Total rainbow connection of digraphs
Hui Lei, Henry Liu, Colton Magnant, Yongtang Shi

TL;DR
This paper investigates the total rainbow connection number in directed graphs, extending existing concepts from undirected graphs to digraphs, and provides results for various classes like tournaments and cactus digraphs.
Contribution
It introduces the concept of total rainbow connection for digraphs and presents new results for biorientations, tournaments, and cactus digraphs.
Findings
Results on total rainbow connection number for biorientations
Findings for tournaments and cactus digraphs
Extension of rainbow connection concepts to digraphs
Abstract
An edge-coloured path is rainbow if its edges have distinct colours. For a connected graph , the rainbow connection number (resp. strong rainbow connection number) of is the minimum number of colours required to colour the edges of so that, any two vertices of are connected by a rainbow path (resp. rainbow geodesic). These two graph parameters were introduced by Chartrand, Johns, McKeon and Zhang in 2008. Krivelevich and Yuster generalised this concept to the vertex-coloured setting. Similarly, Liu, Mestre and Sousa introduced the version which involves total-colourings. Dorbec, Schiermeyer, Sidorowicz and Sopena extended the concept of the rainbow connection to digraphs. In this paper, we consider the (strong) total rainbow connection number of digraphs. Results on the (strong) total rainbow connection number of biorientations of graphs, tournaments and cactus digraphsβ¦
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Total rainbow connection of digraphs
Hui Lei1, Henry Liu2111Corresponding authorββ , Colton Magnant3, Yongtang Shi1
1Center for Combinatorics and LPMC
Nankai University, Tianjin 300071, China
[email protected], [email protected]
2School of Mathematics
Sun Yat-sen University, Guangzhou 510275, China
3Department of Mathematical Sciences
Georgia Southern University, Statesboro, GA 30460-8093, USA
(2 November 2017)
Abstract
An edge-coloured path is rainbow if its edges have distinct colours. For a connected graph , the rainbow connection number (resp.Β strong rainbow connection number) of is the minimum number of colours required to colour the edges of so that any two vertices of are connected by a rainbow path (resp.Β rainbow geodesic). These two graph parameters were introduced by Chartrand, Johns, McKeon, and Zhang in 2008. Krivelevich and Yuster generalised this concept to the vertex-coloured setting. Similarly, Liu, Mestre, and Sousa introduced the version which involves total-colourings.
Dorbec, Schiermeyer, Sidorowicz, and Sopena extended the concept of the rainbow connection to digraphs. In this paper, we consider the (strong) total rainbow connection number of digraphs. Results on the (strong) total rainbow connection number of biorientations of graphs, tournaments, and cactus digraphs are presented.
Keywords: Total rainbow connection; digraph; tournament; cactus digraph; biorientation
1 Introduction
All graphs and digraphs considered in this paper are finite and simple. That is, we do not allow the existence of loops, multiple edges (for graphs), and multiple directed arcs (for digraphs). We follow the terminology and notation of BollobΓ‘s [3] for those not defined here.
The concept of rainbow connection in graphs was introduced by Chartrand, Johns, McKeon, and Zhang [4]. An edge-coloured path is rainbow if its edges have distinct colours. An edge-colouring of a connected graph is rainbow connected if any two vertices of are connected by a rainbow path. The rainbow connection number of , denoted by , is the minimum number of colours in a rainbow connected edge-colouring of . An edge-colouring of is strongly rainbow connected if for every pair of vertices and , there exists a rainbow geodesic, i.e., a path of length equal to the distance between and . The minimum number of colours in a strongly rainbow connected edge-colouring of is the strong rainbow connection number of , denoted by .
As a natural counterpart to the rainbow connection of edge-coloured graphs, Krivelevich and Yuster [8]; and Li, Mao, and Shi [10], proposed the concept of (strong) rainbow vertex-connection. A vertex-coloured path is vertex-rainbow if its internal vertices have distinct colours. A vertex-colouring of a connected graph is rainbow vertex-connected (resp.Β strongly rainbow vertex-connected) if any two vertices of are connected by a vertex-rainbow path (resp.Β geodesic). The rainbow vertex-connection number of , denoted by , is the minimum number of colours in a rainbow vertex-connected vertex-colouring of . The strong rainbow vertex-connection number of , denoted by , is the minimum number of colours in a strongly rainbow vertex-connected vertex-colouring of . We refer the reader to the survey [11] and the monograph [12] on the subject of rainbow connection in graphs.
Liu, Mestre, and Sousa [13]; and Chen, Li, Liu, and Liu [5], proposed the concept of (strong) total rainbow connection. A total-coloured path is total-rainbow if its edges and internal vertices have distinct colours. A total-colouring of a connected graph is total rainbow connected (resp.Β strongly total rainbow connected) if any two vertices are connected by a total-rainbow path (resp.Β geodesic). The total rainbow connection number of , denoted by , is the minimum number of colours in a total rainbow connected total-colouring of . The strong total rainbow connection number of , denoted by , is the minimum number of colours in a strongly total rainbow connected total-colouring of .
In [6], Dorbec, Schiermeyer, Sidorowicz, and Sopena introduced the concept of rainbow connection of digraphs. A directed path, or simply a path , is a digraph consisting of a sequence of vertices and arcs for . We also say that is a path, and its length is the number of arcs . A digraph is strongly connected if for any ordered pair of vertices in , there exists a path. An arc-coloured path is rainbow if its arcs have distinct colours. Let be a strongly connected digraph. An arc-colouring of is rainbow connected if for any ordered pair of vertices in , there is a rainbow path. The rainbow connection number of , denoted by , is the minimum number of colours in a rainbow connected arc-colouring of . Alva-Samos and Montellano-Ballesteros [1] then introduced the notion of strong rainbow connection of digraphs. An arc-colouring of is strongly rainbow connected if for any ordered pair of vertices , there is a rainbow geodesic, i.e., a rainbow path of minimum length. The strong rainbow connection number of , denoted by , is the smallest possible number of colours in a strongly rainbow connected arc-colouring of . We have , where denotes the diameter of . Subsequently, there have been some results on this topic, which considered many different classes of digraphs [2, 7, 14]. Very recently, Lei, Li, Liu, and Shi [9] introduced the (strong) rainbow vertex-connection of digraphs. A vertex-coloured directed path is vertex-rainbow if its internal vertices have distinct colours. A vertex-colouring of is rainbow vertex-connected (resp.Β strongly rainbow vertex-connected) if for any ordered pair of vertices in , there exists a vertex-rainbow path (resp.Β geodesic). The rainbow vertex-connection number of , denoted by , is the minimum number of colours in a rainbow vertex-connected vertex-colouring of . The strong rainbow vertex-connection number of , denoted by , is the minimum number of colours in a strongly rainbow vertex-connected vertex-colouring of . We have .
In this paper, we introduce the concept of total rainbow connection of digraphs. Let be a strongly connected digraph. A total-coloured directed path is total-rainbow if its arcs and internal vertices have distinct colours. A total-colouring of is total rainbow connected if for any ordered pair of vertices in , there exists a total-rainbow path. The total rainbow connection number , denoted by , is the minimum number of colours in a total rainbow connected total-colouring of . Likewise, a total-colouring of is strongly total rainbow connected if for any ordered pair of vertices , there exists a total-rainbow geodesic. The strong total rainbow connection number of , denoted by , is the minimum number of colours in a strongly total rainbow connected total-colouring of .
This paper is organised as follows. In Section 2, we present several general results about the parameters and , as well as their relationships to the parameters , and . In Section 3, we compute the parameters and for some specific digraphs . In Section 4, we study the parameters and for tournaments . Finally in Section 5, we consider the parameters and for cactus digraphs .
