On Pathos Lict Subdivision of a Tree

Transcription

1 International J.Math. Combin. Vol.4 (010), On Pathos Lict Subivision of a Tree Keerthi G.Mirajkar an Iramma M.Kaakol (Department of Mathematics, Karnatak Arts College, Dharwa , Karnataka, Inia) keerthi gm@yahoo.co.in, iramma.k@gmail.com Abstract: Let G be a graph an E 1 E(G). A Smaranachely E 1-lict graph n E 1 (G) of a graph G is the graph whose point set is the union of the set of lines in E 1 an the set of cutpoints of G in which two points are ajacent if an only if the corresponing lines of G are ajacent or the corresponing members of G are incient.here the lines an cutpoints of G are member of G. Particularly, if E 1 = E(G), a Smaranachely E(G)-lict graph n E(G) (G) is abbreviate to lict graph of G an enote by n(g). In this paper, the concept of pathos lict sub-ivision graph P n[s(t)] is introuce. Its stuy is concentrate only on trees. We present a characterization of those graphs, whose lict sub-ivision graph is planar, outerplanar, maximal outerplanar an minimally nonouterplanar. Further, we also establish the characterization for P n[s(t)] to be eulerian an hamiltonian. Key Wors: pathos, path number, Smaranachely lict graph, lict graph, pathos lict subivision graphs, Smaranache path k-cover, pathos point. AMS(010): 05C10, 05C99 1. Introuction The concept of pathos of a graph G was introuce by Harary [1] as a collection of minimum number of line isjoint open paths whose union is G. The path number of a graph G is the number of paths in a pathos. Stanton [7] an Harary [3] have calculate the path number for certain classes of graphs like trees an complete graphs. The subivision of a graph G is obtaine by inserting a point of egree in each line of G an is enote by S(G). The path number of a subivision of a tree S(T) is equal to K, where K is the number of o egree point of S(T). Also, the en points of each path of any pathos of S(T) are o points. The lict graph n(g) of a graph G is the graph whose point set is the union of the set of lines an the set of cutpoints of G in which two points are ajacent if an only if the corresponing lines of G are ajacent or the corresponing members of G are incient.here the lines an cutpoints of G are member of G. For any integer k 1, a Smaranache path k-cover of a graph G is a collection ψ of paths in G such that each ege of G is in at least one path of ψ an two paths of ψ have at most 1 Receive July 18, 010. Accepte December 5, 010.

2 On Pathos Lict Subivision of a Tree 101 k vertices in common. Thus if k = 1 an every ege of G is in exactly one path in ψ, then a Smaranache path k-cover of G is a simple path cover of G. See [8]. By a graph we mean a finite, unirecte graph without loops or multiple lines. We refer to the terminology of [1]. The pathos lict subivision of a tree T is enote as P n [S(T)] an is efine as the graph, whose point set is the union of set of lines, set of paths of pathos an set of cutpoints of S(T) in which two points are ajacent if an only if the corresponing lines of S(T) are ajacent an the line lies on the corresponing path P i of pathos an the lines are incient to the cutpoints. Since the system of path of pathos for a S(T) is not unique, the corresponing pathos lict subivision graph is also not unique. The pathos lict subivision graph is efine for a tree having at least one cutpoint. In Figure 1, a tree T an its subivision graph S(T), an their pathos lict subivision graphs P n [S(T)] are shown. e 1 e 4 c 1 e 3 c P 1 e e 5 P 3 P 1 P 3 e 1 e 4 c 1 e 3 c e e 5 P 1 P 1 P 3 e 1 e 4 c 1 e 3 c P 1 P 1 P 3 e e 5 T S(T) P n [S(T)] Figure 1 The line egree of a line uv in S(T) is the sum of the egrees of u an v. The pathos length is the number of lines which lies on a particular path P i of pathos of S(T). A penant pathos is a path P i of pathos having unit length which correspons to a penant line in S(T). A pathos point is a point in P n [S(T)] corresponing to a path of pathos of S(T). If G is planar graph,the innerpoint number i(g) of a graph G is the minimum number of vertices not belonging to the bounary of the exterior region in any embeing of the plane. A graph is sai to be minimally nonouterplanar if i(g) = 1 was given by [4].

