Supereulerian Graphs and the Petersen Graph

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1 Acta Mathematica Sinica, English Series Feb., 04, Vol. 0, No., pp Published online: January 5, 04 DOI: 0.007/s y Acta Mathematica Sinica, English Series Springer-Verlag Berlin Heidelberg & The Editorial Office of AMS 04 Supereulerian Graphs and the Petersen Graph Xiao Min LI Lan LEI Department of Mathematics and Statistics, Chongqing Technology and Business Uniersity, Chongqing , P.R.China lxm@ctbu.edu.cn leilan@ctbu.edu.cn Hong-Jian LAI Department of Mathematics, West Virginia Uniersity, Morganton, WV , USA and College of Mathematics and System Sciences, Xinjiang Uniersity, Urumqi 80046, P.R.China hjlai@math.u.edu Meng ZHANG Department of Mathematics, West Virginia Uniersity, Morganton, WV , USA mzhang@math.u.edu Abstract AgraphG is supereulerian if G has a spanning eulerian subgraph. Boesch et al. [J. Graph Theory,, (977)] proposed the problem of characterizing supereulerian graphs. In this paper, e proe that any -edge-connected graph ith at most edge-cuts of size is supereulerian if and only if it cannot be contractible to the Petersen graph. This extends a former result of Catlin and Lai [J. Combin. Theory, Ser. B, 66, 9 (996)]. Keyords Supereulerian graphs, petersen graph, edge-cut, reduction, contraction MR(00) Subject Classification O5C5, O5C8, O5C45 Introduction We consider finite, undirected and loopless graphs. Undefined terms and notaions ill follo Bondy and Murty [4]. In particular, κ(g) andκ (G) denote the connectiity and the edgeconnectiity of a graph G, respectiely. A graph G is nontriial if E(G) > 0, and e rite H G to mean that H is a subgraph of G. LetO(G) denote the set of all odd degree ertices of a graph G, andg(g) (called the girth of G) be the length of a shortest cycle in G. A graph G is een if O(G) =, andiseulerian if it is both een and connected. If G has a spanning eulerian subgraph, then G is supereulerian. The supereulerian graph problem, raised by Boesch et al. [], seeks to characterize supereulerian graphs. Pulleyblank [6] shoed that determining if a graph is supereulerian, een hen restricted to planar graphs, is NP-complete. For more in the literature on supereulerian graphs, see Catlin s surey [6] and its update by Chen and Lai []. Receied May 4, 0, accepted March, 0 Supported by National Natural Science Foundation of China (Grant No. 0087) and Science Foundation Chongqing Education Committee (Grant Nos. KJ0075 and KJ07 )

2 9 Li X. M., et al. For X E(G), the contraction G/X is obtained from G by contracting each edge of X and deleting the resulting loops. If H G, eriteg/h for G/E(H). If H is connected, let H denote the ertex in G/H to hich H is contracted, in this case, H is called the preimage of H. AgraphG is collapsible if for eery een subset R V (G), G has a spanning connected subgraph H R ith O(H R )=R. In particular, K is both supereulerian and collapsible, and any collapsible graph G is supereulerian. In [5], Catlin shoed that eery graph G has a unique collection of pairise disjoint maximal collapsible subgraphs H,H,...,H c. The graph obtained from G by contracting each H i into asingleertex( i c), is called the reduction of G. A graph is reduced if it is the reduction of some other graph. Since eery 4-edge-connected graph is collapsible [5], and so supereulerian [5], efforts to characterize supereulerian graphs hae been ithin families of -edge-connected graphs. Chen et al. [0,, ] inestigated conditions under hich a -edge-connected graph G is supereulerian if and only if G cannot be contracted to the Petersen graph. These settled the -edge-connected case of a conjecture by Benhocine et al. []. Caltin et