Long Cycles in 3-Connected Graphs

Transcription

1 Long Cycles in 3-Connected Graphs Guantao Chen Department of Mathematics & Statistics Georgia State University Atlanta, GA Xingxing Yu School of Mathematics Georgia Institute of Technology Atlanta, Georgia Abstract Moon and Moser in 1963 conjectured that if G is a 3-connected planar graph on n vertices, then G contains a cycle of length at least Ω(n log 3 2 ). In this paper, this conjecture is proved. In addition, the same result is proved for 3-connected graphs embeddable in the projective plane, or the torus, or the Klein bottle. MSC Primary 05C38 and 05C50 Secondary 57M15 Partially supported by NSA grant MDA Partially supported by NSF grant DMS

2 1 Introduction and notation A graph is Hamiltonian if it contains a cycle using all vertices, and such a cycle is called a Hamilton cycle. A planar graph is a graph which can be embedded in the plane without crossing edges, and such an embedding is called a plane graph. In 1931, Whitney [11] proved that every 4-connected triangulation of the plane contains a Hamilton cycle. In 1956, Tutte [10] proved a more general result: every 4-connected planar graph contains a Hamilton cycle. However, 3-connected planar graphs need not contain Hamilton cycles. For such examples, see Holton and McKay [7]. The circumference of a graph G, denoted by circ(g), is the length of a longest cycle in G. In 1963, Moon and Moser [9] implicitely made the following conjecture by giving 3-connected planar graphs G with circ(g) 9 V (G) log 3 2. Conjecture 1.1 If G is a 3-connected planar graph on n vertices, then circ(g) Ω(n log 3 2 ). We mention here that Grünbaum and Walther [6] made the same conjecture for a family of 3-connected cubic planar graphs. Barnette [1] showed that every 3-connected planar graph with n vertices contains a cycle of length at least lg n and Clark [3] later improved this lower bound to e lg n. In [8], Jackson and Wormald obtained a polynomial lower bound βn α, where β is some constant and α Recently, Gao and Yu [5] improved α to 0.4 and extended the result to 3-connected graphs embeddable in the projective plane, or the torus, or the Klein bottle. The main result of this paper is the following. Theorem 1.2 Let G be a 3-connected graph with n vertices, and suppose that G is embeddable in the sphere, or the projective plane, or the torus, or the Klein bottle. Then circ(g) Ω(n log 3 2 ). Throughout this paper, we consider finite graphs with no loops or multiple edges. For a graph G, we use V (G) and E(G) to denote its vertex set and edge set, respectively. A graph H is a subgraph of G, denoted by H G, if V (H) V (G) and E(H) E(G). We shall use to denote the empty graph (as well as the empty set). 2

3 For S V (G), the subgraph of G induced by S, denoted by G[S], is the graph whose vertex set is S and whose edge set consists of the edges in G with both incident vertices in S. For S E(G), the subgraph of G induced by S, denoted by G[S], is the graph whose edge set is S and whose vertex set consists of the vertices in G incident with edges in S. Let e be an edge of G with incident vertices x and y; then we write e = xy, and write G[xy] or G[e] instead of G[{xy}] or G[{e}]. Let X V (G), or X E(G), or X G; then G X denotes the graph obtained from G by deleting X and the edges of G incident with a vertex in X. If X = {x}, then we write G x instead of G {x}. Let H G; then G/H denotes the graph with V (G/H) = V (G H) {h} (where h / V (G)) and E(G/H) = E(G H) {hy : y V (G H) and yy E(G) for some y V (H)}. We say that G/H is obtained from G by contracting H to the vertex h. If G is a graph, and x, y V (G), then G + xy = G if xy E(G); otherwise, G + xy denotes the graph obtained from G by adding the edge xy. Let G and H be subgraphs of a graph. Then G H (respectively, G H) is the graph with vertex set V (G) V (H) (respectively, V (G) V (H)) and edge set E(G) E(H) (respectively, E(G) E(H)). We shall use G H instead of G (H G). A block of a graph G is a maximal 2-connected subgraph of G. (The complete graph on two vertices is 2-connected.) Let G be a connected graph and X V (G), where X = k and k is a positive integer; then X is called a k-cut of G if G X is not connected. If X = {x} is a cut set of G, then x is a cut vertex of G. Let G be a plane graph, a plane subgraph of G is a subgraph of G inheriting the embedding of G. The faces of G are the connected components (in topological sense) of the complement of G in the plane. The outer face of a plane graph G is the unbounded face; the bounded faces are inner faces. The boundary of the outer face is called the outer walk of the graph, or the outer cycle if it is a cycle. A cycle is a facial cycle in a plane graph if it bounds a face of the graph. An open disc (respectively, closed disc) in the plane is a homeomorphic image of {(x, y) : x 2 + y 2 < 1} (respectively, {(x, y) : x 2 + y 2 1}). For u, v V (G), a u v path in a graph G is a path with end vertices u and v. For any path P and x, y V (P ), xp y denotes the subpath of P between x and y. Given two vertices x and y on a cycle C in a plane graph, 3

