THE DIAMETER OF A SCALE-FREE RANDOM GRAPH

8/13/2019 THE DIAMETER OF A SCALE-FREE RANDOM GRAPH

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COMBINATORIC ABolyai Society Springer-Verlag

02099683/104/$6.00 c2004 Janos Bolyai Mathematical Society

Combinatorica 24 (1) (2004) 534

THE DIAMETER OF A SCALE-FREE RANDOM GRAPH

BELA BOLLOBAS*,OLIVER RIORDAN

Received August 1, 2000Revised July 24, 2001

We consider a random graph process in which vertices are added to the graph one at atime and joined to a fixed number m of earlier vertices, where each earlier vertex is chosenwith probability proportional to its degree. This process was introduced by Barab asi andAlbert [3], as a simple model of the growth of real-world graphs such as the world-wideweb. Computer experiments presented by Barabasi, Albert and Jeong [1,5] and heuristicarguments given by Newman, Strogatz and Watts [23] suggest that after n steps the

resulting graph should have diameter approximately logn. We show that while this holdsform= 1, for m2 the diameter is asymptotically logn/ loglogn.

1. Introduction

Recently there has been considerable interest in studying complex real-world networks and attempting to understand their properties using randomgraphs as models. Attention has been focused particularly on the small-

world phenomenon, that graphs with a very large number n of vertices of-ten have diameter around log n. In the context of various standard randomgraph models, this phenomenon has of course been known for a long time;see for example [13,14,6]. That such a small diameter can be expected evenwhen the vertex degrees are constant was shown in [9]. Related results in dif-ferent contexts include [19,18,10]; see also chapter X of [7]. A less standardexample showing that when the degrees are constant even a small amount ofrandomness produces this phenomenon is given in [8]. However, real-world

Mathematics Subject Classification (2000): 05C80

*Research supported in part by NSF grant no. DSM 9971788
http://goback/http://goback/http://goback/http://goback/http://goback/


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6 BELA BOLLOBAS, OLIVER RIORDAN

graphs are often not well approximated by these models, as shown by theirdegree sequences, for example.

Barabasi and Albert [3], as well as several other groups (see [5] and the

references therein), noticed that in many real-world examples the fractionP(k) of vertices with degree k follows a power law over a large range, withP(k) proportional to k for some constant independent of the scale ofthe network. Graphs with this behaviour have been studied in several pa-pers, in particular with respect to how well interconnected they are, as mea-sured by the diameter, or by the average distance between pairs of vertices.Such study has been from one of two points of view. In [1,5,23] graphs areconstructed at random subject to the degree sequence having the desireddistribution, and the diameter is analyzed (by computer experiments in [1,

5], and heuristically in [23]), obtaining an answer of the form A + B log n.This small-world phenomenon is often seen as surprising, as it is in thereal world. However, the examples cited in the first paragraph show that inthe random graph context if anything is surprising it is that the diametershould be so large; one would expect such a skewed degree sequence to leadto a diameter no larger than, and perhaps smaller than, the log n obtainedfor all degrees roughly equal. This is discussed briefly insection 10.

The above point of view takes the form of the degree sequence as anexperimental fact. Putting only this information into the model, one inves-tigates which other properties of the graph (such as small diameter) follownaturally from such a degree sequence, without considering how the degreedistribution arises. A different approach is to try to model the growth ofthe graph in such a way as to explain the distribution of the degrees. Inparticular, Barabasi and Albert suggested modeling such networks using arandom graph process with the following description, taken from [3]:

... starting with a small number (m0) of vertices, at every time stepwe add a new vertex with m( m0) edges that link the new vertexto m different vertices already present in the system. To incorporate

preferential attachment, we assume that the probability that a newvertex will be connected to a vertex i depends on the connectivity kiof that vertex, so that (ki) = ki/

j kj . Aftert steps the model leads

to a random network with t + m0 vertices and mt edges.

Such a process provides a highly simplified model of the growth of the world-wide web, for example, the vertices representing web sites or web pages, andthe edges links from sites to earlier sites, the idea being that a new site ismore likely to link to existing sites which are popular (are often linked to)at the time the site is added. It is easy to see heuristically that this process

leads to a degree distributionP(k) approximately of the formP(k) k3

[4],


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THE DIAMETER OF A SCALE-FREE RANDOM GRAPH 7

so that the number of vertices with degree at least k falls off as ck2 forlargek. A precise version of this statement will be proved in a forthcomingpaper [12]. (Note added: although written later, [12] has appeared first.)

Here we shall study the diameter of the resulting graph, showing that it isasymptotically log n/ loglog n if m2. In contrast, for m= 1 a result ofPittel [24] states essentially that the diameter is (log n). The relationshipof these results to previous heuristics is discussed briefly in the final section.

Note that a different type of model leading to the small-world phe-nomenon was introduced earlier by Watts and Strogatz [28], involving regu-larly arranged local connections together with random global connections.As the graphs produced have all degrees about the same, the study of thediameter in this case is a separate topic to that of this paper.

2. The model

When making the model described in the preceding section precise we havesome choice as to how to proceed, since the distribution of a random m-element set is not specified by giving the marginal probability that eachelement is contained in this set. When we add a new vertex to the graph weshall add m new edges one at a time, allowing multiple edges between thesame pair of vertices. Also, it will be convenient to allow loops; in terms of

the interpretation there is no reason to exclude multiple links from one siteto another, or links between different parts of a site or even page.

For precise definitions we start with the case m= 1. Consider a fixedsequence of vertices v1, v2, . . . . (Most of the time we shall take vi = i tosimplify the notation.) We write dG(v) for the degree of the vertex v in thegraph G. We shall inductively define a random graph process (Gt1)t0 sothat Gt1 is a graph on{vi: 1 i t}, as follows. Start with G01, the emptygraph with no vertices, or with G11the graph with one vertex and one loop.GivenGt11 , we formG

t1 by adding the vertex vt together with a single edge

between vt andvi, where i is chosen randomly with

P(i= s) =

dGt1

1

(vs)/(2t 1) 1st 1,1/(2t 1) s= t.