al. considered -edge-connected graphs ith limited number of -edge cuts. They proed the folloing: Theorem. ([9, Theorem.]) Let G be a -edge-connected graph. If G has at most 0 edge-cuts of size, then exactly one of these holds : (i) G is supereulerian ; (ii) The reduction of G is the Petersen graph. Theorem. ([9, Theorem.4]) Let G be a -edge-connected graph. If G has at most edge-cuts of size, then exactly one of these holds : (i) G is supereulerian ; (ii) The reduction of G is the Petersen graph ; (iii) The reduction of G is a nonsupereulerian graph of order beteen 7 and 9, ith girth at least 5, ithexactly ertices of degree and ertex of degree 5, and ith the remaining ertices independent and of degree 4. It has been a question hether graphs stated in Theorem. (iii) exist or not. In this paper, e settle this problem by shoing that no such graphs exist. Theorem. Let G be a -edge-connected graph. If G has at most edge-cuts of size, then the folloing are equialent : (i) G is supereulerian ; (ii) The reduction of G is not the Petersen graph. The folloing notations ill be used throughout this paper. For a graph G and integer i, let D i (G) ={ d G () =i, V (G)} and d i (G) = D i (G). WhenG is understood, e rite d i for d i (G). Let F (G) denote the minimum number of extra edges that must be added to G so that the resulting graph has to edge-disjoint spanning trees. Let E G () ={u u E(G),u V (G)} and N G () ={u u E(G),u V (G)}. When G is understood, e rite N() forn G () ande() fore G (). Our proof depends on a ne sufficient condition for a graph to be supereulerian. Let F

3 Supereulerian Graphs and the Petersen Graph 9 denote the collection of all connected graphs satisfying each of the folloing: (F) d 5 (G) =,d (G) =, (F) δ(g) Δ(G) 5, (F) g(g) 5, and (F4) no edge of G joins to ertices of een degree in G. The folloing associate result plays an important role in our proof of Theorem.. Theorem.4 Let G F be a graph. Then G is supereulerian. The paper ill be organized as follos. In the next section, e present the preliminaries of Catlin s reduction method and the related theory that ill be used in the proofs. We then proe Theorem. assuming the alidity of Theorem.4. The last section ill be deoted to the proof of Theorem.4. Prerequisites In this section, e present Catlin s reduction method to be used in our proofs. Theorem. ([5, Theorems, 5 and 8]) Let G be a connected graph. (i) Let H be a collapsible subgraph of G. Then G is supereulerian if and only if G/H is supereulerian. (ii) G is reduced if and only if G has no nontriial collapsible subgraphs. (iii) Let G be the reduction of G. ThenG is supereulerian if and only if G is supereulerian, and G is collapsible if and only if G = K. (i) If G is reduced, then eery subgraph of G is also reduced. Theorem. ([8, Theorem.5]) Let G be a reduced graph. If F (G), then G {K,K,K,t (t )}. Theorem. ([7, Theorem 7]) E(G). If G is a connected reduced graph, then F (G) = V (G) Corollary.4 4)d j 4. If G is a connected reduced graph, then F (G) =d +d + d j 5 (j Proof As V (G) = j d j and E(G) = j jd j, by Theorem., e hae F (G) = d +d + d j 5 (j 4)d j 4. Theorem.5 ([9, Theorem.]) Let G be a -edge-connected graph ith F (G) =. If G is nonsupereulerian and reduced, then each of the folloing holds : (i) G has no edge joining to ertices of een degree ; (ii) G has girth at least 5; (iii) G has no subgraph H ith κ (H) and F (H) =. For a graph G, an edge-cut X E(G) is called essential edge-cut, if each component of G X has at least one edge. Lemma.6 Let G be a -edge-connected nonsupereulerian reduced graph ith F (G) =. Then eery edge-cut of size is not an essential edge-cut (i.e., the number of edge-cut of size is equal to d (G)).