2 Definitions, remarks, and results for general digraphs
We begin with some definitions about digraphs. For a digraph , its vertex and arc sets are denoted by and . For an arc , we say that is an out-neighbour of , and is an in-neighbour of . Moreover, we call an in-arc of and an out-arc of . We denote the set of out-neighbours (resp.Β in-neighbours) of in by (resp.Β ). Let . For a strongly connected digraph , and , the distance from to (i.e., the length of a shortest path) in is denoted by , or if we wish to emphasise that the distance is taken in the digraph . Let denote the diameter of .
If , then we say that and are symmetric arcs. If and , then is an asymmetric arc. The digraph is an oriented graph if every arc of is asymmetric. A tournament is an oriented graph where every two vertices have one asymmetric arc joining them. A cactus is a strongly connected oriented graph where each arc belongs to exactly one directed cycle. Given a graph , its biorientation is the digraph \overset{\textup{\hskip 0.56917pt\tiny\leftrightarrow}}{\phantom{\in}}G obtained by replacing each edge of by the pair of symmetric arcs and . Let \overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}P_{n} and \overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{n} denote the directed path and directed cycle of order , respectively (where for \overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{n}), i.e., we may let V(\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}P_{n})=V(\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{n})=\{v_{0},\ldots,v_{n-1}\}, and A(\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}P_{n})=\{v_{0}v_{1},v_{1}v_{2}\ldots,v_{n-2}v_{n-1}\} and A(\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{n})=A(\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}P_{n})\cup v_{n-1}v_{0}. If is a directed cycle and , we write for the unique directed path in .
For a subset , we denote by the subdigraph of induced by . Given two digraphs and , and , we define to be the digraph obtained from and by replacing the vertex by a copy of , and replacing each arc (resp.Β ) in by all the arcs (resp.Β ) for . We say that is obtained from by expanding into . Note that the digraph obtained from by expanding every vertex into is also known as the lexicographic product .
Now, we shall present some remarks and basic results for the total rainbow connection and strong total rainbow connection numbers, for general digraphs and biorientations of graphs. We first note that in a total rainbow connected colouring of a strongly connected digraph , there must be a path between some two vertices with at least colours. Thus, we have the following proposition.
Proposition 1**.**
Let be a strongly connected digraph with vertices and arcs. Then
[TABLE]
It is easy to see that the bioriented paths \overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}P_{n}, for , form an infinite family of graphs where we have equalities in the first two inequalities in (1). Also, it is not difficult to see that for every directed cycle \overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{n} with , we have equalities in the last two inequalities in (1). These two results will be included in Theorems 11 and 12.
In the next result, we give equivalences and implications between several conditions, when the rainbow connection parameters are small.
Theorem 2**.**
Let be a non-trivial, strongly connected digraph.
- (a)
The following are equivalent.
- (i)
* is a bioriented complete graph.* 2. (ii)
diam. 3. (iii)
. 4. (iv)
. 5. (v)
. 6. (vi)
. 7. (vii)
. 8. (viii)
. 2. (b)
* if and only if is not a bioriented complete graph.* 3. (c)
- (i)
* if and only if .* 2. (ii)
, if and only if , if and only if diam. 3. (iii)
* if and only if .* 4. (iv)
* if and only if .* 5. (v)
* if and only if .*
Moreover, any of the conditions in (i) implies any of the conditions in (iv), and any of the conditions in (i), (iv) and (v) implies any of the conditions in (ii).
Proof.
(a) In [1], Theorem 2; and [9], Theorem 2, it was proved that (i) to (vi) are equivalent. Now clearly, we have (i) (viii), and by (1), it is easy to see that (viii) (vii) (ii).
(b) If is not a bioriented complete graph, then , and follows from (1). The converse clearly holds by (a).
(c) Part (i) was proved in [1], Theorem 2; and parts (ii) and (iii) were proved in [9], Theorem 2. We prove parts (iv) and (v). Suppose first that . Then (1) implies . Also, there exists a total rainbow connected colouring for , using colours. In such a total-colouring, for any , either , or and there is a total-rainbow path of length , which is also a total-rainbow geodesic. Thus , so that , and the first condition of (iv) implies the second. The same argument, with the role of ββ being replaced by ββ, gives that the first condition of (v) implies the second.
Next, suppose that . Then by (a), we know that is not a bioriented complete graph. By (b), we have , and hence , and (iv) is proved. Finally, suppose that . By (a) and (b), we have . By (c)(iv), we have , so that . This completes the proof of (v).
Now, we consider the final part of (c). Firstly, suppose that either condition in (i) holds, so that . Then (a) implies , and (1) implies . Moreover, there exists a rainbow connected arc-colouring for , using colours. Clearly by colouring all vertices of with a third colour, we have a total rainbow connected colouring for , using colours. Thus, . We have , and thus both conditions of (iv) hold. Secondly, suppose that either of the conditions in (i) holds. Then we have , and (a) implies that . Similarly, if any of the conditions in (iv) or (v) holds, then it is easy to use (1) and (a) to again obtain . Thus, any of the conditions in (i), (iv) and (v) implies any of the conditions in (ii). β
We remark that in Theorem 2(c), no other implication exists between the conditions of (i) to (v). Obviously, the conditions of (ii) and those of (iii) are mutually exclusive. Thus by the last part of (c), the conditions of (iii) are mutually exclusive to those of (i), (iv) and (v). Similarly, the conditions of (iv) and those of (v) are mutually exclusive, and thus the conditions of (i) and those of (v) are also mutually exclusive, since the conditions of (i) imply those of (iv). Clearly, the example of the bioriented stars \overset{\textup{\hskip 2.27626pt\tiny\leftrightarrow}}{\phantom{\in}}K_{1,n} shows that there are infinitely many digraphs where the conditions of (ii) hold, but those of (v) do not hold. Indeed, for , we have \overset{\rightarrow}{rvc}(\overset{\textup{\hskip 2.27626pt\tiny\leftrightarrow}}{\phantom{\in}}K_{1,n})=1, while \overset{\rightarrow}{\smash{t}rc}(\overset{\textup{\hskip 2.27626pt\tiny\leftrightarrow}}{\phantom{\in}}K_{1,n})=3. Also, there are infinitely many digraphs such that the conditions of (ii) and (iv) hold, but those of (i) do not hold. For example, let be a vertex of the directed cycle \overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{3}, and let be a digraph obtained by expanding into a bioriented clique . That is, D=(\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{3})_{u\to K}. Then, we have and , but . In the following lemma, we will see that there are infinitely many examples of digraphs where the conditions of (ii) hold, but the conditions of (iv) do not hold.
Lemma 3**.**
There exist infinitely many digraphs with and .
Proof.
We will construct the digraphs from the Petersen graph . We first consider the digraph \overset{\textup{\tiny\leftrightarrow}}{\phantom{\in}}{\sf P}_{10}. See Figure 1(a), where we have V({\sf P}_{10})=V(\overset{\textup{\tiny\leftrightarrow}}{\phantom{\in}}{\sf P}_{10})=\{u_{i},v_{i}:0\leq i\leq 4\}. Clearly, we have \textup{diam}(\overset{\textup{\tiny\leftrightarrow}}{\phantom{\in}}{\sf P}_{10})=2. We will show that \overset{\rightarrow}{\smash{t}rc}(\overset{\textup{\tiny\leftrightarrow}}{\phantom{\in}}{\sf P}_{10})=\overset{\rightarrow}{s\smash{t}rc}(\overset{\textup{\tiny\leftrightarrow}}{\phantom{\in}}{\sf P}_{10})=4. By (1), it suffices to prove that \overset{\rightarrow}{\smash{t}rc}(\overset{\textup{\tiny\leftrightarrow}}{\phantom{\in}}{\sf P}_{10})\geq 4 and \overset{\rightarrow}{s\smash{t}rc}(\overset{\textup{\tiny\leftrightarrow}}{\phantom{\in}}{\sf P}_{10})\leq 4.
