A note on the rainbow connection of random regular graphs

We prove that a random 3-regular graph has rainbow connection number O(log n). This completes the remaining open case from Rainbow connection of random regular graphs, by Dudek, Frieze and Tsourakakis.


Introduction
A rainbow path in an edge-coloured graph is a path in which every edge has a different colour.The rainbow connection number of a graph G, denoted rc(G), is the minimum number of colours required to colour the edges of G such that every pair of vertices is connected by a rainbow path.(The colouring is not required to be proper, although the edge-colourings in this paper will be.)This was introduced by Chartrand et al. in [2]; see [5] for a survey and motivations.
G n,r is the random r-regular graph on n vertices, where every such graph is selected with equal probability.(We assume that rn is even.)The diameter of G n,r is approximately log r−1 n with high probability (w.h.p.)1 [1]; and clearly this is a lower bound on rc(G n,r ).Dudek, Frieze and Tsourakakis [3] proved that rc(G n,r ) = O(log n) for r 4. Kamcev, Krivelevich and Sudakov [4] provided a short elegant proof for r 5 and extended the result to expander graphs and to vertex colourings (for r 28).Here we briefly note that a small modification to the arguments in [3] proves their result for r = 3.
Let T 1 , T 2 be two copies of a binary tree of height , where the roots x 1 , x 2 have degree three.We allow an adversary to colour the edges of T 1 , T 2 so that both colourings are rainbow; i.e. each colour appears at most once in each tree.L i is the set of leaves of T i .We say a pair of leaves u 1 ∈ L 1 , u 2 ∈ L 2 is compatible if no colour appears both in the path from x 1 to u 1 and in the path from x 2 to u 2 .We define M to be the number of compatible pairs.Lemma 1.For any two rainbow edge colourings of 3 of the pairs of leaves are compatible.Lemma 2 of [3] proves that the same holds for d-ary trees for all d 3 (with a root of degree d), albeit with a different constant multiple.
The proof goes exactly as in [3].There they first obtain an edge-colouring such that for every vertex v, the set of edges within distance = K log log n of v is rainbow for a particular constant K. (That set of edges forms a tree for almost every v.)So consider any two vertices x 1 , x 2 , and let T 1 , T 2 be the trees formed by their distance neighbourhoods.Applying their Lemma 2 (the analogue of our Lemma 1) they prove that at least one of the compatible pairs of leaves is joined by a path which contains none of the colours joining either leaf to its root.For the sake of brevity, we refer the reader to [3] for all the details.
Proof of Lemma 1.For each colour c we let ρ(c) be the number of pairs where the sum is taken over all colours c appearing in both trees.For each such c, let λ i (c) denote the level of the edge in T i coloured c, where the edges adjacent to the leaves are at level 0 and those adjacent to the root are at level − 1.So ρ(c) = 2 λ 1 (c)+λ 2 (c) .Now c ρ(c) is clearly maximized when no colour appears in exactly one tree; so assume that each tree contains the colours c 1 , c 2 , . . ., c 3×2 −1 .Because the trees are isomorphic, the sequences (2 λ 1 (c 1 ) , 2 λ 1 (c 2 ) , . . . ) and (2 λ 2 (c 1 ) , 2 λ 2 (c 2 ) , . . . ) are both permutations of the same multiset.c ρ(c) is the sum of the products of the corresponding elements in those permutations, which is maximized when the permutations are identical.So noting that each tree has 3 × 2 −λ−1 edges at level λ, we obtain c ρ(c) This yields the lemma as Lemma 1 is tight.We will demonstrate this with a recursive edge-colouring of two trees of height .In our construction, s(u) denotes the sibling of u, i.e. the other vertex with the same parent as u (which is unique if u is distance at least 2 from the root).For each u ∈ T 1 , the corresponding vertex in T 2 (i.e.same level and same place in left-to-right order) is labelled u .
For = 1, we colour the edges, left-to-right, 1,2,3 for T 1 and 3, 1, 2 for T 2 .To extend from height to + 1, we use a new set of colours on the new edges of T 1 , and then use the same colours on T 2 , this time exchanging the colours of the edges of each pair of siblings; i.e. u, s(u ) are joined to their parents by edges of the same colour and so are s(u), u .
It is not hard to note first that (u, u ) is always compatible, and then that each leaf u ∈ L 1 lies in 2 −1 + 1 non-compatible pairs: 2 × (2 −2 + 1) arising from children of nodes who were not compatible with its parent in the previous colouring, plus s(u ).
Remark: Lemma 2 of [3] is stated for d-ary trees, d 3, where the root has degree d.Their lemma does not extend to binary trees, as they show with a counterexample (due to Alon) in their section 3.So the fact that the root has degree 3 in our Lemma 1 is crucial.Proving Lemma 1 when the root has degree 3 is sufficient for the remainder of the argument from [3] to apply.Indeed, [3] uses their Lemma 2 to derive Corollary 4 which applies to d-ary trees with a root of degree d + 1, and then use Corollary 4 for the remainder of the paper.There is one exception -they use Lemma 3 (which is stated only for trees where the root has degree d) in section 2.6.3, but Corollary 4 applies there just as well.