4 94 Li X. M., et al. Proof Let X E(G) be an edge-cut of size, and H and H the to components of G X. By (i) of Theorem., H and H are both reduced. Then by Theorem., F (G) = V (G) E(G) =( V (H ) + V (H ) ) ( E(H ) + E(H ) + X ) = V (H ) E(H ) + V (H ) E(H ) = F (H )+F (H ), and so F (G)+ = F (H )+F(H ). Since F (G) =,min{f (H ),F(H )} (sayf(h ) ). By Theorem., H {K,K,K,t (t )}. IfH = K,thenX is not an essential edge-cut. If H = K or H = K,, then ertex of degree ill appear, contrary to κ (G). Hence H = K,t (t ). Since K,t (t ) contains C 4, this is contrary to (ii) of Theorem.5. This completes the proof of the lemma. Proof of Theorem. Let G be the reduction of G. By Theorem. (iii), it suffices to sho that G either is supereulerian or is the Petersen graph. We shall sho that G is contractible to the Petersen graph ith the folloing assumption: G is not supereulerian. (.) Since G has at most edge cut of size, G has at most edge cut of size. Thus d (G ). Since κ (G ) κ (G), d (G )=d (G ) = 0. By Corollary.4, e hae F (G )=d (G )+d (G )+d (G ) (j 4)d j (G ) 4 =d (G ) j (G j 5 j 5(j 4)d ) 4. (.) By (.) and by d (G ), F (G ). If F (G ), then by Theorem., G {K,K,K,t (t )}. By (.), G K,andsoG {K,K,t (t )}, contrary to the fact that κ (G ). Hence F (G )=. In the rest of the proof, e ill rite d j for d j (G ), j. By (.) and by F (G )=, 0 = d j 5(j 4)d j. (.) Thus d 0. If d = 0, by Lemma.6, G has exactly 0 edge-cuts of size. Hence by Theorem., G is the Petersen graph. If d =, then by (.), d 5 =,d j =0,j 6. Thus V (G )=D (G ) D 4 (G ) D 5 (G ). Then by Theorem.5, G F. Thus by Theorem.4, G is supereulerian, contrary to (.). This completes the proof of Theorem.. 4 Proof of Theorem.4 Let G F be a graph. Throughout this section, e alays use V (G) to denote the unique ertex of degree 5. Let H be the subgraph induced by the ertices of distance at least from in G and G 0 = G E(H). Define S = N() D 4 (G), T = N() D (G), S = u S N(u), T =( T N()) D (G) andt =( T N()) D 4(G). Let W = V (H) (S T T ), and let a = D (G) W and b = D 4 (G) W.

5 Supereulerian Graphs and the Petersen Graph 95 Lemma 4. (i) N() =S T. With the notations aboe, each of the folloing holds. (ii) V (G 0 )=V (G) and E(G 0 )= u S T E(u). (iii) u, S T ith u, N(u) N() =. (i) G 0 is acyclic. () (S T T ) V (H) and S D (G). (i) S = S and T + T = T. (ii) d (G) = S + T + T + a and d 4 (G) = S + T + b. (iii) E(H[V (H) D (G)]) = ((a +( S + T )) (4b + T )), and4b + T a +( S + T ). Proof (i) follos from (F) and (F). The definition of H implies (ii). (iii) and (i) follo from (F) and () follos from (F4). Since S D 4 (G) andt D (G), for eery u S, N(u) V (H) = and for eery T, N() V (H) =. These imply (i). By the definitions of S,T and T and by (F), S,T and T are mutually disjoint. Then direct computation yields (ii). By the definition of H, V (H) = a + b + S + T + T.Let H = H[V (H) D (G)]. Then counting V (H ) d G() in to different ays, e obtain a +( S + T )= d G () = E(H ) + S + T +4b + T, and so (iii) follos. V (H ) By (F), = d (G) = S + T + T + a S + T = S +5 S, andso Throughout this section, let S, here S = only if T + a =0. (4.) S = {u,u,...,u S }, (4.) N(u i ) V (H) ={ i, i, i }, here i S, T = {,,..., 5 S }, N( i ) V (H) ={ S +j, S +j }, here j 5 S = T. As S, S +(5 S ). By (F), Lemma 4. Proof i j if and only if i j for i, j. (4.) G must be one of 8 possible graphs. By (4.), S and so e can analyze cases hen S takes different alues. Case S =. Then T =. By(4), T +a =0. Asd (G) =, D (G) =T S and T = T T =4. By Lemma 4. (iii), 0 b. If b =,thenv (G) W D 4 (G) hasaertexz. Since T =0 and by (F4), N(z) S. Since d G (z) = 4, for some i {,, }, N(z) N(u i ), hence G[(N(z) N(u i )) {z, u }] induces a C 4, contrary to (F). Therefore in Case, b =0,andso there is only one possible graph, called G, as presented in Table belo. Case S =.

