The number of edges in a bipartite graph of given radius

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1 The number of edges in a bipartite graph of given radius P. Dankelmann, Henda C. Swart, P. van den Berg University of KwaZulu-Natal, Durban, South Africa Abstract Vizing established an upper bound on the size of a graph of given order and radius. We find a sharp upper bound on the size of a bipartite graph of given order and radius. 1 Introduction Let G be a connected graph with vertex set V (G) and edge set E(G). The order of G is n = V (G), and the size is m = E(G). The distance d G (u, v) between two vertices u and v is the number of edges in a shortest u v path in G. The eccentricity e G (v) of v is the distance from v to a vertex farthest away from it in G. The radius of G, rad(g), is the minimum eccentricity of G, that is rad(g) = min v V (G) e G (v), and the diameter of G, diam(g), is the maximum eccentricity of G, that is diam(g) = max v V (G) e G (v). Several bounds on the radius in terms of other graph parameters are known. Erdös, Pach, Pollack and Tuza [4] proved that if G is a connected graph of order n and minimum degree δ, then rad(g) 3(n 3) (δ + 1) + 5, and also constructed graphs that, apart from the additive constant, attain the bound and, moreover, they gave improved bounds for K 3 -free and C 4 -free Support by the South African National Research Foundation is gratefully acknowledged. 1

2 graphs. Using different methods, Dankelmann, Dlamini and Swart [1,, 3] obtained slight improvements of their bounds. Dankelmann, Mukwembi and Swart [9] proved that if G is a 3 edgeconnected graph of order n, then rad(g) 1 3 n In [10], Mukwembi proved that if G is also bipartite, then rad(g) 3 10 n , and both bounds are sharp, apart from an additive constant. Definition 1 Let n and r be any natural numbers such that n r. Define f(n, r) to be the maximum number of edges in a a graph of order n and radius r, and C(n, r) to be the set of all graphs of order n, radius r, and size f(n, r). Vizing [13] gave the following bound on the size of a connected graph in terms of order and radius. Theorem 1 [13] For any natural numbers n and r such that n r, a) f(n, 1) = 1 n(n 1) b) f(n, ) = 1n(n 1) 1n = 1 n(n ) c) f(n, r) = 1 (n 4rn + 5n + 4r 6r) for n r 6. The graph with radius 1 and the maximum number of edges is the complete graph. C(n, ) consists of all graphs obtained from K n by removing 1 n edges covering V (K n). Examples for graphs in C(n, r), n r 6, consist of a complete graph K n r and a cycle C r, where every vertex of the K n r is joined to the same three consecutive vertices of C r. The goal of this paper is to establish a similar sharp upper bound on the size of a connected, bipartite graph of given radius and order (see Theorem ), and to determine all graphs attaining that bound. The notation we use is as follows. The degree of a vertex v of G, denoted by deg G (v), is the number of vertices adjacent to v. The maximum degree

3 and the minimum degree of G are denoted by (G) and δ(g), respectively. The neighbourhood N G (v) of a vertex v is the set of vertices adjacent to v in G. A set S of vertices is called a cutset if its deletion increases the number of components. A vertex v is called a cut-vertex if {v} is a cutset, and a non-cut vertex or ncv otherwise. A vertex x is said to be separated from a vertex y by a vertex v if v lies on every x y path (i.e., if x and y are in different components of G v). If S V (G), then S G denotes the subgraph of G induced by S. When the graph is understood, then we sometimes drop the argument or subscript G. The join G 1 + G of two vertex disjoint graphs G 1 and G is the graph consisting of the union G 1 G, together with all edges of the type xy, where x V (G 1 ) and y V (G ). For k 3 vertex disjoint graphs G 1, G,..., G k, the sequential join G 1 + G G k is the graph (G 1 + G ) (G + G 3 )... (G k 1 + G k ). The sequential join of k disjoint copies of a graph G will be denoted by [k]g, the union of k disjoint copies of G will be denoted by kg, while [k 1 ]G 1 + G + [k 3 ]G 3 will denote the sequential join G 1 + G G 1 + G + G 3 + G G 3. We write K n and C n for the complete graph and the cycle of order n, respectively. A vertex c of G is called central if e G (c) = rad(g). The centre C(G) is the set of all central vertices in G. An eccentric vertex of a vertex v is a vertex farthest away from v. If there is only one such vertex u, then u is called the unique eccentric point (or uep) of v. A conjugate vertex v of a vertex v is a central vertex which has v as its uep. (So a vertex might have more than one conjugate vertex, or none.) A conjugate pair is a pair of central vertices, each of which is the uep of the other. A spanning tree T of G is said to be radius-preserving if rad(g) = rad(t ). We define a non-trivial graph G to be vertex-radius-decreasing if rad(g v) < rad(g) for every ncv v of G. A graph G is called edge-radius-decreasing or erd if rad(g + e) < rad(g) for every e / E(G). Clearly, the graphs described after Theorem 1 are erd. Erd graphs have been studied by Nishanov [11, 1], Harary and Thomassen [8] and Gliviak, Knor and Soltés [7], but no simple characterizaton is known. Preliminary Results Definition The set B(n, r) consists of all graphs G obtained from C r with three consecutive vertices replaced by ak 1, bk 1, ck 1, where a + c = n r+3, b = n r+3, or a + c = n r+3, b = n r+3. We use the 3

