Volume 18 Figure 1. Notation 1. Notation 2. Observation 1. Remark 1. Remark 2. Remark 3. Remark 4. Remark 5. Remark 6. Theorem A [2]. Theorem B [2].

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1 Bulletn of Mathematcal Scences and Applcatons Submtted: ISSN: , Vol. 18, pp 1-10 Revsed: do: / Accepted: ScPress Ltd., Swtzerland Onlne: On the Lne Degree Splttng Graph of a Graph B. Basavanagoud * and Roopa S. Kusugal Department of Mathematcs, Karnatak Unversty, Dharwad , Inda * b.basavanagoud@gmal.com Keywords: lne degree, lne graph, lne degree splttng graph. Abstract: In ths paper, we ntroduce the concept of the lne degree splttng graph of a graph. We obtan some propertes of ths graph. We fnd the grth of the lne degree splttng graphs. Further, we establsh the characterzaton of graphs whose lne degree splttng graphs are euleran, complete bpartte graphs and complete graphs. 1. Introducton By a graph, we mean a fnte, undrected graph wthout loops or multple lnes. For a graph G, let V(G), E(G) and L(G) denote ts pont set, lne set and lne graph respectvely. We refer the termnology of []. The degree of a pont v V (G) s the number of ponts adjacent to v and s denoted by deg(v). If u and v are two adjacent ponts of G, then the lne connectng them wll be denoted by uv. The degree of a lne e n G s denoted by deg(e), and s defned by deg(e) = deg(u) + deg(v)- wth e = uv. A graph G s sad to be the lne-regular f all lnes have same degree.the open-neghborhood N(e ) of a lne e n E(G) s the set of lnes adjacent to e. For each lne e of G, a new pont e s taken and the resultng set of ponts s denoted by A(G). The lne splttng graph L s (G) of a graph G s defned as the graph havng pont set E(G) A(G) wth two ponts are adjacent f they correspond to adjacent lnes of G or one corresponds to an element e of A(G) and the other to an element e of E(G) where e s n N( e j j ). Ths concept was ntroduced by Kull and Bradar n [4]. Mscellaneous propertes of lne splttng graphs are studed by Basavanagoud and Mathad n [1]. Let G = (V, E) be a graph wth E = S 1 S S t T where each S s a set of lnes of G havng at least two lnes and havng the same lne degree and T = E \ S. For each set S of lnes of G, a new pont w s taken and the resultng set of ponts w s denoted by E 1 (G). The lne degree splttng graph DL s (G) of a graph G s defned as the graph havng pont set E(G) E 1 (G) wth two ponts are adjacent f they correspond to adjacent lnes of G or one correspond to a pont w of E 1 (G) and the other to a lne e of G and e s n S. In Fg. 1, graph G and ts lne degree splttng graph are shown. j j Here S 1 = {e 4, e 6 }, S = {e, e 5, e 7 }, T = {e 1, e 3, e 8 } ScPress apples the CC-BY 4.0 lcense to works we publsh:

2 Volume 18 Fgure 1. Notaton 1. deg*(e) denoted the degree of a pont e n DL s (G). Clearly deg(e) deg*(e) for all e n E(G). Notaton. S, T and E 1 (G) are defned as n the defnton of DL s (G). Observaton 1. For any graph, L(G) s an nduced subgraph of DL s (G). Remark 1. For any graph G wthout solated ponts, L(G) s an nduced subgraph of DL s (G). Clearly f G contans at least two lnes, then G contans at least two lnes of the same degree. Hence G = K s the only graph such that L(G) = DL s (G). Remark. If G s connected then DL s (G) s connected. But not conversely. For example, f G = nk then DL s (G) = K 1,n. Remark 3. A graph G s a cycle C n, n 3 f and only f DL s (G) s a wheel W n+1. Remark 4. If G s a star K 1,n, n then DL s (G) = K n+1. Remark 5. Let e uv be a lne of a graph G. ) If e belongs to S, then deg*(e) = deg(u) + deg(v) 1. ) If e belongs to T, then deg*(e) = deg(u) + deg(v). Remark 6. Let w be a pont belongng to E 1 (G). Then deg*(w ) = E 1 (G). Theorem A []. Unless p = 8, a graph G s the lne graph of K p f and only f p 1) G has ponts, ) G s regular of degree (p ), 3) Every two nonadjacent ponts are mutually adjacent to exactly four ponts, 4) Every two adjacent ponts are mutually adjacent to exactly p ponts. When p = 8, there are exactly three exceptonal graphs satsfyng the condtons. Theorem B []. Unless m = n = 4, a graph G s the lne graph of K m,n f and only f 1) G has mn ponts, ) G s regular of degree m + n, 3) Every two nonadjacent ponts are mutually adjacent to exactly two ponts,