6 96 Li X. M., et al. As d (G) =, T = T S a = a and T =6 ( a) =4+a. Thenby Lemma 4. (iii), 4b +(4+a) a +(6+ a), and so, a +b. Therefore, there ill be 4 different possible graphs in this case. Let G, G, G 4, G 5 denote such a graph hen a = and b =0,orhena =0andb =,orhena =andb =0,ora =0andb = 0, respectiely, as presented in Table belo. Case S =. In this case, T = T S a =4 aand T =8 (4 a) =4+a. By Lemma 4. (iii), 4b +(4+a) a +(+4 a), and so a +b. Let G 6, G 7 denote such agraphhena =andb =0,orhena =0andb = 0, respectiely, as presented in Table belo. Case 4 S =0. Then S = S =. Again by d (G) =, T = T S a =6 a and T = 0 (6 a) =4+a. Then by Lemma 4. (iii), 4b +(4+a) a +(0+6 a), and so a =0 and b = 0. Thus there is one such graph, denoted by G 8, as presented in Table belo. Summing up, e list the 8 possibilities of G in the folloing Table, ith n = V (G). G n S S T T T T a b G 0 {u,u } {,,..., 6 } {,, } 0 { 7, 8,..., } 0 G 9 {u,u,u } {,,..., 9 } {, } 0 { 0,,, } 0 0 G 9 {u,u } {,,..., 6 } {,, } { 7, 8,..., } 0 G 4 9 {u,u } {,,..., 6 } {,, } { 7, 8,..., } 0 G 5 8 {u,u } {,,..., 6 } {,, } { 7, 8,..., } 0 0 G 6 8 {u } {,, } {,,, 4 } { 4, 5,..., } 0 G 7 7 {u } {,, } {,,, 4 } 4 { 4, 5,..., } 0 0 G 8 6 {,,, 4, 5 } 6 {,,..., 0 } 0 0 Table The graphs G i ( i 8) This proes the lemma. Throughout the rest of this section, the graphs G i ( i 8), ill be these graphs defined in Table. Lemma 4. If G {G,G,G 6,G 8 },then E(H[V(H) D (G)]) =0and 4b + T = a +( S + T ). Proof By Lemma 4. (iii), it suffices to sho that 4b + T =a +( S + T ). If G = G,thena =,b =0, T =0and S = 6. By Lemma 4. (i), T = T =6. Thus 4b + T =8=a +( S + T ). If G = G,thena =0,b =, T =, T =4and S =6. Thus4b + T =6=a +( S + T ). If G = G 6,thena =,b =0, T =, T =5and S =. Thus4b + T =5=a +( S + T ). If G = G 8,thena = b =0, T =6, T =4and S =0. Thus4b + T ==a +( S + T ). Lemma 4.4 G G. Proof Suppose G = G. Then as shon in Table, G 0 is isomorphic to G in Figure (see Section 6). Thus S = {u,u }, S = {,,, 4, 5, 6 }, T = {,, }, a =0,b =,

7 Supereulerian Graphs and the Petersen Graph 97 T =and T = 4. Denote the ertex of degree 4 in V (G ) W by x. If the to ertices in T hae one common neighbor in T (say N( 7 ) N( 8 ), and so T = { 7, 8 }), then by (F4), N(x) S T.Since N(x) =4,eitherT N(x), hence G[{,x} T ] contains a 4-cycle, contrary to (F); or for some i =,, N(x) N(u i ), hence G[(N(x) N(u i )) {x, u i }] has a 4-cycle, contrary to (F). Hence by symmetry, e may assume that T = { 7, 9 }. By (F) and (F4), 7, 9 N(x), 8 N( 9 )and 0 N( 7 ), and so N H ( ) S = N(u ) N(u ). Since D (H), then for some i {, }, N H ( ) N(u i ), and so G{,u i } (N H ( ) N(u i )) contains a 4-cycle, contrary to (F). Lemma 4.5 G G 6. Proof Suppose G = G 6. Then as shon in Table, G 0 is isomorphic to G 6 in Figure. Thus e hae S = {u }, S = {,, }, T = {,,, 4 }, a =andb =0. Then T =and T = 5. Denote the ertex of degree in V (G 6 ) W by x. By Lemma 4., E(H[V (H) D (G)]) = E(G 6 [S T ]) =0,andsoN H ( ) N H ( ) N H ( ) T. By (F), N H ( i ) N H ( j )= for all i j, i, j, and so N H ( ) N H ( ) N H ( ) = 6, contrary to the fact that T =5. Lemma 4.6 G