In other words, we send an edgeefromvt to a random vertex vi, where theprobability that a vertex is chosen as vi is proportional to its degree at thetime, counting e as already contributing one to the degree ofvt. (We shallsee why this is convenient later.) For m > 1 we add m edges from vt one ata time, counting the previous edges as well as the outward half of the edge

being added as already contributing to the degrees. Equivalently, we define


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the process (Gtm)t0 by running the process (Gt1) on a sequence v1, v2, . . . ;the graph Gtm is formed from G

mt1 by identifying the vertices v

1,v

2, . . . , v

m

to form v1, identifying vm+1, v

m+2, . . . , v

2m to form v2, and so on.

From now on we shall takevi= i, soGtmis a graph on [t] ={1, 2, . . . , t}.Wehave defined Gtm as an undirected graph, and it is as an undirected graphthat we shall measure its diameter. We note, however, that Gtm has a verynatural orientation: direct each edge ij with i > j from i to j, so each non-loop edge is directed from a later vertex to an earlier one. One can checkthat if any possible graph Gtm is given withoutthe vertex labels, the edgeorientations can be uniquely reconstructed. As we shall always treat Gtm asa graph with labelled vertices, this is of no relevance here.

Often we shall consider not the whole process, but just the graph obtainedat one particular time: we shall writeGnm for the probability space of graphson [n] where a random GnmGnmhas the distribution derived from the processabove.

As the process (Gtm) is defined in terms of (Gt1) we can deduce the prop-

erties of a random Gnm Gnm from those of a random Gmn1 Gmn1 . For themoment we shall thus restrict our attention to the casem =1. The reason forallowing loops is that this allows us to give a precise alternative descriptionof the distribution ofGn1 , in terms of pairings.

Ann-pairingis a partition of the set{1, 2, . . . ,2n}into pairs, so there are(2n)!/(n!2n)n-pairings. These objects are sometimes thought of aslinearizedchord diagrams(or LCDs)[26,11], where an LCD with n chords consists of2n distinct points on the x-axis paired off by semi-circular chords in theupper half plane. Two LCDs are considered to be the same when one can beturned into the other by moving the points on the x-axis without changingtheir order. Thinking of pairings as LCDs, we shall talk of chords and theirleft and right endpoints. We form a graph (L) from an LCD L as follows:starting from the left, identify all endpoints up to and including the first

right endpoint reached to form vertex 1. Then identify all further endpointsup to the next right endpoint to form vertex 2, and so on. For the edges,replace each chord by an edge joining the vertex corresponding to its rightendpoint to that corresponding to its left endpoint. We claim that ifL ischosen uniformly at random from all (2n)!/(n!2n) LCDs withn chords (i.e.,n-pairings), then (L) has the same distribution as a random Gn1Gn1 . Tosee this note thatLcan be obtained by taking a random LCDL withn1chords and adding a new chord whose right endpoint is to the right of alln1 chords, and whose left endpoint lies in one of the 2n1 possible places,each chosen with equal probability. This corresponds to adding a new vertex


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to(L) and joining it to another vertex with probabilities according to thedegrees, exactly as in the description of (Gn1 ).

We now study the diameter of Gnm as an undirected graph, using this

description from pairings.

3. Results

Our aim is to prove the following result, where Gnm Gnm is the randomgraph described above. Following standard terminology we shall say thatalmost everyor a.e. GnmGnm has a certain property if the probability thata randomGnmGnm has this property tends to 1 as n, keepingmfixed.

Theorem 1. Fix an integerm2 and a positive real number. Then a.e.GnmGnm is connected and has diameterdiam(Gnm) satisfying

(1 )log n/ log log ndiam(Gnm)(1 + )log n/ log log n.The lower bound is relatively easy to prove since we can proceed directly

from the definition of the process (Gtm). This bound is proved in section 4,in a tighter form than that given above. Our proof of the upper bound ismuch more complicated; for this we use an alternative method for generatinga random Gn

m

based on generating a random mn-pairing in a certain way,described insection 6.

In the rest of the paper we shall use the following standard notation.Given a graph G we shall write (G) for its maximum degree, E(G) forthe set of its edges, and e(G) = |E(G)| for the number of edges. We writeBi(n, p) for the binomial distribution with parametersnandp, and IAfor theindicator function of an event A. All logarithms are natural unless otherwiseindicated.

4. The lower bound

To prove the lower bound inTheorem 1we shall consider GN1 withN= mn.We shall compareGN1 with a random graph in which every edgeij is presentwith probabilityC/

ij independently of the other possible edges, for some

constant C. It turns out that small graphs are not much more likely tooccur in GN1 than in this random graph, despite the dependence present inGN1 .

Recall thatGN1 is defined as a graph on [N] ={1, 2, . . . , N }. We shall writegj for the vertex to which vertex j sends an edge, soE(G

j1) = E(G

j1

1 ){gjj}.


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Now given Gj11 , the probability thatgj= iis proportional to the degree ofi in Gj11 . Thus, given thatgj= i, the vertex i is likely to have a relativelylarge degree in Gj11 , and hence in G

t1 for all t. This suggests that the

events gj =i and gj =i are positively correlated if i=i, and negativelycorrelated otherwise. However, as we shall now see, for i=i any positivecorrelation is not too strong. Later we shall prove the negative correlation ina more general context. All constants implicit in the O() terms are absoluteconstants.

Lemma 2. If1 i < j, thenP(gj =i) =O

(ij)1/2

.(1)

Also, if1 i < j < k, thenP(gj =i, gk =i) =O

i1(jk)1/2

.(2)

Proof. We shall consider the process (Gt1), applying repeatedly the factthat E(A) =E(E(A |B)), withB the random variable Gt11 . Letdt,i= dGt1(i)be the degree of the vertex i in the graph Gt1. We start by evaluating theexpectation ofdt,i for allt. From the definition of the process (G

t1) we have

Pgt=i|Gt11

= dt1,i/(2t 1)t > i,1/(2i 1) t= i.

(3)

For t > iwe have dt,i= dt1,i+I{gt=i}, so

E

dt,i|Gt11

=dt1,i+

dt1,i2t 1=

1 +

1

2t 1

dt1,i.