3 Bulletn of Mathematcal Scences and Applcatons Vol m 4) Among the adjacent pars of ponts, exactly n pars are mutually adjacent to exactly n m ponts, and the other m pars to n ponts. There s only one exceptonal graph satsfyng these condtons. It has 16 ponts, s not L(K 4,4 ), and was found by Shrkhande [7] when he proved Theorem B for the case m = n.. Results Theorem 1. Let G be a graph wth p ponts and q lnes whose ponts have degree d and let r be the number of lnes n T and t be the number of ponts n E 1 (G). Then V(DL s (G)) = q + t and 1 p E(DL s (G)) = d r. 1 Proof. DL s (G) of a graph G s obtaned from L(G) by addng t new ponts whch corresponds to the set S, 1 t, where S s as n the defnton of DL s (G). Therefore, the number of ponts n DL s (G) s the sum of ponts of L(G) and t. Hence V(DL s (G)) = q + t. It s known n [, pp 7], that L(G) has q 1 p d 1 lnes. By the defnton of DL s (G), the number of lnes n DL s (G) s the sum of number of lnes n L(G) and the lnes due to t new ponts w 1, w,, w t. Hence E(DL s (G)) = E(L(G)) + deg *(w ) t 1 1 q p 1 d q r (1) 1 p d r 1 Theorem. Let G be a connected graph of sze q > 1 or a dsconnected graph wth at least one component havng sze q > 1. Then DL s (G) contans a cycle. Proof. By the defnton of DL s (G), the proof s trval. Corollary.1. Let G be any graph. Then DL s (G) s acyclc f and only f the number of lnes n every component of G s 1, that s, every component of G s ether K 1 or K. Further, DL s (G) = K 1,n when every component of G s K. Theorem 3. Let G be a graph wthout solated ponts. Then DL s (G) s a cycle f and only f G s P 3 or P 4. Proof. Suppose DL s (G) s a cycle. By Remark 1, L(G) s an nduced subgraph of DL s (G). It s clear that L(G) s acyclc and hence G s acyclc. Assume G has at least one component contanng a pont of degree 3. Then K 1,3 s a subgraph of G and hence K 3 s subgraph of L(G), a contradcton. Hence, t follows that every component of G s a tree whose ponts are of degree. It s clear that every component of G s a path. Further f G s a path P n, n > 4 then L(G) s P n-1. DL s (G) contans (n 3) number of cycles, a contradcton. Assume every component of G s K, then by Corollary.1, DL s (G) = K 1,n, a contradcton. Hence, every component of G s a path P n, n 3. Now assume G

4 4 Volume 18 has at least two such paths as ts components. Then by Theorem, DL s (G) contans more than one cycle. Hence, G s connected and s ether P 3 or P 4. Conversely, suppose G = P 3 or P 4. Then t s easy to see that DL s (G) s C 3 or C 4 respectvely. Hence DL s (G) s a cycle. Theorem 4. For any connected graph G of sze q, g(dl s (G)) = 4 f G P4 3 otherwse Proof. Let G be a connected graph of sze q. If G = P 4, then by Theorem 3, DL s (G) s C 4 and hence ts grth s 4. Otherwse () If ether K 3 or K 1,3 s a subgraph of G, then DL s (G) contans a trangle. () If G s any cycle, then by Remark 3, DL s (G) s a wheel. () If G s a path P n, n 5, then DL s (G) contans (n 4) number of trangles. And f G = P 3, then by Theorem 3, DL s (G) s K 3. Hence (), () and () mples that g(dl s (G)) = 3 for all G other than P 4. Theorem 5. DL s (G) s bpartte f and only f every component of G s ether K or P 4. Proof. Suppose DL s (G) s bpartte. Assume G has at least one pont of degree 3. Then L(G) contans K 3 as a subgraph and by Remark 1, DL s (G) contans an odd cycle, a contradcton. Hence every pont of G s of degree. Clearly every component of G s ether a cycle or a path. We consder the followng cases. Case 1. Assume one of the component of G s a cycle. By Remark 3, DL s (G) has a wheel as ts subgraph, a contradcton. Case. Suppose one of the component of G s a path P n, n 5. Then L(G) has P n-1 as a component and DL s (G) has at least (n 4) number of trangles, a contradcton. Hence, every component of G s a