A degree condition for spanning eulerian subgraphs

A Degree Condition for Spanning Eulerian

Zhi-Hong Chen Subgraphs

DEPARTMENT OF MATHEMATICS WAYNE STATE UNIVERSITY

DETROIT, MICHIGAN

ABSTRACTLet p L 2 be a fixed integer. Let G be a simple and 2-edge-connected graph on n vertices, and let g be the girth of G. If d(u) + d(u) 2 (2/(g - Z))((n/p) - 4 + g) holds whenever uv @ €(G), and if n is sufficiently large compared to p, then either G has a spanning eulerian subgraph or G can be contracted to a graph GI of order at most p without a spanning eulerian subgraph. Furthermore, we characterize the graphs that satisfy the conditions above such that GI has order p and does not have any spanning eulerian subgraph. o 1993 John Wiley & Sons, Inc.

1. INTRODUCTION

We shall use the notation of Bondy and Murty [2], unless otherwise stated. A graph may have multiple edges but no loops. A graph is eulerian if it is connect and every vertex has even degree. An eulerian subgraph C of G is called a spanning eulerian subgraph of G if V(C) = V(G). A graph G is called supereulerian if it contains a spanning eulerian subgraph. Denoted by G2. the family of all supereulerian graphs. For u E V(G), we define the neighborhood NG(u) of u in G to be the set of vertices adjacent to u in G. We use K'(G) to denote the edge-connectivity of a graph G.

In this paper, we shall generalize some results of [l], [3], [5], [7], and [ B ] by using the reduction method, which was introduced by Catlin [3].

Catlin [3] introduced the following concept: A graph G is called collapsible if, for every subset S of V(G) of even

cardinality, there is a subgraph of G such that

(i) G - E ( T ) is connected, and (ii) S is the set of vertices of odd degree in I'.

The graph r will be called an S-subgraph of G. Denote by Ck the family of all collapsible graphs. Note that 2-cycle C2 and K3 are collapsible, and we regard K1 as being collapsible.

Journal of Graph Theory, Vol. 17, No. 1, 5-21 (1993) 0 1993 J o h n Wiley & Sons, Inc. CCC 0364-9024/93/010005-I 7

6 JOURNAL OF GRAPH THEORY

One can see that if S is the set of odd degree vertices in a connected graph G and there exists a subgraph r of G satisfying (i) and (ii) above, then G is supereulerian. Thus, a collapsible graph is supereulerian, i.e., CIL SL.

For a graph G with a connected subgraph H , the contraction G/H is the graph obtained from G by contracting all edges of H and deleting any resulting loops.

Catlin proved the following theorems:

Theorem 1 (Catlin [3]). Let H be a subgraph of G . If H E CIL, then

(i) G E ZA if and only if G /H E SL, and (ii) G E CL if and only if G/H E CL. I

Theorem 2 (Catlin [ 3 ] ) . Let HI and H2 be subgraphs of G . If H I , H2 E CIL and if V(H1) f l V(H2) f 0, then HI U H2 E CL. I

In [ 3 ] , Catlin showed that every graph G has a unique collection of maximal collapsible subgraphs H I , . . . , H,. It follows from Theorem 2 that the His are disjoint. Denote by GI the graph of order c obtained from G by contracting the subgraphs HI,. . . , H, to distinct vertices. The graph GI is uniquely determined [3] and is called the reduction of G. The graph Hi is called trivial if IV(Hi)l = 1; otherwise it is nontrivial. Catlin [ 3 ] proved that G1 is simple and has no nontrivial collapsible subgraph. A graph G is said to be reduced if it is the reduction of some graph.

Let E’ = E(G) - u ,?(Hi). Write V(G1) = (u1, 212,. . . , uc}. By the def-

inition of GI, one can define the contraction-mapping 0 on G such that for each i with 1 5 i 5 c,

C

i=l

@(Hi) = v i and 0(E’) = E(G1).

Hence, 8(G) = GI and IE’I = (E(Gl)(. We call each Hi (1 5 i 5 c) of G thepreimage (under 0) of the vertex ui

of G1. Throughout this paper we shall let d ( u ) and dl(u) denote the degree of a vertex u in G and in GI, respectively. We let E(G) denote the edge set of the graph G , and we let E (or E(G1)) denote the edge set of GI. By the definition of GI, we also use E in place of E‘, i.e., E = E’ E(G) .