4 ak 1 bk 1 ck 1 Figure 1: An example of a graph in B(n, r). notation V 1(G) = V (ak 1 ck 1 ) and V (G) = V (bk 1 ). (See Figure 1.) Let h(n, r) := n 4 nr + r + n r for n r 8. In the proof of Lemma 1, we denote, for a vertex v of G, the star induced by v and its neighbours by S G (v). Lemma 1 Let G be a connected bipartite graph of order n and radius at least r 4. If u, v V (G) with d(u, v), then deg G (u) + deg G (v) n r + 4. If deg G (u) + deg G (v) = n r + 4 then m(g) h(n, r). If deg G (u) + deg G (v) = n r + 4 and m(g) = h(n, r), then G is one of the graphs in the family B(n, r). Proof Let F be the union of the two stars S G (u) and S G (v). Since u and v have no common neighbours, F contains no cycle. Hence there exists a spanning tree T of G containing F. Let P be a diametral path of T. By rad(t ) rad(g) r, we have diam(t ) r 1; so P has at least r vertices. Since P contains at most two neighbours of u and v, respectively, we have Hence V (T ) V (P ) deg T (u) + deg T (v) 4 = deg G (u) + deg G (v) 4. deg G (u) + deg G (v) n V (P ) + 4 n r + 4, as desired. Now assume that deg G (u) + deg G (v) = n r + 4. Then P has exactly r vertices, say, P = w 0, w 1,... w r 1, rad(t ) = r, and u and v are internal 4

5 vertices of P, say u = w a and v = w b ; where (say) a < b. Moreover, T has the following properties: (a) each vertex not on a diametral path is an end-vertex of T and adjacent to u or to v, (b) all vertices other than u or v have degree at most in T. To see that these two properties hold observe that, if one of them is violated, then a diametral path of T misses more than deg G (u) + deg G (v) 4 vertices, and thus has fewer than r vertices, hence T has radius less than r, a contradiction. It is clear that every spanning tree of G containing F has properties (a) and (b). We can choose T to also have the property of preserving the distance between u and v. This can be achieved by considering the union F of F and a u v geodesic in G. Clearly F is a (not necessarily spanning) subtree of G, so there exists a spanning tree T of G containing F which has the desired property. We now consider which edges G can contain, in addition to those of T. We show that, if e E(G) E(T ), then either (i) e = w 0 w r 1, or (ii) e joins a vertex in N G (u) to a vertex in N G (v), or (iii) e = xw a+ or e = xw a for some vertex x N G (w a ) V (P ), or (iv) e = xw b+ or e = xw b for some vertex x N G (w b ) V (P ). Note that the indices are taken modulo r, so if a = r then a vertex x N G (w a ) can be joined to w 0. First assume that e joins two vertices of P. Suppose that e = w i w j with w i w j w 0 w r 1. Then at least one of the end points of e, say w i, has degree at least 3 in T + e. Let w i be such a vertex. Clearly, e is not incident with u or v since u and v have the same degree in G and in T, so w i w a, w b. Consider the union of three stars S G (u), S G (v) and S T +e (w i ), which we denote by F 1. First we show that F 1 contains a cycle. Suppose to the contrary that F 1 is a forest. Then there exists a spanning tree T 1 of G containing F 1. In T 1, vertices u and v have degree deg G (u) + deg G (v), respectively, but v i has degree at least 3, so T 1 does not have property (b), a contradiction. This shows that F 1 contains a cycle C 1. Clearly, C 1 must contain w i and either w a and its two neighbours on P or w b and its two neighbours on P. Without loss of generality, we assume the former, so C 1 contains w a, w a+1, w i, w a 1. So i = a + and e = w a 1 w a+ or i = a and e = w a w a+1. If e = w a 1 w a+ consider the tree T = T w a+1 w a+ + w a 1 w a+. Clearly, u and v have full degree in T, but w a 1 has degree 3, contradicting property (b). Similarly, if 5