4 we use xcy to denote the path in C from x to y in clockwise order. Let H be a subgraph of a graph G. An H-bridge of G is a subgraph of G which either (1) is induced by an edge of E(G) E(H) with both incident vertices in H or (2) is induced by the edges in a component of G H and the edges of G from H to that component. For any H-bridge B of G, the attachments of B (on H) are the vertices in V (B) V (H). Although Theorem 1.2 is stated for 3-connected graphs, we need to work with certain 2-connected graphs. The following concepts will serve this purpose. A circuit graph is a pair (G, C), where G is a 2-connected plane graph and C is a facial cycle of G, such that, for any 2-cut S of G, every component of G S contains a vertex of C. An annulus graph is a triple (G, C 1, C 2 ), where G is a 2-connected plane graph and C 1 and C 2 are facial cycles of G, such that, for any 2-cut S of G, every component of G S contains a vertex of C 1 C 2. The rest of the paper is organized as follows. In Section 2, we give the Moon-Moser example which shows that the exponent log 3 2 in Theorem 1.2 cannot be improved. (In fact, we slightly improve their bound of 9 V (G) log 3 2.) In Section 3, we prove a result for weighted graphs, from which Conjecture 1.1 follows as a consequence. In Section 4, we prove Theorem 1.2 for graphs embeddable in the projective plane, or the torus, or the Klein bottle. 2 An example In this section, we use the Moon-Moser example to illustrate that the bound in Conjecture 1.1 (and hence, Theorem 1.2) is in a sense best possible. First, we define a sequence of 3-connected plane graphs {T k } as follows. Let T 1 be a plane graph isomorphic to K 4. Further, let V (T 1 ) = {w, x 0, x 1, x 2 } and let x 0 x 1 x 2 x 0 be the outer cycle of T 1. Suppose that T k is defined for some k 1. Let T k+1 be the graph obtained from T k as follows: in each inner face of T k, add a new vertex and join the new vertex to the vertices of T k incident with that face. The graphs T 1 and T 2 are shown in Figure 1. 4

5 x 0 x 0 w w x 2 T 1 x 1 x 2 T 2 x 1 Figure 1: By the above construction, for any k 1, T k is a 3-connected plane graph with outer cycle x 0 x 1 x 2 x 0. We shall show that circ(t k ) < 7 2 nlog 3 2 ). Let α k be the length of a longest x 1 x 2 path in T k and β k be the length of a longest x 1 x 2 path in T k x 0. By the construction of T k, for i, j {0, 1, 2} and i j, the length of a longest x i x j path in T k is α k and the length of a longest x i x j path in T k ({x 0, x 1, x 2 } {x i, x j }) is β k. Proposition 2.1 For k 1, α k = 3 2 k 1 and β k = 2 k. Proof: It is easy to check that α 1 = 3 and β 1 = 2. Assume that α k = 3 2 k 1 and β k = 2 k. We need to show that α k+1 = 3 2 k and β k+1 = 2 k+1. For i {0, 1, 2}, let D i denote the open disc in the plane bounded by the triangle in T k+1 induced by {w, x 0, x 1, x 2 } {x i }. Let V i denote the set of vertices in T k+1 contained in D i, and let T i be the plane subgraph of T k+1 induced by V i ({w, x 0, x 1, x 2 } {x i }). Clearly, T i is a 3-connected plane graph isomorphic to T k. We shall proceed with claims (a) - (d). (a) α k k. Let P be an x 1 x 2 path in T k+1. Since the outer cycle of each T i is a triangle, P T i consists of a path, or a path and an isolated vertex in {w, x 0, x 1, x 2 } {x i }, or isolated vertices which are contained in {w, x 0, x 1, x 2 } {x i }. 5

6 If P T 0 is a path, then P T 0. In this case, E(P ) α k = 3 2 k k. If P T 1 consists of a path and {w, x 0, x 2 } P T 1, then either V (P ) V 2 = or V (P ) V 0 =. In this case, P T 1 T j for some j {0, 2}, and P T j consists of a path and an isolated vertex. By induction, E(P T 1 ) α k and E(P T j ) β k. Hence, E(P ) α k + β k = 3 2 k k 3 2 k. Similarly, if P T 2 consists of a path and {w, x 0, x 1 } P T 2, then E(P ) 3 2 k. Thus, we may assume that, for i {0, 1, 2}, P T i is not a path containing {w, x 0, x 1, x 2 } {x i }. Hence, for i {0, 1, 2}, P T i consists of a single vertex, or a vertex and a path, or two isolated vertices. Therefore, by induction, E(P T i ) β k, and so, E(P ) 3β k = 3 2 k. Hence, α k k. (b) α k k. By induction, T 0 x 2 contains an x 1 w path P 0 with E(P 0 ) = β k, T 2 x 1 contains a w x 0 path P 1 with E(P 1 ) = β k, and T 1 w contains an x 0 x 2 path with E(P 2 ) = β k. Hence, P = P 0 P 1 P 2 is an x 1 x 2 path in T k+1 with length 3β k = 3 2 k. Hence, α k k By (a) and (b), α k+1 = 3 2 k. (c) β k+1 2 k+1. Let Q be an x 1 x 2 path in T k+1 x 0. If V (Q) V 0 =, then V (Q) (T 1 T 2 ) {x 0 }. In this case, by induction, E(Q T i ) 2 k, and so, E(Q) 2 2 k = 2 k+1. So assume that Q V 0. If Q T 0, then E(Q) α k = 3 2 k 1 2 k+1. So assume that Q T 0. Since x 0 / Q, either V (Q) V 1 = or V (Q) V 2 =. By symmetry assume that V (Q) V 1 =. Then Q T 2 is a w x 1 path in T 2 x 0, and Q T 0 consists of a w x 2 path in T 0 x 1 and the isolated vertex x 1. Hence, E(Q) = E(Q T 0 ) + E(Q T 2 ) 2β k = 2 k+1. Thus, β k+1 2 k+1. (d) β k+1 2 k+1. For i {1, 2}, T i x 0 has an x i w path P i of length β k = 2 k. Hence, P 1 P 2 is an x 1 x 2 path in T k+1 x 0 such that E(P 1 P 2 ) = 2 k+1. Thus, β k+1 2 k+1. By (c) and (d) β k+1 = 2 k+1. 6

