Average Distance and Vertex-Connectivity

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1 Average Distace ad Vertex-Coectivity Peter Dakelma, Simo Mukwembi, Heda C. Swart School of Mathematical Scieces Uiversity of KwaZulu-Natal, Durba, 4041 South Africa March 17, 013 Abstract The average distace µ(g) of a coected graph G of order is the average of the distaces betwee all pairs of vertices of G, i.e., µ(g) = ( ) 1 dg(x, y), where V (G) deotes the vertex {x,y} V (G) set of G ad d G(x, y) is the distace betwee x ad y. We prove that if G is a κ-vertex-coected graph, κ 3 a odd iteger, of order, the µ(g) + O(1). Our boud is show to be (κ+1) best possible ad our results, apart from aswerig a questio of Plesík [1], Favaro, Kouider ad Mahéo [8], ca be used to deduce a theorem that is essetially equivalet to a theorem by Egawa ad Ioue [6] o the radius of a κ-vertex-coected graph of give order, where κ is odd. 1 Itroductio Let G = (V, E) be a coected graph of order. The average distace µ(g) of G is defied as µ(g) = ( ) 1 {x,y} V d G(x, y), where d G (x, y) is the distace betwee vertices x ad y i G. The vertex-coectivity κ(g) (edge-coectivity λ(g)) of G is defied as the miimum umber of vertices (edges, respectively) whose deletio from G results i a discoected or trivial graph. G is k-vertex-coected (k-edge-coected) if κ(g) k (λ(g) k, respectively). It has bee proved by several authors [11, 7, 5] that the average distace of a coected graph with vertices is at most (+1)/3 ad that this boud is attaied oly by a path of order. Plesík The results i this paper are part of the secod author s PhD thesis. Fiacial support by the Natioal Research Foudatio ad the Uiversity of KwaZulu-Natal is gratefully ackowledged. 1

2 [1] showed that this boud ca be improved for a -vertex-coected (- edge-coected) graph G of order to µ(g) 4 /( 1) ad that this boud is attaied if ad oly if G is a cycle. Plesík [1] posed the problem of fidig sharp upper bouds o µ(g) for k-vertex-coected (k-edgecoected) graphs, where k 3. The preset authors [, 3] established asymptotically sharp upper bouds o the average distace of a k-edgecoected graph of give order, where k 3. Favaro, Kouider ad Mahéo [8] geeralised Plesík s boud ad proved that if G is a κ-vertex-coected graph of order, the + κ 1 ( 1 κ µ(g) 1 κ ) κ ( 1) ad that this boud is attaied by the κ/-th power of a cycle whe κ is eve. For odd κ, however, this boud is ot best possible. The goal of this paper is to prove that if G is a κ-vertex-coected graph, κ 3 odd, of (κ+1) order, the µ(g) + 30 ad that this boud is, apart from a additive costat, best possible. The otatio we use is as follows. Let G = (V, E) be a coected simple graph. For a subset S V, G[S] deotes the subgraph iduced by S i G ad diam(s) is the maximum value of d G (x, y), where x, y S. Let v V. The eccetricity ex G (v) of v is the distace from v to a vertex furthest from v. The distace σ G (v) of v is the sum x V d G(v, x) whereas the distace σ(g) of G is u V σ G(u) = (x,y) V V d G(x, y). Thus µ(g) = σ(g)/( 1), where is the order of G. For subsets M 1, M V, the distace σ M1 (v) of v with respect to M 1 is the sum x M 1 d G (v, x) ad the distace σ M1 (M ) of M with respect to M 1 is x M σ M1 (x). The distace betwee M 1 ad M is defied as mi{d G (u, v) : u M 1, v M } ad is deoted by d(m 1, M ). We deote the set of edges with oe ed i M 1 ad the other ed i M by E(M 1, M ). N G (v) deotes the set of all vertices adjacet to v i G; its cardiality is the degree of v i G ad is deoted by deg G (v). The miimum degree δ(g) is the umber mi{deg G (v) : v V }. For a o-egative iteger i, the i-th distace layer N i (v) of v is the set of vertices at distace exactly i from v. We deote the cardiality of N i (v) by k i (v) ad if the vertex v is uderstood we omit the argumet v ad simply write N i, k i. We deote the set 0 j i N j (v) by N i (v) ad the set j i N j (v) by N i (v). N(S) deotes the eighbourhood of subset S V, amely u S N G (u). N(S) deotes the closed eighbourhood of S, i.e. N(S) = N(S) S, whereas N i (S) deotes the i-th eighbourhood of S V, amely u S N i (u). A separatig set of G is a set of vertices of G whose removal icreases the umber of compoets of G. A vertex cutset of G is a separatig set, S say, such that o proper subset of S is a

3 separatig set of G. By V 1 V, we mea the disjoit uio of V 1 ad V. The compositio G 1 [G ] of graphs G 1 ad G is the graph with vertex set V (G 1 ) V (G ) i which the vertices u = (u 1, u ) ad v = (v 1, v ) of G 1 [G ] are adjacet if ad oly if (1) u 1 = v 1 ad u v E(G ) or () u 1 v 1 E(G 1 ). For vertex disjoit graphs G 1, G,, G k, the sequetial joi G 1 + G + + G k is the graph obtaied from the uio of G 1,..., G k by joiig every vertex of G i to every vertex of G i+1 for i = 1,,, k 1. Prelimiary Results Here ad i the sequel, give a κ-vertex-coected graph G, the umber 0 is defied as 0 := 59κ(κ + 1). The followig lemma is eeded for later estimates. Lemma 1 Let G be a κ-vertex-coected graph, κ, of order ad let v be a vertex of G. The (i) σ G (v) 1 8κ ( + κ ). (ii) If 0, the µ(g) (κ+1) (iii) If v is a vertex of largest distace ad ex G (v) 40κ, the µ(g) (κ+1) Proof: (i) Let e be the eccetricity of v. Note that σ G (v) = 1k 1 + k + + ek e. (1) Sice N i is a separatig set of G ad G is κ-vertex-coected, we have k i κ for i = 1,,..., e 1. Clearly, (1) is maximized subject to the ecessary costraits for k 1 = k = = k e 1 = κ. Hece σ G (v) κ( (e 1)) + e( 1 κ(e 1)) = κ e(e 1) + e( κe + κ 1). () By differetiatio with respect to e, () is maximized for e = + κ 1 κ. Thus we obtai σ G (v) κ ( + κ 1 ) ( + κ 1 ) 1 κ κ ( + κ + 1 ) ( ( + κ κ 1 ) ) + κ 1 κ κ = 1 8κ ( + κ ), 3

