Kragujevac Journal of MathematicsVolume 34 (2010), Pages 161–170.**RADIAL** **DIGRAPHS**KUMARAPPAN KATHIRESAN AND R. SUMATHIAbstract. The Radial **graph** of a **graph** G, denoted by R(G), has the same vertexset as G with an edge joining vertices u and v if d(u, v) is equal to the radius of G.This definition is extended to a di**graph** D where the arc (u, v) is included in R(D)if d(u, v) is the radius of D. A di**graph** D is called a Radial di**graph** if R(H) = Df**or** some di**graph** H. In this paper, we shown that if D is a radial di**graph** of type 2then D is the radial di**graph** of itself **or** the radial di**graph** of its complement. Thisgeneralizes a known characterization f**or** radial **graph**s and provides an improvedproof. Also, we characterize self complementary self radial di**graph**s.**1.** **Introduction**A **directed** **graph** **or** di**graph** D consists of a finite nonempty set V (D) of objectscalled vertices and a set E(D) of **or**dered pairs of vertices called arcs. If (u, v) is anarc, it is said that u is adjacent to v and also that v is adjacent from u. The outdegreeod v of a vertex v of a di**graph** D is the number of vertices of D that are adjacentfrom v. The indegree id v of a vertex v of a di**graph** D is the number of vertices ofD that are adjacent to v. The set of vertices which are adjacent from [to] a givenvertex v is denoted by N + D(v) [N − D(v)]. A Di**graph** D is symmetric if whenever uv isan arc, vu is also an arc. As in Chartrand and Oellermann [7], we use D ∗ to denotethe symmetric di**graph** whose underlying **graph** is D. Thus, D ∗ is the di**graph** that isobtained from D by replacing each edge of D by a symmetric pair of arcs. F**or** other**graph** the**or**etic notations and terminology, we follow [3] and [5].F**or** a pair u, v of vertices in a strong di**graph** D the distance d(u, v) is the lengthof a sh**or**test **directed** u − v path. We can extend this definition to all di**graph**s DKey w**or**ds and phrases. Radial **graph**s, Radial di**graph**s, Self-radial **graph**s, Self-radial di**graph**s.2010 Mathematics Subject Classification. 05C12, 05C20.Received: March 20, 2010.161

162 KUMARAPPAN KATHIRESAN AND R. SUMATHIby defining d(u, v) = ∞ if there is no **directed** u − v path in D. The eccentricityof a vertex u, denoted by e(u), is the maximum distance from u to any vertex inD. The radius of D, rad(D), is the minimum eccentricity of the vertices in D;the diameter, diam(D), is the maximium eccentricity of the vertices in D. F**or** adi**graph** D, the Radial di**graph** R(D) of D is the di**graph** with V (R(D)) = V (D)and E(R(D)) = {(u, v)/u, v ∈ V (D) and d D (u, v) = rad(D)}. A di**graph** D is calleda Radial di**graph** if R(H) = D f**or** some di**graph** H. If there exist a di**graph** H withfinite radius and infinite diameter, such that R(H) = D, then the di**graph** D is saidto be a Radial di**graph** of type **1.** Otherwise, D is said to be a Radial di**graph** oftype 2. F**or** the purpose of this paper, a **graph** is a symmetric di**graph**; that is, adi**graph** D such that (u, v) ∈ E(D) implies (v, u) ∈ E(D). Our first result gives auseful property of radial di**graph**s. The proof is straightf**or**ward, so we omit it.Lemma **1.****1.** If D is a symmetric di**graph**, then R(D) is also symmetric.The converse of Lemma **1.**1 need not be true. Figure 1 shows an asymmetric strongdi**graph** D of **or**der p = 4 with rad(D) = 2 and diam(D) = 3 and the c**or**respondingsymmetric radial di**graph** R(D).3✉ ✲ ✉3✻✉♦✉✼D :R(D) :✉✛✲ ✉❄2 3Figure **1.**✴✉✇✉The convention of representing the symmetric pair of arcs (u, v) and (v, u) bythe single edge uv induces a one-to-one c**or**respondence φ from the set of symmetricdi**graph**s to the set of **graph**s. F**or** example in Figure 1, we have φ(R(D)) =⋃K 2 K2 . Theref**or**e, by Lemma **1.**1, it is natural to define, f**or** a **graph** G, theRadial **graph** R(G) of G as the **graph** with V (R(G)) = V (G) and E(R(G)) ={uv/u, v ∈ V (G) and d G (u, v) = rad(G)}.The concept of antipodal **graph** was initially introduced by [13]. The antipodal**graph** of a **graph** G, denoted by A(G), is the **graph** on the same vertices as of G, twovertices being adjacent if the distance between them is equal to the diameter of G.A **graph** is said to be antipodal if it is the antipodal **graph** A(H) of some **graph** H.