[TABLE]
Suppose first that there is a total rainbow connected colouring of \overset{\textup{\tiny\leftrightarrow}}{\phantom{\in}}{\sf P}_{10}, using at most three colours, say . Then whenever we have x,y\in V(\overset{\textup{\tiny\leftrightarrow}}{\phantom{\in}}{\sf P}_{10}) and xy\not\in A(\overset{\textup{\tiny\leftrightarrow}}{\phantom{\in}}{\sf P}_{10}), there must be a total-rainbow path of length in \overset{\textup{\tiny\leftrightarrow}}{\phantom{\in}}{\sf P}_{10}. By considering the unique paths of length between the vertices , we may assume that , , and . Then by considering the unique paths of length between , we must have , , and . Similarly by considering the pairs , then , we must have , , and . But then, we do not have total-rainbow paths between , which is a contradiction. Therefore, \overset{\rightarrow}{\smash{t}rc}(\overset{\textup{\tiny\leftrightarrow}}{\phantom{\in}}{\sf P}_{10})\geq 4.
Now we define a total-colouring of as follows. We assign strongly total rainbow connected colourings for the two cycles and with colours , and then colour to the edges , for . For example, see Figure 1(b). We can then extend this to a total-colouring of \overset{\textup{\tiny\leftrightarrow}}{\phantom{\in}}{\sf P}_{10} where two symmetric arcs of \overset{\textup{\tiny\leftrightarrow}}{\phantom{\in}}{\sf P}_{10} both receive the same colour as the corresponding edge in . Then it is easy to check that is a strongly total rainbow connected colouring for \overset{\textup{\tiny\leftrightarrow}}{\phantom{\in}}{\sf P}_{10}. Therefore, \overset{\rightarrow}{s\smash{t}rc}(\overset{\textup{\tiny\leftrightarrow}}{\phantom{\in}}{\sf P}_{10})\leq 4.
Finally for every , we can obtain the digraph on vertices from \overset{\textup{\tiny\leftrightarrow}}{\phantom{\in}}{\sf P}_{10} by expanding into K\cong\overset{\textup{\hskip 2.27626pt\tiny\leftrightarrow}}{\phantom{\in}}K_{n-9}. That is, D_{n}=(\overset{\textup{\tiny\leftrightarrow}}{\phantom{\in}}{\sf P}_{10})_{v_{0}\to K}. Then note that . We may again apply the first argument above to obtain . Moreover, we can extend the total-colouring on \overset{\textup{\tiny\leftrightarrow}}{\phantom{\in}}{\sf P}_{10} to a total-colouring on where if ; if and ; if and ; if ; if ; and if . Then it is easy to see that is a strongly total rainbow connected colouring for . Indeed, if and , then at most one of is in , and correspond to distinct vertices in \overset{\textup{\tiny\leftrightarrow}}{\phantom{\in}}{\sf P}_{10}. Then a total-rainbow geodesic of length in corresponds to a total-rainbow geodesic of length in \overset{\textup{\tiny\leftrightarrow}}{\phantom{\in}}{\sf P}_{10}. Therefore, , and . β
We propose the following problem.
Problem 4**.**
Among all digraphs with diameter , are the parameters , , , and unbounded?
Alva-Samos and Montellano-Ballesteros [1] showed that for a connected graph ,
[TABLE]
Furthermore, for each inequality, there is an infinite family of graphs where equality holds, and also with the difference between the two parameters unbounded. For example, from [1, 4], we have rc(C_{n})=\overset{\rightarrow}{rc}(\overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}C_{n})=src(C_{n})=\overset{\rightarrow}{src}(\overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}C_{n})=\lceil\frac{n}{2}\rceil for . Also, we have \overset{\rightarrow}{rc}(\overset{\textup{\hskip 2.27626pt\tiny\leftrightarrow}}{\phantom{\in}}K_{1,n})=\overset{\rightarrow}{src}(\overset{\textup{\hskip 2.27626pt\tiny\leftrightarrow}}{\phantom{\in}}K_{1,n})=2 and for , where is the star with edges. On the other hand, for rainbow vertex-connection, Lei, Li, Liu, and Shi [9] showed that
[TABLE]
For total rainbow connection, we have the analogous inequalities to (2).
Proposition 5**.**
For a connected graph , we have
[TABLE]
Proof.
Given a (strongly) total rainbow connected colouring of , it is not hard to see that the total-colouring of \overset{\textup{\hskip 0.56917pt\tiny\leftrightarrow}}{\phantom{\in}}G, obtained by assigning the colour of the edge to both arcs , and the colour of the vertex of to the corresponding vertex of \overset{\textup{\hskip 0.56917pt\tiny\leftrightarrow}}{\phantom{\in}}G, is a (strongly) total rainbow connected colouring of \overset{\textup{\hskip 0.56917pt\tiny\leftrightarrow}}{\phantom{\in}}G. β
As in the case for rainbow connection, for each inequality of (4), there is an infinite family of graphs where equality holds, and also with the difference between the two parameters unbounded. We simply use the same examples. For , we have trc(C_{n})=\overset{\rightarrow}{\smash{t}rc}(\overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}C_{n})=strc(C_{n})=\overset{\rightarrow}{s\smash{t}rc}(\overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}C_{n}), which we will see in Theorem 11 later. Also, for , we have \overset{\rightarrow}{\smash{t}rc}(\overset{\textup{\hskip 2.27626pt\tiny\leftrightarrow}}{\phantom{\in}}K_{1,n})=\overset{\rightarrow}{s\smash{t}rc}(\overset{\textup{\hskip 2.27626pt\tiny\leftrightarrow}}{\phantom{\in}}K_{1,n})=3 and .
For the relationship between the total rainbow connection numbers of a digraph and its spanning subdigraphs, it is not hard to see that the following holds.
Proposition 6**.**
Let and be strongly connected digraphs such that, is a spanning subdigraph of . Then .
However, this is not true for the strong total rainbow connection number, as we will see in the next lemma.
Lemma 7**.**
There are strongly connected digraphs and such that, is a spanning subdigraph of , and .
[TABLE]
Proof.
Let be the digraph consisting of the solid arcs as in Figure 2, and be the digraph obtained from by adding the dotted arc . It is not hard to see that the colouring for as shown in Figure 2, is strongly total rainbow connected, where for the eight pairs of symmetric arcs in the middle, the two arcs in each pair have the same colour. Thus, .
Now, we will show that . Suppose that we have a strongly total rainbow connected colouring for , using at most colours. Notice that any two cut-vertices of must receive distinct colours, and thus without loss of generality, we may assume that have colours . Next, for , we see that the arcs must have distinct colours, and different from colours . Otherwise we can find two vertices such that any geodesic in connecting them is not total-rainbow. Hence, we may assume that , for . For the same reason, the arcs and vertices for , and the arcs for , and for , cannot use the colours . Since the unique geodesic in is , we may assume that , , , , , and . Finally, the unique geodesic in is , and the arcs and internal vertex in this geodesic are not yet coloured. These elements cannot use the colours . Thus the remaining possible colours are , and this is insufficient to make the geodesic total-rainbow. We have a contradiction. Hence and the result follows. β
Our final aim in this section is to compare the rainbow connection parameters. In [8], Krivelevich and Yuster observed that for and , we cannot generally find an upper bound for one of the parameters in terms of the other. Indeed, by taking , we have and . On the other hand, let the graph be constructed as follows. Take vertex-disjoint triangles and, by designating a vertex from each triangle, add a complete graph on the designated vertices. Then and . In [9], similar results were obtained for and , and for and .
When considering the total connection number in addition, we have the following trivial inequalities.
[TABLE]
In the following result, we see that there are infinitely many digraphs where the inequalities (5) and (6) are best possible.
Theorem 8**.**
- (a)
There exist infinitely many strongly connected digraphs with . 2. (b)
Given , there exists a strongly connected digraph with . 3. (c)
Given , there exists a strongly connected digraph with .
