Some Remarks on Rainbow Connectivity

0.11 MB·English

by  Nina Kamčev

Some Remarks on Rainbow Connectivity
NinaKamcˇev ∗ MichaelKrivelevich† BennySudakov‡
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Abstract

p
e An edge (vertex) coloured graph is rainbow-connected if there is a rainbow path between any two
S vertices, i.e. a path all of whose edges (internal vertices) carry distinct colours. Rainbow edge (vertex)
0 connectivityofagraphGisthesmallestnumberofcoloursneededforarainbowedge(vertex)colouring
1 ofG. Inthispaperweproposeaverysimpleapproachtostudyingrainbowconnectivityingraphs. Using
thisidea,wegiveaunifiedproofofseveralknownresults,aswellassomenewones.
]
O
C
1 Introduction
.
h
t
a Anedgecolouring ofagraphG israinbow ifthereisarainbow pathbetweenanytwovertices, thatisapath
m
onwhichalledges havedistinct colours. Anyconnected graphG oforder ncanbemaderainbow-connected
[ using n−1colours bychoosing aspanning tree andgiving each edgeofthespanning tree adifferent colour.

2 Hencewecandefinetherainbowconnectivity, rc(G),astheminimalnumberofcoloursneededforarainbow
v
colouring ofG.
1
2 Rainbow connectivity is introduced in 2008 by Chartrand et al. [9] as a way of strengthening the notion
8
of connectivity, see for example [6], [8], [10], [13], [17], and the survey [18]. The concept has attracted a
0
0 considerable amount of attention in recent years. It is also of interest in applied settings, such as securing
. sensitiveinformation transferandnetworking. Forinstance, [7]describe thefollowingsettinginnetworking:
1
0 we want to route messages in a cellular network such that each link on the route between two vertices is
5
assigned with a distinct channel. Then, the minimum number of channels to use is equal to the rainbow
1
: connectivity oftheunderlying network.
v
i We are interested in upper bounds for rainbow connectivity, first studied by Caro et al. [6]. The trivial
X
lower bound is rc(G) ≥ diam(G), and it turns out that for many classes of graphs, this is a reasonable guess
r
a forthevalueofrainbowconnectivity. Caroetal. [6]showedthataconnected graphofordernandminimum
degreeδ ≥ 3hasrainbowconnectivity atmost 5n. Sincethediameterofsuchagraphisatmost 3n (see,e.g.,
6 δ+1
[11]), itis natural toask whether the rainbow connectivity ofG is ofthe same order. Krivelevich and Yuster
[17]showedthatindeedrc(G) ≤ 20n. ThenChandranetal.[8]settledthisquestionbyprovingrc(G) ≤ 3n +3,
δ δ+1
whichisasymptotically tight.
Arandomr-regulargraphofordernisagraphsampledfromG ,whichdenotestheuniformprobability
n,r
spaceofallr-regulargraphsonnlabelledvertices. Thesegraphswereextensivelystudiedinthelast30years,
see,e.g.,[20]. InthispaperweconsiderG forrconstant andn → ∞. Wesaythataneventholdswithhigh
n,r
∗DepartmentofMathematics,ETH,8092Zurich.Email:[email protected].
†School of Mathematical Sciences, Raymond and Beverly Sackler Faculty of Exact Sciences, Tel Aviv University, Tel Aviv
6997801,Israel.Email:[e-mail]yaUSA-IsraelBSFgrantandbyagrantfromtheIsrael
ScienceFoundation.
‡DepartmentofMathematics,ETH,8092Zurich.Email:[email protected]. ResearchsupportedinpartbySNSF
grant200021-149111.
1
probability (whp)ifitsprobability tends to1asntends toinfinity, butonly overthevalues ofnforwhichnr
iseven(sothatG isnon-empty).
n,r
A random r-regular graph has quite strong connectivity properties, for example, the diameter of G is
n,r
logn
whp asymptotic to , see [5]. The natural question of rainbow connectivity of random regular graphs
log(r−1)
wasfirststudied byFriezeandTsourakakis [13],whoshowedthatwhprc G = O logφrn ,foraconstant
n,r
φr > 2. Dudek etal. [10]improved this bound torc Gn,r = O(logn)whp(cid:0), wh(cid:1)ichis(cid:16)the corr(cid:17)ect dependence
onn. Wewillreturntothisresultlater. (cid:0) (cid:1)
Theaimofthisnoteistopresentasimpleapproachwhichimmediatelyimpliesresultsonrainbowcolour-
ing of several classes of graphs. It provides a unified approach to various settings, yields new theorems,
strengthens someoftheearlierresultsandsimplifiestheproofs. Itisbasedonedge-andvertex-splitting.