path P n, n 4. Assume G has at least one component as P 3. Then by Theorem 3, DL s (G) contans a trangle, a contradcton. Hence, every component of G s ether P 4 or K. Conversely, suppose G s a graph such that each of ts component s ether K or P 4. We consder the followng cases. Case 1. Assume every component of G s K. Then by Corollary.1, DL s (G) s K 1,n, whch s a bpartte graph. Case. Assume every component of G s P 4. Suppose G has exactly one component. Then by Theorem, DL s (G) s C 4, whch s a bpartte graph. Now suppose G has more than one component. Then L(G) contans more than one components each of whch s K 1,. We construct DL s (G) as follows. Snce each P 4 contans one lne of lne degree 4 and the other lne of lne degree 3. Hence DL s (G) contans two ponts w 1 and w correspondng to the set S 1 of lnes of lne degree 4 and set S of lnes of lne degree 3 respectvely. Now partton the ponts of DL s (G) such that V 1 (DL s (G)) contans w 1 and all those ponts of L(G) of degree 1 and V (DL s (G)) contans w and all those ponts of L(G) of degree. And w 1 s adjacent to all ponts of V (DL s (G)) except w and w s adjacent to all ponts of V 1 (DL s (G)) except w 1. Clearly the resultng graph DL s (G) s bpartte. Case 3. Assume G has at least two components one of whch s P 4 and the other s K. Then lne graph of each P 4 s K 1, and each K s K 1. It s easy to see that lne degree of lnes of P 4 s dfferent from the lne degree of lnes of K. Hence no pont of E 1 (G) of DL s (G) whch corresponds to the set of lnes of same lne degree s adjacent to the ponts of both K 1, and K 1 n L(G). By Case 1 and Case of the converse part of ths theorem, DL s (G) contans two components each of whch s bpartte. Hence DL s (G) s bpartte.

5 Bulletn of Mathematcal Scences and Applcatons Vol Theorem 6. For any graph G, (DL s (G)) (G), where (G) denotes the number of components of G. Proof. Let V(L(G)) = {e 1, e,, e q } and V(DL s (G)) \ V(L(G)) = {w 1, w,, w t }. Case 1. Let G be connected. Then by Remark, DL s (G) s connected. Therefore, (G) = 1 = (DL s (G)). Case. Let G be dsconnected. Then L(G) s dsconnected. Let G 1, G,, G k, k be the components of G. If G and G j have lnes of same degree for some, j. Let e 1 E(G 1 ) and e E(G ) be such that deg(e 1 ) = deg(e ). By the defnton of DL s (G), DL s (G) contans a new pont w such that w s adjacent to e 1 and e of L(G) n DL s (G). Hence, (DL s (G)) k 1 < k = (G). If there s no par and j such that G and G j have lnes of same degree, then (DL s (G)) = k = (G). Hence (DL s (G)) (G). Theorem 7. The DL s (G) of a nontrval graph G s complete f and only f G s ether K 3 or K 1,n, n 1. Proof. Suppose DL s (G) s complete graph of order. Then every par of ponts of DL s (G) are adjacent. Thus all the lnes of G are mutually adjacent and hence have same lne degree. Clearly there exsts exactly one new pont n DL s (G) whch s adjacent to all the ponts of L(G). Suppose G K 3 and G has at least one cycle. Then t s easy to observe that there exst at least two nonadjacent lnes. It mples that L(G) s not complete. By Remark 1, DL s (G) s not complete, whch s a contradcton. Suppose next G K 1,n, n 1 and G s a tree wth at least two nonadjacent lnes n G. Then agan L(G) s not complete. By Remark 1, DL s (G) s not complete a contradcton. Conversely, suppose G = K 3. It has three mutually adjacent lnes whch are of same lne degree. Then these three lnes are the ponts of DL s (G) together wth a new pont correspondng to the lnes of same lne degree of G n DL s (G). These four ponts are mutually adjacent ponts n DL s (G). Thus DL s (G) s K 4. Now suppose G s K 1,n, n. By Remark 4, DL s (G) s K n+1. Thus DL s (G) s complete. Theorem 8. If G ( K and K 3 ) s regular graph wth p ponts and q lnes, then DL s (G) s not regular. Proof. If G s regular of degree k, then L(G) s also regular of degree (k 1) wth q ponts and pk q lnes. Let e V(L(G)). Then deg(e) = (k 1) q 1. Clearly (k 1) < q 1, except for G = K and K 3 () Let V(DL s (G)) \ V(L(G)) = {w}. Snce w s adjacent