aa r X i v : . [ m a t h . C O ] N ov On the pancyclicity of digraphs with largesemi-degrees
S. Kh. Darbinyan
Institute for Informatics and Automation Problems, Armenian National Academy of Sciences,P. Sevak 1, Yerevan 0014, ArmeniaEmail: samdarbin @ ipia.sci.am
Abstract
Let D be an directed graph on p ≥
10 vertices with minimum degree at least p − p/ −
1. We present a detailed proof of the following result [13]: The digraph D ispancyclic, unless some extremal cases (which are characterized).Keywords: Digraphs; semi-degrees; cycles; Hamiltonian cycles; pancyclic digraphs. Introduction and Notation
Ghouila-Houri [18] proved that every strong digraph on p vertices with minimum degree at least p is hamiltonian. There are many extentions of this theorem for digraphs and orgraphs. In particular, inmany papers, various degree conditions have been obtained for digraphs (orgraphs) to be hamiltonianor pancyclic or vertex pancyclic (see e.g. [2]-[33]). C. Thomassen [31] proved that any digraph on p = 2 m + 1 vertices with minimum semi-degree at least m is hamiltonian unless some extremal cases,which are characterized. In [9], we proved that if a digraph D satisfies the conditions of this Tomassen’stheorem, then D also is pancyclic (the extremal cases are characterized). For additional information onhamiltonian and pancyclic digraphs see the book [1] by B. Jenssen and G. Gutin.In this paper we present a detailed proof of the following result.Every digraph D (unless some extremal cases) on p ≥
10 vertices with minimum degree at least p − p/ − D contains cycles of length 3, 4, p − p = 2 m , then D also is hamiltonian.In this paper we shall consider finite digraphs without loops and multiple arcs. For a digraph D ,we denote by V ( D ) the vertex set of D and by A ( D ) the set of arcs in D . Sometimes we will write D instead of A ( D ) and V ( D ). If xy is an arc of D , then we say that x dominates y and y is dominatedby x . For subsets A and B ⊂ V ( D ) we define A ( A → B ) as the set { xy ∈ A ( D ) /x ∈ A, y ∈ B } and A ( A, B ) = A ( A → B ) ∪ A ( B → A ). If x ∈ V ( D ) and A = { x } we write x instead of { x } . Fordisjoint subsets A and B of V ( D ) A → B means that every vertex of A dominates every vertex of B .If C ⊂ V ( D ), A → B and B → C , then we write A → B → C . The outset of vertex x is the set O ( x ) = { y ∈ V ( D ) /xy ∈ A ( D ) } and I ( x ) = { y ∈ V ( D ) /yx ∈ A ( D ) } is the inset of x . Similarly, if A ⊆ V ( D ) then O ( x, A ) = { y ∈ A/xy ∈ A ( D ) } and I ( x, A ) = { y ∈ A/yx ∈ A ( D ) } . The out-degreeof x is od ( x ) = | O ( x ) | and id ( x ) = | I ( x ) | is the in-degree of x . Similarly, od ( x, A ) = | O ( x, A ) | and id ( x, A ) = | I ( x, A ). The degree of the vertex x in D defined as d ( x ) = id ( x ) + od ( x ). The subdigraphof D induced by a subset A of V ( D ) is denoted by h A i . The path ( respectively, the cycle ) consistingof the distinct vertices x , x , . . . , x n ( n ≥
2) and the arcs x i x i +1 , i ∈ [1 , n −
1] ( respectively, x i x i +1 , i ∈ [1 , n − x n x ), is denoted x x . . . x n (respectively, x x . . . x n x ). The cycle on k vertices isdenoted C k . For a cycle C k = x x . . . x k x , the indices considered modulo k , i.e., x s = x i for every s and i such that i ≡ s mod k , and we denote by C k [ x i , x j ] := x i x i +1 . . . x j ( C k [ x i , x j ] is a path for x i = x j ).1wo distinct vertices x and y are adjacent if xy ∈ A ( D ) or yx ∈ A ( D ) (or both), i.e. x is adjacentwith y and y is adjacent with x . Notation A ( x, y ) = ∅ (respectively, A ( x, y ) = ∅ ) denote that