Taking expectations of both sides,

E(dt,i) =

1 +

1

2t 1E(dt1,i).(4)

Let us write t,i for E(dt,i). Then, as i,i= 2i/(2i 1), we have for all t ithat

t,i=ts=i

1 +

1

2s 1

=O

t/i

,

where the estimate follows by taking logarithms and comparing the resultingsum with an integral. Taking the expectation of both sides of (3) fort =j > iwe obtain

P(gj =i) =j1,i/(2j 1) =O (ij)1/2 ,


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proving the first part of the lemma.For the second statement it turns out that we need to consider second

moments. Suppose that t > i. Then from(3) we have

E

d2t,i|Gt11

=d2t1,i

1 dt1,i

2t 1

+ (dt1,i+ 1)2dt1,i2t 1

=d2t1,i

1 + 2

2t 1

+ dt1,i2t 1 .

Taking the expectation of both sides we obtain

E(d2t,i) =

1 +

2

2t 1E(d2t1,i) +

E(dt1,i)2t 1 .(5)

Adding (4) and(5), and writing(2)t,i for E(d2t,i) +E(dt,i), we obtain

(2)t,i =

1 +

2

2t 1

(2)t1,i,

while(2)i,i = O(1) as di,i is either 1 or 2. Hence

(2)t,i =

ts=i+1

1 +

2

2s 1

(2)i,i =

2t + 1

2i + 1(2)i,i =O(t/i).

Now from (3) we have

E

dj,iI{gj=i}|Gj11

= (dj1,i+ 1)

dj1,i2j 1 .

Taking the expectation of both sides,

E

dj,iI{gj=i}

=

(2)j1,i

2j 1=O

i1

.

Arguing as in the proof of (1) above, for all t > j we have

E

dt,iI{gj=i}

=O

t/j

E

dj,iI{gj=i}

=O

i1

t/j

.

Now

P

gj =i, gk =i|Gk11

= I{gj=i}

dk1,i2k 1 .

Taking expectations gives

P(gj

=i, gk

=i) =O (ik)1k/j =O i1(jk)1/2 ,


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completing the proof of the lemma.

Note that a result similar to the above can be proved for the probabilitythat gj1= gj2=

= gjr= i, where r is fixed and i < j1 j.

Lemma 4. LetSbe a loopless graph on [N]in which each vertex is joinedto at most one earlier vertex and at most two later vertices. Then

P

SGN1

Ce(S)

ijE(S)1ij

,

for some absolute constant C.

Proof. Grouping the edges ofSaccording to the earlier vertex of each, theevent SGN1 is the intersection of events

Ek ={gjk,1 = = gjk,nk =ik},where the ik are distinct and nk= 1,2 for all k. ByLemma 2 there is anabsolute constant C such that

P(Ek)Cnknkl=1

(ikjk,l)1/2

for every k. Also, byLemma 3, the events Ek andE1 Ek1 are nega-tively correlated for every k, so

P

SGN1

= P

k

Ek

k

P(Ek).

Combined with the bound on P(Ek) above this completes the proof.Note that we did not use the condition that Sis loopless; this was taken

for simplicity. A more general version ofLemma 2 would of course give amore general version ofLemma 4.

Recalling thatGnmis obtained by identifying the vertices ofGmn1 in groups

ofm, we can deduce the lower bound inTheorem 1in the following moreprecise form.

Theorem 5. Let m 1 be fixed. Then diam(Gnm) > log n/ log(3Cm2 log n)for a.e. G

nmG

nm, whereC is the constant inLemma 4.


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Proof. We shall prove slightly more, namely that with probability tendingto 1 the distance(n, n1) between the vertices n andn1 ofGnm is greaterthanL =log n/ log(3Cm2 log n), withCas above.

Consider a particular path P= v0v1 vl on [n] of length l L. A graphSon [mn] is a realizationofP ifSconsists of edgesxtyt+1,t = 0,1, . . . , l1,withxt/m=yt/m= vt. AsPand henceShas maximum degree at mosttwo, Lemma 4 tells us that one such realization S is present in Gmn1 withprobability at most

Cll1t=0

1xtyt+1

Cll1t=0

1vtvt+1

= Cl

v0vl

l1t=1

1

vt.

As there are exactlym2l realizations ofP, andP

Gn

m

if and only if at leastone is present in Gmn1 , we have

P(P Gnm) (Cm2)l

v0vl

l1t=1

1

vt.

Thus, ifn is large enough, the expected number of paths of lengthl betweenthe vertices n andn1 ofGnm is bounded by

(Cm2)ln(n 1) 1v1,...,vl1n

l1

t=1

1

vt=

(Cm2)ln(n 1)

n

v=1

1

vl1

(2Cm2)l

n (log n)l1

(2/3)l(log n)1,asl Limplies that (3Cm2 log n)ln. Summing over 1 lL, we find thatthe expected number of paths of length at most L joining the vertices n andn1 tends to zero, so (n, n1) > Lin a.e. Gnm.

This proves the lower bound in Theorem 1. In the following sections weprove the upper bound for m2.

5. The upper bound

As the proof is rather lengthy, we give an outline before turning to thedetails. The basic idea is to use the random pairing model from section 2.Considering the pairing as an LCD pairing of independent uniformly chosenrandom points in [0, 1], we shall choose the right endpoint of each chord first,

according to the appropriate (non-uniform) distribution. Given these right


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endpoints, the left endpoints are independent of one another, each beingconstrained to come before its corresponding right endpoint. (We may ofcourse ignore the probability 0 event that two endpoints coincide.) We shall

first show that the distribution of right endpoints is likely to be nice, i.e.,to have certain uniformity properties. For the rest of the proof we then fixa nice distribution of right endpoints and work with this.

The main idea of the proof is to start with a given vertex v, and con-sider the set Nk(v) of vertices within distance k ofv, aiming to show thatthis increases in size at a certain rate. Rather than size, it turns out thatone should consider weight, assigning weight i1/2 to the ith vertex. Es-sentially, given the weightw(Nk1) ofNk1, the expected value ofw(Nk) is(log n)w(Nk1), as long as this is not too large. Thus in (log n/ loglog n)

steps from v one can reach many vertices. We complete the proof by show-ing that from these many vertices one can almost certainly reach the firstvertex in one step. Unfortunately there are several complications.