4 as desired. (ii) Assume that 0. By (i), ad summig σ G (v) for all v, we obtai σ(g) 1 8κ (+κ ). Hece ( it suffices ) to show that for 0, we have 1 8κ ( + κ ) ( 1) (κ+1) + 30 or equivaletly that f κ () 0, ( ) where f κ () = ( 1) (κ+1) κ ( + κ ). For costat κ, the graph of f κ () is a parabola that is cocave dow. Hece, o the iterval κ + 1 0, f κ () attais its miimum at a ed poit. At the ed poits we have f κ (κ+1) = 35 8 κ > 0 ad f κ( 0 ) = 1 8κ (35κ +36κ 4) > 0. This establishes (ii). (iii) If 0, the (iii) follows from (ii). So assume that Sice ex G (v) 40κ, the right had side of () is icreasig. Hece by (), σ G (v) κ (40κ)(40κ 1) + 40κ( 40κ + κ 1) = 40κ 800κ 3 + 0κ 40κ. Therefore, σ(g) (40κ 800κ 3 + 0κ 40κ). A log but straight forward calculatio ow shows that, subject to 0 +1, we have (40κ 800κ 3 + 0κ 40κ) ( 1)( + 30), as desired. (κ+1) I view of Lemma 1 (iii), from ow owards if v is a vertex of G of largest distace we may assume, uless otherwise stated, that ex G (v) 40κ + 1 ad so = e i=0 k i 1 + κ(e 1) κ(40κ) + 1 = 40κ +. 3 Sequeces Let G be a κ-vertex-coected graph, κ 3 a odd iteger, ad v a vertex of G of largest distace. I this subsectio, we assume the followig. Assumptio 1 There is o l {0κ,..., e 0κ } for which (k l, k l+1, k l+ ) = (κ, κ, κ). Assumptio Wheever k l = κ = k l+e1, where l 0κ ad e 1 are itegers with l + e 1 e 0κ ad k i κ + 1 for i = l + 1, l +,..., l + e 1 1, the l+e 1 +1 k i (κ + 1)e Our aim i this subsectio is to fid a upper boud o σ G (v) (ad cosequetly o µ(g)) give that Assumptios 1 ad hold for v. We show that if a sequece a 0, a 1, a,..., a e has certai properties, the the sum ia i 4

5 ca be bouded. This will be applied to the sequece k 0 (v), k 1 (v),..., k e (v) to obtai a boud o ik i (v) = σ G (v). For this purpose, let, κ, e be positive itegers with e 40κ + 1 ad 40κ +. Defiitio 1 We say that a sequece (a i ) = (a 0, a 1,, a e ) of umbers is (, κ, e)-realizable if (a i ) satisfies the followig properties: (A) a 0 = 1, a i κ for i = 1,..., e 1 ad a e 1. (B) There is o l {0κ,..., e 0κ } for which (a l, a l+1, a l+ ) = (κ, κ, κ). (C) Wheever a l = κ = a l+e1, where l 0κ ad e 1 are itegers with l + e 1 e 0κ ad a i κ + 1 for i = l + 1, l +,..., l + e 1 1, the l+e 1 +1 a i (κ + 1)e (D) e i=0 a i =. The idex τ[(a i )] of the sequece (a i ) is the umber of values r {0κ,..., e 0κ} for which a r = κ. Clearly, if G is a κ-vertex-coected graph ad v is a vertex of G with ex G (v) = e 40κ + 1 satisfyig Assumptios 1 ad, the (k 0, k 1,, k e ) is (, κ, e)-realizable. The followig sequece, which is (, κ, e)-realizable, is importat. Let (b i ) = (b 0, b 1,, b e ) be the sequece whose terms are defied by the rule, b 0 = 1, { κ if i {1,,..., 0κ + 1} {e 0κ + 1,..., e 1}, b i = κ + 1 if i {0κ +,..., e 0κ} ad b e = + 41κ e(κ + 1). Lemma Assume the otatio for (b i ) above ad let (a i ) be a sequece which is (, κ, e)-realizable. The e ia i i=1 e ib i. Proof: Assume that (a i ) is (, κ, e)-realizable ad let the idex of (a i ) be τ[(a i )] = s. We prove the lemma by iductio o s. First assume that s = 0. The a i κ + 1 for all i {0κ,..., e 0κ}. Clearly, e i=1 ia i is maximized, subject to this coditio ad subject to properties (A) ad (D) (see Defiitio 1), for (a 0, a 1,, a 0κ 1 ) = (1, κ,, κ), (a 0κ,, a e 0κ ) = (κ + 1,, κ + 1), (a e (0κ 1),, a e 1 ) = (κ,, κ) i=1 5

6 ad a e = + 41κ e(κ + 1). Hece sice e 40κ + 1, we have e ia i κ( κ 1) + (κ + 1)(0κ + + e 0κ) i=1 as desired. = +κ(e (0κ 1) + + e 1) + e( + 41 e(κ + 1) ) e e ib i + 40κ + 1 e < ib i, i=1 Now assume that s = 1. The a i κ + 1 for all i {0κ,..., e 0κ} with the exceptio of oe value of i. Clearly, e i=1 ia i is maximized, subject to this coditio ad the properties (A) ad (D), for (a 0, a 1,, a 0κ ) = (1, κ,, κ), (a 0κ+1,, a e 0κ ) = (κ+1,, κ+1), (a e 0κ+1,, a e 1 ) = e (κ,, κ) ad a e = + 41κ e(κ + 1) 1. As above, i=1 ia i e i=1 ib i + 0κ + 1 e < e i=1 ib i, as desired. If s =, the a i κ + 1 for all i {0κ,..., e 0κ} with the exceptio of two values of i. Clearly, e i=1 ia i is maximized, subject to this coditio ad the properties (A) ad (D), for a i = b i for all i. Hece e i=1 ia i e i=1 ib i, as required. Now assume that s > ad that result holds for ay sequece (d i ) which is (, κ, e)-realizable satisfyig τ[(d i )] < s. Cosider the sequece (a i ) ad let l be the largest iteger i {0κ,..., e 0κ} for which a l = κ. We cosider two cases separately. If, o oe had, a l 1 > κ, the let h l be the largest iteger from {0κ,..., e 0κ} for which a h = κ. By property (C), let r {h + 1,..., l 1} be such that a r κ +. Defie a ew sequece (c i ) from (a i ) as follows: a r 1 if i = r, c i = κ + 1 if i = l, a i otherwise. The c i κ+1 for all i {h+1,..., e 0κ}. This, i cojuctio with the fact that (a i ) is (, κ, e)-realizable, yields that (c i ) is (, κ, e)-realizable ad, moreover, τ[(c i )] = s 1. Thus by iductio, e i=1 ic i e i=1 ib i. Clearly e i=1 ia i = e i=1 ic i + (r l) ad thus e i=1 ia i < e i=1 ic i e i=1 ib i, as required. If, o the other had, a l 1 = κ, the by property (B) we have a l κ+1. Let h l 3 be the largest iteger from {0κ,..., e 0κ} for which a h = κ. Applyig property (C) yields l 1 i=h+1 a i (κ + 1)(l h 1) + 1. The we have the followig two subcases. (i) there exist r, t {h + 1,..., l }, r < t, such that a r, a t κ + or (ii) there exists f {h + 1,..., l } such that a f κ + 3. i=1 6