x and y areadjacent (respectively, are not adjacent).For an undirected graph G , we denote by G ∗ symmetric digraph obtained from G by replacing everyedge xy with the pair xy , yx of arcs. K n (respectively, K n,m ) denotes the complete undirected graph on n vertices (respectively, undirected complete bipartite graph, with partite sets of cardinalities n and m ),and K n denotes the complement of K n .If G and G are undirected graphs, then G ∪ G is the disjoint union of G and G . The join of G and G , denoted by G + G , is the union of G ∪ G and of all the edges between G and G .For integers a and b , let [ a, b ] denote the set of all integers which are not less than a and are notgreater than b . If I = [ a, b ] then we denote by a := lef t { I } and b := right { I } .We refer the reader to J.Bang-Jensens and G.Gutin’s book [1] for notations and terminology notdefined here. Preliminary Results . Lemma 1 ([21]). Let D be a digraph on p ≥ C n , n ∈ [2 , p −
1] and let x / ∈ C n . If d ( x, C n ) ≥ n + 1, then D contains a cycle C k for every k ∈ [2 , n + 1].The following Lemma will be used often in the proofs our results. Lemma 2 ([6]). Let D be a digraph on p ≥ P := x x . . . x n , n ∈ [2 , p − x be a vertex not contained in this path. If one of the following holds:(i) d ( x, P ) ≥ n + 2;(ii) d ( x, P ) ≥ n + 1 and xx / ∈ D or x n x / ∈ D ;(iii) d ( x, P ) ≥ n , xx / ∈ D and x n x / ∈ D ;then there is an i ∈ [1 , n −
1] such that x i x, xx i +1 ∈ D , i. e., D contains a path x x . . . x i xx i +1 . . . x n oflength n (we say that the vertex x can be inserted into P or the path x x . . . x i xx i +1 . . . x n is extendedfrom P with x ).Note that the proof of Lemma 1 (see [21]) implies the following: Lemma 3.
Let D be a digraph on p ≥ C m = x x . . . x m x , m ∈ [2 , p − x be a vertex not contained in this cycle. If d ( x, C m ) = m and for some n ∈ [2 , m + 1] the digraph D contains no cycle of length n , then xx i ∈ D if and only if x i + n − x / ∈ D for every i ∈ [1 , m ].Using Lemma 2 it is not difficult to prove the following: Lemma 4.
Let D be a digraph on p vertices containing a path P := x x . . . x n and let x be a vertexnot contained in this path.a). Suppose that xx / ∈ D , x n x / ∈ D and x cannot be inserted into P . Then the following hold:(i) If n ≥ x x, x x, xx n ∈ D and d ( x, P ) ≥ n −
1, then there is an l ∈ [1 , n −
3] such that x l x, xx l +3 ∈ D .(ii) If n ≥ xx n ∈ D , A ( x → { x , x , x } ) = ∅ , d ( x, P ) ≥ n − | A ( x i → x ) | + | A ( x → x i +3 ) | ≤ i ∈ [1 , n − l ∈ [1 , n −
4] such that x l x, xx l +4 ∈ D .b). If n ≥ d ( x, P ) = n + 1 and x is adjacent with at most one vertex of two consecutive vertices of P , then n is odd and O ( x, P ) = I ( x, P ) = { x , x , . . . , x n } .2 otation. Let C n = x x . . . x n x be a cycle. For any pair of integers i, j ∈ [1 , n ]: if i ≤ j we denote by C ( i, j ) := { x i , x i +1 , . . . , x j } , and if i > j let C ( i, j ) := ∅ . Let f ( i, j := | C ( i, j ) | . Lemma 5.
Let D be a strongly connected digraph on p ≥
10 vertices with minimum degree at least p − p/ −
1. Let C := C p − := x x . . . x p − x be an arbitrarycycle of length p − D and let x be the vertex not contained in this cycle. Suppose that x is adjacentwith all vertices of cycle C . Then D contains a cycle C n for all n ∈ [3 , p − Proof.
Suppose, on the contrary, that for some n ∈ [3 , p −
2] the digraph D contains no cycle C n . It isesay to see that n ≥
5. Applying Lemmas 1 and 3 we find that d ( x ) = p − i ∈ [1 , p − | A ( x, x i ) | = 1 and xx i ∈ D if and only if x i + n − x / ∈ D. ( ∗ ) Notation.
