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Indian Journal of Mathematics and Mathematical Sciences Vol 8 No 1 (June 2012) 61-75

TRIPLE CONNECTED GRAPHSJ Paulraj Joseph M K Angel JebithaP Chithra Devi amp G SudhanaAbstractThe concept of connectedness plays an important role in any network A varietyof connectedness have been studied in the literature by considering theexistence of a path between any two vertices In transportation networks thisenables a traveller to have a route from one city to any other city If a travellercan finish some work enroute in any one of the third cities then it will minimizemoney distance time etc A communication network in which a transmittingnode can send a message to two stations at one stretch will be more effectiveand economic Such an optimization leads to the concept of triple connectedgraphs Hence in this paper we introduce triple connected graph byconsidering the existence of a path containing any three vertices of G andstudy their propertiesAMS Subject Classification 05C40Keywords Connected graphs Triple connected graphs1 INTRODUCTIONLet G = (V E) be a finite simple undirected graph of order n and size m If v 1048577V (G)then the neighbourhood of v is the set N (v) consisting of all vertices which areadjacent to v The closed neighbourhood is N[v] = N(v) v The degree of v inG is | N (v) | and is denoted by d (v) A vertex v is said to be pendant vertex ifd (v) = 1 A vertex u is called support if u is adjacent to a pendant vertex G is saidto be connected if any two vertices are joined by a path in G otherwise the graphis disconnected A maximal connected subgraph of a graph G is called a componentof G The number of components of a graph G is denoted by 1048577(G) A vertex v of aconnected graph G is a cut vertex if G ndash v is disconnected An edge e of a connectedgraph G is a cut edge if G ndash e is disconnected A connected graph with no cutvertices is called a block A block of a graph is a block and is maximal with respectto this property Connectivity of a graph G is the minimum number of verticeswhose removal results in a disconnected graph and is denoted by 1048577(G) A graph isr-connected if 1048577(G) 1048577rTree is a connected acyclic graph A path on n vertices is denoted by Pn A cycleon n vertices is denoted by Cn A complete graph on n vertices is denoted by Kn62 J Paulraj Joseph amp G SudhanaA complete bipartite graph G with bipartition (X Y) such that |X| = r and |Y | = sis denoted by Kr s A set of vertices in a graph G is said to be an independent set ifno two of them are adjacent in G The cardinality of a maximum independent set inG is called the independence number of G and is denoted by 1048577(G) Terms notdefined here are used in the sense of [3]Different types of connectedness have been defined and studied in the literatureA graph G is said to be traceable if there exists a Hamilton path in G A graph G isHamiltonian connected [2] if any two vertices of G are connected by a Hamiltonpath A graph is said to be a s-Hamiltonian connected [4] if for any S 1048577V(G) oforder at most s G ndash S is hamiltonian connected graph A graph G = (V E) is said tobe panconnected [1] if for every pair of vertices u and v of G and for each integerk satisfying d (u v) 1048577k 1048577|V| ndash 1 there is a u ndash v path of length k contained in G Inthis paper we consider connectedness involving three vertices of a graph2 PRELIMINARIESDefinition 21 A graph is said to be triple connected if any three vertices lie on a

path in GExample 22 All paths and cycles complete graphs and wheels are somestandard examples of triple connected graphs Petersen graph Kr s where r s 10485772and Km1 m1 mk

mi 10485771 and k 10485773 are also triple connected graphsRemark 23 A triple connected graph may or may not have cut vertex or cutedge The graph in Fig 21(a) is a triple connected graph without cut vertex whereasthe graph in Fig 21(b) is a triple connected graph with cut edgeFigure 21Remark 24 Hamiltonian graphs and traceable graphs are triple connectedbut not conversely The graph in Fig 22 is triple connected but neither traceablenor hamiltonianFigure 22Triple Connected Graphs 63Remark 25 Clearly every triple connected graph is a connected graph If Ghas a spanning subgraph which is triple connected then G is triple connectedTheorem 26 Every 2-connected graph is triple connectedProof Let G be a 2-connected graph and let x y z 1048577G Since G is 2-connectedit has no cut vertex Hence any two vertices are joined by two internally disjointpaths Let P1 and P2 be two internally disjoint paths from x to y Since G is connectedthere is a y ndash z path say Q If Q is internally disjoint with P1 or P2 then x y z lie onpath P1 Q or P2 Q Otherwise let w be the last vertex common to P1 and QThen z ndash w section of Q together with w ndash y section of P1 and P2

ndash 1 is a path containingx y and z Hence G is triple connectedObservation 27 Three vertices u v and w lie on a path P if and only if P isthe union of two paths with exactly one common vertexTheorem 28 If G has a cut vertex v such that 1048577(G ndash v) 10485773 then G is not tripleconnectedProof Let v be a cut vertex of G such that 1048577(G ndash v) 10485773 Let G1 G2 G3 be anythree components of G ndash v Let x 1048577V(G1) y 1048577V(G2) z 1048577V(G3) Then any pathconnecting two vertices of x y and z must pass through v and hence any two suchpaths have at least two vertices in common By Observation 27 x y and z do notlie on a path Hence G is not triple connectedTheorem 29 Given any two positive numbers a and b where b 10485772a ndash 1 thereexists a triple connected graph G such that 1048577(G) = a and |V(G) | = bProof If a = 1 then b 10485771 and the complete graph on b vertices is a tripleconnected graph with 1048577(G) = 1 Now let a gt 1 If b = 2a ndash 1 then the path on bvertices labelled as v1 v2 v2a ndash 1 is a triple connected graph with 1048577(G) = a If b gt2a ndash 1 then b = (2a ndash 1) + c where c 10485771 Construct the graph G as followsConsider a path on 2a ndash 1 vertices v1 v2 v2a ndash 1 and a complete graph on cvertices w1 w2 wc Let G be the graph obtained by identifying v1 and w1 ClearlyG is a triple connected graph with b vertices Also v1 v3 v5 v2a ndash 1 is a maximumindependent set so that 1048577(G) = a3 CHARACTERIZATION THEOREMSTheorem 31 A tree T is triple connected if and only if T ~=Pn n 10485772Proof Let T be a tree Assume that T is triple connected Suppose T ~= Pn n 10485772Then there exists a vertex v such that d (v) 10485773 Clearly v is a cut vertex of T and1048577(T ndash v) 10485773 Hence by Theorem 28 T is not triple connected which is a

contradiction Thus T ~=Pn n 10485772 Converse is obviousDefinition 32 A connected subgraph H of a connected graph G is called aH-cut if 1048577(G ndash H) 1048577264 J Paulraj Joseph amp G SudhanaTheorem 33 A connected graph G is not triple connected if and only if thereexists a H-cut with 1048577(G ndash H) 10485773 such that |V(H) N (Ci) | = 1 for at least threecomponents C1 C2 and C3 of G ndash HProof Let G be a connected graph Assume that G is not triple connectedThen there exists at least vertices u v w 1048577V(G) such that there is no path containingu v and w in G Let P1 P2 and P3 be arbitrary u ndash v path v ndash w path and u ndash w pathrespectively If Pi and Pj are internally disjoint then Pi Pj is a path containingu v and w Otherwise let x1 x2 xk 1048577V (P1) V (P2) 1 1048577k 1048577n ndash 3 andxi 1048577u v w Clearly at least one vertex of x1 x2 xk must be a cut vertexOtherwise we have another two vertexdisjoint paths say P4 and P5 whose unionform a path on which u v and w lie in G Let xi (1 1048577i 1048577k) be a cut vertex of GThen G ndash xi is disconnected Let C1 be a component of G ndash xi which contains vSince V(P2) V(P3) ndash u v w 10485771048577 there exists a cut vertex yj 1048577V(P2) V(P3)(yj = xi or yj 1048577xi) Thus G ndash yj is disconnected Let C2 be a component of G ndash yj

which contains w Also since V(P1) V(P3) ndash u v w 10485771048577 there exists a cut vertexzk 1048577V(P1) V(P3) (where zk may or may not be equal to xi and yj) Thus G ndash zk isdisconnected Let C3 be a component of G ndash zk which contains u Now we define aH-cut as follows Let H be any connected subgraph such that xi yj zk V (H)and V(H) V(Ci) = 1048577 i = 1 2 3 Clearly V(H) N(C1) = xi V(H) N(C2) = yjV(H) N (C3) = zk Thus |V (H) N (Ci) | = 1 for every i = 1 2 3 Also sincexi yj zk ndash V(H) 1048577(G ndash H) 10485773Conversely assume that there exists a H-cut with 1048577(G ndash H) 10485773 such that|V(H) N (Ci) | = 1 for at least three components C1 C2 and C3 of G ndash H Since|V(H) N(Ci) | = 1 let V(H) N(Ci) = xi i = 1 2 3 Clearly xi is a cut vertex of GLet u 1048577V(C1) v 1048577V(C2) and w 1048577V(C3)Claim No path in G contains u v and wSince G is connected any two vertices of G are connected by a path Let P1 beany u ndash v path P2 be any v ndash w path and P3 be any u ndash w path in GCase (i) x1 1048577x2 1048577x3Since u 1048577V(C1) and v 1048577V(G) ndash V(C1) x1 lies on every u ndash v path in G Thusthe path P1 must pass through x1 Also v 1048577V(C2) and u 1048577V(G) ndash V(C2) Thus x2

lies on every v ndash u path in G Thus the path P1 must contain the cut vertex x2 Thusx1 x2 V (P1) Similarly we can prove that x1 and x3 lie on every u ndash w pathThus x1 x3 V(P3) Also x2 and x3 lie on every v ndash w path Thus x2 x3 V(P2)Clearly x1 1048577P1 P3 x2 1048577P1 P2 and x3 1048577P2 P3 Since P1 P2 and P3 are arbitrarypaths there is no path in G containing u v and wCase (ii) x1 = x2 = x3Clearly x1 lies on every u ndash v path v ndash w path and u ndash w path This implies thatx1 1048577V(Pi) V(Pj) for i 1048577j i j 10485771 2 3 Thus there is no path containing u vand wTriple Connected Graphs 65The above theorem can be rewritten as followsA connected graph G is triple connected if and only if G has no triple cutDefinition 35 A block graph B (G) of a graph G is the graph in which thevertex set is the set of all blocks of G and two vertices of B(G) are adjacent if andonly if the vertex set of the corresponding blocks of G have non-empty intersection