6 e = w a w a+1 the tree T = T w a w a 1 +w a w a+1 does not have property (b), a contradiction. Hence w 0 w r 1 is the only edge between two vertices of P present in G but not in T. Now let e E(G) E(T ) be an edge joining a vertex x N G (w a ) V (P ) to a vertex w i on P. Suppose that e is not of type (iii), i.e., that i a, a+. Then either i a + 4 or i a 4. (Note that in this part of the proof, subscripts are not taken modulo r.) Case 1: w i is not a neighbour of w b on P. So i b 1, b + 1. If i a + 3 consider the graph T + xw i, which has the unique cycle w a w a+1 w a+..., w i xw a. Clearly, all edges in the set E := {w a+1 w a+, w a+ w a+3,..., w i w i 1 } are on this cycle, so T +xw i e =: T (e ) is a spanning tree of G for all e E. Since vertex w i has degree 3 in T (e ), and vertex w a has full degree, property (b) implies that in T (e ) vertex w b does not have full degree. So each edge in E is incident with vertex w b. Since only two edges of E can be incident with w b, we have E = {w a+1 w a+, w a+ w a+3 } and w b = w a+. But then w a and w b are at distance, contradicting our hypothesis. If i a 3 then similar arguments lead to the same conclusion. Case : w i is a neighbour of w b on P. So i = b 1 or i = b + 1. Then w a xw i w b is a (w a w b )-path of length 3, so w a and w b are at distance 1 or 3 in T (and in G). First consider the case that w a and w b are at distance 1, so b = a + 1. Then i = b + 1 (since i = b 1 = a is not possible) and thus i = a + ; so e = xw a+, as desired. Now consider the case that w a and w b are at distance 3; hence b = a + 3. But then i {b 1, b + 1} = {a +, a + 4}. If i = a + then e = xw a+, so e is of type (iii). That leaves the case i = b + 1 = a + 4. We show that a + 4 = r 1, i.e., that w a+4 is an end-vertex of P. Suppose to the contrary that r 1 > a + 4. In the tree T w a+1 w a+ + xw a+4 =: T, vertices u and v have full degree and vertex w a+4 has three neighbours, contradicting property (b). Hence a + 4 = r 1. We now show that not all vertices in N G (w b ) are adjacent to a vertex in N G (w a ). Suppose to the contrary that each vertex y N G (w b ) has a neighbour y N G (w a ). Then we can reduce the distance from w a to the end-vertices in N T (w b ) as follows. Consider the tree T = T {yw b y N T (w b ), y w b 1 } + {yy y N T (w b ), y w b 1 }. Since every end-vertex of T, except possibly w 0, is within distance 3 of w a, the distance from w 0 to any end-vertex of T is at most d T (w 0, w a ) + 3 = 6