We denote by M , M , . . . , M k , N , N , . . . , N k the maximal subpaths (sets) on cycle C forwhich both of the following hold ( we take the indices of M i and N i modulo k ):(i) Every vertex of M i (respectively, N i ) is dominated by x (respectively, dominates x );(ii) The subpaths M i and N i are labeled in such way that on the cycle C the subpath M i precedingof N i and N i preceding of M i +1 .Let m i := | M i | and n i := | N i | . Without loss of generality, we may assume that m = max { m i / ≤ i ≤ k } ≥ max { n i / ≤ i ≤ k } (1)( for otherwise we consider the digraph ←− D ). Let I := [3 , m + n + 1] and for all l ∈ [2 , k ] let I l = " l X i =2 m i + l − X i =1 n i + 3 , l X i =1 ( m i + n i ) + 1 . From the definitions of the sets M i , N i and from (*) it is easy to see that k ≥ n / ∈ ∪ ki =1 I i . From(1) it follows that for each j ∈ [2 , k ], right { I j − } + 1 ≥ lef t { I j } − right { I j } > right { I j − } . Hence, since n / ∈ ∪ ki =1 I i , for some s ∈ [2 , k ] we have right { I s − } + 1 ≤ n ≤ lef t { I s } −
1, i.e., s − X i =1 ( m i + n i ) + 2 ≤ n ≤ s X i =2 m i + s − X i =1 n i + 2 . This implies that m ≤ m s . Hence by (1) we have m = m s and n = s − X i =1 ( m i + n i ) + 2 . From (*) it follows that for all l ∈ [1 , k ], m l = m l + s − , n l = n l + s − and n = l + s − X i = l ( m i + n i ) + 2 . (2)For any t ∈ [2 , k + 1] denote by q t := P t − i =1 ( m i + n i ), in particular, q s = n − q k +1 = p −
1. Note that xx q t +1 ∈ D, xx q t / ∈ D and x q t + n − x ∈ D by (*) . (3)To be definite, assume that M := { x , x , . . . , x m } . We first prove the following Claims 1-5.3 laim 1. If j ∈ [ s + 1 , k + 1], then(i) x n − x q j +1 / ∈ D , in particular, x n − x / ∈ D ;(ii) | A ( x q j +1 → x n − ) | + | A ( x n − → x q j +2 ) | ≤
1, in particular, d ( x n − , { x q j +1 , x q j +2 } ) ≤ Proof.
Assume that Claim 1 is not true. Then(i) x n − x q j +1 ∈ D and C n = x n − x q j +1 x q j +2 . . . x q j + n − x x n − by (3);(ii) x q j +1 x n − , x n − x q j +2 ∈ D and C n = xx q j +1 x n − x q j +2 . . . x q j + n − x by (3). In both cases we havea contradiction. Claim 2. If m ≥
2, then L := A ( x n − → ∪ kj = s N j ) = ∅ . Proof.
In the converse case, if x n − z ∈ L , then C n = xx x . . . x n − zx , a contradiction. Claim 3. If m ≥ j ∈ [ s, k ], then L ( j ) := A ( x n − → ( M j − { x q j +1 , x q j +2 } )) = ∅ . Proof.
In the converse case, if x n − x i ∈ L ( j ), then from the maximality of m it follows that d − |{ x i , x i +1 , . . . , x q j + m j +1 }| ≤ m − C n = xx d x d +1 . . . x n − x i x i +1 . . . x q j + m j +1 x , a con-tradiction. Claim 4. x n − x p − / ∈ D . Proof.
In the converse case, x n − x p − ∈ D and C n = x x . . . x n − x p − x , a contradiction.From the maximality of m and Claims 2 and 4 it follows that d ( x n − , N k ) ≤ n k . (4)Using Claims 1-3, we get d ( x n − , C ( n, p − n k − ≤ p − n − n k + 1 . (5)Since the vertex x n − cannot be inserted into the path x x . . . x n − and x n − x / ∈ D (Claim 1), usingLemma 2(ii), we get d ( x n − , C (1 , n − ≤ n − . (6)Hence, by (4) and (5), we conclude that n k ≥ d ( x n − , N k ) ≥ n k − ,d ( x n − , C (1 , n − ( = n − , if d ( x n − , N k ) = n k − , ≥ n − , if d ( x n − , N k ) = n k . (7) Claim 5. If t ∈ [2 , n − | A ( x t − → x n − ) | + | A ( x n − → x t ) | ≤ Proof.