**RADIAL** **DIGRAPHS** 163Aravamudhan and Rajendran [1] and [2] gave the characterization of antipodal **graph**s.After that, Johns [8] gave a simple proof f**or** the characterization of antipodal **graph**s.Motivated by the above concepts, Kathiresan and Marimuthu [9], [10], [11] and [12]introduced a new type of **graph** called Radial **graph**s and the following properties ofradial **graph**s have been verified.Proposition **1.****1.** [12] If rad(G) > 1, then R(G) ⊆ G.The**or**em **1.****1.** [12] Let G be a **graph** of **or**der n. Then R(G) = G if and only ifrad(G) = **1.**i.Let S i (G) be the subset of the vertex set of G consisting of vertices with eccentricityLemma **1.**2. [12] Let G be a **graph** of **or**der n. Then R(G) = G if and only ifS 2 (G) = V (G) **or** G is disconnected in which each component is complete.The**or**em **1.**2. [12] A **graph** G is a radial **graph** if and only if it is the radial **graph**of itself **or** the radial **graph** of its complement.2. A Characterization of Radial Graphsusing Radial Di**graph**sWe begin with some properties of Radial di**graph**s.Lemma 2.**1.** If rad(D) > 1, then R(D) ⊆ D.Proof. By the definition of R(D) and D, we have V (R(D)) = V (D) = V (D). If (u, v)is an arc of R(D), then d D (u, v) = rad(D) > 1 in D and hence uv /∈ E(D). Theref**or**e,uv ∈ E(D). Thus, E(R(D)) ⊆ E(D). Hence R(D) ⊆ D.□As a special case, we have Proposition **1.****1.**The**or**em 2.**1.** Let D be a di**graph** of **or**der p.rad(D) = **1.**Then R(D) = D if and only ifProof. Let D be a di**graph** of **or**der p. Suppose rad(D) = **1.** Then, by the definitionof radial di**graph**, R(D) = D.Conversely, assume that R(D) = D. Suppose rad(D) ≠ **1.** Then, by Lemma 2.1we have R(D) ⊆ D, a contradiction. Hence, rad(D) = **1.**□

164 KUMARAPPAN KATHIRESAN AND R. SUMATHIIf D is a symmetric di**graph** of radius 1, then φ(D) is a **graph** of radius **1.** Thisimplies The**or**em **1.****1.**We now present a result that will be useful in our characterization of radial di**graph**s.Let S i (D) be the subset of the vertex set of D consisting of vertices with eccentricityi.The**or**em 2.2. Let D be a di**graph** of **or**der p. Then R(D) = D if and only if anyone of the following holds.(a) S 2 (D) = V (D),(b) D is not strongly connected such that f**or** any v ∈ V (D), od v < p − 1 and f**or**every pair u, v of vertices of D, the distance d D (u, v) = 1 **or** d D (u, v) = ∞.Proof. If S 2 (D) = V (D), then (u, v) ∈ E(R(D)) if and only if (u, v) /∈ E(D). Also,there are no vertices u and v in D such that d D (u, v) > 2. Hence, R(D) = D. Now,suppose that b holds. If d D (u, v) = ∞ f**or** every pair u, v of vertices, then D = K ∗ pf**or** some positive integer p and R(D) = R(K ∗ p) = K ∗ p = D. Since f**or** any v ∈ V (D),od v < p − 1, rad(D) ≠ **1.** Hence rad(D) = ∞. In this case, if (u, v) ∈ E(D), then(u, v) /∈ E(R(D)). If (u, v) /∈ E(D), then d D (u, v) = ∞ and so (u, v) ∈ E(R(D)).Hence, R(D) = D.Conversely, assume that R(D) = D.Case **1.