7 r, while any two end-vertices of T, other than w 0, are within distance at most 5. Hence the diameter of T is at most r 1, which implies rad(t ) r 1, a contradiction to rad(g) r. This proves that there exist a vertex y N G (w b ) not adjacent to any vertex in N G (w a ). Hence, we can obtain, if necessary by renaming y and w r 1, that no vertex in N G (w a ) is adjacent to to vertex w a+4. Hence property (iv) holds. We now show that in addition to properties (i)-(iv) the following holds: (v) if x N G (w a ), then at most one of the edges xw a, xw a+ is present in G, (vi) if y N G (w b ), then at most one of the edges xw b, xw b+ is present in G, (vii) if xy E(G) for some x N G (w a ), y N G (w b ), then b = a + 1 or b = a + 3. To prove (v), suppose that a vertex x N G (a) is adjacent to w a and to w a+. Then the tree T := T {w a w a 1, w a+1 w a+ } + {xw a, xw a+ } preserves the degrees of w a and w b, but has another vertex, namely x of degree 3. This contradicts property (b), and so (v) holds. Similarly, (vi) holds. Property (vii) follows directly from the fact that T preserves the distance between w a and w b in G. Now the bound on the size of G follows easily. In addition to the edges of T, G can only have edges satisfying (i)-(vii). There is only one edge satisfying (i), namely the edge w 0 w r 1. The graph G has at most (deg G (w a ) )(deg G (w b ) ) (n r) edges of the form xy, where x N 4 G (w a ) V (P ) and y N G (w b ) V (P ), that are not in T. Finally, each vertex not on P has at most one edge, not in T joining it to a vertex on P. Hence m(g) m(t ) (deg G (w a ) )(deg G (w b ) ) + (n V (P ) ) n + (n r) + n r 4 = h(n, r), as desired. From the above proof it follows that, if m(g) = h(n, r), then (N G (w a ) N G (w b )) V (P ) G is a balanced, complete bipartite graph of order n r, w 0 w r 1 E(G) and every vertex in N G (w a ) V (P ) (or in N G (w b ) V (P )) is adjacent to either w a+ or w a (or to either w b or w b+, respectively.) We show next that if x N G (w a ) V (P ) and y N G (w b ) V (P ), then it is impossible that both xw a and yw b+ are edges in G. Suppose to the 7

8 contrary that xw a, yw b+ E(G). Then b = a + 3 as otherwise rad(g) < r and consider the spanning tree T of G, where T =: T {w b+1 w b+, w a+1 w a+, w a 1 w a } + {yw b+, xy, xw a }. In T the vertices w a and w b have full degree, while x and y are both of degree 3, which contradicts (b). Consequently, it follows that G B(n, r). We now present propositions that will be needed in the proof of our main result. Proposition 1 [13] For any connected graph G of order n, (G) n rad(g) +. Definition 3 Given integers n, d with 3 d n, define a path-complete bipartite graph as follows: G(n, d) = [d 1 t]k 1 + n d + 1 where 1 t d. K1 + n d + 1 K1 + [t]k 1, Proposition [3] Let G be a bipartite graph of order n and diameter d 3. Then n m(g) 4 nd + 3n + d 4 d 7, 4 with equality if and only if G is a path-complete bipartite graph G(n, d). Proposition 3 [5] Let {v, v } be any conjugate pair in a graph G = K. If G {v, v } is connected, then rad(g {v, v }) rad(g). Proposition 4 [6, 5] Let v be an ncv of a graph G. Then rad(g v) < rad(g) if and only if v has a conjugate vertex, and in this case rad(g v) = rad(g) 1. Proposition 5 [5] Let G be a vertex-radius-decreasing graph, and v a ncv of G. If v is not central, then all its conjugate vertices are cut-vertices. If v is central, then it has exactly one conjugate vertex v, and v is a ncv (so v and v form a conjugate pair). 8

9 Proposition 6 [6, 5] A graph G of order n is a vertex-radius-decreasing block if and only if G is self-centered, n is even, and V (G) can be partitioned into conjugate pairs. Proposition 7 [5] In any vertex-radius-decreasing graph containing at least one cut-vertex, every ncv has degree 1. Proposition 8 Let G be a bipartite graph and let v be a vertex in a partite set V i, i = 1,. Then deg G (v) V 3 i rad(g) +. Proof Let T v be a distance-preserving spanning tree of G with v as its root; so deg Tv (v) = deg G (v). Let P be a diametral path of T v. Then P has length diam(t v ) rad(t v ) 1 rad(g) 1. So P contains at least rad(g) vertices, with at least rad(g) of them in V 3 i. Moreover, at most two of them can be neighbours of v on P. So there are at least deg G (v) neighbours of v which are not on P. So and Proposition 8 follows. 3 The Main Result V 3 i rad(g) + deg G (v), In this section we obtain a bound on the size of a bipartite graph of order n and radius r. The following lemma deals with the case r = 4 of our main theorem. Lemma Let G be a bipartite graph of order n 8 and radius 4. Then n m(g) n + 8, 4 Moreover, if m(g) = n 4 n + 8, then G B(n, 4). Proof Since rad(g) = 4, there exists a vertex x V (G) such that e G (x) = 4. Moreover, there is a vertex x 4 V (G) such that d(x, x 4 ) = 4, having xx 1 x x 3 x 4 as a shortest x x 4 path in G. For 1 i 4, let N i be the ith distance layer of x. So x i N i for 1 i 4. Since e G (x 1 ) 4, there is a vertex x 1 V (G) such that d(x 1, x 1 ) = 4. Thus x 1 N 3 and 9

