Minimizing the number of edges in C ≥r -saturated graphs
MMinimizing the number of edges in C ≥ r -saturated graphs ∗ Yue Ma a , Xinmin Hou b , Jun Gao c a,b,c Key Laboratory of Wu Wen-Tsun MathematicsSchool of Mathematical SciencesUniversity of Science and Technology of ChinaHefei, Anhui 230026, China.
Abstract
Given a family of graphs F , a graph G is said to be F -saturated if G does not containa copy of F as a subgraph for any F ∈ F but the addition of any edge e / ∈ E ( G ) createsat least one copy of some F ∈ F within G . The minimum and maximum size of an F -saturated graph on n vertices are called the saturation number and the Tur´an numberof F , denoted by sat( n, F ) and ex( n, F ), respectively. Let C ≥ r be the family of cycles oflength at least r . Erd˝os and Gallai (1959) proved that ex( n, C ≥ r ) ≤ ( r − n − , where n ≥ r ≥
3. In this paper, we determine the exact values of sat( n, C ≥ r ) for r ∈ { , , } and n ≤ r ≤ n and give upper and lower bounds for the other cases. Let G = ( V, E ) be a graph. We call | V | the order of G and | E | the size of it. If | V | = n , wecall G an n -vertex graph. Given a family F of graphs, a graph G is said to be F -saturated if G does not contain a subgraph isomorphic to any member F ∈ F but G + e containsat least one copy of some F ∈ F for any edge e / ∈ E ( G ). The Tur´an number ex( n, F )of F is the maximum number of edges in an n -vertex F -saturated graph. The minimumnumber of edges in an n -vertex F -saturated graph is called the saturation number , denotedby sat( n, F ), i.e.sat( n, F ) = min {| E ( G ) | : G is an n -vertex F -saturated graph } .Let Sat( n, F ) = { G : G is an n -vertex F -saturated graph } .Let C r denote the cycle of length r and C ≥ r be the family of cycles of length at least r .Erd˝os and Gallai (1959) proved the following celebrated theorem on Tur´an number of C ≥ r . ∗ The work was supported by NNSF of China (No. 11671376) and Anhui Initiative in Quantum Informa-tion Technologies (AHY150200).. a r X i v : . [ m a t h . C O ] F e b heorem 1.1 (The Erd˝os-Gallai Theorem, [10]) . Let n ≥ r ,ex ( n, C ≥ r ) ≤ ( r − n − . For a single cycle C r , there are many results for ex( n, C r ) and sat( n, C r ) have beenknown, we review some of them in the following. • (Simonovits [17]) ex( n, C k +1 ) = (cid:100) n (cid:101) for sufficiently large n ; • (Erd˝os-Bondy-Simonovits [4], The Even Cycle Theorem) ex( n, C k ) = O ( n k ); • (Erd˝os, Hajnal, and Moon [11]) sat( n, C ) = n − n ≥ • (Ollmann [16], Tuza [19], Fisher et al [12]) sat( n, C ) = (cid:98) n − (cid:99) for n ≥ • (Chen [5, 6]) sat( n, C ) = (cid:100) ( n − (cid:101) for n ≥ • (Barefoot et al [1]) sat( n, C ) ≤ n for n ≥ • (F¨uredi and Kim [13]) (1 + r +2 ) n − < sat( n, C r ) < (1 + r − ) n + (cid:0) r − (cid:1) for all r ≥ n ≥ r − • (Clark, Entringer, and Shapiro [7, 8], Lin et al [15]) sat( n, C n ) = (cid:100) n (cid:101) for n = 17 or n ≥ n, C ≥ r ) for n ≥ r ≥
3. In this paper, we determinethe exact values of sat( n, C ≥ r ) for r = 3 , , n ≤ r ≤ n , and give lower and upperbounds of sat( n, C ≥ r ) for 6 ≤ r ≤ n . The exact values of sat( n, C ≥ r ) for 3 ≤ r ≤ Theorem 1.2. (1) sat ( n, C ≥ ) = n − for n ≥ ;(2) sat ( n, C ≥ ) = (cid:100) n − (cid:101) for n ≥ ;(3) sat ( n, C ≥ ) = (cid:100) ( n − (cid:101) for n ≥ . The following theorem gives the lower bounds of sat( n, C ≥ r ) for n ≥ r ≥ Theorem 1.3. (1) sat ( n, C ≥ r ) ≥ n for n ≥ r ≥ ;(2) sat ( n, C ≥ r ) ≥ n + r for r ≥ n ≥ r ≥ . The last one gives the upper bounds of sat( n, C ≥ r ) and the exact values of sat( n, C ≥ r )for n ≤ r ≤ n . To state it, we define a function g ( x )(see Figure 1) on x ∈ (0 , ∩ Q : g ( x ) = x, if x ∈ [ , , k x, if x ∈ ( k , k − ] , − k − x, if x ∈ [ k − , k − ) , for k ≥ . (1)2igure 1: The image of g ( x ) Theorem 1.4. sat ( n, C ≥ r ) ≤ g ( rn ) n + O ( nr ) for n ≥ r ≥ . Moreover, sat ( n, C ≥ r ) = n + (cid:100) r (cid:101) for n ≥ r ≥ n . The rest of the article is arranged as follows. We give the proof of Theorem 1.2 inSection 2. The proof of Theorems 1.3 and 1.4 will be given in Sections 3 and 4. We givesome remarks in Section 5 and some additional definitions in Appendix.
The following result is due to Dirac.
Theorem 2.1 (Dirac 1952, Theorem 4 in [9]) . Let G be a connected graph with δ ( G ) ≥ d .If n ≥ d then G contains a path of length at least d . Note that a cycle is 2-connected. From the definition of the C ≥ r -saturated graph, wehave the following two facts. Fact 1. A C ≥ r -saturated graph G on n vertices must be connected and e ( G ) ≥ n − . Fact 2.
Let G be a C ≥ r -saturated graph. Then any pair of nonadjacent vertices in G mustbe connected by a path of length at least r − in G . It is obviously true that any tree T on n vertices is a C ≥ -saturated graph. So sat( n, C ≥ ) = n − n, C ≥ ) = { T n } , where T n is a tree on n vertices. Let B ( G ) be the set of blocks of G isomorphic to K and b ( G ) = | B ( G ) | . A ( x, y )-pathis a path connecting x and y . 3 emma 2.2. Let G be a C ≥ r -saturated graph for r ≥ . Then the following holds.(a) Every block B of G is C ≥ r -saturated. Specifically, each block B with | V ( B ) | < r is acomplete graph.(b) B ( G ) forms a matching of G .Proof. (a) Let B be a block of G . Since B is a maximal 2-connected subgraph of G , anycycle containing edges of B and any path connecting two nonadjacent vertices in B mustbe totally contained in B . Since G is C ≥ r -saturated, B contains no cycle of length at least r and any pair of nonadjacent vertices in B is connected by a path of length r − B , i.e. B is C ≥ r -saturated too.Specifically, if | V ( B ) | < r then the longest path in B has length no more than r − B contains no nonadjacent vertices, i.e., B is a complete graph.(b) Suppose there is a vertex u incident with two blocks of B ( G ), say uv , uv . Then v v / ∈ E ( G ), otherwise uv , uv is contained in the triangle uv v u , a contradiction to thefact that uv , uv ∈ B ( G ). So there exists a ( v , v )-path P in G on at least r vertices.However, both uv and uv are cut edges, which forces that uv , uv ∈ E ( P ), i.e., P = v uv ,a contradiction to | V ( P ) | = r ≥ t -triangle , denoted by T t , is a connected graph consisting of t blocks each of which isisomorphic to a triangle. It is easy to show by induction on t that T t has 2 t + 1 vertices.If the t triangles of a T t intersect in exactly one common vertex, we call it a t -fan , denotedby F t . Lemma 2.3.