We shall use the following theorem:

Theorem 3 (Catlin [3 ] ) . G . Then

Let G be a graph and GI be the reduction of

(i) GI is simple and K3-free; (ii) G has a spanning eulerian subgraph if and only if GI has a spanning

eulerian subgraph; and

DEGREE CONDITION FOR EULERIAN SUBGRAPHS 7

(iii) For any subgraph H of GI, either H E (K1, K2} or

IE(H)I 5 21V(H)( - 4.

In particular, if GI (K1, Kz}, then

IE(Gi)I 5 ~ I V ( G I ) I - 4. I

2. MAIN RESULTS

Theorem 4. Let G be a 2-edge-connected simple graph with girth g, where g E {3,4}. Let p 1 2 be an integer. If

d(u) + d ( v ) 2 - - - 4 + g 8 - 2 2 ( n P 1

whenever uu 4 E(G), and if

then exactly one of the following holds:

(a) G is collapsible; (b) G can be contracted to a noncollapsible graph GI of order c, where

c 5 p and GI is the reduction of G. Further, if c = p , then either

(bl) p = 4, and GI = Cq, or (b2) p 2 5, n = (g - 2)ps for some integer s, and S(G) =

(l/(g - 2 ) ) ( ( n / p ) - 4 + g), and either

(i) g = 3, and the preimage Hi of each vertex ui of GI is at

(ii) g = 4, and the preimage Hi of each vertex ui of GI is at most :dl(ui) edges short of being K,, or

most 3dl (ui ) edges short of being K,,s. 1

Note; In Theorem 4 g E {3,4} is necessary. In fact, we have the following proposition.

Proposition 1. be a simple graph of order n. If

Let m and n be two given integers with n 1 m2. Let G

d(u) + d(u) I 2 - + 1 (: ) (4)

whenever uu 4 E(G) , then the girth g of G is at most 4.


Proof. By contradiction, suppose g 2 5. By (4) there exists a vertex u E V ( G ) such that

n m

d(v) 2 - + 1. (5)

Let N ( u ) = { x I , x 2 , ... ,q}, where k = d ( u ) 2 ( n / m ) + 1. Since the girth of G is at least 5, for i # j , x i x j @ E(G) and

N ( x ~ ) n N ( x ~ ) = {u},

d(Xi) + d ( X j ) 2 - + 1 . (: ) Therefore, except for at most one vertex in N(u) , say xl, all vertices in N(u) have degree at least (n/m) + 1, i.e.,

Since

by the definition of k , and by (5) and (6),

contrary to n 2 m2.

By Proposition 1 we can see that there is no graph G of order n with girth g 2 5 satisfying the condition (2) of Theorem 4, if n is sufficiently large compared to g and p . Thus g E {3,4} is necessary in Theorem 4. But (3) is not best possible.

To prove Theorem 4 we need the following lemmas:

Lemma 1 for each x E V ( H ) , define

Catlin [4]). Let H be a triangle-free graph, and not a star, and

B ( x ) = {w E V(H)lwx E E(H')} .

Then the family {B(x) lx E V ( H ) } has a complete system of distinct representatives. I


Lemma 2. Let G be a simple graph of girth g E {3,4}. Let G1 be the reduction of G, and let E be the edge set of GI. Let H I , H 2 , . . . , H , be the maximal collapsible subgraphs of G. Then for each i E {1,2,. . . , c } there is a ui E V(Hi) such that

If IV(H)I > 1, H E { H I , H 2 , . . . , Hc} , then there are at least two vertices in H satisfying the inequality (7).

Proof. If g = 3, then (7) is trivial. Thus, we may assume g = 4.

Case 1. If IV(Hi)l = 1, then V(Hi) = {vi}, so N G - E ( u ~ ) = 0, and (7) follows.

Case 2. If IV(Hi)l > 1, then since Hi is collapsible and K3-free, IV(Hi)l 2 6. Let ulvz and u3v4 be two nonadjacent edges of Hi. Since G is K3-free, we have

and so

and

By (8) and (9), we can see that two of the vertices in Hi satisfy (7). Hence the lemma holds. I

Lemma 3. Let G be a 2-edge-connected simple graph, and let GI be the reduction of G and IV(G,)l > 1. Let V(G1) = ( ~ 1 ~ x 2 , ..., xc}, S =

{1,2,. . . , c } . Then there is a permutation IT on S such that V(G1) =


with no edge of G joining a vertex of 6 - ' ( x i ) to a vertex of 13-1(xr(i)), where 13 is the contraction mapping defining G I .