** Suppose that the radius is finite. Assume rad(D) ≠ 2. If rad(D) = 1, thenby The**or**em 2.1, R(D) = D, which is a contradiction to R(D) = D. Thus, we assumethat 2 < rad(D) < ∞. Let u and v be vertices of D such that d D (u, v) = 2. Notethat (u, v) /∈ E(D) and (u, v) /∈ E(R(D)); so R(D) ≠ D, which is a contradiction.Case 2. Assume that the radius is infinite. Then there exist vertices u and v suchthat 1 < d D (u, v) < ∞.Then (u, v) /∈ E(D) and (u, v) /∈ E(R(D)) and againR(D) ≠ D, which is a contradiction. Hence, rad(D) = 2.There are two possibilities rad(D) = diam(D) = 2 and rad(D) = 2, diam(D) > 2.It is well known that rad(D) ≤ diam(D). Suppose that rad(D) < diam(D). Let xand y be vertices in D such that d D (x, y) = diam(D). Now (x, y) /∈ E(D) implies(x, y) ∈ E(D). But (x, y) /∈ E(R(D)), a contradiction to R(D) = D. Hence, the onlypossibility is rad(D) = diam(D) = 2.If D is a self centered symmetric di**graph** of radius 2, then φ(D) is a self centered**graph** of radius 2. On the otherhand, if D is symmetric but not strongly connected□

**RADIAL** **DIGRAPHS** 165such that f**or** any v ∈ V (D), od v < p − 1 and f**or** every pair u and v of vertices of D,the distance d D (u, v) = 1 **or** d D (u, v) = ∞, then φ(D) is a disconnected **graph** whereeach component is complete. This implies Lemma **1.**2.We now give a characterization of radial **graph**s using radial di**graph**s of type 2.The**or**em 2.3. If D is a radial di**graph** of type 2, then D is the radial di**graph** of itself**or** the radial di**graph** of its complement.Proof. Suppose that D is a radial di**graph** of type 2 and let H be a di**graph** such thatR(H) = D. We consider three cases based on H.Case **1.** Suppose that rad(H) = **1.** Then by The**or**em 2.1, R(H) = H.Case 2. Suppose that 1 < rad(H) < ∞. Then the diameter of H may be finite **or**infinite. Since D is a radial di**graph** of type 2, diameter of H is finite. Then H isstrongly connected and f**or** every pair u, v of vertices of H, the distance d H (u, v) < ∞.Define H ′ as the di**graph** f**or**med from H by adding the arc (u, v) to E(H) if d H (u, v) ≠rad(H). Note that d H ′(u, v) = 1 when d H (u, v) ≠ rad(H) and d H ′(u, v) = 2 whend H (u, v) = rad(H). Hence f**or** every vertex v in H ′ , there exist a vertex which areat distance rad(H). Thus, D = R(H) = R(H ′ ). Since rad(H ′ ) = diam(H ′ ) = 2,by The**or**em 2.2 we have R(H ′ ) = H ′ . Theref**or**e, D = H ′ and D = H ′ which givesD = R(D) as desired.Case 3. Suppose that rad(H) = ∞. Define H ′ as the di**graph** f**or**med from H byadding the arc (u, v) to E(H) if d H (u, v) ≠ rad(H). Now, if d H (u, v) < ∞, thend H ′(u, v) = 1 and if d H (u, v) = ∞, then d H ′(u, v) = ∞. Thus, D = R(H) = R(H ′ ).Since f**or** every pair u, v of vertices of H ′ , the distance d H ′(u, v) = 1 **or** d H ′(u, v) = ∞,by The**or**em 2.2 we have R(H ′ ) = H ′ . Theref**or**e, D = H ′ and D = H ′ which givesD = R(D) as desired.□If D is a symmetric radial di**graph** of type 2, then φ(D) is a radial **graph**. Also byThe**or**em 2.3, G is the radial **graph** of itself **or** the radial **graph** of its complement.This implies the characterization of radial **graph**s in The**or**em **1.**2 follows immediately.Figure 2 shows that D is a di**graph** on four vertices whose radius is one and diameteris infinite such that R(D) = D, but D is not a radial di**graph** of type 2.