A graph G is C ≥ -saturated if and only if(1) B ( G ) forms a matching of G ;(2) every component of G − B ( G ) is isomorphic to K or T t for some t ≥ .Proof. Necessity: (1) is a corollary of Lemma 2.2 (b).Now, let B be a block of G . If | V ( B ) | ≤ B ∼ = K or K by Lemma 2.2 (a). Nowsuppose | V ( B ) | ≥
4. Since B is 2-connected, δ ( B ) ≥
2. By Theorem 2.1, B has a cycle oflength at least 4, a contradiction to G is C ≥ -free. Sufficient:
Since every nontrivial component of G − B ( G ) is isomorphic to T t and B ( G )forms a matching of G , adding any edge e / ∈ E ( G ) will give rise to a 2-connected subgraph H on at least four vertices in G + e . By Theorem 2.1, H contains a cycle of length at leastfour. So G truly is C ≥ -saturated.By Lemma 2.3, we have Sat ( n, C ≥ ) = { G : | V ( G ) | = n and G satisfies (1) and (2) of Lemma 2.3 } . emma 2.4. For n ≥ , sat ( n, C ≥ ) ≥ (cid:100) n − (cid:101) .Proof. Let G be a C ≥ -saturated graph on n ≥ G is connected byFact 1 and each component of G − B ( G ) is isomorphic to either a K or a T t for some t ≥ n t ( G ) and n ( G ) be the number of components isomorphic to T t and K of G − B ( G ), respectively. By Lemma 2.3, B ( G ) is a matching and so b ( G ) ≤ n . So n = (cid:80) t ≥ (2 t + 1) n t ( G ) and e ( G ) = b ( G ) + (cid:88) t ≥ t · n t ( G ).By induction on b ( G ), we get the number of components of G − B ( G ) is b ( G ) + 1. Hence (cid:88) t ≥ n t ( G ) = b ( G ) + 1.Since n − b ( G ) ≥
0, we have (cid:80) t ≥ (2 t + 1) n t ( G ) − (cid:32) (cid:80) t ≥ n t ( G ) − (cid:33) ≥
0, i.e. (cid:88) t ≥ (2 t − n t ( G ) ≥ − e ( G ) = b ( G ) + (cid:88) t ≥ t · n t ( G )= (cid:88) t ≥ (3 t + 1) n t ( G ) −
1= 54 (cid:88) t ≥ (2 t + 1) n t ( G ) + 14 (cid:88) t ≥ (2 t − n t ( G ) − ≥ n −
32= 5 n −
64 .To complete the proof for r = 4, we need only construct a family of C ≥ -saturated graphson n vertices and (cid:100) n − (cid:101) edges. Define M , k +2 be the graph on the vertex set V ( M , k +2 ) = { a ij : 1 ≤ i ≤ k, ≤ j ≤ } ∪ { u , u } and the edge set E ( M , k +2 ) = { u a , u a k } ∪ ( ∪ ki =1 { a i a i , a i a i , a i a i , a i a i } ) ∪ ( ∪ k − i =1 { a i a i +1 , } ).Let M , k +1 = M , k +2 − { u } , M , k = M , k +2 − { u , u } and M , k +3 be the graphobtained from M , k +2 by adding a new vertex v and two new edges vu , va k . M ,n for n = 12 , , ,
15 are shown in Figure 2. Obviously, | V ( M ,n ) | = n , e ( M ,n ) = (cid:100) n − (cid:101) and M ,n ∈ Sat ( n, C r ≥ ). 5igure 2: M ,n for n = 12 , , , Given integers n ≥ k ≥ r , let H ( n, k, r ) be the graph obtained from the complete graph K k − r by connecting each vertex of the empty graph K n − k + r to the same r vertices choosingfrom K k − r . When k = 2 r or 2 r + 1, write S n,r for H ( n, k, r ), i.e., S n,r = K r ∨ K n − r , thejoin graph of K r and the empty graph K n − r , we call the vertices of K r the center of S n,r .Define f ( n, k, r ) = (cid:18) k − r (cid:19) + r ( n − k + r ) . Clearly, f ( n, k, r ) is the number of edges of H ( n, k, r ). The following result is due toKopylov [14] Theorem 2.5 (Kopylov [14]) . Let n ≥ k ≥ and let r = (cid:98) k − (cid:99) . If G is a 2-connected n -vertex graph with e ( G ) ≥ max { f ( n, k, , f ( n, k, r ) } , then either G has a cycle of lengthat least k , or G = H ( n, k, or G = H ( n, k, r ) . Note that when k = 5, f ( n, k,
2) = f ( n, k, r ) = 2 n −
3. So we have the followingcorollary.
Corollary 2.6.
Let n ≥ k ≥ . If G is a 2-connected n -vertex graph with e ( G ) ≥ n − ,then either G has a cycle of length at least , or G = S n, . The following theorem due to Whitney [20] characterizes the structure of 2-connectedgraphs. Given a graph H , we call P an H -path if P is nontrivial and meets H exactly inits ends. Theorem 2.7 (Whitney, 1932) . A graph is 2-connected if and only if it can be constructedfrom a cycle by successively adding H -paths to graph H already constructed. orollary 2.8. Let G be a 2-connected graph on n ≥ vertices. If G is C ≥ -saturated then G = S n, .Proof. By Theorem 2.7, G can be constructed from a cycle C by successively adding H -paths to graph H already constructed. Since G is 2-connected and contains no cycle oflength more than four, | V ( C ) | = 3 or 4. If C is of length three, denote C = u u u u ,then every H -path P must have two fixed ends, say u , u , and has length two because G is C ≥ -free. That is G = S n, with center u , u . Now suppose G contains no cycle of lengththree. Then C is an induced cycle of length four. Denote C = u u u u u . Also since G is C ≥ -free, every H -path P must have two fixed ends, say u , u , and has length two. Thatis G = K n − , with { u , u } being a part. But this contradicts to the C ≥ -saturation of G since adding the non-edge u u to G can not give rise to a cycle of length at length five.A ( r, s, t ) -cactus , denoted by T ( r, s, t ), is a connected graph such that its blocks consistof r triangles, s copies of K , and t members of { S k, : k ≥ } . Lemma 2.9.