to all the lnes of G n L(G), deg*(w) = q > (k 1) + 1 = k 1. (by ()). But deg*(e) = deg(e) + 1 = (k 1) + 1 = k 1. Hence DL s (G) s not regular. Corollary 8.1. If G s k-regular graph, then DL s (G) = L(G) + K 1. Corollary 8.. DL s (G) s lne-regular f and only f G s K 1,n or K 3. Theorem 9. Let G be any graph. A necessary and suffcent condton for DL s (G) to be euleran s that each of the followng holds: 1) Each even degree lne of G occurs exactly once. ) Each odd degree lne of G occurs even number of tmes. Proof. Suppose DL s (G) s euleran. Let x be a pont of DL s (G). Then x s a lne of G or a pont of E 1 (G). Suppose x = uv s a lne of G. Remark 5 mples that f x belongs to T, then deg*(x) = deg(u) + deg(v). If x belongs to S, then deg*(x) = deg(u) + deg(v) 1. Suppose x belongs to T. Then both deg u and deg v must be even or odd. Ths mples that x s adjacent to even number of lnes of G.

6 6 Volume 18 Hence all such x whch belongs to T are even degree lnes. Thus (1) holds. Suppose next x belongs to S. Then ether deg(u) or deg(v) s odd. Ths mples that x s adjacent to odd number of lnes. Hence all such x are odd degree lnes and snce they belong to S, the pont w belongng to E 1 (G) n DL s (G) contrbutes one to the degree of x n DL s (G). Thus degree of x n DL s (G) s even. If x s a pont of E 1 (G), then Remark 6 mples that deg*(x) = E 1 (G). Snce deg*(x) s even, E 1 (G) s even. Thus, each S contans even number of lnes of odd degree. Hence () holds. Conversely, assume the condton (1) and () on G hold. Suppose x s a pont of DL s (G). Then x s a lne of G or x belongs to E 1 (G). Assume x s a lne of G. If x belongs to T, then by (1), deg*(x) s even. And f x belongs S, then by () and defnton of DL s (G), deg*(x) s even. Now f x belongs to E 1 (G), then by condton (), each S contans even number of lnes of odd degree and hence deg*(x) s even. Hence DL s (G) s euleran. Theorem 10. Let G be a nontrval graph. If DL s (G) s euleran, then G s not euleran. Proof. Suppose DL s (G) s euleran. Theorem mples that there exsts at least one component n G of sze q > 1. Then by Remark 1, G has at least two lnes x and y of same lne degree. Let x and y be the ponts of DL s (G) correspondng to the lnes x and y of G respectvely. Snce DL s (G) s euleran deg*(x) and deg*(y) are even. Then by Remark 5, one of the end ponts of both the lnes x and y n G s odd. Thus, G s not euleran. Theorem 11. If G s euleran, then DL s (G) s hamltonan. The converse of the Theorem 11 s need not be true. Ths s easly seen n Fg.. Fgure. In the next results we present characterzaton of lne degree splttng graph of complete bpartte graphs and complete graphs. The lne degree splttng graph of complete bpartte graphs and complete graphs are characterzed by mmedate observatons nvolvng adjacences and degree of lnes n K m,n and K p. In the followng theorem, we establsh the characterzaton for lne degree splttng graphs of complete bpartte graphs. Propertes of DL s (K m,n ): If we suppose that m n 1 t can be seen that DL s (K m,n ) has the followng four propertes. 1) The graph has mn + 1 ponts, ) Of the mn + 1 ponts, mn ponts are of degree m + n 1 and one pont s of degree mn, 3) Every two nonadjacent ponts are mutually adjacent to three ponts, m 4) n of the pars of adjacent ponts of same degree are mutually adjacent to m 1 other n ponts, m of the remanng pars of adjacent ponts of same degree are mutually adjacent to n 1 ponts and every two adjacent ponts of dfferent degree are mutually adjacent to exactly m + n ponts. The object of ths note s to show that f any graph satsfes the four condtons (1) (4), then t s somorphc to DL s (K m,n ) except possbly when (m, n) = (4, 4). Here we have tred to gve a generalzaton for lne degree splttng graphs, whch s smlar to the characterzaton gven for lne graphs whch was studed by Moon [5] and Hoffman [3]. Before provng that f any graph satsfes the condtons (1) (4) then t s somorphc to DL s (K m,n ) except for (m, n) = (4, 4), we state and prove the followng lemma.