The main idea of the edge-splitting lemma is simple: we decompose G into two edge-disjoint spanning
trees T and T with a common root vertex and small diameters. We use different palettes for edges of T
1 2 1
and T , ensuring that each tree contains arainbow path from any vertex to the root. Hence if wecan get the
2
diameters of T and T ‘close’ to the diameter of G (say within a constant factor), then we have obtained a
1 2
strongresult.
Weexhibitafewapplications ofthelemma. Firstweuseittogiveastraightforward proofoftheresultof
KrivelevichandYuster[17],thatis
16n
Theorem1.1. Foraconnected n-vertex graphGofminimumdegreeδ≥ 4, rc(G) ≤ .
δ
Next we turn to random regular graphs. The rainbow colouring of G of Dudek et al. [10] typically
n,r
usesΩ(rlogn)colours, whichforlargerissignificantly biggerthanthediameterofG . Usingoursplitting
n,r
lemmawecanimproveittoanasymptotically tightbound.
clogn
Theorem1.2. Thereisanabsolute constant csuchthatforr ≥ 5,rc(G ) ≤ whp.
n,r
logr
For r ≥ 6, the theorem is an immediate consequence of the contiguity of different models of random
regular graphs. With little extra work, our approach also works for 5-regular graphs. Wewould like topoint
outthattheproofofDudeketal. worksstartingfromr = 4.
ThequestionofwhichcharacteristicsofG ensuresmallrainbowconnectivityarisesnaturally. Recalling
n,r
thatexpandergraphsalsohavediameterlogarithmicinn,itmakessensetolookatexpansionproperties. The
followingtheorem canbeviewedasageneralisation ofthepreviousresultonG .
n,r
Theorem 1.3. Let ǫ > 0. Let G be a graph of order n and degree r whose edge expansion is at least ǫr.
Furthermore, assumethatr ≥ max 64ǫ−1log 64ǫ−1 , 324 . Thenrc(G) = O ǫ−1logn .
n (cid:16) (cid:17) o (cid:16) (cid:17)
Inparticular, thistheorem applies to(n, r, λ)-graphs withλ ≤ r(1−2ǫ), i.e. n-vertex r-regular graphs whose
alleigenvalues exceptthelargest oneareatmostλinabsolutevalue.
Krivelevich and Yuster [17] have introduced the corresponding concept of rainbow vertex connectivity
rvc(G), the minimal number of colours needed for a rainbow colouring of vertices of G. The only point to
clarify is that a path is said to be rainbow if its internal vertices carry distinct colours. The easy bounds
diam(G) − 1 ≤ rvc(G) ≤ n also hold in this setting. Krivelevich and Yuster have demonstrated that it is
impossible to bound the rainbow connectivity of G in terms of its vertex rainbow connectivity, or the other
wayaround. Theyalsoboundrvc(G)intermsoftheminimaldegree.
Ourapproach essentially works for vertex colouring as well. InSection 3wepresent the vertex-splitting
lemma. Itisthenusedtoprovethevertex-colouring analogue ofTheorem1.2onrandom regulargraphs.
2
clogn
Theorem1.4. Thereisanabsolute constant csuchthatwhprvc(G )≤ forallr ≥ 28.
n,r
logr
2 Edge rainbow connectivity
2.1 The edge-splitting lemma
We state and prove the main lemma. The rest of the section uses the same notation for spanning subgraphs
G andG .
1 2
Lemma2.1. LetG = (V, E)beagraph. SupposeGhastwoconnectedspanningsubgraphsG = (V,E )and
1 1
G = (V, E )suchthat|E ∩E |≤ c. Thenrc(G) ≤ diam(G )+diam(G )+c.
2 2 1 2 1 2
Proof. LetB= E ∩E . Colourtheedgesof Bindistinctcolours. Thesecolourswillremainunchanged, and
1 2
the remaining edges get coloured according to graph distances inG andG , denoted by d and d . Choose
1 2 1 2
an arbitrary v ∈ V and define distance sets U = {u ∈ V : d (v,u) = j}andW = {u ∈ V : d (v,u) = j}. For
j 1 j 2
1 ≤ j ≤ diam(G ),colour theedgesbetweenU andU withcolour a . Similarly, usinganewpalette (b ),
1 j−1 j j j
colourtheedgesbetweenW andW withcolourb foreach1 ≤ j ≤ diam(G ). Thecolouring indeeduses
j−1 j j 2
atmost diam(G )+diam(G )+c colours.