7 We form a ew sequece (c i ) from (a i ) as follows. If (i) holds, set a r 1 if i = r, a c i = t 1 if i = t, κ + 1 if i {l 1, l}, otherwise. Whereas, if (ii) holds, set a f if i = f, c i = κ + 1 if i {l 1, l}, otherwise. a i a i Clearly, i either case (c i ) is (, κ, e)-realizable, ad, moreover, τ[(c i )] = s. Simple calculatios, as above, i cojuctio with iductio give e i=1 ia i e i=1 ic i e i=1 ib i. Propositio 1 Let G be a κ-vertex-coected graph, κ 3 odd, with vertices. Let v be a vertex of G of largest distace. If v satisfies Assumptios 1 ad above, the µ(g) (κ + 1) Proof: Deote the eccetricity of v by e. Let (b i ) be the sequece defied i Sectio 3. The fact that G is κ-vertex-coected, i cojuctio with Assumptios 1 ad, guaratees that the sequece (k 0, k 1,, k e ) of the umber of vertices i the distace layers of v is (, κ, e)-realizable. By Lemma, ad lettig e(κ + 1) 41κ 1 =: g, we obtai e e σ G (v) = ik i ib i i=1 i=1 = κ[ κ + (0κ + 1)] + (κ + 1)[(0κ + ) + +(e 0κ)] + κ[(e 0κ + 1) + + (e 1)] + e[ 1 g] = e 1 e (κ + 1) + 1 e (41κ + 1) 40κ 1. (3) A simple differetiatio shows that (3) is maximized for e = + 41κ+1 κ + 1 ad we obtai σ G (v) = ( + 41κ+1 κ ( + 41κ+1 κ (κ + 1) ) 1 ( + 41κ+1 ) (κ + 1) κ + 1 ) (41κ + 1) 40κ 1 ( + (41κ + 1) κ 13 κ ).,

8 Therefore, sice v is a vertex of largest distace, ad summig σ G (v) for all v V (G), we obtai ( σ(g) σ G (v) + (41κ + 1) (κ + 1) 4 κ 13 κ 7 ). 4 Recall that 40κ +. It suffices to show that ( (κ+1) + (41κ + 1) κ 13 κ ) ( ) 7 4 ( 1) (κ+1) + 30, that is, that + (41κ + 1) κ 13 κ 7 4 ( 1)( + 60(κ + 1)). We may equivaletly show that g κ () 0 for all 40κ +, where g κ () = ( 1)( + 60(κ + 1)) ( + (41κ + 1) κ 13 κ 4) 7. But this follows from g κ () = (19κ + 58) κ + 3 κ 33 4 g κ(40κ + ) = 760κ κ + 79 κ > 0, as desired. 4 Extedig the Result Whereas the previous subsectio bouds the average distace of a κ-vertexcoected graph G whose vertex of largest distace v satisfies Assumptios 1 ad, this sectio exteds this boud to ay κ-vertex-coected graph. We will show i the ext subsectio (Lemmas 3 ad 5) that if at least oe of the Assumptios 1 ad is violated with respect to v, the there exists N α, 0κ + 1 α ex G (v) 0κ 1, such that diam(n α ) 9. The distace layer N α is cetral to our proof; it will allow us to boud the average distace roughly as follows: N α separates G ito two subgraphs, G[N α ] ad G[N α ], each of which ca be made κ-vertex-coected by attachig a few vertices ad edges to the vertices i N α without chagig the distaces i G. The fact that distaces i G betwee vertices of N α are small allows us to do this. Let the resultig κ-vertex-coected subgraphs obtaied from G[N α ] ad G[N α ] be G 1 ad G. We will boud σ(g) i terms of upper bouds o σ(g 1 ) ad σ(g ) ad use iductio o the order of G to establish the result. 4.1 Distace Layers of a Arbitrary Vertex Throughout this subsectio, G is a κ-vertex-coected graph, κ 3 is odd, ad v is a fixed vertex (ot ecessarily a cetre vertex) of G. We deote the eccetricity of v by e. Sice at least oe of the Assumptios 1 ad is violated, we cosider itegers l, e 1 where 1 l < l + e 1 e 1, with k l = κ = k l+e1 ad l+e k i (κ + 1)(e 1 1) + 1. We show that, uder this hypothesis, there exists a distace layer N l+a, a {1,..., e 1 1} such that diam(n l+a ) 9. This result, a improvemet o correspodig results i [6, 10, 9], is more geeral as it applies to distace layers of a 8