Proof. Since G1 is simple triangle-free graph and G1 # K1, by the definition of contraction, K ' ( G ~ ) 2 K'(G) 2 2, and so G1 is not a star. By lemma 1, there is a system of distinct representatives y i E B ( x i ) = {w E V(G1)Jwxi E E(GE)} ( 1 Cr i 5 c). Therefore xiyi E E(Gf ) . It follows that

@(xi) n e - l ( y i ) = 0 , no edge of G joins 13-'(xi) and 0 - ' ( y i ) , and { y l , . . . . yc} = V(G1). Thus there is a permutation T on S such that yi = Hence the lemma holds. I

Proof of Theorem 4. Suppose (a) fails for G. By (ii) of Theorem 1, G1 is not collapsible and so G1 # K1 and K ' ( G ~ ) 1 K'(G) 2 2. Obviously, (V(G1)I = c 2 4. Let V(G1) = {xl,xz, . . . ,xc} and S = {1,2,. . . , c}. By Lemma 3, there is a permutation T on S such that for 1 5 i I c ,

e-l(xi) n ~ - ~ ( x . ( ~ , ) = 0,

and no edge of G joins 8 - ' ( x i ) and 13-1(xr(i)). Let Hi = 8 - ' ( x i ) for

1 I i I c. We can pick ui E @-'(x i ) and vi E B-'(x,+I) such that ui and ui satisfy the inequality (7) in Lemma 2, and uivi E E(G"). Then ui E V ( H i ) and v i E V(H,+)) for 1 5 i I c . Therefore

and so


= -{n 2 + c(g - 4)). g - 2

Let U = {ul, u2,. . . u,) and V = {vl, v2,. . .,vc). By E = E(G1), and by (iii) of Theorem 3, there are at most 2(EI 5 4c - 8 incidences in G of edges of E with U and at most 21EI I 4 c - 8 incidences in G of edges of E with V. Hence

By (2), (10) implies

2 L c ( E - 4 + g) 5 8c - 16 + -{n + c(g - 4)), 8 - 2 P g - 2

and so by (ll), c < p + 1. Hence

If c < p, then (b) holds. Next we consider the case when c = p. If p = 4, then (bl) of Theorem 4 holds. In the following, we suppose that

Arrange the components HI, Hz, . . . , H, of G - E such that c = p > 4. We then show that we have case (bz).

IV(Hi)I 5 IV(H2)I 5 * . . 5 I V W C I I . (12)


The claim is proved by contradiction. Suppose that the claim is false. = IV(H,>I = 1 for some s and Then we havelV(H1)I = IV(H& =

I v ( H ~ ) ~ > 1 for j > s.

Case 1. Suppose s 1 3. Then V(Hi) = {x i } for 1 5 i I 3. Since G1 is K3-free, two of {x1,x2,xg}, say x1 and x2, are not adjacent in G . By c = p , and by (2) and (l),

If g = 3, then by (13), we have n 5 c(c - l), contrary to (3). If g = 4, then (13) implies that n I 2c(c - 2), contrary to (3) again.

Case 2. 1 5 s 5 2. Then V(H1) = {x l } , and IV(Hj)l > 1 for j 2 3.

Since G I is simple, K3-free and K ’ ( G ~ ) L 2, obviously, d ( x l ) 5 c - 2 and at most one edge of E(G) joins V(H1) = {XI} and V(H,) for 3 5 j I c. Since IV(Hj)l > 1 for 3 5 j 5 c, by the last part of lemma 2, there exists a z j E V ( H j ) - N ( x l ) , such that

and there are at most 2(IE) - d(x1)) incidences in G of edges of E with (23, 24,. . . , zc}. Hence,

and so

{ n - 2 + (c - 2)(g - 4)) 1 + -

8 - 2 5 (c - 4)(c - 2) + 2(2c - 4)


By (2), we have

(c - N g - 4) (n - 2 ) + 1 + - ) ( c - 2)-(C - 4 + g I ( c - 4)(c - 2 ) + 2(2c - 4)

2 g - 2 c

8 - 2 8 - 2

(14) ( C - 2)(4 - g ) - 2 -- (" ; 4) 5 c(c - 2) +

8 - 2 8 - 2

If g = 3, then by c = p inequality (3) gives n 2 4 2 . But by (14) and c > 4, n < 4c2, a contradiction.

If g = 4, then by (3) and c = p, n 2 8c2. But by (14) and c > 4, n < 6 2 , a contradiction again. Hence Claim 1 holds.