7 By the definition of T k+1, we have n = V (T k+1 ) = k = 3 + (3 k+1 1)/2. Next, we show that circ(t k+1 ) α k + 2β k. Let C be a longest cycle in T k+1, and let T i be defined as in the proof of Proposition 2.1. If C uses an edge of the outer cycle of T k+1, then E(C) α k+1 +1 α k +2β k. If C T i for some i, then by induction, E(C) α k 1 + 2β k 1 α k + 2β k. So assume that C T i for any i {0, 1, 2}. Note that E(C T i ) induces a path in T i between two vertices on the outer cycle of T i, and at most one of these C T i contains all vertices of the outer cycle of T i. Hence, E(C) α k +2β k. Since n = 3 + (3 k+1 1)/2 and by a simple calculation, E(C) < 7 2 nlog 3 2 ). Therefore, the bound in Conjecture 1.1 is best possible. 3 Planar graphs In this section, we prove Conjecture 1.1. The following elementary result is needed. Lemma 3.1 Let m, n, k be non-negative real numbers. Then (1) m r + n r (m + n) r for 0 < r < 1. (2) m r + n r (m + n + k) r if 0 r log 3 2 and k = min{m, n, k}. Proof: To prove (1), it is sufficient to show that f(x) = x r + (1 x) r 1, where 0 < r < 1 and 0 < x < 1. This can be done by showing that f(x) has a unique critical point x = 1/2 in [0, 1] and f(1/2) > 1 (and so, the absolute minimum of f(x) on [0,1] is 1). Next, we prove (2). Without loss of generality, assume that m n k. If m n (m+n+k)/3, then m r +n r (2/3 r )(m+n+k) r (m+n+k) r (since 2/3 r 1 when 0 r log 3 2). Hence, we may assume that n = ( 1 t)(m + n + k), where 0 < t < 1/3. 3 Since m n k, m (m + n + k) 2n = ( 1 + 2t)(m + n + k). Thus, 3 m r + n r [( t)r + ( 1 3 t)r ](m + n + k) r. Hence, it is sufficient to show that f(t) = ( t)r + ( 1 3 t)r 1 for 0 t 1. 3 Note that f(1/3) = 1. Since 0 r log 3 2, f(0) 1. Also, a simple calculation shows that f(1/6) > 1. Since f (t) = 2r( t)r 1 r( 1 3 t)r 1 = 0 has a unique solution, f(t) 1 when 0 t 1/3. 7

8 In order to prove Conjecture 1.1, we need to work with a class of 2- connected planar graphs which includes all 3-connected planar graphs. Definition 3.2 Let (G, C) be a circuit graph, and let x, y V (C). We say that (G, xcy) is a strong circuit graph if, for any 2-cut S of G, S V (ycx {x, y}). We also need to work with graphs in which vertices are assigned nonnegative weights. Let R + denote the set of non-negative real numbers. Let G be a graph, and w : V (G) R +. For H G, we write w(h) = Σ v V (H) w(v). Define w( ) = 0. Theorem 3.3 Let (G, xcy) be a strong circuit graph, and let w : V (G) R +. Then G contains an x y path P such that Σ v V (P y) [w(v)] log 3 2 [w(g y)] log 3 2. Remark: In the inequality of Theorem 3.3, y is not included. This is for a technical reason which facilitates counting. Also note the symmetry between x and y (we can always re-embed G so that the clockwise direction of C becomes the counter clockwise direction). Proof: We use induction on V (G) + E(G). Since G is 2-connected and since G contains the cycle C, V (G) + E(G) 6. If V (G) + E(G) = 6, then G is the complete graph on three vertices, and the inequality in Theorem 3.3 follows from (1) of Lemma 3.1. So assume that V (G) + E(G) > 6. For convenience, let r = log 3 2, and assume that C is the outer cycle of G. We shall proceed with claims (a) - (g). (a) We may assume that xy / E(G). Suppose that xy E(G). Since (G, xcy) is a strong circuit graph, {x, y} is not a cut set of G, and so, xy E(C). Since G is 2-connected, we can label the cut vertices of G xy as v 1,, v m 1 and blocks of G xy as B 1,, B m such that B i B i+1 = {v i } for i {1, 2,, m 1}, B i B j = for i j 2 and i, j {1,, m}, and v 0 = x B 1 v 1 and v m = y B m v m 1. See Figure 2. In the left figure xcy = G[xy] and v 0, v 1,, v m occur on C in counter clockwise order, and in the right figure ycx = G[xy] and 8

