is denoted V + f ( G ). If v is a broadcast vertex and ∗ Faculty of Mathematics, Laboratory L’IFORCE, University of Sciences and Technology Houari Boumediene(USTHB), B.P. 32 El-Alia, Bab-Ezzouar, 16111 Algiers, Algeria. † Univ. Bordeaux, CNRS, Bordeaux INP, LaBRI, UMR5800, F-33400 Talence, France. a vertex such that d ( u, v ) ≤ f ( u ), then the vertex v f -dominates the vertex u . The cost of abroadcast f on G is the value σ ( f ) = P v ∈ V + f f ( v ).A broadcast f is independent if no broadcast vertex f -dominates another broadcast vertex,or, equivalently, if d ( u, v ) > max { f ( u ) , f ( v ) } for every two distinct broadcast vertices u and v .The maximum cost of an independent broadcast on G is the broadcast independence number of G , denoted β b ( G ). An independent broadcast with cost β b ( G ) is referred to as a β b -broadcast .A subset S of V ( G ) is an independent set if no two vertices in S are adjacent in G . The independence number of G , denoted α ( G ), is then the maximum cardinality of an independentset in G . Note that the characteristic function f S of every maximal independent set S in agraph G is an independent broadcast and, therefore, α ( G ) ≤ β b ( G ) for every graph G .Broadcast independence and broadcast domination were introduced by Erwin [9] in hisPh.D. dissertation, using the terms cost independence and cost domination, respectively. Healso discussed several other types of broadcast parameters and gave relationships between them.Most of the corresponding results are published in [8, 10]. Since then, several papers have beendevoted to the study of these broadcast parameters, but there were not so many results concern-ing the broadcast independence number [6, 8] until recently (see [1–5, 7]). In particular, Bessyand Rautenbach discussed the algorithmic complexity of broadcast independence in [3] andthe links between girth, minimum degree, independence number and broadcast independencenumber in [4, 5].In this paper, we study the broadcast independence number of circulant graphs. Recallthat for every integer n ≥
3, and every sequence of integers a , . . . , a k , k ≥
1, satisfying1 ≤ a ≤ · · · ≤ a k ≤ (cid:4) n (cid:5) , the circulant graph G = C ( n ; a , . . . , a k ) is the graph defined by V ( G ) = { v , v , . . . , v n − } and E ( G ) = (cid:8) v i v i + a j | a j ∈ { a , . . . , a k } (cid:9) (subscripts are taken modulo n ). Note that, in particular, C ( n ; a , . . . , a k ) is 2 k -regular andvertex-transitive (see [12] for a survey on properties of undirected circulant graphs).Our paper is organized as follows. In Section 2, we give some preliminary results and de-termine the broadcast independence number of circulant graphs of the form C (2 a ; 1 , a ) and C (3 a ; 1 , a ). In Section 3 we prove that almost all circulant graphs of the form C ( n ; 1 , a ) admitan optimal independent broadcast all whose broadcast values are at most 2. General upper andlower bounds on the cost of independent broadcasts on circulant graphs of the form C ( n ; 1 , a )are proposed in Section 4. We then determine the value of the broadcast independence numberof various classes of circulant graphs in Section 5. We finally propose a few concluding remarksin Section 6. Let µ ( G ) denote the maximum cardinality of a set of pairwise antipodal vertices in G . Dunbar et al. proved the following lower bound on the broadcast independence number of a graph. Proposition 1 (Dunbar et al. [8]) . For every graph G , β b ( G ) ≥ µ ( G )(diam( G ) − ≥ G ) − . Moreover, this bound is sharp.
In addition to grid graphs G m,n = P m (cid:3) P n with m ∈ { , , } and m ≤ n [6] and paths [9],the relation β b ( G ) = 2(diam( G ) −
1) also holds for cycles of order at least 4 [7].2 v v v v k v k +1 v a − v a − v a − v a v a +1 v a +2 v a +3 v a + k v a + k +1 v a − v a − v a − Figure 1: The circulant graph C(2a;1,a).It can also be observed that the value 2(diam( G ) −
1) is an upper bound on the cost of someindependent broadcasts.In order to compare the values of the independence number and of the broadcast indepen-dence number of the graphs we will consider in Section 5, the following observation will beuseful.
Observation 2.
For every graph G , β b ( G ) ≥ α ( G ) . Moreover, β b ( G ) = α ( G ) if and only ifthere exists a β b -broadcast f on G such that f ( v ) = 1 for every broadcast vertex v ∈ V + f . Indeed, the fact that the characteristic function f S of every maximal independent set S in a graph G is an independent broadcast on G , as noticed in the previous section, gives theinequality and the necessity of the condition for the second part of the statement, while thesufficiency follows from the fact that V + f is always an independent set.Before considering general cases in the next sections, we will determine in the following theindependence number and the broadcast independence number of circulant graphs of the form C ( n ; 1 , a ) for two particular cases, namely when n = 2 a or n = 3 a . Lemma 3.
For every integer a ≥ , α ( C (2 a ; 1 , a )) = ( a, if a is odd, a − , if a is even. Proof.
Since C (4; 1 ,
2) = K and α ( K ) = 1, we can assume a ≥ a is odd, then the set S = { v i | i is even } is an independent set of C (2 a ; 1 , a ), and thus α ( C (2 a ; 1 , a )) ≥ | S | = a . Since C a is a subgraph of C (2 a ; 1 , a ), we get α ( C (2 a ; 1 , a ) ≤ α ( C a ) = a and the result follows.If a is even, then the set S ′ = { v i | ≤ i ≤ a − , i is even }∪{ v i | a +1 ≤ i ≤ a − , i is odd } isan independent set of C (2 a ; 1 , a ), and thus α ( C (2 a ; 1 , a )) ≥ | S ′ | = a −
1. Note that the odd cycle C = v v . . . v a v , with α ( C ) = a , is a subgraph of C (2 a ; 1 , a ). Therefore, for every independentset I of C (2 a ; 1 , a ), there are at least two consecutive vertices v i , v i +1 with v i , v i +1 / ∈ I . Thisimplies α ( C (2 a ; 1 , a )) ≤ a − , which gives α ( C (2 a ; 1 , a )) = a − (cid:3) Theorem 4.
For every integer a ≥ , β b ( C (2 a ; 1 , a )) = α ( C (2 a ; 1 , a )) = a, if a is odd ,α ( C (2 a ; 1 , a )) = a − , if a = 2 p for some integer p ≥ ,a otherwise. Proof.
The case a = 2 directly follows from Lemma 3. We can thus assume a ≥
3. The graph C (2 a ; 1 , a ) can be viewed as the Cartesian product graph P a (cid:3) K with two additional edges (see3igure 1, where the two additional edges are drawn as dashed lines). Recall that α ( C (2 a ; 1 , a ))is given by Lemma 3, and let f be any independent β b -broadcast on C (2 a ; 1 , a ). If | V + f | = 1,then σ ( f ) = diam( C (2 a ; 1 , a )) = (cid:22) a + 12 (cid:23) < α ( C (2 a ; 1 , a )) , which gives σ ( f ) < β b ( C (2 a ; 1 , a )) by Observation 2, a contradiction. Therefore, | V + f | ≥ v ∈ V + f f -dominates exactly 4 f ( v ) vertices, each f -broadcast vertex v is f -dominated exactly once, each non-broadcast vertex is f -dominated at most three times, andat most | V + f | vertices can be dominated three times (namely the vertices v i + a when v i ∈ V + f ),we get 4 f ( V + f ) ≤ | V + f | + | V + f | + 2 (cid:0) a − | V + f | (cid:1) = 4 a, (1)and thus β b ( C (2 a ; 1 , a )) = σ ( f ) = X v ∈ V + f f ( v ) = f ( V + f ) ≤ a. (2)We now consider the three cases of the statement of the theorem separately.1. a is odd.Let g be the mapping from V ( C (2 a ; 1 , a )) to { , } defined by g ( v i ) = 1 if and onlyif i is even. Since a is odd, g is an independent broadcast on C (2 a ; 1 , a ). This gives β b ( C (2 a ; 1 , a )) ≥ σ ( g ) = a and, since g satisfies (2), β b ( C (2 a ; 1 , a )) = a . By Observation 2,we then get β b ( C (2 a ; 1 , a )) = α ( C (2 a ; 1 , a )) = a. a = 2 p for some integer p ≥ β b ( C (2 a ; 1 , a )) ≥ α ( C (2 a ; 1 , a )) = a −
1, since a is even. Let g be any independent β b -broadcast on C (2 a ; 1 , a ).Suppose first that not all g -broadcast vertices have the same g -value, and let v i and v j be any two vertices with g ( v i ) < g ( v j ) such that the distance d ( v i , v j ) is minimum amongall g -broadcast vertices with distinct g -values. Without loss of generality, we can assume i < j . We consider two subcases, depending on whether v i and v j are on the same side ofthe “ladder” (refer to Figure 1) or not.(a) j − i > a ( v i and v j are not on the same side of the ladder). Since no g -broadcast vertex lies on a shortest path linking v i and v j , v j + a is not g -dominated by v i and is thus g -dominated at most twice. Therefore, the inequality (1)becomes 4 g ( V + g ) ≤ | V + g | −
