TThe diameter and radius of radially maximal graphs ∗ Pu Qiao a , Xingzhi Zhan b † a Department of Mathematics, East China University of Science and Technology, Shanghai 200237, China b Department of Mathematics, East China Normal University, Shanghai 200241, China
Abstract
A graph is called radially maximal if it is not complete and the addition of anynew edge decreases its radius. In 1976 Harary and Thomassen proved that theradius r and diameter d of any radially maximal graph satisfy r ≤ d ≤ r − . Dutton, Medidi and Brigham rediscovered this result with a different proof in 1995and they posed the conjecture that the converse is true, that is, if r and d arepositive integers satisfying r ≤ d ≤ r − , then there exists a radially maximalgraph with radius r and diameter d. We prove this conjecture and a little more.
Key words.
Radially maximal; diameter; radius; eccentricity
We consider finite simple graphs. Denote by V ( G ) and E ( G ) the vertex set and edgeset of a graph G respectively. The complement of G is denoted by ¯ G. The radius anddiameter of G are denoted by rad( G ) and diam( G ) respectively. Definition.
A graph G is said to be radially maximal if it is not complete andrad( G + e ) < rad( G ) for any e ∈ E ( ¯ G ) . Thus a radially maximal graph is a non-complete graph in which the addition of anynew edge decreases its radius. Since adding edges in a graph cannot increase its radius, ∗ E-mail addresses: [email protected] (P.Qiao), [email protected] (X.Zhan). † Corresponding author. a r X i v : . [ m a t h . C O ] O c t very graph is a spanning subgraph of some radially maximal graph with the same radius.It is well-known that the radius r and diameter d of a general graph satisfy r ≤ d ≤ r [4, p.78]. In 1976 Harary and Thomassen [3, p.15] proved that the radius r and diameter d of any radially maximal graph satisfy r ≤ d ≤ r − . (1)Dutton, Medidi and Brigham [1, p.75] rediscovered this result with a different proof in1995 and they [1, p.76] posed the conjecture that the converse is true, that is, if r and d are positive integers satisfying (1) then there exists a radially maximal graph with radius r and diameter d. We prove this conjecture and a little more.We denote by d G ( u, v ) the distance between two vertices u and v in a graph G. The eccentricity , denoted by e G ( v ) , of a vertex v in G is the distance to a vertex farthestfrom v. The subscript G might be omitted if the graph is clear from the context. Thus e ( v ) = max { d ( v, u ) | u ∈ V ( G ) } . If e ( v ) = d ( v, x ) , then the vertex x is called an eccentricvertex of v. By definition the radius of a graph G is the minimum eccentricity of allthe vertices in V ( G ) , whereas the diameter of G is the maximum eccentricity. A vertex v is a central vertex of G if e ( v ) = rad( G ) . A graph G is said to be self-centered ifrad( G ) = diam( G ) . Thus self-centered graphs are those graphs in which every vertex isa central vertex. N G ( v ) will denote the neighborhood of a vertex v in G. The order of agraph is the number of its vertices. The symbol C k denotes a cycle of order k. We will need the following operation on a graph. The extension of a graph G at a vertex v, denoted by G { v } , is the graph with V ( G { v } ) = V ( G ) ∪ { v (cid:48) } and E ( G { v } ) = E ( G ) ∪{ vv (cid:48) } ∪ { v (cid:48) x | vx ∈ E ( G ) } where v (cid:48) (cid:54)∈ V ( G ) . Clearly, if G is a connected graph of order atleast 2 , then e G { v } ( u ) = e G ( u ) for every u ∈ V ( G ) and e G { v } ( v (cid:48) ) = e G { v } ( v ) = e G ( v ) . Inparticular, rad( G { v } ) = rad( G ) and diam( G { v } ) = diam( G ) . Gliviak, Knor and ˇSolt´es [2, Lemma 5] proved the following result.
Lemma 1.
Let G be a radially maximal graph. If v ∈ V ( G ) is not an eccentric vertexof any central vertex of G, then the extension of G at v is radially maximal. Now we are ready to state and prove the main result.
Theorem 2.
Let r, d and n be positive integers. If r ≥ and n ≥ r, then there exists self-centered radially maximal graph of radius r and order n. If r < d ≤ r − and n ≥ r − , then there exists a radially maximal graph of radius r, diameter d and order n. Proof.
