On the Line Degree Splitting Graph of a Graph

In this paper, we introduce the concept of the line degree splitting graph of a graph. We obtain some properties of this graph. We find the girth of the line degree splitting graphs. Further, we establish the characterization of graphs whose line degree splitting graphs are eulerian, complete bipartite graphs and complete graphs.


Introduction
By a graph, we mean a finite, undirected graph without loops or multiple lines. For a graph G, let V(G), E(G) and L(G) denote its point set, line set and line graph respectively. We refer the terminology of [2].

Remark 1.
For any graph G without isolated points, L(G) is an induced subgraph of DL s (G). Clearly if G contains at least two lines, then G contains at least two lines of the same degree. Hence G = K 2 is the only graph such that L(G) = DL s (G).

Remark 2.
If G is connected then DL s (G) is connected. But not conversely.
For example, if G = nK 2 then DL s (G) = K 1,n .

Remark 3.
A graph G is a cycle C n , n  3 if and only if DL s (G) is a wheel W n+1 .

Remark 4.
If G is a star K 1,n , n  2 then DL s (G) = K n+1 .

Remark 5.
Let e  uv be a line of a graph G.
Theorem A [2]. Unless p = 8, a graph G is the line graph of K p if and only if 3) Every two nonadjacent points are mutually adjacent to exactly four points, 4) Every two adjacent points are mutually adjacent to exactly p -2 points.
When p = 8, there are exactly three exceptional graphs satisfying the conditions.
Theorem B [2]. Unless m = n = 4, a graph G is the line graph of K m,n if and only if 1) G has mn points, 2) G is regular of degree m + n -2, 3) Every two nonadjacent points are mutually adjacent to exactly two points, There is only one exceptional graph satisfying these conditions. It has 16 points, is not L(K 4,4 ), and was found by Shrikhande [7] when he proved Theorem B for the case m = n.

Results
Theorem 1. Let G be a graph with p points and q lines whose points have degree d i and let r be the number of lines in T and t be the number of points in E 1 (G).
Then | V(DL s (G)) | = q + t and | E(DL s (G)) | = r d 2 Proof. DL s (G) of a graph G is obtained from L(G) by adding t new points which corresponds to the set S i , 1  i  t, where S i is as in the definition of DL s (G). Therefore, the number of points in DL s (G) is the sum of points of L(G) and t. Hence | V(DL s (G)) | = q + t.
It is known in [2, pp 72], that L(G) has      Proof. Suppose DL s (G) is a cycle. By Remark 1, L(G) is an induced subgraph of DL s (G). It is clear that L(G) is acyclic and hence G is acyclic. Assume G has at least one component containing a point of degree 3. Then K 1,3 is a subgraph of G and hence K 3 is subgraph of L(G), a contradiction. Hence, it follows that every component of G is a tree whose points are of degree  2. It is clear that every component of G is a path. Further if G is a path P n , n > 4 then L(G) is P n-1 . DL s (G) contains (n -3) number of cycles, a contradiction. Assume every component of G is K 2 , then by Corollary 2.1, DL s (G) = K 1,n , a contradiction. Hence, every component of G is a path P n , 2  n  3. Now assume G

Bulletin of Mathematical Sciences and Applications Vol. 18
has at least two such paths as its components. Then by Theorem 2, DL s (G) contains more than one cycle. Hence, G is connected and is either P 3 or P 4 . Conversely, suppose G = P 3 or P 4 . Then it is easy to see that DL s (G) is C 3 or C 4 respectively. Hence DL s (G) is a cycle. □ Theorem 4. For any connected graph G of size q  2, Proof. Let G be a connected graph of size q  2. If G = P 4 , then by Theorem 3, DL s (G) is C 4 and hence its girth is 4. Otherwise (i) If either K 3 or K 1,3 is a subgraph of G, then DL s (G) contains a triangle.
(ii) If G is any cycle, then by Remark 3, DL s (G) is a wheel.
(iii) If G is a path P n , n  5, then DL s (G) contains (n -4) number of triangles. And if G = P 3 , then by Theorem 3, DL s (G) is K 3 .