Assume that the claim is false, that is t ∈ [2 , n −
2] and x t − x n − , x n − x t ∈ D . Let the integer t with these properties be the smallest. If t ≤ n − n s − −
1, then C n = xx q k + m k . . . x t − x n − x t . . .x n − n s − − x since n k = n s − by (2), a contradiction. Thus we may assume that t ≥ n − n s − (inparticular, x t ∈ N s − ). Hence from t ≤ n − n s − ≥
2. Therefore m ≥ t ≥
4. It is not difficult to see that for all i ∈ [1 , n − d := |{ x i +1 , x i +2 , . . . , x n − }| ≤ m , then x i x n − / ∈ D (8)(otherwise x i x n − ∈ D and C n = xx m − d +1 x m − d +2 . . . x i x n − . . . x q s + m s +1 x since m = m s ). Togetherwith t ≥ n − n s − , m ≥ n s − and the fact that x n − cannot be inserted into the path x x . . . x n − thisimplies that t = n − n s − , m = n s − and A ( { x t − , x t − } → x n − } ) = A ( x n − , x t − ) = ∅ . (9)4ote that n k = n s − ≥ i ∈ [3 , t ], then | A ( x i − → x n − ) | + | A ( x n − → x i ) | ≤ , (10)(otherwise C n = xx q k + m k . . . x i − x n − x i . . . x t x ). From (10), in particular, we have x t − x n − / ∈ D .Suppose first that A ( x n − , x t − ) = ∅ . Then, since x n − x / ∈ D and the vertex x n − cannot be insertedinto the path x x . . . x n − , using Lemma 2 and (9), we obtain d ( x n − , C (1 , n − ≤ n −
3. From this,(7) and Claim 2, we get N k → x n − and d ( x n − , C (1 , t − t −
3. Since x t − x n − / ∈ D by (10), weobtain that t ≥
5, and by Lemma 2 there is an i ∈ [2 , t −
3] such that x n − x i and x i − x n − ∈ D , whichcontradicts the minimality of t .Suppose next that A ( x n − , x t − ) = ∅ . Then x t − x n − / ∈ D and x n − x t − ∈ D by (9). From (8) wehave d ( x n − , C ( t, n − ≤ n − t . Then it follows from (7) that d ( x n − , C (1 , t − ≥ ( t − , if d ( x n − , N k ) = n k ,t − , if d ( x n − , N k ) = n k − . In both cases it is easy to see that d ( x n − , { x p − , x p − , x , x , . . . , x t − } ) ≥ t − . Since A ( x n − → { x p − , x p − , x } ) = ∅ , x n − x t − ∈ D and x t − x n − / ∈ D , by Lemma 4(ii) there is an j ∈ [2 , t −
1] such that x j − x n − , x n − x j ∈ D or x j − x n − , x n − x j ∈ D ( j ≥ t or inequality (10). This completes the proof of Claim 5. Claim 6. If x n − x ∈ D , then L := A ( ∪ kj = s M j \ { x n − } → x n − ) = ∅ . Proof.
Otherwise x n − x ∈ D , zx n − ∈ L and C n = xzx n − x x . . . x n − x , a contradiction. Claim 7.
The vertex x n − dominates at most ( p − n − / C ( n, p − Proof.
