Total Rainbow k-connection in Graphs

Transcription

1 Total Rainbow k-connection in Graphs Henry Liu, Ângela Mestre, and Teresa Sousa, Departamento de Matemática and Centro de Matemática e Aplicações Faculdade de Ciências e Tecnologia Universidade Nova de Lisboa Quinta da Torre, 89- Caparica, Portugal {hliu tmjs}@fctunlpt Universidade de Lisboa Centro de Estruturas Lineares e Combinatórias Avenida Prof Gama Pinto, 9- Lisboa, Portugal mestre@ciifculpt March Abstract Let k be a positive integer and G be a k-connected graph In 9, Chartrand, Johns, McKeon, and Zhang introduced the rainbow k-connection number rc k (G) of G An edge-coloured path is rainbow if its edges have distinct colours Then, rc k (G) is the minimum number of colours required to colour the edges of G so that any two vertices of G are connected by k internally vertex-disjoint rainbow paths The function rc k (G) has since been studied by numerous researchers An analogue of the function rc k (G) involving vertex colourings, the rainbow vertex k-connection number rvc k (G), was subsequently introduced In this paper, we introduce a version which involves total colourings A total-coloured path is total-rainbow if its edges and internal vertices have distinct colours The total rainbow k-connection number of G, denoted by trc k (G), is the minimum number of colours required to colour the edges and vertices of G, so that any two vertices of G are connected by k internally vertex-disjoint total-rainbow paths We study the function trc k (G) when G is a cycle, a wheel, and a complete multipartite graph We also compare the functions rc k (G), rvc k (G), and trc k (G), by considering how close and how far apart trc k (G) can be from rc k (G) and rvc k (G) Keywords: Graph colouring, rainbow (vertex) connection number, k-connected AMS subject classifications: C, C Research partially supported by FCT - Fundação para a Ciência e a Tecnologia (Portugal), through the projects PEst-OE/MAT/UI9/ (CMA) and PTDC/MAT//9 Research supported by FCT - Fundação para a Ciência e a Tecnologia (Portugal), reference SFRH/BPD/8/8

2 Introduction In this paper, all graphs are finite, simple, and undirected For undefined terms in graph theory, we refer the reader to the book by Bollobás [] A set of internally vertex-disjoint paths will be called disjoint An edge-coloured path is rainbow if its edges have distinct colours For a positive integer k, an edge-colouring of a k-connected graph G, not necessarily proper, is rainbow k-connected if any two vertices of G are connected by k disjoint rainbow paths The rainbow k-connection number of G, denoted by rc k (G), is the minimum integer t such that there exists a rainbow k-connected colouring of G using t colours By Menger s theorem [], a graph is k-connected if and only if any two vertices are connected by k disjoint paths Hence, rc k (G) is well-defined if and only if G is k-connected The function rc k (G) was first introduced by Chartrand et al ([] for k = in 8, and [] for general k in 9) and has since been very well-studied For an overview of the rainbow connection subject, we refer the reader to the survey by Li et al [8], and the book by Li and Sun [] A vertex-coloured path is vertex-rainbow if its internal vertices have distinct colours A vertex-colouring of a k-connected graph G, not necessarily proper, is rainbow vertex k- connected if any two vertices of G are connected by k disjoint vertex-rainbow paths The rainbow vertex k-connection number of G, denoted by rvc k (G), is the minimum integer t such that there exists a rainbow vertex k-connected colouring of G using t colours Again, rvc k (G) is well-defined if and only if G is k-connected The function rvc k (G) was introduced by Krivelevich and Yuster [] (for k = in ), and by the authors [] (for general k in ) Here, we consider an analogous function using total colourings A total-coloured path is total-rainbow if its edges and internal vertices have distinct colours A total colouring of a k-connected graph G, not necessarily proper, is total-rainbow k-connected if any two vertices of G are connected by k disjoint total-rainbow paths The total rainbow k-connection number of G, denoted by trc k (G), is the minimum integer t such that, there exists a total-rainbow k-connected colouring of G using t colours We have trc k (G) is well-defined if and only if G is k-connected We write trc(g) for trc (G), and similarly for rc(g) and rvc(g) For a non-trivial connected graph G, many observations about the function trc k (G) can be made We have trc(g) = if and only if G is a complete graph, and trc(g) if G is not complete A simple upper bound is trc(g) n + q, where V (G) = n and q is the number of vertices of G with degree at least, and equality holds if and only if G is a tree (see Proposition ) If G is k-connected, then trc k (G) if k, and trc k (G) diam(g) for k, where diam(g) denotes the diameter of G In relation to rc k (G) and rvc k (G), we have trc k (G) max(rc k (G), rvc k (G)) Also, if rc k (G) =, then trc k (G) = If rvc k (G), then trc k (G) In the rest of this paper, we will study the function trc k (G) when G is a cycle, a wheel, a complete bipartite graph, and a complete multipartite graph We will also compare the functions trc k (G), rc k (G), and rvc k (G), by considering how close and how far apart trc k (G) can be from each of rc k (G), rvc k (G), and max(rc k (G), rvc k (G)) Total Rainbow k-connection Numbers of some Graphs In this section, we first derive the upper bound on trc(g) as mentioned in Introduction Then, we study the function trc k (G) for some specific graphs G