We first treat the easier case of self-centered graphs. Suppose r ≥ n ≥ r. The even cycle C r is a self-centered radially maximal graph of radius r and order2 r. Choose any but fixed vertex v of C r . For n > r, successively performing extensionsat vertex v starting from C r we obtain a graph G ( r, n ) of order n. G (4 ,
11) is depictedin Figure 1.Denote G ( r, r ) = C r . Since G ( r, n ) has the same diameter and radius as C r , it isself-centered with radius r. Let xy be an edge of the complement of G ( r, n ) . Denote by S the set consisting of v and the vertices outside C r . Then S is a clique. If one endof xy, say, x lies in S, then y (cid:54)∈ N [ v ] , the closed neighborhood of v in G ( r, n ) . We have e ( x ) < r. Otherwise x, y ∈ V ( C r ) \ S. We then have e ( x ) < r and e ( y ) < r. In both cases,rad( G ( r, n ) + xy ) < rad( G ( r, n )) . Hence G ( r, n ) is radially maximal.Next suppose r < d ≤ r − n ≥ r − . We define a graph H = H ( r, d, r − r − V ( H ) = { x , x , . . . , x r − } ∪ { y , y , . . . , y r } and E ( H ) = { x i x i +1 | i = 1 , , . . . , r − } ∪ { x r − y } ∪ { x r − j +2 y j | j = 1 , , . . . , r − d }∪ { x d − r +1 y r − d +1 } ∪ { y t y t +1 | t = 2 r − d + 1 , . . . , r − d ≥ r + 2 } where x r = x . H is obtained from the odd cycle C r − by attaching edges and one path.A sketch of H is depicted in Figure 2, and H (6 , d,
17) with d = 7 , , ,
10 are depicted inFigure 3. 3learly, H has radius r, diameter d and order 3 r − . To see this, verify that x d − r +1 isa central vertex and e H ( y r ) = d. Now we show that H is radially maximal. Let C be the cycle of length 2 r −
1; i.e., C = x x . . . x r − x . We specify two orientations of C. Call the orientation x , x , . . . , x r − , x clockwise and call the orientation x r − , x r − , . . . , x , x r − counterclockwise. For twovertices a, b ∈ V ( C ) , we denote by −→ C ( a, b ) the clockwise ( a, b )-path on C and by ←− C ( a, b )the counterclockwise ( a, b )-path on C. For uv ∈ E ( ¯ H ) , denote T = H + uv. To show rad( T ) < r, it suffices to find a vertex z such that e T ( z ) < r. Denote A = V ( C ) = { x , x , . . . , x r − } and B = V ( H ) \ V ( C ) = { y , y , . . . , y r } . We distinguish three cases.
Case 1. u, v ∈ A. Let u = x i and v = x j with i > j. d − r + 1 ≤ r − , the vertex y is a leaf whose only neighbor is x r − . Note thatin H, the three vertices x r , x r − and x r − are central vertices, y is the unique eccentricvertex of x r , and y is the unique eccentric vertex of x r − and x r − . If j ≥ r or i ≤ r, then e T ( x r ) < r. Indeed, in the former case −→ C ( x r , v ) ∪ vu ∪ −→ C ( u, x r − ) ∪ x r − y is an( x r , y )-path of length less than r and in the latter case, ←− C ( x r , u ) ∪ uv ∪ ←− C ( v, x ) ∪ x y is an ( x r , y )-path of length less than r. Next suppose i > r > j. If | ( i − r ) − ( r − j ) | ≥ , then in T there is an ( x r , y )-path of length less than r, which implies that e T ( x r ) < r. It remains to consider the case | ( i − r ) − ( r − j ) | ≤ . If ( i − r ) − ( r − j ) = 0 or 1 , then in T, there is an ( x r − , y )-pathof length less than r and hence e T ( x r − ) < r. If ( r − j ) − ( i − r ) = 1 , then in H, there isan ( x r − , y )-path of