Proof.
(a) We may simply consider the digraphs as described after the proof of Theorem 2. That is, is the digraph on vertices, obtained by expanding a vertex of \overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{3} into \overset{\textup{\hskip 2.27626pt\tiny\leftrightarrow}}{\phantom{\in}}K_{n-2}. We have .
(b) For , let be the simple graph with disjoint triangles attached to , as described above. Let D_{s}=\overset{\textup{\hskip 0.56917pt\tiny\leftrightarrow}}{\phantom{\in}}G_{s}. Then by (4), (5), and (3), we have . In [13], Theorem 11, it was proved that . Thus we have .
(c) For , we construct the simple graph as follows. First, we take the graph , and let be the vertices of the , and the remaining vertices are , where is a triangle, for . We then add new vertices , and add the edges , for . Throughout, all indices are taken modulo . See Figure 3 for the case .
[TABLE]
Now let D_{s}=\overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}H_{s}. Then by (4), (6), and (3), we have . Thus it suffices to prove that and , which would imply .
We first prove that . Suppose that we have a vertex-colouring of , using fewer than colours. Then some two vertices of have the same colour. We may assume that and have the same colour, for some . If , then the unique geodesic is . If , then the unique geodesic is . If , then the unique geodesic is . In each case, we have two vertices in such that there is no vertex-rainbow geodesic connecting them. Thus .
Now we prove that . We define a total-colouring of , using the colours , as follows. For , let , , , , , , all modulo . For , let be a colour not in . Such a colour exists since . We will show that is a strongly total rainbow connected colouring for , which implies that . Let . We show that there always exists a total-rainbow geodesic . If at least one of belongs to , or if for some , then clearly such an geodesic exists, with length at most . Thus, it suffices to consider and the six cases as shown in the following table, where in each case.
In each case, we have a desired geodesic , where is some colour different from the remaining four colours. For example, in the very first case, we have . The proof is thus complete. β
We may also consider how far from equality we can be in (5) and (6). In the following result, we see that there is an infinite family of digraphs such that is unbounded on , while is bounded. Similar results also hold for in comparison with , and for in comparison with each of and .
Theorem 9**.**
- (a)
Given , there exists a strongly connected digraph such that and . 2. (b)
Given , there exists a strongly connected digraph such that and .
Proof.
(a) Let be the simple graph consisting of with a pendent edge attached to each of the vertices of , and D_{s}=\overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}F_{s}. Let be the vertices of the , and be the pendent vertices, where for . Then in any total rainbow connected colouring of , the vertices must receive distinct colours, and thus by (1), . Also, the arc-colouring of where for all ; and , for , is strongly rainbow connected, and thus . Since , we have , and thus .
(b) Let be the digraph consisting of copies of the triangle \overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{3}, all having one vertex in common. Let , and let the arcs of be , for . It was proved in Lemma 6 of [9] that , and . Now, note that for any two vertices , there is a unique path, and thus . The assertion then follows from . β
Finally, we consider how far apart the quantities and in (5) can be, and similarly for and in (6). For example, by considering the bioriented path \overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}P_{n}, we have \overset{\rightarrow}{\smash{t}rc}(\overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}P_{n})=\overset{\rightarrow}{s\smash{t}rc}(\overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}P_{n})=2n-3 and \max(\overset{\rightarrow}{rc}(\overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}P_{n}),\overset{\rightarrow}{rvc}(\overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}P_{n}))=\max(\overset{\rightarrow}{src}(\overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}P_{n}),\overset{\rightarrow}{srvc}(\overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}P_{n}))=n-1, so that
[TABLE]
and each difference can be arbitrarily large. However, in this example, all four quantities in consideration are unbounded in . Thus, we propose the following problem.
Problem 10**.**
Does there exist an infinite family of digraphs such that is unbounded on , while is bounded? Similarly, does there exist an infinite family of digraphs such that is unbounded on , while is bounded?
3 Total rainbow connection of some specific digraphs
In this section, we determine the (strong) total rainbow connection number of some specific digraphs. For , the wheel is the graph obtained by taking the cycle , and joining a new vertex to every vertex of . The vertex is the centre of . For , let denote the complete -partite graph with class-sizes . The following theorem determines the two parameters for the biorientations of paths, cycles, wheels, and complete multipartite graphs.
Theorem 11**.**
- (a)
For , \overset{\rightarrow}{\smash{t}rc}(\overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}P_{n})=\overset{\rightarrow}{s\smash{t}rc}(\overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}P_{n})=2n-3. 2. (b)
For , \overset{\rightarrow}{\smash{t}rc}(\overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}C_{n})=\overset{\rightarrow}{s\smash{t}rc}(\overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}C_{n})=g(n), where
[TABLE] 3. (c)
For , \overset{\rightarrow}{\smash{t}rc}(\overset{\textup{\hskip 2.56064pt\tiny\leftrightarrow}}{\phantom{\in}}W_{n})=\overset{\rightarrow}{s\smash{t}rc}(\overset{\textup{\hskip 2.56064pt\tiny\leftrightarrow}}{\phantom{\in}}W_{n})=3. 4. (d)
Let , and let \overset{\textup{\hskip 2.27626pt\tiny\leftrightarrow}}{\phantom{\in}}K_{n_{1},n_{2},\ldots,n_{t}} be a complete -partite digraph with for some . Then, \overset{\rightarrow}{\smash{t}rc}(\overset{\textup{\hskip 2.27626pt\tiny\leftrightarrow}}{\phantom{\in}}K_{n_{1},n_{2},\ldots,n_{t}})=\overset{\rightarrow}{s\smash{t}rc}(\overset{\textup{\hskip 2.27626pt\tiny\leftrightarrow}}{\phantom{\in}}K_{n_{1},n_{2},\ldots,n_{t}})=3.
Proof.
(a) In [5], Proposition 6, it was shown that for . Since \textup{diam}(\overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}P_{n})=n-1, it follows from (1) and (4) that
[TABLE]
and thus part (a) holds.
(b) For , it was shown in [13], Theorem 2, that ; and in [5], Theorem 8, that . Therefore, \overset{\rightarrow}{\smash{t}rc}(\overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}C_{n})\leq\overset{\rightarrow}{s\smash{t}rc}(\overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}C_{n})\leq strc(C_{n})=g(n). It suffices to prove that \overset{\rightarrow}{\smash{t}rc}(\overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}C_{n})\geq g(n). We have \overset{\rightarrow}{\smash{t}rc}(\overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}C_{n})\geq 2\,\textup{diam}(\overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}C_{n})-1=2\lfloor\frac{n}{2}\rfloor-1, and this gives \overset{\rightarrow}{\smash{t}rc}(\overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}C_{n})\geq g(n) for . To see that \overset{\rightarrow}{\smash{t}rc}(\overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}C_{7})\geq 6 (resp.Β \overset{\rightarrow}{\smash{t}rc}(\overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}C_{9})\geq 8), one can check that in any total-colouring of \overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}C_{7} (resp.Β \overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}C_{9}) with at most (resp.Β ) colours, there exist vertices and with (resp.Β ) such that the path with length (resp.Β ) is not total-rainbow. Since the other path has length (resp.Β ), it is also not total-rainbow. Hence, there does not exist a total-rainbow path, and the results for follow. Finally, let or . Let . Suppose that we have a total-colouring of \overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}C_{n} with fewer than colours. Let . Then for some , we have arcs of and vertices with the same colour. Without loss of generality, for some , either and , or and , or and , have the same colour. Consider the two paths in \overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}C_{n}. The path is not total-rainbow. The path has at least arcs and internal vertices, and hence is also not total-rainbow. Therefore \overset{\rightarrow}{\smash{t}rc}(\overset{\textup{\hskip 1.42271pt\tiny\leftrightarrow}}{\phantom{\in}}C_{n})\geq n.