G 7. Proof Suppose G = G 7. Then as shon in Table, G 0 is isomorphic to G 7 in Figure. Thus e hae S = {u }, S = {,, }, T = {,,, 4 } and a = b =0. Then T =4 and T = 4. By (F4) and Lemma 4. (iii), E(G 7 [S ] T ) = E(H[V (H) D (G)]) = ((a +( S + T )) (4b + T )) =. By (F), for any i j ith i, j {,, }, N H ( i ) N H ( j )=. As in this case, {,, } D (H), and so N H ( ) N H ( ) N H ( ) =6. Since E(G 7 [S ] T ) =,ehae (N H ( ) N H ( ) N H ( )) T, and so (N H ( ) N H ( ) N H ( )) T 5byN H ( ) N H ( ) N H ( ) T T, contrary to the fact that T =4. Lemma 4.7 If G = G,thenG is supereulerian. Proof Suppose G = G. We use the notation in Table for G.Asa =,letd (G) W = {x, y}. By Lemma 4., E(G[S {x, y}]) =. Hence N(x) N(y) T = { 7, 8, 9, 0,, }. If N(x) N(y), then there is a ertex in T (say 7 ) hich is adjacent to neither x nor y. Hence N H ( 7 ) S. Since ertex 7 has degree in H, C 4 must be induced. Therefore N(x) N(y) =. Without loss of generality, by (F) e may assume that x N( 7 ) N( 9 ) N( ), and y N( 8 ) N( 0 ) N( ). Thus N( 7 ) S = N( 8 ) S =. By (F), ithout loss of generality, e may assume that 7 N( ) N( 4 )and 8 N( ) N( 5 ). Hence N( ) { 9, 0,, } = N( 6 ) { 9, 0,, } =. By symmetry and by (F), e may also assume that N( 9 ) N( )and 6 N( 0 ) N( ). By the assumptions aboe, e got a graph G = G[E(G 0 ) {x 7,x 9,x,y 8,y 0, y, 7, 4 7, 8, 5 8, 9,, 6 0, 6 }](seefigure). ThenG is a spanning subgraph of G. SinceG {,,,, 6,x 9,y 0 } is a spanning eulerian subgraph of G, G is supereulerian.

8 98 Li X. M., et al. Lemma 4.8 If G = G,thenG is supereulerian. Proof Suppose G = G. We use the notation in Table for G. Then T =, andsoby Lemma 4. (i), T = { 0,,, }. As a = b =0,a +( S + T ) 4b + T = 8 = 6, and so by Lemma 4. (iii) and by (F4), E(G[S ]) =. LetH = H E(G[S ]). By (F), g(g) 5, and so N H ( 0 ) N H ( )= and N H ( ) N H ( )=. Let P = N H ( 0 ) N H ( )andq = N H ( ) N H ( ). Then by (F4), P Q S. (4.4) As { 0,,. } D (H ), N H ( 0 ) = N H ( ) = N H ( ) = N H ( ) =. Thus P = Q =6. If P Q 5, then N H ( 0 ) (P Q) orn H ( ) (P Q). We suppose N H ( 0 ) (P Q). By N H ( 0 ) =, 0 has to neighbors in some member of {N H ( ),N H ( )}, sayinn H ( ). Thus the to neighbors and { 0, } together induce a 4-cycle in G, contrary to (F). If P Q, then P Q 0 > 9= S, contrary to (4.4). Hence P Q 4. Case P Q =4. Since P Q = 4 and S =, for some u i S, (P Q) N(u i ). Hence e may assume that, (P Q) N(u ). By (F), N H ( ) N H ( ) =. As {, } (P Q) D (H), e hae N H ( ) { 0, } = N H ( ) {, } = and N H ( ) { 0, } = N H ( ) {, } =. Hence by N H ( ) N H ( )=, { 0,,, } N H ( ) N H ( ). Without loss of generality, assume that { 0,,, } E(G ). By symmetry and (F), e may further assume { 0 4, 0 7, 5, 8 } E(G ). As P Q = P + Q P Q = 8 < 9 = S and by (4.4), S P Q =. If P Q, thenby{ 0,,, } N H ( ) N H ( ), for some i {0,,, }, N( i ) N H (u ), say N( 0 ) N H (u ). Then G[{u, 0 } (N( 0 ) N H (u ))] contains a 4-cycle, contrary to (F). Therefore, / P Q. It follos that either 6 N( )and 9 N( )or 6 N( )and 9 N( ). By symmetry, e assume 6 N( )and 9 N( ). Thus must be adjacent to one of ertices 4, 5 and 6. The proofs for each of