Theorem 36 G is triple connected if and only if the block graph B(G) of G isa pathProof Assume that B(G) is a path Let B(G) = x1 x2 x3 xn where xi is a vertexin B(G) corresponding to the block Bi in G Let u v w 1048577V(G) be arbitraryCase (i) u v and w belong to the same blockLet u v w 1048577Bi Since Bi is a block it is 2-connected Hence by Theorem 26there exists a path in Bi containing u v and wCase (ii) Exactly two of u v and w belong to the same blockLet u v 1048577Bi and w 1048577Bj with i lt j Let x be the vertex common to Bi and Bi + 1Since Bi is triple connected and u v x 1048577V (Bi) there exists a path P containingu v and x in Bi Since Bi 1048577Bj and Bj is connected there exists a x ndash w path P1048577in Gwhich is internally disjoint from P Clearly P P1048577is a path containing u v and wCase (iii) u v and w belong to three different blocks of GCase (iii) Exactly two vertices of x1 x2 and x3 are equalWithout loss of generality assume that x1 = x2 Clearly x1 lies on every u ndash vpath v ndash w path and u ndash w path Thus x1 1048577P1 P2 P3 and x3 1048577P2 P3 and hencethere is no path containing u v w Thus G is not triple connectedRemark 34 The H-cut mentioned in the above theorem is called a triple cutSuch a triple cut need not be unique For the non-triple connected graph G in Fig 31H1 = 1048577u1048577 H2 = 1048577u v1048577 H3 = 1048577u v w1048577 H4 = 1048577u v w x1048577 H5 = 1048577u v w x y1048577are triple cuts of GFigure 3166 J Paulraj Joseph amp G SudhanaLet u 1048577Bi v 1048577Bj and w 1048577Bk Without loss of generality assume that i lt j lt kLet V (Bi) V (Bi + 1) = x V (Bj ndash 1) V(Bj) = y V (Bj) V(Bj + 1) = z andV(Bk ndash 1) V(Bk) = h Now u x 1048577V(Bi) y 1048577V(Bj) and i lt j As in case (ii) thereexists a u ndash x ndash y path P1 in G Also z 1048577V(Bj) and h w 1048577V(Bk) and j lt k Hencethere exists a z ndash h ndash w path P2 in G Since Bj is a block and y v z 1048577V(Bj) thereexists a path P3 containing y v and z in G Since B(G) is a path these three paths P1P2 and P3 are internally disjoint Thus P1 P2 P3 is a path containing u v and win G Thus G is triple connectedConversely assume that G is triple connected Suppose that there exists a vertexx 1048577B(G) such that d (x) 10485773 Let B be the block in G corresponding to the vertex xin B(G) Clearly 1048577(G ndash B) 10485773 and |N(Ci) V(B) | = 1 for at least three componentsC1 C2 and C3 of G ndash B Thus by Theorem 33 G is not triple connected which is acontradiction Hence B (G) is a pathDefinition 37 An edge e = uv is said to be subdivided if it is deleted andreplaced by a u ndash v path of length two with a new internal vertex w A subdivisiongraph S (G) of a graph G is obtained from G by applying a finite number ofsubdivisions of edges in successionObservation 38 B (G) is a path if and only if B(S (G)) is a pathTheorem 39 G is triple connected if and only if S (G) is triple connectedProof The theorem follows by repeated application of Theorm 364 COMPLEMENTARY GRAPHSTheorem 41 Let T be a tree Then Tmdash is triple connected if and only if T ~= K1 rProof Let T be a tree Assume that T ~= K1 r Let u v and w be any three verticesin V(Tmdash) and let S = u v wCase (i) 1048577S1048577= K

mdash

3 in TThen 1048577S1048577= K3 in Tmdash and uvw is a path in TmdashCase (ii) 1048577S1048577= K2 K1 in TWithout loss of generality let u and v be adjacent in T Thus uw vw 1048577E (Tmdash)and hence uwv is a path in TmdashCase (iii) 1048577S1048577= P3 in TWithout loss of generality let u be adjacent to both v and w in T Thus vw 1048577E(Tmdash)Since T ~= K1 r there exists another vertex x which is not adjacent to u in T Thusxu 1048577E(Tmdash) Since T is a tree x can not be adjacent to both v and w in T Without lossof generality assume that x is not adjacent to w in T Then xw 1048577E(Tmdash) and uxwv isa path in TmdashTriple Connected Graphs 67Thus any three vertices lie on a path in Tmdash Hence Tmdash is triple connectedConversely assume that Tmdash is triple connected Suppose Tmdash ~=K1 r This impliesthat Tmdash~=K1 Kr r 10485772 which is disconnected Thus Tmdash is not triple connected whichis a contradiction Thus T ~= K1 rProposition 42 Let G be a connected graph Then Gmdash is disconnected with 1048577components if and only if G contains a complete 1048577-partite graph (104857710485772) as aspanning subgraphProof Let Gmdash be disconnected with 1048577components C1 C2 C1048577 Let V(Ci) = Vi

and |Vi | = ni i = 1 2 1048577Claim Kn1 n2 n1048577is a spanning subgraph of GIt is enough if we prove that any two vertices in different partite sets are adjacentin G Let u 1048577Vi and v 1048577Vj Since u and v are the vertices of Ci and Cj in Gmdash respectivelyuv 1048577E (Gmdash) Hence uv 1048577E (G) and hence the claimConversely assume that G contains a complete 1048577-partite graph as a spanningsubgraph say Kn1 n2 n1048577where 1048577is as large as possible Then V(G) can be partitionedinto 1048577subsets V1 V2 V1048577such that every two vertices in different partite sets are

joined by an edge where | Vi | = ni We claim that each 1048577Vi1048577is connected in GmdashSuppose that there exists a j (1 1048577j 10485771048577) such that 1048577Vj1048577is disconnected with at leasttwo components in Gmdash As in previous part there exists a complete 10485771048577-partite graphas a spanning subgraph of 1048577Vj1048577in G where 1048577104857710485772 Then Vj can be partitioned into 10485771048577subsets Vj1 Vj2 Vj10485771048577such that any two vertices in different partite sets Vj1 Vj2 Vj10485771048577

are joined by an edge in G Then V1 V2 Vj ndash 1 Vj + 1 V1048577 Vj1 Vj2 Vj10485771048577are thepartite sets of V such that any two vertices in different partite sets are adjacent in GHence G contains a complete (1048577ndash 1) + 10485771048577partite graph as a spanning subgraphwhere (1048577ndash 1) + 10485771048577gt 1048577which is a contradiction to the choice of 1048577 Hence each 1048577Vi 1048577is connected in Gmdash Further by hypothesis if uv 1048577E (Gmdash) then u and v belong tosame partite set in Gmdash Hence Gmdash is disconnected with 1048577componentsCorollary 43 Let G be a connected graph Then Gmdash is disconnected if andonly if G contains a complete bipartite graph Kr s (r s 10485771) as a spanning subgraphDefinition 44 A graph G satisfying Proposition 42 is called a 1048577-complementgraphTheorem 45 Let G be a disconnected graph Then Gmdash is triple connected ifand only if G ~= K1 H where H is a 1048577-complement graph 104857710485773Proof Assume that Gmdash is triple connected Suppose G ~=K1 H where K1 = vThen v is a triple cut for G and hence by Theorem 33 G is not triple connectedwhich is a contradiction Thus G ~= K1 H68 J Paulraj Joseph amp G SudhanaConversely assume that G ~= K1 H If 1048577(G) 10485773 then Gmdash contains acomplete 1048577-partite graph as a spanning subgraph which is triple connected ByRemark 25 Gmdash is triple connectedNow assume that 1048577(G) = 2 If G ~=G1 G2 such that |V(G1) | |V(G2) | 10485772 thenGmdash contains a complete bipartite graph as a spanning subgraph which is a tripleconnected By Remark 25 Gmdash is triple connectedNow let G ~=K1 H where H is a 2-complement graph Then B(Gmdash) = P2 andhence by Theorem 36 Gmdash is triple connectedLemma 46 Let G be a connected graph with a cut vertex v and d (v) = 3 ThenG