3 10 Keerthi G.Mirajkar an Iramma M.Kaakol We nee the following for immeiate use. Remark 1.1 For any tree T, n[s(t)] is a subgraph of P n [S(T)]. Remark 1. For any tree T, T S(T). Remark 1.3 If the line egree of a nonpenant line in S(T) is o(even), the corresponig point in P n [S(T)] is of even(o) egree. Remark 1.4 The penant line in S(T) is always o egree an the corresponing point in P n [S(T)] is of o egree. Remark 1.5 For any tree T with C cutpoints, the number of cutpoints in n[s(t)] is equal to sum of the lines incient to C in T. Remark 1.6 For any tree T, the number of blocks in n[s(t)] is equal to the sum of the cutpoints an lines of T. Remark 1.7 n[s(t)] is connecte if an only if T is connecte. Theorem 1.1([5]) If G is a non trivial connecte (p, q) graph whose points have egree i an l i be the number of lines to which cutpoint C i belongs in G, then lict graph n(g) has q + C i points an q + [ i + l i ] lines. Theorem 1.([5]) The lict graph n(g) of a graph G is planar if an only if G is planar an the egree of each point is atmost 3. Theorem 1.3([]) Every maximal outerplanar graph G with p points has p 3 lines. Theorem 1.4([6]) A graph is a nonempty path if an only if it is a connecte graph with p points an i 4p + 6 = 0. Theorem 1.5([]) A graph G is eulerian if an only if every point of G is of even egree.. Pathos Lict Subivision Graph In the following Theorem we obtain the number of points an lines of P n [S(T)]. Theorem.1 For any (p, q) graph T, whose points have egree i an cutpoints C have egree C j, then the pathos lict sub-ivision graph P n [S(T)] has (3q +C +P i ) points an 1 i +4q + Cj lines. Proof By Theorem 1.1, n(t) has q + c points by subivision of T n(s(t))contains q + q + c points an by Remark 1.1, P n S(T) will contain 3q + c + P i points, where P i is the path number. By the efinition of n(t), it follows that L(T) is a subgraph of n(t). Also, subgraphs of L(T) are line-isjoint subgraphs of n[s(t)] whose union is L(T) an the cutpoints c of T having egree C j are also the members of n[s(t)]. Hence this implies that n[s(t)] contains q + 1 i + c j lines. Apart from these lines every subivision of T generates

4 On Pathos Lict Subivision of a Tree 103 a line an a cutpoint c of egree. This creates q + q lines in n[s(t)]. Thus n[s(t)] has 1 i + c j + q lines. Further, the pathos contribute q lines to P n S(T). Hence P n [S(T)] contains 1 i + c j + 4qlines. Corollary.1 For any (p, q) graph T, the number of regions in P n [S(T)] is (p + q) Planar Pathos Lict Sub-ivision Graph In this section we obtain the conition for planarity of pathos. Theorem 3.1 P n [S(T)] of a tree T is planar if an only if (T) 3. Proof Suppose P n [S(T)] is planar. Assume (T) 4. Let v be a point of egree 4 in T. By Remark 1.1, n(s(t)) is a subgraph of P n [S(T)] an by Theorem 1., P n [S(T)] is non-planar. Clearly, P n [S(T)] is non-planar, a contraiction. Conversely, suppose (T) 3. By Theorem 1., n[s(t)] is planar. Further each block of n[s(t)] is either K 3 or K 4. The pathos point is ajacent to atmost two vertices of each block of n[s(t)]. This gives a planar P n [S(T)]. We next give a characterization of trees whose pathos lict subivision of trees are outerplanar an maximal outerplanar. Theorem 3. The pathos lict sub-ivision graph P n [S(T)] of a tree T is outerplanar if an only if (T). Proof Suppose P n [S(T)] is outerplanar. Assume T has a point v of egree 3. The lines incient to v an the cut-point v form K 4 as a subgraph in n[s(t)]. Hence P n [S(T)] is non-outerplanar, a contraiction. Conversely, suppose T is a path P m of length m 1, by efinition each block of n[s(t)] is K 3 an n[s(t)] has m 1 blocks. Also, S(T) has exactly one path of pathos an the pathos point is ajacent to atmost two points of each block of n[s(t)]. The pathos point together with each block form m 1 number of K 4 x subgraphs in P n [S(T)]. Hence P n [S(T)] is outerplanar. Theorem 3.3 The pathos lict sub-ivision graph P n [S(T)] of a tree T is maximal outerplanar if an only if. Proof Suppose P n [S(T)] is maximal outerplanar. Then P n [S(T)] is connecte. Hence by Remark 1.7, T is connecte. Suppose P n [S(T)] is K 4 x, then clearly, T is K. Let T be any connecte tree with p > points, q lines an having path number k an C cut-points. Then clearly, P n [S(T)] has 3q+k+C points an 1 i +4q+ C j lines. Since P n [S(T)] is maximal