10 x x 1 / E(G). But x 1 must have a neighbour in N, say x, where x x and x 1 x / E(G). Moreover, x must have a neighbour in N 1 that is not x 1, say x 1. Since e G (x ) 4, there is a vertex x V (G) such that d(x, x ) = 4, where x / {x, x 4 }. Suppose, without loss of generality, that x V 1. Then certainly {x, x 4 } and {x, x } are disjoint pairs of vertices in V 1 that are distance 4 apart. Since e G (x 1) 4, there is a vertex x 1 V such that d(x 1, x 1) = 4, where x 1 / {x 1, x 1 }. Then certainly {x 1, x 1 } and {x 1, x 1} are disjoint pairs of vertices in V that are distance 4 apart. So there exist four disjoint pairs of vertices, say u i and v i, such that d(u i, v i ) = 4 for 1 i 4, where u i, v i V 1 for i = 1, and u i, v i V for i = 3, 4. Denote by G, the bipartite complement of G; that is the graph with bipartition (V 1, V ) such that for u V 1, v V, uv E(G) if and only if uv / E(G). Let V 1 = V 1 {u 1, v 1, u, v } and V = V {u 3, v 3, u 4, v 4 }. We show that m(g) n 8. For each vertex w V, there exist edges e 1 (w) and e (w) joining w to a vertex in {u 1, v 1 } and {u, v }, respectively, in G since otherwise d G (u 1, v 1 ) =. Similarly, for each vertex w V 1, there exist edges e 3 (w) and e 4 (w) joining w to a vertex in {u 3, v 3 } and {u 4, v 4 }. Clearly, the subsets of E(G) are disjoint. Hence, A = {e 1 (w) w V } {e (w) w V }, B = {e 3 (w) w V 1} {e 4 (w) w V 1} m(g) A + B = V + ( V 1 4) = n 8. We have m(g)+m(g) n 4 since the maximum size of a complete bipartite graph is n 4. Hence m(g) n n m(g) n + 8, 4 4 as required. We now show that if m(g) = n 4 n + 8, then G B(n, 4). Suppose that m(g) = n 4 n + 8. Then, m(g) = n 8, and hence, m(g) = A + B = V + ( V 1 4). Hence, in G, every vertex in V is adjacent to exactly one vertex in {u 1, v 1 } and exactly one vertex in {u, v }, 10

11 and every vertex in V 1. Every vertex in V 1 is adjacent to exactly one vertex in {u 3, v 3 } and exactly one vertex in {u 4, v 4 }. Let x, y be an arbitrary adjacent pair of vertices in V 1 V. Then deg G (x)+deg G (y) = V 1 + V = n 4. Hence, by Lemma 1, the result follows. We now present our main theorem. Theorem For natural numbers n and r such that n r, the maximum number of edges in a bipartite graph of order n and radius at least r is b(n, r), where a) b(n, 1) = n 1, b) b(n, ) = n 4, c) b(n, 3) = n 4 n, d) b(n, r) = n 4 nr + r + (n r) for n r 8. The bipartite graph with radius 1 and the maximum number of edges is the star K 1,n 1. The bipartite graph with radius and the maximum number of edges is the complete bipartite graph K. The bipartite graph with n n, radius 3 and the maximum number of edges is obtained from the complete,, by the removal of a minimum edge cover. If G is a bipartite graph K n n graph with radius r 4 and the maximum number of edges, then G B(n, r). Proof a) The only bipartite graph with radius 1 and order n is the star K 1,n 1, which has n 1 edges. b) The bipartite graph with radius and the maximum number of edges is the complete bipartite graph K n, n which has n n = n 4 edges. c) Let G be a bipartite graph of order n, radius 3 and partite sets V 1 and V. Since rad(g) = 3, every vertex in V 1 must be non-adjacent to at least one vertex in V, and vice versa. Thus, m(g) n, and since the maximum size of a complete bipartite graph is n n, we have m(g) m(g), 4 4 and thus m(g) n n. Clearly, equality holds if G is obtained from 4 the complete graph K n, n, by the removal of a minimum edge cover. 11