Theorem 5. DL s (G) is bipartite if and only if every component of G is either K 2 or P 4 .
Proof. Suppose DL s (G) is bipartite. Assume G has at least one point of degree  3. Then L(G) contains K 3 as a subgraph and by Remark 1, DL s (G) contains an odd cycle, a contradiction. Hence every point of G is of degree  2. Clearly every component of G is either a cycle or a path. We consider the following cases.
Case 1. Assume one of the component of G is a cycle. By Remark 3, DL s (G) has a wheel as its subgraph, a contradiction.

Case 2.
Suppose one of the component of G is 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 triangles, a contradiction. Hence, every component of G is a path P n , n  4. Assume G has at least one component as P 3 . Then by Theorem 3, DL s (G) contains a triangle, a contradiction. Hence, every component of G is either P 4 or K 2 . Conversely, suppose G is a graph such that each of its component is either K 2 or P 4 . We consider the following cases.
Case 1. Assume every component of G is K 2 . Then by Corollary 2.1, DL s (G) is K 1,n , which is a bipartite graph.

Case 2.
Assume every component of G is P 4 . Suppose G has exactly one component. Then by Theorem 2, DL s (G) is C 4 , which is a bipartite graph. Now suppose G has more than one component. Then L(G) contains more than one components each of which is K 1,2 . We construct DL s (G) as follows. Since each P 4 contains one line of line degree 4 and the other line of line degree 3. Hence DL s (G) contains two points w 1 and w 2 corresponding to the set S 1 of lines of line degree 4 and set S 2 of lines of line degree 3 respectively. Now partition the points of DL s (G) such that V 1 (DL s (G)) contains w 1 and all those points of L(G) of degree 1 and V 2 (DL s (G)) contains w 2 and all those points of L(G) of degree 2. And w 1 is adjacent to all points of V 2 (DL s (G)) except w 2 and w 2 is adjacent to all points of V 1 (DL s (G)) except w 1 . Clearly the resulting graph DL s (G) is bipartite. Proof. If G is regular of degree k, then L(G) is also regular of degree 2(k -1) with q points and q 2 pk 2  lines. Let e  V(L(G)). Then deg(e) = 2(k -1)  q -1.

Corollary 8.2. DL s (G) is line-regular if and only if G is K 1,n or K 3 .
Theorem 9. Let G be any graph. A necessary and sufficient condition for DL s (G) to be eulerian is that each of the following holds: 1) Each even degree line of G occurs exactly once.
2) Each odd degree line of G occurs even number of times.

Bulletin of Mathematical Sciences and Applications Vol. 18
Hence all such x which belongs to T are even degree lines. Thus (1) holds. Suppose next x belongs to S i . Then either deg(u) or deg(v) is odd. This implies that x is adjacent to odd number of lines. Hence all such x are odd degree lines and since they belong to S i , the point w i belonging to E 1 (G) in DL s (G) contributes one to the degree of x in DL s (G). Thus degree of x in DL s (G) is even. If x is a point of E 1 (G), then Remark 6 implies that deg*(x) = | E 1 (G) |. Since deg*(x) is even, | E 1 (G) | is even. Thus, each S i contains even number of lines of odd degree. Hence (2) holds. Conversely, assume the condition (1) and (2)  The converse of the Theorem 11 is need not be true. This is easily seen in Fig. 2.

Figure 2.
In the next results we present characterization of line degree splitting graph of complete bipartite graphs and complete graphs. The line degree splitting graph of complete bipartite graphs and complete graphs are characterized by immediate observations involving adjacencies and degree of lines in K m,n and K p .
In the following theorem, we establish the characterization for line degree splitting graphs of complete bipartite graphs.

Properties of DL s (K m,n ):
If we suppose that m ≥n ≥1 it can be seen that DL s (K m,n ) has the following four properties.
1) The graph has mn + 1 points, 2) Of the mn + 1 points, mn points are of degree m + n -1 and one point is of degree mn, 3) Every two nonadjacent points are mutually adjacent to three points,  4). Here we have tried to give a generalization for line degree splitting graphs, which is similar to the characterization given for line graphs which was studied by Moon [5] and Hoffman [3]. Before proving that if any graph satisfies the conditions (1) -(4) then it is isomorphic to DL s (K m,n ) except for (m, n) = (4, 4), we state and prove the following lemma.