5 104 Keerthi G.Mirajkar an Iramma M.Kaakol outerplanar, by Theorem 1.3, it has [(3q + k + C) 3] lines. Hence 1 i + 4q + C j = [(3q + k + C) 3] = [(3(p 1) + k + C) 3] = 6p 6 + k + C 3 = 6p + k + C 9. But for k = 1, i + 8q + C j = 1p + 4C , i + C j = 4p + 4C 6, i + C j 4p 4C + 6 = 0. Since every cut-point is of egree two in a path, we have, Cj = C. Therefore i + 6 4p = 4C xc = 0. Hence i + 6 4p = 0. By Theorem 1.4, it follows that T is a non-empty path. Conversely, Suppose T is a non-empty path. We now prove that P n [S(T)] is maximal outerplanar by inuction on the number of points ( ). Suppose T is K. Then P n [S(T)] = K 4 x. Hence it is maximal outerplanar. As the inuctive hypothesis, let the pathos lict subivision of a non-empty path P with n points be maximal outerplanar. We now show that P n [S(T)] of a path P with n + 1 points is maximal outerplanar. First we prove that it is outerplanar. Let the point an line sequence of the path P be v 1, e 1, v, e, v 3, e 3,..., v n, e n, v n+1. P, S(P ) an P n [S(P )] are shown in Figure. Without loss of generality, P v n+1 = P. By inuctive hypothesis P n [S(P)] is maximal outerplanar. Now the point v n+1 is one point more in P n [S(P )] than in P n [S(P)]. Also there are only eight lines (e n 1, e n), (e n 1, e n 1), (e n 1, e n ), (e n, R), (e n, e n), (e n, C n), (C n, e n), (e n, R) more in P n [S(P )]. Clearly, the inuce subgraph on the points e n 1, C n 1, e n, e n, C n, R is not K 4. Hence P n [S(P )] is outerplanar. We now prove P n [S(P )] is maximal outerplanar. Since P n [S(P)] is maximal outerplanar, it has (3q + C + 1) 3 lines. The outerplanar graph P n [S(P )] has (3q + C + 1) lines = [3(q + 1) + (C+) + 1] 3 lines. By Theorem 1.3, P n [S(P )] is maximal outerplanar. Theorem 3.4 For any tree T, P n [S(T)] is minimally nonouterplanar if an only if (T) 3 an T has a unique point of egree 3. Proof Suppose P n [S(T)] is minimally non-outerplanar. Assume (T) > 3. By Theorem 3.1, P n [S(T)] is nonplanar, a contraiction. Hence (T) 3. Assume (T) < 3. By Theorem 3., P n [S(T)] is outerplanar, a contraiction. Thus (T) = 3. Assume there exist two points of egree 3 in T. Then n[s(t)] has at least two blocks as K 4. Any pathos point of S(T) is ajacent to atmost two points of each block in n[s(t)] which gives i(p n [S(T)]) > 1, a contraiction. Hence T has exactly one point point of egree 3.

6 On Pathos Lict Subivision of a Tree 105 Conversely, suppose every point of T has egree 3 an has a unique point of egree 3, then n[s(t)] has exactly one block as K 4 an remaining blocks are K 3 s. Each pathos point is ajacent to atmost two points of each block. Hence i(p n [S(T)]) = 1. e 1 e e n 1 e n P : v1 v v 3 v n 1 v n v n+1 S(P ) : e 1 c 1 e c 1 e c 1 e c c n c n 1 v 1 v v 1 v v 3 v n 1 v n 1 c n 1 c n en 1 e n 1 e n e n v n v n v n+1 c 1 c 1 c c n 1 c n 1 c n P n [S(P )] : e 1 e 1 e e e n 1 e n 1 e n e n R Figure 4. Traversability in Pathos Lict Subivision of a Tree In this section, we characterize the trees whose P n [S(T)] is eulerian an hamiltonian. Theorem 4.1 For any non-trivial tree T, the pathos lict subivision of a tree is non-eulerian. Proof Let T be a non-trivial tree. Remark 1.4 implies P n [S(T)] always contains a point of o egree. Hence by Theorem 1.5, the result follows. Theorem 4. The pathos lict subivision P n [S(T)] of a tree T is hamiltonian if an only if every cut-point of T is even of egree.