Let m = 1. Then by (1), m i = n i = 1 for all i ∈ [1 , k ] and n ≤ p −
3. Observe that p − n are even. Using Claims 1(i) and 4, we obtain O ( x n − , C ( n, p − ⊆ { x n , x n +2 , x n +4 , . . . , x p − } . Hence the claim is true for m = 1 since |{ x n , x n +2 , x n +4 , . . . , x p − }| = ( p − n − / . Let now m ≥
2. Then m = m s ≥ n = p −
2, then s = k , m k = 2 and n k = 1. Togetherwith (2) and (*) this implies that m i = 2 and n i = 1 for all i ∈ [1 , k ] . Therefore id ( x ) ≤ p/ −
2, acontradiction. Thus we may assume that n ≤ p −
3. According to Claims 1-4 we haveif m s ≥ , then od ( x n − , C ( n, p − ≤ ( m s − m s +1 + · · · + m k ) / ≤ ( p − n − / m s = 2 , then od ( x n − , C ( n, p − ≤ m s +1 + m s +2 + · · · + m k ) / ≤ ( p − n − / n ≤ p −
3. Claim 7 is proved.Now we shall complete the proof of Lemma 5.From Claim 7, od ( x n − ) ≥ ( p − / A ( x n − → { x, x } ) = ∅ it follows that thereis an l ∈ [2 , n −
3] such that x n − → { x l , x l +1 } . Choose l with these properties is as small as possible.5ote that x n − cannot be inserted into the path x x . . . x n − . Now using the minimality property of l ,Lemma 2 and Claim 5, we see that A ( x n − , x l − ) = A ( { x l − , x l − } → x n − ) = ∅ . (11)First we prove that l ≥
3. Assume that l = 2. Then A ( x n − , x p − ) = ∅ by (11) and Claim 4. Henceit is easy to see that d ( x n − , C ( n, p − ≤ p − n − . (12)Indeed, if m ≥
2, then (12) immediately follows from Claims 2 and 6 and if m = 1, then M j = { x q j +1 } and (12) follows from Claims 1(i) and 6. By (12), p − ≤ d ( x n − ) = d ( x n − , C (1 , n − d ( x n − , C ( n, p − ≤ d ( x n − , C (1 , n − p − n. Hence d ( x n − , C (1 , n − ≥ n −
1, which contradicts (6). This proves that l ≥ A ( x l − , x n − ) = ∅ . If l = 3, then A ( x n − , { x p − , x , x } ) = ∅ , d ( x n − , N k ) ≤ n k − d ( x n − , C (1 , n − d ( x n − , C (3 , n − ≤ n − , which contradicts (7). Thus we may assume that l ≥
4. Since A ( x n − , { x l − , x l − } ) = ∅ , using Lemma2 and (11), we obtain d ( x n − , C (1 , n − d ( x n − , C (1 , l − d ( x n − , C ( l, n − ≤ n − , which also contradicts (7).Suppose next that A ( x l − , x n − ) = ∅ . Then x n − x l − ∈ D by (11). Since A ( x n − → { x p − , x } ) = ∅ it follows that l ≥
4. Therefore, since A ( x l − , x n − ) = ∅ by (11), from (7) and Lemma 2 it follows that d ( x n − , C (1 , n − n − d ( x n − , N k ) = n k . From this and Claims 2, 4 it is easy to see that N k → x n − . If n k ≥ x x n − ∈ D , then, since d ( x n − , N k ∪ C (1 , l − ≥ n k + l −
3, by Lemma 4(i) there is an i ∈ [2 , l − x i − x n − , x n − x i ∈ D , which contradicts Claim 5. So, we may assume that n k = 1 and A ( x , x n − ) = ∅ . Because of this and (13), by Lemma 2 we have x n − x ∈ D . Therefore Claim 6 holds(i.e., L = ∅ ). Now using Claims 1 and 2, we see that d ( x n − , C ( n, p − ≤ p − n . Together with (13)this implies that d ( x n − ) ≤ p −
2, a contradiction. This completes the proof of Lemma 5.
Notation.
In the following, for any integer k , k denotes the element of [1 , p −
1] ( p ≥
3) which iscongruent to k modulo p − k ≡ k mod ( p − Lemma 6.
Let D be a digraph on p ≥
10 vertices with minimum degree at least p − p/ −
1. Let D contains a cycle C := C p − := x x . . . x p − x of length p − n ∈ [5 , p −
2] the digraph D contains no cycle of length n . Suppose that the vertex x / ∈ V ( C ) isadjacent with the vertex x , is not adjacent with the vertex x p − and there are positive integers k and a with k + a ≤ p − xx a , x p − k − x ∈ D and A ( C ( p − k, p − → x ) = A ( x → C (1 , a − ∅ . ( a ≥
2) (14)Then the following statements hold: 6i) n ≤ p − k − a + 2;(ii) If i ∈ [ p − k − , p − j ∈ [1 , a ] and f ( j, i ) ≥ n −
1, then x i x j / ∈ D ;(iii) If p − k − ≤ i < j ≤ p −
1, then x i x j ∈ D if and only if j = i + 1 (i.e., f ( i, j ) = 2). Proof.