1) + | V + g | + 2 (cid:0) a − | V + g | + 1 (cid:1) = 4 a − , which gives β b ( C (2 a ; 1 , a )) = σ ( g ) = X v ∈ V + g g ( v ) = g ( V + g ) ≤ (cid:22) a − (cid:23) = a − . (b) j − i < a ( v i and v j are on the same side of the ladder). If there exists a g -broadcast vertex v k with i + a < k < j + a then, since no g -broadcast vertex lies on a shortest path linking v i and v j , we necessarily have v k ∈ { v i + a +1 , v j + a − } . By considering either v i and v k , or v j and v k , instead of v i and v j , we are back to the previous subcase.If no such vertex exists, then both v i + a and v j + a are g -dominated at most twice andthus, using the same argument as before, we get β b ( C (2 a ; 1 , a )) ≤ a − β b ( C (2 a ; 1 , a )) = a − g ( v ) = k for every vertex v ∈ V + g and let v i be any such vertex.If v i + a + k / ∈ V + g , then the vertex v i + a is g -dominated at most twice and thus, as previously,we get β b ( C (2 a ; 1 , a )) = σ ( g ) = a −
1. The same conclusion arises if v i − a − k / ∈ V + g .Suppose finally v i + a + k , v i − a − k ∈ V + g for every vertex v i ∈ V + g and assume, without loss ofgenerality, v ∈ V + g . We thus have (recall that indices are taken modulo 2 a ) V + g = { v , v a + k , v k , v a +3 k , . . . } , = { v , v k , v k , . . . , v a + k , v a +3 k , v a +5 k , . . . } . Hence, every g -broadcast vertex v i with 0 ≤ i < a satisfies i = 2 kt for some t , 0 ≤ t < a k .Since v − a − k = v a − k is a g -broadcast vertex and 0 < a − k < a , we get a − k = 2 kt ′ , forsome t ′ , 0 < t ′ < a k . This gives a = (2 t ′ + 1) k , contradicting the assumption a = 2 p , sothat this last case cannot appear.3. a is even and a = 2 p for every p > a = (2 ℓ + 1)2 k for some positive integers k and ℓ . Let g be the mapping from V ( C (2 a ; 1 , a )) to { , k } defined by g ( v i ) = 2 k if and only if i ≡ k +1 ), whichgives V + g = { v p k +1 | ≤ p ≤ ℓ } .For any two g -broadcast vertices v i = v p k +1 and v j = v q k +1 , 0 ≤ p < q ≤ ℓ , we have d ( v i , v j ) = min { ( q − p )2 k +1 , | ( p − q )2 k +1 + a | + 1 , ( p − q )2 k +1 + 2 a } , which gives, since a = (2 ℓ + 1)2 k , d ( v i , v j ) = min { ( q − p )2 k +1 , | p − q + ℓ ) + 1 | k + 1 , ( p − q + 2 ℓ + 1)2 k +1 } ≥ k + 1 . Therefore, g is an independent broadcast on C (2 a ; 1 , a ), with cost σ ( g ) = 2 k (cid:0) a k +1 (cid:1) = a ,which gives β b ( C (2 a ; 1 , a )) ≥ σ ( g ) = a and thus, since g satisfies (2), β b ( C (2 a ; 1 , a )) = a .This completes the proof. (cid:3) We finally consider the case n = 3 a . Theorem 5.
For every integer a ≥ , β b ( C (3 a ; 1 , a )) = α ( C (3 a ; 1 , a )) = a. Proof.
Let f be an independent β b -broadcast on C (3 a ; 1 , a )). For each vertex v i ∈ V + f , we let C if = { v i , . . . , v i + f ( v i ) − } ∪ { v i + a , . . . , v i + a + f ( v i ) − } ∪ { v i +2 a , . . . , v i +2 a + f ( v i ) − } . We clearly have | C if | = 3 f ( v i ) for every v i ∈ V + f , and C if ∩ C i ′ f = ∅ for every two distinct vertices v i and v i ′ in V + f , since otherwise we would have d ( v i , v i ′ ) ≤ max { f ( v i ) , f ( v i ′ ) } , contradictingthe fact that f is an independent broadcast. This gives3 f ( V + f ) = X v i ∈ V + f | C if | ≤ a, and thus β b ( C (3 a ; 1 , a )) = σ ( f ) = X v ∈ V + f f ( v ) = f ( V + f ) ≤ a a. Consider now the mapping f from V ( C (3 a ; 1 , a )) to { , } defined as follows, depending onthe parity of a . 5 v i Figure 2: The set D f ( v i ) (black vertex and grey vertices), with a = 6 and f ( v i ) = 2.1. If a is odd, then f ( v i ) = 1 if and only if i is even and i < a .2. If a is even, then f ( v i ) = 1 if and only if ( i mod a + 1) is odd and i ≤ a .In both cases, f is clearly an independent broadcast on V ( C (3 a ; 1 , a ) with σ ( f ) = a Thisimplies β b ( C (3 a ; 1 , a )) ≥ a and thus, thanks to Observation 2, β b ( C (3 a ; 1 , a )) = α ( C (3 a ; 1 , a )) = a . (cid:3) Recall that we denote by v , v , . . . , v n − the vertices of C ( n ; 1 , a ) and that subscripts are alwaysconsidered modulo n . We will say that an edge v i v j is a k -edge for some integer k , 1 ≤ k ≤ (cid:4) n (cid:5) ,if j = i + k or i = j + k . Therefore, every edge in C ( n ; 1 , a ) is either a 1-edge or an a -edge.Let f be an independent broadcast on C ( n ; 1 , a ). For every f -broadcast vertex v i ∈ V + f , wedenote by D f ( v i ) the set of vertices that are f -dominated by v i , that is D f ( v i ) = S f ( v i ) k =0 { v j | i − ( f ( v i ) − k ) a − k ≤ j ≤ i − ( f ( v i ) − k ) a + k }∪ S f ( v i ) k =0 { v j | i + ( f ( v i ) − k ) a − k ≤ j ≤ i + ( f ( v i ) − k ) a + k } . Figure 2 illustrates this definition on a circulant graph of the form C ( n ; 1 ,
6) (with n ≥ v i with f ( v i ) = 2.Let us say that an independent broadcast f is ℓ -bounded , for some integer ℓ ≥
1, if f ( v ) ≤ ℓ for every vertex v . In particular, a 1-bounded independent broadcast is the characteristicfunction of an independent set. This implies that such a broadcast always exists for everygraph, and thus that every graph admits an ℓ -bounded independent broadcast for every ℓ ≥ C ( n ; 1 , a ),2 ≤ a ≤ n , admit a 2-bounded optimal independent broadcast. Considering the β b -broadcastsused in the proofs of Theorems 4 and 5, we already have the following result. Proposition 6.
For every integer a ≥ , the following holds.1. C (2 a ; 1 , a ) admits a -bounded β b -broadcast if a is odd or a = 2 p for some p ≥ .2. C (3 a ; 1 , a ) admits a -bounded β b -broadcast. We will now prove that every circulant graph of the form C ( n ; 1 , a ), a ≥ n ≥ a + 2,admits a 2-bounded β b -broadcast. We first consider the case when 2 a + 2 ≤ n < a . Lemma 7. If n , a and r are three integers such that n = 2 a + r , ≤ a ≤ (cid:4) n (cid:5) and ≤ r < a ,then C ( n ; 1 , a ) admits a -bounded β b -broadcast. i (3 , v i + a v i − a (a) Item 1: f ( v i ) ∈ { , } , a = 10 and r = 5 (so that p = 0 and n = 25) v i (6)1 1 11 1 1 1 v i + a v i − a (b) Item 1: f ( v i ) = 6, a = 8 and r = 7 (so that p = 2 and n = 23) v i (5)1 1 11 1 1 v i + a v i − a (c) Item 2: f ( v i ) = 5, a = 12 and r = 2 (so that p = 2 and n = 26) v i (6)1 1 1 11 1 1 v i + a v i − a (d) Item 3: f ( v i ) = 6, a = 12 and r = 3 (so that d = 2 and n = 27) v i (5)1 1 11 1 1 v i + a v i − a (e) Item 3: f ( v i ) = 5, a = 12 and r = 3 (so that d = 1 and n = 27)Figure 3: Construction of the mapping g in the proof of Lemma 7.7 roof. Note that it is enough to prove that for every independent broadcast f on C ( n ; 1 , a ),there exists an independent broadcast g on C ( n ; 1 , a ) such that σ ( g ) ≥ σ ( f ) and g ( v ) ≤ v ∈ V + g .Let f be any independent broadcast on C ( n ; 1 , a ), and g be the mapping from V ( C ( n ; 1 , a ))to { , , } defined as follows (the construction of the mapping g is illustrated in Figure 3,where the value of f ( v i ) is indicated in brackets; not all a -edges are drawn, but the missing a -edges are parallel to the drawn ones; note also that v i − a = v i + a + r , and thus v i + a and v i − a areseparated by r − v i is an f -broadcast vertex such that 2 < f ( v i ) ≤ r , then we let g ( v j ) = j = i, j = i − a , or i − ≤ j ≤ i + p + 1 and j − i + 1 is even , or i + a ≤ j ≤ i + a + p and j − i − a is even , where p = f ( v i ) − f ( v i ) is odd, and p = f ( v i ) − f ( v i ) is even (see Figure 3(a,b)).2. If v i is an f -broadcast vertex such that l a m > f ( v i ) ≥ r + 1 and r is even, then we let g ( v j ) = j = i, i − ≤ j ≤ i + p + 1 and j − i + 1 is even , or i + a ≤ j ≤ i + a + r + p and j − i − a is even , where p = f ( v i ) − f ( v i ) is odd, and p = f ( v i ) − f ( v i ) is even (see Figure 3(c)).3. If v i is an f -broadcast vertex such that f ( v i ) ≥ r + 1 and r is odd, then we let g ( v j ) = j = i, i − ≤ j ≤ i + (cid:22) d + 22 (cid:23) r + ( d mod 2)and ( j − i + 2) mod ( r + 2) is odd , or i + a ≤ j ≤ i + a + (cid:24) d + 22 (cid:25) r + 1 − ( d mod 2)and ( j − i − a + 3) mod ( r + 2) is odd , where d = f ( v i ) − ( r + 1) (see Figure 3(d,e)).4. For every other vertex v k , we let g ( v k ) = f ( v k ).Note that, in particular, g ( v i ) = f ( v i ) for every f -broadcast vertex v i with f ( v i ) ≤ v i and at distance notgreater than f ( v i ) from v i , which means that their f -value was 0.We now prove that g is an independent broadcast on C ( n ; 1 , a ). For that, we first prove thefollowing claim. Claim A.