9 v 1 x=v y=v 0 m B 1 B m v m-1 v i-1 B i v i B v i i-1 v i v 1 B 1 x=v 0 B m y=v m v m-1 Figure 2: v 0, v 1,, v m occur on C in clockwise order. We view each B i as a plane subgraph of G. Next, we shall find a v i 1 v i path P i in B i such that m i=1 P i gives the desired path P. If V (B i ) = 2, then let P i = B i. It is easy to see that Σ v V (Pi v i )[w(v)] r [w(b i v i )] r. So assume that V (B i ) 3. Then V (B i ) + E(B i ) 6. Let C i denote the outer cycle of B i. We claim that (B i, v i 1 C i v i ) is a strong circuit graph. Let S be an arbitrary 2-cut of B i. If B i S contains a component T with T C i =, then T is also a component of G S with T C =, and so, (G, C) is not a circuit graph, a contradiction. Hence, any component of B i S contains a vertex of C i, and so, (B i, C i ) is a circuit graph. Now assume that S (v i C i v i 1 {v i, v i 1 }) =. Then S v i 1 C i v i. If xcy = G[xy], then G S contains component T with T C =, a contradiction. So ycx = G[xy]. Then S is a 2-cut of G such that S (ycx {x, y}) =, and so, (G, xcy) is not a strong circuit graph, a contradiction. Hence (B i, v i 1 C i v i ) is a strong circuit graph. By induction, B i contains a v i 1 v i path P i such that Σ v V (Pi v i )[w(v)] r [w(b i v i )] r. 9

10 Now let P = m i=1 P i. Then P is an x y path in G. Moreover, Σ v V (P y) [w(v)] r = Σ m i=1 (Σ v V (P i v i )[w(v)] r ) Σ m i=1[w(b i v i )] r [w(g y)] r. The first inequality follows from the previous inequalities about P i. Since V (G y) is the disjoint union of V (B i v i ) for i {1,, m}, the second inequality follows from (1) of Lemma 3.1. This completes the proof of (a). Since (G, xcy) is a strong circuit graph, x is not a cut vertex of G y. Hence, G y contains a unique block, say H, containing x. Since xy / E(G) and since H is a block of G y, V (H) + E(H) 6. Let D denote the outer cycle of H. Let y 1, y 2,, y m V (D) denote the attachments of (H {y})-bridges of G, and assume that x, y 1,, y m occur on D in this clockwise order. See Figure 3. y y 2 y 1 y m H x Figure 3: (b) For i {1,, m}, there are exactly two {y, y i }-bridges of G; and for i {1,, m 1}, yy i E(G) and the {y, y i }-bridge of G not containing x is G[yy i ]. Since (G, xcy) is a strong circuit graph, for any 2-cut S of G, S (ycx {x, y}). Hence, for i {1,, m 1}, {y, y i } is not a cut set of G, and 10

11 so, G has exactly two {y, y i }-bridges, yy i E(G), and the {y, y i }-bridge of G not containing x is G[yy i ]. If G has at least three {y, y m }-bridges, then G {y, y m } contains a component T with T C =, a contradiction. Thus, (b) follows. Let X = ycx ycy m. Let x = x if X = {x}; otherwise, let x be the end vertex of X other than x. Let B denote the minimal subgraph of H y 1 Dy m such that X B and B is a union of blocks of H y 1 Dy m. Let w 1,, w n V (B) be the attachments of (B y 1 Dy m )-bridges of H. For k {1,, n}, define s k, t k V (y 1 Dy m ) as follows: y 1, s k, t k, y m occur on D in this clockwise order, {s k, w k } and {t k, w k } are contained in (B y 1 Dy m )-bridges of H, and subject to these conditions, s k Dt k is maximal. We can choose the notation of w 1,, w n so that y 1, s 1, t 1, s 2, t 2,, s k, t k, s k+1, t k+1,, s n, t n, y m occur on D in this clockwise order. Then y 1 = s 1, y m = t n, and w n = x. See Figure 4. y sk tk s 1 = y 1 M y m = t n w 1 w x =w k n x B Figure 4: (c) There is some k {1,, n} such that B contains an x w k path P B and Σ v V (PB )[w(v)] r [w(b)] r. Let Y = G B and B = G/Y, and let y denote the unique vertex in B B. 11

12 We may assume that B {x}. Otherwise, V (B ) = {x, y }, E(B ) = {xy }, n = 1 and w 1 = x = x. Let P B = {x}. Clearly, P B is an x w 1 path in B such that Σ v V (PB )[w(v)] r [w(b)] r. Hence, V (B ) + E(B ) 6. Let C denote the outer cycle of B, where E(C ) = E(w n Dw 1 ) {w 1 y, y w n }. We claim that (B, xc y ) is a strong circuit graph. Suppose that S is an arbitrary 2-cut of B. By the construction of B, if y S, then S y X x y C x {y, x} and the vertex in S y is a cut vertex of B, and so, each component of B S contains a vertex of C. Hence, we may assume that y / S. If T is a component of B S, then T C ; otherwise, T is a component of G S with T C =, a contradiction. Thus, since y / S, S V (y C x {y, x}). Hence (B, xc y ) is a strong circuit graph. Let w : V (B ) R + be defined as follows: w (v) = w(v) if v V (B), and w (y ) is an arbitrary non-negative integer. By induction, B contains an x y path P such that Σ v V (P y )[w (v)] r [w (B y )] r. Let P B = P y. Then P B is a path in B from x to w k for some k {1,, n}. Since B y = B, and by the definition of w, we have This completes (c). Σ v V (PB )[w(v)] r [w(b)] r. Let M be the union of s k Dt k and those (B y 1 Dy m )-bridges of H whose attachments are contained in s k Dt k {w k }, where k is given in (c). See Figure 4. Let L = if s k = y 1 ; otherwise, let L denote the component of H (B M) containing y 1. Let R = G (B L M {y}) (possibly R = ). Note that V (G y) is the disjoint union of V (B), V (L), V (R), and V (M w k ). (d) We may assume that w(b) > min{w(l), w(r)}. Suppose that w(b) min{w(l), w(r)}. Let L = G[V (L M B)]/B, and let x denote the unique vertex of L (L M). Thus x is the result of the contraction of B. Let R = G[V (R) {y, t k }]. We shall find an x t k 12