A graph G is C ≥ -saturated if and only if(i) B ( G ) forms a matching of G ;(ii) every component of G − B ( G ) is isomorphic to K or T ( r, s, t ) for either r ≥ , s + t ≥ or r ≥ , s + t ≥ ;(iii) the center of S k, ( k ≥ ) and the vertices of K can not incident with a cut edge.Proof. (Necessity): (i) follows from Lemma 2.2 (b).To prove (ii), it is sufficient to show that every block of G is isomorphic to one of { K , K , K , S k, } . Let B be a block of G . If | V ( B ) | ≤ B is isomorphic to K , K or K by Lemma 2.2 (a), we are done. Now suppose | V ( B ) | ≥
5. Since adding a non-edge in B can only give new cycles in B , B is C ≥ -saturated too. By Corollary 2.8, B is isomorphicto S k, for some k ≥ u , u be the center of a block of S k, or a triangle and uu be a cut edge of G incidentto u . Clearly, by adding the non-edge uu just gives rise to a new block isomorphic to S k +1 , or S , , which contains no cycles of length five by Corollary 2.6, a contradiction tothe C ≥ -saturation of G .( Sufficient ): It can be checked that T ( r, s, t ) of G − B ( G ) is C ≥ -saturated. Since thecenter of S k, ( k ≥
5) and the vertices of K does not incident with a cut edge, it also canbe checked that adding a non-edge between two components also gives rise to a cycle oflength at least 5.By Lemma 2.9, we have Sat ( n, C ≥ ) = { G : | V ( G ) | = n and G satisfies (i), (ii) and (iii) of Lemma 2.9 } . G , a leaf of G is a vertex of degree one. Lemma 2.10. sat ( n, C ≥ ) ≥ (cid:100) ( n − (cid:101) for n ≥ .Proof. Let G be a C ≥ -saturated graph on n vertices. Let b ( G ) , b ( G ) , b ( G ) and b ( G )denote the number of blocks isomorphic to K , K , K and S k, ( k ≥
5) in G . By Lemma 2.3,each component of G − B ( G ) is isomorphic to K or T ( r, s, t ). Let C ( G ) be the set of allcomponents of G − B ( G ). For a component H ∈ C ( G ) isomorphic to some T ( r, s, t ), wehave | V ( H ) | = 1 + 2 r + 3 s + t (cid:80) i =1 k i , where r = b ( H ), s = b ( H ) and t = b ( H ). Since B ( G )is a matching, we also have the number of components in G − B ( G ) is b ( G ) + 1. So n = | V ( G ) | = (cid:88) H ∈ C ( G ) | V ( H ) | = b ( G ) + 1 + 2 b ( G ) + 3 b ( G ) + b ( G ) (cid:88) i =1 ( k i − | E ( G ) | = b ( G ) + 3 b ( G ) + 6 b ( G ) + b ( G ) (cid:88) i =1 (2 k i − b ( G ) ≤ b ( G ) + b ( G ) (cid:88) i =1 ( k i − . Therefore, | E ( G ) | = b ( G ) + 3 b ( G ) + 6 b ( G ) + b ( G ) (cid:88) i =1 (2 k i − b ( G ) + 2 b ( G ) + 3 b ( G ) + b ( G ) (cid:88) i =1 ( k i − + 17 − b ( G ) + b ( G ) + 12 b ( G ) + b ( G ) (cid:88) i =1 (4 k i − ≥
107 ( n −
1) + 17 b ( G ) + b ( G ) (cid:88) i =1 ( k i − ≥
107 ( n − C ≥ -saturated graph in Sat ( n, C ≥ )with exactly (cid:100) ( n − (cid:101) edges for any n ≥
5. For n = 5 ,
6, let M , = H (2 , ,
0) and8 , be the graph obtained from H (0 , , ∼ = K by pending two leaves to two distinctvertices of K . For n = 7 p + 2 s + 1 with p ≥ , ≤ s ≤
3, let M ,n be the graph obtainedfrom H (0 , , p ) by pending a leaf to each vertex of its independent set, where H (0 , , p )consists of p − S , and one isomorphic to S s, . Note that H (0 , , p ) has 4( p −
1) + (4 + s ) + 1 = 4 p + s + 1 vertices and its independent set has size3( p −
1) + 3 + s = 3 p + s . Hence | V ( M ,n ) | = 4 p + s + 1 + 3 p + s = 7 p + 2 s + 1. For n = 7 p + 2 s with p ≥ , ≤ s ≤
3, let M ,n be obtained from M ,n +1 by deleting one leaf.Figure 3 shows M ,n for n = 5 , , , Sat ,n for n = 5 , , , | E ( M ,n ) | = (cid:100) n − (cid:101) and by Lemma 2.9, M ,n ∈ Sat ( n, C ≥ ). For a graph G and a subset X ⊆ V ( G ), let δ G ( X ) = min { d G ( v ) : v ∈ X } and ∆ G ( X ) =max { d G ( v ) : v ∈ X } . We write d G ( X ) = d for short if δ G ( X ) = ∆ G ( X ) = d . Let d G ( X ) = | X | (cid:80) v ∈ X d G ( v ) be the average degree of X . Let N G ( X ) be the set of neighborsof X out of X . For a graph G and two disjoint vertex sets U, W ⊂ V ( G ), let G [ U ] be thesubgraph induced by U , and G [ U, W ] be the bipartite subgraph of G with vertex classes U, W and edge set E G [ U, W ] = { uv ∈ E ( G ) : u ∈ U and v ∈ W } .The following theorem gives the structure of a C ≥ r -saturated graph for r ≥ Theorem 3.1.
Let G be a C ≥ r -saturated graph on n vertices for n ≥ r ≥ . Let X bethe set of leaves in G and X = { v ∈ V ( G ) : d G ( v ) = 3 and v ∈ N G ( X ) } and X ≥ = { v ∈ V ( G ) : d G ( v ) ≥ and v ∈ N G ( X ) } . Let X (cid:48) be the set of vertices of degree two withat least one neighbor of degree two and X be the rest of the vertices of degree two. Let C ≥ r -saturated graph for r ≥ Y = N G ( X (cid:48) ∪ X ∪ X ) \ X and Z be the set of remaining vertices in G . Then the followinghold.(i) G [ X ] , G [ X , X ∪ X (cid:48) ∪ Y ∪ Z ] , G [ X ∪ X ] , G [ X ∪ X , X (cid:48) ] , G [ X (cid:48) ∪ X ∪ X , X ≥ ∪ Z ] are all empty graphs;(ii) Both G [ X (cid:48) ] and G [ X , X ∪ X ≥ ] are perfect matchings;(iii) For each uv ∈ G [ X (cid:48) ] , there is a w ∈ Y such that w ∈ N G ( u ) ∩ N G ( v ) ;(iv) If Y, Z ∪ X ≥ (cid:54) = ∅ then E G [ Y, Z ∪ X ≥ ] (cid:54) = ∅ ;(v) For each vertex of X ∪ X , its two neighbors in Y are adjacent.(vi) Let Y be the set of isolated vertices in G [ Y ] and Y = Y \ Y . Let H = G [ Y ∪ Z ∪ X ≥ ] .Then δ H ( Y ) ≥ and d H ( Y ) ≥ .