1 2
To see that it is a rainbow colouring, look at two vertices x and x in V. Let P be a shortest path inG
1 2 i i
from x to v. By our definition of colouring on distance sets, both paths P and P are rainbow. If they are
i 1 2
edge-disjoint, theconcatenation isarainbowpathbetween x and x . Otherwise,P andP canonlyintersect
1 2 1 2
in edges of B. If this occurs, we walk from x along P to the earliest common edge. We use this edge to
1 1
switchto P andwalkto x . (cid:3)
2 2
2.2 Rainbow connectivity andminimum degree
Inthissetting,thebestpossibleresulthasbeenshownbyChandranetal[8]. Namely,aconnectedgraphGof
ordernandminimumdegreeδsatisfiesrc(G) ≤ 3n +3.Weshowhowthesplitting lemmacanbeusedwith
δ+1
basicgraphtheorytoobtainagoodupperbound, rc(G) ≤ 16n.
δ
ProofofTheorem1.1. Let G = (V,E) be as in the statement. We split G into two spanning subgraphs of
minimumdegreeatleast δ−1. FirstassumethatallverticesofGhaveevendegree. Then,usingconnectedness
2
ofG,orderitsedgesalonganEuleriancyclee , e ... e ,anddefine
1 2 m
F = {e : j∈ [m]even} and F = {e : j∈ [m]odd}.
1 j 2 j
Edges around each vertex are coupled into adjacent pairs ejej+1, so this is indeed a balanced split. Let
H = (V,F)betheassociated graphs.
i i
Toapplythissplitting togeneralG,notethatthenumberofvertices ofodddegreeiseven,sowecanadd
amatching M betweenthosevertices. EvenifG′ = (V,E∪ M)contains double edges, itstillhasanEulerian
cycle. Weapply the above procedure toG′, and then remove the auxiliary edges M. The end result is that a
vertexofodddegreed inG hasdegree d±1 in H,soindeed subgraphs H haveminimumdegreeatleast δ−1.
2 i i 2
Thegraph H maynotbeconnected. Butsince theminimum degreeofthisgraphis δ−1,eachconnected
1 2
component has order at least δ. Hence the number of components of H is at most 2n, so we can add a
2 1 δ
set B ⊂ E such that G = (V,F ∪ B )isconnected, and|B | ≤ 2n. We define the set B analogously. An
1 1 1 1 1 δ 2
elementary graph-theoretic result (mentioned in the introduction, see also [11]) shows that subgraphs G
1
3
and G of G have diameters at most 3n ≤ 6n. Applying the edge-splitting lemma to G and G gives
2 δ−1+1 δ 1 2
2
rc(G) ≤ 6n + 6n + 4n ≤ 16n. (cid:3)
δ δ δ δ
2.3 Expanders
Weadoptaweakdefinition ofanexpander. Asbefore,G = (V, E),thedegreer isfixedandtheorderntends
toinfinity. ForS ⊂ V,wedefineout(S)tobethesetofedgeswithexactlyoneendpoint inS . AgraphGhas
edgeexpansion ΦifeverysetS ⊂ V with|S| ≤ n satisfies|out(S)| ≥ Φ|S|.
2
FriezeandMolloy[12]haveshownusingtheLova´szLocalLemmathatthenaturalrandomk-splitting of
E giveskexpandergraphswithpositiveprobability. Westatetheirtheorem fork = 2.
Theorem 2.2. Letr beanatural number, λ > 0arealnumber, andG = (V,E)anr-regular graph withedge
expansion Φ. Suppose
Φ r
≥ 8λ−2 and ≥ 14λ−2.
logr logr
Then there is a partition E = E ∪ E such that both subgraphs G = (V,E) have edge expansion at least
1 2 i i
Φ
(1−λ) .
2
Under stronger conditions on expansion, they also give a randomised polynomial-time algorithm for the
splitting, whichimmediately givesarainbowcolouring.
ProofofTheorem1.3. LetG beanr-regular graph withedgeexpansion ǫr. Wewillapply Theorem2.2with
λ = 1. The hypothesis r ≥ 64ǫ−1log 64ǫ−1 ensures that ǫr ≥ 32, and the second inequality follows from
2 logr
r ≥ 324. We get a partition E = E ∪(cid:16) E su(cid:17)ch that each graph G = (V,E) has edge expansion at least ǫr.