9 arbitrary vertex ad ot oly of a cetre vertex. We employ the followig otatio. Notatio: l( 1) ad e 1 ( ) are itegers satisfyig l + e 1 e 1 ad k l = κ = k l+e1. Let N l = {u 1, u,..., u κ }, N l+e1 = {w 1, w,..., w κ }. Defie I := {1,,..., κ}. For x N i, T (v, x) deotes a shortest v-x path. The collectio of all shortest v-x paths form the set T S (v, x). For k I defie M k := {x l+e 1 N i there exists T (v, x) T S (v, x) with u k V (T (v, x))}. That is, M k = {x N i d(x, u k ) = i l ad l i l + e 1 }. It is evidet that l+e 1 M k = N i. (4) k I Let G M be the graph whose vertex set V (G M ) is the family of sets {M i i I}, where M t M p E(G M ) if ad oly if d G (M t, M p ) 1, that is, if M t ad M p have a vertex i commo or there exists a edge e = ab E(G), where a M t ad b M p. Propositio G[ l+e 1 N i ] is coected. Proof: Cosider the graph G M. By (4), it is sufficiet to prove that G M is coected. Suppose to the cotrary that G M is discoected. Let C 1 be a compoet of G M with V (C 1 ) = {M i i I 1 I} ad let C be the uio of the remaiig compoets of G M, hece V (C ) = {M i i I I 1 }. Let M = i I 1 M i ad M = i I I 1 M i. The sice o vertex i C 1 is adjacet to a vertex i C, we have from (4), that, i G, l+e 1 N i = M M is a disjoit uio ad E(M, M ) =. A path from N 0 to N e thus cotais vertices from at least two of the sets M N l, M N l+e1, M N l ad M N l+e1. Hece the uio of ay two of the four sets M N l, M N l+e1, M N l, M N l+e1 is a separatig set of G. Sice κ is odd, we have M N l M N l ad M N l+e1 M N l+e1. Assume that, say, M N l < M N l ad M N l+e1 < M N l+e1. (The other cases follow aalogously.) The M N l κ 1 ad M N l+e1 κ 1, hece (M N l ) (M N l+e1 ) is a separatig set of cardiality at most κ 1, a cotradictio. Hece G M is coected. Lemma 3 If e 1 = ad k l+1 κ +, the diam(n l+1 ) 7. Proof: Let S = N l N l+1 N l+. The, by Propositio, G[S] is coected. Now suppose to the cotrary that there exist x, y N l+1 with d G (x, y) 8. The N 3 (x) ad N 3 (y) are disjoit ad o-adjacet. 9

10 We first show that every w N l+1 is i either N (x) or N (y). If this were ot true, the N(x), N(w), N(y) would be pairwise disjoit ad cosequetly, from N(x) N(w) N(y) S, we have κ + (κ + ) + κ k l + k l+1 + k l+ N(x) + N(w) + N(y) (κ + 1) + (κ + 1) + (κ + 1) = 3κ + 3, a cotradictio. It follows that N l+1 = (N (x) N l+1 ) (N (y) N l+1 ). Hece, sice each vertex i N l+ is adjacet to some vertex i N l+1, we have that N l+ = (N 3 (x) N l+ ) (N 3 (y) N l+ ). Sice N l is a miimum separatig set, every vertex t N l has a eighbour i N l+1. Thus N l = (N 3 (x) N l ) (N 3 (y) N l ). We coclude that S = (N 3 (x) S) (N 3 (y) S). Cosequetly it follows from the fact that N 3 (x) ad N 3 (y) are disjoit ad o-adjacet that G[S] is discoected, a cotradictio to Propositio. Lemma 4 Assume that k l+1 + k l+ κ + 3 ad let S = l+ N i. If G[S] is coected, the diam(n l+1 ) 9. Proof: Let B = N l+1 ad A = N l+. The N l + B + A κ + κ + 3 = 3κ + 3. (5) We prove that diam(b) 9 by cotradictio. Suppose that d G (x, y) 10 for some x, y B. The the sets N 4 (x) ad N 4 (y) are disjoit ad oadjacet. We first show that every vertex of B is i N 3 (x) or i N 3 (y). Note that if this were false, the for some w B, the sets N(w), N(x) ad N(y) would be pairwise disjoit ad pairwise o-adjacet. Thus from (5) ad δ(g) κ, 3(κ + 1) N(w) + N(x) + N(y) N l + B + A 3κ + 3. Therefore, we have S = N(w) N(x) N(y) ad so G[S] is discoected with three compoets G[N(w)], G[N(x)] ad G[N(y)], a cotradictio to the hypothesis of the lemma. Thus we have proved that B = (B N 3 (x)) (B N 3 (y)). As i the proof of Lemma 3, we get S = (S N 4 (x)) (S N 4 (y)). Thus, sice N 4 (x) ad N 4 (y) are o-adjacet, G[S] is discoected, a cotradictio. We remark that by the coditio o the miimum degree (of vertices i N l+1 ), it ca easily be verified that if the hypothesis of Lemma 3 (Lemma 4) is satisfied, the N l+1 ca be partitioed ito two oempty sets S 1 ad S with diam(s i ) 3 for i = 1,. I order to prove the mai result of this subsectio, we eed more otatio. Cosider the graph G M above. Let us assig weights to the 10

11 edges of G M. For a edge yz E(G), with y N l+α ad z N l+β, let w(yz) = α + β. Let M t M p be a edge of G M. We first show that there exists ab E(G) with a M t, b M p, (a b). (6) Note that (6) immediately follows by defiitio of adjacecy betwee M t ad M p uless M t M p. If M t M p, let a M t M p. The a N i, l < i l + e 1. Hece, there exists a vertex b N i 1 M t (or b N i 1 M p ) which is adjacet to a as desired. Now let f : E(G M ) Z be defied by f(m t M p ) = mi w(ab). ab E(M t,m p) Let T M be a spaig tree of (G M, f) of miimum total edge weight with respect to f. For every edge M t M p of T M choose a edge ab E(G) with w(ab) = f(m t M p ). Deote the set of these κ 1 edges of G by E 1 = {e 1,..., e κ 1 } ad let e i = a i b i. Further, assume the otatio a i N l+αi ad b i N l+βi for i = 1,,..., κ 1. Let M r M s be a edge of T M with largest weight. Let, without loss of geerality, e 1 = a 1 b 1 E(M r, M s ), with a 1 M r N l+α1 ad b 1 M s N l+β1. Now assume, without loss of geerality, that α 1 β 1. We show that α β 1 for all α {α i, β i i = 1,,..., κ 1}. (7) Cosider α i. Note that α i β i + 1. Let M t M p be the edge of T M that iduces a i b i. The α i 1 α i + β i = w(a i b i ) = f(m t M p ) f(m r M s ) = w(a 1 b 1 ) = α 1 + β 1 β 1. Hece α i β 1. Similarly, β i β 1. Here ad i the sequel, we assume the give otatio for M r, M s, a 1, b 1, β 1, uless otherwise stated. We ow prove the mai result of this subsectio. Lemma 5 If k i κ + 1 for i = l + 1, l +,..., l + e 1 1 ad l+e k i (κ+1)(e 1 1)+1, the there exists a {1,..., e 1 1} such that diam(n l+a ) 9. Moreover, N l+a ca be partitioed ito two oempty sets S 1 ad S with diam(s i ) 3 for i = 1,. Proof: With the otatio above, we cosider β 1. If β 1, the by (7), G[ l+ N i] is coected ad the result follows by a applicatio of Lemma 4. Now assume that β 1 3. By the coditio k i κ + 1 for i = l + 1, l +,..., l + e 1 1 ad l+e k i (κ + 1)(e 1 1) + 1, it follows 11