For every vertex v j whose g -value is set to 1 in Item 1, 2 or 3 above, we have d ( v i , v j ) ≤ f ( v i ) − . roof. In Item 1, every vertex whose g -value is set to 1 is at distance at most p + 1 ≤ f ( v i ) − v i . In Item 2, every vertex whose g -value is set to 1 is at distance at most max { p +1 , r − } ≤ f ( v i ) − v i .Consider now a vertex v j whose g -value is set to 1 in Item 3 and whose distance to v i ismaximal (see Figure 3(d,e)), and suppose first that i − ≤ j ≤ i + (cid:4) d +22 (cid:5) r + ( d mod 2). Since r ≥ r is odd, v j is at distance d ( v i , v j ) = 2 (cid:4) d +22 (cid:5) + (cid:4) r (cid:5) − v i (going to v i + ⌊ d +22 ⌋ r using (2 (cid:4) d +22 (cid:5) ) a -edges, and then back to v j using ( (cid:4) r (cid:5) −
1) 1-edges). Since d = f ( v i ) − r − r ≥
3, this gives d ( v i , v j ) ≤ f ( v i ) − r + 1 + r − − f ( v i ) − r + 12 ≤ f ( v i ) − . Suppose finally that i + a ≤ j ≤ i + a + (cid:6) d +22 (cid:7) r + 1 − ( d mod 2). In that case, v j is at distance d ( v i , v j ) = 1 + 2( (cid:4) d +22 (cid:5) −
1) + (cid:4) r (cid:5) − v i (going to v i + a + ⌈ d +22 ⌉ r using (1 + 2( (cid:4) d +22 (cid:5) − a -edges, and then back to v j using ( (cid:4) r (cid:5) −
1) 1-edges). As before, since d = f ( v i ) − r − r ≥
3, this gives d ( v i , v j ) ≤ (cid:18)(cid:22) d + 22 (cid:23) − (cid:19) + r − − (cid:18)(cid:22) f ( v i ) − r + 12 (cid:23) − (cid:19) + r − ≤ (cid:18) f ( v i ) − r + 12 (cid:19) + r − − f ( v i ) − r − − ≤ f ( v i ) − , which concludes the proof of the claim. (cid:3) Thanks to this claim, and since f was an independent broadcast on C ( n ; 1 , a ), no g -broadcastvertex v i with g ( v i ) = f ( v i ) ∈ { , } g -dominates a vertex whose g -value has been set to 1.Therefore, in order to prove that g is indeed an independent broadcast on C ( n ; 1 , a ), it remainsto prove that the set of vertices whose g -value has been set to 1 is an independent set. Moreover,thanks to Claim A, Items 1, 2 and 3 can be considered separately. This is readily the case forvertices whose g -value has been set to 1 in Item 1 and 2. In Item 3, thanks to the parity oftheir subscript, no two such vertices are linked by a 1-edge. Moreover, any two such verticescannot be linked by an a -edge since ( j − i + 2) − ( j − i − a + 3) = a − σ ( g ) ≥ σ ( f ). Indeed, inItem 1, the number of vertices set to 1 is n = 1 + p +42 + p +22 = p + 4, which gives n = f ( v i ) if f ( v i ) is even, or n = f ( v i ) + 1 > f ( v i ) if f ( v i ) is odd.In Item 2, the number of vertices set to 1 is n = p +42 + p +22 + r = p +6+ r , which gives n = f ( v i ) − r ≥ f ( v i ) (recall that r ≥
2) if f ( v i ) is even, or n = f ( v i )+ r > f ( v i ) if f ( v i ) isodd.Finally consider Item 3. Observe that, since r +2 is odd, in every sequence of r +2 consecutivevertices lying between v i − and v i + ⌊ d +22 ⌋ r +( d mod 2) , or between v i + a and v i + a + ⌈ d +22 ⌉ r +1 − ( d mod 2) ,exactly r +12 vertices are set to 1. Therefore, the number of vertices set to 1 in Item 3 is either n = (cid:24) r + 12( r + 2) (cid:18) d + 22 r + 3 (cid:19)(cid:25) + (cid:24) r + 12( r + 2) (cid:18) d + 22 r + 2 (cid:19)(cid:25) , if d is even, or n = (cid:24) r + 12( r + 2) (cid:18) d + 12 r + 4 (cid:19)(cid:25) + (cid:24) r + 12( r + 2) (cid:18) d + 32 r + 1 (cid:19)(cid:25) , g f ( v i ) ∈ { , } and a = 7 (so that p = 2) fg f ( v i ) = 7 and a = 5 (so that d = 1) fg f ( v i ) = 7 and a = 4 (so that d = 2)Figure 4: Construction of the mapping g in the proof of Lemma 8.if d is odd. In both cases, we get n ≥ r + 12( r + 2) (cid:0) ( d + 2) r + 5 (cid:1) = r + 1 + ( dr + 1)( r + 1)2( r + 2) ≥ r + 1 + dr ( r + 1)2( r + 2) . Since r ≥
3, we have r ( r + 1) ≥ r + 2), and thus n ≥ r + 1 + d = f ( v i ) . We thus have σ ( g ) ≥ σ ( f ), as required. This completes the proof. (cid:3) We now consider the case n > a . Lemma 8. If n and a are two integers such that ≤ a ≤ (cid:4) n (cid:5) and n > a , then C ( n ; 1 , a ) admits a -bounded β b -broadcast. Proof.
Again, it is enough to prove that for every independent broadcast f on C ( n ; 1 , a ), thereexists an independent broadcast g on C ( n ; 1 , a ) such that σ ( g ) ≥ σ ( f ) and g ( v ) ≤ v ∈ V + g .Let f be any independent broadcast on C ( n ; 1 , a ), and g be the mapping from V ( C ( n ; 1 , a ))to { , , } defined as follows (the construction of the mapping g is illustrated in Figure 4, notall a -edges being drawn).1. If v i is an f -broadcast vertex such that 2 < f ( v i ) ≤ a , then we let g ( v j ) = j = i, i − ≤ j ≤ i + p + 1 and j − i + 1 is even,or i − a ≤ j ≤ i − a + p and j − i + a is even,or i + a ≤ j ≤ i + a + p and j − i − a is even , where p = f ( v i ) − f ( v i ) is odd, and p = f ( v i ) − f ( v i ) is even (see Figure 4(a)).10. If v i is an f -broadcast vertex such that f ( v i ) ≥ a + 1 and a is odd, then we let g ( v j ) = ( j = i, i − a ≤ j ≤ i + (1 + d ) a and j − i + a is even , where d = f ( v i ) − ( a + 1) (see Figure 4(b)).3. If v i is an f -broadcast vertex such that f ( v i ) ≥ a + 1 and a is even, then we let g ( v j ) = ( j = i i − a − ≤ j ≤ i + (2 + d ) a and ( j − i + a + 1) mod ( a + 1) is odd , where d = f ( v i ) − ( a + 1) (see Figure 4(c)).4. For every other vertex v k , we let g ( v k ) = f ( v k ).Note that, as in the proof of the previous lemma, g ( v i ) = f ( v i ) for every f -broadcast vertex v i with f ( v i ) ≤
2. Moreover, all vertices set to 1 in the above items are also distinct from v i and at distance not greater than f ( v i ) from v i , which means that their f -value was 0.We now prove that g is an independent broadcast on C ( n ; 1 , a ). For that, we first prove thefollowing claim. Claim B.
For every vertex v j whose g -value is set to 1 in Item 1, 2 or 3 above, we have d ( v i , v j ) ≤ f ( v i ) − . Proof.