(c) Let be the centre of \overset{\textup{\hskip 2.56064pt\tiny\leftrightarrow}}{\phantom{\in}}W_{n}. Note that any total-colouring of \overset{\textup{\hskip 2.56064pt\tiny\leftrightarrow}}{\phantom{\in}}W_{n} where every in-arc (resp.Β out-arc) of has colour (resp.Β ), and has colour , is strongly total rainbow connected. Since , we have \textup{diam}(\overset{\textup{\hskip 2.56064pt\tiny\leftrightarrow}}{\phantom{\in}}W_{n})=2, and 3\leq\overset{\rightarrow}{\smash{t}rc}(\overset{\textup{\hskip 2.56064pt\tiny\leftrightarrow}}{\phantom{\in}}W_{n})\leq\overset{\rightarrow}{s\smash{t}rc}(\overset{\textup{\hskip 2.56064pt\tiny\leftrightarrow}}{\phantom{\in}}W_{n})\leq 3 by (1). Thus the result holds.
(d) Let be the partition classes of \overset{\textup{\hskip 2.27626pt\tiny\leftrightarrow}}{\phantom{\in}}K_{n_{1},n_{2},\ldots,n_{t}}. For each arc with and , assign colour to if and colour if , and assign colour to each vertex of \overset{\textup{\hskip 2.27626pt\tiny\leftrightarrow}}{\phantom{\in}}K_{n_{1},n_{2},\ldots,n_{k}}. Then it is easy to check that this is a strongly total rainbow connected colouring. Also, since for some , we have \textup{diam}(\overset{\textup{\hskip 2.27626pt\tiny\leftrightarrow}}{\phantom{\in}}K_{n_{1},n_{2},\ldots,n_{t}})=2. Hence by (1), we have 3\leq\overset{\rightarrow}{\smash{t}rc}(\overset{\textup{\hskip 2.27626pt\tiny\leftrightarrow}}{\phantom{\in}}K_{n_{1},n_{2},\ldots,n_{k}})\leq\overset{\rightarrow}{s\smash{t}rc}(\overset{\textup{\hskip 2.27626pt\tiny\leftrightarrow}}{\phantom{\in}}K_{n_{1},n_{2},\ldots,n_{k}})\leq 3, and the result follows. β
In the next result, we determine the (strong) total rainbow connection numbers for directed cycles.
Theorem 12**.**
Let . Then,
[TABLE]
Proof.
Let \overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{n}=v_{0}v_{1}\cdots v_{n-1}v_{0}. One can easily verify that \overset{\rightarrow}{\smash{t}rc}(\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{3})=\overset{\rightarrow}{s\smash{t}rc}(\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{3})=3, and \overset{\rightarrow}{\smash{t}rc}(\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{4})=\overset{\rightarrow}{s\smash{t}rc}(\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{4})=6. Now let . By (1), we have \overset{\rightarrow}{\smash{t}rc}(\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{n})\leq\overset{\rightarrow}{s\smash{t}rc}(\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{n})\leq 2n, and hence it suffices to prove that \overset{\rightarrow}{\smash{t}rc}(\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{n})\geq 2n. Suppose that we have a total rainbow connected colouring for \overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{n}, using fewer than colours. Then, there exist two elements of V(\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{n})\cup A(\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{n}) with the same colour. We recall that for , we have \overset{\rightarrow}{rc}(\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{n})=n from [1], Theorem 4; and \overset{\rightarrow}{rvc}(\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{n})=n from [9], Proposition 8. It follows that we must have a vertex and an arc with the same colour. We may assume that and have the same colour, for some . If , then the unique path is not total-rainbow. Otherwise, the unique path is not total-rainbow. We have a contradiction, and therefore, \overset{\rightarrow}{\smash{t}rc}(\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{n})\geq 2n. β
4 Tournaments
We now study the (strong) total rainbow connection number of tournaments. Our first aim is to consider the range of values that the (strong) total rainbow connection number can take, over all strongly connected tournaments of a given order . Clearly, {\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C}_{3} is the only such tournament of order . For , there is also a unique tournament, namely , which is the union of {\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C}_{4}=v_{0}v_{1}v_{2}v_{3}v_{0} and the arcs .
For the rainbow connection analogue, we have \overset{\rightarrow}{rc}({\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C}_{3})=\overset{\rightarrow}{src}({\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C}_{3})=\overset{\rightarrow}{rc}(T_{4})=\overset{\rightarrow}{src}(T_{4})=3. Dorbec, Schiermeyer, Sidorowicz, and Sopena [6]; and Sidorowicz and Sopena [14], proved the following results.
Theorem 13**.**
[6, 14]* If is a strongly connected tournament with vertices, then .*
Theorem 14**.**
[6, 14]* For every and such that , there exists a tournament on vertices such that .*
It was also remarked in [6] that there does not exist a tournament on or vertices with rainbow connection number , and that such a tournament exists if the order is (mod . In response, Alva-Samos and Montellano-Ballesteros [2] proved the following result.
Theorem 15**.**
[2]* For every , there exists a tournament on vertices such that .*
Lei, Li, Liu, and Shi [9] considered both the rainbow vertex-connection and strong rainbow vertex-connection analogues. We have \overset{\rightarrow}{rvc}({\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C}_{3})=\overset{\rightarrow}{srvc}({\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C}_{3})=1, and . For general values of , they proved the following results.
Theorem 16**.**
[9]* If is a strongly connected tournament on vertices, then .*
Theorem 17**.**
[9]* For and , there exists a tournament on vertices such that .*
Here, we consider the same situation for the (strong) total rainbow connection number. Clearly, we have \overset{\rightarrow}{\smash{t}rc}({\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C}_{3})=\overset{\rightarrow}{s\smash{t}rc}({\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C}_{3})=3, and . Now, Theorem 18 below is the analogous result to Theorems 13 and 16.
Theorem 18**.**
If is a strongly connected tournament on vertices, then
[TABLE]
Proof.
The theorem holds for . Now, let . Clearly by (1), we have , so it remains to prove that .
Let . If , then by Theorem 13, there exists a strongly rainbow connected arc-colouring of , using at most colours. By assigning a new colour to all vertices of , we obtain a strongly total rainbow connected total-colouring of , using at most colours. Indeed, if , then either ; or and there exists a total-rainbow path of length , which is also an geodesic.
Therefore, we may assume that . We shall define a total-colouring of with colours, and show that is a strongly total rainbow connected colouring. Our approach will use ideas in the proofs of Theorems 13 and 16 (see [14], Theorem 16; and [9], Theorem 20). Define an arc-colouring and a vertex-colouring as follows. Let be two vertices with , and let be an geodesic. Note that we have . Let be a bijective mapping such that . We let
[TABLE]
Also, note that are distinct vertices. For , we let , , and the remaining vertices are given the distinct colours .
Now let be the total-colouring of obtained by combining and . That is,
[TABLE]
Let . From the proofs of Theorems 13 and 16, the following facts can be seen.
- (i)
In the arc-colouring , every geodesic is rainbow. 2. (ii)
In the vertex-colouring , every geodesic is vertex-rainbow, if . 3. (iii)
In the vertex-colouring , there exists a vertex-rainbow geodesic, if .
By (ii) and (iii), we can choose a vertex-rainbow geodesic , with respect to . Then by (i), is also a rainbow geodesic, with respect to . Therefore, is a total-rainbow geodesic, with respect to . It follows that is a strongly total rainbow connected colouring for . This completes the proof. β
Next, Theorem 19 below is an analogue to Theorems 14, 15, and 17.
Theorem 19**.**
For and with odd, there exists a tournament on vertices such that
[TABLE]
Proof.
We first consider the case . By Theorem 15, we know that for every , there exists a tournament with . Thus by Theorem 2(c), we have . For , we let be the tournament which is the union of two copies of {\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C}_{5}: and . Then it is easy to check that .