12 d) Let G be a bipartite graph of order n, radius at least r 4 and maximum size with partite sets V 1 and V. By double induction, we prove that if G has order n and rad(g) r, then m(g) b(n, r) for n r 8, and m(g) = b(n, r) if and only if G B(n, r). We first show the inequality for the case n = r, i.e., we show that m(g) b(r, r) for r 4. Let G be a graph of radius r and order r. By Proposition 1, (G) n r + =. It follows that m(g) 1 n (G) n = r = b(r, r). Moreover, G must be a cycle of length r and thus G B(r, r). For the case r = 4, it has been shown in Lemma that, for n 8, m(g) b(n, 4) and if m(g) = b(n, 4), G B(n, 4). Now let n and r be natural numbers such that r 5 and n r + 1 and assume validity of the theorem for all bipartite graphs of order n and radius at least r, where either 4 r r 1 or else r = r and r n n 1. Let G be any bipartite graph of order n and radius at least r. Claim 1 If {x, x } is a conjugate pair of vertices in G, and the graph G {x, x } is disconnected, then m(g) b(n, r) and if m(g) = b(n, r), then G B(n, r). Let S = {x, x }. Let G 1, G,..., G k be the components of G S. Let G x = V (G 1 ) S G and G y = V (G )... V (G k ) S G. Note that G x and G y are connected for otherwise either x or x is not central. Suppose n(g x ) = t and thus n(g y ) = n t +. Moreover diam(g x ), diam(g y ) r and thus r + 1 t n r + 1. By m(g) = m(g x ) + m(g y ) and Proposition we have m(g) n nt + 5n t nr r + r t 4 = n nr + 4 r + n r + 1 (t r 1)(t n + r 1) b(n, r) since r + 1 t n r + 1 and therefore 1 (t r 1)(t n + r 1) 0. If m(g) = b(n, r), then equality holds throughout the above inequalities, and it follows that G x and G y are both graphs of diameter r and maximum size, given their orders. Moreover, t = r + 1 or t = n r + 1. Without loss of generality, say n(g x ) = n r + 1 and thus n(g y ) = r + 1. Since diam(g x ) = r, Proposition implies G x = G(n r + 1, r) = [r ]K1 + n r K 1 + n r K 1 + K 1. 1

13 So G x contains partite sets X and Y where X = n r + 1, and Y = n r + 1, where every vertex in X has degree n r = n r +, and every vertex in Y has degree n r +1+1 = n r +. So G contains adjacent vertices, x X and y Y, such that deg G (x)+deg G (y) = n r+4. It follows from Lemma 1 that G B(n, r). Claim If G contains a conjugate pair of vertices then m(g) b(n, r). If m(g) = b(n, r), then G B(n, r). Let {x, x } be a conjugate pair of vertices in G. By Claim 1, we may assume that G = G {x, x } is connected. Then by Proposition 3, rad(g ) r. By Lemma 1, we need only consider the case where deg G (x) + deg G (x ) < n r + 4. Moreover, by the induction hypothesis, we know that m(g ) b(n, r). Hence, m(g) m(g ) + deg G (x) + deg G (x ) b(n, r) + n r + 3 = ( ) n (n )r + r + (n r) + n r + 3 = n 4 nr + r + n r = b(n, r), as required. If m(g) = b(n, r), then we have equality throughout i.e., m(g ) = b(n, r) and deg G (x) + deg G (x ) = n r + 3. Without loss of generality, say deg G (x) deg G (x ). Then, deg G (x) n r +. By the induction hypothesis, G B(n, r) and so in G, V 1(G ) = n r+3 = n r + 1 and V (G ) = n r+3 = n r or V 1(G ) = n r, V (G ) = n r + 1. Since n(g ) r and n(g ) + = n, n r +. Thus deg G (x) n r + r + r + = 3. Note that x can be adjacent to at most vertices in V (G ) (V 1(G ) V (G )) as otherwise rad(g) < r. However, as rad(g) r, it then follows that x cannot be adjacent to a vertex in V 1(G ) V (G ) and to two vertices in V (G ) (V 1(G ) V (G )). So x is adjacent to at most one vertex in V (G ) (V 1(G ) V (G )), and thus x is adjacent to at least n r+ 1 = n r+1 vertices in V 1(G ) V (G ), i.e., x is adjacent to every vertex in V 1(G ) or x 13