3 Proposition Let G be a connected graph on n vertices, with q vertices having degree at least Then, trc(g) n + q, with equality if and only if G is a tree Proof The proposition is trivial for n =, Now, let n First, we show that trc(g) n + q Take a spanning tree T of G, and note that T has at most q non-leaves Then, take a total colouring of G with at most n + q colours such that the edges and non-leaves of T have distinct colours We have a total-rainbow connected colouring for G Now, let G be a tree Suppose that we have a total colouring for G with fewer than n + q colours Then, either there are two edges, or two non-leaves, or an edge and a non-leaf, with the same colour In each case, we can find two vertices u, v V (G) such that the unique u v path in G is not total-rainbow Hence, trc(g) n + q Conversely, if G is not a tree, then G contains a spanning unicyclic subgraph G Let C be the unique cycle of G We recall from Chartrand et al [, Proposition ] that rc(c ) = and rc(c p ) = p for p, where C p is the cycle of order p Now, consider a total colouring of G where C is given a rainbow connected colouring with rc(c) colours, and the vertices of degree at least in G and the edges of E(G ) \ E(C) are given distinct colours Then, we have a total-rainbow connected colouring for G, and hence for G, with at most e(g ) e(c) + rc(c) + q n + q colours Therefore, trc(g) < n + q Now, we study the function trc k (G) for some specific graphs G Let κ(g) = max{k : G is k-connected} denote the vertex connectivity of G Note that trc k (G) is well-defined if and only if k κ(g) We first consider the case when G is a cycle Let C n denote the cycle of order n, and note that κ(c n ) = Theorem (a) For n, the values of trc(c n ) are given in the following table For n, we have trc(c n ) = n n 8 9 trc(c n ) 8 9 (b) trc (C ) =, trc (C ) =, and trc (C n ) = n for n Proof Throughout, let the vertices of C n be v,, v n, with the indices taken cyclically modulo n (a) One can easily verify that trc(c ) = and trc(c ) = trc(c ) = Now, let n We first prove the upper bounds Define a total colouring c n of C n, using colours,,,, as follows For n or n =, let c n (v ) = c n (v n/ ) = c n (v n/ ) =, and c n (v ) = c n (v n/ + ) = c n (v n/ + ) = Then, colour v v, v v, v, v v, v,, v n, v n v with the colours,,, n,,,, n, omitting v, v, v n/, v n/ +, v n/ and v n/ + Note that each of the colours,,, n appear exactly twice See Figure Then, every path of length at most n is total-rainbow, and when n is even, any two opposite vertices v i, v i+n/ ( i n ) are connected by a total-rainbow path Hence, c n is total-rainbow connected, and trc(c n ) n

4 v v v v v v v v v v v v v 8 v v v v 8 v v v v v v 8 v v v v v v v 8 v 9 v v v v v v v v v 9 v v 8 v v v v v 9 v v v 8 8 v v 9 Figure Total-rainbow connected colourings of C n, for n and n = For n = or n, and i n, let c n (v i v i+ ) = i, and c n (v i ) = i + n (mod n) Then, for i n, consider the path Q i = v i v i+ v i+ n/ The edges of Q i have colours i, i +,, i + n (mod n), and the internal vertices have colours i + n +,, i + n (mod n) Hence, Q i is a total-rainbow path Since any two vertices v j and v l are contained in some Q i, there exists a total-rainbow v j v l path Hence, c n is total-rainbow connected, and trc(c n ) n Now we prove the lower bounds We have trc(c n ) diam(c n ) = n, and this gives the matching lower bound for n =, 8,, To see that trc(c ) (resp trc(c 9 ) 8), one can check that in any total colouring of C (resp C 9 ) with at most five (resp seven) colours, there exists a path of length (resp ) which is not total-rainbow, so that the end-vertices of such a path are not connected by a total-rainbow path Finally, for n = or n, take a total colouring of C n with fewer than n colours Then for some m {,,, }, we have m edges and m vertices with the same colour Without loss of generality, for some < i n m +, either v and v i, or v and v i v i+, or v v and v i v i+, have the same colour Then, the path v v v i+ is not total-rainbow, and the path v i+ v i+ v n v has at least n n m > n edges and internal vertices, and hence is also not total-rainbow Therefore, trc(c n ) n (b) One can easily verify that trc (C ) = and trc (C ) = For n, we clearly have trc (C n ) n If we have a total colouring of C n with fewer than n colours, then some a, b E(C n ) V (C n ) have the same colour There exist vertices v i and v j such that one of the two v i v j paths, say P, satisfies a, b E(P ) V (P {v i, v j }), so that P is not total-rainbow Hence, trc(c n ) n A graph closely related to the cycle C n is the wheel W n This is the graph obtained from C n by joining a new vertex v to every vertex of C n The vertex v is the centre of W n We now determine trc k (W n ) Observe that κ(w n ) =