Since C n D , it follows from Lemmas 1 and 3 that d ( x ) = p − id ( x ) , od ( x ) ≤ m , where m := ⌊ p/ ⌋ , and for each i ∈ [1 , p − | A ( x → x i ) | + | A ( x i + n − → x ) | = 1 , in particular, x a + n − x ∈ D. ( ∗ )Proof of statements (i) and (ii) immediately follows from d ( x ) = p −
1, (14), (*) and Lemma 1.
Proof of (iii).
Suppose that statement (iii) is false. Then there are integers s and t with p − k − ≤ s < t − ≤ p − x s x t ∈ D . Choose the vertices x s and x t so that f ( s, t ) is as small as possible.Let d := f ( s + 1 , t − ≤ d + 2 = f ( s, t ) ≤ k + 2 . (15)From A ( C ( p − k, p − → x ) = ∅ by (14) and (*) it follows that x → C ( p − k − n + 2 , p − n + 1) , in particular , xx p − k − n +2 ∈ D. (16)Note that p − k − n + 2 ≥ a by Lemma 6(i). Together with (16) and A ( x, x p − ) = ∅ this implies thatthere is a vertex x q , q ∈ [ p − n + 1 , p − xx q +1 / ∈ D and x → C ( p − k − n + 2 , q ) . (17)Remark that from xx q +1 / ∈ D and (*) we have x q + n − x ∈ D . Case 1. q ≤ p − k − f ( q + n, q ) ≥ d , then by (15), (17) and x q + n − x ∈ D we have xx q − d +1 ∈ D and C n = xx q − d +1 x q − d +2 . . . x s x t . . . x q + n − x , a contradiction. Therefore, we may assume that f ( q + n, q ) < d. From this,(15), Lemma 6(i) and f ( q + n, q ) + n = p it is easy to see that p − k + 1 ≤ n = p − f ( q + n, q ) ≤ p − k − a + 2 . This implies that a = 1, f ( q + n, q ) = k − n = p − k + 1 and d = k . Therefore, xx ∈ D , s = p − k − t = p − x p − k − x p − ∈ D ). From 5 ≤ n ≤ p − n = p − k + 1 it follows that p − k ≥ k ≥
3. Hence by the minimality of f ( s, t ) we have A ( C ( p − k − , p − → x p − ) = A ( C ( p − k − , p − → x p − ) = ∅ . (18)Since n = p − k + 1, using A ( C ( p − k, p − → x ) = ∅ by (14) and (*) we see that x → C (1 , k ) (19)We have a cycle C n − := xx x . . . x p − k − x of length n − x p − cannot be inserted into this cycle C n − . Hence, since x p − k − x p − ∈ D , we have x p − x p − k − / ∈ D . Therefore, A ( x p − , x p − k − ) = ∅ by (18),and d ( x p − , C (1 , p − k − ≤ p − k − d ( x p − ) ≥ p − x p − → C ( p − k, p − . (20)7f x i x p − k − ∈ D , where i ∈ [ p − k, p − C n = xx x . . . x p − k − x p − x i x p − k − x by (19) and (20),a contradiction. So we may assume that A ( C ( p − k, p − → x p − k − ) = ∅ . (21)Therefore id ( x p − k − , C (1 , p − k − ≥ p/ − . (22)It is easy to see that if i ∈ [1 , p − k − | A ( x i → x p − k − ) | + | A ( x p − k → x i +1 ) | ≤ , for otherwise x i x p − k − , x p − k x i +1 ∈ D and C n = x p − x x . . . x i x p − k − x p − k x i +1 . . . x p − k − x p − , acontradiction. From this and (22) it follows that x p − k does not dominate at least p/ − C (2 , p − k − f ( s, t ) we also have that x p − k does not dominate k − C ( p − k + 1 , p − A ( x p − k → { x, x p − k − , x } ) = ∅ . Hence, by our arguments above we have that x p − k does not dominate atleast p/ k − od ( x p − k ) ≤ p/ − k + 1 ≤ p/ − k ≥
3, a contradiction.The discussion of Case 1 is completed.