One essential complication is that whenw(Nk1) is fairly small, althoughw(Nk) has expectation (log n)w(Nk1), it has a very skew distribution, andis almost always much smaller than its mean. This means that we have towork with a blow-up factor which starts off as a constant and increases aswe proceed.

Another complication is that the main argument we use does not workstarting from a single vertex near the end. To deal with this we show sep-arately that all such vertices are likely to be joined by short paths to suffi-ciently early vertices.

Finally, the nice distributions of right endpoints are not really all thatnice; globally they behave exactly as we would like, but it is impossible toavoid some local variation. This complicates the details of the proof.

The main probabilistic tool we shall use in the rest of the paper is thefollowing lemma given by Janson [17], which may be deduced from the Cher-noff bounds [15].

Lemma 6. Let X= X1 + + Xk where theXi are independent Bernoullirandom variables with P(Xi= 1) =pi. Let =EX=p1 + +pk. Then fort0we have

P(X + t)exp

t2

2( + t/3)

and

P(X t)exp t2

2 .


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6. Pairings on [0,1]

We start by describing the method we shall use to generate the randomgraphGn

m

. Using the results ofsection 2, this is equivalent to generating arandom pairing of the integers{1, 2, . . . ,2N}, where N= mn.

Letx1, . . . , x2Nbe 2Nindependent samples from the uniform distributionon [0, 1]. Assuming that these 2Npoints are distinct, our LCD (linearizedchord diagram) is given by pairing x2i1 withx2i for eachi. Relabelling thexi as 1, 2, . . . in ascending order, this gives a random pairing with the correctdistribution. To see this note that, loosely speaking, for any set{x1, . . . , x2N}of 2Ndistinct elements of [0, 1], all (2N)! possibilities for the order in whichx1, . . . , x2Ntake these values are equally likely.

We now consider generating a pairing starting with the right end-

points. We call a random variable with density function 2x, 0 < x < 1,an M2(0, 1) random variable. Let{li,ri} ={x2i1, x2i} with li < ri. AsP(max{x2i1, x2i} t) =t2, the ri are independent M2(0,1) random vari-ables. Also, givenr1, . . . , rN, the random variablesl1, . . . , lNare independentwithliuniformly distributed on [0, ri]. To obtain the pairing we must sort allthe endpoints together. We shall first sort the right endpoints, and then con-sider between which right endpoint each left endpoint lies. Our first lemmaconcerns properties of the sorted right endpoints.

For the rest of this section set N = mn. Let r1, . . . , rN be indepen-

dent M2(0, 1) random variables, and let R1, . . . , RN be obtained by sortingr1, . . . , rN into ascending order. When it comes to constructing the actualgraphGnm, we shall only be interested in every m

th endpoint and the spac-ings between them, so for 1 i n let Wi= Rmi, and let wi= Wi Wi1,takingW0 =0. We shall show that, in a certain range, the Wi have the val-ues one might expect. For thewi we can only say that this holds on averageover certain intervals.

For the rest of the paper we writes = 2a for the smallest power of 2 largerthan (log n)7, and 2b for the largest power of 2 smaller than 2n/3. Note that

a < b ifn is large enough. We shall consider the intervals It= [2t+ 1, 2t+1],fora t < b.Lemma 7. Let m2 be fixed. Using the definitions above, each of thefollowing five events has probability tending to1 asn:

E1=

Wi

i

n

1

10

i

n forsin

,

E2

= Itcontains at least 2t

1 verticesi with w

i 1

10in,a

t < b ,


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E3 =

w1 4

log n

n

,

E4=

wi (log n)

2

n fori < n1/5

,

E5 ={win4/5 fori > n/(log n)5}.The events E1 andE2 describe the main properties of the Wi andwi we

shall use. The remaining properties are simple technicalities we shall needin section 7. In the following proof, and in the rest of the paper, we shall

always assume that n is larger than some sufficiently large constant, evenwhen this is not explicitly stated.

Proof ofLemma 7. Let us write R(x) for|{i : ri x}| = |{i : Ri x}|. For0 x 1 we have R(x) Bi(mn,x2). As Wi= Rmi, we have that Wi x ifand only ifR(x)mi.

Let x = 910

i/n. Then as ER(x) = mnx2 = 81mi/100, Lemma 6impliesthat

P(Wix) = P(R(x)mi)emi/60.Similarly, withx =1110

i/neitherx 1 or ER(x)=121mi/100, and in either

case

P(Wi > x) = P(R(x)< mi)emi/60.Thus

P(Ec1)2i=s

ei/60 =o(1),

sinces asn.It now suffices to show that P(EcrE1) = o(1) for r = 2, . . . , 5.GivenRmiand the setJof indicesj for whichrj > Rmi, each of them(n

i) variablesrj,j J, is independently distributed with density 2x/(1R2mi),Rmi < x < 1. For Rm(i+1)= Rmi+m to be less than Rmi+ y it must happenthat at least mof theserj fall in the interval [Rmi,Rmi+y]. Conditional onRmi, the expected number ofrj in this interval is exactly

m(n

i)(Rmi+ y)

2 R2mi1 R2mi

.


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Thus

P(Rm(i+1)Rmi+ y|Rmi)(n i)(Rmi+ y)

2 R2mi1

R2mi

.

Translating this to a statement about the Wi, and replacing i by i 1 forconvenience later,

P(wiy|Wi1)(n i + 1)(Wi1+ y)2 W2i1

1 W2i1.(7)

For s + 1 i2n/3, provided Wi1 1110i1n this probability is at most

2n(2.2yi/n + y2). Thus, for i in this range,P(wi 2t1)exp(2t/400).

AsEc2E1 is exactly the event that for some a t < b we have Zt> 2t1,

P(Ec2 E1)b1t=a

exp(2t/400) =o(1),

asa.


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Turning toE3, note that w1 = W1. TakingW0 to be identically zero, theargument giving (7) applies with i = 1 to give P(w1 x) nx2 and henceP(E3) = 1o(1).