In Item 1, every vertex whose g -value is set to 1 is at distance at most p + 1 ≤ f ( v i ) − v i .Among the vertices whose g -value might be set to 1 in Item 2, the vertex whose distance to v i is maximal is, since a is odd, the vertex v j with j = da + a +12 , which gives d ( v i , v j ) = d + a + 12 = f ( v i ) − a + 12 ≤ f ( v i ) − a ≥ g -value might be set to 1 in Item 3, the vertex whosedistance to v i is maximal is, since a is even, the vertex v j with j = ( d + 1) a + a , which gives d ( v i , v j ) = d + 1 + a f ( v i ) − a ≤ f ( v i ) − a ≥ (cid:3) Thanks to this claim, and since f was an independent broadcast on C ( n ; 1 , a ), no g -broadcastvertex v i with g ( v i ) = f ( v i ) ∈ { , } g -dominates a vertex whose g -value has been set to 1.Therefore, in order to prove that g is indeed an independent broadcast on C ( n ; 1 , a ), it remainsto prove that the set of vertices whose g -value has been set to 1 is an independent set. Moreover,thanks to Claim B, Items 1, 2 and 3 can be considered separately. This is readily the case forvertices whose g -value has been set to 1 in Item 1. It follows from the parity of their subscriptin Item 2 (neither a 1-edge nor an a -edge, since a is odd, can link any two such vertices), andfrom the value modulo ( a + 1) of their subscript in Items 3 (which, again, implies that neithera 1-edge nor an a -edge, since a is even, can link any two such vertices).11n order to finish the proof, we only need to show that we have σ ( g ) ≥ σ ( f ). Indeed,in Item 1, the number of vertices set to 1 is n = p +42 + p +22 + p +22 = p +82 , which gives n = f ( v i ) − ≥ f ( v i ) if f ( v i ) is even (in that case, f ( v i ) ≥
4, and equality holds only when f ( v i ) = 4), or n = f ( v i ) − > f ( v i ) if f ( v i ) is odd (in that case, f ( v i ) ≥ n = l (2+ d ) a +12 m = (cid:6) a +1+ da (cid:7) . If d = 0, since a is odd, we get n = (cid:6) a +12 (cid:7) = a + 1 = f ( v i ). Otherwise, that is, if d ≥
1, since a ≥
3, we get n = (cid:6) a +1+ da (cid:7) ≥ a + + ad ≥ a + d + 1 = f ( v i ).Finally, in Item 3, note that, for every sequence of a + 1 consecutive vertices, a of them areset to 1. Therefore, the total number of vertices set to 1 is n = (cid:24) a a + 1) (( d + 3) a + 2) (cid:25) ≥ a + da + 2 a a + 1) = a + da a + 1) + a a + 1) . Since a ≥
4, we have a a +1) >
1, which gives n > a + d + 1 = f ( v i ).We thus have σ ( g ) ≥ σ ( f ), as required. This completes the proof. (cid:3) From Proposition 6 and Lemmas 7 and 8, we directly get the following theorem.
Theorem 9.
Every circulant graph of the form C ( n ; 1 , a ) , ≤ a ≤ (cid:4) n (cid:5) , admits a -bounded β b -broadcast if none of the following conditions is satisfied: (i) n = 2 a and a is even, or (ii) n = 2 a + 1 . The following example will show that when a = 2 or n = 2 a + 1, not all circulant graphs ofthe form C ( n ; 1 , a ) admit a 2-bounded β b -broadcast. Consider the circulant graph C (21; 1 , f be the mapping from V ( C (21; 1 , { , } defined by f ( v ) = f ( v ) = f ( v ) = 3and f ( v i ) = 0 otherwise. Since 2 is even, f is clearly an independent broadcast on C (21; 1 , σ ( f ) = 9, and thus 9 ≤ β b ( C (21; 1 , β b -broadcast g on C (21; 1 , | V + g | ≤
4, we immediately get σ ( g ) ≤
8, since g is 2-bounded. Sup-pose now | V + g | >
4. Each vertex v i ∈ V + g dominates at most three vertices among { v i , v i +1 , v i +2 } (subscripts are taken modulo 21), and none of these vertices is dominated more than once.Therefore, since g is an independent broadcast, we get X v i ∈ V + g (1 + 2 g ( v i )) = | V + g | + 2 g ( V + g ) ≤ , which gives (recall that | V + g | > σ ( g ) = g ( V + g ) ≤ − | V + g | ≤ . In both cases, we get a contradiction to the optimality of g . Finally, since C (21; 1 ,
10) is iso-morphic to C (21; 1 , C (2 a + 1; 1 , a )that do not admit any 2-bounded β b - broadcast. C ( n ; 1 , a ) In this section, we will provide some general upper and lower bounds on the cost of independentbroadcasts on circulant graphs of the form C ( n ; 1 , a ), 2 ≤ a ≤ (cid:4) n (cid:5) , that will be useful in thenext section. 120 0 v i A if The vertices of the set A if , a = 7 v j B jf (the black vertices) , a = 7Figure 5: The sets A if and B jf .We first introduce some notation and a useful lemma. Let f be an independent broadcaston C ( n ; 1 , a ). We then let V f = { v i ∈ V + f | f ( v i ) = 1 } , V f = { v j ∈ V + f | f ( v j ) = 2 } and V ≥ f = { v j ∈ V + f | f ( v j ) ≥ } . In particular, if f is 2-bounded, we then have V + f = V f ∪ V f .Consider now a 2-bounded independent broadcast f and any vertex v i ∈ V f such that f ( v i − ) = f ( v i − ) = 0. Since f is an independent broadcast, we necessarily have f ( v i +1 ) = 0.Moreover, we then have either f ( v i +2 ) = 0 or f ( v i +2 ) = 1. Therefore, the broadcast values ofthe sequence of vertices v i v i +1 v i +2 . . . is of the form either 100, 10100 or 1010 . . . v i ∈ V f such that f ( v i − ) = f ( v i − ) = 0, we then let A if = { v i + ℓ , ≤ ℓ ≤ p + 2 } be the set of vertices satisfying (i) f ( v i +2 k ) = 1 and f ( v i +2 k +1 ) = 0 for every k , 0 ≤ k ≤ p , and(ii) f ( v i +2 p +2 ) = 0.Now, for each vertex v j ∈ V f , we let B jf = { v j − a +1 } ∪ { v j , v j +1 , v j +2 } ∪ { v j + a +1 } . The definition of these two sets is illustrated in Figure 5. These sets have the followingproperties.
Lemma 10.
For every -bounded independent broadcast f on C ( n ; 1 , a ) , ≤ a ≤ (cid:4) n (cid:5) , thefollowing holds.1. For every vertex v i ∈ V f , | A if | = 2 f ( A if ) + 1 .2. For every vertex v j ∈ V f , | B jf | = 5 .3. P v i ∈ V f | A if | + P v j ∈ V f | B jf | ≤ n . Proof.
The first two items directly follow from the definition of the sets A if and B jf . It alsofollows from the definition that A if ∩ A i ′ f = ∅ for every two distinct vertices v i and v i ′ in V f .13imilarly, we necessarily have B jf ∩ B j ′ f = ∅ for every two distinct vertices v j and v j ′ in V f ,since otherwise we would have d ( v j , v j ′ ) ≤ max { f ( v j ) , f ( v j ′ ) } = 2, contradicting the fact that f is an independent broadcast. The same argument gives A if ∩ B jf = ∅ for every two vertices v i ∈ V f and v j ∈ V f . All together, these three properties imply that Item 3 also holds. (cid:3) The next result provides a general upper bound on the broadcast independence number ofcirculant graphs of the form C ( n ; 1 , a ), with 3 ≤ a ≤ (cid:4) n (cid:5) and 3 a ≤ n . Proposition 11. If n and a are two integers such that ≤ a ≤ (cid:4) n (cid:5) and a ≤ n , then, forevery -bounded independent broadcast f on C ( n ; 1 , a ) , we have σ ( f ) ≤ $ n − (cid:12)(cid:12) V f (cid:12)(cid:12) % . Proof.
Let f be any 2-bounded independent broadcast on C ( n ; 1 , a ) (it follows from Theorem 9that such broadcasts exist). From Lemma 10, we get X v i ∈ V f | A if | + X v j ∈ V f | B jf | = 2 f ( V f ) + (cid:12)(cid:12) V f (cid:12)(cid:12) + 3 f ( V f ) − (cid:12)(cid:12) V f (cid:12)(cid:12) ≤ n, which gives2 f ( V + f ) = 2 f ( V f ) + 2 f ( V f ) ≤ n − (cid:12)(cid:12) V f (cid:12)(cid:12) + (cid:12)(cid:12) V f (cid:12)(cid:12) − f ( V f ) ≤ n + (cid:12)(cid:12) V f (cid:12)(cid:12) − f ( V f ) . Now, since f ( v j ) = 2 for every v j ∈ V f , we have f ( V f ) = 2 (cid:12)(cid:12) V f (cid:12)(cid:12) , and thus σ ( f ) = f ( V + f ) ≤ $ n − (cid:12)(cid:12) V f (cid:12)(cid:12) % . This completes the proof. (cid:3)
When a is even, the upper bound given in Proposition 11 can be improved as follows. Proposition 12. If n and a are two integers such that ≤ a ≤ (cid:4) n (cid:5) and a is even, then, forevery -bounded independent broadcast f on C ( n ; 1 , a ) , we have σ ( f ) ≤ (cid:22) a a + 1) (cid:18) n − a − a (cid:12)(cid:12) V f (cid:12)(cid:12)(cid:19)(cid:23) . Proof.
