Color Code Techniques In Rainbow Connection

Let G be a graph with an edge k-coloring γ : E(G) → {1, …, k} (not necessarily proper). A path is called a rainbow path if all of its edges have different colors. The map γ is called a rainbow coloring if any two vertices can be connected by a rainbow path. The map γ is called a strong rainbow coloring if any two vertices can be connected by a rainbow geodesic. The smallest k for which there is a rainbow k-coloring (resp. strong rainbow k-coloring) on G is called the rainbow connection number (resp. strong rainbow connection number) of G, denoted rc(G) (resp. src(G)). In this paper we generalize the notion of “color codes” that was originally used by Chartrand et al. in their study of the rc and src of complete bipartite graphs, so that it now applies to any connected graph. Using color codes, we prove a new class of lower bounds depending on the existence of sets with common neighbours. Tight examples are discussed, involving the amalgamation of complete graphs, generalized wheel graphs, and a special class of sequential join of graphs.


Introduction
In 2008, Chartrand et al. introduced rainbow colorings, as a way to strengthen connectedness.A coloring on a graph G refers to any map γ : E(G) → {1, . . ., k}, which is also called edge-with color sequence c 1 , c 2 , • • • , c n−1 is said to be a rainbow path if all of its edges have different colors.A coloring is called a rainbow coloring if any two vertices can be connected by a rainbow path.A trivial way to produce a rainbow coloring on any connected graph is using |E(G)| colors to give each individual edge its own color.This may not be efficient, e.g. two colors are enough to rainbow-color C 4 (put 1 and 2 alternately).The smallest k for which there is a rainbow k-coloring on G is called the rainbow connection number of G, denoted by rc(G).
These notions can be developed further by considering the notion of a geodesic, which is a shortest path between two vertices x, y in a graph G.The distance d G (x, y) is defined as the length of any such shortest path.The diameter of G, denoted by diam(G), is defined as the largest distance between pairs of vertices of G.A coloring is called a strong rainbow coloring if any two vertices can be connected by a rainbow geodesic.The smallest k for which there is a strong rainbow k-coloring on G is called the strong rainbow connection number of G, denoted by src(G).
Chartrand et al. [1] noted the following chain, Some equality cases are known.The complete graphs are the only graphs whose rc and src are equal to 1, and trees are the only graphs whose rc and src are equal to the number of edges in those graphs (see [1]).However, it remains open to characterize when rc(G) = diam(G).
Li and Sun [5] tightened the upper bound to src(G) ≤ |E(G)| − 2t, where t is the number of edge-disjoint triangles.Schiermeyer [7] observed a different lower bound rc(G) ≥ n 1 (G), where n 1 is the number of vertices of degree one.The reader is referred to [6] for a more detailed survey.
In this paper, we prove some lower bounds based on the presence of sets with common neighbours.For a non-empty Q ⊆ V (G), its common neighborhood is denoted where β 0 is the vertex-independence number, and ω is the clique number.These parameters are described e.g. in [4].We also prove a version of (1.4) for multiple sets.In Section 2.2 we prove similar bounds for rc.In Section 2.3 we discuss some miscellaneous bounds that will be useful in our discussion of tight examples involving the amalgamation of complete graphs, generalized wheel graphs, and a class of sequential join.
We use color codes.This notion was used in [1] as a tool to study the rc and src of complete bipartite graphs.Now we adapt it to any connected graph.Given a coloring γ : E(G) → {1, . . ., k} (not necessarily rainbow) and a non-empty set Q ⊆ V (G) with non-empty common neighborhood CN (Q) = {t 1 , . . ., t b }, we define the color code of a vertex v ∈ Q as follows, (1.5) The tuple code(v) depends on the set Q that we consider v a member of, as illustrated in Figure 1.(1.6) Lemma 1.1.Let γ be a coloring on G, and Then there is a rainbow geodesic between two non-adjacent vertices in Q * if and only if their color codes are different.
So there is a rainbow v−w geodesic if and only if there is a t ∈ CN (Q) with γ(vt) = γ(wt).
A set is called co-neighboring if any two of its vertices have precisely the same (non-empty) neighborhood.An independent set has any two of its vertices non-adjacent.
By Lemma 1.1 there are no rainbow v−w geodesics.Let L : v−x− • • • −y−w be a rainbow path with x ∈ N (v) and y ∈ N (w).Then x, y ∈ CN (Q) and x = y (since L is not geodesic).So, the length of L is at least 2 + d G (x, y) ≥ 4 because x, y are non-adjacent.Lemma 1.3.Let γ be a coloring on G, and then there are non-adjacent vertices in Q * with the same color code.
, then |code(Q)|ω(Q * ) < |Q| so at least ω(Q * ) + 1 vertices in Q have the same code; if X is a set of such vertices, then some v, w ∈ X are non-adjacent in Q * .
Later we deal with multiple subsets.The problem is how to compare the codes in different subsets.Let us call two disjoint sets Q 1 , Q 2 ⊆ V (G) CN-bridged if for every v ∈ Q 1 and w ∈ Q 2 we have v and w non-adjacent in G, and any geodesic between them has the form v−x− • • • −y−w with x ∈ CN (Q 1 ) and y ∈ CN (Q 2 ).A diagonal tuple has the form (i, i, . . ., i).
for all i ∈ {1, . . ., p}, and let γ be a k-coloring on G. Suppose there is a natural number r that satisfies the following condition, . (1.8) Then one of the following must hold: (1) For some i ∈ {1, . . ., p}, there are non-adjacent vertices in Q * i with the same code.(2) For some i, j ∈ {1, . . ., p} with i = j, there is a diagonal tuple in code(Q i ) ∩ code(Q j ).
Proof.Suppose (1) fails to hold.Let A and B be the set of diagonal and non-diagonal b-tuples of numbers from {1, . . ., k}, respectively.Then |A| = k and |B| = k b − k.We need to show code(Q i ) ∩ code(Q j ) ∩ A = ∅ for some i = j.Assuming otherwise, for all i = j we have (1.9) Since (1) fails, we have |code where the rightmost inequality in (1.10) follows from the rightmost inequality in the hypothesis (1.8).Since p ≥ 2, we get r 2 ≤ r−1 p ≤ r−1 2 , a contradiction.