mdash is not triple connected if and only if the degree set of N(v) is 1 n ndash 2 n ndash 2Proof Let G be a connected graph with a cut vertex v and d (v) = 3 Assumethat Gmdash is not triple connected Then G ndash v has exactly two components in whichone of them is trivial Let C be a component of G ndash v where V(C) = x1 x2 xn ndash 2Clearly v is adjacent to an end vertex x and two more vertices say xi and xj It isenough if we prove that d (xi) = n ndash 2 and d (xj) = n ndash 2Suppose this is not true we consider two casesCase (i) d (xi) d (xj) lt n ndash 2Clearly x is adjacent to all the vertices x1 x2 xn ndash 2 of C and v is adjacent toall the vertices of x1 x2 xn ndash 2 ndash xi xj of C in Gmdash Since d (xi) d (xj) lt n ndash 2 in Gxi and xj are adjacent to at least one vertex of C in Gmdash If xi is adjacent to either xj orx1 x2 xn ndash 2 ndash xj in Gmdash then Gmdash is triple connected Similarly if xj is adjacent toeither xi or x1 x2 xn ndash 2 ndash xi in Gmdash then we have at most two componentswhich are adjacent to x Thus Gmdash is triple connected which is a contradictionCase (ii) d (xi) lt n ndash 2 and d (xj) = n ndash 2In this case xj is an end vertex which is adjacent to x in Gmdash Since d (xi) lt n ndash 2in G xi is adjacent to at least one vertex of x1 x2 xn ndash 2 ndash xj in Gmdash Thus thesubgraph induced by the vertices x1 x2 xn ndash 2 ndash xj v forms a connectedcomponent which is adjacent to x Thus we have exactly two components whichare adjacent with x which is a contradiction Hence d (xi) = n ndash 2 and d (xj) = n ndash 2Conversely assume that the degree set of N(v) is 1 n ndash 2 n ndash 2 Then in Gmdashboth xi and xj are end vertices that are adjacent to x and the subgraph induced byx1 x2 xn ndash 2 ndash xi xj v is connected with d (x) = n ndash 2 Thus Gmdash ndash x hasexactly three components such that all their vertices are adjacent to x Thus x isa triple cut and hence Gmdash is not triple connectedTheorem 47 Let G be a connected but not triple connected graph with aunique vertex v of degree n ndash 2 Then v is a triple cut in GTriple Connected Graphs 69Proof Let G be a connected graph which is not triple connected Then byTheorem 33 there exists a triple cut H in G such that G ndash H has at least threecomponents Cirsquos with |N(Ci) V(H) | = 1 Let N(Ci) V(H) = xi for all i wherexi need not be distinct Let x1 x2 xp be the set of distinct vertices of HClaim 1 v 1048577V(H)Suppose v 1048577V(H) then there exists a component Ci in V ndash H such that v 1048577V(Ci)and v 1048577xi Thus there exist at least two vertices x 1048577Cj y 1048577Ck i 1048577j 1048577k andx 1048577xj y 1048577xk such that v is not adjacent to both x and y Then d (v) 1048577n ndash 3 which isa contradiction Hence v 1048577V(H)Claim 2 v 1048577x1 x2 xpSuppose v 1048577x1 x2 xp then we can find at least three vertices x 1048577Ci y 1048577Cj

z 1048577Ck i 1048577j 1048577k and x 1048577xi y 1048577xj z 1048577xk such that v is not adjacent to x y and z Thend (v) 1048577n ndash 4 which is a contradiction Hence v 1048577x1 x2 xpClaim 3 p 10485772Suppose p 10485773 Then there exist at least three elements x1 x2 and x3 such thatx1 x2 x3 1048577i

(N(Ci) V(H)) and i 10485773 Let x1 = v Since N(C1) V(H) = x1 = vwe can find at least two vertices x 1048577V(C2) x 1048577x2 and y 1048577V(C3) y 1048577x3 such that vis not adjacent to both x and y Thus d (v) 1048577n ndash 3 which is a contradiction Hencep 10485772 If p = 1 then obviously v is a triple cut in G If p = 2 then leti

(N(Ci)V(H)) = x1 x2 Without loss of generality we may assume that x1 = v Clearly bythe previous argument there exists exactly one component say Cj such thatN(Cj) V(H) = x2 and the remaining (l ndash 1) components are C1 C2 Cj ndash 1 Cj + 1 Cl such that N (Ci) V (H) = x1 = v i = 1 2 j ndash 1 j + 1 Also thecomponent Cj contains exactly one vertex which is not adjacent to v in G Inparticular x1 and x2 are adjacent and form a cut edge of G Thus x1 = v is atriple cut of GTheorem 48 Let G be a connected graph with a unique cut vertex v such thatd (v) lt n ndash 1 and v be a support with pendant vertex x Then Gmdash is not triple connectedif and only if G ndash v x contains a complete k-partite graph (k 10485773) as a spanningsubgraph with V(G) ndash N [v] belonging to the same partite setProof Let G1048577= 1048577N(v) ndash x1048577and G10485771048577= 1048577V(G) ndash N[v]1048577 Assume that Gmdash is not tripleconnected Since v is a unique cut vertex of G d (v) 10485773Case (i) d (v) = 3Let V(G1048577) = N(v) ndash x = u w Then by Lemma 46 d (u) = n ndash 2 d (w) = n ndash 2and hence u and w are adjacent in G and G1048577= K1 1 Thus u and w are therequired partite sets in G1048577 Since every vertex in G10485771048577is adjacent to both u and wu w V (G10485771048577) gives the partite sets of G ndash v x and form a completek-partite graph (k 10485773) as a spanning subgraph70 J Paulraj Joseph amp G SudhanaCase (ii) d (v) gt 3Since d (v) 1048577n ndash 1 in G |V(G10485771048577) | 10485771048577 Since v is adjacent to all the vertices of G10485771048577in Gmdash 1048577G10485771048577v1048577is a connected subgraph in Gmdash Since Gmdash is not triple connected andd (x) = n ndash 2 by Theorem 47 x is a triple cut of Gmdash Then there exist at least threecomponents C1048577is in Gmdash ndash x Let C1 be a component of Gmdash ndash x which contains thesubgraph 1048577G10485771048577v1048577 Then there exist at least two components in Gmdash ndash x otherthan C1 Let C2 C3 Cp be the components of Gmdash ndash x p 10485773 Clearly the verticesof Ci (i = 2 3 p) are in N(v) But C1 may or may not contain vertices of N (v)Hence we distinguish into two casesSubcase (a) NG (v) V(C1) = 1048577

Then 1048577NG (v)1048577is a disconnected subgraph of Gmdash with components C2 C3 CpHence by Proposition 42 there exists a complete (p ndash 1) partite graph as a spanningsubgraph of G1048577 Also since every vertex of C1 ndash v = G10485771048577is adjacent to everyvertex in Ci (i = 1 2 p) V(G10485771048577) V(C2) V(Cp) are the partite sets of a completep-partite graph as a spanning subgraph of G ndash v x where V(G10485771048577) belongs to thesame partite setSubcase (b) NG (v) V(C1) = 1048577Let A = V(C1) ndash V(G10485771048577) ndash v Then in G every vertex in A is adjacent to all thevertices of Ci i = 2 3 p Clearly N(v) = V(C2) V(C3) V(Cp) V(A)Hence by Proposition 42 V(C2) V(C3) V(Cp) and V(A) are the partite sets ofa complete p-partite graph as a spanning subgraph of G1048577 In particular every verexin C1 ndash v is adjacent to all the vertices of Ci i = 2 3 p Clearly G ndash v x =V(C1 ndash v) V(C2) V(Cp) and by Proposition 42 V(C1 ndash v) V(C2) V(Cp)are the partite sets of a complete p-partite graph as a spanning subgraph of G ndash x vwhere V(G10485771048577) belongs to the same partite setConversely assume that G ndash v x contains a complete k-partite graph (k 10485773)as a spanning subgraph with V (G10485771048577) belonging to the same partite set In Gmdash thevertex x is adjacent to all vertices except v Thus d(x) = n ndash 2 in Gmdash Clearly Gmdash ndash xis disconnected Since d (v) 10485771 v is adjacent to at least one vertex in Gmdash Thus1048577G10485771048577v1048577is connected subgraph of Gmdash If 1048577G10485771048577v1048577is maximal then C1 = 1048577G10485771048577v1048577is a component in G ndash x Otherwise we can find a component C1 of Gmdash ndash xwhich contains 1048577G10485771048577v1048577 By assumption V(C1) ndash v is the required partite setwhich contains V (G10485771048577) Also we can find at least two partite sets other thanV(C1) ndash v Let V (C2) V(C3) be the partite sets of G ndash v x Clearly V (C2)V(C3) 1048577V(G1048577) Thus by Proposition 42 we have 1048577Gmdash ndash v x1048577has at least threecomponents C1 ndash v C2 C3 Hence Gmdash ndash x has at least three components C1 C2 C3Thus x is a triple cut and hence Gmdash is not triple connectedTriple Connected Graphs 715 DERIVED GRAPHSDefinition 51 The line graph L (G) of a graph G is the graph in which the vertexset is the edge set of G and two vertices of L (G) are adjacent if and only if thecorresponding edges are adjacent in GTheorem 52 Let G be a connected graph Then L (G) is not triple connectedif and only if G has triple cut H such that G ndash H has at least three componentsC1 C2 and C3 with |V(Ci) | 10485772 and |V(Ci) N(H) | = 1 for i = 1 2 3Proof Assume that G has a triple cut H as in hypothesis Let V(Ci) N(H) = yiand N(Ci) V(H) = xi for every i Therefore xi yi 1048577E(G) and so xi yi 1048577V(L(G))Let H1048577= 1048577E (H) x1 y1 x2 y2 xk yk10485771048577L (G) Since H is connected H1048577isconnected Since 1048577V(Ci) xi1048577is connected L (1048577V(Ci) xi1048577) is connected Thusin L (G) 1048577E(Ci) xi yi1048577 say Ai is a connected subgraph and N(Ai) V(H1048577) = xi yifor all i Therefore by Theorem 33 H1048577is a triple cut in L (G) and hence L (G) is not