13 path P L in L and a t k y path P R in R, and use P L and P R to construct the desired path P. (d1) L contains an x t k path P L such that Σ v V (PL {x,t k })[w(v)] r [w((l M) {w k, t k })] r. First, assume that t k = y 1. Since (G, xcy) is a strong circuit graph, L =, t k w k E(G), M = G[t k w k ], V (L ) = {x, t k }, and E(L ) = {x t k }. Let P L = L. Clearly, Σ v V (PL {x,t k })[w(v)] r = [w( )] r [w((l M) {w k, t k })] r. So assume that t k y 1. Then L is a 2-connected plane graph with at least three vertices. Hence, V (L ) + E(L ) 6. Let C L denote the outer cycle of L such that w 1 Dt k is the clockwise segment of C L from x = w 1 to t k. Next, we show that (L, t k C L x ) is a strong circuit graph. Let S be an arbitrary 2-cut of L. First, assume that L S has a component T with T C L =. Then x S; otherwise, T is also a component of G S with T C =, a contradiction. Thus, by the construction of L, there is some w i, where 1 i k, such that S = (S {x }) {w i } is a 2-cut of G, and T is a component of G S with T C =, a contradiction. Hence, (L, C L ) is a circuit graph. Now, assume that S (x C L t k {x, t k }) =. Then S t k C L x. Thus, G S contains a component U with U (t k C L x {t k, x }). Hence, w k x ; otherwise, t k = y m and t k C L x {t k, x } =, a contradiction. Let S = S if x / S; otherwise, let S = (S {x }) {w k }. Thus, U is a component of G S with U C =, a contradiction. Hence, (L, t k C L x ) is a strong circuit graph. Let w : V (L ) R + be defined as follows: w (v) = w(v) for v L x, and w (x ) = 0. By induction, L contains an x t k path P L such that By the definition of w, we have This proves (d1). Σ v V (PL t k )[w (v)] r [w (L t k )] r. Σ v V (PL {x,t k })[w(v)] r [w((l M) {w k, t k })] r. 13

14 (d2) R contains a t k y path P R such that Σ v V (PR y)[w(v)] r [w(r {t k })] r. If V (R ) = 2, then t k = y m, t k y E(G), and the {y, y m }-bridge of G not containing x is G[t k y]. In this case, R =. Let P R = R. Clearly, Σ v V (PR y)[w(v)] r [w(r y)] r = [w(r {t k })] r. So assume that V (R ) 3. Hence, V (R + t k y) + E(R + t k y) 3. Without loss of generality, assume that R + t k y is a plane graph with outer cycle C R such that {t k y} E(yCy m ) E(C R ), and E(t k C R y) = {t k y}. Next we show that (R + t k y, t k C k y) is a strong circuit graph. Suppose that S is an arbitrary 2-cut of R + t k y. If (R + t k y) S has a component T with T C R =, then T is a component of G S with T C =, a contradiction. Hence, (R + t k y, C R ) is a circuit graph, and so, S C R. Since E(t k C R y) = {t k y}, S (yc R t k {y, t k }). Thus, (R + t k y, t k C k y) is a strong circuit graph. By induction, R contains at k y path P R such that Σ v V (PR y)[w(v)] r [w(r y)] r = [w(r {t k })] r. Note that we can always select P R so that E(P R ) {t k y}. Hence, P R is a path in R. This proves (d2). Finally, we find the desired path P as follows. Without loss of generality, assume that the edge of P L incident with x is x v, and v is incident with w l, where 1 l k. Let Q be a path in B from x to w l, and let P = ((P L x ) Q P R ) + vw l. Then P is an x y path in G such that Σ v V (P y) [w(v)] r Σ v V (PL {x,t k })[w(v)] r + Σ v V (PR y)[w(v)] r [w((l M) {w k, t k })] r + [w(r {t k })] r [w(g y)] r. Here, the first inequality is obvious, and the second inequality follows from (d1) and (d2). By the assumption that w(b) min{w(l), w(r)} and since V (G y) is the disjoint union of V ((L M) {w k, t k }), V (B), and V (R {t k }), the third inequality follows from (2) of Lemma 3.1. This completes the proof of (d). 14

15 (e) We may assume that w(l) < w(r). Otherwise, assume that w(l) w(r). Let L = G[V (L) {y, s k }]. We shall extend P B in (c) to the desired path P by finding an s k y path P L in L and a w k s k path P M in M. (e1) L contains an s k y path P L such that Σ v V (PL {s k,y})[w(v)] r [w(l)] r. If y 1 = s k, then L = and L = G[yy 1 ]. In this case, let P L = L. Clearly, Σ v V (PL {s k,y})[w(v)] r = [w(l {s k, y})] r = [w( )] r = [w(l)] r. So assume that y 1 s k. Then L + ys k is a 2-connected graph with at least three vertices. Hence, V (L + ys k ) + E(L + ys k ) 6. Without loss of generality, we may assume that L + ys k is embedded in the plane with outer cycle C L such that y 1 C L s k = y 1 ys k (and hence, E(yC L s k ) = {ys k }). Next we show that (L + ys k, yc L s k ) is a strong circuit graph. Let S be an arbitrary 2-cut of L + ys k. If (L + ys k ) S has a component T with T C L =, then T is a component of G S with T C =, a contradiction. Hence, (L + ys k, C L ) is a circuit graph, and so, S C L. Since E(yC L s k ) = ys k, (L + ys k, yc L s k ) is a strong circuit graph. Let w : V (L + ys k ) R + be defined as follows: w (v) = w(v) for v L s k, and w (s k ) = 0. By induction and by the definition of w, L + ys k contains an s k y path P L such that Σ v V (PL {s k,y})[w(v)] r = Σ v V (PL y)[w (v)] r [w (L y)] r = [w(l)] r. Note that we can always select P L so that E(P L ) {ys k }. Hence, P L is a path in L. This proves (e1). (e2) M contains a path P M from w k to s k such that Σ v V (PM w k )[w(v)] r [w(m w k )] r. 15