7 Bulletn of Mathematcal Scences and Applcatons Vol Lemma 1. Let there be gven a graph G satsfyng condtons (1), (), (3) and (4) where m n 1, but (m, n) (4, 4), (5, 4) or (4, 3). Let p 11 and p 1 be two adjacent ponts of same degree of G whch are mutually adjacent to each of the m 1 ponts n A = {p 13,, p 1m, w}. Let C 1 = {p 1,, p n1 } be the set of n 1 ponts whch are adjacent to p 11 but not to p 1. Furthermore let there be at least m 1 ponts n A C 1 such that each of these ponts and p 11 are mutually adjacent to m 1 other ponts. Then A p 11 p 1 and C 1 p 11 are the pont set of complete graphs of m + 1 and n ponts respectvely, and no pont of A s adjacent to any of C 1 whose degree s same as that of the pont of A. Proof. We consder frst the case n whch m n 5. No pont n C 1 can be adjacent to more than two ponts n A wthout volatng condton (3) wth respect to the pont p 1. Therefore, each pont n C 1 s adjacent to at least n 3 of the remanng n ponts n C 1 n order to satsfy condton (4) wth respect to the pont p 11. Then, f there exsts two nonadjacent ponts n C 1 they must be mutually adjacent to the remanng n 3 ponts n C 1 as well as to p 11 and w. Ths contradcts condton (3) snce n 3 n the case beng consdered. Hence, every pont n C 1 s adjacent to every other pont n C 1. If some pont n C 1, p 1 say, s adjacent to some pont n A whose degree s same as p 1 then there are n ponts adjacent to both p 1 and p 11. Thus n must equal m 1, by (4), or m = n + 1. If m = n + 1 suppose that some pont n A, p 13 say, s adjacent to some ponts of C 1. It s easly seen that p 13 cannot be adjacent to more than one pont of C 1 but not to all ponts of C 1 wthout volatng condton (3). But f p 13 s adjacent to all ponts of C 1 then the number of ponts whch are adjacent to both p 11 and p 13 s, ncludng p 1, at least n whch contradcts condton (4). Hence, p 13 can be adjacent to at most one pont n C 1. From condton (4) t follows that there s at least one other pont n A, p 14 say, whch s not adjacent to p 13. But p 13 and p 14 are each adjacent to at least m 4 of the remanng m 3 ponts of A and f m > 6 there wll be at least two of these ponts whch s adjacent to both p 13 and p 14. Ths, however, contradcts condton (3) snce p 13 and p 14 are both adjacent to p 11 and p 1. The only alternatve remanng to be treated, under the assumpton that m = n + 1 and that some lnes jon ponts n A to ponts n C 1, s when m = n + 1 = 6. In ths case t s not dffcult to see that the only confguraton whch can satsfy condton (4) wthout mplyng a contradcton of the type just descrbed s one n whch each pont of A except w s adjacent to p 15, say, and p 14 s adjacent to p 16. Suppose that p 1 and p 31 are the dfferent ponts n C 1 whch are adjacent to p 13 and p 15, respectvely. Then p 1 and p 15 are not adjacent to each other but are mutually adjacent to w, p 31, p 13 and p 11, contradctng condton (3). Hence, no pont of A s adjacent to any pont of C 1 under the gven assumptons. Ths and the fourth sentence of the hypothess of the lemma mples that each pont n A s adjacent to every other pont n A whch suffces to complete the proof of the lemma when m n 5. Next consder the case n whch n = 4 and m 6. No pont n C 1 can be adjacent to m 1 other ponts of A C 1 by an earler remark and the fact that m 1 5. Hence, from the hypothess, each of the m 1 ponts of A must be adjacent to m other ponts of A C 1. Usng agan the fact that no pont n C 1 can be adjacent to more than two