triple connectedConversely assume that L (G) is not triple connected Then there exists a triple cutH1048577 Therfore L (G) ndash H1048577has components C11048577 C21048577 Ck1048577 k 10485773 with |N(Ci) V(H1048577) | = 1Let N (Ci1048577) V (H1048577) = xi1048577 for every i Then xi1048577are edges of G Therefore takexi1048577= xi yi for all i Let A1048577= x11048577 x21048577 xk1048577Claim 1 |V(H1048577) | 10485771Suppose V(H1048577) = x1048577 Then x11048577= x21048577= = xk1048577= x1048577 Let ui10485771048577N (x1048577) V (Ci1048577)i = 1 2 k Since ui1048577s are adjacent to x1048577in L (G) k-edges u11048577 uk1048577are adjacent toan edge x1048577in G Then at least two edges uj1048577and uk1048577are incident with the same endvertex of x1048577 Hence uj1048577and uk1048577belong to the same component which is a contradictionThus the claimClaim 2 All the vertices of A1048577are distinctSuppose that there exist two components Ci and Cj1048577 i 1048577j such that V(H1048577) N(Ci1048577)= xi1048577 = xj1048577 = V(H1048577) N(Cj) Clearly N(xi1048577) V(Ci1048577) 10485771048577and N(xi1048577) V(Cj1048577) 10485771048577By Claim 1 and since H1048577is connected N(xi1048577) V(H1048577) 10485771048577 Let ui10485771048577N(xi1048577) V(Ci1048577)uj10485771048577N (xi1048577) V(Cj1048577) and xk10485771048577N (xi1048577) V (H1048577) Then ui1048577 uj1048577and xk1048577are adjacent to acommon vertex xi1048577in L (G) If ui1048577 uj1048577and xi1048577have a common end vertex in G then ui1048577and uj1048577lie in same component in L (G) which is a contradiction If ui1048577 xi1048577and xk1048577havea common end vertex in G then | N (Ci) V (H1048577) | 10485772 which is a contradictionHence the claimSince k 10485773 by Claim 2 |A1048577| 10485773 and hence |V(H1048577) | 10485773 Since V(H1048577) N(Ci1048577)= xi1048577 there exists at least one ui10485771048577V(Ci1048577) such that ui1048577xi10485771048577E(L (G)) Then ui1048577andxi1048577are two edges with common vertex in G Without loss of generality let it be yiSince H1048577is a triple cut of L (G) and Ci1048577is a component of L (G) ndash H1048577such that| N (Ci) V (H1048577) | = 1 no edge of V (Ci1048577) is incident with xi in G and no edge of72 J Paulraj Joseph amp G SudhanaH1048577ndash xi1048577 is incident with yi in G Therefore for every i xi yi is a bridge and1048577V(Ci1048577) yi1048577(= Ci say) is connected Since V(Ci1048577) 10485771048577in L (G) Ci has at least oneedge in G Hence |V(Ci)| 10485772 in G for all iCase (i) V(H1048577) 1048577A1048577= 1048577By definition of L (G)1048577NH1048577[xi1048577]1048577is a complete subgraph of L (G) for all i (51)Let xi1048577 xj10485771048577V (H1048577) Since H1048577is connected xi1048577and xj1048577are connected by a pathP1048577= xi1048577xi1048577+ 1 xj1048577ndash 1 xj1048577 By (51) xi1048577xi1048577+ 1 1048577E (H1048577) and by repeated application of(51) xi1048577xj10485771048577E(H1048577) Hence H1048577is a complete subgraph of L (G) Thus G has a vertexv with which all the vertices of H1048577are incident and hence v = x1 = x2 = = xk ThusH = v is a triple cut for G such that V(Ci) N(v) = yi for all iCase (ii) V(H1048577) ndash A104857710485771048577Let H = 1048577V(H1048577)1048577ndash y1 y2 yk Since V(H1048577) 1048577E(G) 1048577V(H1048577)1048577is a subgraph ofG and hence H is a subgraph of G Since H1048577is connected and no edge of H1048577is incidentwith yirsquos H is connected Hence H is a required triple cut with V(Ci) N(H) = yifor all iDefinition 53 The closure of a graph G is the graph obtained from G byrecursively joining pairs of non-adjacent vertices whose degree sum is at least nuntil no such pair remainsTheorem 54 If G is not triple connected then c (G) is also not triple connectedProof Let G be not triple connected Then there exists a triple cut H in GThen G ndash H has at least three components C1 C2 C3 with N(Ci) V(H) = xi LetC1 C2 Ck be the components of G ndash H with |N(Ci) V(H) | = 1 1 1048577i 1048577k k 10485773

Let c (G) = G + e1 e2 el where each ei joins two vertices in G whose degreesum 1048577nClaim The ends of each ei belong to either 1048577N (Ci)1048577or HCase (i) u 1048577V(Ci) and v 1048577V(Cj) i 1048577jLet |N(Ci) | = a + 1 and N (Ci) V(H) = xi Thus d (u) 1048577a Since G is nottriple connected there exists at least one component Cr other than Ci and Cj inG ndash H such that we can find at least one vertex in Cr which is not adjacent to vAlso v is adjacent to at most one vertex xi in N(Ci) Thus d (v) 1048577(n ndash 1) ndash a ndash 1 =n ndash a ndash 2 Now d (u) + d(v) 1048577a + n ndash a ndash 2 = n ndash 2 lt n Thus u and v are not adjacentin c (G) Thus no edge in e1 e2 em joins two components of G ndash H in c (H)Case (ii) u 1048577V(Ci) and v 1048577HIf | V (H) | = 1 then by Case (i) v is a triple cut in c (G) Hence assume that|V(H) | 10485772 Now d (v) 1048577a Now v can be adjacent to at most one vertex xi in N(Ci)Triple Connected Graphs 73Thus d (v) 1048577(n ndash 1) ndash a Thus d (u) + d(v) 1048577(n ndash 1 ndash a) + a lt n Thus u and v are notadjacent in c (G) Hence the claim and H satisfies the hypothesis of the theoremThus c (G) is not triple connected graphCorollary 55 G is triple connected if and only if c (G) is triple connectedProof If G is triple connected then obviously c (G) is also triple connectedConversely assume that c (G) is triple connected Suppose that G is not tripleconnected Then by Theorem 54 c (G) is not triple connected which is acontradiction Hence G is triple connectedDefinition 56 Let G and H be any two graphs Then G + H is the graphobtained from G H by joining each vertex of G to every vertex of HTheorem 57 If G and H are any two nontrivial connected graphs then G + His triple connectedProof Let G and H be any two nontrivial connected graphs Let | V(G) | = rand |V(H) | = s where r s 10485772 Clearly G + H contains a complete bipartite graphKr s as a spanning subgraph which is triple connected Hence by Remark 25 G + His triple connectedTheorem 58 Let G and H be any two graphs Then G + H is not tripleconnected if and only if G ~=K1 and 1048577(H) 10485773Proof Let G and H be any two graphs Assume that G + H is not tripleconnected Suppose that G ~= K1 or 1048577(H) 10485772 If G ~= K1 then | V(G) | 10485772 NowG + H contains a complete bipartite graph as a spanning subgraph Hence G + H istriple connected which is a contradiction Now let 1048577(H) 10485772 If 1048577(H) = 1 andG ~=K1 G + H is 2-connected If 1048577(H) = 2 and G ~=K1 = v then G +H is aconnected graph having exactly two blocks intersecting at a unique cut vertex offull degree Hence B (G + H) = P2 Thus in both cases G + H is triple connectedwhich is a contradiction Hence G ~=K1 and 1048577(H) 10485773Conversely assume that G ~=K1 and 1048577(H) 10485773 Let V(G) = v Now G + H isa connected graph with v as a cut vertex and 1048577(G ndash v) = | V (H) | 10485773 Thus byTheorem 28 G + H is not triple connected which is a contradictionDefinition 59 The corona of two graphs G1 and G2 is the graph G = G1

1048577G2

formed from one copy of G1 and |V(G1)| copies of G2 where the i th vertex of G1 is

adjacent to every vertex in the i th copy of G2If both G1 and G2 are disconnected then G1

1048577G2 and G2

1048577G1 are disconnectedIf G1 is disconnected then G1

1048577G2 is disconnected If G1 is connected then G1

1048577G2

is always connected but need not be triple connected For example if G1

~=P2 andG2

~=K2 K1 then G1

1048577G2 is connected but not triple connected as shown in theFig 5174 J Paulraj Joseph amp G SudhanaTheorem 510 Let G1 and G2 be any two connected graphs Then G1

1048577G2 istriple connected if and only if |V(G1) | = 1 or 2Proof If |V(G1) | = 1 then G1

1048577G2 has no cut vertex and hence by Theorem 26it is triple connectedIf |V(G1) | = 2 then G1

1048577G2 has a cut edge whose ends are the only cut verticesClearly 1048577V(G2) x1048577and 1048577V(G2) y1048577are isomorphic blocks of G1

1048577G2 Nowlet u v and w be any three vertices of G1

1048577G2 If all the three lie in any one blockthen by Theorem 26 they lie on a path in G1

1048577G2 Otherwise without loss ofgenerality we assume that u v 10485771048577V (G2) x1048577and w 10485771048577G2 y1048577(w may beequal to y also) Since 1048577G2 x1048577is a block there is a u ndash v path P1 in which x isa not an internal vertex Since v is adjacent to x and w is adjacent to y in G1

1048577G2vxyw is a v ndash w path P2 in G2 Then P1 P2 is a u ndash v ndash w path in G1

1048577G2If |V(G1)| 10485773 then w(G1

1048577G2 ndash V(G1)) 10485773 and every vertex in the i th copy ofG2 is adjacent to only the i th vertex of G1 Thus G1 is a triple cut and hence byTheorem 33 G1

1048577G2 is not triple connectedRemark 511 It is well known that G1

1048577G2 need not be isomorphic to G2

1048577G1Similarly G1

1048577G2 is triple connected need not imply that G2

1048577G1 is triple connectedFor example K2

1048577P3 is triple connected but P3

1048577K2 is not triple connectedsince u v and w do not lie on a path (See Fig 52(b))

Figure 51Figure 52Theorem 512 Let G1 be a connected graph and G2 be a disconnected graphThen G1

1048577G2 is triple connected if and only if G1

~=K1 and 1048577(G2) = 2Triple Connected Graphs 75Proof Assume that G1

1048577G2 is triple connected Suppose that G1

~=K1 or1048577(G2) gt 2 Then in both cases G1 is a triple cut of G1

1048577G2 and by Theorem 33 G11048577G2 is not triple connected If 1048577(G2) gt 2 and G1 is a trivial graph then clearlyG1