Let f be any 2-bounded independent broadcast on C ( n ; 1 , a ). Observe first that wenecessarily have | A if | ≤ a + 1 for every vertex v i ∈ V f , since otherwise this would give f ( v i ) = f ( v i + a ) = 1, contradicting the fact that f is an independent broadcast. This implies f ( A if ) ≤ a .Using item 1 of Lemma 10, we then get | A if | f ( A if ) = 2 f ( A if ) + 1 f ( A if ) = 2 + 1 f ( A if ) ≥ a = 2( a + 1) a , and thus | A if | ≥ a + 1) a f ( A if ) . From Lemma 10, we then get n ≥ X v i ∈ V f | A if | + X v j ∈ V f | B jf | ≥ a + 1) a f ( V f ) + 3 f ( V f ) − (cid:12)(cid:12) V f (cid:12)(cid:12) , i v i + a v i +2 a ( a + 1 vertices) ( a + 1 vertices)(a) two consecutive sequences of a + 1 vertices v i v i + a v i +2 a ( a + 1 vertices) ( a − v i v i + a v i +2 a ( a − a − a − f in the proof of Proposition 13 ( a = 12) .which gives n ≥ a + 1) a f ( V ) + a − a f ( V f ) − (cid:12)(cid:12) V f (cid:12)(cid:12) . Finally, since f ( V f ) = 2 (cid:12)(cid:12) V f (cid:12)(cid:12) , we get2( a + 1) a f ( V ) ≤ n + (cid:12)(cid:12) V f (cid:12)(cid:12) − a − a (cid:12)(cid:12) V f (cid:12)(cid:12) = n − a − a (cid:12)(cid:12) V f (cid:12)(cid:12) , and thus σ ( f ) = f ( V ) ≤ (cid:22) a a + 1) (cid:18) n − a − a (cid:12)(cid:12) V f (cid:12)(cid:12)(cid:19)(cid:23) . This completes the proof. (cid:3)
Proposition 13. If n , a , k and k are four integers such that n = k ( a + 1) + k ( a − , ≤ a ≤ (cid:4) n (cid:5) , and a is even, then, for every independent broadcast f on C ( n ; 1 , a ) , we have σ ( f ) ≥ k (cid:16) a (cid:17) + k (cid:16) a − (cid:17) . Proof.
Let n = k ( a + 1) + k ( a − C ( n ; 1 , a ) consists of k sequencesof a + 1 vertices and k sequences of a − f be a mapping from V ( C ( n ; 1 , a ) to { , } , defined as follows (see Figure 6 for the case a = 12). For every sequence of a + 1 or a − . . . a is even, for everytwo consecutive sequences, the f -broadcast vertices are pairwise non adjacent and then, f isan independent broadcast on C ( n ; 1 , a ), with cost σ ( f ) = k (cid:16) a (cid:17) + k (cid:16) a − (cid:17) . Hence, σ ( f ) ≥ k (cid:16) a (cid:17) + k (cid:16) a − (cid:17) . This completes the proof. (cid:3) Some exact values
We determine in this section the broadcast independence number of circulant graphs of the form C ( n ; 1 , a ), for various values of n and a . In several cases, we prove, thanks to Observation 2, thatthe independence number and the broadcast independence number of these graphs coincide.In [11], Liancheng, Zunquan and Yuansheng determined the exact value of the independencenumber of some circulant graphs of the form C ( n ; 1 , a ). Proposition 14 (Liancheng et al. [11]) . For every two integers n and a with ≤ a ≤ (cid:4) n (cid:5) , wehave1. α ( C ( n ; 1 , a )) = n , for even n and odd a ,2. α ( C ( n ; 1 , a )) = n − k , for odd n and a ∈ { , } ,3. α ( C ( n ; 1 , (cid:4) n (cid:5) ,4. α ( C ( n ; 1 , (cid:4) n (cid:5) . Several of our results in this section will thus extend the results of Proposition 14.We first consider the case of circulant graphs of the form C ( n ; 1 , n ≥
4. It is not difficultto check that, for every n ≥
4, antipodal vertices in C ( n ; 1 ,
2) are at distance (cid:6) n − (cid:7) apart fromeach other. We thus have the following. Observation 15.
For every integer n , n ≥ , diam( C ( n ; 1 , (cid:24) n − (cid:25) . The broadcast independence number of circulant graphs of the form C ( n ; 1 ,
2) is given bythe following result.
Theorem 16.
For every integer n ≥ , β b ( C ( n ; 1 , α ( C ( n ; 1 , , if n ∈ { , } , n − , if n ≡ , C ( n ; 1 , −
1) = 2 (cid:18)(cid:24) n − (cid:25) − (cid:19) , otherwise. Proof.
Since C (4; 1 ,
2) and C (5; 1 ,
2) are both complete graphs, the result obviously holds for n ∈ { , } .Suppose now n ≥
6. By Proposition 1, β b ( C ( n ; 1 , ≥ C ( n ; 1 , −
1) holds forevery n . We will prove that we have β b ( C ( n ; 1 , ≤ C ( n ; 1 , −
1) if n β b ( C ( n ; 1 , n − otherwise.Let f be an independent β b -broadcast on C ( n ; 1 , v ∈ V + f f -dominates4 f ( v ) + 1 vertices. Moreover, each f -broadcast vertex is f -dominated exactly once, and eachnon-broadcast vertex is f -dominated at most twice. This gives4 f ( V + f ) + | V + f | ≤ (cid:0) n − | V + f | (cid:1) + | V + f | , and thus β b ( C ( n ; 1 , σ ( f ) = X v ∈ V + f f ( v ) = f ( V + f ) ≤ n − | V + f | . We now consider three cases, depending on the value of | V + f | .16. | V + f | ≤ | V + f | = 1, then V + f = { v i } for some vertex v i , and thus σ ( f ) = f ( v i ) ≤ e ( v i ) = diam( C ( n ; 1 , ≤ C ( n ; 1 , − . If | V + f | = 2, then V + f = { v i , v j } for some distinct vertices v i and v j , and thus σ ( f ) = f ( v i ) + f ( v j ) ≤ C ( n ; 1 , − . | V + f | ≥ σ ( f ) ≤ $ n − | V + f | % ≤ (cid:22) n − (cid:23) ≤ (cid:18)(cid:24) n − (cid:25) − (cid:19) and thus σ ( f ) ≤ (cid:0)(cid:6) n − (cid:7) − (cid:1) = 2(diam( C ( n ; 1 , −
1) by Observation 15.3. | V + f | = 3.Let V + f = { v i , v i , v i } , with 0 ≤ i < i < i < n −
1. We consider two subcases,depending in the parity of n .(a) n is even.Since f is a β b -broadcast, we have f ( v i j ) = min (cid:8) d ( v i j , v i j − ) − , d ( v i j , v i j +1 ) − (cid:9) for every j , 0 ≤ j ≤ { x, y } ≤ x + y for every two integers x and y , we get σ ( f ) = f ( v i ) + f ( v i ) + f ( v i ) ≤ d ( v i , v i ) + d ( v i , v i ) + d ( v i , v i ) − . Now, since d ( v i j , v i j ′ ) = (cid:24) | i j − i j ′ | (cid:25) ≤ | i j − i j ′ | + 12for every two distinct vertices v i j and v i j ′ , we get σ ( f ) ≤ (cid:22) | i − i | + | i − i | + | i − i | + 32 (cid:23) − (cid:22) n − (cid:23) . Finally, since n is even, we get σ ( f ) ≤ (cid:22) n − (cid:23) = n − ≤ (cid:18)(cid:24) n − (cid:25) − (cid:19) = 2(diam( C ( n ; 1 , − . (b) n is odd.If every non-broadcast vertex is f -dominated exactly twice, then we necessarily have f ( v i ) = f ( v i ) = f ( v i ) = ℓ for some value ℓ . Moreover, since each vertex v i j ,0 ≤ j ≤ f -dominates 4 f ( v i j ) + 1 = 4 ℓ + 1 vertices, we get 12 ℓ + 3 = 2( n −
3) + 3(each vertex in V + f is f -dominated only once), and thus ℓ = n − . This implies n ≡ σ ( f ) = n − . 17ow, we have σ ( f ) = n −
32 = 3 ℓ = 2 (cid:18)(cid:24) n − (cid:25) − (cid:19) = 2(diam( C ( n ; 1 , − ℓ is even, that is n ≡ σ ( f ) = n −
32 = 3 ℓ > ℓ − (cid:18)(cid:24) n − (cid:25) − (cid:19) = 2(diam( C ( n ; 1 , − ℓ is odd, that is n ≡ f -dominated only once, whichimplies 4 f ( V + f ) + 3 ≤ n −
4) + 4 = 2 n − σ ( f ) = f ( V + f ) ≤ (cid:22) n − (cid:23) . Since n is odd, we get σ ( f ) ≤ (cid:22) n − (cid:23) = (cid:22) n − (cid:23) ≤ (cid:18)(cid:24) n − (cid:25) − (cid:19) = 2(diam( C ( n ; 1 , − . In all cases, we thus get β b ( C ( n ; 1 , σ ( f ) ≤ f ( V + f ) ≤ C ( n ; 1 , −
1) if n β b ( C ( n ; 1 , σ ( f ) = n − if n ≡ (cid:3) Comparing the value of α ( C ( n ; 1 , β b ( C ( n ; 1 , α ( C ( n ; 1 , < β b ( C ( n ; 1 , C (2 a + 1; 1 , a ) and C (2 a + 1; 1 ,
2) are isomorphic for every integer a , a ≥
2, Theorem 16 admits the following corollary.