12 that at most oe of the umbers k l+1, k l+,..., k l+e1 1 is equal to κ +. Let C = N l+β1 1, A = N l+β1 ad B = N l+β1 3. We are goig to show that diam(a) 9. Note first that B + A + C = { κ + (κ + 1) + (κ + 1) + 1 if β1 = 3, (κ + 1) + (κ + 1) + (κ + 1) + 1 if β 1 4 { 3κ + 3 if β1 = 3, 3κ + 4 if β 1 4. (8) Fact 1 l+β1 1 N i is partitioed ito two disjoit o-adjacet sets. Proof of Fact 1: Cosider the tree T M. The T M M r M s has exactly two compoets, C 1 ad C say. Let C 1 (C ) be the compoet that cotais M s (M r ). Let V (C 1 ) = {M i i I 1 }. Put M = i I 1 M i ad M = i I I 1 M i. The ( l+β 1 1 N i = M ( ) ( l+β 1 1 N i ) M ( ) l+β 1 1 N i ). It remais to show that M ( l+β 1 1 N i ) ad M ( l+β 1 1 N i ) are disjoit ad o-adjacet. As i the proof of (6), it is sufficiet to show that there ca be o edge ab E(G) with a M ( l+β 1 1 N i ) ad b M ( l+β 1 1 N i ). Suppose to the cotrary that ab E(G) is such a edge. The there exist M t ad M p such that a M t V (C 1 ) ad b M p V (C ). Thus M t M p E(G M ). Note that w(ab) (β 1 1) < w(a 1 b 1 ), ad hece f(m t M p ) w(ab) < w(a 1 b 1 ) = f(m r M s ). Therefore, the spaig tree of G M, (T M M r M s ) + M t M p has a total edge weight which is less tha the total edge weight of T M, cotradictig the choice of T M. This establishes Fact 1. Fact With the otatio above, { M N l, M N l } = { κ 1, } κ+1. Proof of Fact : If o oe had M N l κ 3, the sice by Fact 1, (M N l ) (C M ) is a separatig set of G, we have that C M κ+3. Further, sice by Fact 1, (M N l ) (C M ) is a separatig set of G, we have that C M κ+3. Thus, by the same fact, C = C M + C M κ κ + 3 = κ + 3, a cotradictio to (8) ad the fact that A κ + 1 ad B κ + 1 if β 1 4. If o the other had, M N l κ+3, the M N l κ κ+3 = κ 3, ad as above, we derive a cotradictio. Therefore, κ 3 < M N l < κ+3, i.e., M N l { κ 1, } κ+1. This, i cojuctio with Nl = κ ad Fact 1, yields Fact. Fact ad the fact that (M N l ) (C M ), (M N l ) (C M ), 1

13 (M N l ) (C M ) ad (M N l ) (C M ) are separatig sets of G, give Fact 3 C M κ+1 ad C M κ+1. Now assume that κ 5. (For κ = 3 aalogous cosideratios yield a similar result.) We first obtai a upper boud o B M + A M + C M. As i Fact 3, if β 1 4, the A M κ+1 ad A M κ+1, whereas sice (M N l ) (B M ) ad (M N l ) (B M ) are separatig sets, by Fact, B M κ 1 ad B M κ 1. If, however, β 1 = 3, the sice for D {B, A}, (M N l ) (D M ) ad (M N l ) (D M ) are separatig sets, we have D M κ 1 ad D M κ 1. We have established B M + A M + C M hece, from (8) ad Fact 1, we get B M + A M + C M = κ 1 + κ 1 + κ+1 if β 1 = 3, κ 1 + κ+1 + κ+1 if β 1 4 3κ 1 if β 1 = 3, 3κ+1 if β 1 4, Similarly, B M + A M + C M 3κ+7. 3κ + 3 3κ 1 if β 1 = 3, 3κ + 4 3κ+1 if β 1 4 = 3κ + 7. (9) Fact 4 (a) diam(a M ), (b) diam(a M ). Proof of Fact 4: We show that (a) holds as (b) follows aalogously. Suppose that (a) is false ad assume that d(x, y) 3 for some x, y A M. The N(x) ad N(y) are disjoit. Sice by Fact 1, x (y) ca oly be adjacet to elemets i (B C A) M, we have from (9) that (κ + 1) + (κ + 1) N(x) + N(y) B M + A M + C M 3κ + 7, i.e., κ 3, a cotradictio to κ 5. This establishes (a). We ca ow complete the proof of Lemma 5. Let h A M, z 13

14 A M. Recall that a 1 M r M ad b 1 M s M. Let a V (T (u r, a 1 )) A ad b V (T (u s, b 1 )) A. The by Fact 4, d(h, a ) ad d(z, b ). Hece d(h, z) d(h, a ) + d(a, a 1 ) + d(a 1, b 1 ) + d(b 1, b ) + d(b, z) = 9, as desired. This, i cojuctio with Fact 4, establishes diam(a) 9 with S 1 = A M ad S = A M. The graph G s,t, defied below, will play a importat role i the proof of our mai result. Its distace is computed below. Defiitio For positive itegers κ, s, t the graph G s,t is the sequetial joi G s,t = 8 i=1 H i where H 1 = K sκ, H 8 = K tκ ad, for i =, 3,, 7, H i = K κ. Lemma 6 Let κ, s, t, c be itegers, s, t 1, satisfyig s + t = κ + c. The σ(g s,t ) = κ[κ 3 + 4κ + κ c + κc + 4κc + 75κ c 6] + 1κ ts, ad moreover, σ(g s,t ) κ[4κ 3 + 4κ + 8κ c + 4κc + 4κc + 75κ c 6]. Proof: A tedious but straight forward calculatio yields the exact value ( ) ( ) ( ) s t of σ(g s,t ). Now 0 = s+t st, hece st s+t = ( ). κ+c This, substituted i the exact value of σ(gs,t ), implies the secod iequality. 4. A Upper Boud o the Average Distace We ow state ad prove the mai result of this paper. Theorem 1 Let G be a κ-vertex-coected graph, κ 3 odd, of order. The µ(g) (10) (κ + 1) Moreover, apart from a additive costat, this iequality is best possible. Proof: We prove the equivalet statemet ( ) σ(g) ( 1) (κ + 1) + 30 by iductio o the order of G. For small values of say < 0, the result follows by the secod part of Lemma 1. (Recall that 0 was defied 14