1 2 i i 4
The maximum degree ofG is at most r, so every set S of order |S| ≤ n has a neighbourhood Γ(S) of order
i 2
|Γ(S)| ≥ 1+ ǫ |S|. Thus the number of vertices within distance at most l from any vertex in G is at least
4 i
min (1+(cid:16)ǫ/4)l,(cid:17)n/2 andtherefore diam(G )= O ǫ−1logn .
i
AnpplyingLemmoa2.1givesrc(G) ≤ diam(G )(cid:16)+diam(G(cid:17) ) = O ǫ−1logn . (cid:3)
1 2
(cid:16) (cid:17)
2.4 Random regular graphs
TwosequencesofprobabilityspacesF andG onthesameunderlyingmeasurablespacesarecalledcontigu-
n n
ous,writtenF ≈ G ,ifasequence ofevents(A )occurs whpinF ifandonlyifitoccurswhpinG . LetG
n n n n n
andG′ betwomodelsofrandom graphsonthesamevertexset. Wegetanewrandom graphG bytaking the
unionofindependently chosengraphsG ∈ GandG ∈ G′,conditional ontheeventE(G )∩E(G ) = ∅. The
1 2 1 2
probability spaceofsuchdisjoint unionsisdenoted byG⊕G′.
It is known that G is contiguous with any other model which builds an r-regular graph as an edge-
n,r
disjoint union of random regular graphs and Hamiltonian cycles. Thisgoes back tothe workof Janson [14],
Robinson and Wormald [19], and is also laid out in the survey [20]. The specific results we use in proving
Theorem 1.2areGn,r+r′ ≈ Gn,r ⊕Gn,r′ andGn,r+2 ≈ Gn,r ⊕Hn,where Hn isarandom Hamiltonian cycleon
[n]. RecallthatTheorem1.2saysthatforr ≥ 5, rc(G ) ≤ clogn whp.
n,r logr
ProofofTheorem1.2forr ≥ 6. As usually, we assume that rn is even, and define r so that G are non-
i n,ri
empty for i = 1, 2. If r is odd, then n is even and we can set r = r±1. Otherwise, we set r = r = r or
i 2 1 2 2
r = r ±1asappropriate. Theobservation attheendoftheproofresolvesthecaser = 6.
i 2
4
LetG bea random r-regular graph, r ≥ 3. Then withhigh probability diam(G ) ≤ (1+o(1))logn ≤ clogn,
i i i i log(ri−1) 2logr
wherecisasuitable constant. LetG betheunion oftwosuch edge-disjoint graphsG andG . Thesplitting
1 2
lemmagivesrc(G) ≤ clogn.
logr
SinceGwasarandomelementofG ⊕G ,therandomr-regular graphhasthesamepropertywhp.
n,r1 n,r2
For r = 6 and odd n, we take G to be sampled from H ⊕ H ⊕ H . The first two Hamiltonian cycles
n n n
belongtoG ,resp. G . WesplittheedgesofthethirdHamiltoniancycleH alternately, sothat n−1 edgesare
1 2 n 2
assignedtoG and n+1 toG . ThenwecanquoteProposition 2.4,aresultofBolloba´sandChungwhichsays
1 2 2
thattheunionofaHamiltonian cycleandarandomperfectmatchinghaswhplogarithmicdiameter[4]. (cid:3)
The remainder of the section deals with the case r = 5. Since G ≈ G ⊕ H ⊕ H , we can model
n,5 n,1 n n
our 5-regular graph as aunion of two random graphsG andG , where eachG is an edge-disjoint union of
1 2 i
a Hamiltonian cycle and a matching of size n . The following theorem says that whp eachG has diameter
4 i
O(logn),sorc(G) = O(logn)whpfollowsfrjomk thesplitting lemma.
Theorem2.3. LetGbearandomgraphon[n],theunionofthecycle(1, 2,...,n, 1)andarandommatching
on[n]consisting of n edges. ThenGhasdiameterO(logn)whp.
4
j k
Thetheorem canbeprovedbyadapting theargument ofKrivelevich etal. [16],whoshowedthatstarting
fromaconnectedn-vertexgraphC andinaddition, turningeachpairofverticesintoanedgewithprobability
ǫ, the resulting graph typically has logarithmic diameter. This is very similar to what we need when C is
n
a Hamiltonian cycle. However, since we are adding a random matching rather than independent edges, our
modelisslightlydifferent. Insteadofreprovingtheresultof[16]inoursetting,wedecidedtogiveadifferent
(very short) proof relying on the following result (see [20]), which by contiguity simply says that G has
n,3
logarithmic diameter whp. Without assuming that the cycle and matching are edge disjoint this was proved
earlierbyBolloba´sandChung[4].