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Color code techniques in rainbow connection | F. Septyanto and K. A. Sugeng

Lower bounds for src
(2.1) . So Lemma 1.3 applies, and we get a contradiction with Lemma 1.1.
If we have several subsets Q 1 , . . ., Q p ⊆ V (G), then an application of Theorem 2.1 to each individual set gives p lower bounds, which can be averaged to The following is a better bound that incorporates all the subsets simultaneously, under the additional assumption that the sets are pairwise CN-bridged.Moreover, the bound can also make use of a previously known lower bound for src and possibly improve it to a sharper bound.
Theorem 2.2.Let G be a connected graph, p ≥ 2, and , so this geodesic is not rainbow.
Remark 2.1.Even if we start with the trivial estimate r = 1, this is already stronger than the average bound (2.2), due to Jensen's inequality for the concave function f (x) = b √ x on x ≥ 0 and the fact that 1 + ⌊x⌋ > x.

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Color code techniques in rainbow connection | F. Septyanto and K. A. Sugeng

Lower bounds for rc
We consider analogous version of the previous bounds for rainbow connection number.
Theorem 2.3.Let G be a connected graph and . (2.4) Theorem 2.4.Let G be a connected graph, p ≥ 2, and since Q i and Q j are co-neighboring sets.Since v, w are non-adjacent, the length of this path is at least two.But γ(vx) = u = γ(wy), so the path is not rainbow.

Miscellaneous Bounds
Now we prove some additional bounds that will be useful in our discussion in Section 3. We call G an s-strong graph if G is connected and every rainbow s-coloring on G is strong rainbow.For example, any connected graph is 1-strong, and any tree is s-strong for every s ∈ N. Later we need 2-strong and 3-strong graphs.
Theorem 2.6.Any connected graph is 2-strong.Therefore, for any connected graph G, with equality if and only if rc(G) ≤ 3.
Proof.Any path of length two between non-adjacent vertices must be a geodesic.So, any rainbow 2-coloring is strong rainbow.
Theorem 2.7.Any connected (C 3 , C 5 )-free graph is 3-strong.Therefore, if G is connected and with equality if and only if rc(G) ≤ 4.
Proof.Suppose there is a rainbow 3-coloring on G that is not strong rainbow.Let v, w ∈ V (G) be non-adjacent vertices without any rainbow geodesics.Let L be a rainbow v−w path.If the length of L is two or d G (v, w) = 3, then L will be a geodesic.So the length of L is three and

Tight Examples
Our examples involve the notion of graph join.Recall that the join of two graphs A and B is a new graph obtained from their disjoint union by adding a new edge between every vertex of A and every vertex of B. The resulting graph is denoted by A + B. For example, the complete bipartite graph K s,t is a join K s + K t of two edgeless graphs.
In the case of a graph join, the CN-graph construction becomes the well-known construction of graph square.Recall that the square graph of a graph A, denoted by A 2 , is a new graph with the same vertex-set as A, but with edge-set E(A 2 ) = {vw : 1 ≤ d A (v, w) ≤ 2}.