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Volume 18 Lemma 1. Let there be given a graph G satisfying conditions (1), (2), (3) and (4) where m ≥ n ≥1, but (m, n)  (4, 4), (5,4) or (4,3). Let p 11 and p 12 be two adjacent points of same degree of G which are mutually adjacent to each of the m -1 points in A = {p 13 , …, p 1m , w}. Let C 1 = {p 21 , …, p n1 } be the set of n -1 points which are adjacent to p 11 but not to p 12 . Furthermore let there be at least m -1 points in A  C 1 such that each of these points and p 11 are mutually adjacent to m -1 other points. Then A  p 11  p 12 and C 1  p 11 are the point set of complete graphs of m + 1 and n points respectively, and no point of A is adjacent to any of C 1 whose degree is same as that of the point of A.
Proof. We consider first the case in which m ≥ n ≥5. No point in C 1 can be adjacent to more than two points in A without violating condition (3) with respect to the point p 12 . Therefore, each point in C 1 is adjacent to at least n -3 of the remaining n -2 points in C 1 in order to satisfy condition (4) with respect to the point p 11 . Then, if there exists two nonadjacent points in C 1 they must be mutually adjacent to the remaining n -3 points in C 1 as well as to p 11 and w. This contradicts condition (3) since n -2 ≥ 3 in the case being considered. Hence, every point in C 1 is adjacent to every other point in C 1 . If some point in C 1 , p 21 say, is adjacent to some point in A whose degree is same as p 21 then there are n points adjacent to both p 21 and p 11 . Thus n must equal m -1, by (4), or m = n + 1. If m = n + 1 suppose that some point in A, p 13 say, is adjacent to some points of C 1 . It is easily seen that p 13 cannot be adjacent to more than one point of C 1 but not to all points of C 1 without violating condition (3). But if p 13 is adjacent to all points of C 1 then the number of points which are adjacent to both p 11 and p 13 is, including p 12 , at least n which contradicts condition (4). Hence, p 13 can be adjacent to at most one point in C 1 . From condition (4) it follows that there is at least one other point in A, p 14 say, which is not adjacent to p 13 . But p 13 and p 14 are each adjacent to at least m -4 of the remaining m -3 points of A and if m > 6 there will be at least two of these points which is adjacent to both p 13 and p 14 . This, however, contradicts condition (3) since p 13 and p 14 are both adjacent to p 11 and p 12 .
The only alternative remaining to be treated, under the assumption that m = n + 1 and that some lines join points in A to points in C 1 , is when m = n + 1 = 6. In this case it is not difficult to see that the only configuration which can satisfy condition (4) without implying a contradiction of the type just described is one in which each point of A except w is adjacent to p 15 , say, and p 14 is adjacent to p 16 . Suppose that p 21 and p 31 are the different points in C 1 which are adjacent to p 13 and p 15 , respectively. Then p 21 and p 15 are not adjacent to each other but are mutually adjacent to w, p 31 , p 13 and p 11 , contradicting condition (3). Hence, no point of A is adjacent to any point of C 1 under the given assumptions. This and the fourth sentence of the hypothesis of the lemma implies that each point in A is adjacent to every other point in A which suffices to complete the proof of the lemma when m≥ n ≥5.
Next consider the case in which n = 4 and m ≥6. No point in C 1 can be adjacent to m -1 other points of A  C 1 by an earlier remark and the fact that m 1 ≥5. Hence, from the hypothesis, each of the m -1 points of A must be adjacent to m -2 other points of A  C 1 . Using again the fact that no point in C 1 can be adjacent to more than two point in A it follows that there is at least one point of A which is not adjacent to any point in C 1 and hence is adjacent to each of the remaining m -2 points of A. To avoid contradicting condition (4) it must be that A  p 11  p 12 is the point set of a complete graph of m + 1 points. Condition (4) now implies that no point of A is adjacent to any point of C 1 and that C 1  p 11 is the point set of complete graph of 4 points, which completes the proof of the lemma for this case.
An entirely analogous argument proves the lemma when n = 3 and m ≥ 5 (see Figure 3).
By the construction of DL s (G) we have considered that its validity when n = m = 3 follows from the similar result to that of Shrikhande [8] for line graphs and the remaining cases, when n = 1 or 2, are also easily established.