length r − e T ( x r − ) < r. Case 2. u, v ∈ B. Let u = y i and v = y j with 1 ≤ i < j ≤ r. Subcase 2.1. i = 1 and j ≤ r − d. In the sequel the subscript arithmetic for x k istaken modulo 2 r − . x r − j +2 is a central vertex of H whose unique eccentric vertex is y j . To see this, note that if r − j + 2 ≤ d − r + 1 then d H ( x r − j +2 , y r ) ≤ d − r + 1 − ( r − j +2) + r − (2 r − d ) = 2 d − r + 2 j − ≤ r − j ≤ r − d, and if r − j + 2 > d − r + 1then d H ( x r − j +2 , y r ) ≤ r − j + 2 − ( d − r + 1) + r − (2 r − d ) = r − j + 1 ≤ r − j ≥ . If r − j + 2 ≥ , in T there is the ( x r − j +2 , y j )-path ←− C ( x r − j +2 , x ) ∪ x y ∪ y y j . Hence d T ( x r − j +2 , y j ) ≤ r − j + 2 − r − j + 3 ≤ r − j ≥ , implying e T ( x r − j +2 ) < r. If r − j + 2 ≤ , in T there is the path −→ C ( x r − j +2 , x r − ) ∪ x r − y ∪ y y j . Hence d T ( x r − j +2 , y j ) ≤ − ( r − j + 2) + 2 = 2 j − r ≤ r − j ≤ r − d and d ≥ r + 1 , implying e T ( x r − j +2 ) < r. Subcase 2.2. i = 1 and 2 r − d + 1 ≤ j ≤ r. First suppose j = r. Observe that x d − r +1 is a central vertex of H whose unique eccentric vertex is y r . Also the condition d ≤ r − d − r +1 < d − r +1 . If 2 d − r +1 ≥ , then d T ( x d − r +1 , y r ) ≤ d − r +1 − ≤ r − . If 2 d − r + 1 ≤ , then d T ( x d − r +1 , y r ) ≤ − (2 d − r + 1) + 2 ≤ r − , where wehave used the fact that d ≥ r + 1 . Hence e T ( x d − r +1 ) < r. Next suppose 2 r − d + 1 ≤ j ≤ r − . Observe that x r is a central vertex of H whoseunique eccentric vertex is y . Note also that r > d − r + 1 . Now in T, there is the ( x r , y )-path ←− C ( x r , x d − r +1 ) ∪ x d − r +1 y r − d +1 . . . y j ∪ y j y . Hence d T ( x r , y ) ≤ r − ( d − r + 1) + j − (2 r − d ) + 1 = j ≤ r − , implying e T ( x r ) < r. i ≥ j ≤ r − d. First suppose 2( j − i ) ≤ r − . Then 2 r − j + 2 ≥ r − i + 3 . Clearly x r − j +2 is the unique neighbor of y j in H. By considering the twopossible cases r − i + 3 ≤ d − r + 1 and r − i + 3 > d − r + 1 , it is easy to verifythat x r − i +3 is a central vertex of H whose unique eccentric vertex is y i . In T thereis the ( x r − i +3 , y i )-path −→ C ( x r − i +3 , x r − j +2 ) ∪ x r − j +2 y j ∪ y j y i . Hence d T ( x r − i +3 , y i ) ≤ r − j + 2 − ( r − i + 3) + 1 + 1 = r − j − i ) + 1 ≤ r − , implying e T ( x r − i +3 ) < r. Next suppose 2( j − i ) ≥ r. Then r − i + 2 ≥ r − j + 2 . Observe that x r − i +2 is acentral vertex of H whose unique eccentric vertex is y i . Also j − i ≤ r − d − . Similarlywe have d T ( x r − i +2 , y i ) ≤ r − i + 2 − (2 r − j + 2) + 1 + 1= 2 − r + 2( j − i ) ≤ − r + 2(2 r − d − ≤ r − , implying e T ( x r − i +2 ) < r. Subcase 2.4. 2 ≤ i ≤ r − d and 2 r − d + 1 ≤ j ≤ r. First suppose 2 r + 2 ≤ i + d. Then d − r +1 ≥ r − i +3 . Note that x r − i +3 is a central vertex of H whose unique eccentric vertexis y i . In T we have the ( x r − i +3 , y i )-path −→ C ( x r − i +3 , x d − r +1 ) ∪ x d − r +1 y r − d +1 . . . y j ∪ y j y i . Thus d T ( x r − i +3 , y i ) ≤ d − r + 1 − ( r − i + 3) + j − (2 r − d ) + 1 ≤ d − r + 1 − ( r − i + 3) + r − (2 r − d ) + 1= 2 d − r + 2 i − ≤ r − , implying e T ( x r − i +3 ) < r. Next suppose 2 r + 2 ≥ i + d + 1 . Then r − i + 2 ≥ d − r + 1 . Observe that x r − i +2 is a central vertex of H whose unique eccentric vertex is y i . Similarly we have d T ( x r − i +2 , y i ) ≤ r − i + 2 − ( d − r + 1) + j − (2 r − d ) + 1 ≤ r − i + 2 − ( d − r + 1) + r − (2 r − d ) + 1= r − i + 2 ≤ r − , e T ( x r − i +2 ) < r. Subcase 2.5. 