(The structure of G is shown in Figure 4.)Proof. (i). By definition of X , a component of G [ X ] is either an edge or an isolated vertex.Since G is connected and n ≥ r ≥ X must be an independent set of G .By definition of Y and Z , G [ X , Y ∪ Z ] is an empty graph. Clearly, every vertex of X is contained in a block isomorphic to K . If there exists a vertex v ∈ N G ( X ) ∩ ( X ∪ X (cid:48) ),then v is a cut vertex of G and so v is contained in two adjacent blocks each of which isisomorphic to K , a contradiction to (b) of Lemma 2.2. Therefore, G [ X , X ∪ X (cid:48) ] is empty.By definition, G [ X ] is empty. Suppose there is an edge uv ∈ E ( G [ X ∪ X ]) and u ∈ X .Let u (cid:48) be the leaf adjacent to u . Since v ∈ X ∪ X and G [ X , X ] is empty, v must havea non-leaf neighbor, say w . Then u (cid:48) w / ∈ E ( G ). Thus there is a path P of length at least r − ≥ u (cid:48) and w in G . Clearly, v / ∈ V ( P ), otherwise P = wvuu (cid:48) is of lengththree, a contradiction. So P − u (cid:48) u + uvw is a cycle of length at least r in G , a contradiction to G is C ≥ r -saturated. Therefore, G [ X ∪ X ] is an empty graph. With a similar discussion, wehave that there is no edge uv with u ∈ X (or u ∈ X ≥ ) and v ∈ X (cid:48) (or v ∈ X (cid:48) ∪ X ∪ X ).That is G [ X , X (cid:48) ] (or G [ X (cid:48) ∪ X ∪ X , X ≥ ]) is empty. Since E ( G [ X , X (cid:48) ]) is empty by10efinition, we have G [ X ∪ X , X (cid:48) ] is an empty graph. By definition, G [ X (cid:48) ∪ X ∪ X , Z ] isempty. So G [ X (cid:48) ∪ X ∪ X , X ≥ ∪ Z ] is empty too. The proof of (i) is completed.By (i), we have X ∪ X ≥ = N G ( X ) and X , X , X (cid:48) , X , X ≥ , Y, Z form a partition of V ( G ).(ii). By the definition and (b) Lemma 2.2, G [ X , X ∪ X ≥ ] is a matching. To complete(ii), we prove that ∆( G [ X (cid:48) ]) = δ ( G [ X (cid:48) ]) = 1. By definition, δ ( G [ X (cid:48) ]) ≥
1. Supposethere exists a vertex v ∈ X (cid:48) having two neighbors in X (cid:48) , say u , u . Then u u / ∈ E ( G ),otherwise, G [ { v, u , u } ] forms a component of G , a contradiction to the connectivity of G .So G contains a path P connecting u and u of length at least r − ≥
5. Clearly, v / ∈ V ( P ),otherwise, P = u vu is of length three, a contradiction. So P + u vu is a cycle in G oflength at least r + 1, a contradiction.(iii). Let uv be a component in G [ X (cid:48) ] and u (cid:48) (resp. v (cid:48) ) be the second neighbor of u (resp. v ). Then u (cid:48) , v (cid:48) ∈ Y . To complete (iii), we show that u (cid:48) = v (cid:48) . If not, then u (cid:48) v / ∈ E ( G ).Hence G contains a ( u (cid:48) , v )-path of length at least r − ≥ G . With a similar discussionas in (ii), we have u / ∈ V ( P ) and so P + u (cid:48) uv is a cycle of length at least r + 1 in G , acontradiction.(iv). If not, then G [ Z ∪ X ≥ ∪ N G ( X ≥ )] forms a component of G , which is a contradictionto Fact 1.(v). If not, then there is a w ∈ X ∪ X with N G ( w ) ∩ Y = { u, v } but uv / ∈ E ( G ).So there is a ( u, v )-path P of length at least r − G . With the same reason as in (ii), w / ∈ V ( P ). Therefore, P + uwv is a cycle of length at least r + 1 in G , a contradiction.(vi). Recall that H = G [ Y ∪ Z ∪ X ≥ ]. We first prove δ H ( Y ) ≥
2. Suppose there existsa vertex v ∈ Y with d H ( v ) ≤ d H ( v ) = 0, then by (v), E G [ v, X ∪ X ] = ∅ . So N G ( v ) ⊆ X (cid:48) . By (iii), the componentcontaining v is isomorphic to a fan F t centered at v for some t >
0. By Fact 1, G isconnected. This forces that G is isomorphic to F t . Clearly, F t is not C ≥ r -saturated for r ≥
6, a contradiction.Now suppose d H ( v ) = 1 and let N H ( v ) = { u } . By (v), N G ( v ) ∩ ( X ∪ X ) ⊆ N G ( u ) ∩ ( X ∪ X ). By (iii), N G ( v ) ∩ X (cid:48) is disjoint with N G ( u ) ∩ X (cid:48) . We first claim that N G ( v ) ∩ X (cid:48) = ∅ .If not, choose w ∈ N G ( v ) ∩ X (cid:48) . Then wu / ∈ E ( G ) because u and v have no commonneighbor in X (cid:48) . So there is a path of length at least r − ≥ u and w in G .Since the edge containing w in G [ X (cid:48) ] only connect to v in G , any path connecting w and u must pass through v . But the longest ( u, v )-path in G has length at most two (equalityholds when N G ( v ) ∩ ( X ∪ X ) (cid:54) = ∅ ) and the longest ( v, w )-path has length two, so thelongest ( u, w )-path has length at most four, a contradiction. With similar discussion, wehave N G ( u ) ∩ X (cid:48) = ∅ . Therefore, the block B containing v is isomorphic to S k, centeredat { u, v } , where k = | N G ( v ) ∩ ( X ∪ X ) | + 2. If | N G ( v ) ∩ ( X ∪ X ) | ≥ B is not11 ≥ r -saturated because adding any edge in X ∪ X gives rise to a longest cycle of length atmost 5 ≤ r − B , a contradiction to the C ≥ r -saturation of B . So | N G ( v ) ∩ ( X ∪ X ) | ≤ d G ( v ) ≤
2, which is a contradiction to d G ( v ) ≥
3. Therefore, we have δ H ( Y ) ≥ d H ( Y ) ≥ using a discharging argument. Recall that every vertexof Y has at least one neighbor in Y . Claim 1.
For any v ∈ Y with d H ( v ) = 2 , the two neighbors of v are adjacent. If not, denote N H ( v ) = { v , v } , then there is a ( v , v )-path P of length r − ≥ G . If v ∈ V ( P ), by (i), G [ X (cid:48) ∪ X ∪ X , X ≥ ∪ Z ] is empty, then the only vertices used by P are v , v , v and at most two vertices in X ∪ X , i.e. P has length at most 4 < r −
1, acontradiction. Hence, v / ∈ V ( P ). Then P + v vv is a cycle of length at least r + 1 in G , acontradiction, too. Claim 2.
For any pair of vertices u, v ∈ Y with d H ( u ) = d H ( v ) = 2 , uv / ∈ E ( G ) . Otherwise, let w be the other neighbor of v in H , then wu ∈ E ( G ) by Claim 1. Hencethe triangle T = uvwu forms a block of H . Let B be the block of G containing T . Then B is obtained from T by adding T -paths of length exactly two, each of which has ends in { u, v, w } and the internal vertex in X ∪ X . If B ∩ ( X ∪ X ) (cid:54) = ∅ , we claim that B is not C ≥ r -saturated, so we have a contradiction to Lemma 2.3. In fact, let P = uxv be a T -pathwith x ∈ X ∪ X . Then wx / ∈ E ( G ). But the longest path connecting w and x is at mostfour by the structure of B . So B is not C ≥ r -saturated. Now assume B ∩ ( X ∪ X ) = ∅ .That is B = T = uvwu . Since d G ( u ) ≥
3, by (ii), there must be an edge u u ∈ G [ X (cid:48) ] suchthat the triangle T (cid:48) = uu u u forms a block of G . Clearly, the longest path connecting anypair of nonadjacent vertices in V ( T ) ∪ V ( T (cid:48) ) has length at most 4 < r −
1, a contradiction.A vertex v ∈ Y with d H ( v ) = r (or d H ( v ) ≥ r ) is called an r -vertex (or an r + -vertex).From Claims 1 and 2, we have that for each 2-vertex v ∈ Y , either v has two adjacent3 + -neighbors in Y (we call v an inner vertex), or v has two adjacent neighbors such thatone is a 3 + -vertex in Y and the other in Z ∪ X ≥ (we call v a boundary vertex). Claim 3.