Corollary 17.
For every integer a ≥ , β b ( C (2 a + 1; 1 , a )) = a − , if a = 2 , or a ≡ , (cid:16)l a m − (cid:17) , otherwise. We now determine the broadcast independence number of circulant graphs of the form C ( n ; 1 , a ) when n is even and a is odd. Theorem 18. If n and a are two integers such that n is even, n ≥ , a is odd and ≤ a ≤ (cid:4) n (cid:5) ,then β b ( C ( n ; 1 , a )) = α ( C ( n ; 1 , a )) = n . Proof.
From Proposition 11, we get that σ ( g ) ≤ j n − | V g | k for every 2-bounded independentbroadcast g on C ( n ; 1 , a ), which implies β b ( C ( n ; 1 , a )) ≤ n . Consider now the mapping f from V ( C ( n ; 1 , a )) to { , } defined by f ( v i ) = 1 if and only if i is even. Since a is odd, f is clearly anindependent broadcast on C ( n ; 1 , a ). This implies β b ( C ( n ; 1 , a )) ≥ σ ( f ) = n and thus, thanksto Observation 2, β b ( C ( n ; 1 , a )) = α ( C ( n ; 1 , a )) = n . This completes the proof. (cid:3) We are now able to determine the broadcast independence number of circulant graphs ofthe form C ( n ; 1 , heorem 19. For every integer n ≥ , β b ( C ( n ; 1 , α ( C ( n ; 1 , n , if n is even ,n − , otherwise. Proof. If n is even, the result directly follows from Theorem 18.Suppose now that n is odd and consider the mapping f from V ( C ( n ; 1 , { , } definedby f ( v i ) = 1 if and only if i is even and i ≤ n −
5. Since all broadcast vertices have an evenindex not greater than n − f is clearly a 1-bounded independent broadcaston C ( n ; 1 ,
3) with σ ( f ) = n − and V f = ∅ . We thus get β b ( C ( n ; 1 , ≥ n − and, thanks toObservation 2, β b ( C ( n ; 1 , α ( C ( n ; 1 , σ ( g ) ≤ j n − | V g | k for every 2-bounded independent broad-cast g on C ( n ; 1 , (cid:12)(cid:12) V g (cid:12)(cid:12) ≥
2, then σ ( g ) ≤ (cid:4) n − (cid:5) = n − = σ ( f ). If (cid:12)(cid:12) V g (cid:12)(cid:12) = 1, say V g = { v j } ,then we necessarily have g ( v j − ) = g ( v j − ) = g ( v j − ) = g ( v j − ) = 0, and thus v j − and v j − donot belong to any set A if for any v i ∈ V g . Using this remark together with Lemma 10, we thenget X v i ∈ V g | A ig | + X v j ∈ V g | B jg | = 2 f ( V g ) + (cid:12)(cid:12) V g (cid:12)(cid:12) + 3 f ( v j ) − ≤ n − , which gives, since f ( v j ) ≥ σ ( g ) = f ( V g ) + f ( v j ) ≤ n − − (cid:12)(cid:12) V g (cid:12)(cid:12) − f ( v j ) + 12 ≤ n −
32 = σ ( f ) . Finally, if (cid:12)(cid:12) V g (cid:12)(cid:12) = 0 then, since n is odd, there necessarily exists a vertex v i ∈ V g suchthat g ( v i +2 ) = 0, which implies g ( v i +1 ) = g ( v i +3 ) = 0. This implies that v i +3 does not belongto any set A i ′ g for any v i ′ ∈ V g . Using this remark together with Lemma 10, we then have2 f ( V g ) + (cid:12)(cid:12) V g (cid:12)(cid:12) ≤ n −
1, and thus σ ( g ) = f ( V g ) ≤ n − − (cid:12)(cid:12) V g (cid:12)(cid:12) ≤ (cid:22) n − (cid:23) = n −
32 = σ ( f ) . Hence, in all the previous cases, we have σ ( g ) ≤ σ ( f ) = n − , which completes the proof. (cid:3) We now determine the broadcast independence number of circulant graphs of the form C ( n ; 1 , Theorem 20.
For every integer n ≥ , β b ( C ( n ; 1 , α ( C ( n ; 1 , (cid:22) n (cid:23) . Proof.
From Proposition 12, we get that σ ( g ) ≤ (cid:22) a a + 1) (cid:18) n − a − a (cid:12)(cid:12) V g (cid:12)(cid:12)(cid:19)(cid:23) for every 2-bounded independent broadcast g on C ( n ; 1 , a ), which gives σ ( g ) ≤ (cid:4) n (cid:5) , and thus β b ( C ( n ; 1 , ≤ (cid:4) n (cid:5) .We now construct a mapping f from V ( C ( n ; 1 , { , } . Let n = 5 k + r with 0 ≤ r ≤ r .19. r = 0.We let f ( v i ) = 1 if ( i mod 5) is odd, and f ( v i ) = 0 otherwise.2. r = 1.We let f ( v i ) = 1 if ( i mod 5) is odd and i ≤ n − f ( v n − ) = f ( v n − ) = 1, and f ( v i ) = 0otherwise.3. r = 2.We let f ( v i ) = 1 if ( i mod 5) is odd and i ≤ n −
3, and f ( v i ) = 0 otherwise.4. r = 3.We let f ( v i ) = 1 if ( i mod 5) is odd, and f ( v i ) = 0 otherwise.5. r = 4.We let f ( v i ) = 1 if ( i mod 5) is odd and i ≤ n − f ( v n − ) = f ( v n − ) = f ( v n − ) = 1,and f ( v i ) = 0 otherwise.Clearly, in each of the previous cases, f is a 1-bounded independent broadcast on C ( n ; 1 , σ ( f ) = (cid:4) n (cid:5) and V f = ∅ . Hence, β b ( C ( n ; 1 , (cid:4) n (cid:5) and, thanks to Observation 2, β b ( C ( n ; 1 , α ( C ( n ; 1 , (cid:3) Thanks to Proposition 12, we are now able to determine the broadcast independence numberof circulant graphs of the form C (( a + 1) k ; 1 , a ) with a ≥ k ≥ a = 2, 3 and4 are already covered by Theorems 16, 19 and 20, respectively). Theorem 21. If a and k are two integers such that a ≥ and k ≥ , then we have β b ( C (( a + 1) k ; 1 , a )) = α ( C (( a + 1) k ; 1 , a )) = ak , if a is even , ( a + 1) k , otherwise. Proof. If a is odd, then ( a + 1) k is even and the result directly follows from Theorem 18.Suppose now that a is even, which implies a ≥
6. From Proposition 12, we get that σ ( g ) ≤ (cid:22) a a + 1) (cid:18) ( a + 1) k − a − a (cid:12)(cid:12) V g (cid:12)(cid:12)(cid:19)(cid:23) ≤ (cid:22) a a + 1) ( a + 1) k (cid:23) = ak g on C (( a + 1) k ; 1 , a ), which implies β b ( C (( a +1) k ; 1 , a )) ≤ ak . Consider now the mapping f from V ( C (( a + 1) k ; 1 , a )) to { , } defined by f ( v i ) = 1 if andonly if ( i mod a + 1) is odd. Since a is even, f is clearly a 1-bounded independent broadcaston C (( a + 1) k ; 1 , a ) with σ ( f ) = ak and V f = ∅ . This implies β b ( C (( a + 1) k ; 1 , a )) ≥ ak andthus, thanks to Observation 2, β b ( C (( a + 1) k ; 1 , a )) = α ( C (( a + 1) k ; 1 , a )) = ak . This completesthe proof. (cid:3) We now consider the case of circulant graphs C ( n ; 1 , a ) when a divides n . We first introducetwo new sets of vertices, slightly modifying the definition of the sets A if and B jf defined inSection 2, using a -edges instead of 1-edges. Let f be any 2-bounded independent broadcaston C ( n ; 1 , a ). Now consider any vertex v i ∈ V f such that f ( v i − a ) = f ( v i − a ) = 0. Since f is an independent broadcast, we necessarily have f ( v i + a ) = 0. Moreover, we then have either f ( v i +2 a ) = 0 or f ( v i +2 a ) = 1. Therefore, the broadcast values of the sequence of vertices v i v i + a v i +2 a . . . is of the form either 100, 10100 or 1010 . . . v i ∈ V f such that f ( v i − a ) = f ( v i − a ) = 0, we then let A ′ if = { v i + ℓa , ≤ ℓ ≤ p + 2 } be the set of vertices satisfying (i) f ( v i +2 ka ) = 1 and f ( v i +(2 k +1) a ) = 0 for every k , 0 ≤ k ≤ p ,and (ii) f ( v i +(2 p +2) a ) = 0.Now, for each vertex v j ∈ V f , we let B ′ jf = { v j } ∪ { v j + a − , v j + a , v j + a +1 } ∪ { v j +2 a } . These sets satisfy the same properties as those of the sets A if and B jf given in Lemma 10.The proof is similar to the proof of Lemma 10 and is omitted. Lemma 22.