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Color code techniques in rainbow connection | F. Septyanto and K. A. Sugeng

Amalgamation of Complete Graphs
Our first example is one in which the β 0 lower bound in Theorem 2.1 is stronger than the ω lower bound.The amalgamation of (disjoint) complete graphs K m 1 , . . ., K mt , denoted by is a new graph obtained by choosing one vertex from each K m i and identifying those vertices as a single vertex (called the central vertex).The rainbow connection number of Amal(K m 1 , . . ., K mt ) when m 1 = • • • = m t ≥ 3 was studied by Fitriani and Salman [2].Now we settle the general case.
It remains to compute the rc.Since G is not a complete graph, rc(G) ≥ 2. If t = 2, then rc(G) ≤ src(G) = t = 2 and so rc(G) = 2.If t ≥ 3 and u = t, then G is a tree and rc(G) = |E(G)| = u = max{3, u}.Now let t ≥ 3 and u < t.By Theorem 2.6, rc(G) ≥ min{3, t} = 3.By Schiermeyer's lower bound, rc(G) ≥ n 1 (G) = u.So rc(G) ≥ max{3, u}.A rainbow max{3, u}-coloring γ on G can be produced as follows.First, give all u vertices of degree 1 in G different colors.For all i such that m i ≥ 3, put γ(e) = 3 for all e ∈ E(K m i −1 ), put color 1 on half the edges from K m i −1 to K 1 , and put color 2 on the remaining edges from K m i −1 to K 1 .This way, any two non-adjacent vertices in A can be connected by a rainbow path whose color sequence is i, j, or i, 1, 3, or i, 2, 3, or i, j where i, j ∈ {1, . . ., u} and j = i (see Figure 2 below).

Generalized Wheel Graphs
This is an example in which the ω lower bound in Theorem 2.1 is sharper than β 0 .The join of a cycle with any graph, i.e.C n + H, is called the generalized wheel graphs.This class of graph has been studied under various labelling schemes [3].Now we consider the rc and src.Theorem 3.2.Let n ≥ 3 and H be any graph.Then Proof.
(1) By Theorem 2.6 it is enough to prove rc(G) ≤ 3. A rainbow 3-coloring on G = C n +H can be produced as follows.Put the color 3 on all edges in C n .Let the cycle be v in this order.If i is odd, assign color 1 to all v i −H edges.If i is even, assign color 2 to all v i −H edges.In this way, any two vertices in H can be connected by a rainbow path with color sequence 1, 3, 2, and any two non-adjacent vertices in C n can be connected by a rainbow path with color sequence 1, 2 or 3, 1, 2. So this coloring is rainbow, and rc(G) ≤ 3. ( so that G is not a complete graph and src(G) ≥ 2. The following claim simplifies computation.
For the upper bound, we quote Theorem 2.3 in [8] stating that where i(A) is the independent domination number of A, which is the smallest cardinality of a set of independent (pairwise non-adjacent) vertices that are also dominating (i.e.any other vertex is adjacent to at least one of them).We apply this with A = C n and B = H.The following figure shows that i(C n ) ≤ k.