Every + -vertex v ∈ Y has at most d H ( v ) − neighbors of degree two in Y . Suppose v ∈ Y is a 3 + -vertex adjacent to r vertices of degree two in Y . Let v , . . . , v r be the 2-vertices in Y adjacent to v and u , . . . , u r be their other neighbors so that u i isadjacent to v i for i = 1 , . . . , r . By Claims 1 and 2, u , . . . , u r ∈ N H ( v ) and v , . . . , v r areindependent in H . Hence d H ( v ) ≥ r + 1, the equality holds if and only if u = · · · = u r . Claim 4. No -vertex in Y is adjacent to two boundary vertices in Y . If not, suppose that there is a 3-vertex adjacent to two boundary vertices v , v ∈ Y .By Claim 3, v, v , v have a common neighbor u ∈ Z ∪ X ≥ . Hence u is a cut vertex12eparating v, v , v and the other vertices of Z ∪ X ≥ (if Z ∪ X ≥ (cid:54) = ∅ ). By definition, v , v are 2-vertices. By Claim 2, v v / ∈ E ( G ). Hence there is a ( v , v )-path P of length at least r − G . Let B be the block of G containing { v, v , v , u } . By (i) and (v), v ∈ V ( P ) andthe length of P is at most 4 < r −
1, a contradiction.For each v ∈ Y , define its initial charge as ch ( v ) = d H ( v ) − . Then (cid:88) v ∈ Y ch ( v ) = (cid:88) v ∈ Y d H ( v ) − | Y | . Hence to show d H ( Y ) ≥ , it is sufficient to show (cid:80) v ∈ Y ch ( v ) ≥
0. Now we redistributethe charges according to the following rules.(R1) Every 3 + -vertex v ∈ Y gives to each of its incident inner vertex in Y .(R2) Every 3 + -vertex v ∈ Y gives to each of its incident boundary vertex in Y .We proceed to derive that each vertex v ∈ Y ends up with a nonnegative final charge ch (cid:48) ( v ).For a 2-vertex v ∈ Y , if v is an inner vertex, by Claim 2, v has two 3 + -neighbors in Y .Hence by (R1), v receives at least 2 × = from its 3 + -neighbors. If v is a boundary vertex,by (R2), v receives at least from its 3 + -neighbor. So the final charge ch (cid:48) ( v ) = 2 − + = 0.For a 3-vertex v ∈ Y , by Claim 4, if v is adjacent to a boundary vertex then v has noother neighbor of degree two, so v gives to its boundary neighbor. If v is not adjacent toboundary vertex then, by Claim 3, v has at most two neighbors of degree two, so v gives atmost 2 × to its neighbors. Therefore, the final charge ch (cid:48) ( v ) = 3 − − = 0.For a 4 + -vertex v ∈ Y , by Claim 3, v has at most d H ( v ) − v gives at most ( d H ( v ) −
1) to its neighbors of degree two. So the finalcharge ch (cid:48) ( v ) = d H ( v ) − − ( d H ( v ) −
1) = d H ( v ) − ≥ (cid:88) v ∈ Y ch ( v ) = (cid:88) v ∈ Y ch (cid:48) ( v ) ≥ . This completes the proof of (vi).
Corollary 3.2.
Let G be a C ≥ r -saturated graph on n vertices for n ≥ r ≥ . X , X , X (cid:48) , X , X ≥ , Y , Y , Z are defined the same as in Theorem 3.1 and let x = | X | , x = | X | , x (cid:48) = | X (cid:48) | , x = | X | , x = | X ≥ | , y = | Y | , z = | Z | and y = | Y | . We have(a) x = x + x and n = x + x (cid:48) + 2 x + 2 x + y + z ;(b) y ≤ x (cid:48) and y ≤ x + 2 x + x (cid:48) ;(c) if x + x = 0 and G [ Y ∪ Z ∪ X ≥ ] is a complete graph, then z + x + y = r − ;otherwise, x + x + x (cid:48) ≤ n − r and x + 2 x + z − x (cid:48) ≥ r − n .Proof. (a) follows directly from (ii) of Theorem 3.1.13b) By (v) of Theorem 3.1, N G ( X ∪ X ) ∩ Y ⊆ Y . Hence N G ( Y ) ⊆ X (cid:48) . By (ii), (iii)of Theorem 3.1 and the double-counting method, 2 y = 2 | Y | ≤ | E G ( Y , X (cid:48) ) | ≤ | X (cid:48) | = x (cid:48) .Similarly, we have y − y = | Y | ≤ | E G ( X ∪ X , Y ) | = 2 | X ∪ X | = 2 x + 2 x . So y ≤ x + 2 x + x (cid:48) .(c) If x + x = 0 and G [ Y ∪ Z ∪ X ≥ ] is a complete graph, then G is obtained fromthe complete graph K y + z + x by attaching leaves to X ≥ and K ’s to Y . It is easy to checkthat this graph G is C y + z + x +1 -saturated, which implies y + z + x = r − x + x = 0 but G [ Y ∪ Z ∪ X ≥ ] is not a complete graph, then any pair of nonadjacentvertices in Y ∪ Z ∪ X ≥ are connected by a path of length at least r − G . Obviously,all of the vertices in this path are in Y ∪ Z ∪ X ≥ , which implies y + z + x ≥ r . Note that n = z + y + x (cid:48) + 2 x . So x + x (cid:48) + x = x (cid:48) + x ≤ n − r .Now suppose x + x (cid:54) = 0. Denote H = G [ Y ∪ Z ∪ X ≥ ∪ X ∪ X ]. Since every vertex in X ∪ X has degree exactly two in H , H is not a complete graph if y + z + x + x + x ≥ y + z + x + x + x ≤
3, since each vertex in X ∪ X has two neighbors in Y , y ≥ y = 2, x + x = 1, and z + x = 0. Therefore, G is isomorphic to a t -triangle T t for some t ≥ x = 1) or is the graph obtained from T t by attaching one leaf to thevertex in X (for x = 1). By Lemma 2.3, G is C ≥ -saturated but not C ≥ -saturated, acontradiction to r ≥
6. Hence G [ Y ∪ Z ∪ X ≥ ∪ X ∪ X ] is not a complete graph. So anypair of nonadjacent vertices is connected by a path P of length at least r − G . By (i)and (iii) of Theorem3.1, V ( P ) ⊆ Y ∪ Z ∪ X ≥ ∪ X ∪ X . Therefore, y + z + x + x + x ≥ r .Note that n = ( y + z + x + x + x ) + x + x + x (cid:48) . So x + x + x (cid:48) ≤ n − r . By (b), wehave 3 x + 2 x + z − x (cid:48) ≥ y + z + x − x (cid:48) = n − x + x + x (cid:48) ) ≥ r − n .Now we prove (1) of Theorem 1.3. Corollary 3.3.
Let G be a C ≥ r -saturated graph on n vertices for n ≥ r ≥ . Then e ( G ) ≥ n .Proof. Let X , X , X (cid:48) , X , X ≥ , Y , Y , Z are defined the same as in Theorem 3.1 and let x , x , x (cid:48) , x , x , y, z and y defined as in Corollary 3.2. Let H = G [ Z ∪ Y ∪ X ≥ ]. By (i)and (vi) of Theorem 3.1, δ H ( X ≥ ) ≥ − δ H ( Z ) ≥ δ H ( Y ) ≥ d H ( Y ) ≥ . So e ( H ) = 12 (cid:88) v ∈ V ( H ) d H ( v ) ≥ (cid:18) | X ≥ | + 3 | Z | + 2 | Y | + 52 | Y | (cid:19) = 12 (cid:18) x + 3 z + 2 y + 5( y − y )2 (cid:19) = 3 x z y − y .