pont n A t follows that there s at least one pont of A whch s not adjacent to any pont n C 1 and hence s adjacent to each of the remanng m ponts of A. To avod contradctng condton (4) t must be that A p 11 p 1 s the pont set of a complete graph of m + 1 ponts. Condton (4) now mples that no pont of A s adjacent to any pont of C 1 and that C 1 p 11 s the pont set of complete graph of 4 ponts, whch completes the proof of the lemma for ths case. An entrely analogous argument proves the lemma when n = 3 and m 5 (see Fgure 3). By the constructon of DL s (G) we have consdered that ts valdty when n = m = 3 follows from the smlar result to that of Shrkhande [8] for lne graphs and the remanng cases, when n = 1 or, are also easly establshed.

8 8 Volume 18 Theorem 1. Let there be gven a graph G satsfyng condtons (1), (), (3) and (4), where m n 1 but (m, n) (4, 4). Then G s somorphc to DL s (K m,n ). L(K 5,3 ) : p 11 p 1 p 13 p 14 p 15 p 1 p 31 Fgure 3. Proof. We prove the theorem by consderng the followng cases. Case 1. When (m, n) (5, 4) or (4, 3). m Condton (4) mples that there are n ponts p of same degree n G, countng multplctes, for whch there exsts another pont q of G such that p s adjacent to q, and p 1 and q are m mutually adjacent to m 1 other ponts of G. Snce n / mn m 1t follows that there exsts two ponts, p 11 and p 1 say, whch satsfy the hypothess of the lemma. Retanng the notaton of the lemma let C = {p,, p n } be the set of n 1 ponts whch are adjacent to p 1 but not to p 11 ; by symmetry t follows that C p 1 s the pont set of a complete graph of n ponts and no pont n C s adjacent to any pont n A whose degree s same as the degree of the consdered pont n C. By applyng condton (3) to the ponts of C 1 wth respect to p 1 and to the ponts of C wth respect to p 11 we see that each pont of C 1 s adjacent to one, and only one, pont of C and vce versa. We may assume that the ponts are labeled n such a way that p j1 s adjacent to p j, for j =,, n. The hypothess of the lemma are now satsfed wth any par of dstnct ponts, p 1 and p 1j, playng the roles of the ponts earler labeled as p 11 and p 1. Hence we may assert that for each pont p 1j, j = 1,, m, there exsts a set of n 1 ponts, C j = {p j,, p nj }, such that C j p 1j s the pont set of a complete graph of n ponts and no pont of C j s adjacent to any pont p l, where j. Also there are

9 Bulletn of Mathematcal Scences and Applcatons Vol no ponts common to C and C j and each pont of C s adjacent to one and only one pont of C j, for, j = 1,, m, j. Ths exhausts the ponts and lnes of G. Let p 3 be that pont of C 3 whch s adjacent to p. If p 3 s not adjacent to p 1 suppose that p 33 s the pont of C 3 whch s adjacent to p 1 and that p 31 s the pont of C 1 whch s adjacent to p 3. Then the nonadjacent ponts p 1 and p 3 are mutually adjacent to the dstnct ponts p, p 33, p 31 and w whch contradcts condton (3). Hence p 3 and p 1 are adjacent. Lettng p j be that pont of C j whch s adjacent to p j-1, for j = 4,, m, and repeatng ths argument t can be seen that the ponts p j, j = 1,, m, form the pont set of a complete graph of m + 1 ponts. Next let p 33 be the pont of C 3 whch s adjacent to p 3 and repeat the above argument. Carryng through ths procedure n 1 tmes t s seen that the mn ponts of G may be labeled p j, = 1,, n, j = 1,, m, n such a way that p j s adjacent to p rs f, and only f, = r or j = s, but not both and one pont w s adjacent to all other mn ponts of G. Ths