166 KUMARAPPAN KATHIRESAN AND R. SUMATHID :✉✻✉✣R(D) :✉✻✉✼✉✲ ✉✉✲ ✉Figure 2.Figure 3 shows that D is a radial di**graph** since there exist only one di**graph** H onfour vertices whose radius is two and diameter is infinite. Also, H is neither D n**or**D and hence D is a radial di**graph** of type **1.** Since there is no relationship betweenradial di**graph**s of type 1 and **graph**s, we have not considered such di**graph**s in thispaper. So, we propose the following problem.Problem: Characterize Radial di**graph**s of type **1.**H :✉✻❪✲ ✉✣D :✉✛✉✻❄✉✢✉✉✛✉Figure 3.In view of the The**or**em 2.3, it is natural to ask whether there exist a di**graph** whichare not radial di**graph** of type 2. The next the**or**em answers this question.The**or**em 2.4. A disconnected di**graph** D is a radial di**graph** of type 2 if and only ifeach vertex in D has outdegree at least one.Proof. Let D be a radial di**graph** of type 2. Then there exist a **graph** H such thatR(H) = D where H is either D **or** D. Since D is disconnected, R(D) and D areconnected. Hence R(D) ≠ D. Assume that the components of D contains a vertex(say u) whose outdegree is zero. Then by definition of R(D), there is an arc from u toevery other vertex in D. Hence rad(R(D)) = 1 and so R(D) ≠ D. Also rad(D) = 1,by The**or**em 2.1 we have R(D) = D. Hence R(D) ≠ D. Hence each vertex in D hasoutdegree at least one.F**or** the converse, suppose each vertex in D has outdegree at least one. Since f**or**every vertex (say u) in D there is an arc from u to at least one vertex in the samecomponent, d D(u, v) = 1 if u ∈ D i , v ∈ D j , i ≠ j and d D(u, v) = 2 if u, v ∈ D i .