14 is adjacent to every vertex in V (G ). Moreover, deg G (x) = n r +, and thus deg G (x ) = n r + 1. Since rad(g) 5, d G (x, x ) 5 and thus x cannot be adjacent to any vertex in V 1 V as otherwise rad(g) < r, and thus deg G (x ) =. Hence, n = r + since n r + 1 = and n r +. Moreover, n(g ) = r and so G = Cr. Hence, deg G (x) = r+ r + = 3, and thus x must be adjacent to three vertices on G = Cr, which is a contradiction as then rad(g) < r. Hence, equality cannot be attained in this case. Claim 3 If G is a vertex-radius-decreasing graph then m(g) b(n, r), and if m(g) = b(n, r) then G B(n, r). By Claim, we need only consider the case where G has no conjugate pairs. Then, by Proposition 6, G must contain at least one cut-vertex and by Proposition 7, any ncv of G must have degree 1. Hence, G contains two end vertices x 1 and x. Let G = G {x 1, x }, and note that if rad(g ) r, then any central vertex c of G is within distance r from every vertex in V (G) {x 1, x }, including the neighbours of x 1 and x. But then c is within distance r 1 from x 1 and x, contradicting rad(g) = r. Hence rad(g ) r 1. So, by the induction hypothesis, m(g ) b(n, r 1). Hence, m(g) = + m(g ) + b(n, r 1) = b(n, r), If m(g) = b(n, r), we have equality throughout. So m(g ) = b(n, r 1) and thus by our induction hypothesis, G B(n, r 1). If V 1(G) 3 or V (G), then G is not a vertex-radius-decreasing graph; thus V 1(G) = and V (G) = 1. Hence, n r + 3 = 3, and thus n = r which is a contradiction as n > r. Hence, equality cannot be attained in this case. Claim 4 If v is a ncv of G with rad(g v) r and deg G (v) n r +, then m(g) b(n, r). If m(g) = b(n, r), then G B(n, r). By the induction hypothesis, m(g v) b(n 1, r), and hence, m(g) = m(g v) + deg G (v) b(n 1, r) + n r + 14

15 = (n 1) (n 1)r + r + (n 1 r) + n r + 4 = n nr + r + n r 4 = b(n, r), as required. If m(g) = b(n, r), we have equality throughout; so m(g v) = b(n 1, r) and deg G (v) = n r +. By the induction hypothesis, G v B(n 1, r). If n(g v) = r, then G v is a cycle of length r, and moreover every vertex in G v has degree. Hence, any neighbour of v in G, say z, has degree 3 and thus G contains adjacent vertices v and z such that deg G (z) + deg G (x) = 5 = n r + 4. Hence G B(n, r) by Lemma 1. Since n(g v) r + 1 and n = n(g v) 1, n r +. Hence deg G (v) = n r + r+ r + 3. Note that v can be adjacent to at most one vertex in G {v} (V 1(G v) V (G v)) as otherwise rad(g) < r. Thus v is adjacent to at least n r + 1 = n r + 1 vertices in V 1(G v) V (G v). Let w V i (G v), i = 1, such that vw E(G), and let y V 3 i(g v) such that wy E(G v). Then and thus deg G v (w) + deg G v (y) = V 1(G v) + V (G v) + 1, deg G (w) +deg G (y) = V 1(G v) + V (G v) + = (n 1) r+3+ = n r+4, and hence G B(n, r) by Lemma 1. Claim 5 If w is a ncv of G with deg G (w) n r+ and rad(g w) r 1, then every neighbour of w is a ncv. By Proposition 4, w has a conjugate vertex w such that d G (w, w) = r and d G (w, u) r 1 for every u V (G) {w}. Let s and t be neighbours of w. It follows that if u is any vertex in V (G) {w, s}, then no shortest w u path can contain s. In particular, G s contains a w t path and hence a w w path. So G s is connected. Since s N G (w) was chosen arbitrarily, it follows that no neighbour of w is a cut-vertex. Claim 6 If v is a ncv of G with rad(g v) r and deg G (v) > n r +, then m(g) b(n, r). If m(g) = b(n, r), then G B(n, r). 15