7 106 Keerthi G.Mirajkar an Iramma M.Kaakol Proof If T =, then P n [S(T)] is K 4 x. If T is a tree with p 3 points. Suppose P n [S(T)] is hamiltonian. Assume that T has at least one cut-point v of o egree m. Then G = K 1,m is a subgraph of T. Clearly, n(s(k 1,m )) = K m+1, together with each point of K m incient to a line of K 3. In number of path of pathos of S(T) there exist at least one path of pathos P i such that it begins with the cut-point v of S(T). In P n [S(T)] each pathos point is ajacent to exactly two points of K m. Further the pathos beginning with the cut-point v of S(T) is ajacent to exactly one point of K m in n(s(t)). Hence this creates a cut-point in P n [S(T)], a contraiction. Conversely, suppose every cut-point of T is even. Then every path of pathos starts an ens at penant points of T. We consier the following cases. Case 1 If T has only cut-points of egree two. Clearly, T is a path. Further S(T) is also a path with p + q points an has exactly one path of pathos. Let T = P l, v 1, v,, v l is a path. Now S(T):v 1, v 1, v, v,, v l 1, v l for all v i V [S(P l )] such that v i v i = e i, v i v i+1 = e i are consecutive lines an for all e i, e i E[S(P n)]. Further V [n(s(t))]={e 1, e 1, e, e,, e i, e i } {C 1, C 1, C, C,, C i } where,(c 1, C 1, C, C,, C i ) are cut-points of S(T). Since each block is a triangle in n(s(t)) an each block consist of points as B 1 = (e 1 C 1 e 1 ), B = (e C e ),, B m = (e i C i e i ). In P n[s(t)], the pathos point w is ajacent to e 1, e 1, e, e,, e i, e i. Hence, P n [S(T)]= e 1, e 1, e, e,, e i, e i (C 1, C 1, C, C,, C i ) w form a cycle as we 1C 1 e 1 C 1e C e e i w containing all the points of P n[s(t)].hence P n [S(T)] is hamiltonian. Case If T has all cut-points of even egree an is not a path. we consier the following subcases of this case. Subcase.1. If T has exactly one cut-point v of even egree m, v = (T) an is K 1,m. Clearly, S(K 1,m ) = F, such that E(F)={e 1, e 1, e, e,, e q, e q}. Now n(f) contains point set as {e 1, e 1, e, e,, e q, e q } {v, C 1, C, C 3,, C q }. For S[K 1,m], it has m paths of pathos with pathos point as P 1,,, P m. By efinition of n[s(t)], each pathos point is ajacent to exactly two points of n(s(t)). Also, V [P n [S(T)]]={e 1, e 1, e, e,, e q, e q } {v, C 1, C, C 3,, C q } {P 1,,, P m }. Then there exist a cycle containing all the points of n[s(t)] as P 1, e 1, C 1, e 1, v, e, C, e,,, P m, e q 1, C q 1, e q 1, e q, C q, e q, P 1. Subcase.. Assume T has more than one cut-point of even egree. Then in n(s(t)) each block is complete an every cut-point lies on exactly two blocks of n(s(t)). Let V [n(s(t))]={e 1, e 1, e, e,, e q, e q } {C 1, C,, C i } {C 1, C, C 3,, C q } {P 1,,, P j }. But each P j is ajacent to exactly two point of the block B j except {C 1, C,, C i } {C 1, C, C 3,, C q} an all these points together form a hamiltonian cycle of the type, {P 1, e 1, C 1, e 1, v, e, C, e,,, P r, e k, C k, e k, e k+1, C k+1, e k+1, P r+1,, P j, e q 1, C q 1, e q 1, e q, C q, e q, P 1 }. Hence P n [S(T)] is hamiltonian. References [1] Harary. F, Annals of New York, Acaemy of Sciences,(1974)175.

8 On Pathos Lict Subivision of a Tree 107 [] Harary. F, Graph Theory, Aison-Wesley, Reaing, Mass,1969. [3] Harary F an Schwenik A.J, Graph Theory an Computing, E.Rea R.C, Acaemic press, New York,(197). [4] Kulli V.R, Proceeings of the Inian National Science Acaemy, 61(A),(1975), 75. [5] Kulli V.R an Muebihal M.H, J. of Analysis an computation,vol, No.1(006), 33. [6] Kulli V.R, The Maths Eucation, (1975),9. [7] Stanton R.G, Cowan D.D an James L.O, Proceeings of the louisiana conference on combinatorics, Graph Theory an Computation (1970), 11. [8] S. Arumugam an I. Sahul Hami, International J.Math.Combin. Vol.3,(008),