14y (ii), (iii) and (v) of Theorem 3.1, e G ( X ∪ X ≥ , X ) = x + x , e ( G [ x (cid:48) ]) = x (cid:48) , e G ( Y, X ∪ X ) = 2( x + x ) and e G ( Y, X (cid:48) ) = x (cid:48) . By Theorem 3.1, we have e ( G ) = e ( H ) + e G ( X ∪ X ≥ , X ) + e G ( Y, X ∪ X ) + e ( G [ x (cid:48) ]) + e G ( Y, X (cid:48) ) (2)= e ( H ) + ( x + x ) + 2( x + x ) + 32 x (cid:48) ≥ z + 54 y − y + 52 x + 3 x + 2 x + 32 x (cid:48) = 54 n + 14 ( z + 3 x + x (cid:48) + 2 x − y ) ≥ n + 18 (2 z + 6 x + x (cid:48) + 4 x ) (3) ≥ n, the fifth inequality holds because y ≤ x (cid:48) by (b) of Corollary 3.2.The following corollary complete (2) of Theorem 1.3. Corollary 3.4.
Let G be a C ≥ r -saturated graph on n vertices for some r ≥ and n ≤ r ≤ n .Then e ( G ) ≥ n + r .Proof. If x + x = 0 and G [ Y ∪ Z ∪ X ≥ ] is a complete graph, then y + z + x = r − ≥ e ( G ) = | E ( G [ Y ∪ Z ∪ X ≥ ]) | + ( x + x ) + 32 x (cid:48) = (cid:18) r − (cid:19) + x + 32 x (cid:48) = (cid:18) r − (cid:19) + 12 x (cid:48) + n − ( r − n + 12 ( r − r + 4) + 12 x (cid:48) ≥ n + r x (cid:48) ≥ n + r , where the third equality holds since n = ( z + x + y ) + x + x (cid:48) = r − x + x (cid:48) and thefifth inequality holds since r ≥ x + x (cid:54) = 0 or G [ Y ∪ Z ∪ X ≥ ] is not a complete graph. Then x + x + x (cid:48) ≤ n − r and 3 x + 2 x + z − x (cid:48) ≥ r − n by (c) of Corollary 3.2. Let A = y − (2 x + 2 x + x (cid:48) ), B = ( x + x + x (cid:48) ) − ( n − r ) and C = (2 r − n ) − (3 x + 2 x + z − x (cid:48) ). Then B, C ≤ e G ( Y, X (cid:48) ∪ X ∪ X ) by the double-counting argument, we have A ≤
0. Thus, by n ≤ r ≤ n , we get (cid:18) r n − (cid:19) A + (cid:18) rn − (cid:19) B + (cid:18) − r n (cid:19) C ≤ e ( G ) ≥ e ( G ) + (cid:18) r n − (cid:19) A + (cid:18) rn − (cid:19) B + (cid:18) − r n (cid:19) C = e ( G ) + (cid:18) r n − (cid:19) (cid:0) z + y + x (cid:48) + x + 2 x + 2 x (cid:1) + x (cid:48) − x − x − z ≥ n + 18 (cid:0) z + 6 x + x (cid:48) + 4 x (cid:1) + (cid:18) r n − (cid:19) n + 18 (cid:0) x (cid:48) − x − x − z (cid:1) ≥ n + r x (cid:48) ≥ n + r In this section, we construct maximally C ≥ r -saturated graphs that achieve the bounds statedin Theorem 1.4. Our constructions are based on the constructions of the least maximallynonhamiltonian graphs given in [7, 8, 15, 18]. Bollob´as [3] posed the problem of findingsat( n, C n ). Bondy [3] has shown that sat( n, C n ) ≥ (cid:100) n (cid:101) for n >
7. In [7, 8, 15], theauthors completely determined that sat( n, C n ) = (cid:100) n (cid:101) by constructing the least maximallynonhamiltonian graphs. These constructions came from appropriate modifications of afamily of well-known snarks, Isaacs’ flower snarks. Let J k be the Isaacs’ flower snark on 4 k vertices with k = 2 p + 1 and p ≥
7, and for a vertex v ∈ V ( J k ), J k ( v ) denotes the graphobtained from J k by expanding v to a triangle and for an edge uv ∈ E ( J k ), J k ( uv ) denotesthe graph obtained from J k by replacing the edge uv by a bowtie (i.e. an F in this paper),detailed definitions can be found in [18] (Definitions 1, 2 and 3) and the appendix of thispaper). The following table lists the optimal C n -saturated graphs for all n , where Clark etal [7, 8] gave the construction for n = 8 p, p + 2 , p + 4 and 8 p + 6, and the optimality of theother cases can be verified from Proposition 2 in [18] (we also include this in Appendix).We define an almost 3-regular graph is a graph with all vertices of degree three but one,say u , of degree four with the property that the neighborhood N G ( u ) induces a perfectmatching in G , say { u u , v v } , such that u , u (resp. v , v ) have distinct neighbors outof { u } ∪ N G ( u ). Note that G [ N G ( u ) ∪ { u } ] ∼ = F by the definition. A barbell is a graphobtained from two disjoint triangles by adding a new edge connecting them. For simplify,16rder construction order construction8 p J k − ( v , v ) 8 p + 1 J k − ( v )( v v )8 p + 2 J k − ( v , v , v ) 8 p + 3 J k − ( v , v )( v v )8 p + 4 J k p + 5 J k − ( v , v , v )( v v )8 p + 6 J k ( v ) 8 p + 7 J k ( v v )Table 1: The optimal C n -saturated graphs of order n .we call a 3-regular (or an almost 3-regular) graph containing no barbell as a subgraph agood graph . By the definitions of J k , J k ( v ) and J k ( uv ), we can check that all optimal graphsconstructed in the above table are good. So we have Lemma 4.1.
For any r ≥ , there exists a C r -saturated good graph G on r vertices and e ( G ) = (cid:100) r (cid:101) . We also need the following property of C r -saturated graph on r vertices. Lemma 4.2.