~=K1 is a triple cut of G1

1048577G2 and by Theorem 33 G1

1048577G2 is not triple connectedConversely assume that G1

~=K1 and 1048577(G2) = 2 Let V (G1) = x ClearlyG1

1048577G2 has exactly two blocks say B1 and B2 with common cut vertex x ThenB(G1

1048577G2) = P2 and hence by Theorem 36 G1

1048577G2 is triple connectedACKNOWLEDGMENTThe research of the second author is supported by the University Grants Commission NewDelhi through the Basic Science Research Fellowship (Grant No F4-12006(BSR)7-201-2007)REFERENCES[1] Y Alavi and J E Williamson (1975) Panconnected Graphs Studio ScientianemMathematicarum Hungarica 10 19-22[2] J A Bondy and U S R Murty (2008) Graph Theory Springer[3] Gary Chartrand and Ping Zhang (2006) Introduction to Graph Theory TataMcGraw-Hill Edition[4] Yan Jin Zhao Kewen Hong-Jin Lai and Ju Zhou (2008) New Sufficient Conditionsfor s-Hamiltonian Connected Graphs ARS Combinatoria 88 217-227J Paulraj Joseph M K Angel Jebitha ampP Chithra DeviDepartment of MathematicsManonmaniam Sundaranar UniversityTirunelveli-627 012 Tamil Nadu IndiaE-mail jpaulraj_2003yahoocoinG SudhanaDepartment of MathematicsNesamony Memorial Christian CollegeMarthandam-629 165 Tamil Nadu India

JP Journal of Mathematical SciencesVolume 7 Issues 1 amp 2 2013 Pages 13-39copy 2013 Ishaan Publishing House

This paper is available online at httpwwwiphscicom2010 Mathematics Subject Classification05C70Keywords and phrases graph decompositions path decompositionsReceived June 13 2013- 4 P DECOMPOSITION OF PRODUCT GRAPHSP CHITHRA DEVI and J PAULRAJ JOSEPHDepartment of MathematicsManonmaniam Sundaranar UniversityTirunelveli - 627 012 Tamil NaduIndiae-mail chithradevi095gmailcomjpaulraj_2003yahoocoinAbstractA decomposition of a graph G is a family of edge-disjoint subgraphsG1 G2hellip Gksuch that EGEG1_ EG2 ___ EGk If eachGi is isomorphic to H for some subgraph H of G then the decomposition iscalled an H- decomposition of G In this paper we give necessary andsufficient conditions for the decomposition of some product graphs into pathsof length three1 IntroductionLet G V Ebe a simple undirected graph without loops and multiple edgesPath on n vertices is denoted by Pn Cycle on n vertices is denoted by Cn Completegraph on n vertices is denoted by Kn The set Nvconsists of all vertices that areadjacent to v and is called the neighbourhood of v G is the graph with vertex setVGand two vertices are adjacent in G if and only if they are non adjacent in GkG denotes the graph consisting of k components each of which is isomorphic to GAn m- centipede is a path Pm together with a leaf at each vertex of the path Termsnot defined here are used in the sense of [1]P CHITHRA DEVI and J PAULRAJ JOSEPH14A decomposition of a graph G is a family of edge-disjoint subgraphsG1 G2hellip Gksuch that EGEG1_ EG2 ___ EGk If each Gi isisomorphic to H for some subgraph H of G then the decomposition is called anH- decomposition of G If H has at least three edges then the problem of deciding ifa graph G has an H- decomposition is NP- complete [3] Heinrich et al [7] provedthat a connected 4-regular graph G admits a P4 - decomposition if and only ifEG0mod 3by characterizing graphs of maximum degree 4 that admit atriangle-free Eulerian tour Haggkvist and Johansson [4] proved that every maximalplanar graph with at least 4 vertices has a P4 - decomposition Sunil Kumar [9]proved that a complete r- partite graph is P4 - decomposable if and only if its size isa multiple of 3 In [2] we gave necessary and sufficient conditions for theP4 - decomposition of some biregular and triregular triple connected graphs In thispaper we give necessary and sufficient conditions for the decomposition of someproduct graphs into paths of length three2 Building BlocksIn this section we prove some lemmas and collect certain results which are usedin the subsequent sections These are the building blocks in the construction of themain theoremsDefinition 21 The Cartesian product G1 G2 of two graphs G1 and G2 is thesimple graph with V1 V2 as its vertex set and two vertices u1 v1 and u2 v2are adjacent if and only if either u1 u2 and v1 is adjacent to v2 in G2 or u1 isadjacent to u2 in G1 and v1 v2Definition 22 The corona of two graphs G1 and G2 is the graphG G1 _ G2 formed from one copy of G1 and V G1 copies of G2 where the ithvertex of G1 is adjacent to every vertex in the ith copy of G2Remark 23 Pk Kn is P4 - decomposable if and only if k 1mod 3

Remark 24 Ck Kn is P4 - decomposable if and only if k 0mod 3Remark 25 Let G be an m- centipede m 2 Then G Kn is notP4 - decomposableP4 - DECOMPOSITION OF PRODUCT GRAPHS15Remark 26 Let G be any graph Then G Kn is P4 - decomposable if and onlyif G itself is P4 - decomposableLemma 27 K2 Cn is P4 - decomposable for all n 3Proof Assume that n 3Let V K2 x1 x2 and VCn y1 y2hellip yn Then VK2 Cn x1 y1 x1 y2 hellip x1 yn x2 y1 x2 y2 hellip x2 yn Rename x1 yiui for all i 1 2hellip n and x2 yi vi for all i 1 2hellip nThen P4 - decomposition of K2 Cn is given by uiui1vi1vi unu1v1vn1 i n minus1 Hence K2 Cn is P4 - decomposable ıLemma 28 Let G be the graph K2 Cn with a pendant vertex attached to thefirst vertex of each copy of K2 Then G is P4 - decomposable if and only ifn 0mod 3Proof Assume that n 0mod 3Let V K2 x1 x2 and VCn y1 y2hellip yn ThenVK2 Cn x1 y1 x1 y2 hellip x1 yn x2 y1 x2 y2 hellip x2 yn Rename x1 yi ui and x2 yi vi for all i 1 2hellip n and letw1 w2hellip wn be the pendant vertices attached to v1 v2hellip vn respectivelyThen for n 3 P4 - decomposition of G is given by uiui1vi1wi1 unu1v1w1_ECn h1 i n minus1 where Cn is v1v2_vnv1 and Cn is P4 - decomposableWhen n 3 P4 - decomposition of G is w1v1u1u2 w2v2u2u3 u1u3v3v2v2v1v3w3Hence G is P4 - decomposableConverse is obvious Lemma 29 Let G be the graph Cn _ K1 together with a pendant vertexattached to each of its pendant vertex Then G is P4 - decomposable for all n 3P CHITHRA DEVI and J PAULRAJ JOSEPH16Proof Let VCn v1 v2hellip vn Let N vi minusV Cn ui and N ui minusviwi Then wiuivivi1 wnunvnv1 1 i n minus1 is a P4 - decomposition of GHence G is P4 - decomposable Lemma 210 The graph Cn _ K2 is P4 - decomposable for all n 3Proof Let VCn v1 v2hellip vn Let wi and vi be the pendant vertices atvi 1 i n Then wivivi1ui1 wnvnv1u1 1 i n minus1 is a P4 - decompositionof Cn _ K2 Hence Cn _ K2 is P4 - decomposable Theorem 211 [9] Kr s is P4 - decomposable if and only if r 2 s 2 andrs 0mod 3Theorem 212 [9] Kn is P4 - decomposable if and only if n 3 andn 2mod 3Lemma 213 [8] The graph K2n1 can be decomposed into Hamilton cyclesfor every natural number nLemma 214 [8] The graph K2n minusI can be decomposed into Hamilton cyclesfor every natural number n where I is a 1-factor of K2n3 P4 -Decomposition of Cartesian Product of GraphsIn this section we give necessary and sufficient conditions for theP4 - decomposition of Cartesian product of some graphsTheorem 31 The graph Pk Cn is P4 - decomposable if and only if n 0mod 3or 2k 1mod 3Proof Let G be the graph Pk Cn Let V Cn v1 v2hellip vn and V Pku1 u2hellip uk Then VPk Cn ui v j 1 i k 1 j n Rename ui v j v ji for all 1 i k and 1 j n

Suppose that n 0mod 3P4 - DECOMPOSITION OF PRODUCT GRAPHS17Case (i) k is oddWhen k 3 and n 3 P3 C3 is P4 - decomposableAssume that n 3 and k 3Now v1 j v2 j hellip vnj Cn for all 1 j 3 AlsoEv11 vn1v11v12 vn1vn2 v12v13 vn2vn3G hellip _ hellip hellip where Gis the graph Cn _ K1 together with a pendant vertex attached to each of itspendant vertexThus EGEG_ ECn _ ECn Since n 0mod 3 Cn is P4 - decomposable Also by Lemma 29 Gis P4 -decomposable Hence G is P4 - decomposableAssume that k 3Now k minus1 is even and v1 j v2 j hellip vnj Cn for all j 2 4hellip k minus1and v1k v2k hellip vnk CnAlsoEv1 j v2 j hellip vnj _ v1 jv1j1hellip vnjvnj1v1 j1 v1 j2 hellip vn j1 vn j2G j 1 3hellip k minus2where Gis the graph Cn _ K1 together with a pendant vertex attached to each of itspendant vertexThus1 times21times21__________ _____________________minus______minuskn nkE G E G E G E C E CSince n 0mod 3 Cn is P4 - decomposable Also by Lemma 29 GisP4 - decomposable Hence G is P4 - decomposableP CHITHRA DEVI and J PAULRAJ JOSEPH18Case (ii) k is evenThen k minus1 is oddNowv1 j v2 j hellip vnj Cn for all j 2 4hellip k minus2andEv1 j v2 j hellip vnj _ v1 jv1j1hellip vnjvnj1

v1 j1 v1 j2 hellip vn j1 vn j2G j 1 3hellip k minus3where Gis the graph Cn _ K1 together with a pendant vertex attached to each of itspendant vertexAlso v1k minus1hellip vnk minus1 v1k hellip vnk K2 CnThus2 times22times22nkn nkE G E GE GE C E C E K C______ minus______ minus___________ _______________Since n 0mod 3 Cn is P4 - decomposableAlso by Lemmas 27 and 29 both K2 Cn and Gare P4 - decomposableHence G is P4 - decomposableSuppose that n 0mod 3 Then 2k 1mod 3and hence k 3t 2 t NNowv1 j v2 j hellip vnj v1 j1 hellip vn j1 K2 Cn for all j 1 4hellip 3t 1andE v1 j hellip vnj _ v1 jv1 jminus1 hellip vnjvn jminus1 v1 jv1 j1 hellip vnjvn j1 Cn _ K2for all j 3 6hellip 3tP4 - DECOMPOSITION OF PRODUCT GRAPHS19Thustimes2 21 times2 2 ________________ _ ___ ___________________tn ntE G E K Cn E K Cn E C K E C KBy Lemmas 27 and 210 both K2 Cn and Cn _ K2 are P4 - decomposableand hence G is P4 - decomposable