5 Theorem (a) trc(w ) =, trc(w n ) = for n, trc(w n ) = for n 9, and trc(w n ) = for n (b) trc (W ) =, trc (W ) =, trc (W ) =, and trc (W n ) = trc(c n ) for n (hence, trc (W n ) is determined and given by Theorem (a) for n ) (c) trc (W ) =, trc (W ) =, and trc (W n ) = n for n Proof Let v be the centre of W n, and the vertices of the cycle C n be v,, v n Throughout this proof, indices of the vertices are taken cyclically modulo n (a) Clearly, we have trc(w ) = Now, let n We first prove the upper bounds Define a total colouring f n of W n, using colours,,,, as follows For n 9 and i n, let f n (vv i ) = i, f n(v i v i+ ) = i (mod ); and f n (v) = f n (v i ) = if n ; and f n (v) = f n (v i ) = if n 9 For n and i n, let f n (vv i ) = i (mod ), f n (v i v i+ ) =, f n (v i ) =, and f n (v) = See Figure for the cases n =,, Then, f n is total-rainbow connected Hence, trc(w n ) for n, trc(w n ) for n 9, and trc(w n ) for n Figure Total-rainbow connected colourings of W, W and W Now, we prove the lower bounds Clearly, we have trc(w n ) for n Suppose that we have a total-rainbow connected colouring f n of W n with colours,,, b, where b = for n 9, and b {, } for n Assume that f n(v) = If f n(vv i ) = for some i n, then there is no total-rainbow v i v i+ path Hence, some four of the edges vv,, vv n have the same colour Assume that for some i n, we have f n(vv ) = f n(vv i ) = But then, we do not have a total-rainbow v v i path, a contradiction Hence, trc(w n ) for n 9, and trc(w n ) for n (b) We first prove the upper bounds For n, we define a total colouring g n of W n, using colours,,,, as follows For g, g and g, we take the total-rainbow -connected colourings as shown in Figure Hence, trc (W ), trc (W ) and trc (W ) Figure Total-rainbow -connected colourings of W, W and W

6 For n and n =, let g n c n when we restrict g n to the cycle C n, where c n is the total colouring on C n as defined in Theorem (a) Let g n (v) =, and colour the edges vv i, for i n, as in Figure Note that g n uses trc(c n ) = n colours In each case, g n is total-rainbow -connected Indeed, for v and a vertex v i, we have vv i and vv i v i are total-rainbow For two vertices v i and v j, assume that v i v i+ v j is total-rainbow (by the choice of g n c n on C n ) If v i vv j is total-rainbow, then this is the second suitable v i v j path Otherwise, we can take the second path to be v j v j+ v i if n =,, 8, or v i v i vv j if n =, 9,, Hence, we have trc (W n ) trc(c n ) = n Figure Total-rainbow -connected colourings of W n, for n and n = For n = or n, let g n c n on the cycle C n as before Let g n (vv i ) = i for i n, g n (vv n ) = n and g n(v) = n Note that g n uses trc(c n ) = n colours We claim that g n is total-rainbow -connected For two vertices v i and v j, two of v i v i+ v j, v j v j+ v i and v i vv j are total-rainbow For v and a vertex v i, we have vv i and vv i+ v i are total-rainbow, except for i = n and n even, in which case we take vv n and vv v n v n Hence, we have trc (W n ) trc(c n ) = n Now, we prove the lower bounds Clearly, we have trc (W ) and trc (W ) If we have a total colouring of W with fewer than five colours, then without loss of generality, vv and vv i have the same colour for i = or i =, and we do not have two disjoint total-rainbow v v i paths Hence, trc (W ) Now let n If we have a total colouring of W n with fewer than trc(c n ) colours, then there exist v i and v j where both of the v i v j paths in the C n, say P and P, are not total-rainbow Since any two disjoint v i v j paths in W n must include at least one of P and P, we cannot have two disjoint total-rainbow v i v j paths in W n Hence, trc (W n ) trc(c n ) (c) We first prove the upper bounds For n, we define a total colouring h n of W n, using colours,,,, as follows For h and h, we take the total-rainbow -connected colourings as shown in Figure Hence, trc (W ) and trc (W )