Let G be a C r -saturated good graph on r ≥ vertices. Then every edge e ∈ E ( G ) is contained in a cycle of length r − in G .Proof. Suppose there exists an edge e = u v ∈ E ( G ) which is not contained in any cycleof length r − Case 1: d G ( u ) = d G ( v ) = 3.Let N G ( u ) = { v , a , a } and N G ( v ) = { u , b , b } .Suppose a a , b b ∈ E ( G ). If |{ a , a } ∪ { b , b }| = 2, then { a , a } = { b , b } and G [ { v , u , a , a } ] is isomorphic to K . Since G is connected and r ≥ G must be analmost 3-regular graph and the unique 4-vertex is in { a , a } . But this is impossible since G [ N G ( a i ) ∪ { a i } ] can not be a 2-fan for i = 1 ,
2. If |{ a , a } ∪ { b , b }| = 3, without lossof generality, let a = b and a (cid:54) = b , then a is a 4-vertex in G and so G must be analmost 3-regular graph. Note that N G ( a ) = { a , b , v , u } . So G [ N G ( a ) ∪ { a } ] (cid:29) F , acontradiction. So |{ a , a } ∪ { b , b }| = 4. But G [ { u , a , a , v , b , b } ] induces a barbell in G , a contradiction.Now suppose one of a a , b b is not an edge in G . Without loss of generality, assume a a / ∈ E ( G ). Then G contains a Hamiltonian ( a , a )-path P . So u is an internal vertexof P . We claim that u v ∈ E ( P ). If not, then u a , u a ∈ E ( P ) and so P = a u a ,a contradiction. Thus u v ∈ E ( P ). Since one of u a , u a is contained in P , withoutloss of generality, assume u a ∈ E ( P ). Hence P − u a is a ( u , a )-path on vertex set V ( G ) \ { a } . Since a u / ∈ E ( P ), P − u a + u a is a cycle of length r − u v ,a contradiction. 17 ase 2: One of { u , v } is a 4-vertex.Without loss of generality, assume d G ( u ) = 4 and N G ( u ) = { u , u , v , v } with u u , v v ∈ E ( G ). Let N G ( v ) = { u , v , a } . By the definition of the almost 3-regulargraph, av / ∈ E ( G ). Hence G contains a Hamiltonian path connecting a and v . If u v / ∈ E ( P ), then P = av v is of length 2, a contradiction. Thus u v ∈ E ( P ). Since thereis another one of v v , v a contained in P , without loss of generality, assume v v ∈ E ( P ).Then v a / ∈ E ( P ). Hence P − v v + v a is a cycle on r − u v in G ,a contradiction.Let G and H be two distinct graphs and v ∈ V ( G ). We attach H to v means thatwe identify a vertex of H and v to obtain a new graph. Let U, W be two disjoint subsetsof V ( G ). We define L ( G ; U, W ) be the graph obtained from G by attaching a K to eachvertex of U and attaching a K to each vertex of W . For two graphs G , G , let v i ∈ V ( G i )be a leaf and u i ∈ V ( G i ) be its support vertex in G i for i = 1 ,
2, define C ( G , G ; u u )be the graph obtained from G , G by deleting v , v and adding a new edge u u . Let G , G , . . . , G k be a sequence of graphs. We recursively define C ( G , ..., G k ; u v , . . . , u k v k ) = C ( C ( G , ..., G k − ; u v , . . . , u k − v k − ) , G k ; u k v k ) . Let M r,r be a C r -saturated good graph. We define M r,n as follows: • If r ≤ n ≤ r , define M r,n = L ( M r,r ; U, ∅ ), where U ⊂ V ( M r,r ) and | U | = n − r ; • if 2( k − r − k − < n < k − r for some k ≥
2, define G = C ( G , ..., G k − ; u v , . . . , u k − v k − ) , where G i = L ( M ir,r ; U i , V i ) and M ir,r are pairwise disjoint copies of M r,r , U i ( (cid:54) = ∅ )and V i form a partition of V ( M ir,r ) with (cid:80) k − i =1 | V i | = n − k − r + 2( k − v i − , u i ∈ U i and v i − (cid:54) = u i for 1 ≤ i ≤ k − • if k − r ≤ n ≤ kr − k −
1) for some k ≥
2, define M r,n = C ( G , ..., G k ; u v , . . . , u k − v k − ) , where G i = L ( M ir,r ; U i , ∅ ) and M ir,r are pairwise disjoint copies of M r,r , U i (cid:54) = ∅ aresubsets of V ( M ir,r ) with (cid:80) k − i =0 | U i | = n − kr + 2( k − v i − , u i ∈ U i and v i − (cid:54) = u i for 1 ≤ i ≤ k . Remark.
In the following proof, u i will be viewed as a leaf of G i +1 with support v i and v i a leaf of G i with support vertex u i for 1 ≤ i ≤ k . Proposition 1.
For n ≥ r ≥ , M r,n is C ≥ r -saturated graph. roof. It is sufficient to show that C ( G , . . . , G k ; u v , . . . , u k − v k − ) is C ≥ r -saturated for k ≥
1, where G i = L ( M ir,r ; U i , V i ) and M ir,r are pairwise disjoint copies of M r,r , U i ( (cid:54) = ∅ ) and V i are disjoint subsets of V ( M ir,r ), and v i − , u i ∈ U i with v i − (cid:54) = u i for 1 ≤ i ≤ k − H = C ( G , . . . , G k ; u v , . . . , u k − v k − ). By definition, the blocks of H are isomor-phic to M r,r , K , or K . So H is C ≥ r -free since M r,r is C r -saturated. Now we prove thatfor any a, b ∈ V ( H ) with ab / ∈ E ( H ), H contains an ( a, b )-path on at least r vertices. Case 1: a, b ∈ V ( G i ) for some 1 ≤ i ≤ k .Without loss of generality, assume i = 1. If a, b ∈ V ( M r,r ), we are done since M r,r is C r -saturated. If a ∈ V ( M r,r ) but b is not, then b has a neighbor, say b (cid:48) , in V ( M r,r ). If ab (cid:48) ∈ E ( M r,r ) then ab (cid:48) is contained in a cycle C on r − M r,r by Lemma 4.2.Thus, C − ab (cid:48) + bb (cid:48) is an ( a, b )-path on r vertices in L ( M r,r ; U , V ). If ab (cid:48) / ∈ E ( M r,r ) then M r,r contains a ( a, b (cid:48) )-path P on r vertices. Thus P + bb (cid:48) is an ( a, b )-path on r + 1 verticesin L ( M r,r ; U , V ). If a, b / ∈ V ( M r,r ) then a, b have two different neighbors in V ( M r,r ) (thisis because ab / ∈ E ( G )). If a (cid:48) b (cid:48) ∈ E ( M r,r ) then M r,r contains a cycle C on r − a (cid:48) b (cid:48) . Thus C − a (cid:48) b (cid:48) + a (cid:48) a + b (cid:48) b is an ( a, b )-path on r + 1 vertices in L ( M r,r ; U , V ),we are done. If a (cid:48) b (cid:48) / ∈ E ( M r,r ) then M r,r contains an ( a (cid:48) , b (cid:48) )-path P on r vertices. So P + a (cid:48) a + b (cid:48) b is an ( a, b )-path on r + 2 vertices in L ( M r,r ; U , V ). Case 2: a ∈ V ( G i ) and b ∈ V ( G j +1 ) for some 1 ≤ i ≤ j ≤ k − H is connected and u j v j is a cut edge, there exists an ( a, u j )-path P from a to u j containing no vertices in V ( M j +1 r,r ). If b (cid:54) = v j , then u j b / ∈ E ( H ) ∩ E ( G j +1 ). Hence G j +1 contains a ( u j , b )-path P on at least r vertices by Case 1. Hence P + P (cid:48) is an ( a, b )-pathon at least r vertices in H . If b = v j , we may also assume a = u i by symmetry, whichimplies i < j . Let P be a ( u i , v j − )-path P containing no vertices in V ( M jr,r ) \ { v j − } .Since v j − (cid:54) = u j , v j − v j / ∈ E ( G j ). Again from Case 1, G j contains a ( v j − , v j )-path P onat least r vertices. Hence P + P is an ( a, b )-path on at least r vertices in H . Proposition 2.