**RADIAL** **DIGRAPHS** 167Hence, e D(u) = 2, f**or** all u ∈ V (D) and so R(D) is disconnected which is equal to D.Hence, D is a radial di**graph**.□If each vertex in symmetric di**graph** D has outdegree at least one, then φ(D) is a**graph** which has no K 1 component. Theref**or**e, we have the following result.The**or**em 2.5. A disconnected **graph** G is a radial **graph** if and only if G has no K 1component.3. Self-Radial Di**graph**s and GraphsIn the previous section, we proved, f**or** a di**graph** D of **or**der p, that the radialdi**graph** R(D) is identical to D if and only if rad(D) = **1.** Similarly, f**or** a **graph** G of**or**der p, the radial **graph** R(G) is identical to G if and only if rad(G) = **1.** A m**or**einteresting question can also be asked. When is R(D) isom**or**phic to D **or** when isR(G) isom**or**phic to G? If R(D) ∼ = D, then we will call D a self radial di**graph** and ifR(G) ∼ = G, we will call G a self radial **graph**.F**or** a class of self radial di**graph**s D that are strongly connected, given a positiveinteger p ≥ 3, the **directed** cycle C ′ p where V (C p) ′ = {v 1 , v 2 , . . . , v p } and E(C p) ′ ={(v i , v i+1 )/1 ≤ i ≤ p − 1} ⋃ {(v p , v 1 )} is self radial.✉✲✉✼ ✻❄✛ ✴✉✉Figure 4.The self radial di**graph** D in Figure 4 is an example of minimum **or**der that isweakly connected but not unilaterally connected.Proposition 3.**1.** There exist a family of self radial di**graph**s which are unilaterallyconnected but not strongly connected.Proof. F**or** a di**graph** D ′ on p ≥ 4 vertices, where V (D ′ ) = {v 1 , v 2 , . . . , v p } andE(D ′ ) = {(v i , v i+1 )/1 ≤ i ≤ p − 2} ⋃ {(v p−1 , v 1 )} ⋃ {(v 1 , v p )}. Then e(v 1 ) = p − 2,e(v 2 ) = p − 1, e(v i ) = p − 2, 3 ≤ i ≤ p − 1 and e(v p ) = ∞. Thus, rad(D ′ ) = p − 2 anddiam(D ′ ) = ∞. We define a mapping f between D ′ and R(D ′ ) as follows: f(v 1 ) = 3,

168 KUMARAPPAN KATHIRESAN AND R. SUMATHIf(v 2 ) = 2, f(v 3 ) = 1, f(v p ) = p and f(v i ) = p + 3 − i, 4 ≤ i ≤ p − **1.** The mapping fis an isom**or**phism between D ′ and R(D ′ ) and so D ′ is self radial.□Proposition 3.2. If D is a disconnected di**graph**, then D is not self radial.Proof. Let u and v be vertices of D. If u and v are in different components of D, thend D (u, v) = ∞ = rad(D). Thus, (u, v) ∈ E(R(D)) and d R(D) (u, v) = **1.** If u and v arein the same component of D, then there exists a vertex w in the second componentof D. Now, d D (u, w) = d D (w, v) = ∞, so (u, w) ∈ E(R(D)) and (w, v) ∈ E(R(D))and d R(D) (u, v) ≤ 2. Theref**or**e, R(D) is strongly connected and rad(R(D)) ≤ 2. □Proposition 3.3. A self centered di**graph** of radius 2 is self radial if and only if Dis self complementary.Proof. Let D be a self centered di**graph** of radius 2. Then R(D) = D. Since D is selfradial, R(D) ∼ = D. Hence, D ∼ = D.Conversely, let D be self complementary. Since D is a self centered di**graph** ofradius 2, R(D) = D. Hence, R(D) ∼ = D and so D is self radial.□The**or**em 3.**1.** A self complementary di**graph** D is self radial if and only if D is selfcentered di**graph** of radius 2.Proof. Since D is self complementary and self radial, R(D) ∼ = D. Suppose D is a selfcentered di**graph** with rad(D) ≥ 3. Then D is a self centered di**graph** of radius 2.Hence, rad(R(D)) ≠ rad(D), which is a contradiction. Hence, D is a self centereddi**graph** of radius 2.Conversely, if D is a self centered di**graph** of radius 2, then by The**or**em 2.2 we haveR(D) = D. Since D is self complementary, R(D) ∼ = D. Hence, D is self radial. □The**or**em 3.2. If D is a self radial di**graph** with R(D) ≠ D, then p ≤ q(D) ≤p(p − 1)/2.Proof. If D is disconnected, then by Proposition 3.2 we have D is not self radial. Theminimum number of arcs in a connected di**graph** is p − **1.