Claim 2. IV(Hi)l > 3 p - 1 = 3c - 1 (1 I i I c).

Let r = IV(H1)I. By Claim 1 and (12), 1 < r I n/c. Since G1 is simple, any vertex in V ( H i ) is adjacent to at most one vertex in V ( H j ) i # j . Since IV(Hj)l > 1 (1 I j 5 c), by Lemma 2, there exist at least two vertices in V ( H l ) , say y 1 and y i , satisfying the inequality (7). Let O(H1) = v1. There are at most d l ( v l ) incidences in G of edges of E with {y l , y : } . Therefore, one of { y l , y i } , say y1, is incident with at most dl(v1)/2 edges of E. Since dl(v1)/2 5 ( E ( / 2 and y 1 satisfies the inequality (7),

For j 2 2, by Lemma 2, there exists a y j E V ( H j ) - N( y l ) such that

Obviously, there are at most 2()EI - ( d ( y l ) - ~ N G - E ( Y I ) ~ ) ) incidences in G of edges of E with { yz , y3,. . . , yc}. Hence,


Simplifying the inequality above, we have

5 (g - 2)(c - 2)(c + 1)

+ (c - 2)r + 2(c - I)(g - 4) + n,

.: - (g - 2)(c + 1) 5 r .

By this and (3) and c = p ,

(18) r 2 (g - 2 ) ( 3 ~ - 1) 2 3~ - 1.

By (12) and (18), for 1 I i I c,

Now we can show that the case (b2) of Theorem 4 holds.

Case A. g = 3. By IV(G1)I = c, (E(G1)I 5 2c - 4 and so by (19), there is a vertex u E V ( H I ) that is not incident with any edges of E(G1). Thus N ( v ) C V ( H l ) . By (12), IV(H1)I 5 n/c, and so d ( v ) 5 (n/c) - 1 . Hence, there is a real number t 2 1 such that

n d(U) = - - t .

C


Therefore,

By (19) and (l), we can choose ui E V(Hi ) (c 2 i 2 2) such that ui’s are not incident with any edges of E(GI). Hence vivj 6Z E(G) if i # j.

Let d(u i ) = (n /c ) + ti for some real number ti, (2 I i I c ) . Then for i 2 2,

By (2), for 2 5 i 5 c,

n n I d(u1) + d ( U j ) = - - t + - + ti,

C C

t - 2 2 t i

By (20), (21), and t - 2 2 t i ,

2 ( c - l ) - + t - l + - - t + l (a ) a = n + (c - 2)(t - 1).

Then

(c - 2)(t - 1) I 0,

Since c > 4, it follows that t I 1, and so t = 1. Hence, by (20) and (12) we have IV(H1)I = n/c . By (12), for i with 1 I i I c ,

Let s = n / c , which shall be an integer. Since g = 3 in this case, we have n = cs = (g - 2)ps.

Write V(G1) = {u1,u2,. . . , uc}, where v i = t9(Hi) (1 5 i I c). By (2), (22), and IE(G1)I I 2c - 4, we can see that for any u E V(G) , d(u) 2 ( n / c ) - 1, and that those vertices that are not incident with the edges of E ( G I ) must have degree ( n / c ) - 1. Since only dl(ui) edges of G have exactly one end in V(Hi) , it follows that Hi (1 I i I c ) is at most ~ d l ( u j ) edges short of being K , for s = ( n / c ) = ( n / p ) . Hence (i) of case (b2) of Theorem 4 holds.

1


Case B. g = 4. By IV(G,)l = c, IE(G1)I 5 2c - 4, (19) and c > 4, there are at least c + 1 > 3 vertices in V ( H i ) (1 I i 5 c ) , say { x f , x i , . . . , x f + l } , which are not incident with any edges of E(Gl ) . Since G is K3-free in this case, we can choose two vertices in {xi, x i , . . . , ~ f + ~ } , say xf and x i , which are not incident with any edges of ,?(GI) and xix i @ E(G). Then N ( x j ) C V ( H i ) ( j = 1,2) . By (2) and g = 4,

n n P C

d ( x l ) + d ( x i ) 2 - = - ,

We may assume

BY (317

IN(xf)l z n/2c > 4(c - 1) = 2c + (2c - 4). (25)

By (1) and (25), there are yi, yi in N ( x f ) that are not incident with any edges of GI and so N(yj) V ( H i ) ( j = 1,2) . Since G is K3-free, we have that yfyi B: E(G) . By (2),

la d ( y l ) + d(yi) 2 - .