Conversely suppose that G is P4 - decomposable Then EG0mod 3That is 2k minus1n 0mod 3That is 2k 1mod 3or n 0mod 3 ıTheorem 32 The graph Ck Cn is P4 - decomposable if and only ifk 0mod 3or n 0mod 3Proof Let G be the graph Ck Cn Let V Cn v1 v2hellip vn and V Cku1 u2hellip uk Then VCk Cn ui v j 1 i k 1 j nRename ui v j v ji for all 1 i k and 1 j nSuppose that n 0mod 3Case (i) k nAssume that k n 3 ThenEC3 C3 EK2 C3 _ C3 _ K2 where both K2 C3 and C3 _ K2 are P4 - decomposableHence C3 C3 is P4 - decomposableAssume that n 3Nowv1 j v2 j hellip vnj Cn for all 1 j kandvi1 vi2hellip vik Ck for all 1 i nP CHITHRA DEVI and J PAULRAJ JOSEPH20Thus2 times______________ ___nE G E Cn E Cn E CnSince n 0mod 3 Cn is P4 - decomposable and hence Cn Cn is P4 -decomposableCase (ii) k nIf k 0mod 3and n 0mod 3 then Cn and Ck are P4 - decomposableNowtimes times__________ _______________nk kkE G E Cn E Cn E C E CHence G is P4 - decomposableIf k 0mod 3 then n 0mod 3Suppose k is odd Then k 2t 1 t NNowE v1 j v2 j hellip vnj v1 j1 v2 j1 hellip vn j1_ v1 j1 v1 j2 hellip vn j1 vn j2G j 1 3hellip 2t minus3E v1 2t1 v2 2t1 hellip vn 2t1_ v1 2t v1 2t1 hellip vn 2t vn 2t1 v1 2t1 v11hellip vn 2t1 vn1Cn K2andv1 2tminus1 v2 2tminus1 hellip vn 2tminus1 v1 2t v2 2t hellip vn 2t

K2 _ Cn P4 - DECOMPOSITION OF PRODUCT GRAPHS21where Gis the graph K2 Cn with a pendant vertex attached to the first vertex ofeach copy of K2Thus2 2 1 timesE G E G E G E K Cn E Cn Kt_ _ _____________ minusBy Lemmas 27 28 and 210 K2 Cn Gand Cn _ K2 are P4 -decomposable and hence G is P4 - decomposableSuppose k is even Then k 2t t 1NowE v1 j v2 j hellip vnj v1 j1 v2 j1 hellip vn j1_ v1 j1 v1 j2 hellip vn j1 vn j2G j 1 3hellip 2t minus3andE v1 2tminus1 hellip vn 2tminus1 v1 2t hellip vn 2t_ v1 2t v11hellip vn 2t vn1Gwhere Gis the graph K2 Cn together with a pendant vertex attached to the firstvertex of each copy of K2Thustimes____________tE G E GE GBy Lemma 28 Gis P4 - decomposableHence G is P4 - decomposableIf n 0mod 3 then the case is similar as aboveP CHITHRA DEVI and J PAULRAJ JOSEPH22Conversely suppose that G is P4 - decomposable Then EG0mod 3That is 2nk 0mod 3 That is n 0mod 3or k 0mod 3 ıTheorem 33 Let G be an m- centipede Then G Cn is P4 - decomposable ifand only if n 0mod 3or 4m 1mod 3Proof Let V Cn v1 v2hellip vn and VGu1 u2hellip um w1 w2hellip wmwhere u1u2_um is the path in G and wi rsquos are leaves at ui 1 i mRename ui v j u ji and wi v j wji for all 1 i m 1 j nSuppose that n 0mod 3NowE w1 j w2 j hellip wnj u1 j u2 j hellip unj_ u1 ju1 j1 hellip unjun j1G 1 j m minus1w1m w2mhellip wnm u1m u2mhellip unm K2 Cn where Gis the graph K2 Cn together with a pendant vertex attached to the firstvertex of each copy of K2Thus

2 1 timesnmE G Cn E GE GE K Cminus_____________By Lemmas 27 and 28 K2 Cn and Gare P4 - decomposable and henceG Cn is P4 - decomposableSuppose that 4m 1mod 3 Then m 3t 1 t NNoww1 j w2 j hellip wnj u1 j u2 j hellip unj K2 Cn for all 1 j mP4 - DECOMPOSITION OF PRODUCT GRAPHS23andu1i u2i hellip umi P3t1 for all 1 i nThustimes3 1 3 1times2 2 ____________ _____________________nt tmE G Cn E K Cn E K Cn E P E P By Lemma 27 K2 Cn is P4 - decomposable Also P3t1 is P4 -decomposable Hence G Cn is P4 - decomposableConversely suppose that G Cn is P4 - decomposableThus EG Cn 0mod 3That is 4m minus1n 0mod 3That is n 0mod 3or 4m 1mod 3 ıNow we investigate the P4 - decomposition in the Cartesian product of anm- centipede G with Kn If G Kn is P4 - decomposable thenEG Kn 0mod 3That is2 10mod 3212 minusminusn mnmnThat is nmn 1minus10mod 3That is n 0mod 3or mn 11mod 3Theorem 34 Let G be an m- centipede and let n 0mod 3 Then G Kn isP4 - decomposableProof Let V Cn v1 v2hellip vn and VGu1 u2hellip um w1 w2hellip wmwhere u1u2_um is the path in G and wi rsquos are leaves at ui 1 i m Then

VG Kn ui v j wi v j 1 i m 1 j n Rename ui v j u ji and wi v j wji for all 1 i m 1 j nP CHITHRA DEVI and J PAULRAJ JOSEPH24Suppose that n 0mod 3 ThenClaim 1 K2 Kn is P4 - decomposableLet VK2 x1 x2 Rename x1 vi vi1 and x2 vi vi2 for all1 i nWhen n 3 K2 K3 is P4 - decomposable Hence suppose that n 3Assume that n is odd Then by Lemma 213 K2n1 can be decomposed into nHamilton cycles C2n1Nowv11 v21hellip vn1 Knandv12 v22hellip vn2 KnThus v11 v21hellip vn1 can be decomposed into2n minus1Hamilton cyclesWithout loss of generality let C1 v11v21hellipvn1v11 be a Hamiltonian cycle of itSimilarly let C2 v12v22hellipvn2v12 be a Hamiltonian cycle of v12 v22hellip vn2 NowEC1_ EC2 _ v11v12 v21v22hellip vn1vn2 K2 CnThus__________ _______________times23times232______ minus______ minusnn nnE K Kn E Cn E Cn E C E C_EK2 Cn Since n 0mod 3 Cn is P4 - decomposable and by Lemma 27 K2 Cn isP4 - decomposableHence K2 Kn is P4 - decomposableAssume that n is evenP4 - DECOMPOSITION OF PRODUCT GRAPHS25By Lemma 214 Kn minusI can be decomposed into2n minus2

Hamilton cycles forevery natural number n where I is a 1- factor of KnNow v11 v21hellip vn1 Kn and v11v21 v31v41hellip v nminus1 1vn1 is a 1- factorof KnThus v11 v21hellip vn1 minusv11v21 v31v41hellip v nminus1 1vn1 can be decomposedinto2n minus2Hamilton cycles Cn Similarly v12 v22hellip vn2 minusv22v32 v42v52hellipv nminus2 1v nminus1 1 vn2v12 can be decomposed into2n minus2Hamilton cyclesAlsov11v21 v31v41hellip vnminus11vn1 v22v32 v42v52hellip vnminus21vnminus11vn2v12 v11v12 v21v22hellip vn1vn2C2nThustimes222times222 __________ _ _______________minusminusnn n nnE K Kn E Cn E Cn E C E C E CSince n 0mod 3 Cn and C2n are P4 - decomposable and hence K2 Kn isP4 - decomposableClaim 2 G1 Kn _ K1 together with a pendant vertex attached to each of itspendant vertex is P4 - decomposable if n is odd and n 0mod 3Let xi be the pendant vertex at vi in Kn _ K1 and let yi be the pendant vertexat xi in G1When n 3 clearly G1 is P4 - decomposable Hence suppose that n 3Since n is odd Kn can be decomposed into2n minus1Hamilton cyclesP CHITHRA DEVI and J PAULRAJ JOSEPH26Without loss of generality let C1 v1v2_vnv1 be a Hamiltonian cycle of G1NowEC1viuiuiwiG _ _ 1 i nwhere Gis the graph Cn _ K1 together with a pendant vertex attached to each of itspendant vertexThustimes2