7 Figure Total-rainbow -connected colourings of W and W For n, let h n (v i ) = i, h n (v i v i+ ) = i + n and h n (vv i ) = i + for every i n, and h n (v) = Then h n is total-rainbow -connected, and trc (W n ) n Now, we prove the lower bounds Clearly, for a total-rainbow -connected colouring of W, we need v vv, v vv and v vv to be total-rainbow This implies that trc (W ) For n, suppose that we have a total colouring of W n, using fewer than trc (C n ) colours Then, there exist v i and v j such that one of the v i v j paths along the C n is not totalrainbow Hence, we cannot have three disjoint total-rainbow v i v j paths in W n, and by Theorem (b), trc (W ) and trc (W n ) n for n Next, we consider the function trc k (G) when G is a complete bipartite graph K m,n, where m n Note that κ(k m,n ) = m We first determine trc(k m,n ) exactly Clearly, trc(k, ) = and trc(k,n ) = n + if n For m, we have the following result Theorem For m n, we have trc(k m,n ) = min( m n +, ) Proof Let the classes of K m,n be U and V, where U = {u,, u m } and V = n Let b = m n, and note that m n m if and only if b + Hence, we need to prove that trc(k m,n ) = b + if m n m, and trc(k m,n ) = if n > m We first prove the upper bound Note that whenever we have constructed a total colouring of K m,n and want to show that it is total-rainbow connected, it suffices to verify that any two vertices in the same class of K m,n are connected by a total-rainbow path, since any two vertices in different classes are adjacent For m n m, we define a total colouring f of K m,n with b + colours as follows Assign to the vertices of V distinct vectors of length m, with entries from {,, b}, such that the vectors (,,, ), (,,,, ),, (,,, ) are all present For v V, let v denote the vector assigned to v For u i U and v V, let f(u i v) = v i (where v i denotes the ith coordinate of v ) Let f(w) = b + for all w V (K m,n ) Now for u i, u j U, the path u i zu j is total-rainbow, where z V is assigned the vector z = (,,,,,, ), with the in the ith position For x, y V, there exists i m such that x i y i, and the path xu i y is total-rainbow Hence, f is total-rainbow connected, and trc(k m,n ) b + For n > m, we define a total colouring f of K m,n with seven colours as follows Let V = V V, where V = m Assign the m distinct vectors of length m with entries from {,, } to the vertices of V, and the vector (,,, ) (with length m) to every vertex of V Again for v V, let v be the vector assigned to v For u i U and v V, let f (u i v) = v i Let f (u ) =, f (u i ) = for i m, and f (w) = for all w V Then as before, any two vertices u i, u j U, and any two vertices x, y V, are connected by a totalrainbow path Now, for x V and y V, the path xu y is total-rainbow For x, y V, the path xu zu y is total-rainbow, where z V is assigned the vector z = (,,, ) Hence, f is total-rainbow connected, and trc(k m,n )

8 Now, we prove the lower bound If m n m, then trc(k m,n ) = b + Next, let m < n m Then (b ) m < n b m, with b {,,, } Let g be a total colouring of K m,n, using colours from {,, b} For v V, assign v with the vector v of length m, where v i = g(u iv) for i m For two partitions P and P of V, we say that P refines P, written as P P, if for all A P, we have A B for some B P In other words, P can be obtained from P by partitioning some of the sets of P We define a sequence of refining partitions P P P m of V, with P i (b ) i for i m, as follows Initially, set P = {V } Now, for i m, suppose that we have defined P i with P i (b ) i Let P i = {A,, A l }, where l (b ) i Define P i as follows For q l and A q P i, let B q = {v A q : v i = g(u i ) or g(u i ) + (mod b)}, B q r = {v A q : v i = g(u i ) + r (mod b)}, for r b Let P i = {Br q : q l, r b and Br q }, so that P i is a partition of V with P i (b ) i and P i P i Proceeding inductively, we obtain the partitions P P P m of V, with P i (b ) i for i m Now, observe that for every i m, and any two vertices y and z in the same set in P i, the path yu i z is not total-rainbow, since g(yu i ) = y i and g(zu i) = z i are either in {g(u i), g(u i ) + } (mod b), or they are both g(u i ) + r (mod b) for some r b Since n > (b ) m P m, there exists a set in P m with at least two vertices w and x, and since P P m, this means that w and x are in the same set in P i for every i m Therefore, wu i x is not a total-rainbow path for every i m Since any other w x path has length at least, and g uses only at most b colours, we cannot have a total-rainbow w x path Hence, g is not total-rainbow connected, and trc(k m,n ) b + Finally, let n > m Let g be a total colouring of K m,n, using colours from {,, } For v V, assign to v the vector v of length m, where v i = g (u i v) for i m Then, there exist x, y V with x = y, and hence the path xu i y is not total-rainbow for any i m Since any other x y path has length at least, and g uses only at most six colours, we cannot have a total-rainbow x y path Hence, g is not total-rainbow connected, and trc(k m,n ) Now, we consider trc k (K m,n ) for k m n We may consider the related problems of finding rc k (K m,n ) and rvc k (K m,n ) for k m n It is easy to see that rvc(k, ) =, rvc(k m,n ) = if n, and rvc k (K m,n ) = for k Also, we have rc(k,n ) = n, and Chartrand et al [] proved that rc(k m,n ) = min( m n, ) if m n For k, the determination of rc k (K m,n ) has been well-studied and remains an open problem Partial solutions for the balanced case rc k (K n,n ) have been obtained by Chartrand et al [], Li and Sun [], and Fujita et al [] The result of Fujita et al [, Theorem ] says that if < ε < and k (θ )( ε) +, where θ = θ(ε) is the largest solution of x e ε (x ) =, then we have rc k (K n,n ) = for n k ε + From this result, we have the following corollary Corollary Let < ε < and k (θ )( ε) +, where θ = θ(ε) is the largest solution of x e ε (x ) = If n k ε +, then trc k(k n,n ) = Proof By the result of Fujita et al, we have rc k (K n,n ) = We can take a rainbow k- connected colouring of K n,n with three colours, and colour the vertices with two further colours, with the vertices within each class having the same colour This gives trc k (K n,n ) 8