Suppose that Ec4E1 holds, and let =(log n)

2

/n. Then as E1 holds wehave Wn1/52n2/5. As E4 does not hold it follows that some interval[x, x + ] with 0 x < 2n2/5 contains two of the Wi, and hence two of theri. Each such interval is contained in one of the intervals Jt= [t,(t + 2)],0 t n3/5. The probability that a particular ri lies in one such Jt isexactly (4t+4)2 , so the probability that at least two ri lie in Jt is at mostmn(mn1)(4t + 4)24/2 < (log n)9(t + 1)2/n2. Thus

P(Ec

4E1

)

n3/5

t=0

(log n)9(t + 1)2/n2 =o(1),

so P(E4) = 1o(1).Finally, suppose that Ec5 E1 holds. This time let =n4/5. Then for

somex, (log n)3 < x < 1, the interval [x, x+] contains noWi, and hencecontains at most m 1 of the ri. Setting = /(m +1), each such intervalcontains mdisjoint intervals of the form [t, (t+1)) withtan integer and(log n)3 < t< 1 , one of which must contain no ri. For a given t, thenumber ofri in [t

, (t + 1)) has a Bi(mn,pt) distribution with

pt= (2t+ 1)2 >(log n)3 > n5/6.

The probability that no ri lies in this interval is thus

(1 pt)mn emnpt < en1/6 =o(n1).Summing over the O(n1/5) values of t shows that P(Ec5 E1) =o(1), andhence that P(E5) = 1o(1), completing the proof of the lemma.

From now on we shall assume that the right endpoints R1, . . . , Rmn are

given, writing Wi for Rmi. The remaining randomness is given by takingindependent random variables Li, 1 i mn, uniformly distributed on[0,Ri], to obtain an LCD given by chords{Li,Ri} already sorted by theirright endpoints. Recall that the graph Gmn1 is constructed from this LCDby taking an edge from each vertex i to a vertex k, where Rk1 < Li < Rk(taking R0= 0). Thus the graph G

nm we shall study is given by taking m

edges from each vertex i, 1 i n, joining i to li,j, 1 j m, where li,j= kifWk1 < Lm(i1)+j < Wk.

Speaking somewhat imprecisely, we would like to forget the Ri and con-

sider only the Wi. The only problem is that the value of li,j is determined


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by a variable Lm(i1)+j distributed uniformly on [0,Rm(i1)+j ]. However,as Wi1< Rm(i1)+jWi, if we take Lm(i1)+j uniformly distributed on[0,Wi] we increase the probability that the diameter is large, by increasing

the probability of loops ati. After making this change we can essentially re-strict our attention to the case m =2. More precisely, we shall only considertwo edges from each vertex i, to the vertices li,1 and li,2. This change justinvolves deleting edges from the graph, which again can only increase thediameter.

Fixm 2. A precise description of the model we shall use from now onis as follows. Suppose that n distinct real numbers 0 < W1< < Wn 0 be fixed. For the rest of the paper we concentrate onthe Wi rather than the Ri. We shall assume the Wi are given and have the

properties E1, . . . , E 5 above, and form a random graph G = G(W1, . . . , W n)as above, writing PL for the corresponding probability measure. To simplifythe notation in the following results and proofs, the implicit bound in eachoccurrence of o(.) should be taken as a function ofn only, not dependingon the Wi or any other parameters, tending to 0 as n . (Of coursethese implicit functions depend on our fixed parameters m and ; they arefunctions ofn,mandwhich tend to 0 asn withmandfixed.) In thislight, the results are most simply seen as proving an (implicit) bound on someprobability that holds for every sufficiently largen, and every sequence0 j from i toj.

Let v be fixed, and consider the vertices reached from v after k steps.More precisely, letD0 ={v}, and fork 1 letDk be the set of those verticesoutsideD0 Dk1 incident with an edge directed from some vertex ofDk1. Our first lemma shows that|Dk|is likely to be not much smaller than2|Dk1|, as long as neither Dk1 norDk contains any useful vertices.

Letnkbe |Dk|ifD0Dkcontains no useful vertices, andotherwise.

Lemma 9. For1v n, 1k8loglog n and1c|Dk1| we have

PL(nk2nk1 c|D0, D1, . . . , Dk1)n3c/5+o(1).

Proof. Let t=|Dk1|, and let i1, . . . , it be the vertices of Dk1 in someorder. We may assume that no vertex ofD0 Dk1is useful as otherwisenk1= and there is nothing to prove. Thus nk1= t, and for each s wehave isn1/5.

Suppose that is sends its first and second edges to j2s

1 and j2s. Then

j1, . . . , j2t are independent, and for 1s2t and 1j is/2 we havePL(js =j

) =wj/Wis/2 .(8)

Now|Dk| is the number of distinct js which lie outside D0 Dk1, a setof at most 2k1 vertices none of which is useful. Considering thejsin order,we say that a possible value ofjs is repetitiveif it coincides with a previous

js which is not useful, or lies in D0 Dk1. Recalling thatnk=if anyvertex inDk is useful, we see thatnk is at least 2nk1 minus the number of

sfor which a repetitive choice is made. For a fixeds the number of possible


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repetitive choices is at most 2k+1. Consider one such choice j. Asj is notuseful, either wj < (log n)

2/norj> n/(log n)5. In the first case, using E1,

Wis/2Wn1/5 n1/5/n/2 =n2/5/2,so

PL(js =j) =wj/Wis/2n3/5+o(1).

In the second case we have wj n4/5 asE5 holds. Also, asis/2jn4/5we have Wis/2 n1/10/2, and again PL(js=j)n3/5+o(1).

Thus at each stage the probability of making a repetitive choice is atmost 2k+1n3/5+o(1). Since there are only 2

|Dk1

| n/(log n)5. In the first case 2b i/2, so,using E1 twice, W2b /Wi1/8. In the second case 2b

n/(log n)5/2, soW2b (log n)5/2/2100(log n)3. As we always have Wi 1, in eithercase W2b /Wi 100(log n)3. Combined with (9) and(10) this shows that99(log n)

3

.