these subcases ill be similar, and so e shall only proe the case hen 4 E(G) and omit the others. Let G = G 0 +{ 0,,,, 0 4, 0 7, 5, 8, 6, 9, 4 } (see Figure ). Then G is a spanning subgraph of G. As G {, 0,, } is a spanning eulerian subgraph of G, G is supereulrian. Case P Q =. By (4.4) and P Q = P + Q P Q =9= S, P Q = S,andsoΔ(G [S ]) =. Let P Q = {z,z,z }. Hence {N H (z ),N H (z ),N H (z )} {{ 0, }, { 0, }, {, }, {, }}. By symmetry, e may assume N H (z )={ 0, }, N H (z )= { 0, } and N H (z )={, }.LetG = G 0 + E(H ). Then G is a spanning subgraph of G. (Anexampleithz =,z = 4,z = 7 is shon in Figure.) By E(G [S ]) = and Δ(G [S ]) =, O(G ) ={,,,z,z,z }. It follos that G {,z 0, z 0, z, } is a spanning eulerian subgraph of G, andsogis supereulrian. Lemma 4.9 If G = G 4,thenG is supereulerian.

9 Supereulerian Graphs and the Petersen Graph 99 Proof Suppose G = G 4. We use the notation in Table for G 4.Asa =,letd (G) W = {x}. Since T =, by Lemma 4. (i), T = T T = 5. Without loss of generality, let T = { 7 } and so T = { 8, 9, 0,, }. By Lemma 4. (iii), E(G 4 [S { 7,x}]) = a +( S + T ) 4b + T =. LetE(G 4 [S { 7,x}]) = {e}. Case x is not incident ith e. Since E(G 4 [S { 7,x}]) = {e}, x is an isolated ertex in G 4 [S { 7,x}]andsoN(x) T. If N(x) T { 8 },thenby N(x) =,forsomei {, }, N( i ) N(x), and so G[{x, i } (N( i ) N(x))] has a 4-cycle, contrary to (F). Hence x N( 8 ). Without loss of generality, e may assume x N( 9 ) N( ). Thus by (F4), N H ( 0 ) S { 7 }. If N H ( 0 ) S, then as N H ( 0 ) =, for some i {, }, N(u i ) N H ( 0 ), and so G[{u i, 0 } (N(u i ) N H ( 0 ))] has a 4- cycle, contrary to (F). Hence 0 N( 7 ). Similarly, N( 7 ). Since N H ( 8 ) =, 8 N( 7 )andx N H ( 8 ), e hae N H ( 8 ) S =. Then by (F), 8 must be adjacent to one ertex in {,, } and to one ertex in { 4, 5, 6 }. Thus e may assume 8 N( ) N( 4 ). Since e cannot be incident ith to ertices in {,, }, ith 8 N( ), one of {, } must be adjacent to to ertices in { 9, 0,, }. Similarly, one of { 5, 6 } must be adjacent to to ertices in { 9, 0,, }. Without loss of generality, let N( ) { 9, 0,, } = N( 5 ) { 9, 0,, } =. By (F), {N H ( ),N H ( 5 )} = {{ 9, }, { 0, }} and N H ( ) N H ( 5 ). By symmetry, e assume { 9,, 5 0, 5 } E(G 4 ). Note that N H ( ) { 7, 8 } = and N H ( 6 ) { 7, 8 } =. Under these assumptions, e shall sho e = 6. If N H ( ) { 9, 0,, },then N H ( ) {{ 9, 0 }, { 9, }, { 9, }, { 0, }, { 0, }, {, }} by N H ( ) =. In any case, G ould hae a 4-cycle (see Table ), contrary to (F). N H ( )isin G has a 4-cycle in 9, 0 G[{, 9, 0, }] 9, G[{, 9,,x}] 9, G[{, 9,, }] 0, G[{, 0,, 5 }] 0, G[{, 0,, 7 }], G[{,,, }] Table Possible 4-cycles in G Hence, by N H ( ) { 7, 8 } =, N H ( ) { 4, 5, 6 }. If N H ( ) { 4, 5, 6 }, then G[N( ) N(u ) { }] contains a 4-cycle, contrary to (F). Hence N H ( ) { 4, 5, 6 } =. By symmetry, N H ( 6 ) {,, } =. As{e} = E(G [S { 7,x}]), e hae e = 6. Let G 4 = G 0 + {x 8,x 9,x, 7 0, 7, 8, 4 8, 9,, 5 0, 5, 6 }. Thus e obtained a spanning subgraph G 4 of G 4 (see Figure ). Since G 4 {,,, 7 0, 5,,x 9 } is a spanning eulerian subgraph of G 4, G 4 is supereulerian. Case x is incident ith e.