** Then D can contain no**directed** cycles and hence D is not strongly connected. If D is unilaterally connected,then D can contain a **directed** walk that passes through each vertex of D. This canonly be done with p − 1 arcs if D itself is a **directed** path P ′ : v 1 , v 2 , . . . , v p . Howeverin R(P ′ ), there is only one arc from v 1 to v p and so R(P ′ ) is disconnected. Hence D

**RADIAL** **DIGRAPHS** 169is not self radial. Finally, if D is a weakly connected but not unilaterally connected,then the radius may be finite **or** infinite. If the radius of D is finite and e(v) = rad(D),then there exist a vertex u ∈ N D(v) + and there is no path from u to v and so e(u) = ∞.If we take any vertex v ′ ∈ D, the **directed** distance from u to v ′ (v ′ to u) is either lessthan the radius of D **or** infinity, then in R(D), u must be an isolated vertex. SinceD is connected and R(D) contains an isolated vertex, D is not self radial. Supposethe radius of D is infinite, then there exist two vertices u and v in D such that nou − v **directed** path and no v − u **directed** path exist in D. Theref**or**e, both the arcs(u, v) and (v, u) are in R(D). Since D contains no **directed** cycles and R(D) containsa **directed** 2-cycles, D is not self radial. Hence, q(D) ≥ p.F**or** the upperbound, since R(D) ≠ D, R(D) ⊂ D. Now, D ∼ = R(D) ⊂ D impliesthat q(D) ≤ q(D). Hence, q(D) ≤ 1 2 q(K∗ p) = p(p − 1)/2.□✉✻✲ ✉✣✢✉✛✉❄Figure 5.The bounds in The**or**em 3.2 are sharp. The lower bound is sharp f**or** the class of**directed** cycles and Figure 5 is an example of minimum **or**der self complementary selfcentered di**graph** of radius 2 which satisfies the sharpness of the upperbound.References[1] R. Aravamuthan and B. Rajendran, Graph equations involving antipodal **graph**s, Presented atthe seminar on Combinat**or**ics and applications held at ISI, Calcutta during 14-17 December1982, 40-43.[2] R. Aravamuthan and B. Rajendran, On antipodal **graph**s, Discrete Math. 49 (1984), 193-195.[3] F. Buckley and F. Harary, Distance in **graph**s, Addison-Wesley Redwood City, CA 1990.[4] F. Buckley, The Eccentric di**graph** of a **graph**, Cong. Numer. 149 (2001), 65-76.[5] G. Chartrand and L. Lesniak, Graphs and Di**graph**s, 3rd Edition, Chapman and Hall, London,1996.[6] G. Chartrand and P. Zhang, Distance in **graph**s-Taking the long View, AKCE J. Graphs. Combin.1 (2004), 1-13.[7] G. Chartrand and O. R. Oellermann, Applied and Alg**or**ithmic Graph The**or**y, McGraw-Hill,New Y**or**k, 1993.[8] G. Johns, A simple proof of the characterization of antipodal **graph**s, Discrete Math. 128 (1994),399-400.

170 KUMARAPPAN KATHIRESAN AND R. SUMATHI[9] KM. Kathiresan, G. Marimuthu and S. Arockiaraj, Dynamics of radial **graph**s, Bull. Inst. Combin.Appl. 57 (2009), 21-28.[10] KM. Kathiresan and G. Marimuthu, A study on radial **graph**s, Ars Combin., to appear.[11] KM. Kathiresan and G. Marimuthu, Further results on radial **graph**s, Discussiones Mathematica,to appear.[12] G. Marimuthu, A study on radius and diameter of **graph**s, Ph. D Thesis, Madurai KamarajUniversity, Madurai, 2006.[13] R. R. Singleton, There is no irregular mo**or**e **graph**, Amer. Math. Monthly 7 (1968), 42-43.Centre f**or** Research and PostGraduate Studies in Mathematics,Ayya Nadar Janaki Ammal College (Autonomous),Sivakasi West - 626 124,Tamil Nadu, IndiaE-mail address: kathir2esan@yahoo.comE-mail address: sumathiramint@yahoo.com