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Let us say that a vertex i is goodifwi1/(10

in). ThenE2says exactlythat fora t < beach intervalItcontains at least 2t1 good vertices. As eachgood vertex i in It has weight wi 1/(10

2t+1n), the total weight of good

vertices within It is at least 2t1/(102t+1n) 2t/n/30. On the otherhand, the total weight ofIt is W2t+1 W2t W2t+1 3

2t/n. For a t b,

the whole ofIt lies to the left ofi, and the probability thatli,1 is good given

that li,1 It is thus at least 1/90. As 2b n/(log n)5, any good vertex ofItis also useful. This shows that the probability that li,1 is useful is at least/90, from which the lemma follows.

Using the two lemmas above, it is now easy to prove Lemma 8.

Proof of Lemma 8. Fix a vertex v, and let K=8loglog n 1. FromLemma 9, with probability at least 1 n

6/5+o(1)

= 1 o(n1

) either v iswithin distance Kof a useful vertex, or we have|Dk| = 2|Dk1| for all butat most one value ofk, 1kK, and|Dk|2|Dk1|1 for allk, 1kK.Suppose that the second case holds. Then|DK|2K12(log n)4, providedn is sufficiently large. Given the sequence D0, . . . , DK, the verticesli,1, iDK,are independent with their original distribution. IfDKdoes not contain auseful vertex then, fromLemma 10, the probability that no i in DKhas auseful neighbour is at most

(1

(log n)3)|DK |

exp(

|DK

|/(log n)3)

n2.

Thus with probability at least 1o(n1) the given vertexv is within distance8loglogn of a useful vertex. As there are only n vertices v to consider,Lemma 8follows.

8. Expanding neighbourhoods

The lemma below gives the heart of the argument. As before, o(.) notation

refers to an implicit function depending on nand only. The statement of thelemma then applies for all sufficiently large n and all 0 < W1 < < Wn < 1satisfying the stated conditions.

Lemma 11. Assume that E1, . . . , E 5 hold, let > 0 be fixed, and let v,1vn, be a useful vertex. With PL probability1o(n1)there is a pathin G between v and1 of length at most (1/2+ )log n/ loglog n.

The basic strategy of the proof is to divide [s + 1,n] into intervals of theformIt= [2

t+ 1, 2t+1]. Given the set of vertices k at distance k from v, we

consider the expected size ofk+1It. Whenever this is bigger than log nwe


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will have enough independence to show that the actual size ofk+1Itwill beclose to its expected value. This will show that the weight fk=

ik1/

in

almost certainly grows at a certain rate, depending on its current value. In

fact we only consider the intervals It for a t < bwhere, as before, s = 2a

isthe smallest power of 2 larger than (log n)7, and 2b is the largest power of 2smaller than 2n/3. Thusa7log2 log n andb log2 n.

As the proof of Lemma 11 is a little complicated, we state part of theestimation as separate lemma. What this lemma claims is that if the fksatisfy a certain condition then they grow fast enough.

Lemma 12. Suppose that > 0, and setK= (1/2+)log n/ loglog n1. Letf0, f1, . . . be a sequence of real numbers with f0(log n)2/n and

fk+1min{2log2(fkn/ log n) 31, b a 1}fk/1000,(11)fork 0. Then, providednis sufficiently large, =min{k :fk(log n)2/

n}

exists and is at most K.

Proof. Provided n is sufficiently large we have log2(f0n/ log n) log2(log n)1016. Thus (11) implies that f12f0. It follows inductivelythatfk+12fk andfk2kf0 hold for all k0. This shows that exists.

As b a 1 log2 n 8log2 log n, the minimum in (11) is different fromthe first term only when f

k(log n)

3/

n. If this first happens at k = k

0,

say, then for kk0 we havefk+1(b a 1)fk/1000 = (log2 n)1+o(1)fk,

which implies that k0 +6. Thus

fk+1 log2(fkn/ log n) 16500

fk(12)

for 0

k <

6. Asfk

2kf0 and log2(f0n/ log n)

17 (ifnis large enough),

we have log2(fkn/ log n) k+ 17. Combined with (12) this implies thatfk+1fk(k + 1)/500 for 0k < 6, and hence that

f6 ( 6)!5006

f0

6500e

6f0,

using Stirlings formula. As f6


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Proof of Lemma 11. Roughly speaking, we consider the set of verticeswithin distance k ofv, showing that the weight of this set increases rapidlywith k. In fact, to keep independence, when working outwards from v we

only allow ourselves to use an edge ij from a vertexiat distancek to a newvertexj ifi> j andij is the first edge from ior ifi< j andij is the secondedge from j. As before we say that a vertex i is goodif

wi 110

in

.

Let 0= {v}, and for k 1 let k consist of those j in [s + 1, 2b] \ (0 k1) such that j is good and either lj,2 k1 or there is an i k1withli,1 =j. We shall write Nk for0 k.

Rather than the actual weight (sum of the wi) we consider the idealweight ofk, given by

fk =ik

1in

.

Note that for k1 the set k contains only good vertices, so its actualweight is at least fk/10. For the sake of simplicity we shall assume that vitself is good, as well as useful. The case that v is useful but not good canbe dealt with by redefining f0 to be (log n)

2/n. It is easy to check that the

first step in the argument that follows requires only that 1/vn f0 andthat wv f0/10. At the end of the argument we only use f0 (log n)2/n,and nowhere else is the goodness ofv relevant.

Given0, . . . , k, for each i k the random variable li,1 has its originalunconditioned distribution, since we have looked at li,2 but not li,1. Simi-larly, for i / Nk all we know about li,2 is that li,2/ Nk1, and the variables{li,1 :ik}{li,2 : i /Nk} are independent.