Sequential Join
This example shows that Theorem 2.2 can be strictly stronger than Theorem 2.1 alone.We will also see some constructive use of color codes.The sequential join of vertex-disjoint graphs G 1 , G 2 , . . ., G t is defined as the union [3]).We consider on a sequential join of the form mK 1 + bK 1 + bK 1 + mK 1 .When b = 1 this graph is a tree so its rc and src are already known.So, we assume b ≥ 2.
By construction, there is a rainbow path from x to y.
−y is rainbow.This completes the proof of (1).
To prove (2) and the remaining statements, we need the following claim.
We prove this by constructing a strong rainbow . First, we define γ as a coloring on H. Later, we will erase the new vertices and restrict γ to G m,b .
We begin by coloring the middle part, i.e. bK 1 + bK 1 whose src is 2. If d = 2, put a strong rainbow 2-coloring on this part with the colors c+1 and c+2.If d = 1, then we put γ(t i u j ) = c+1 instead for all i, j ∈ {1, 2}.Now we color the left wing.Including v 1 , choose any c b − c vertices in Q ′ 1 to form a set Q 11 .The edges adjacent to Q 11 are colored in such a way so that, with respect to {t 1 , . . ., t b }, the set code(Q 11 ) consists of all non-diagonal b-tuples with entries from {1, . . ., c}.If c ≥ b, we also put code(v 1 ) = (1, 2, 3, . . ., b).Analogously, we form Q 21 ⊆ Q ′ 2 and put the coloring in the same way.Next, for each i ∈ {1, 2}, choose any c 2 vertices from Q ′ i \Q i1 and let them form a set Q i2 .Put the coloring on edges adjacent to Q i2 so that code(Q 12 ) and code(Q 22 ) are disjoint and their union consists of all diagonal tuples with entries taken from {1, . . ., 2 c 2 } ⊆ {1, . . ., c}.Finally, for each i ∈ {1, 2}, let we are done.Otherwise, put the coloring on edges incident to Q i3 in a way so that code(Q i3 ) consists of permutations of (a, a, . . ., a, c+1, c+1, . . ., c+1), where a ∈ {1, . . ., c} with (a, a, . . ., a) ∈ code(Q i2 ) is repeated j times, for some j ∈ {1, . . ., b}.The number of such a tuple (a 1 , a 2 , . . ., a b ) is precisely Therefore, all vertices of Q i3 can be allocated such tuples.
After erasing all the new vertices, we end the definition of γ.Now we prove that γ is strong rainbow.Let x, y ∈ V (G m,b ) be non-adjacent.
For each i ∈ {1, 2}, all vertices in Q i have distinct codes.We are done by Lemma 1.1.
There is i ∈ {1, . . ., b} such that γ(xt i ) ≤ c.Then x−t i −y is a rainbow geodesic.In this subcase there is a rainbow geodesic between x and y in the middle part (bK 1 + bK 1 ).
In the remaining cases we consider x ∈ Q 1 and y ∈ Q 2 .
Case 4: x ∈ Q 11 and y ∈ Q 21 .If code(x) with respect to {t 1 , . . ., t b } is equal to code(y) with respect to {u 1 , . . ., u b }, choose i, j ∈ {1, . . ., b} with i = j and γ(xt i ) = γ(xt j ) = γ(yu j ).Then the geodesic x−t i −u j −y is rainbow.Now suppose that code(x) = code(y), say they differ at the i'th component.Then the geodesic x−t i −u i −y is rainbow.

Figure 1 .
Figure 1.If we consider a ∈ {a, d}, code(a) is a 3-tuple.It is a 2-tuple if we consider a ∈ {a, d, f }.To avoid ambiguity, we also refer to the tuple (γ(vt 1 ), γ(vt 2 ), • • • , γ(vt b )) as the code of v with respect to {t 1 , . . ., t b }.Let code(Q) = {code(v)|v ∈ Q}.Since every code is a b-tuple, we have |code(Q)| ≤ k b .(1.6) where k is the right hand side minus 1.Let γ be a strong rainbow k-coloring on G.Note that (1.8) holds, so one of the options (1) or (2) in Lemma 1.4 holds.If (1) holds, Lemma 1.1 is contradicted.So (2) holds.Let v ∈ Q i and w ∈ Q j have the same diagonal tuple as their code.By CN-bridging, any v−w geodesic has the form

. 5 )
Proof.Suppose rc(G) ≤ k, where is the right hand side minus 1.Let γ be a rainbow k-coloring on G.Note that (1.8) holds, so one of the options (1) or (2) in Lemma 1.4 holds.If (1) holds, Lemma 1.2 is contradicted because k ≤ 3.So (2) holds.Let v ∈ Q i and w ∈ Q j have the same diagonal tuple as their code, with i = j.In any path v−x− • • • −y−w, we have x ∈ N

Figure 3 .
Figure 3.The marked vertices form an independent dominating set of cardinality k.

Case 3 :v 1 j
x, y ∈ CN (Q 1 ) (or by symmetry x, y ∈ CN (Q 2 )).Say x = t i and y = t j with 1 ≤ i < j ≤ b.Subcase 3.1: m ≥ c b − c.In this subcase the set code(Q 11 ) contains all off-diagonal tuples with entries from {1, . . ., c}, so there is v ∈ Q 11 such that the i'th component of code(v) is different than the j'th component.Then x−v−y is a rainbow geodesic.Subcase 3.2: c ≥ b.In this subcase code(v 1 ) = (1, 2, . . ., b), so x i −−y is a rainbow geodesic.Subcase 3.3: d = 2.