Now, let be odd. We first consider the case , and construct a tournament on vertices such that (7) holds. Let , and and . Note that we have . Since is the only path in , we have diam, and thus by (1). Now, consider the total-colouring of where for ; if and ; for ; and . Then uses colours. Let . If , then is the unique path, and if , then . If , then , and a geodesic is if , and if . In each case, we always have a total-rainbow geodesic, so that . Therefore, (7) holds for , by (1).
Finally, let . We construct a tournament on vertices such that (7) holds. Let be a tournament on vertices, where is a single arc if ; T=\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{3} if ; if ; and if . All of these tournaments exist from our previous arguments. Let , which is obtained from by expanding into . Then, note that we have diam, which again gives . Now, we extend the colouring on to a colouring on by letting if ; if and ; if and ; if ; and
- β’
if , let and , where is the only arc of ;
- β’
if (resp.Β ), let be a strongly total rainbow connected colouring when restricted to , using three (resp.Β five) colours that are already used.
From previous arguments and remarks, the strongly total rainbow connected colourings on exist for . We claim that is a strongly total rainbow connected colouring for . Let . A similar argument as for shows that, there exists a total-rainbow geodesic in , if we do not have . Now, let . Then, we have exactly one of the following three situations.
- β’
.
- β’
, and there exists such that is a path in . Then there exists a total-rainbow path of length in , which is also a total-rainbow geodesic in .
- β’
, and there does not exist such that is a path in . Then, is a total-rainbow geodesic in , since .
In each case, we have a total-rainbow geodesic in , and the claim holds. Therefore, . Again (7) holds, by (1). β
In Theorem 19, we have not been able to consider the case when is even. Thus we pose the following problem.
Problem 20**.**
Do there exist with and even such that, there exists a tournament on vertices with ? Similarly, what happens for ?
In [6], Dorbec, Schiermeyer, Sidorowicz, and Sopena also proved the following result, which shows that the rainbow connection number of a tournament is in fact closely related to its diameter.
Theorem 21**.**
[6]* Let be a tournament of diameter . We have .*
Lei, Li, Liu, and Shi [9] then proved an analogous result for the rainbow vertex-connection number, as follows.
Theorem 22**.**
[9]* Let be a tournament of diameter . We have .*
Here, we shall prove the following analogue for the total rainbow connection number.
Theorem 23**.**
Let be a tournament of diameter .
- (a)
If , then . 2. (b)
If , then .
Proof.
Clearly for both parts, the lower bound follows from (1). We prove the upper bounds.
(a) Let . Let , and , . Note that ; that are the vertices at distance from ; and that for all . Also, note that for every vertex , there must be an out-neighbour of in , since there must exist a path in with length .
We define a total-colouring of as follows. Let if either , or is an arc in or ; if one of is in and the other is in ; for all ; ; and for all . We claim that is a total rainbow connected colouring for . Let , and suppose that . We show that a total-rainbow path always exists. Clearly this is true if . Next, let , and let be an out-neighbour of . If then we take , and if , then we take . If , then since , there must exist a suitable path for some . Finally, let . If then we take , and if , then we can take for some . Thus the claim holds, and we have .
(b) Let . We prove the upper bound by constructing a total-colouring of , using colours for the arcs, and for the vertices. As in the proofs of Theorems 21 and 22 (see [6], Theorem 24; and [9], Theorem 23), we consider the following decomposition of . Let be a vertex with eccentricity , i.e., there exists a vertex of at distance from . For , let denote the set of vertices at distance from . Then, every is non-empty. Note that for every with , and whenever , , and . Let have maximum in-degree in , and have maximum out-degree in . Now, we define the total-colouring as follows.
- β’
if , and if and , for .
- β’
if and , where .
- β’
if , and for every other arc in .
- β’
if , and for every other arc in .
- β’
if , for .
- β’
, and for .
- β’
, and for .
Next, we update by considering a geodesic . Note that we have or . Thus, we consider these two cases.
- (i)
If , then must contain one arc from to , for every . Let be the vertex of in . We recolour by letting , and the out-arc of in with colour . 2. (ii)
If , then must contain one arc from to , for every , and one arc from to , for some . If , then let be the vertex of in , and be the arc of in . If , then let be the arc of in . In either case, we recolour and by letting and , the out-arc of in with colour , and the arc in (either or ) with colour .
The situation in (ii) is shown in Figure 4.
[TABLE]
Now, we prove that is a total rainbow connected colouring for , which implies the required upper bound . The proof from here onwards is the same as in [9], but we shall provide it for the sake of completeness. Let . We show that there always exists a total-rainbow path. This is true if . Let . If , then . Otherwise, for some . Let be an geodesic, which contains one vertex in each of . If , then contains a total-rainbow path. Otherwise, , and is a desired path.
Hence, suppose that . Note that by the choice of , if , then either , or there exists such that is a path. Indeed, if and no such exists in , then in , the set of in-neighbours of contains , and all in-neighbours of . Thus, contradicts the choice of . Hence in , there is an path of length at most two, which is total-rainbow. Now, consider . If , then contains a total-rainbow path. Otherwise, is a desired path. Next, if , then by a similar argument as for , we have a total-rainbow path of length at most two in , by the choice of . We have is a desired path. Finally, let . If in case (i) or case (ii), or in case (ii), then or contains a total-rainbow path. Otherwise, we can find an in-neighbour of in , say . Note that exists by considering an geodesic, and that in case (i), or in case (ii). Then, is a desired path. β
5 Cactus digraphs
In [2], Alva-Samos and Montellano-Ballesteros studied the rainbow connection number of cactus digraphs. In this section, we will consider both the rainbow vertex-connection and the total rainbow connection numbers of cactus digraphs.
Recall that a cactus is a strongly connected oriented graph where every arc belongs to exactly one directed cycle. We begin by describing the structure of a cactus. Recall that for a digraph , a block is a maximal subdigraph without a cut-vertex. An end-block is a block which is incident with at most one cut-vertex, and there must exist at least one end-block in . Moreover, if itself is not a block, then there must exist at least two end-blocks in . The block graph of , denoted , is the graph with is block of and if and share a vertex in . From the definition of cactus, it is not hard to obtain the following characterisation, as remarked in [2].
Lemma 24**.**
[2]* Let be a digraph with vertices and arcs. Then the following statements are equivalent.*
- (i)
* is a cactus.* 2. (ii)
* is a strongly connected digraph in which every block is a directed cycle.* 3. (iii)
Let be the number of blocks in . Then has a decomposition into directed cycles such that, for each we have
[TABLE]
and . 4. (iv)
There is exactly one directed path between each pair of vertices of .
Thus, (iv) shows that a cactus may be considered as a sort of analogue in a digraphs setting to trees in a simple graphs setting (although, the block graph of a cactus is not always a tree). Lemma 24 allows us to define some terms about a cactus . For , we denote by the unique directed path in . Let denote the set formed by all the cut-vertices of . If and belongs to the cycle , then we write and for the out-neighbour and in-neighbour of in . In particular, if , then belongs to exactly one cycle, and we may simply write and . We say that a cactus on vertices is an -cactus when it has a decomposition into cycles. Also, we always consider a cactus along with its cycle decomposition as given in (iii) of Lemma 24. It is clear that such a decomposition is unique (up to the ordering of the cycles).
From the equation in (iii), we see that if for , then
[TABLE]
Since for all , we have . It is also easy to obtain the following lemma.
Lemma 25**.**
Let be a cactus with blocks. Let and , where are two distinct cycles of . Then the unique and paths in must be of the form and , where and . Furthermore, no arc or internal vertex of belongs to or , and likewise for .
Proof (sketch).