16 If s k = t k, then w k s k E(G) and M = G[w k s k ]. In this case, let P M = M. Clearly, Σ v V (PM w k )[w(v)] r [w(m w k )] r. So assume that s k t k. Then M + w k s k is a 2-connected graph with at least three vertices. Hence, V (M + w k s k ) + E(M + w k s k ) 6. Without loss of generality, assume that M +w k s k is embedded in the plane with outer cycle C M such that E(s k Dt k ) {w k s k } C M, and E(w k C M s k ) = {w k s k }. Next we show that (M + w k s k, w k C M s k ) is a strong circuit graph. Let S be an arbitrary 2-cut of M + w k s k. If (M + t k s k ) S has a component T with T C M =, then T is a component of G S with T C =, a contradiction. Thus, (M + w k s k, C M ) is a circuit graph, and so, S C M. Since E(w k C M s k ) = w k s k, (M + w k s k, w k C M s k ) is a strong circuit graph. By induction, M + w k s k contains a w k s k path P M such that Σ v V (PM w k )[w(v)] r [w(m w k )] r. Note that we can always select P M so that E(P M ) {w k s k }. Hence, P M is a path in M. This proves (e2). Now let P = P B P M P L. Then P is an x y path in G. Moreover, Σ v V (P y) [w(v)] r = Σ v V (PB )[w(v)] r + Σ v V (PM w k )[w(v)] r + Σ v V (PL {s k,y})[w(v)] r [w(b)] r + [w(m w k )] r + [w(l)] r [w(b L R)] r + [w(m w k )] r [w(g y)] r. Here, the first inequality follows from (c), (e1) and (e2). By (d) and the assumption that w(l) w(r), the second inequality follows from (2) of Lemma 3.1. The third inequality follows from (1) of Lemma 3.1. This proves (e). By (e), R, and so t k y m. We shall extend P B in (c) to the desired path P by finding a t k y path P R and a w k t k path P M. The argument is similar to that for (e). Since it is not too long and is not completely symmetric to (e), we provide the details. Let R = G[V (R) {y, t k }]. (f) R contains a t k y path P R such that Σ v V (PR {t k,y})[w(v)] r [w(r)] r. 16

17 If V (R ) = 2, then R =, t k = y m, yy m E(G), and the {y, y m }-bridge of G not containing x is G[yy m ]. In this case, let P R = R. Clearly, Σ v V (PR {t k,y})[w(v)] r = [w( )] r = [w(r)] r. So assume that V (R ) 3. Then, V (R + t k y) + E(R + t k y) 6. Without loss of generality, assume that R +t k y is embedded in the plane with outer cycle C R such that {t k y} E(yCy m ) E(C R ) and E(t k C R y) = {t k y}. Now we show that (R + t k y, t k C R y) is a strong circuit graph. Let S be an arbitrary 2-cut of R + t k y. If (R + t k y) S contains a component T with T C R =, then T is a component of G S with T C =, a contradiction. Hence, (R + t k y, C R ) is a circuit graph, and so, S C R. Since E(t k C R y) = {t k y}, (R + t k y, t k C R y) is a strong circuit graph. Let w : V (R ) R + be defined as follows: w (v) = w(v) for v V (R t k ), and w (t k ) = 0. By induction and by the definition of w, R + t k y contains a t k y path P R such that Σ v V (PR {t k,y})[w(v)] r = Σ v V (PR y)[w (v)] r [w (R y)] r = [w(r)] r. Note that we can always select P R so that E(P R ) {t k y}. Hence, P R R. This completes (f). (g) M contains a w k t k path P M such that Σ v V (PM w k )[w(v)] r [w(m w k )] r. If s k = t k, then t k w k E(G) and M = G[t k w k ]. In this case, let P M = M. Clearly, Σ v V (PM w k )[w(v)] r [w(m w k )] r. So assume that s k t k. Then M + t k w k is a 2-connected graph with at least three vertices. Hence, V (M + t k w k ) + E(M + t k w k ) 6. Without loss of generality, assume that M + t k w k is embedded in the plane with outer cycle C M such that E(s k Dt k ) {t k w k } E(C M ) and E(t k C M w k ) = {t k w k }. Next, we show that (M + t k w k, t k C M w k ) is a strong circuit graph. Let S be an arbitrary 2-cut of M + t k w k. If (M + t k w k ) S has a component T with T C M =, then T is a component of G S with T C =, 17