10 00 Li X. M., et al. If e = x 7,thenas E(G [S { 7,x}]) =,N H ( ) N H ( ) N H ( ) { 8, 9, 0,, } and by g(g 4 ) 5, N H ( i ) N H ( j ) = (i j, i =,,,j =,, ). Hence N H ( ) N H ( ) N H ( ) = 6, contrary to N H ( ) N H ( ) N H ( ) { 8, 9, 0,, }. Therefore x / N( 7 )andso N(x) S =. Thus by (F), for eery { 8, 9, 0,, }, N H () {x, 7 }. Therefore, and so { 8, 9, 0,, } N H (x) N H ( 7 ), N H (x) N H ( 7 ) { 8, 9, 0,, } 5. But as d H (x) =,d H ( 7 )=and N H (x) S =, (N H ( 7 ) N H (x)) { 8, 9, 0,, } 4, contrary to N H (x) N H ( 7 ) { 8, 9, 0,, } 5. Lemma 4.0 If G = G 5,thenG is supereulerian. Proof Suppose G = G 5. We use the notation in Table for G 5, and so S = {u,u }, S = {,,, 4, 5, 6 }, T = {,, } and a = b =0. Since T =, by Lemma 4. (i), T = T T = 4. By Lemma 4. (iii), Denote E(G 5 [S T ]) =a +( S + T ) 4b + T =. E(G 5 [S T ]) = {e,e }. As T =, e may assume that T = { 7, } for some { 8, 9,..., }. Case N( ). Then = 8. Without loss of generality, e may assume that, 4 and 7 N( 9 ), and that, 5 and 8 N( 0 ). Then each of and must be adjacent to one in { 7, 8 }. By symmetry, assume 7, 8 E(G 5 ). As 9, N( 7 )andas 0, N( 8 ), both of e and e can only be adjacent to ertices in S.By(F),g(G 5 ) 5, and so e is not adjacent to e.sincen H ( 9 )={, 4, 7 } and N H ( 0 )={, 5, 8 },eachof and 6 is adjacent to at least one in {, }. Thus e may assume that, 6 E(G 5 ) (the proofs for the other cases, 6 E(G 5 )or, 6 E(G 5 )or, 6 E(G 5 )are similar). Let G 5 = G 0 + { 9, 4 9, 7 9, 0, 5 0, 8 0, 7, 8,, 6 }. Then G 5 is a spanning subgraph of G 5 (see Figure ). Since G 5 {,,, 7, 8 } is a spanning eulerian subgraph of G 5, G 5 is supeuelerian. Case / N( ). Thus e may assume that = 9. Then by (F), 8 9, 0 7 E(G 5 ). By symmetry, each of and must be adjacent to one in { 7, 9 },toonein{,, } and one in { 4, 5, 6 }. Without loss of generality, e assume ertex, 4 and 7 N( ), and, 5 and 9 N( ). Let G 5 = G 0 + { 8 9, 7 0,, 4, 7,, 5, 9 }. Thus G 5 is a spanning subgraph G 5 (see Figure ). As N H ( 7 )={ 0, } and N H ( 9 )={ 8, }, E(G 5 [S ] T )=E(G 5 [S ]). By (F), Δ(G 5 [S ]) =. Since N H ( ) = {, 4, 7 } and N H ( ) = {, 5, 9 }, each of