shows that G s somorphc to DL s (K m,n ) under the gven condtons. Case. When (m, n) = (5, 4) or (4, 3). By Theorem B t s clear that, the graph H = L(K m,n ) s a regular graph of degree m + n. Hence n [3] Hoffman has gven the spectral characterzaton for graph H to be L(K m,n ) when (m, n) = (5, 4) or (4, 3). G = DL s (K m,n ) s obtaned from H by ntroducng a new pont n H and makng ths pont adjacent to all the ponts of H. Ths s because, by Observaton 1, H s an nduced subgraph of DL s (K m,n ) and snce H s a regular graph of degree m + n, every lne of K m,n s of lne degree m + n. Hence to obtan G = DL s (K m,n ) a unque pont w correspondng to the set of lnes of K m,n s ntroduced n H and ths pont s made adjacent to all the ponts of H. Thus degree of every pont n G whch corresponds to the ponts of H are ncreased by one, and the degree of w s equal to the number of ponts n H. Hence, G s a rregular graph. So we felt t dffcult to gve the spectral characterzaton for G = DL s (K m,n ). But by mmedate observaton of adjacences of lnes and degree of lnes of G, t s clear that G = DL s (K m,n ) satsfes condtons (1) (4). Case 3. When (m, n) = (4, 4). Ths theorem does not hold when (m, n) = (4, 4). Ths s because there exsts just one counter example (see Fg. 4). Ths graph contans 17 ponts, but t s not somorphc to DL s (K 4,4 ). Concluson: Studyng the above theorem, we conclude that unless m = n = 4, a graph G s the lne degree splttng graph of K m,n f and only f t satsfes the condtons (1) (4) of propertes of DL s (K m,n ). Fgure 4. In the followng theorem, we establsh the characterzaton for lne degree splttng graphs of complete graphs.

10 10 Volume 18 Theorem 13. Unless p = 8, a graph G s the lne degree splttng graph of K p f and only f p 1) G has 1 ponts, p p ) G has ponts of degree p 3 and one pont of degree, 3) Every two nonadjacent ponts are mutually adjacent to exactly fve ponts, 4) Every two adjacent ponts of same degree are mutually adjacent to exactly p 1 ponts and every two adjacent ponts of dfferent degree are mutually adjacent to exactly (p ) ponts. Proof. The proof s smlar to the proof of DL s (K m,n ). It s evdent that DL s (K p ) has the above four propertes. It s not at all obvous that when p = 8, there are exactly three exceptonal graphs satsfyng the above condtons. By Observaton 1, L(G) s an nduced subgraph of DL s (G). Hence, these three exceptonal graphs are obtaned by consderng the exceptonal graphs of Theorem A, and ntroducng a new pont n t and makng t adjacent to every other pont n t. The graphs obtaned thus are the exceptonal graphs for Theorem 13. References [1] B. Basavanagoud, V.N. Mathad, Mscellaneous propertes of lne splttng graphs, Proc. Nat. Conf. On Graphs, Comb., Algor., and Appl. (eds. S. Arumugam, B.D. Acharya, S. B. Rao) Narosa Publshng House, New Delh, Inda, 005, pp [] F. Harary, Graph Theory, Addson-Wesley, Readng, Mass, [3] A. J. Hoffman, On the lne-graph of the complete bpartte graph, Ann. Math. Statst. 35 (1964) [4] V.R. Kull, M.S. Bradar, The lne-splttng graph of graph, Acta Cenca Indca. 8(3) (00) [5] J. Moon, On the lne-graph of the complete bgraph, Ann. Math. Statst. 34 (1963) [6] R. Ponraj, S. Somasundaram, On the degree splttng graph of a graph, Natl. Acad. Sc. Lett. 7(7-8) (004) [7] S.S. Shrkhande, On a characterzaton of the trangular assocaton scheme, Ann. Math. Statst. 30 (1959) [8] S. S. Shrkhande, The unqueness of the L assocaton scheme, Ann. Math. Statst. 30 (1959)