9 Also, we have trc k (K n,n ), since to have at least two disjoint total-rainbow paths connecting two vertices in different classes, at least one path must have length at least For example, if we set ε = in Corollary, then for k 9 and n k, we have trc k (K n,n ) = Fujita et al [, Problem ] also asked the following question: For k m n, is it true that rc k (K m,n ) {, } for sufficiently large m? Here, we may ask a similar question Problem For k m n, is it true that trc k (K m,n ) {,, } for sufficiently large m? Also, for which values of m and n does trc k (K m,n ) take a particular value in {,, }? Next, we consider the more general problem of finding trc k (G) when G is a complete multipartite graph For t, let K n,,n t be the complete multipartite graph with classsizes n n t Let m = t i= n i and n = n t Observe that κ(k n,,n t ) = m We first determine trc(k n,,n t ) Theorem Let t, n n t, m = t i= n i and n = n t Then, if n =, trc(k n,,n t ) = if n and m > n, min( m n +, ) if m n Proof Write G for K n,,n t, and let V i be the ith class (with n i vertices) for i t If n =, then G = K t and trc(g) = For the case n and m > n, Chartrand et al [, Theorem ] proved that rc(g) = This implies that trc(g) = Now, let m n Let b = m n, and note that m n m if and only if b+ Hence, we need to prove that trc(g) = b + if m n m, and trc(g) = if n > m We first prove the upper bound For the case m n m, observe that K m,n is a spanning subgraph of G with classes V V t and V t Hence, by Theorem we have trc(g) trc(k m,n ) = b + For the case n > m, we define a total colouring f of G with five colours as follows Let V t = V V, where V = m Assign the m distinct vectors of length m with entries from {,, }, except for (,,, ), to the vertices of V Assign the vector (,,, ) to all vertices of V For v V t, let v denote the vector assigned to v Let V V t = {u,, u m }, where u V and u V For i m and v V t, let f(u i v) = v i Let f(e) = for every edge e inside V V t (note that such edges exist, since t ), f(u ) =, and f(w) = for all w V (G) \ {u } As in the proof of Theorem, we only need to verify that any two vertices in the same class of G are connected by a totalrainbow path Any two vertices in V (G) \ V, and one vertex in each of V (G) \ V and V, are connected by a total-rainbow path If x, y V, then the path xu u y is total-rainbow Therefore, f is total-rainbow connected, and trc(g) Now, we prove the lower bound If m n m, then trc(g) = b + Next, let m < n m Then (b ) m < n b m, with b {, } Suppose that we have a total colouring of G, using at most b colours Again, K m,n is a spanning subgraph of G with classes V V t and V t Apply the same argument involving the refining partitions as in Theorem We have vertices u, v V t which are not connected by a total-rainbow path of length Since we have used at most b colours, we do not have a total-rainbow u v path of length at least in G Therefore, trc(g) b+ Finally for n > m, consider a total 9

10 colouring of G with at most four colours We obtain K m,n as before and apply the same argument as in the end of Theorem Then there are u, v V t which are not connected by a total-rainbow path of length Since we have used only at most four colours, again we cannot have a total-rainbow u v path in G Hence, trc(g) Now, we consider trc k (K n,,n t ) for k m Again, we have the related problems of finding rc k (K n,,n t ) and rvc k (K n,,n t ) for k m, and these have been well-studied The function rvc k (K n,,n t ) has been completely determined by Liu et al [] Roughly speaking, for s t, if n i = for i t s and n i = for i > t s, then we have rvc k (K n,,n t ) = s Otherwise, we have rvc k (K n,,n t ) {,,, } Also, Chartrand et al [] proved that if n =, rc(k n,,n t ) = if n and m > n, min( m n, ) if m n The problem of determining rc k (K n,,n t ) remains open for k For the balanced case, let K t n denote the graph K n,,n with t classes Fujita et al [, Theorem 8] proved that if < ε < and k θ(t )( ε) +, where θ = θ(ε, t) is the largest solution of t x e (t )εx k =, then we have rc k (K t n ) = for n (t )( ε) From this, we instantly have the following corollary Corollary 8 Let t, < ε <, and k θ(t )( ε) +, where θ = θ(ε, t) is the largest solution of t x e (t )εx k = If n (t )( ε), then trc k(k t n ) = For example, if we set t = and ε = in Corollary 8, then for k 9 and n k, we have trc k (K n ) = For general complete multipartite graphs, Fujita et al [, Problem ] asked the following question for rc k (K n,,n t ): For t, k and sufficiently large n, is it true that rc k (K n,,n t ) {, }? Here, we ask a similar question for trc k (K n,,n t ) Problem 9 For t, n n t and k m, where m = t i= n i, is it true that trc k (K n,,n t ) {,, } for sufficiently large n? Also, for which values of n,, n t does trc k (K n,,n t ) take a particular value in {,, }? To end this section, we mention the analogous situation for complete graphs Obviously, we have rc(k n ) =, rvc(k n ) =, and rvc k (K n ) = if k The problem of determining rc k (K n ) for k has also been well-studied Chartrand et al [] proved that if k and n (k +), then we have rc k (K n ) = The bound n (k +) was subsequently improved to n ck / +o(k / ) (for some constant c) by Li and Sun [9], and then to n (+o())k by Dellamonica et al [, Theorem ] This latter bound for n is asymptotically best possible, and we have the following corollary Corollary For k, we have trc k (K n ) = for n ( + o())k Comparing rc k (G), rvc k (G), and trc k (G) In [], Krivelevich and Yuster observed that we cannot upper-bound one of the functions rc(g) and rvc(g) in terms of the other, by providing examples of graphs G where rc(g) is