15 as 59κ(κ + 1).) So assume that 0 ad that the result holds for ay κ-vertex-coected graph, κ 3 odd, of order less tha. Let v be a vertex of G of largest distace. If v satisfies Assumptios 1 ad of the previous sectio, the the result follows by Propositio 1. So assume that at least oe of the two assumptios is violated. The by Lemmas 3 ad 5, there exists a distace layer N α, 0κ + 1 α ex G (v) 0κ 1, such that diam(n α ) 9, κ k α κ+ ad further N α = S 1 S, where S 1, S ad diam(s 1 ), diam(s ) 3. Let S 1 = {u 1, u,, u s }, S = {v 1, v,, v t }. Cosider the graph G s,t (see Defiitio ). Form a ew graph H from the disjoit uio of G ad G s,t by attachig G s,t to N α as follows. Partitio the vertices of H 1 = K sκ (H 8 = K tκ ) ito s (t) disjoit subsets U 1, U,, U s (W 1, W,, W t ) of cardiality κ each. The for each vertex u i S 1 (v j S ) joi u i (v j ) to every vertex i U i (W j ). Formally, H is the graph with vertex set V (H) = V (G) V (G s,t ), ad edge set E(H) = E(G) E(G s,t ) {u i x x U i, i = 1,, s} {v i x x W i, i = 1,, t}. Deote the vertex set of G s,t by A. The order of G s,t is A = κs + 6κ + κt = κ(s + t + 6) = κ(k α + 6). Recall that N α = {x V (G) d G (x, v) α}, N α = α i N i ad let G 1 = H[N α A], G = H[N α A], ad i = V (G i ), i = 1,. The followig claim is easy to verify. Claim 1 G 1 ad G are κ-vertex-coected. Let N α = k α = κ + c, where c {0, 1, }. The Claim (a) 1 + = + κ + 13κ + c(κ + 1). (b) 1κ + 7κ + c(κ + 1) + 1 i 19κ + 6κ + cκ 1 for i = 1,. Proof of Claim : Cosider the graph H. Note that the oly vertices i commo betwee G 1 ad G are vertices i A N α. Therefore, V (G 1 ) + V (G ) A N α = V (H). Recall that V (H) = V (G) + V (G s,t ) = + κ(k α + 6). Thus 1 + κ(k α + 6) k α = + κ(k α + 6), hece from k α = κ + c we obtai (a). Also, 1 0κ k i + k α + A i=0 1 + κ(0κ) + κ + c + κ(κ + c + 6) = 1κ + 7κ + c(κ + 1) + 1. Similarly, 1κ + 7κ + c(κ + 1) + 1. I cojuctio with (a), we obtai (b). 15

16 Claim 3 For all x, y V (G), d G (x, y) = d H (x, y) that is, G is isometric i H. Proof of Claim 3: Let E 1 = E(H) E(G). First ote that by our costructio of H, for x, y S 1 (S ), d H[E1](x, y) = 3 ad for x S 1, y S, d H[E1](x, y) = 9. To prove the claim, it is sufficiet to show that for all x, y V (G) V (H[E 1 ]), d G (x, y) d H[E1](x, y). Now V (G) V (H[E 1 ]) = N α = S 1 S. If x, y S 1 (x, y S ), it follows from diam(s 1 ) 3 (diam(s ) 3) that d G (x, y) 3 = d H[E1](x, y). Also, if x S 1 ad y S, it follows from diam(n α ) 9 that d G (x, y) 9 = d H[E1](x, y), as desired. By Claim 3 we ca verify the formula Claim 4 Let X = x A σ G 1 (x), Y = x A σ G (x) ad T = x N <α y N >α d G (x, y). The σ(g) = σ(g 1 ) + σ(g ) (X + Y ) + σ A (A) + T (x,y) N α N α d H (x, y). We use Claim 4 to obtai a upper boud o σ(g) by fidig correspodig upper bouds o the positive terms ad correspodig lower bouds o the egative terms. Note that σ A (A) σ(g s,t ). It follows by Lemma 6, that Also σ A (A) κ[4κ 3 + 4κ + 8κ c + 4κc + 4κc + 75κ c 6]. (11) (x,y) N α N α d H (x, y) k α (k α 1) = (κ + c)(κ + c 1). (1) We ow fid a upper boud o T usig a vertex of N α, u 1 say: T = d G (x, y) x N <α y N >α [d G (x, u 1 ) + d G (u 1, y)] x N <α y N >α = N >α d G (x, u 1 ) + [σ G (u 1 ) σ Nα A(u 1 )] x N <α = N >α [σ G1 (u 1 ) σ Nα A(u 1 )] + N <α [σ G (u 1 ) σ Nα A(u 1 )]. The distace betwee u 1 ad each other vertex of N α is at least 1. Hece σ Nα A(u 1 ) 1(κ + c 1 + κ) + (tκ + κ + (s 1)κ) + 3(κ) + 4(κ) + 5κ = 1κ + κ(t + s) + c 1 = κ + 1κ + c(κ + 1) 1. 16

17 Therefore, T [ (κ + 7κ + c(κ + 1))][σ G1 (u 1 ) (κ + 1κ + c(κ + 1) 1)] +[ 1 (κ + 7κ + c(κ + 1))][σ G (u 1 ) (κ + 1κ + c(κ + 1) 1)]. We ow fid a lower boud o X i terms of σ G1 (u 1 ). Recall that X = x A σ G 1 (x) ad i particular A = 8 i=1 V (H i), where V (H 1 ) = s i=1 U i. For each u U 1, we have, by the triagle iequality σ G1 (u 1 ) [d(u 1, u) + d(u, x)] = [1 + d(u, x)] = 1 + σ G1 (u), x V (G 1) x V (G 1) hece σ G1 (u) σ G1 (u 1 ) 1. Similarly for u s i= U i, because d H (u 1, u) =, we have σ G1 (u) σ G1 (u 1 ) 1. Thus σ G1 (V (H 1 )) κ[σ G1 (u 1 ) 1 ] + (s 1)κ[σ G1 (u 1 ) 1 ] = κ[sσ G1 (u 1 ) (s 1) 1 ]. For i =, 3,, 7, we have σ G1 (V (H i )) κ[σ G1 (u 1 ) i 1 ] ad σ G1 (V (H 8 )) tκ[σ G1 (u 1 ) 8 1 ]. Summig for i = 1,, 8 yields X κ[(s + t + 6)σ G1 (u 1 ) 1 (6 + (s + t) + 6t)] = (κ(κ + c + 6))σ G1 (u 1 ) κ 1 ((κ + c) t). Sice s + t = κ + c ad 1 s, we obtai Aalogously, Thus X (κ + 6κ + cκ)σ G1 (u 1 ) 1 (8κ + 8κc + 0κ). Y (κ + 6κ + cκ)σ G (u 1 ) (8κ + 8κc + 0κ). (X +Y ) (κ +6κ+cκ)[σ G1 (u 1 )+σ G (u 1 )] (8κ +8κc+0κ)[ 1 + ]. (14) Combiig Claim 4 ad (11)-(14), we obtai σ(g) σ(g 1 ) + σ(g ) + f 1 σ G1 (u 1 ) + f σ G (u 1 ) + g, where f 1, f ad g are the fuctios, (13) f 1 = ( κ cκ 13κ c), f = ( 1 κ cκ 13κ c), 17