C

We may assume

(N(yl)( = d ( y i ) 2 n/2c . (26)

Since G is K3-free and yi E N(ni ) , N ( x f ) fl N ( y i ) = 0. Hence,

Therefore, by (24) and (26),

C C

i = l i = l

This implies that IN(xi)I = IN(yi)l = n/2c and V ( H i ) = N ( x i ) U N ( y i ) (1 5 i 5 c). Let s = n/2c, which shall be an integer. Since g = 4 in this case, we have n = 2cs = (g - 2 ) p s . Since there always exist some vertices in V ( H i ) that are not incident with any edges of E ( G l ) , and by (2) we can


see that d(u) 2 n/2c for every u E V(G), and S(G) = n / 2 c . Since G is K3-free, this implies that Hi is a spanning subgraph of K,,,, where s = n/2c . Since 6(G) 2 n/2c and only dl(vi) edges of G have exactly one end in Hi, it follows that Hi is at most idl(ui) edges short of being K,,,. Hence part (ii) of case (bz) of theorem 4 holds.

This completes the proof of theorem 4. I

Remark. Theorem 4 still holds if we replace the inequality (3) by n > 4(g - 2 ) p ( p - l), but it is still not quite best possible.

We express Theorem 4 in terms of spanning eulerian subgraph in the following theorem:

Theorem 5. Let G be a 2-edge-connected simple graph with girth g, where g E {3,4}. Let p L 2 be an integer. If

d(u) + d ( u ) 2 - (1. g - 2 P

whenever uu @ E(G), and if

4 + g ) (2')

(3')

then exactly one of the following holds:

(a) G has a spanning eulerian subgraph; (b) G can be contracted to a graph G1 of order c, where c 5

p and G1 contains no spanning eulerian subgraphs. Further, if c = p , then n = (g - 2)ps , for some integer s, and S(G) = (l/(g - 2 ) ) ( ( n / p ) - 4 + g), and either

(i) g = 3, and the preimage Hi of each vertex ui of G1 is at most ~ d l ( u i ) edges short of being K,, or

(ii) g = 4, and the preimage H i of each vertex ui of GI is at most 5dl(ui) short of being K,,,.

1

1 I

Proof. Let GI be the reduction of G. If G is collapsible then G1 = K1. By Theorem 4, no matter whether G is collapsible or not, the reduction G1 of G has order c, where c 5 p . By Theorem 3, G is supereulerian if and only if G1 is supereulerian. Thus, if G1 is supereulerian, then (a) of Theorem 5 holds. Otherwise, G1 is a nonsupereulerian graph of order c, and (b) of Theorem 4 holds. Hence GI is not C4, and so (b) of Theorem 5 holds. Furthermore, if c = p , then (i) and (ii) of Theorem 4 imply that (i) and (ii) of Theorem 5 hold. I


3. COROLLARIES

In Theorem 5, the case g = 3, p = 2 gives a result of Lesniak-Foster and Williamson [8] (except for the bound on n). The case g = p = 3 of Theorem 5 is related to a result of Benhocine, Clark, Kohler, and Veldman [l]. The case g = 3 of Theorems 4 and 5 was done by Catlin [4] when the inequality (2) is strict.

Corollary 1 (Catlin [4]). order n, and let p L 2 be an integer. If

Let G be a 2-edge-connected simple graph of

d(u) + d(u) > 2(; - 1)

whenever uu @ E(G), and if n 2 4p2, then exactly one of the following conclusions holds:

(a) G is collapsible; (b) G is contractible to a noncollapsible graph G1 of order less than p; (c) p = 4, G is contractible to C,.

Proof. Let g = 3 in Theorem 4. Then (27) is a special case of (2). To prove this corollary, we only need to show that (i) of case (b2) of Theorem 4 does not hold in this case.

Suppose (i) of (d) of theorem 4 holds in this case. Then there are some vertices ui E V ( H i ) 1 5 i 5 c with d(ui) = (n /c ) - 1 and N ( v i ) C V ( H i ) , and so uiuj 4 E(G) if i # j. Hence

d ( v J + d ( U 2 ) = 2 - - (: 7

contrary to (27). The corollary holds. I

For triangle-free graphs we have

Corollary 2. Let Gbe a 2-edge-connected and K3-free graph of order n, and let p 2 2 be an integer. If

whenever uu @ E(G), and n 2 8p2, then exactly one of the following conclusions holds:

(a) G is collapsible; (b) G can be contracted to a noncollapsible graph GI of order less than p;


(c) p = 4, G is contractible to C,.