3E G1 E C E C E Gnn n ______ minus_____________Since n 0mod 3 Cn is P4 - decomposable Also by Lemma 29 GisP4 - decomposable and hence G1 is P4 - decomposableCase (i) n is oddAssume that n 3NowE w1 j w2 j w3 j u1 j u2 j u3 j _ u1 ju1 j1 u2 ju2 j1 u3 ju3 j1Gfor all 1 j m minus1where Gis the graph K2 K3 together with a pendant edge attached to the firstvertex of each copy of K2Thus2 3 1 timesE G K3 E G E G E K Kmminus_____________Clearly Gand K2 K3 are P4 - decomposableHence G K3 is P4 - decomposableAssume that n 3P4 - DECOMPOSITION OF PRODUCT GRAPHS27NowE w1 j w2 j hellip wnj _ w1 ju1 j hellip wnjunj u1 ju1 j1 hellip unjun j1G1 for all 1 j m minus1u1 j u2 j hellip unj Kn for all 1 j m minus1andw1m w2mhellip wnm u1m u2mhellip unmK2 Kn where G1 is the graph Kn _ K1 together with a pendant vertex attached to each of itspendant vertexThus1 times21 times1 1 __________ _ _______________minusminus

mn n nmE G Kn E G E G E K K E K E KBy Claims 1 and 2 both K2 Kn and G1 are P4 - decomposable Also sincen 0mod 3 by Theorem 212 Kn is P4 - decomposableHence G Kn is P4 - decomposableCase (ii) n is evenClaim 3 Kn _ K1 is P4 - decomposable if n mod 3and n is evenBy Lemma 214 the graph Kn minusI can be decomposed into2n minus2Hamiltoncycles Cn where I is a 1- factor of KnNow I together with the n pendant edges in Kn _ K1 form 2 4 Pn______Thus times24 4times221 __________ _____________ _______________ minusn nE Kn K E Cn E Cn E P E PSince n 0mod 3 Cn is P4 - decomposableP CHITHRA DEVI and J PAULRAJ JOSEPH28Thus Kn _ K1 is P4 - decomposableNoww1 j w2 j hellip wnj u1 j u2 j hellip unj minusE u1 j u2 j hellip unjKn _ K1 for all 1 j m minus1andu1 j u2 j hellip unj u1 j1 u2 j1 hellip un j1 minusE u1 j1 u2 j2 hellip un j1Kn _ K1 for all 1 j m minus1andw1m w2mhellip wnm u1m u2mhellip unm K2 KnThusn

mE G Kn E Kn K E Kn K E K Kminus21 times1 1 _______________ ___ _1 times_____1 ________1_ _ ___ _mminusE Kn K E Kn KSince n 0mod 3 by Claim 1 K2 Kn is P4 - decomposable Also since nis even by Claim 3 Kn _ K1 is P4 - decomposableHence G Kn is P4 - decomposable ıConjecture 35 If G is an m- centipede and mn 11mod 3 then G Knis P4 - decomposableTheorem 36 Pk Kn is P4 - decomposable if and only if n 0mod 3orkn 12mod 3Proof Let V Pk u1 u2hellip uk and VKn v1 v2hellip vn ThenVPk Kn ui v j 1 i k and 1 j nRename ui v j v ji for all 1 i k 1 j nP4 - DECOMPOSITION OF PRODUCT GRAPHS29Suppose that n 0mod 3or kn 12mod 3Case (i) n 0mod 3Subcase (i) n is evenNowv1 j v2 j hellip vnj v1 j1 v2 j1 hellip vn j1minusE v1 j1 v2 j1 hellip vn j1Kn _ K1 for all 1 j k minus1andv1k v2k hellip vnk KnThus1 times1 1 nkE Pk Kn E Kn K E Kn K _ E K______________ ___ _minusSince n 0mod 3 by Theorem 212 Kn is P4 - decomposable Also byClaim 3 of Theorem 34 Kn _ K1 is P4 - decomposableHence Pk Kn is P4 - decomposableSubcase (ii) n is oddClaim 1 Kn _ K2 is P4 - decomposableWhen n 3 clearly K3 _ K2 is P4 - decomposableHence assume that n 3Since n is odd Kn can be decomposed into2

n minus1Hamilton cycles Cn Now2 times23E K K1 E C E C E Cn Knn n n _ ___________ ___minusP CHITHRA DEVI and J PAULRAJ JOSEPH30Since n 0mod 3 Cn is P4 - decomposable Also by Lemma 210 Cn _ K2 isP4 - decomposableHence Kn _ K2 is P4 - decomposableIf k 0mod 3 then k 3lWhen n 3v1 j v2 j v3 j v1 j1 v2 j1 v3 j1K2 K3 for all j 1 4hellip 3l minus8v1 jminus1 v2 jminus1 v3 jminus1 v1 j v2 j v3 j v1 j1 v2 j1 v3 j1minusE v1 jminus1 v2 jminus1 v3 jminus1 minusE v1 j1 v2 j1 v3 j1K3 _ K2 for all j 3 6hellip 3l minus6v1 jminus1 v2 jminus1 v3 jminus1 v1 j v2 j v3 j v1 j1 v2 j1 v3 j1minusE v1 j1 v2 j1 v3 j1G j 3l minus2 3l minus4andv13lminus1 v23lminus1 v33lminus1 v13l v23l v33l K2 K3where Gis the graph K2 K3 together with a pendant edge attached to the firstvertex of each copy of K2Thus__________________2 times3 2 3 2 3minuslE Pk K E K K E K K________________ _ ___ _2 times3 2 3 2lminusE K K E K K_EG_ EG_ EK2 K3 P4 - DECOMPOSITION OF PRODUCT GRAPHS31Clearly K2 K3 K3 _ K2 and Gare P4 - decomposableHence Pk K3 is P4 - decomposableNow assume that n 3 Thenv1 j v2 j hellip vnj v1 j1 v2 j1 hellip vn j1K2 Kn for all j 1 4hellip 3l minus5

v1 jminus1 v2 jminus1 hellip vn jminus1 v1 j v2 j hellip vnj v1 j1 v2 j1 hellip vn j1minusE v1 jminus1 v2 jminus1 hellip vn jminus1minusE v1 j1 v2 j1 hellip vn j1Kn _ K2 for all j 3 6hellip 3l minus3v13l v23l hellip vn3l Kn v13lminus1 v23lminus1hellip vn3lminus1Kn v13lminus2 v23lminus2hellip vn3lminus2v1 3lminus1 v2 3lminus1 hellip vn 3lminus1 v1 3l v2 3l hellip vn 3lminusE v1 3lminus1 v2 3lminus1 hellip vn 3lminus1 minusE v1 3l v2 3l hellip vn 3lGwhere Gis Kn _ K1 together with a pendant edge attached to each of its pendantvertexThus________________1 times2 2minuslE Pk Kn E P Kn E P Kn________________ _ ___ _1 times2 2lminusE Kn K E Kn KEKn EKn EG _ _ _ P CHITHRA DEVI and J PAULRAJ JOSEPH32Since n 0mod 3 by Theorem 212 by Claims 1 and 2 of Theorem 34 andby Claim 1 Kn K2 Kn Gand Kn _ K2 are P4 - decomposableHence Pk Kn is P4 - decomposableIf k 1mod 3 then k 3l 1When n 3v1 j v2 j v3 j v1 j1 v2 j1 v3 j1K2 K3 for all j 1 4hellip 3l minus2v1 jminus1 v2 jminus1 v3 jminus1 v1 j v2 j v3 j v1 j1 v2 j1 v3 j1minusE v1 jminus1 v2 jminus1 v3 jminus1 minusE v1 j1 v2 j1 v3 j1K3 _ K2 for all j 3 6hellip 3l minus3andv1 3lminus1 v2 3lminus1 v3 3lminus1 v1 3l v2 3l v3 3l v1 3l1 v2 3l1 v3 3l1minusE v1 3lminus1 v2 3lminus1 v3 3lminus1Gwhere Gis the graph K2 K3 together with a pendant edge attached to the firstvertex of each copy of K2Thus__________________times3 2 3 2 3l

E Pk K E K K E K K1 timesE K3 K2 E K3 K2 E Glminus_________________ _ ___ _Clearly K2 K3 K3 _ K2 and Gare P4 - decomposableHence Pk K3 is P4 - decomposableP4 - DECOMPOSITION OF PRODUCT GRAPHS33Now assume that n 3 Thenv1 j v2 j hellip vnj v1 j1 v2 j1 hellip vn j1K2 Kn for all j 1 4hellip 3l minus2v1 jminus1 v2 jminus1 hellip vn jminus1 v1 j v2 j hellip vnj v1 j1 v2 j1 hellip vn j1minusE v1 jminus1 v2 jminus1 hellip vn jminus1 minusE v1 j1 v2 j1 hellip vn j1Kn _ K2 for all j 3 6hellip 3l minus3v13l v23l hellip vn3l Kn v13lminus1 v23lminus1hellip vn3lminus1 v13l v23l hellipvn 3l v1 3l1 v2 3l1 hellip vn 3l1minusE v1 3lminus1 v2 3lminus1 hellip vn 3lminus1minusE v1 3l v2 3l hellip vn 3lGwhere Gis the graph Kn _ K1 together with a pendant vertex attached to each of itspendant vertexThus__________________times2 2lE Pk Kn E K Kn E K Kn1 timesE K K2 E K K2 E Kn E Gln n minus_ _________________ _ ___ _Since n 0mod 3 by Theorem 212 Claims 1 and 2 of Theorem 34 and byClaim 1 Kn K2 Kn Gand Kn _ K2 are P4 - decomposableHence Pk Kn is P4 - decomposableIf k 2mod 3 then k 3l 2P CHITHRA DEVI and J PAULRAJ JOSEPH34Now assume that n 3 Thenv1 j v2 j hellip vnj v1 j1 v2 j1 hellip vn j1K2 Kn for all j 1 4hellip 3l 1

v1 jminus1 v2 jminus1 hellip vn jminus1 v1 j v2 j hellip vnj v1 j1 v2 j1 hellip vn j1minusE v1 jminus1 v2 jminus1 hellip vn jminus1 minusE v1 j1 v2 j1 hellip vn j1Kn _ K2 for all j 3 6hellip 3lThus__________________1 times2 2lE Pk Kn E K Kn E K Kntimes_____2 __________2_ _ ___ _lE Kn K E Kn KSince n 0mod 3 by Claim 1 of Theorem 34 and by Claim 1 K2 Kn andKn _ K2 are P4 - decomposableHence Pk Kn is P4 - decomposableCase (ii) n 1mod 3Thus n 3t 1 and hence kn 12mod 3_ k 1mod 3_ k 3l 1Nowv1 j v2 j hellip vnj Kn for all 1 j kandvi1 vi2hellip vin Pk for all 1 i nP4 - DECOMPOSITION OF PRODUCT GRAPHS35Thus____________k timesE Pk Kn E Kn E Kntimes__________ ___nE Pk E PkSince n 1mod 3 by Theorem 212 Kn is P4 - decomposableAlso since k 3l 1 Pk is P4 - decomposableHence Pk Kn is P4 - decomposableCase (iii) n 2mod 3Thus n 3t 2Now kn 1k3t 30mod 3 which is a contradictionHence this case does not ariseHence Pk Kn is P4 - decomposableConversely suppose that Pk Kn is P4 - decomposable ThenEPk Kn 0mod 3That is10mod 3