Let us fix at witha t < b, and consider the intervalIt= [2t+1, 2t+1]. LetXbe the number of vertices ofNkIt (these vertices are excluded for ourpresent purpose), and letSbe the set of good vertices ofIt\Nk. NowE2statesexactly that It contains at least 2

t1 good vertices, so|S| 2t1 X. Leti1, . . . , id be the vertices ofk [2t+1+1, n], listed in any order. We considerexamining each i in turn to see whether li,1 lies in S. More precisely, letCr be the number of distinct elements ofS equal to li,1 for i {i1, . . . , ir},so|k+1It|Cd. Conditioning on everything so far, i.e., on 0, . . . , k andli1,1, . . . , lir1,1, providedCr12t2Xthe probabilityPr thatCr= Cr1+1satisfies

Pr

2t2 1

102t+1nW1ir

,


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as there are at least|S|Cr12t2 vertices j ofSunused, each of weightwj 110jn 1102t+1n . Using Wi

2i/nfor is,

Pr 2t2

20

2tn

nir

= 2t80

ir

=pr,(13)

say. LetYbe a random variable given by the sum ofdindependent Bernoullirandom variables Bi(1, pr) with means p1, . . . , pd. Then Cd stochasticallydominates min{Y, 2t2 X}. (Roughly speaking, (13) implies by inductiononr that Cr

rr=1 Bi(1, pr) until Cr= 2

t2X.)Let

1=

ik, i>2t+1

2t

80i =d

r=1pr.

Provided110log nwe have fromLemma 6that P(Y1/2)n5/4, andhence that

PL(Cdmin{1/2, 2t2 X} |0, . . . , k)n5/4.(14)In other words, if1 is large enough, then 1/2 vertices ofIt are likely tobe hit by edges coming out ofk [2t+1+1, n], unless this would mean thatmore than 2t2 vertices ofIt would be contained inNk+1.

We now consider edges from It into k [s + 1, 2t].Let X =X+ Cd and let S be the set of good vertices of It \ Nk not

counted in Cd, so|S| 2t1 X. We consider the number C ofj S forwhich lj,2 lies in k, noting that|k+1 It| C. Given 0, . . . , k, for each

j S the random variable lj,2 has its original distribution conditioned onlj,2/Nk1. Thus, as k is disjoint from Nk1,

P(lj,2k)

ik, ijwi/Wj

ik, i2t

wi/W2t+1

ik, i2t

1

10

in

n

2t+2

=

ik, i2t

1

20

2ti=p,

say. As the lj,2 are independent, the number C ofj S for which lj,2 is in

k stochastically dominates a Bi(2t1

X, p) distribution. Let 2= 2t2

p.
http://goback/http://goback/http://goback/http://goback/


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Then provided210log n we have fromLemma 6that

PL(C2/2|0, . . . , k, X 1. Then 2 (1 + 2)/2 and 2 10log n.Thus (15) implies that

PL(C(1+ 2)/4|0, . . . , k, X


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Given 0, . . . , k, letT be the set of indices t, a t < b, for which2tnfk/160 20log n. As 2b n/3, considering the smallest element of

T shows that|T | min{2log2(fkn/ log n) 30, b a}.

Let us writeT for the set of t T for which the quantity 1 + 2 definedabove (which depends on t) is at least

2tnfk/160, so|T| |T | 1. We

shall say that an interval It is fullat stage k +1 if|Nk+1 It| 2t2. From(16), for eachtT with probability 1O(n5/4) either It is full inNk+1or|k+1It|

2tnfk/640. In the latter case we have

ik+1It

1

in

2tnfk

6402t+1n fk

1000

.

Thus, with probability 1O((log n)n5/4) = 1O(n6/5) either some It isfull at stage k +1, or

fk+1|T|fk

1000 min{2log2(fkn/ log n) 31, b a 1}

1000 fk.(18)

Starting with 0 ={v} let us construct 1,2, . . . , and let K be theminimum k for which either fk(log n)2/

n or (18) does not hold. As v is

useful we have f0 = wv (log n)2

/n. As (18) holds for 0 k < K, Lemma 12implies that K(1/2 + )log n/ loglog n1.We claim that

fK(log n)2/

n(19)

holds with probability 1o(n1). The only possible problem is that some Itmay become full at some stage.

Suppose first that some It, a t < b, is full at some stage k, 1 k K.Then (leaving out 0 as v is not necessarily good), in NK\{v} we have atleast 2

t

2

1 good vertices in It. This givesf1+ + fK 2

t2 110

2t+1n

2t

160

n

2a

160

n=

s

160

n>

(log n)3n

.

AsK < log nandfk< (log n)2/

nfor k < K, this implies (19). On the other

hand, if noIt becomes full at any stage, then (18) holds with probability atleast 1O(n6/5) at each ofK log nstages, so (19) holds with probability1o(n1), as claimed.

If (19) holds we stop the construction of the sequence 1, 2, . . . at K.

Given the sequence so far, the li,1, i K, are independent. As K >0 we


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have K [s + 1,n] by construction. Thus, since E1 and E3 hold, for eachiKwe have

PL(li,1= 1|0, . . . , K) = w1Wi

4log n

n

n2i

> 2log n

i

.

Conditioning on the sequence 0, . . . , K, if (19) holds the probability thatKdoes not send an edge to vertex 1 is at most

iK

1 2log n

i

exp

iK2

log n

i

= exp

2fK

n

log n

n2.

Since (19) holds with probability 1o(n1), it follows that with probability1o(n1) the vertexv is connected to 1 by a path of length at most K+1(1/2 + )log n/ loglog n, completing the proof ofLemma 11.

Combining the results above proves Theorem 1.

Proof ofTheorem 1. We have already proved the lower bound asTheo-rem 5. Fixm

2 and > 0. TogetherLemmas 8and 11, applied with/4 in

place of, imply that given W1, . . . , W n

PL

diam(G(W1, . . . , W n))> (1 + /2)log n/ log log n + 16 log log n

=o(1),

whenever E1, . . . , E 5 hold. Let R1, . . . , Rmn be obtained by sorting mninde-pendent M2(0, 1) random variables, set Wi= Rmi, and consider the graphG= G(W1, . . . , W n). As each Ei holds with probability 1o(1) byLemma 7,it follows that

P

diam(G)> (1 + /2)log n/ log log n + 16 log log n

=o(1).

Finally, we have shown at the end of section 6 that the random graph Gdefined in this way may be coupled withGnmso thatG can be obtained fromGnm by (perhaps) deleting some edges and adding some loops. Under thiscoupling diam(G)diam(Gnm), completing the proof.

We next turn briefly to the case m = 1, which has been studied earlier

and behaves rather differently.