Proof. The proof is similar to that of Corollary 1. I

In [l], Benhocine, Clark, Kohler, and Veldman conjectured that for a 2- edge-connected simple graph G of order n , if d(u) + d(u) > 2((n/5) - 1) whenever uu # E(G) and n is sufficiently large, then G contains a spanning eulerian subgraph. Catlin proved this conjecture in [4]. The following corollary is related to this conjecture. Also, we discuss the same kind of problems for 3-edge-connected graphs in Corollaries 4 and 5.

Corollary 3. Let G be a 2-edge-connected graph of order n with girth g = 3 or 4. If n is sufficiently large and if

( E - 4 + g ) 2

d(u) + d(u) 2 - 8 - 2 5

whenever uu @ E(G), then exactly one of the following holds:

(a) G has a spanning eulerian subgraph; (b) n = ( g - 2)5s for some integer s, and G can be contracted to the

graph G1 = K2.3, and either

(i) g = 3, the preimage Hi of each vertex of G1 = K2.3 is K, or K, - e for some e E E(K,), or

(ii) g = 4, the preimage Hi of each vertex of G1 = K2,3 is K,,s or Ks,s - e for some e E E(KS,,).

Proof. Suppose G has no spanning eulerian subgraph. By Theorem 5 , G can be contracted to a graph GI with IV(G1)l 5 5. Since K’(GI) 2 K’(G) 2 2, by inspection, there is no 2-edge-connected graph of order less than 5 with no spanning eulerian subgraph. Thus if IV(Gl)l 5 4, then G1 has a spanning eulerian subgraph, and so by Theorem 3, G has one too, a contradiction. Therefore IV(Gl)l = 5. Since G1 has no spanning eulerian subgraph with d(G1) 2 2, by inspection, this implies GI = K2,3, and so dl(u) 5 3 for any u E V(G1). By (b2) of Theorem 5, this implies (b) of Corollary 3. Hence the corollary holds. I

For 3-edge-connected graphs, we have the following:

Corollary 4. Let G be a 3-edge-connected simple graph of order n with girth g = 3 or 4. If n is sufficiently large and if


whenever uu @ E(G), then exactly one of the following holds:

(a) G is collapsible. (b) The reduction graph G I of G is the Petersen graph.

To prove Corollary 4 we need the following lemma:

Lemma 4 (Chen [6] ) . n 5 11 vertices. Then either G is collapsible or G is the Petersen graph.

Let G be a 3-edge-connected simple graph on I

Proof of Corollary 4. The inequality (30) is the case when p = 11 in Theorem 4. By Theorem 4, if n is sufficiently large, then G is either collapsible or contractible to a noncollapsible graph G1 with IV(G1)I I p = 11. By Lemma 4, either G is collapsible or its reduction G I can only be the Petersen graph. I

Corollary 5. girth g E {3,4}. If n is sufficiently large and if

Let G be a 3-edge-connected simple graph of order n with

whenever uu 4 E(G) , then exactly one of the following holds:

(a) G is collapsible; (b) n = (g - 2)10s, for some integer s, and G can be contracted to the

Petersen graph G I , and either

(i) g = 3, the preimage Hi of each vertex of GI is K, or K , - e

(ii) g = 4, the preimage of each vertex of G1 is either K,,s or for some s = n/10, and e E E(KS,,); or

K,, , - e for some s = n/20, and e E E(Ks, , ) .

Proof. The inequality (31) is the case when p = 10 in Theorem 4. By Theorem 4, if n is sufficiently large, then either G is collapsible or G is contractible to a noncollapsible graph G1 with /V(Gl ) / I 10. Suppose G is not collapsible. Since d ( G I ) 2 K’(G) Z 3, G I is 3-edge-connected. By Lemma 4, there is no 3-edge-connected noncollapsible graph of order less than 10. Hence / V ( G , ) / = p = 10, G I Petersen graph and so d l ( v ) = 3, for any u E V(GI). By (b2) of Theorem 4, the corollary follows. I

ACKNOWLEDGMENT

The author wishes to thank Paul A. Catlin, the author’s Ph.D. supervisor, for his many helpful suggestions. He also wants to thank the referees for their helpful comments.


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Documents

A degree condition for spanning eulerian subgraphs