For every -bounded independent broadcast f on C ( n ; 1 , a ) , if any, the followingholds.1. For every vertex v i ∈ V f , | A ′ if | = 2 f ( A ′ if ) + 1 .2. For every vertex v j ∈ V f , | B ′ jf | = 5 .3. P v i ∈ V f | A ′ if | + P v j ∈ V f | B ′ jf | ≤ n . We are now ready to determine the independent broadcast number of circulant graphs of theform C ( qa ; 1 , a ), a ≥ q ≥
4. Recall that the cases a = 2, 3 and 4 are already covered byTheorems 16, 19 and 20, respectively, while the cases q = 2 and q = 3 are covered by Theorems4 and 5, respectively. Theorem 23. If a and q are two integers such that a ≥ and q ≥ , then we have β b ( C ( qa ; 1 , a )) = α ( C ( qa ; 1 , a ))= qa , if a is odd, and q is even, ( q − a , if a and q are odd, (cid:22) qa a + 1) (cid:23) , if a and q are even, min (cid:26)(cid:22) qa a + 1) (cid:23) , ( q − a (cid:27) , otherwise. Proof.
We consider the four cases separately.1. a is odd and q is even.In that case, qa is even and the result directly follows from Theorem 18.2. a and q are odd.In that case, q ≥ C ( qa ; 1 , a ) admits a 2-bounded β b -broadcast. Let f be any 2-bounded independent broadcast on C ( qa ; 1 , a ). Observe firstthat, since q is odd, we necessarily have | A ′ if | ≤ q for every vertex v i ∈ V f , since otherwisethis would give f ( v i ) = f ( v i +( q − a ) = f ( v i − a ) = 1, contradicting the fact that f is an21ndependent broadcast. Therefore f ( A ′ if ) ≤ q − , and thanks to Item 1 of Lemma 22, weget | A ′ if | f ( A ′ if ) = 2 f ( A ′ if ) + 1 f ( A ′ if ) = 2 + 1 f ( A ′ if ) ≥ q − qq − , which gives | A ′ if | ≥ qq − f ( A ′ if ) . Now, using Item 2 and Item 3 of Lemma 22, we get qa ≥ X v i ∈ V f | A ′ if | + X v j ∈ V f | B ′ jf | ≥ qq − f ( V f ) + 52 f ( V f ) , which gives qa ≥ qq − σ ( f ) + q − q − f ( V f ) ≥ qq − σ ( f ) , and then σ ( f ) ≤ ( q − a . Consider now the mapping g from V ( C ( qa ; 1 , a )) to { , } defined by g ( v i ) = 1 if i is evenand i ≤ ( q − a −
1, and g ( v i ) = 0 otherwise. Clearly, g is a 1-bounded independentbroadcast on C ( qa ; 1 , a ). Moreover, β b ( C ( qa ; 1 , a )) ≥ σ ( g ) = ( q − a , and thus, thanks to Observation 2, β b ( C ( qa ; 1 , a )) = α ( C ( qa ; 1 , a )) = ( q − a . a and q are even.Note first that if a + 1 divides q , say q = ℓ ( a + 1) for some integer ℓ ≥
1, which gives qa = ℓa ( a + 1), the result directly follows from Theorem 21 for k = ℓa , since ak ℓa qa a + 1) = (cid:22) qa a + 1) (cid:23) . Assume now that this is not the case, so that k ( a + 1) < q < ( k + 2)( a + 1) for some eveninteger k ≥
2. Let q = k ( a + 1) + 2 ℓ (recall that k and q are even) for some integer ℓ ,1 ≤ ℓ ≤ a . From Proposition 12, we get that σ ( f ) ≤ (cid:22) a a + 1) (cid:18) qa − a − a (cid:12)(cid:12) V f (cid:12)(cid:12)(cid:19)(cid:23) ≤ (cid:22) qa a + 1) (cid:23) for every 2-bounded independent broadcast f on C ( qa ; 1 , a ). Moreover, we have (cid:22) qa a + 1) (cid:23) = ( a − q (cid:22) q a + 1) (cid:23) = ( a − q k , which implies β b ( C ( qa ; 1 , a )) ≤ ( a − q k . a 2a3a4a(q-1)a(q-2)a(q-3)a 1 a+12a+13a+14a+1(q-1)a+1 i a+i2a+i3a+i4a+i(q-1)a+i a-1 2a-13a-14a-15a-1qa-1 0 a 2a 3a4a5a(i+1)a ia(i+2)a C C C i C ( a − C Figure 7: The circulant graph C ( qa ; 1 , a ) (only subscripts of vertices are indicated) v v v v v v v v v v v v v v v v v v v v v v v v C C C C C C C Figure 8: Construction of the sets S i in the proof of Theorem 23 ( a = 6, q = 5)Since qa = ( ka + ℓ )( a + 1) + ℓ ( a − β b ( C ( qa ; 1 , a )) ≥ ( ka + ℓ ) a ℓ ( a − ka ℓ ( a −
1) = ka q − k ( a + 1)2 ( a − , which gives β b ( C ( qa ; 1 , a )) ≥ ( a − q k . Finally, by Observation 2, we get β b ( C ( qa ; 1 , a )) = α ( C ( qa ; 1 , a )) = (cid:22) qa a + 1) (cid:23) . a is even and q is odd.The graph C ( qa ; 1 , a ) can be seen as a copies C , . . . , C a − of a q -cycle, with C k = { v k , v k + a , v k +2 a . . . , v k +( q − a } , for every k , 0 ≤ k ≤ a −
1, cyclically connected as de-picted in Figure 7.We consider three subcases.(a) q = a − q is odd, similarly to Case 2 ( q and a odd), we have σ ( f ) ≤ ( q − a v v v v v v v v v v v v v v v v v v v v v v v v C C q C q +1 C q +2 C q +3 C q +4 C q + 1 cycles 2 ℓ cyclesFigure 9: Construction of the sets S i in the proof of Theorem 23 ( a = 10, q = 5, ℓ = 2)for every independent broadcast f on C ( qa ; 1 , a ). Consider the sets S k , 0 ≤ k ≤ a − a = 6 and q = 5). S = { v , v a , v a , . . . , v ( q − a , v ( q − a } ,S = { v a , v a , v a , . . . , v q − a , v q − a } ,S = { v a , v a , v a , . . . , v q − a , v q − a } , ... S a − = { v a − , v a − , v a − , . . . , v n − a − , v n − a − } . From this definition, we clearly get that S a − k =0 S k is a independent set in V ( G ). Thisgives (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a − [ k =0 S k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = a q − ≤ α ( C ( qa ; 1 , a )) ≤ β b ( C ( qa ; 1 , a )) . Thanks to Observation 2, we then get β b ( C ( qa ; 1 , a )) = α ( C ( qa ; 1 , a )) = ( q − a . (b) q < a − a = q + 1 + 2 ℓ (recall that a and q have different parity) for some integer ℓ , ℓ ≥
1. For every cycle C k , k = 0 , , . . . , q , let S k be the set defined as in the previoussubcase and, for every cycle C k , k = q + 1 , . . . , q + 2 ℓ , let (see Figure 9 for the case a = 10, q = 5 and ℓ = 2) S k = { v k , v k +2 a , v k +4 a , . . . , v k +( q − a , v k +( q − a } , if k is even ,S k = { v k + a , v k +3 a , v k +5 a , . . . , v k +( q − a , v k +( q − a } , if k is odd . From this definition, we clearly get that S a − k =0 S k is an independent set of C ( qa ; 1 , a ).We then have | a − [ k =0 S k | = a q − ≤ α ( C ( qa ; 1 , a )) ≤ β b ( C ( qa ; 1 , a ))and, since q is odd, we get β b ( C ( qa ; 1 , a ) ≤ a (cid:18) q − (cid:19) . β b ( C ( qa ; 1 , a )) = α ( C ( qa ; 1 , a )) = ( q − a . (c) q > a + 1.Note first that if a + 1 divides q , say q = ℓ ( a + 1) for some integer ℓ ≥
1, which gives qa = ℓa ( a + 1), the result directly follows from Theorem 21 for k = ℓa , since ak ℓa qa a + 1) = (cid:22) qa a + 1) (cid:23) . Suppose now that this is not the case, so that k ( a + 1) < q < ( k + 2)( a + 1), forsome odd integer k ≥
1. Let q = k ( a + 1) + 2 ℓ (recall that k and q are odd) for someinteger ℓ , 1 ≤ ℓ ≤ a . From Proposition 12 we get β b ( C ( qa ; 1 , a )) ≤ (cid:22) qa a + 1) (cid:23) = (cid:22) q ( a − a + 1) + q a + 1) (cid:23) = (cid:22) q ( a − a + 1) + ( a + 1)2( a + 1) + q − ( a + 1)2( a + 1) (cid:23) = (cid:22) q ( a − a + 1) + ( a + 1)2( a + 1) + ( k − a + 1) + 2 ℓ a + 1) (cid:23) = ( a − q + 12 + k − . Moreover, we have qa = ( ka + ℓ )( a + 1) + ℓ ( a −
1) and, thanks to Proposition 13, weget β b ( C ( qa ; 1 , a )) ≥ ( ka + ℓ ) a ℓ ( a −
1) = ka ℓ ( a −
1) = ka q − k ( a + 1)2 ( a − , which gives β b ( C ( qa ; 1 , a )) ≥ ( a − q + 12 + k − . Therefore, by Observation 2, we finally get β b ( C ( qa ; 1 , a )) = α ( C ( qa ; 1 , a )) = ( a − q + 12 + k −
12 = (cid:22) qa a + 1) (cid:23) . This completes the proof. (cid:3)
Our last general result is the following.