11 Supereulerian Graphs and the Petersen Graph 0 and 6 is adjacent to 8 or 0. If { 0, 6 8 } E(G 5 ) (or similarly, { 8, 6 0 } E(G 5 )), then G 5 + { 0, 6 8 } {,,, 8 9, 7 0 } is an eulerian subgraph of G 5 + { 0, 6 8 } hich spans G 5,andsoG 5 is supereulerian. If { 8, 6 8 } E(G 5 ) (or similarly, { 0, 6 0 } E(G 5 )), then G 5 +{ 8, 6 8 } {, 7, 9 } is a spanning eulerian subgraph of G 5 + { 8, 6 8 } that spans G 5,and so G 5 must be supereulerian. Lemma 4. If G = G 8,thenG is supereulerian. Proof Suppose G = G 8. We use the notation in Table for G 8, and so S =, T = {,,, 4, 5 }. By Lemma 4., E(H[V (H) D (G)]) =, and so H is a bipartite graph ith a ertex bipartition (T,T ). (4.5) By (F), for any i ith i 5, Without loss of generality, assume that 8, 0 T. Define N H ( i ) N H ( i )=. (4.6) T = { T : N H () T }. If T, as T =6and T = 4, e may assume { 7, 8, 9, 0 } = T. By (F) and (4.5), N H ( 7 ) N H ( 8 )={,,, 4, 5, 6 } = N H ( 9 ) N H ( 0 ), N H ( 7 ) = N H ( 8 ) = N H ( 9 ) = N H ( 0 ) =andn H ( 7 ) N H ( 8 )=N H ( 9 ) N H ( 0 )=. It follos that either N H ( 7 ) N H ( 9 ) or N H ( 7 ) N H ( 0 ), forcing G 8 to has a 4-cycle, contrary to (F). Hence T. Case T =. We may assume that T = { 5 }, and so by symmetry, assume that T = { 6, 8, 9, 0 }. By (4.6) and (F), e hae that N H ( 9 ) N H ( 0 )={,,, 4, 5, 7 }. By symmetry, let { 9, 9, 5 9 } E(G 8 ), it follos { 0, 4 0, 7 0 } E(G 8 ). By (F), 6 7, 5 8 E(G 8 ). Let G 8 = G 0 + { 9, 9, 5 9, 0, 4 0, 7 0, 6 7, 5 8 }. Thus G 8 is a spanning subgraph of G 8 (see Figure ). Since G 8 {,,, 5 9, 9 5, 7 4 } is eulerian, G 8 is supereulerian. Case T =0. Then e may assume that T = { 4, 6, 8, 0 }. By (4.6) and by symmetry, e may assume that { 4, 6, 8, 0 } E(G 8 ). As 8, 0 N( ), by (F), / N( 8 ) N( 0 ), and so 6 E(G 8 ). Similarly, 4 5, 7 0, 8 9 E(G 8 ). Let G 8 = G 0 + { 4, 6, 8, 0, 6, 4 5, 7 0, 8 9 }. Thus G 8 is a spanning subgraph of G 8 (see Figure ). As is a Hamilton cycle of G 8, G 8 is supereulerian. 5 Remarks Remark 5. Both Theorem. (Theorem. of [9]) and Theorem. in this paper raise the folloing a question: if G is a -edge-connected graph and if the number of -edge-cuts of

12 0 Li X. M., et al. G is k, hat is the largest alue of k such that eery -edge-connected graph G ith at most k edge-cuts of size is supereulerian if and only if G cannot be contracted to the Petersen graph? Theorem. says that k 0 and in this paper e proe k. Hoeer, since either of the to Blanusa snarks (see [] or [4]) is -edge-connected and nonsupereulerian, has exactly 8 edge-cuts of size, and cannot be contracted to the Petersen graph, e hae k 7. We conclude this section by conjecturing that k = 7. u u u u u x G y G u u u u u 0 G x G u u u u G 4 x G 5

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