Proposition 2.4. Let H be a graph formed by taking a disjoint union of a random matching of size n and
2
ann-cycle. ThenthediameterofH iswhp(1+o(1))log n. j k
2
Denote m = n . Note that G in Theorem 2.3 can be built in two steps as follows. First we select a
4
random subset B j= k{b , b , ...,b } ⊂ [n] of order 2m, and then independently a random perfect matching
1 2 2m
on {b , b ,...,b }. Throughout the proof we identify the vertices of G with natural numbers up to n and
1 2 2m
assumeb < b < ···< b .
1 2 2m
Given a subset B, define variables Yi = bi+1 −bi for i = 1,...2m−1. Moreover, we define Y0 = b1 and
Y = n−b to record the positions of the first and the last vertex in B. An important observation is that
2m 2m
a random set B of order 2m induces a random sequence (Y , Y ,...,Y ) with Y ≥ 1 for i < 2m, Y ≥ 0
0 1 2m i 2m
and 2m Y = n and, vice versa, given such arandom sequence, wecan uniquely reproduce acorresponding
i=0 i
set BP,which isuniformly distributed over allsubsets of[n] oforder 2m. Tocomplete theproof, weneed the
followingsimplelemmaabout(Y).
i
Lemma2.5. Let(Y , Y ,...,Y )bearandom sequence asdefined above. Fixasetofindices 0 ≤ i < i <
0 1 2m 1 2
··· < i < 2m. ThenP Y > logn = o(1)and
s 2m
(cid:2) (cid:3)
s
P Y > 10s ≤ e−2s.
 ij 
Xj=1 
5
ProofofLemma2.5. SincepermutingthevariablesY,i< 2m,doesnotchangetheprobabilityspace,without
i
lossofgenerality wemayassume(i1, ...is) = (0,..., s−1). RecallthatYi weredefinedbyYi = bi+1−bi,so
that s−1Y > 10smeansexactlythatthereareatmost s−1verticesofBamongthefirst10svertices. Onthe
i=0 i
otherPhand, |B∩[10s]| is a hypergeometric random variable with mean 2m ·10s. Therefore, by the standard
n
tailbounds(see,e.g.,Theorem2.10in[15]).
Ps−1Yi > 10s = P[|B∩[10s]| ≤ s−1] ≤ e−2(20nm10−s1)2s2 ≤ e−2s.
Xi=0 
Similarly, Y > lognmeansthatnovertexof Bisintheinterval [n−logn,n]. Theprobability ofthis event
2m
is n−logn / n = o(1). (cid:3)
2m 2m
(cid:16) (cid:17) (cid:16) (cid:17)
ProofofTheorem2.3. Asweexplained,ourGcanbeconstructedasfollows. Startwithacycleb b ...b b .
1 2 2m 1
Pickarandomperfectmatching MonB= {b , b ,...b }whoseedgesdonotcoincidewithanyedgesofthe
1 2 2m
cycle. LetH = H(M)bethegraph on Bformedastheunionofthecycleb b ...b b andthematching M.
1 2 2m 1
Choose a random sequence (Y , Y ,...,Y ) as above. The graph G on [n] is obtained by subdividing each
0 1 2m
edge bibi+1 into Yi edges. The exception is the edge b2mb1, which is subdivided into Y2m +Y0 edges. Note
that M and(Y)arechosen independently. Since M israndom, byProposition 2.4whp H(M)hasdiameterat
i
most (1+o(1))log (2m) ≤ 1.5logn−1. Condition on this event, and fix an arbitrary M which satisfies the
2
condition.
We will show that for random (Y), whp G will have small diameter. We further condition on the event
i
that Y ≤ logn, which by the previous lemma holds whp. Let s = 1.5logn. Take the vertices u and v in
2m
[n], and single out the segments to which they belong, bi ≤ u < bi+1 and bj ≤ v < bj+1 (i and jare possibly
0 or 2m−1). H contains a path P between b and b of length at most s− 1, which we turn into a path in
i j
G as follows. If an edge on P belongs to the matching M, then it is also an edge of G. Otherwise, if the
edge has form bkbk+1, we replace it by the segment bk, bk +1, bk +2,...,bk+1 in G, whose length is Yk. If
Pcontains theedge b b ,thecorresponding segmenthaslength Y +Y . Attheends ofthepath, wewalk
2m 1 2m 0
from uto b and from b tov. Denote byU theset ofindices k < 2m such that Pcontains avertex b . Since
i j k
Y ≥ 1 for i < 2m, the distance between u and v in G is at most Y +1+ max{1,Y } < s+ Y .
i 2m k∈U k k∈U k
Note also that |U| = |P|+ 1 ≤ s and that P, U do not depend on variablesP(Yk). Thus, by LemmaP2.5, the
probability that this distance exceeds 11s is at most e−2s = n−3. Taking the union bound over all pairs of
vertices, P[diam(G) > 11s | M] = O n−1 . Since weconditioned on the event with probability 1−o(1), the
probability thatdiam(G) > 11sisatm(cid:16)ost(cid:17)o(1),completing theproof. (cid:3)
3 Vertex rainbow connectivity
Wenowstatethevertex-colouring analogue ofLemma2.1.