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9. The case m= 1

One particular step in the proof of the upper bound on diam(Gnm) given

above appears at first sight to be rather wasteful, namely the step where wesplit the edges into two separate classes, considering at every point only edgesfrom one class. Something like this is necessary to achieve independencewhen considering how the (k + 1)-neighbourhood Nk+1 of a given vertexrelates toNk. However, as the neighbourhoods grow quite quickly, one wouldimagine that we only rarely try to use edges already used. This suggests thata version of the proof above should apply with m =1, losing about a factor oftwo in the growth rate, which would have negligible effect on the diameter.This turns out to be not the case for m =1 the diameter is in fact(log n),

as shown by Pittel [24] in a slightly different context described below.For m = 1 the graph Gnm has a rather simple structure. Each vertex

sends an edge either to itself or to an earlier vertex, so the graph consistsof components which are trees, each with a loop attached. It is much morenatural to consider trees rather than forests. Trees on {1,2, . . . , n} whereeach vertex other than the first sends an edge to an earlier vertex are knownas recursivetrees; random recursive trees have been studied for some time,see, for example, [22]. When a random recursive tree is constructed onevertex at a time, each new vertex must be joined to an earlier vertex chosen

according to some rule. The most studied case is that of a uniform choice, butprobabilities essentially proportional to degrees have also been considered.Such objects (where for the first vertex degree plus one is used) are knownasrandom plane-oriented recursive trees, see[27,21], for example. Pittel [24]showed that the height (maximum distance of a vertex from the root) of suchan object is (c + o(1))log n with probability 1o(1), wherec = (2)1 for the solution ofe1+=1. It is easy to see that his method also shows thatthe diameter is (2c+o(1))log n. NowGn1 has on average(log n) loops. Thuswith high probabilityGn1 hasO(log n) loops and henceO(log n) components.

It is easy to see that Pittels method can be applied to prove the followingresult.

Theorem 13. Let be the solution of e1+= 1, and let >0 be fixed.Then for almost everyGn1Gn1 the largest distance between two vertices inthe same component is between (1)log nand(1 + )log n.

Note that this result is given only for completeness for m =1 the random

plane-oriented recursive tree model is actually more natural than Gn1 .


27/30


10. Conclusions

In several papers random graphs modeling complex real-world networks are

studied, in particular with respect to their diameter or average diameter, theaverage distance between two vertices. One model studied is the process Gnmconsidered here. Another model is given by taking a functionP(d),d 1, torepresent the probability that a vertex has degree d, and choosing a graphonn vertices at random from all graphs with this distribution of degrees.

In [1,5,23] results from computer experiments have been published sug-gesting that for the second model the diameter of the random graph willvary as A + B log n when n is large, for a wide variety of functions P(d).In [23] a heuristic argument is given, based on the standard neighbourhood

expansion method; roughly speaking, if one follows a random edge to oneof its endvertices v, the probability P(d) that v will have degree d shouldbe proportional to dP(d). Thus, writingNk(v) for the set of vertices withindistancek of a given vertex v, one would expect|Nk+1(v)| to be about

f=

dP(d) =

d2P(d)

/

dP(d)

times larger than|Nk(v)|. One then expects the diameter of the graph to beabout log n/ log f.

In the context ofGnm, which hasP(d) proportional tod3, this argument

actually predicts an expansion factor of log n, and hence a diameter of aroundlog n/ loglog n. The same expansion factor is predicted by approximatingGnmby a graph in which each edge ij is present with probability proportional to1/

ij, as in the proof ofTheorem 5. As shown byTheorem 1, this prediction

is correct ifm 2. The explanation for the prediction of(log n) given bythe more sophisticated heuristic in [23] is that a cutoff is imposed on thedegree distribution so that the formalism used there applies. In the lightof our results, we believe that the heuristic given above will apply in manysituations, in particular when the graph is chosen at random subject to a

power law distribution being imposed on the degrees. On the other hand,in any given case it can be highly non-trivial to take the actual structureinto account. Furthermore, this structure can cause the heuristic argumentto fail, as shown by the case of Gn1 , for which exactly the same heuristicsapply as for Gn2 .

We finish with two more remarks concerning possible extensions ofTheo-rem 1. For the first, note that the upper bound on the diameter ofGnm givenbyTheorem 1is a pure existence result. An interesting question is whethera short path between two given vertices can be constructed quickly using

only local information (see [20], for example). InGnm we believe that it will


28/30


take (log n) steps to reach vertex 1 from vertex n using local information.In the other direction, given any set Sofo(n) vertices of a typicalGn1m it iseasy to check that the probability of vertex n sending an edge to S is o(1).

It follows that at least (n) steps are needed to locally construct a pathfrom 1 ton inGnm.

The second remark concerns the robustness of the graph Gnm. In [2]the following question is raised, and answered experimentally: how manyvertices can be deleted at random from a scale-free random graph withoutthe diameter increasing too much, ignoring small components if necessary?It is easy to see that the expanding neighbourhoods argument ofsection 8isunaffected if the vertices ofGnm are deleted independently with probability

p, for any constant p < 1. In the terminology ofsection 7, from any useful

vertex there will be a short path to the first surviving vertex of Gn

m. Ingeneral, after deleting vertices in this manner, what remains of Gnm willconsist of many small components, together with a giant component whosediameter is asymptotically log n/ loglog n, for any constant p < 1.

Finally, we finish by mentioning an interesting problem to which one ofthe referees has drawn our attention. Suppose we are given a rooted treeof height n in which each non-leaf node has k2 children, and supposethat the edges are weighted with independent standard exponential randomvariables. What is the asymptotic distribution of the minimum weight of apath from root to leaf as n

with k fixed? For results on this question,

related to the question of which vertices ofGnmcan be reached from an initialvertex by descending paths of a certain length, see [16,25].

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34 BOLLOBAS, RIORDAN: THE DIAMETER OF A SCALE-FREE RANDOM GRAPH

Bela Bollobas

Department of Mathematical Sciences

University of Memphis

Memphis TN 38152

USA

and

Trinity College

Cambridge CB2 1TQ

UK

[email protected]

Oliver Riordan

Trinity College

Cambridge CB2 1TQ

UK

[email protected]
mailto:[email protected]:[email protected]:[email protected]:[email protected]

THE DIAMETER OF A SCALE-FREE RANDOM GRAPH

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