By Lemma 24(iii), in order to connect to in , we connect and by a chain of intermediate cycles. We may think of , and these intermediate cycles as a chain of βkissing cyclesβ, with and at the two ends, and every two consecutive cycles meeting at a cut-vertex of . The vertices where two consecutive cycles meet are distinct. Now to connect to , we start at , traverse along until we reach the cut-vertex in , say . Then we traverse along the chain of intermediate cycles until we reach the cut-vertex in , say . Finally, we traverse along until we reach . Now, note that to connect to , we traverse in the opposite direction, first from , through to , then the intermediate cycles in reverse order, and then through , , and finally to . By Lemma 24(iv), these are the unique and paths in , and they have the forms as described in the lemma. β
By Lemma 24(iv), since the directed path between any pair of vertices in a cactus is unique, we have
[TABLE]
Thus, it suffices to consider the parameters , , and . Moreover, note that again by Lemma 24(iv), we have
[TABLE]
since a total rainbow connected colouring of can be obtained by combining a rainbow connected colouring with colours, and a rainbow vertex-connected colouring with additional colours.
If is an -cactus, then Q=\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{n}. We have \overset{\rightarrow}{rc}(\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{n})=n for . In [9], it was proved that
[TABLE]
Finally, \overset{\rightarrow}{\smash{t}rc}(\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{n}) is given by Theorem 12. From now on, we consider -cactus digraphs where . In [2], Alva-Samos and Montellano-Ballesteros proved the following result.
Theorem 26**.**
[2]* Let be an -cactus with . We have the following.*
- (a)
. 2. (b)
* if and only if is independent.* 3. (c)
* if and only if and Q[K_{Q}]=\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}P_{q-1}.*
They also showed that for every value in the range in (a), there exists an -cactus whose rainbow connection number is equal to .
Here, our aim is to prove similar results for the parameters and . To proceed, we first prove the following lemma. Given a digraph with a vertex-colouring , we say that a vertex is singularly coloured if no other vertex of has colour . Likewise, if is a total-colouring of , we say that an element is singularly coloured if no other element of has colour .
Lemma 27**.**
Let be a cactus with blocks. Let be an end-block of , let be the unique cut-vertex of in , and . Let and be the out-neighbour and in-neighbour of in .
- (a)
If is a rainbow vertex-connected colouring on , then all vertices of must be singularly coloured. 2. (b)
If is a total rainbow connected colouring on , then all elements of must be singularly coloured. 3. (c)
We have
[TABLE]
Proof.
We first prove (b), from which we can easily deduce (a). We then prove (c).
(b) Let be a total rainbow connected colouring on . We will proceed by assuming that there are and with . We then obtain a contradiction by finding a path whose set of arcs and internal vertices contains both and , since the end-vertices of would then not be connected by a total-rainbow path in .
Firstly, let , and suppose that we have with . Suppose that . Then note that either has length at least , or has length and . In either case, we may let or . Now, suppose that . Then belongs to some cycle in . By Lemma 25, the unique and paths in are and , for some . Thus, we may let or .
Next, suppose that we have with . If , then since has length at least , we let or . If , then let be the unique cycle in containing . By Lemma 25, the unique and paths in are and , for some . The same condition holds for in place of . Thus, we may let or .
Secondly, let , and suppose that we have with . If , then we may let or . Now, suppose that . Then belongs to some cycle in . By Lemma 25, the unique and paths in are and , for some . The same condition holds for in place of . Thus, we may let or .
Finally, suppose that we have with . If , then we let or . If , then let be the unique cycle in containing . By Lemma 25, the unique and paths in are and , for some . The same conditions hold for when is replaced by , or is replaced by , or both. Thus, we may let or .
In every case, we have found a desired path , and thus we always have a contradiction.
(a) Let be a rainbow vertex-connected colouring on , and suppose that there exist vertices and with . We can use exactly the same argument as in the first part of (b) to obtain a path which contains and as internal vertices. Thus is not vertex-rainbow, which is a contradiction.
(c) Suppose that is a rainbow vertex-connected colouring of . Then in , we require at least colours for , otherwise there would exist two vertices in which are not connected by a vertex-rainbow path in , and thus also in . Also by (a), the vertices of must provide additional colours for . Therefore, uses at least colours, and . Likewise, if is a total rainbow connected colouring of , then in , we require at least colours for . By (b), the elements of must provide additional colours for . Therefore, . β
We are now ready to present the main results of this section. Firstly, we have the following result, which contains the rainbow vertex-connection and total rainbow connection analogues of Theorem 26.
Theorem 28**.**
Let be an -cactus, with . We have the following.
- (a)
. 2. (b)
.
Proof.
(a) We first prove the lower bound by induction on . For , consists of two cycles meeting at a cut-vertex . Then is a path in with length , and therefore . Now let , and suppose the lower bound holds for any cactus with blocks. Let be an end-block of , let be the unique cut-vertex of in , and let . By Lemma 27(c) and induction, we have
[TABLE]
since . This proves the lower bound of (a).
Now we prove the upper bound. Let be two end-blocks of , and be the cut-vertices of in . Let be the out-neighbour and in-neighbour of in , and similarly for in . We define the vertex-colouring on by setting , , and all remaining vertices are given the distinct colours . It is easy to check that is a rainbow vertex-connected colouring for . Indeed, if or , then clearly is a vertex-rainbow path. Otherwise, we may assume that and . Then does not have and as internal vertices, and thus is also vertex-rainbow. Therefore, .
(b) The lower bound can be proved similarly by induction . For , since we have a path of length in , this implies by (1). Now let , let be an end-block of , and be as defined in (a). By Lemma 27(c) and induction, we have
[TABLE]
and the lower bound of (b) holds.
The upper bound easily follows from (8), Theorem 26(a) and part (a). β
We have the following corollary.
Corollary 29**.**
For every -cactus , we have and .
In the next result, we characterise the cactus digraphs that attain equality in each of the two lower bounds in Theorem 28. It turns out that we have the same characterisation in both cases.
Theorem 30**.**
Let be an -cactus, with . Then the following are equivalent.
- (i)
. 2. (ii)
. 3. (iii)
For all , we have .
Proof.
We first prove that (iii) (i). We use induction on to show that if is an -cactus and for all , then we have . For any such cactus , we may define a vertex-colouring as follows. Let for some . Note that for every , we have . Now, for , set for all , and for all , where the and are some colours. Assign further distinct colours to all remaining vertices. Observe that every cycle of is rainbow coloured.
We use induction on to show that is a rainbow vertex-connected colouring for with colours. The assertion holds for , since this is included in the proof of the upper bound in Theorem 28(a). Now suppose that and the assertion holds for . Let be the vertex-colouring of as described. Let be an end-block of , let be the unique cut-vertex of in , and let . Let be the vertex-colouring of when is restricted to . Then obviously we have for all , and it is easy to see that is the vertex-colouring for of the type as described. Thus by induction, is a rainbow vertex-connected colouring of , using colours. This means that the number of colours used by is , since . It remains to show that is a rainbow vertex-connected colouring for . Let for all , and for all . Let . Then the unique path is vertex-rainbow if (by induction), or (since is rainbow coloured by ). Now let and . Then the unique path has the form . By induction, the path is vertex-rainbow, and furthermore, by the definition of , the vertices of have distinct colours and do not use the colour . Since the path does not have an internal vertex with colour , it follows, again from the definition of , that is a vertex-rainbow path. A very similar argument can be used for the case and . Therefore by induction, is a rainbow vertex-connected colouring of with colours. Hence, , and by Theorem 28(a). This completes the proof that (iii) (i).
Next, we prove that (i) (iii), and (ii) (iii), by using similar arguments for both implications. We use induction on to show that if (resp.Β ), then for all . The assertions trivially hold for , since we have (resp.Β ) by Corollary 29, and has only one cut-vertex. Next, we prove the assertion for . Suppose that has two cut-vertices with . Then consists of three cycles where , , , and . The unique path in is and has length at least , so that (resp.Β , by (1)). Thus the assertions hold for .