Lemma3.1. LetG = (V,E)beagraph. Suppose that V ,V ⊂ V satisfy: 1) V ∪V = V;2) |V ∩V | ≤ c;
1 2 1 2 1 2
3)everyvertexv∈ V hasaneighbour inV andviceversa;4)G[V]isconnected, fori = 1,2. Then
1 2 i
rvc(G) ≤ diam(G[V ])+diam(G[V ])+c+2.
1 2
Proof. Let B = V ∩V . Colour the vertices of B in distinct colours. These colours will remain unchanged,
1 2
and the remaining vertices get coloured according to graph distances d inG = G[V]. Choose root vertices
i i i
6
v ∈ V such that v v is an edge of G. Give each distance set {u ∈ V : d (v ,u) = j} the colour a , for
i i 1 2 1 1 1 j
0 ≤ j≤ diam(G ). Similarly,eachset{u∈ V : d (v ,u) = j}getscolourb .
1 2 2 2 j
Toseethatitisarainbowvertexcolouring, lookattwovertices x ∈ V and x inV. Supposefirstthat x
1 1 2 2
liesinV ,andletP beashortestpathinG from x tov. Byourdefinitionofcolouringondistancesets,both
2 i i i i
paths P and P are rainbow. If they are vertex-disjoint, the concatenation P −v v −P is a rainbow path
1 2 1 1 2 2
between x and x . Otherwise, P and P canonly intersect invertices of B. Ifthisoccurs, wewalkfrom x
1 2 1 2 1
along P totheearliestcommonvertex. WeusethisvertextoswitchtoP andwalkto x .
1 2 2
If x does not lie in V , we replace it with its neighbour in V , which exists by hypothesis, and then
2 2 2
proceedwiththeargument. Thecasewhere x ,x < V istreatedsimilarly. (cid:3)
1 2 1
3.1 Random regular graphs
Lemma3.2. LetG beanr-regular graph,r ≥ 28. ThentheverticesofG canbepartitioned asV = U ∪U
1 2
sothateachv∈ V hasatleast 0.11r neighbours inbothU andU .
1 2
Proof. Thisisastandard application oftheLova´sz LocalLemma. Denote γ = 0.11forthe restofthepaper.
Foreachvertexv,putitintoU randomly andindependently withprobability 1/2. Let E betheeventthatv
1 v
does notsatisfy thestatement ofthelemma. Bythestandard Chernoff bounds the probability ofthiseventis
at most2e−2(21−γ)2r. Twoevents Ev and Eu are adjacent in thedependency graph ifu and vare at distance at
most 2from each other,and otherwise they are independent. Hence, each event has degree at most∆ = r2 in
thedependency graph. Thenforγ = 0.11andr ≥ 28,thecondition
(∆+1)eP[Ev]≤ (r2+1)·2e1−2(12−γ)2r < 1,
issatisfied. Therefore, bytheLocalLemma,withpositiveprobability noevent E occurs. (cid:3)
v
To use such a partition, we need an estimate on the number of edges spanned by subsets of vertices of
G . Similar results have appeared e.g. in [2], but for our purposes weneed a more explicit dependence on
n,r
thedegreer. Toprovetheestimate,weworkinthepairing(configuration) modelforr-regulargraphs. Forrn
even, wetake aset of rn points partitioned into n cells v , v , ...v , each cell containing r points. A perfect
1 2 n
matching (or pairing) P induces amultigraphG(P)inwhich the cells are regarded as vertices and pairs in P
as edges. For fixed degree r and P chosen uniformly from the set of pairings P , G(P) is a simple graph
n,r
with probability bounded away from zero, and each simple graph occurs with equal probability. It is known
(see,e.g.,[20])thatifaneventholdswhpinG(P),thenholdsitholdswhpevenonthecondition thatG(P)is
asimplegraph, andtherefore itholdswhpinG .
n,r
Lemma3.3. Letr ≥ 3beafixed integer. Let Pbeapairing selected uniformly from P . If E ⊂ [n](2) isa
n,r 0
fixedsetofm≤ nr pairsofverticesfromn,then
4
m
2r
P[E ⊂ E(G(P))] ≤ 2 .