18 a contradiction. Recall that, by (e), t k y m, and so, w k x. Hence, (M + t k w k, C M ) is a circuit graph, and so, S C M. Since E(t k C M w k ) = {t k w k }, (M + t k w k, t k C M w k ) is a strong circuit graph. By induction, M + t k w k contains a t k w k path P M such that Σ v V (PM w k )[w(v)] r [w(m w k )] r. Note that we can always select P M so that E(P M ) t k w k. Hence, P M M. This proves (g). Now let P = P B P M P R. Then P is a path in G from x to y. Moreover, Σ v V (P y) [w(v)] r = Σ v V (PB )[w(v)] r + Σ v V (PM w k )[w(v)] r + Σ v V (PR {t k,y})[w(v)] r [w(b)] r + [w(r)] r + [w(m w k )] r [w(b L R)] r + [w(m w k )] r [w(g y)] r. Here, the first inequality follows from (c), (f) and (g). By (d) and by (e), the second inequality follows from (2) of Lemma 3.1. The third inequality follows from (1) of Lemma 3.1. Corollary 3.4 If (G, C) is a circuit graph and e E(C). Then G contains a cycle T through e such that E(T ) V (G) log 3 2. Proof: Let e = xy such that xcy = G[xy]. Then (G, xcy) is a strong circuit graph. Let w : V (G) R + such that w(v) = 1 for all v V (G). By Theorem 3.3, G contains an x y path P such that V (P ) 1 = Σ v V (P y) [w(v)] log 3 2 [w(g y)] log 3 2 = ( V (G) 1) log 3 2. Hence, by (1) of Lemma 3.1, V (P ) 1 + ( V (G) 1) log 3 2 V (G) log 3 2. Thus P + xy gives the desired cycle. 18

19 Corollary 3.5 Let G be a 3-connected planar graph, and let e E(G). Then G contains a cycle C through e such that E(C) V (G) log 3 2. Proof: Without loss of generality, we assume that G is embedded in the plane such that e E(C), where C is a facial cycle of G. Then (G, C) is a circuit graph. Hence, Corollary 3.5 follows from Corollary Graphs on other surfaces In this section, we prove that Conjecture 1.1 also holds for 3-connected graphs embedded in the projective plane, or the torus, or the Klein bottle. Definition 4.1 Given an embedding σ : G Π of a graph G into a surface Π, the representativity of σ is defined to be the number min{ σ(g) Γ : Γ is a non-null homotopic simple closed curve in Π}. For graphs embedded in the projective plane, Fiedler et al proved the following result ([4], Proposition 1). Note that in [4], a cycle in G is called a polygon in G. Lemma 4.2 Let σ : G Π be an embedding of a 3-connected graph G into the projective plane Π, let P 1 be a cycle in G such that σ(p 1 ) is null homotopic, and let D 1 be the open disc in Π bounded by σ(p 1 ). Then there is a cycle P in G such that σ(p ) is null homotopic and bounds a disc D containing D 1, and the closure of D contains σ(v (G)). Note that the condition about P 1 and D 1 in Lemma 4.2 holds if the representativity of σ is at least 2. Also note that the subgraph H of G contained in the closure of D is a 2-connected spanning subgraph of G, and σ(h) can be viewed as a plane graph embedded in the closed disc bounded by D. Lemma 4.3 Let G be a 3-connected graph embeddable in the projective plane. Then G contains a 2-connected spanning planar subgraph H. Moreover, H can be embedded in the plane with a facial cycle C such that (H, C) is a circuit graph. 19

20 Proof: Let σ : G Π be an embedding of G in the projective plane Π. If the representativity of σ is at least 2, then by the remarks following Lemma 4.2, G has a cycle C, such that the subgraph H of G contained in the closed disc in Π bounded by σ(c) is a spanning subgraph of G. Since G is 3-connected, it is easy to see that (H, C) is a circuit graph. So assume that the representativity of σ is at most 1. Then G is a 3- connected planar graph. Let H be a plane embedding of G and let C be facial cycle of H. Then (H, C) is a circuit graph. Now we are ready to prove Conjecture 1.1 for 3-connected graphs embeddable in the projective plane. Theorem 4.4 Let G be a 3-connected graph embeddable in the projective plane. Then circ(g) V (G) log 3 2. Proof: By Lemma 4.3, let H be a 2-connected plane graph with a facial cycle C such that H is a spanning subgraph of G and (H, C) is a circuit graph. By Corollary 3.4, H, and hence G, has a cycle of length at least V (H) log 3 2 = V (G) log 3 2. For graphs embeddable in the torus or the Klein bottle, we need the following result ([2], Theorems 2 and 3). Lemma 4.5 Let σ : G Π be an embedding of a 3-connected graph G in a surface Π, where Π is the torus or the Klein bottle. Suppose σ has representativity at least 1. Then there is a spanning subgraph H of G and either (1) there is a cycle C of H such that (H, C) is a circuit graph or (2) there are two cycles C 1 and C 2 of H such that (H, C 1, C 2 ) is an annulus graph. Theorem 4.6 Let G be a 3-connected graph embeddable in the torus or the Klein bottle. Then circ(g) ( V (G) /2) log 3 2. Proof: Let σ : G Π be an embedding of G into a surface Π, where Π is the torus or the Klein bottle. If the representativity of σ is 0, then G is embeddable in the projective plane, and Theorem 4.6 follows from Theorem 4.4. So we may assume that the representativity of σ is at least 1. Then by Lemma 4.5, there is a spanning subgraph H of G and either (1) there is a 20