Now let , and suppose that the assertions hold for . Suppose that there exist with . Note first that there is a unique cycle containing both and . Indeed, if , then is the unique cycle containing the arc . If , then we may assume that the unique path in is , where (if , then we could consider instead of ). Then is the unique cycle containing the arc , and also contains . Next, note that the set of the remaining cycles of can be partitioned so that each part forms a βbranch of cycles attached to β. More formally, for , let be the component of that contains . Thus the subdigraphs are the βbranchesβ, and every cycle other than belongs to exactly one of the . Now, we choose an end-block of as follows. If has at least two cycles, then we let be an end-block of in , and likewise if has at least two cycles. Otherwise, both and must have exactly one cycle, and since , there exists a vertex in . We let be an end-block of in . Now, let be the only cut-vertex of in , and . By the choice of , we see that and .
For the rainbow vertex-connection, we have by induction. Therefore by Lemma 27(c), we have
[TABLE]
since . By induction, we have proved that (i) (iii).
Similarly, for the total rainbow connection, we have by induction. Therefore by Lemma 27(c), we have
[TABLE]
By induction, we have proved that (ii) (iii).
Finally, we can now easily show that (i) (ii), and this would mean that the three conditions are equivalent. Suppose that . Since we know that (i) (iii), we have for all . In particular, is independent. By Theorem 26(b), we have . Thus by (8), we have . Therefore, by Theorem 28(b), and (ii) holds. β
Similarly in the next result, we characterise the cactus digraphs that attain equality in each of the two upper bounds in Theorem 28.
Theorem 31**.**
Let be an -cactus, with . Then the following are equivalent.
- (i)
. 2. (ii)
. 3. (iii)
* and Q[K_{Q}]=\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}P_{q-1}.*
It is easy to see (as shown in [2]) that an equivalent way of saying and Q[K_{Q}]=\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}P_{q-1} is that has the following structure. consists of cycles and cut-vertices such that, , , and for . Moreover, is a directed path in . We shall identify with this notation and say that such a cactus is a special path cactus.
Proof of Theorem 31.
(iii) (ii). Suppose that and Q[K_{Q}]=\overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}P_{q-1}. In other words, is a special path cactus, with the notation as described above. Let be the out-neighbour of in , and be the in-neighbour of in . Then is a path of length , and thus by (1). We have by Theorem 28(b).
(ii) (i). Let . By Theorem 26(a), we have . By (8), we have . We have by Theorem 28(a).
(i) (iii). We use induction on to show that if , then is a special path cactus. The case holds, since by Corollary 29 and is always a special path cactus. Next, we consider the case . Then consists of three cycles . Suppose that is not a special path cactus. Then either has one cut-vertex, or two cut-vertices with and . If we have the former, or the latter with and , then by Theorem 30, we have . Otherwise, we may assume that , , , and . Define the vertex-colouring on where if , if , if , and all remaining vertices are given further distinct colours. It is easy to check that is a rainbow vertex-connected colouring for with colours, and thus . In fact, we have by Theorems 28(a) and 30.
Now, let , and suppose that the assertion holds for . Let be an end-block of , let be the only cut-vertex of in , and . If is not a special path cactus, then by induction and Theorem 28(a), we have . We have a rainbow vertex-connected colouring for , using colours, since . Therefore, must be a special path cactus with at least three cycles. Using the prescribed notation, the end-blocks of are and . Suppose that . Then we can apply the same argument with instead of , and conclude that is a special path cactus, and we have is the out-neighbour of in . Thus is a special path cactus. A similar argument holds if , where we would apply the same argument with instead of . β
Finally in the following result, we see that, as in the case for , every value in the range in Theorem 28(a) can be attained by . However, the same is not quite true for with respect to the range in Theorem 28(b). We will see that can attain every value in the range, except for the value of .
Theorem 32**.**
Let .
- (a)
Let . For every where
[TABLE]
there is an -cactus with . 2. (b)
Let with . For every where
[TABLE]
there is an -cactus with . 3. (c)
For every -cactus , we have .
Proof.
For (a) and (b), we construct an -cactus as follows. For some , we take copies of \overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{3}, say , and amalgamate them linearly. That is, we may let and for . Then, we attach one copy of \overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{n-2q+2}, say , and further copies of \overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{3}, say , at the same vertex . That is, we identify one vertex from each of with . Let be the resulting -cactus, and note that can be constructed for . We define the path
[TABLE]
(a) Suppose first that is even, and note that . Let . The path as defined in (9) has length . Thus . Now, consider the vertex-colouring of where if or ; if or ; and the remaining vertices are given further distinct colours. It is easy to check that is a rainbow vertex-connected colouring for using colours, since no path can have two internal vertices with both having colour , or both having colour . Thus, , and as required.
Now, let be odd, and note that . We take and replace the arc with the path , and with a copy of \overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{n-2q+1}. Let be the resulting cactus, and note that can be constructed for . Now the path as in (9) has length , so that . We consider the vertex-colouring of where if ; if or ; ; and the remaining vertices are given further distinct colours. Again, it is easy to check that is a rainbow vertex-connected colouring for using colours. Thus, , and as required.
(b) For the case (mod ), let and . We have . Let . Then
[TABLE]
Indeed, (10) can be seen with a similar argument as in the proof of Lemma 27(c). If we have a total rainbow connected colouring for , then must use at least colours. Furthermore, by Lemma 27(b), the arcs in not incident with must be singularly coloured by , and thus they provide an additional colours for . Now, the path in (9) has length in , so that by (1), we have , and thus
[TABLE]
Now we define a total-colouring of as follows. Let for ; if ; if ; if ; and if . All remaining arcs and vertices are given further distinct colours. Then, since the colours are each used twice, and the colours are each used times, and , we have that uses colours. Moreover, it is easy to check that is a total rainbow connected colouring for , since no path can have two arcs or internal vertices with the same colour. Thus, , and as required.
Next, for the case (mod ), let , so that . With such an , we let be the -cactus as in the second part of (a). Let . The path in (9) has length in , so that by (1), we have . Again (10) holds, and thus
[TABLE]
Now we define a total-colouring of as follows. Let for ; ; ; if ; if ; if ; and if . All remaining arcs and vertices are given further distinct colours. Then, since the colours are each used twice, the colours are each used times, and , we have that uses colours. Again, it is easy to check that is a total rainbow connected colouring for . Thus, , and as required.
Finally, we consider the case (mod ) with . Let , and note that . With such an , we take and replace the arcs and with the paths and , and with a copy of \overset{\textup{\hskip 1.42271pt\tiny\rightarrow}}{\phantom{\in}}C_{n-2q}. Let be the resulting -cactus, and note that can be constructed for . Now the path in (9) has length , so that by (1). Again (10) holds, and thus
[TABLE]
Now we define a total-colouring of as follows. Let ; for ; ; ; ; ; if ; if ; if ; and if . All remaining arcs and vertices are given further distinct colours. Then, since the colours are each used twice, the colours are each used times, and , we have that uses colours. Again, it is easy to check that is a total rainbow connected colouring for . Thus, , and as required.
(c) If is a special path cactus, then by Theorem 31, we have . Otherwise, by Theorems 26, 28, and 31, we have and . Thus by (8), we have . β
Acknowledgements
Hui Lei and Yongtang Shi are partially supported by National Natural Science Foundation ofβ Chinaβ (Nos.Β 11371021,β 11771221),β andβ Naturalβ Scienceβ Foundationβ ofβ Tianjinβ (No. 17JCQNJC00300). Henry Liu is partially supported by the Startup Fund of One Hundred Talent Program of SYSU. Henry Liu would also like to thank the Chern Institute of Mathematics, Nankai University, for their generous hospitality. He was able to carry out part of this research during his visit there.
The authors thank the anonymous referees for the careful reading of the manuscript.
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