2 r − d + 1 ≤ i < j ≤ r. Observe that x r +1 is a central vertex of H whoseunique eccentric vertex is y r . Clearly e T ( x r +1 ) < r. Case 3. u ∈ A and v ∈ B. Let u = x i and v = y j . Observe that x r is a central vertex of H whose unique eccentric vertex is y . If j = 1 , then e T ( x r ) < r. Now suppose 2 ≤ j ≤ r − d. Then both x r − j +2 and x r − j +3 are centralvertices of H whose unique eccentric vertex is y j . If u lies on the path −→ C ( x r − j +2 , x r − j +2 ) , then e T ( x r − j +2 ) < r ; if u lies on the path ←− C ( x r − j +2 , x r − j +3 ) , then e T ( x r − j +3 ) < r. Finally suppose 2 r − d + 1 ≤ j ≤ r. We have 2 d − r + 1 < d − r + 1 < r + 1 . Observethat both x r +1 and x d − r +1 are central vertices of H whose unique eccentric vertex is y r . If 2 d − r + 1 ≤ i ≤ d − r + 1 , then d T ( x d − r +1 , y r ) ≤ r − e T ( x d − r +1 ) < r. Similarly, if d − r + 2 ≤ i ≤ r + 1 then e T ( x r +1 ) < r. It remains to consider the case when u = x i lies on the path −→ C ( x r +2 , x d − r ) . Weassert that e T ( u ) < r. First note that if w ∈ { y r − d +1 , y r − d +2 , . . . , y r } then d T ( x i , w ) ≤ d − r ≤ r − . Also if w ∈ V ( C ) we have d T ( x i , w ) ≤ r − C ) = r − . Next suppose w = y s with 1 ≤ s ≤ r − d. Let x k and x k +1 be the two vertices on C with d C ( x i , x k ) = d C ( x i , x k +1 ) = r − . Since x i lies on the path −→ C ( x r +2 , x d − r ) , we have k ≥ k + 1 ≤ d − r < d − r + 1) . It follows that d H ( x i , w ) ≤ r − , since N H ( y ) = { x r − , x } and N H ( y r − d ) = { x d − r +1) } . This completes the proof that H isradially maximal.Note that by the two inequalities in (1), any non-self-centered radially maximal graphhas radius at least 3 . Obviously, the vertex x r − is not an eccentric vertex of any vertexin H. Hence by Lemma 1, the extension of H at x r − , denoted H r , is radially maximal.Also, H r has the same diameter and radius as H, and has order 3 r. Again, the vertex x r − is not an eccentric vertex of any vertex in H r . For any n > r − , performingextensions at the vertex x r − successively, starting from H, we can obtain a radiallymaximal graph of radius r, diameter d and order n. This completes the proof. (cid:50)
Combining the restriction (1) on the diameter and radius of a radially maximal graphand Theorem 2 we obtain the following corollary.
Corollary 3.
There exists a radially maximal graph of radius r and diameter d if andonly if r ≤ d ≤ r − . Final Remarks
Since any graph with radius r has order at least 2 r, Theorem 2 covers all the possibleorders of self-centered radially maximal graphs.Gliviak, Knor and ˇSolt´es [2, p.283] conjectured that the minimum order of a non-self-centered radially maximal graph of radius r is 3 r − . This conjecture is known to be truefor the first three values of r ; i.e., r = 3 , , Acknowledgement.
This research was supported by the NSFC grants 11671148 and11771148 and Science and Technology Commission of Shanghai Municipality (STCSM)grant 18dz2271000.