18 g = [1κ + 1cκ κ c + ]( 1 + ) + 16κ 4 + 3cκ 3 + 4κ 3 +16c κ + 36cκ + 733κ + 1c κ + 104cκ 39κ + 3c 3c. By Claim 1 ad Claim, we ca use our iductio to boud σ(g 1 ) ad σ(g ). Lemma 1 (i) gives us upper bouds o σ G1 (u 1 ) ad o σ G (u 1 ). By Claim (b), we have f 1, f > 0. Thus ( ) ( ) 1 σ(g) 1 ( 1 1) (κ + 1) ( 1) (κ + 1) + 30 = ( 1 +f 1 8κ ( 1 + κ ) 1 8κ(κ + 1) f ( 1, ), ) + f ( 1 8κ ( + κ ) ) + g where f ( 1, ) is the fuctio, f ( 1, ) = 4κ 3 + (8κ κ 3 16cκ + 10κ 4cκ + 13κ 8c)+ (8κ κ 1 16κ κ cκ 3 6κ cκ 110κ + 4cκ 16κ + 16c + 8) 4κ ( 16κ 3 16cκ + 10κ 4cκ + 13κ 8c) + 1 (80κ cκ 3 6κ cκ 110κ + 4cκ 16κ + 16c + 8)+18κ cκ κ c κ cκ κ 4 + 4c κ cκ κ c κ + 80cκ 344κ + 4c κ 56cκ 08κ 16c. We use Claim (a) to elimiate from f. Deote the result by h ( 1 ). The h ( 1 ) = 1(4κ 8 4κ 3 4cκ + 60κ 60cκ + 176κ 4c + 3) + 1 (8 4κ + 16κ cκ 96κ + 7cκ 7κ + 3c κ 5 +96cκ 4 08κ 4 +48c κ 3 64cκ 3 373κ c κ +16cκ 35κ + 108c κ+7cκ 416κ+4c 3c)+4κ 3 + (8κ 3 +8cκ +76κ 1cκ+ 13κ 8c)+( 16κ 5 3cκ κ 4 16c κ cκ κ 3 80c κ + 51cκ + 33κ 68c κ + 80cκ 16κ 16c + 16c + 8) + p c (κ), where p c (κ) is a polyomial i κ, c of degree 7, p c (κ) = 3κ 7 96cκ κ 6 96c κ cκ κ 5 3c 3 κ c κ cκ κ 4 11c 3 κ c κ cκ κ 3 10c 3 κ 1c κ + 310cκ 536κ 5c 3 κ + 4c κ + 15cκ 104κ 8c c 8c. The boud (10) is established if we show that ( ) 1 8κ(κ + 1) h ( 1 ) ( 1) (κ + 1) + 30, i.e., if we show that h ( 1 ) 4κ( 1)( + 60κ + 60), 18

19 or equivaletly that the fuctio t ( 1 ) := 4κ( 1)(+60κ+60) h ( 1 ) is at least 0. For costat, t ( 1 ) = 4κ( 1)( + 60κ + 60) h ( 1 ) gives a cocave dow parabola. So its miimum o a iterval is attaied at a ed poit. Recall from Claim (b) that we miimize t subject to 1κ + 7κ + c(κ + 1) κ + 6κ + cκ 1. It is sufficiet to show that t 0 at 1 = 1κ + 7κ + c(κ + 1) + 1, 19κ + 6κ + cκ 1. Thus we arrive at (10) by the followig claim whose proof will appear at the ed of this sectio. Claim 5 With the otatio above t (1κ + 7κ + c(κ + 1) + 1) = t ( 19κ + 6κ + cκ 1) 0. It remais to show that, apart from the additive costat, the iequality (10) is best possible. For positive itegers, k, κ, with κ odd, ad = k (κ+1), let G be the compositio C k [K κ+1 ]. Clearly, G is κ-vertexcoected ad µ(g ) = 1 ( ( 1) σ(g ) ) = (κ+1) + O(1), as desired. We ow prove the claim made i the above theorem. Proof of Claim 5: Note that t (1κ + 7κ + c(κ + 1) + 1) = t ( 19κ + 6κ + cκ 1). Deote this term by ϕ(). The ϕ() = (76κ 3 4cκ 176κ + 8cκ + 5κ 8) (084κ cκ κ 4 + 4c κ cκ κ 3 + 8c κ + 95cκ κ + 4c κ + 15cκ 180κ + 8c 3) + q c (κ), where q c (κ) = 9608κ cκ κ 6 +4c κ cκ κ 5 +8c 3 κ 4 76c κ 4 880cκ κ 4 + 8c 3 κ c κ cκ 3 666κ c 3 κ + 8c κ 46cκ + 164κ + 8c 3 κ 56c κ 5cκ 104κ + 8c 3 + 8c 3. We have to show that ϕ() is at least 0 for 0 = 59κ(κ + 1) ad c {0, 1, }. We prove that ϕ() 0 by showig that ϕ( 0 ) > 0 ad that the gradiet of ϕ is positive. Now, ϕ( 0 ) = 15108κ cκ κ 6 1c κ cκ κ 5 + 8c 3 κ 4 784c κ cκ κ 4 +8c 3 κ c κ cκ κ c 3 κ 1860c κ 11430cκ 15176κ + 8c 3 κ 58c κ 5cκ κ + 8c 3 +8c κ ()κ κ 6 1() κ ()κ κ 5 + 8(0) 3 κ 4 784() κ ()κ κ 4 19