16 We first show that v has a neighbour that is a ncv. Suppose to the contrary that every neighbour of v is a cut-vertex. Let T v be a distance-preserving spanning tree of G with v as its root; so deg Tv (v) = deg G (v). Let P be a diametral path of T v. Then P has length diam(t v ) rad(t v ) 1 rad(g) 1. So P contains at least rad(g) vertices. Moreover, the (deg G (v) ) neighbours of v not on P cannot be leaves because they are cut-vertices, and so they must be adjacent to a vertex that is non-adjacent to every other neighbour of v. Hence, since deg Tv (v) n r + 3, n r + (deg Tv (v) ) r + ( ) n r = n +, which is a contradiction. Thus, v must have a neighbour, say x, which is a ncv. If deg G (x) n r + 3, then deg G(v) + deg G (x) n r + 6, which is a contradiction by Lemma 1. Hence, deg G (x) n r +. Moreover, since x is a ncv, rad(g x) r 1 by Claim 4. By Proposition 4, x has a conjugate vertex, say x. If rad(g x) r 1, then {x, x} would form a conjugate pair and the result follows by Claim. So rad(g x) r and since d(x, v), deg G (x) n r + by Lemma 1. Hence, x must be a cut-vertex by Claim 4, and so G {x} has at least two components, say G 1 and G. Assume without loss of generality, that v, x V (G 1 ). Let x 1 be a neighbour of x of degree at least in V (G 1 ). Since d(v, x 1 ), deg G (x 1 ) n r + by Lemma 1. Suppose x 1 is a ncv. Then, by Claim 4, rad(g x 1 ) r 1. Applying Claim 5 to x 1 now yields that x is not a cut-vertex, which is a contradiction. Hence, x 1 is a cut-vertex. Let H be the component of G x 1 containing x and denote by N i the ith distance layer of x 1 in H. Since x 1 is a cut-vertex; it follows that every vertex in N 1 is an endvertex or a cut-vertex. By the same argument, if every vertex in N i, i 1, is an end-vertex or a cut-vertex, then so is every vertex in N i+1 (if any exists). Hence, by induction, each vertex in H is either an end-vertex or a cut-vertex. Consider a distance preserving spanning tree T of V (H) {x}. Then either T is a path or T contains at least two end-vertices distinct from x 1. 16

17 In the former case, let x be the end-vertex of T, x x 1, and y the neighbour of x, and in the latter case, let x and y be two end-vertices distinct from x. In both cases G =: G x y has n vertices, rad(g ) r 1 and m(g ) = m(g). Hence, by induction, m(g) = m(g ) + b(n, r 1) + = b(n, r). If m(g) = b(n, r), we have equality throughout; so m(g ) = b(n, r 1). By the induction hypothesis, G B(n, r 1). Hence G contains vertices w, v such that d G (w, v) and deg G (w)+deg G (v) = (n ) (r 1)+4 = n r + 4. By Lemma 1, G B(n, r). Claim 7 If v is a ncv of G with rad(g v) r, then m(g) b(n, r). If m(g) = b(n, r), then G B(n, r). This follows from Claims 4 and 6. References [1] P. Dankelmann, G. Dlamini and H.C. Swart, Upper bounds on distance measures in K,l -free graphs. (submitted) [] P. Dankelmann, G. Dlamini and H.C. Swart, Upper bounds on distance measures in K 3,3 -free graphs. Util. Math. 67 (005) [3] G. Dlamini, Aspects of Distances in Graphs, Ph.D. thesis, University of Natal, Durban, 003. [4] P. Erdös, J. Pach, R. Pollack and Z. Tuza, Radius, diameter and minimum degree. J. Combin. Theory B 47 (1989), [5] S. Fajtlowicz, A characterisation of radius critical graphs. J. Graph Theory 1 (1988), [6] F. Gliviak, On radially critical graphs. Recent Advances in Graph Theory (Proc. Sympos. Prague 1974), Academia Praha, Prague (1975), [7] F. Gliviak, M. Knor and L. Soltes, On radially maximal graphs. Australas. J. Combin. 9 (1994),

18 [8] F Harary and C. Thomassen, Anticritical graphs. Math. Proc. Cambridge Philos. Soc. 79 (1976), [9] S. Mukwembi, Bounds on Distances in Graphs. Ph.D. Thesis, University of KwaZulu-Natal, Durban, 006. [10] S. Mukwembi, The radius of a triangle-free graph with prescribed edgeconnnectivity, Util. Math. (to appear). [11] Y. Nishanov, On radially critical graphs with the maximum diameter (in Russian). Trudy Samarkand. Gos. Univ. 35 (197), [1] Y Nishanov, A lower bound on the number of edges in radially critical graphs (in Russian). Trudy Samarkand. Gos. Univ. 56 (1975), [13] V. Vizing, The number of edges in a graph of given radius. Soviet Math. Dokl. 8 (1967),