21minusminusk nn nkThat is1 0mod 321___minusk n nThat is n 0mod 3or kn 12mod 3 ı4 P4 - decomposition of Lexicographic Product of GraphsIn this section we give sufficient condition for the lexicographic product of anygraph G with Kn Cn and Kn to be P4 - decomposableDefinition 41 Let G VG EGand H V H EHbe two graphsP CHITHRA DEVI and J PAULRAJ JOSEPH36Then the lexicographic product of G and H is the graph G H with vertex setVGVHand two vertices g1 h1 and g2 h2 are adjacent in G H if g1is adjacent to g2 or g1 g2 and h1 is adjacent to h2The other way of viewing G H is by replacing each vertex in G by a copy ofH and two vertices in G are adjacent if and only if there exists a complete bipartitesubgraph with the corresponding vertices of H as partite setsNow we investigate the P4 - decomposition in the lexicographic product of a nontrivial graph G with Kn If G Kn is P4 - decomposable thenEG Kn 0mod 3That is 0mod 3 n2 E G That is n 0mod 3or EG0mod 3Theorem 42 Let G be any non trivial graph If n 0mod 3 then G Kn isP4 - decomposableProof Let V G v1 v2hellip vk and VKn u1 u2hellip un Then V GKnvi u j 1 i k and 1 j nRename vi u j v ji 1 i k and 1 j nNow for each viv j EGv1i v2i hellip vni v1 j v2 j hellip vnj Kn nThustimes ______________E GE G Kn E Kn n E Kn nSince n 0mod 3 by Theorem 211 Kn n is P4 - decomposableHence G Kn is P4 - decomposable ıConjecture 43 If G is a non trivial graph and EG0mod 3 thenG Kn is P4 - decomposable

P4 - DECOMPOSITION OF PRODUCT GRAPHS37Now we investigate the P4 - decomposition in the lexicographic product of a nontrivial graph G with Cn If G Cn is P4 - decomposable thenEG Cn 0mod 3That is 0mod 3 n V G n2 E G That is n 0mod 3or VGn EG0mod 3Theorem 44 Let G be any non trivial graph If n 3 and n 0mod 3 thenG Cn is P4 - decomposableProof Let V G v1 v2hellip vk and VCn u1 u2hellip un ThenVG Cn vi u j 1 i k and 1 j nRename vi u j v ji 1 i k and 1 j nNowv1i v2i hellip vni Cn for all 1 i kAlso for each viv j EGv1i v2i hellip vi v1 j v2 j hellip vnj minusE v1i v2i hellip vniminusE v1 j v2 j hellip vnjKn nThustimes times____________ _______________E Gn n n nkE G Cn E Cn E Cn E K E KSince n 0mod 3 by Theorem 211 Kn n is P4 - decomposableAlso Cn is P4 - decomposableHence G Cn is P4 - decomposable ıP CHITHRA DEVI and J PAULRAJ JOSEPH38Conjecture 45 If G is a non trivial graph and VGn EG0mod 3then G Cn is P4 - decomposableNow we investigate the P4 - decomposition in the lexicographic product of a nontrivial graph G with Kn If G Kn is P4 - decomposable thenEG Kn 0mod 3That is0mod 321 2 minusV G n E GnnThat is n 0mod 3or0mod 321

minusV G n E GnTheorem 46 Let G be any non trivial graph If n 3 and n 0mod 3 thenG Kn is P4 - decomposableProof Let V G v1 v2hellip vk and VKn u1 u2hellip un ThenVG Kn vi u j 1 i k and 1 j nRename vi u j v ji 1 i k and 1 j nNow v1i v2i hellip vni Kn for all 1 i kAlso for each viv j EGv1i v2i hellip vni v1 j v2 j hellip vnj minusE v1i v2i hellip vniminusE v1 j v2 j hellip vnjKn nThustimes times____________ _______________E Gn n n nkE G Kn E Kn E Kn E K E KSince n 0mod 3 by Theorems 211 and 212 Kn n and Kn areP4 - decomposableHence G Kn is P4 - decomposable ıP4 - DECOMPOSITION OF PRODUCT GRAPHS39Conjecture 47 If G is a non trivial graph andVGn EGnminus210mod 3 then G Kn is P4 - decomposableReferences[1] J A Bondy and U S R Murty Graph Theory Springer 2008[2] P Chithra Devi and J Paulraj Joseph On P4 - decomposition of triple connectedgraphs Amer J Appl Math Math Sci (2013) (accepted)[3] D Dor and M Tarsi Graph decomposition is NP-complete a complete proof ofHolyers conjecture SIAM J Comput 26 (1997) 1166-1187[4] R Haggkvist and R Johansson A note on edge-decompositions of planar graphsDiscr Math 283(1-3) (2004) 263-266[5] S L Hakimi On the realizability of a set of integers as degrees of the vertices of agraph SIAM J Appl Math 10 (1962) 496-506[6] V Havel A remark on the existence of finite graphs (Czech) Casopis Pest Mat 80(1995) 477-480[7] K Heinrich J Liu and M Yu P4 - decompositions of regular graphs J Graph Theo31(2) (1999) 135-143[8] E Lucas Recreations Mathematiques Vol 2 Gauthier-Villars Paris 1884[9] C Sunil Kumar On P4 - decompositions of graphs Taiwan J Math 7(4) (2003)

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path in GExample 22 All paths and cycles complete graphs and wheels are somestandard examples of triple connected graphs Petersen graph Kr s where r s 10485772and Km1 m1 mk

mi 10485771 and k 10485773 are also triple connected graphsRemark 23 A triple connected graph may or may not have cut vertex or cutedge The graph in Fig 21(a) is a triple connected graph without cut vertex whereasthe graph in Fig 21(b) is a triple connected graph with cut edgeFigure 21Remark 24 Hamiltonian graphs and traceable graphs are triple connectedbut not conversely The graph in Fig 22 is triple connected but neither traceablenor hamiltonianFigure 22Triple Connected Graphs 63Remark 25 Clearly every triple connected graph is a connected graph If Ghas a spanning subgraph which is triple connected then G is triple connectedTheorem 26 Every 2-connected graph is triple connectedProof Let G be a 2-connected graph and let x y z 1048577G Since G is 2-connectedit has no cut vertex Hence any two vertices are joined by two internally disjointpaths Let P1 and P2 be two internally disjoint paths from x to y Since G is connectedthere is a y ndash z path say Q If Q is internally disjoint with P1 or P2 then x y z lie onpath P1 Q or P2 Q Otherwise let w be the last vertex common to P1 and QThen z ndash w section of Q together with w ndash y section of P1 and P2

ndash 1 is a path containingx y and z Hence G is triple connectedObservation 27 Three vertices u v and w lie on a path P if and only if P isthe union of two paths with exactly one common vertexTheorem 28 If G has a cut vertex v such that 1048577(G ndash v) 10485773 then G is not tripleconnectedProof Let v be a cut vertex of G such that 1048577(G ndash v) 10485773 Let G1 G2 G3 be anythree components of G ndash v Let x 1048577V(G1) y 1048577V(G2) z 1048577V(G3) Then any pathconnecting two vertices of x y and z must pass through v and hence any two suchpaths have at least two vertices in common By Observation 27 x y and z do notlie on a path Hence G is not triple connectedTheorem 29 Given any two positive numbers a and b where b 10485772a ndash 1 thereexists a triple connected graph G such that 1048577(G) = a and |V(G) | = bProof If a = 1 then b 10485771 and the complete graph on b vertices is a tripleconnected graph with 1048577(G) = 1 Now let a gt 1 If b = 2a ndash 1 then the path on bvertices labelled as v1 v2 v2a ndash 1 is a triple connected graph with 1048577(G) = a If b gt2a ndash 1 then b = (2a ndash 1) + c where c 10485771 Construct the graph G as followsConsider a path on 2a ndash 1 vertices v1 v2 v2a ndash 1 and a complete graph on cvertices w1 w2 wc Let G be the graph obtained by identifying v1 and w1 ClearlyG is a triple connected graph with b vertices Also v1 v3 v5 v2a ndash 1 is a maximumindependent set so that 1048577(G) = a3 CHARACTERIZATION THEOREMSTheorem 31 A tree T is triple connected if and only if T ~=Pn n 10485772Proof Let T be a tree Assume that T is triple connected Suppose T ~= Pn n 10485772Then there exists a vertex v such that d (v) 10485773 Clearly v is a cut vertex of T and1048577(T ndash v) 10485773 Hence by Theorem 28 T is not triple connected which is a





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