11 much larger than rvc(g), and vice versa By taking G to be the star K,s, we have rc(g) = s and rvc(g) = On the other hand, let G be constructed as follows Take s vertex-disjoint triangles and, by designating a vertex from each triangle, add a complete graph on the designated vertices Then rc(g) and rvc(g) = s Subsequently, Liu et al [] compared the functions rc k (G) and rvc k (G) by extending these examples of Krivelevich and Yuster, and they obtained similar results Here, we similarly compare trc k (G) with rc k (G) and rvc k (G) Recall that for any k- connected graph G, we have trc k (G) max(rc k (G), rvc k (G)) () The first question we may ask is, how tight are the inequalities trc k (G) rc k (G) and trc k (G) rvc k (G)? Theorem below shows that the second inequality is the best possible in the sense that, for every sufficiently large s, there exists an example of a graph G where trc k (G) = rvc k (G) = s This is somewhat surprising since in total colourings, we colour edges and vertices of graphs, as opposed to only vertices for vertex-colourings, so we could possibly expect that trc k (G) > rvc k (G) if rvc k (G) is sufficiently large Theorem For every s k +, there exists a graph G with trc k (G) = rvc k (G) = s To prove Theorem, we need the result of Chartrand et al [] which was mentioned at the end of Section, as well as an auxiliary lemma Theorem (Chartrand et al [, Theorem ]) For k and n (k + ), we have rc k (K n ) = Lemma Let k, s k +, and u,, u s be the vertices of the complete graph K s For i s, let L(u i ) {,, s}, where L(u i ) = {i, i +,, i + k + } modulo s Then, there exists an edge-colouring of K s, using at most k + colours from {,, s} such that, for every y, z V (K s ), the following property holds (P) There exist k + disjoint rainbow y z paths of length, say yv z,, yv k+ z for some v,, v k+ V (K s ), such that for every j k +, the path yv j z does not use the colours from L(y) L(z) L(v j ) Proof We take a random edge-colouring of K s with k+ colours For y, z V (K s ), let E y,z be the event that the property (P) does not hold For fixed y and z, we can choose a set L {,, s} such that L(y) L(z) L and L = k+, and then a set S V (K s )\{y, z} such that S = s k 8, and L(v) L = for all v S For v S, we say that the path yvz is good if it is rainbow, and does not use the colours of L(v) L The probability that the path yvz is good is p = (8k+) (8k+), and for different v in S, these probabilities (k+) are independent Let X be the number of good y z paths Then, X Bi(s k 8, p) By the Chernoff bound (see for example, [, Theorem ]), we have P(E y,z ) P(X k + 9) P(X ( ε)(s k 8)p) ( exp ) ε (s k 8)p, where ε is a constant such that < ε < k+9 (s k 8)p Note that we have p > so that k + 9 (s k 8)p > = 8 8, ()

12 and we may take ε = 8 8 Applying the union bound, we have ( ) ( ) s ( P E y,z exp ) ε (s k 8)p y,z V (K s) ( ) s < exp( 9(s k 8)) <, if s k + and k Hence, there exists an edge-colouring of K s, using at most k + colours, such that the property (P) holds for all y, z V (K s ) We remark that in Lemma, the constant can be replaced by any constant greater than θ, where θ 8 is the largest solution of the cubic equation (x ) (x ) x = Using any such constant in place of will ensure that the term corresponding to ε can be found, as in the calculation in () The constant will then have to be replaced by another constant Proof of Theorem We shall focus mainly on the case when k, and then briefly consider the case k = at the end We generalise the construction of Krivelevich and Yuster with the disjoint triangles attached to a clique, as described earlier Take s disjoint k-sets of vertices V,, V s Let V i = {v i,, vi k } for every i s For every p k, we add a clique on {v p,, vp} s This gives k disjoint copies of K s Let G p be the copy of K s on {vp,, vp} s ( p k) Take further disjoint sets X,, X s and Y,, Y s, each with (k + ) vertices For each i s, add a clique with vertex set X i Y i, and a complete bipartite graph with classes X i Y i and V i Let G be the resulting graph We claim that rvc k (G) s, and trc k (G) s First, suppose that we have a vertex-colouring of G, using fewer than s colours We may assume that v, v V (G ) have the same colour Let u X and v X Then, in any set of k disjoint u v paths in G, one path must use both v and v ; note that the edges and vertices between v and v in such a path are in G Hence, rvc k (G) s Now, we construct a total colouring c of G, using the colours,, s, as follows Let c be an edge-colouring on K s as given by Lemma, where the vertices of the K s are u,, u s and the sets L(u i ) {,, s} are as defined For w V i and w V j, for some i j, if ww E(G), we let c(ww ) = c (u i u j ) For i s and p k, let c(vp) i = i + p, c(e) = i + k for e E(V i, X i ), c(x) = i + k + for x X i, c(e ) = i + k + for e E(X i, Y i ), c(y) = i + k + for y Y i, and c(e ) = i + k + for e E(Y i, V i ), all modulo s Here, E(A, B) denotes the set of edges in G with one end-vertex in A and the other in B, for disjoint sets of vertices A, B V (G) Note that for i s, the set of colours used within V i X i Y i is L(u i ) = {i, i +,, i + k + } (mod s) By Theorem, we can give rainbow k-connected edge-colourings to the copies of K (k+) induced by X i and Y i with two colours not in L(u i ), for i s We claim that c is total-rainbow k-connected for G Let u, v V (G) We show that there exist k disjoint total-rainbow u v paths It is easy to see that such paths exist if u, v V i X i Y i for some i Note that we apply Theorem if u, v X i or u, v Y i Let u V i and v V j for some i j, with u = vp i and v = vq j for some p and q If p = q, then u, v V (G p ) Applying Lemma on G p, we can find k suitable u v paths of length, all within G p If p q, then applying Lemma on G p and G q, we can find the k suitable u v paths, where each path has the form uvpxyv l qv l for some l {i, j}, x X l and y Y l, with distinct paths having distinct values of l