0
n !
Proof. The total number of pairings P is (nr)! . In order to bound from above the number of pairings P
(nr)!2n2r
inducing E , first for each edge e = (u,v) ∈2E , choose a point in the cell of u and a point in the cell of v in
0 0
at most r2 ways, the total number of such choices is then at most r2m. Theremaining rn−2m points can be
pairedin (nr−2m)! ways. Altogether, usingStirling’sformula,theprobability ofgetting E isatmost
(nr−m)!2n2r−m 0
2
7
(nr−2m)! nr !
P[E ⊂ E(G(P))] ≤ r2m · 2
0 (cid:16) (cid:17)
(nr)! nr −m !2−m
2
nr−2m nr nr(cid:16)−2m −(cid:17)2m nr nr nr−2m m
=(1+o(1))r2m · 2
nr ! e ! (cid:18)nr−2m(cid:19) e !
2m n2r er2 m r m 2r m
=(1+o(1)) 1− ≤ 2 ≤ 2 .
nr ! nr−2m! n− 2m n !
 r 
Hereweusedthatsince1− x≤ e−x,then(1− 2m)n2r ≤ e−m andthat 2m ≤ n. (cid:3)
nr r 2
Lemma3.4. LetPbearandomelementofP ,andG(P)bethecorresponding r-regularmultigraphon[n].
n,r
WeobtainitsmaximalsimplesubgraphG(P)bydeletingtheloopsandidentifyingtheparalleledgesofG(P).
(i) Assumethatγ′r ≥ 3. Thenthereisanabsolute constant α > 0such thatwhpallvertex setsS ⊂ [n]of
e
orderuptoαnspanfewerthan |S|γ′r edgesinG(P).
2
(ii) There isanabsolute constant β > 0such that whpall vertex sets S ⊂ [n]of order upto βn span fewer
e r
than3|S|edgesinG(P).
Proof. Denote theevent tehatG(P)[S]contains atleast |S|d edges by B . Fixtheorder |S| = s. SinceG(P)is
2 S
asubgraph ofG(P),wecanapplytheprevious lemmatoeachsubset E ⊂ S(2) of sd edgestoget
e 0 2 e
s2/2 2r sd/2 2ser sd/2
P[B ] ≤ 2 ≤ 2 .
S
sd/2! n ! nd !
Takingtheunionboundoverallsetsofvertices oforder sgives
d s
n ne s 2er 2
P B ≤ P[B ]≤ 2 · .
 S s! S  s n d ! 
S∈_[n](s)   
For(i)setd = γ′r ≥ 3andchooseαsothattheterminsquarebracketsislessthan 1 for s= αn(notethat
2
thistermisincreasing in s). Wesplittherangeof sintoγ′r ≤ s≤ n14 andn14 < s≤ αntoget
P BS ≤ n14 ·O n−38 + 2−s+1 = o(1),
_S  (cid:18) (cid:19) sX≥n41
asrequired.
For (ii), set d = 6. Take β such that s = β again makes the term in brackets at most 1. The same
n r 2
calculation givestheresult.
(cid:3)
Fromthediscussion above,conditional ontheeventthatG(P)isasimplegraph(whichisexactlyG(P)=
G(P)), G(P) satisfies the statement of Lemma 3.4. Therefore the same holds for the random regular graph
e
G . Wecannowprovethemainresultofthissection, rvc(G )= O logn whpforr ≥ 28.
n,r n,r logr
(cid:16) (cid:17)
8
ProofofTheorem1.4. Let G be a random r-regular graph, γ = 0.11. Use Lemma 3.2 to obtain a partition
V = U ∪U suchthateachv∈ V hasatleastγr neighbours ineachpart.
1 2
All statements about G from now on will hold with high probability. In particular, we assume that G
satisfiesLemma3.4withγ′ = γ ,whereǫ = 0.02ischosensothat γr > 3. Weonlyneedtheextra(1+ǫ)−1
1+ǫ 1+ǫ
factor later, forClaim3. Suchedge distribution implies thateach connected component ofG[U ]contains at
i
leastαnvertices, whereαistheconstant fromLemma3.4.