For n ≥ r ≥ , M r,n is C ≥ r -saturated with e ( M r,n ) = g ( rn ) n + O ( nr ) .Furthermore, if r ≤ n ≤ r , we have sat ( n, C ≥ r ) = n + (cid:100) r (cid:101) .Proof. By Theorem 4.1 and Proposition 1, we know that M r,n is indeed C ≥ r -saturated.In the following, we check the order and the number of edges of M r,n . If r ≤ n ≤ r , | V ( M r,n ) | = r + | U | = n and e ( M r,n ) = (cid:100) r (cid:101) + n − r = n + (cid:100) r (cid:101) . Combining with sat( n, C ≥ r ) ≥ n + r , we have sat( n, C ≥ r ) = n + (cid:100) r (cid:101) . If 2( k − r − k − < n < k − r for some k ≥ | V ( M r,n ) | = ( k − r + k − (cid:88) i =0 | U i | + 2 k − (cid:88) i =0 | V i | − k −
2) = n e ( M r,n ) = ( k − e ( M r,r ) + k − (cid:88) i =0 | U i | + 3 k − (cid:88) i =0 | V i | − ( k − k − (cid:24) r (cid:25) + ( k − r + 2 ( n − k − r + 2( k − − ( k − n − ( k − (cid:22) r (cid:23) + 3( k − g (cid:16) rn (cid:17) n + O (cid:16) nr (cid:17) . If k − r ≤ n ≤ kr − k −
1) for some k ≥
2, by definition, | V ( M r,n ) | = kr + k − (cid:88) i =0 | U i | − k −
1) = n and e ( M r,n ) = ke ( M r,r ) + k − (cid:88) i =0 | U i | − ( k − k (cid:24) r (cid:25) + n − kr + 2( k − − ( k − n + k (cid:108) r (cid:109) + ( k − g (cid:16) rn (cid:17) n + O (cid:16) nr (cid:17) . It is obvious that the Tur´an function has monotonicity, i.e., ex( n, F ) ≥ ex( n, F ) for F ⊆ F . But the saturation number does not have this property. For example, from theresult of F¨uredi and Kim [13] and (1) of Theorem 1.3, for r ≥ n , wehave sat( n, C r ) < (1 + 1 r − n + (cid:18) r − (cid:19) < n ≤ sat( n, C ≥ r ) . In this paper, we determine the exact values of sat( n, C ≥ r ) for r = 3 , , n ≤ r ≤ n .From the image of g ( x ), we guess that sat( n, C ≥ r ) does not have monotonicity with respectto r too. It is also an interesting question to determine the exact values of sat( n, C ≥ r ) forthe other cases of r . 20 eferences [1] C. A. Barefoot, L. H. Clark, R. C. Entringer, T. D. Poter, L. A. Sz´ekely, and Z.Tuza, Cycle-saturated graphs of minimum size. Discrete Math. 150(1-3)(1996), 31-48.Selected papers in honour of Paul Erd˝os on occasion of his 80th birthday (Keszthely,1993).[2] B. Bollob´as, Extremal Graph Theory, Academic Press, Inc, London, New York, 1978.[3] J. A. Bondy, Variations on the hamiltonian theme. Canad. Math. Bull. 15 (1972),57-62.[4] J. A. Bondy and M. Simonovits, Cycles of even length in graphs, J. Combin. TheorySer. B 16 (1974), 97-105.[5] Y.-C. Chen, Minimum C -saturated graphs, J. Graph Theory 61(2) (2009), 111-126.[6] Y.-C. Chen, All minimum C -saturated graphs, J. Graph Theory 67(1)(2011), 9-26.[7] L. H. Clark, R.C. Entringer, Smallest maximally nonhamiltonian graphs, PeriodicaMathematica Hungarica 14 (1983), 57-68.[8] L. H. Clark, R. C. Entringer, H. D. Shapiro, Smallest maximally nonhamiltonian graphsII. Graph Combin. 8(3) (1992), 225-231.[9] G. A. Dirac, Some theorems on abstract graphs, Proc. London Math. Soc. (3) 2 (1952),69-81.[10] P. Erd˝os, T. Gallai, On maximal paths and circuits of graphs, Acta Math. Acad. Sci.Hungar. 10 (1959), 337-356.[11] P. Erd˝os, A. Hajnal, and J. W. Moon, A problem in graph theory, Amer. Math.Monthly, 71 (1964), 1107-1110.[12] D. C. Fisher, K. Fraughnaugh, and L. Langley, P -connected graphs of minimum size,Ars Comin. 47 (1997), 299-306.[13] Z. F¨uredi and Y. Kim, Cycle-saturated graphs with minmum number of edges, J. GraphTheory 73(2) (2013), 203-215.[14] G. N. Kopylov, Maximal paths and cycles in a graph, Dokl. Akad. Nauk SSSR 234(1977), 19-21; English translation: Sov. Math., Dokl. 18(3) (1977), 593-596.[15] X. Lin, W. Jiang, C. Zhang, and Y. Yang, On smallest maximally non-Hamiltoniangraphs, Ars Combin. 45 (1997), 263-270.2116] L. T. Ollmann, K , saturated graphs with minimal number of edges, Proceedingsof Third Southeastern Conference on Combinatorics, Graph Theory, and Computing,Florida Atlantic University, Boca Raton, FL, 1972, pp. 367-392.[17] M. Simonovits, Extremal graph problems with symmetrical extremal graphs, additionalchromatic conditions, Discrete Math. 7 (1974), 349-376.[18] L. Stacho, Non-isomorphic smallest maximally non-Hamiltonian graphs, Ars Combi-natoria, 48 (1998) 307-317.[19] Z. Tuza, C -saturated graphs of minimum size, Acta Univ. Carolin. Math. Phys. 30(2)(1989), 161-167.[20] H. Whitney, Non-separable and planar graphs, Transactions of the American Mathe-matical Society 34 (1932), 339-362. In this section, we revisit the definition of Isaacs’ snarks J k for odd k ≥ J k used in Table 1. Definition 1.
For odd k ≥ , the Issacs’ snark J k is a -regular graph on vertex set V = { v , v , ..., v k − } with edge set E = k − (cid:91) j =0 { v j v j +1 , v j v j +2 , v j v j +3 , v j +1 v j +7 , v j +2 v j +6 , v j +3 v j +5 } in the sense of module k . The first modification is called ”replacing a vertex of degree 3 with a triangle”.
Definition 2.
For a graph G with a vertex v of degree , let N G ( v ) = { x, y, z } . Then G ( v ) is a new graph with vertex set V ( G ( v )) = ( V ( G ) \{ v } ) ∪ { u , u , u } where u , u , u / ∈ V ( G ) and edge set E ( G ( v )) = ( E ( G ) \{ vx, vy, vz } ) ∪ { u u , u u , u u , u x, u y, u z } . If we apply the operations ”replacing a vertex of degree 3 with a triangle” successivelyon vertices v , v , . . . , v k , we write G ( v , v , ..., v k ) for G ( v )( v ) ... ( v k ) for short.The second modification is ”replacing an edge with an F ” (is called a bowtie in [18]).22igure 5: Flower snark J . Definition 3.
For a graph G with uv ∈ E ( G ) where d G ( u ) = d G ( v ) = 3 , let N G ( u ) = { v, x u , y u } and N G ( v ) = { u, x v , y v } . Let F be a copy of F on vertex set { x , x , y , y , z } with edge set { x y , x y , zx , zx , zy , zy } , where V ( F ) ∩ V ( G ) = ∅ . Then G ( uv ) is a newgraph with vertex set V ( G ( uv )) = ( V ( G ) \{ u, v } ) ∪ V ( F ) and edge set E ( G ( uv )) = ( E ( G ) \{ uv, ux u , uy u , vx v , vy v } ) ∪ E ( F ) ∪ { x x u , y y u , x x v , y y v } . With these definitions, it is easy to see that the constructions in Table 1 are all goodgraphs with r vertices and (cid:100) r (cid:101) edges for r ≥
56. We show that the graphs constructedin Figure 1 when n = 8 p + 1 , p + 3 , p + 5 , p + 7 are optimal by Proposition 2 given byStacho [18]. Proposition 3 ([18]) . The graph G = J k ( v i +2 , . . . , v i m +2 )( v i m +1 v i m +1 +2 ) is a C r -saturated for k ≥ odd, where r = | V ( G ) | = 4 k + 2 m + 3 , m ≥ , ≤ i (cid:96) ≤ k − for (cid:96) = 1 , ..., m + 1 and the distance d J k ( v i (cid:96) +2 , v i p +2 ) ≥ for any (cid:96) (cid:54) = p . When p ≥ k = 2 p + 1, it is not difficult to show that the distance of anypair from v , v , v , v in J k − is at least 3. In fact, by induction, we can show that d J k − ( v a +2 , v b +2 ) = min {| b − a | , k − − | b − a |} . Then by Proposition 3, the graphsconstructed in Table 1 are optimal for n = 8 p + 1 , p + 3 , p + 5 , pp