13 Let u X i Y i and v V j for some i j, with v = vq j for some q By Lemma, in the edge-colouring c on K s, there are k + disjoint rainbow u i u j paths of length, where each such path u i wu j does not use the colours of L(u i ) L(u j ) L(w) We may choose k of these paths, say u i u l u j,, u i u lk u j, such that L(u lr ) L(u i ) = for r k Then, we have the k suitable u v paths, where one path is uvqv i l q v, and the other k paths have the form uvpv i pxyv l qv, l for some l {l,, l k }, p q, x X l and y Y l, with distinct paths having distinct values of p and l Let u X i and v X j for some i j For p k, one of the following holds c(uv i p) = i + k, c(v i p) = i + p, c(vv j p) = j + k and c(v j p) = j + p (mod s) are pairwise distinct c(uy) = i + k +, c(y) = i + k +, c(yv i p) = i + k +, c(v i p) = i + p, c(vv j p) = j + k and c(v j p) = j + p (mod s) are pairwise distinct, where y Y i c(uv i p) = i + k, c(v i p) = i + p, c(vy ) = j + k +, c(y ) = j + k +, c(y v j p) = j + k + and c(v j p) = j + p (mod s) are pairwise distinct, where y Y j Hence, there is a u v i p path Q i p and a v v j p path Q j p such that, the colours of the elements of V (Q i p u) E(Q i p) V (Q j p v) E(Q j p) are distinct We may take the paths Q i p and Q j p, for p k, such that the only common vertex of the paths Q i p is u, and the only common vertex of the paths Q j p is v Now, by Lemma, in the edge-colouring c on K s, there are k + disjoint rainbow u i u j paths of length, where each such path u i wu j does not use the colours of L(u i ) L(u j ) L(w) Hence, we may choose k of these paths, say u i u l u j,, u i u lk u j, such that c(vp lp ) L(u i ) L(u j ) for p k Therefore, we have the k suitable u v paths uq i pv pv i p lp vpq j j pv, for p k The cases when u Y i and v Y j, and when u X i and v Y j for some i j, can be considered similarly Therefore, c is total-rainbow k-connected, and trc k (G) s Finally, for the case k =, we may simply use the construction of Krivelevich and Yuster Let G be the graph in their construction, where the clique K s has s Let v,, v s be the vertices of the K s We have already seen that rvc(g ) = s Now, consider a total colouring c on G with s colours as follows For the vertices and edges other than the edges of the K s, we colour these as in the total colouring c on G For i j, let c (v i v j ) be a colour not present in the triangles at v i and v j This can be done since s Then, c is total-rainbow connected for G, and trc(g ) s Next, we consider the tightness of the inequality trc k (G) rc k (G) For s k +, let G be the complete bipartite graph K k,s Then, we have rc k (G) = s and trc k (G) = s + Indeed, the lower bounds rc k (G) s and trc k (G) s + are fairly trivial To see that rc k (G) s, we take an edge-colouring of G with colours,, s as follows Let the classes of G be {x,, x k } and {y,, y s }, and colour x i y j with colour i + j (mod s) Then, this is a rainbow k-connected colouring for G To see that trc k (G) s +, we take a total colouring of G with colours,, s + by taking the above edge-colouring and in addition, colour every x i with colour s +, and y j with colour j + k (mod s) This is a total-rainbow k-connected colouring for G Hence, we see that rc k (G) and trc k (G) can differ by, and attaining all sufficiently large values However, we have not been able to improve this to obtain an analogue of Theorem, and we pose the following problem