21 cycle C of H such that (H, C) is a circuit graph or (2) there are cycles C and D of H such that (H, C, D) is an annulus graph. If there is a cycle C of H such that (H, C) is a circuit graph, then by Corollary 3.4, H, and hence, G, contains a cycle of length at least ( V (G) /2) log 3 2. So assume that there are cycles C and D of H such that (H, C, D) is an annulus graph. Without loss of generality, we may assume that C is the outer cycle of H. We consider two cases: C D = and C D. Case 1. C D =. Since H is 2-connected, H contains vertex disjoint paths P 1, P 2 from v 1, v 2 V (C) to u 1, u 2 V (D), respectively, such that, for i {1, 2}, (P i {u i, v i }) (C D) =. Let C 1 = v 1 Cv 2 P 2 u 1 Du 2 P 1, and let C 2 = v 2 Cv 1 P 1 u 2 Du 1 P 2. For i {1, 2}, let H i be the subgraph of H contained in the closed disc in the plane bounded by C i. Then (H i, C i ), i {1, 2}, are circuit graphs. See Figure 5. H 1 v 1 P 1 u1 u D H 2 P v C Figure 5: Since V (H 1 ) + V (H 2 ) > V (H), we may assume, without loss of generality, that V (H 1 ) > V (H) /2. Since (H 1, C 1 ) is a circuit graph, by Corollary 3.4, circ(h 1 ) V (H 1 ) log 3 2. Hence, circ(g) ( V (G) /2) log 3 2. Case 2. C D. 21

22 Assume that u V (C D). Let u 1,, u n be the neighbors of u in clockwise order around u such that u 1, u n V (C) and u k, u k+1 V (D), where 1 k < n. Let H be the plane graph obtained from H u by adding two vertices u and u, and adding edges u u i for i {1,, k} and edges u u i for i {k + 1,, n}. See Figure 6. Since H is 2-connected, we can label the cut vertices of H as v 1,, v m 1 and blocks of H as B 1,, B m such that u, v 1,, v m 1 occur on C in this clockwise order, B i B i+1 = {v i } for i {1,, m 1}, B i B j = for i, j {1,, m} with i j 2, and v 0 = u B 1 v 1 and v m = u B m v m 1. u n u u 1 u k+1 u k u k+1 u k u =v 0 u n v B i v i-1 i u 1 u =v m Figure 6: Let W C V (H) (respectively, W D V (H)) be defined as follows: For v V (H), v W C (respectively, v W D ) if, and only if, H has a 2-cut S v V (C) (respectively, S v V (D)) such that v and D S v (respectively, C S v ) are contained in different components of H S v. Note that W C W D = and u / W C W D. We may assume that W C and W D. Otherwise, by symmetry assume that W D =. Then (H, C) is a circuit graph. Hence, by Corollary 3.4, circ(h) V (H) log 3 2, and so, circ(g) ( V (G) /2) log 3 2. Without loss of generality, assume that W D W C. Then, W D < V (H) /2 (since u / W C W D ). 22

23 Next, for i {1,, m}, we find a v i 1 v i path P i in B i. If V (B i ) = 2, then let P i = B i. So assume that V (B i ) 3. Then let C i denote the outer cycle of B i. Note that v i 1 C i v i = v i Dv i 1 and v i C i v i 1 = v i Cv i 1 (where we view v 0 and v m as u). We construct a new graph Bi from B i as follows. If W D B i =, then let Bi = B i and let Ci = C i. If W D B i, then W D V (B i ) can be partitioned into sets W1 i,, W m i i with the following properties: (1) for j {1,, m i }, B i has a 2-cut Sj i = {s i j, ti j } V (D B i) and G[Wj i] is the component of B i Sj i not containing C B i, and (2) B i has no 2-cut S V (D B i ) such that the component of B i S not containing C B i properly contains G[Wj i ]. Let Bi denote the graph obtained from B i by deleting G[Wj i] and adding the edges si j ti j for j {1,, m i }. Let Ci be obtained from C i by deleting Wj i D and adding edges s i j ti j. Let the edges si j ti j be added so that C i is the outer cycle of Bi and v i Ci v i 1 = v i C i v i 1. By the above construction, (Bi, v i Ci v i 1 ) is a strong circuit graph. Let w : V (Bi ) R+ with w(v) = 1 for v V (Bi ). By Theorem 3.3, B i contains a v i 1 v i path Pi such that V (P i ) 1 = Σ v V (P i v i )[w(v)] log 3 2 [w(b i v i)] log 3 2 = ( V (B i ) 1)log 3 2. If s i j ti j E(P i ), then we replace si j ti j in P i by a path in G[Wj i {si j, ti j }] from s i j to t i j, and let P i denote the resulting path. Then V (P i ) 1 V (Pi ) 1 ( V (B i ) 1)log 3 2. Now let P = m i=1 P i. Then V (P ) 1 = Σ m i=1( V (P i ) 1) Σ m i=1 ( V (B i ) 1)log 3 2 [ V (H) W D ] log 3 2 ( V (G) /2) log 3 2. Here, the first inequality follows from previous inequalities. Since V (H) W D is the disjoint union of V (B i v i) for i = 1,, m, the second inequality follows from (1) of Lemma 3.1. The third inequality follows from the fact that W D < V (G) /2. 23

24 Hence, V (P ) ( V (G) /2) log Therefore, identifying v and v in P gives the desired cycle. It is easy to see that our main result, Theorem 1.2, follows from Corollary 3.5, Theorem 4.4 and Theorem 4.6. Acknowledgment The authors would like to thank Ron Gould and Robin Thomas for helpful discussions. 24

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