20 +8(0) 3 κ () κ ()κ κ (0) 3 κ 1860() κ 11430()κ 15176κ + 8(0) 3 κ 58() κ 5()κ κ + 8(0) 3 + 8(0) 3 = 15108κ κ κ κ κ κ 43κ 3 (15108κ )κ 6 ( )κ 5. Now, for κ 3, we have (15108κ )κ Hece the last term is positive. It remais to show that dϕ dϕ d > 0. Note that d = (76κ3 4cκ 176κ + 8cκ + 5κ 8) (084κ cκ κ 4 + 4c κ cκ κ 3 + 8c κ + 95cκ κ + 4c κ + 15cκ 180κ + 8c 3). Note that sice 0 c, we have 76κ 3 4cκ 176κ + 8cκ + 5κ 8 76κ 3 4()κ 176κ + 8(0)κ + 5κ 8 = 76κ 3 184κ + 5κ 8 > 0 for all κ 3. Hece, dϕ d dϕ d ( 0) = 6884κ 5 960cκ 4 411κ 4 4c κ cκ κ 3 8c κ 8cκ κ 4c κ 15cκ 764κ 8c κ 5 960()κ 4 411κ 4 4() κ ()κ κ 3 8() κ 8()κ κ 4() κ 15()κ 764κ 8() + 3 = κ(344κ κ 3 889κ κ 58) = κ 3 (344κ 3016κ 889) + κ(1863κ 58). Now, for κ 3, the last term is positive ad this completes the proof of the claim. 5 A Upper Boud o the Radius I this sectio, we show that our lemmas o distace layers proved i Subsectio 4.1 ca be used to deduce a theorem essetially equivalet to the boud o the radius of graphs of give order ad coectivity by Egawa ad Ioue [6]. Let z be a fixed cetre vertex of G, ad ex G (z) = r = rad(g) (= mi{ex G (v) : v V (G)}). For each i = 0, 1,..., r, N i := {x V (G) d G (x, z) = i} ad k i = N i. Sice N r, choose a fixed vertex z r N r ad let T (z, z r ) = zz 1 z z r be a shortest z-z r path. For a vertex x of G, we also employ the otatio T (z, x) for a shortest z-x path. Defiitio 3 Assume the otatio above ad let r 0. Let l {9, 10,..., r 10}. We say (N l, N l+1 ) is a (κ) z -class if (k l, k l+1 ) = (κ, κ). 0

21 Lemma 7 Let r 0 ad (N l, N l+1 ) a (κ) z -class. If 10 l r 11, the k l 1 κ + 1 ad k l+ κ + 1. Proof: Suppose to the cotrary that k l 1 = κ (respectively k l+ = κ). The sice (k l, k l+1 ) = (κ, κ), ad k l κ+ (respectively k l+ κ+), by Lemma 3, diam(n α ) 7 < 9 for some α {10, 11,..., r 10}. Assume the otatio for z r, T (z, z r ) ad T (z, x) give above. We show that ex G (z 10 ) < r, which cotradicts the fact that rad(g) = r. For x N r 11, we have d(z 10, x) d(z 10, z) + d(z, x) 10 + r 11 = r 1 < r. If x N r 10, let z α V (T (z, x)) N α. The sice diam(n α ) 9, d(z 10, x) d(z 10, z α ) + d(z α, z α) + d(z α, x) (α 10) (r α) = r 1 < r, as desired. Deote by Υ p,q the sum q i=p k i. Lemma 8 Let r 0, l( 9) ad e 1 ( ) with l + e 1 r 9 be itegers that satisfy k l = κ = k l+e1, k i κ + 1 for i = l + 1, l +,..., l + e 1 1. The Υ l+1,l+e1 (κ + 1)e Proof: Suppose to the cotrary that Υ l+1,l+e1 1 (κ+1)(e 1 1)+1. The by Lemma 5, there exists N α, 10 α r 10, such that diam(n α ) 9. Hece, as i the proof of Lemma 7, ex G (z 10 ) < r, a cotradictio. Lemmas 7 ad 8 imply the followig lemma which establishes the lower boud o the umber of vertices betwee ay two cosecutive (κ) z -classes. Lemma 9 Let (N l, N l+1 ) ad (N L, N L+1 ) be ay two cosecutive (κ) z - classes. The Υ l+1,l (κ + 1)(L l). Cosiderig Lemmas 7-9 ad after log but straightforward calculatios, we obtai that = rad(g) i=0 k i + 16κ + (κ + 1)(rad(G) 17) ad upo rearragig we arrive at the first part of the followig theorem. Theorem Let G be a κ-vertex-coected graph, κ 3 odd, of order. The rad(g) κ κ + 1. Apart from a additive costat, the above boud is best possible. 1

22 To see that the give boud is, apart from a additive costat, best possible, for positive itegers, k, κ, with κ odd, ad = k (κ + 1), let G be the compositio C k [K κ+1 ]. Clearly, G is κ-vertex-coected ad the radius of G is k = κ+1. We remark that Egawa ad Ioue [6] proved the asymptotically equivalet boud rad(g) κ κ+1 o the radius of a κ-vertex-coected graph G, κ 3 odd, of order. Refereces [1] P. Dakelma ad R.C. Etriger, Average distace, miimum degree ad spaig trees. J. Graph Theory 33 (000), [] P. Dakelma, S. Mukwembi ad H. C. Swart, Average Distace ad Edge-Coectivity I. (Submitted). [3] P. Dakelma, S. Mukwembi ad H. C. Swart, Average Distace ad Edge-Coectivity II. (Submitted). [4] E. DeLaVi a ad B. Waller, Spaig trees with may leaves ad average distace. Preprit. [5] J. K. Doyle ad J. E. Graver, Mea distace i a graph. Discrete Math. 7 (1977), [6] Y. Egawa ad K. Ioue, Radius of (k 1)-coected graphs, Ars Combi. 51 (1999), [7] R. C. Etriger, D. E. Jackso ad D. A. Syder, Distace i graphs, Czech Math J. 6 (1976), [8] O. Favaro, M. Kouider ad M. Mahéo, Edge-vulerability ad mea distace. Networks 19 (1989), [9] J. Harat, A upper boud for the radius of a 3-coected graph. Discrete Math. 1 (1993), [10] J. Harat ad H. Walther, O the radius of graphs. J. Combi. Theory Ser. B 30 (1981), [11] L. Lovasz, Combiatorial problems ad exercises. Akadémiai Kiadó, Budapest (1979). [1] J. Plesík, O the sum of all distaces i a graph or digraph. J. Graph Theory 8 (1984), 1-4.