14 Problem Let k Does there exist an integer N = N(k) such that for all s N, there exists a graph G with trc k (G) = rc k (G) = s? There exist some graphs G such that trc k (G) = rc k (G) Let χ (H) denote the chromatic index of the graph H Chartrand et al [, Proposition ] proved that { k + if k is even, rc k (K k+ ) = χ (K k+ ) = k if k is odd Hence, if k is even, we have a total-rainbow k-connected colouring for K k+ with k + colours as follows Take a proper edge-colouring of K k+ with k + colours, and then for every vertex v V (K k+ ), colour v with the colour not used by the edges at v It follows that trc k (K k+ ) = rc k (K k+ ) = k + Note that this argument does not hold if k is odd Also, we have trc(c ) = rc(c ) = Moreover, there are infinitely many graphs G such that trc(g) = rc(g) = We may take G to be any graph formed by taking C, then replacing a vertex v by a clique and joining all edges from the clique to the two neighbours of v in C For k odd, we have not been able to find a graph G such that trc k (G) = rc k (G) Now, the second question we can ask is, how far from equality can the inequality () be? As we have mentioned at the beginning of this section, Liu et al [] showed that the functions rvc k (G) and rc k (G) can be arbitrarily far apart They proved that, for t < s, there exists a graph G such that rc k (G) s and rvc k (G) = t [, Theorem ] They also proved that for s (k + ), there exists a graph G such that rvc k (G) = s and rc k (G) 9 [, Theorem 9] Hence, we instantly have the following corollaries Corollary Given t < s, there exists a graph G such that trc k (G) s and rvc k (G) = t Corollary Let s (k + ) Then, there exists a graph G such that trc k (G) s and rc k (G) 9 The constructions given by Liu et al were as follows For Corollary, we take the star K,s and identify the centre with one end-vertex of the path of length t This graph is a broom We take a blow-up of this broom by replacing each non-leaf vertex with a clique K k, and joining an edge between two vertices if the corresponding two vertices in the broom are neighbours For Corollary, we take the same construction as that of Theorem, but with the condition s (k + ) instead of s k + We observe that the difference between trc k (G) and max(rc k (G), rvc k (G)) can be arbitrarily large Take a path x x x s of length s Then, take a blow-up by replacing each vertex x i ( i s ) with a clique K (k+) on vertex set X i, and joining all edges between x and X ; x s and X s ; and X i and X i+ for i s Let G be the resulting graph Clearly, trc k (G) s and rvc k (G) = s Using s and Theorem, we have rc k (G) = s, and hence trc k (G) max(rc k (G), rvc k (G)) s can be arbitrarily large However, in this simple construction, we see that max(rc k (G), rvc k (G)) = s is unbounded as s increases We may ask the following question ( ) For fixed k, does there exist an infinite family F of k-connected graphs such that max(rc k (G), rvc k (G)) is bounded on F, but trc k (G) is not? We see that the constructions of Corollaries and do not answer this question In the first construction, we have rc k (G) s Similarly in the second construction, using

15 Theorem, we have rvc k (G) = trc k (G) = s if we also have s k + In both cases, max(rc k (G), rvc k (G)) is unbounded as s increases We make the following conjecture Conjecture For every k, there exists a function f k : N N such that if G is a k-connected graph and max(rc k (G), rvc k (G)) = c, then trc k (G) f k (c) A positive solution to Conjecture would imply that the answer to the question ( ) is no Acknowledgements Henry Liu and Teresa Sousa acknowledge the support from FCT - Fundação para a Ciência e a Tecnologia (Portugal), through the projects PEst-OE/MAT/UI9/ (CMA) and PTDC/MAT//9 Ângela Mestre acknowledges the support through a fellowship provided by FCT - Fundação para a Ciência e a Tecnologia (Portugal), reference SFRH/BPD/8/8 The authors would also like to thank the anonymous referees for reading the paper carefully References [] B Bollobás, Modern Graph Theory, Springer-Verlag, New York, 998, xiv+9 pp [] G Chartrand, G L Johns, K A McKeon, and P Zhang, Rainbow connection in graphs, Math Bohem () (8), 8-98 [] G Chartrand, G L Johns, K A McKeon, and P Zhang, The rainbow connectivity of a graph, Networks () (9), -8 [] D Dellamonica Jr, C Magnant, and D M Martin, Rainbow paths, Discrete Math () (), -8 [] S Fujita, H Liu, and C Magnant, Rainbow k-connection in dense graphs, J Combin Math Combin Comput, in press [] S Janson, T Luczak, and A Ruciński, Random Graphs, Wiley-Interscience Series in Discrete Mathematics and Optimization, New York,, xii+ pp [] M Krivelevich and R Yuster, The rainbow connection of a graph is (at most) reciprocal to its minimum degree, J Graph Theory () (), 8-9 [8] X Li, Y Shi, and Y Sun, Rainbow connections of graphs: a survey, Graphs Combin 9() (), -8 [9] X Li and Y Sun, On the rainbow k-connectivity of complete graphs, Australas J Combin 9 (), - [] X Li and Y Sun, Note on the rainbow k-connectivity of regular complete bipartite graphs, Ars Combin (), -8

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