Theorem 24.
Let n , a , q and r be integers such that n = qa + r , a is even, a ≥ and q ≥ max { , r } . If either q and r have the same parity, or q and r have different parity and q + r ≥ a − , then we have β b ( C ( n ; 1 , a )) = α ( C ( n ; 1 , a )) = (cid:22) a a + 1) n (cid:23) roof. From Proposition 12, we get that σ ( f ) ≤ (cid:22) a a + 1) (cid:18) n − a − a (cid:12)(cid:12) V f (cid:12)(cid:12)(cid:19)(cid:23) ≤ (cid:22) a a + 1) n (cid:23) for every 2-bounded independent broadcast f on C ( n ; 1 , a ). We consider two cases.1. q and r have the same parity.Note first that if a + 1 divides q − r , say q − r = ℓ ( a + 1) for some integer ℓ , which gives qa + r = ( q − ℓ )( a + 1), the result directly follows from Theorem 21 for k = q − ℓ , since ak a ( q − ℓ )2 = ( qa + r ) a a + 1) = (cid:22) ( qa + r ) a a + 1) (cid:23) . Suppose now that this is not the case, so that k ( a + 1) < q − r < ( k + 2)( a + 1), for someeven integer k . Let q − r = k ( a + 1) + 2 ℓ (recall that k and q − r are even) for some integer ℓ , 1 ≤ ℓ ≤ a . We have (cid:22) a a + 1) ( qa + r ) (cid:23) = (cid:22) qa ( a + 1) − ( q − r ) a a + 1) (cid:23) = aq (cid:22) − ( q − r ) a a + 1) (cid:23) . Since − ℓ < − ℓa ( a +1) < − ℓ + 1, we get (cid:22) − ( q − r ) a a + 1) (cid:23) = (cid:22) − ( k ( a + 1) + 2 ℓ ) a a + 1) (cid:23) = − ka (cid:22) − ℓa ( a + 1) (cid:23) = − ka − ℓ, which gives β b ( C ( qa + r ; 1 , a )) ≤ (cid:22) a a + 1) ( qa + r ) (cid:23) = aq − ak − ℓ = ( q − k ) a − ℓ. Furthermore, since qa + r = q + r a + 1) + q − r a −
1) = q + r + k ( a − a + 1) + ℓ ( a − , and thanks to Proposition 13, we get β b ( C ( qa + r ; 1 , a )) ≥ q + r + k ( a − (cid:16) a (cid:17) + ℓ ( a −
1) = q + r + k ( a −
1) + 2 ℓ (cid:16) a (cid:17) − ℓ. Again, since q − r = k ( a + 1) + 2 ℓ , we have β b ( C ( qa ; 1 , a )) ≥ q + r + q − r − k (cid:16) a (cid:17) − ℓ = ( q − k ) a − ℓ, and thus, thanks to Observation 2, we finally get β b ( C ( qa + r ; 1 , a )) = α ( C ( qa + r ; 1 , a )) = ( a − q + 12 + k −
12 = (cid:22) qa a + 1) (cid:23) . q and r have different parity.Similarly to the previous case, if a + 1 divides q − r , then ak (cid:22) ( qa + r ) a a + 1) (cid:23) . Suppose now that this is not the case. We consider two subcases, depending on whether q − r is greater than a + 1 or not.(a) q − r < a + 1.In that case, we have β b ( C ( qa + r ; 1 , a )) ≤ (cid:22) a a + 1) ( qa + r ) (cid:23) = (cid:22) qa ( a + 1) − ( q − r ) a a + 1) (cid:23) = aq (cid:22) − ( q − r ) a a + 1) (cid:23) . Since q − r < a + 1, we get − q − r + 12 < − ( q − r ) a a + 1) < − q − r + 12 + 1 , and thus β b ( C ( qa + r ; 1 , a )) ≤ aq − q − r + 12 . Since qa + r = q + r + 1 − a a + 1) + q − r + 1 + a a − β b ( C ( qa + r ; 1 , a )) ≥ (cid:18) q + r + 1 − a (cid:19) a (cid:18) q − r + 1 + a (cid:19) (cid:16) a − (cid:17) , which gives β b ( C ( qa ; 1 , a )) ≥ aq − q − r + 12 . Finally, thanks to Observation 2, we get β b ( C ( qa ; 1 , a )) = α ( C ( qa ; 1 , a )) = aq − q − r + 12 . (b) q − r > a + 1.In that case, we have k ( a + 1) < q − r < ( k + 2)( a + 1), for some odd integer k ≥ q − r = k ( a + 1) + 2 ℓ (recall that k and q − r are odd) for some integer ℓ ,1 ≤ ℓ ≤ a . Since (cid:22) − ( q − r ) a a + 1) (cid:23) = (cid:22) − ( k ( a + 1) + 2 ℓ ) a a + 1) (cid:23) = − ka (cid:22) − ℓa ( a + 1) (cid:23) and − ℓ < − ℓa ( a +1) < − ℓ + 1 , we get (cid:22) a ( qa + r )2( a + 1) (cid:23) = (cid:22) qa ( a + 1) − ( q − r ) a a + 1) (cid:23) = aq (cid:22) − ( q − r ) a a + 1) (cid:23) = aq − ka − ℓ. β b ( C ( qa + r ; 1 , a )) ≤ (cid:22) a ( qa + r )2( a + 1) (cid:23) = aq − ak − ℓ = ( q − k ) a − ℓ. Moreover, since qa + r = q + r + 1 − a a + 1) + q − r + a + 12 ( a −
1) and q − r = k ( a + 1) + 2 ℓ , we get qa + r = q + r + k ( a − a + 1) + ℓ ( a − . Hence, we have β b ( C ( qa + r ; 1 , a )) ≥ q + r + k ( a − · a ℓ ( a −
1) = q + r + k ( a −
1) + 2 ℓ · a − ℓ, which gives β b ( C ( qa + r ; 1 , a )) ≥ q + r + q − r − k · a − ℓ = ( q − k ) a − ℓ. Finally, thanks to Observation 2, we get β b ( C ( qa + r ; 1 , a )) = α ( C ( qa + r ; 1 , a )) = (cid:22) qa a + 1) (cid:23) . This completes the proof. (cid:3)
We proved that every circulant graph of the form C ( n ; 1 , a ), 3 ≤ a ≤ ⌊ n ⌋ , admits a 2-bounded β b -broadcast,except when n = 2 a + 1, or n = 2 a and a is even. Using this property, wedetermined the exact value of the broadcast independence number of several classes of circulantgraphs of the form β b ( C ( n ; 1 , a )), 2 ≤ a ≤ ⌊ n ⌋ . In several cases, we showed that β b ( C ( n ; 1 , a ))reaches one of its lower bounds, namely α ( C ( n ; 1 , a )) or 2(diam( C ( n ; 1 , a )) − β b ( C ( n ; 1 , a )) = α ( C ( n ; 1 , a )), we get that C ( n ; 1 , a ) admits a 1-bounded β b -broadcast.We finally mention a few open problems that seem worth to be investigated.1. Determine the value of β b ( C ( n ; 1 , a )) for the remaining unsolved cases namely:(a) For odd integers a and n , with a ≥ a n .(b) For integers n , a , q and r , with n = qa + r , a ≥ n is divisible neither by a nor by a + 1, and eitheri. q < r , orii. q > r , q and r have different parity, q + r < a − G for which β b ( G ) = α ( G ) or β b ( G ) = 2(diam( G ) − Acknowledgment.
The first and second authors acknowledge General Directorate of ScientificResearch and Technological Development of the Algerian Ministry of Higher Education andScientific Research (DGRSDT) for support of this work.28 eferences [1] M. Ahmane, I. Bouchemakh, E. Sopena. On the Broadcast Independence Number of Cater-pillars.
Discrete Appl. Math.
244 (2018), 20–35.[2] M. Ahmane, I. Bouchemakh, E. Sopena. On the Broadcast Independence Number of Lo-cally Uniform 2-Lobsters. Available on arXiv:1902.02998 [cs.DM] (2019).[3] S. Bessy, D. Rautenbach. Algorithmic aspects of broadcast independence. Available onarXiv:1809.07248 [math.CO] (2018).[4] S. Bessy, D. Rautenbach. Relating broadcast independence and independence.
DiscreteMath.
Commun. Comb. Optim.
GraphsCombin.
30 (2014), 83–100.[7] S. Bouchouika, I. Bouchemakh, E. Sopena. Broadcasts on Paths and Cycles. Broadcastson paths and cycles.
Discrete Appl. Math.
283 (2020), 375-395.[8] J.E. Dunbar, D.J. Erwin, T.W. Haynes, S.M. Hedetniemi, S.T. Hedetniemi. Broadcasts ingraphs.
Discrete Appl. Math.
154 (2006), 59–75.[9] D. Erwin.
Cost domination in graphs . Ph.D. Dissertation, Western Michigan University,2001.[10] D. Erwin. Dominating broadcasts in graphs.
Bull. Inst. Combin. Appl.
42 (2004), 89–105.[11] X. Liancheng, X. Zunquan, Y. Yuansheng. Some Results on the Independence Number ofCirculant Graphs C ( n ; { , k } ). OR Trans.
13 (2009), no. 4, 65–70.[12] E.A. Monakhova. A survey on undirected circulant graphs.