Claim1. WecanfindW ⊂ V suchthatW = O(1)andG[U ∪W]isconnected.
i i i i
Forasetofvertices A ⊂ V,denoteΓj(A) = {v ∈ V : d (v,A) ≤ j}. Itiswell-knownthatarandom regular
G
graph hasgood expansion properties (see[3]),i.e. thereisaconstant φ > 0such thatwhp|Γ(A)| ≥ (1+φ)|A|
whenever|A| ≤ n. NowsupposethatAhaslinearorder,|A| ≥ αn,andtakeanintegerl > logα−1−log2. Iterating
2 log(1+φ)
the expansion property gives that |Γl(A)| > n. To prove Claim 1, suppose A and B are vertex sets of two
2
connected components of G[U], each of order at least αn. We just showed that Γl(A)∩Γl(B) , ∅, so there
i
isapath of length at most2l from Ato BinG. Adding thevertices of this path toW reduces the number of
i
connected components byone, so repeating this step α−1 timesensures that V = U ∪W spans aconnected
i i i
graphG = G[V]. Choose alarge integer asuch that|W| ≤ aforall nandr. Thevertex sets V andV now
i i i 1 2
satisfy|V ∩V |≤ 2a,soweturntothediametersofG andG .
1 2 1 2
Claim2. Forr ≥ 112(sothatγr ≥ 12),everyT ⊂ V oforderatmost βn satisfies|Γ (T)| ≥ 1+ γr |T|.
i γr2 Gi 12
SupposeT doesnotsatisfytheclaim,andletS = Γ (T). SincealltheedgesinG withanen(cid:16)dpoint(cid:17)inT
Gi i
lieinG[S],wegetthatS spansatleast
i
γr|T| γr|S| 3γr|S|
≥ ≥ = 3|S|
2 2 1+ γr γr
12
(cid:16) (cid:17)
edges. Notethat bythehypothesis |S| ≤ 1+ γr · βn < βn. Hencewecandeduce from Lemma3.4(ii) that
12 γr2 r
S spansfewerthan3|S|edges, whichisa(cid:16)contrad(cid:17)iction.
Claim 3. Letα bethe constant from Lemma3.4(i)and ǫ > 0asabove. Everysubset T ⊂ V oforder at
i
most αn satisfies|Γ (T)| ≥ (1+ǫ)|T|.
1+ǫ Gi
AssumethatT doesnotexpand, anduseLemma3.4forS = Γ (T),γ′ = γ > 3. Sincealltheedgesof
Gi 1+ǫ r
G withanendpoint inT lieinG [S],wegetthatS spansatleast
i i
γr|T| γr|S| γ′r|S|
≥ =
2 2(1+ǫ) 2
edges. Thiscontradicts statement(i)ofLemma3.4.
For r ≥ 112, Claim 2 implies that starting from any vertex v ∈ V, wecan expand inG to a set of order
i i
βn in c1logn steps, where c is a constant independent of r and n. Further O(logr) steps give a set of order
γr2 logr 1
αn ,byClaim3. Forr < 112, weusedirectly Claim3O(logn)times(thus avoiding Claim 2)toexpand toa
1+ǫ
setoforder αn . Inthisrange, logr < log112andhenceO(logn) = O logn .
1+ǫ logr
Denote k = clogn, where c > c is sufficiently large for the describ(cid:16)ed ex(cid:17)pansion to go through. Suppose
logr 1
the diameter of G is larger than 4k, and take x and x such that the shortest path x x ...x is longer
i α 0 R 0 1 R
than 4k (such a path exists since G is connected). Then we can use the procedure above to expand from
α i
vertices x , x , x ... inkstepstoget 4 disjoint (bythechoiceofthepath)neighbourhoods, eachoforder
0 3k 6k 3α
αn , which is a contradiction. Thus applying Lemma 3.1 to subsets V and V gives rvc(G) ≤ 9clogn, as
1+ǫ 1 2 αlogr
required. (cid:3)
Remark. The constants γ = 0.11 and ǫ = 0.02 are chosen so that Theorem 1.4 holds for r ≥ 28. If we are
onlyinterested inlargevaluesofr,wecansetγarbitrarily closeto0.5and,say,ǫ = 0.25
9
Concluding remarks
Inthispaperweproposedasimpleapproachtostudyingrainbowconnectivityandrainbowvertexconnectivity
in graphs. Using it we gave a unified proof of several known results, as well as of some new ones. Two
obvious interesting questions which remain open are to show that rainbow edge connectivity and rainbow
vertexconnectivity ofrandom3-regular graphsonnvertices arelogarithmic inn.
Acknowledgement. PartofthisworkwascarriedoutwhenthethirdauthorvisitedTelAvivUniversity,Israel.
He would like to thank the School of Mathematical Sciences of Tel Aviv University for hospitality and for
creating astimulating researchenvironment.
Wewouldliketothanktherefereesforhelpfulremarks.
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10