Case 2. q ≥ p − k − xx q +1 / ∈ D and (14) it follows that A ( x, x q +1 ) = ∅ . Since δ ( x ) ≥ p/ − d ( x ) = p − od ( x ) ≤ p/
2. Hence from q ≥ p − k − n − ≤ p/ n ≤ p/ Claim 1. L := A ( x q → C ( q + 2 , p − ∪ C (1 , a − ∅ . Proof.
Assume that x q x i ∈ L for some i ∈ [ q + 2 , p − ∪ [1 , a − n ≤ p − k − a + 2by Lemma 6(i). Hence the cycle C ′ := x a x a +1 . . . x q x i . . . x a has length at least n −
1. Then, since d ( x ) = p −
1, (14) and A ( x q +1 , x ) = ∅ , we obtain d ( x, V ( C ′ )) ≥ | V ( C ′ ) | + 1. Therefore C n ⊂ D byLemma 1, a contradiction. Claim 2. If p − k − n ≥ a + n −
4, then A ( x q → C ( a, a + n − ∅ . Proof.
Suppose, on the contrary, that there is an i ∈ [0 , n −
3] such that x q x a + i ∈ D . Then, since x a + n − x ∈ D by (*), f ( p − k − n + 2 , q ) ≥ n − C n = x q x a + i x a + i +1 . . . x a + n − xx q − i x q − i +1 . . . x q , a contradiction. Claim 3. If a < p − k − n + 1 ≤ a + n −
4, then A ( x q → C ( a, p − k − n + 1)) = ∅ . Proof.
Assume that the claim is not true, that there is an i ∈ [0 , p − k − n − a + 1] such that x q x a + i ∈ D .Note that a + n − ≤ p − k −
3. From q ≥ p − k − i ∈ [0 , p − k − n − a + 1] it follows that q − i ≥ n + a −
2. Therefore xx q − i ∈ D by (17), and C n = x q x a + i x a + i +1 . . . x a + n − xx q − i x q − i +1 . . . x q , acontradiction. Claim 4. If q ≥ p − k , then the following hold:(i) f ( p − k, q ) ≤ n − A ( x q → C ( p − k − n + 2 , p − k − n + 2 + f ( p − k, q − ∅ . Proof. (i) Suppose, on the contrary, that f ( p − k, q ) ≥ n −
2. From this, since p − k + f ( p − k, q ) = q + 1,we have q − n + 3 ≥ p − k . Note that I ( x ) ⊆ C (1 , p − k −
1) by (14). It is easy to see that if x i ∈ I ( x ), then x q x i / ∈ D (otherwise x i x , x q x i ∈ D and C n = x q x i xx q − n +3 x q − n +4 . . . x q by xx q − n +3 ∈ D ). Thereforefrom id ( x ), od ( x q ) ≥ p/ − x q x / ∈ D it follows that x q x j ∈ D if and only if x j / ∈ I ( x ). Then,8ince x p − x / ∈ D we have x q x p − ∈ D , and q = p − x q x ∈ D ( x q = x p − ), then C ′ := x x . . . x p − x is a cycle of the length p − d ( x, C ′ ) = p −
1, and hence D contains acycle C n by Lemma 1, a contradiction. So, we may assume that x q x / ∈ D and x x ∈ D . Then, since q − n + 4 ≥ p − k + 1, C n = xx q − n +4 . . . x q x q +1 x x , a contradiction. This completes the proof of inequality f ( p − k, q ) ≤ n − i ∈ [1 , f ( p − k, q )] such that x q x p − k − n +1+ i ∈ D . Then p − k − n + 1 + i ≤ p − k − n + 1 + n − p − k − C n = x q x p − k − n +1+ i . . .x p − k − xx q − i +1 . . . x q , a contradiction. Claim 4 is proved. Claim 5. If a + n ≥ p − k − n + 4, then n ≥ ( p/ , if q ≥ p − k,p/ , if q = p − k − . Proof.
Put B := C ( p − k − n + 2 + f ( p − k, q ) , q + 1) \ { x q } . From Claims 1-4 it follows that O ( x q ) ⊆ ( B, if q ≥ p − k,B ∪ { x } , if q = p − k − . Then p/ − ≤ od ( x q ) ≤ ( n − , if q ≥ p − k,n − , if q = p − k − | B | = n −
2. Claim 5 is proved.We now consider the following two subcases.
Subcase 2.1. a + n ≥ p − k − n + 4.Using (*) and (17), we obtain x q + n − x ∈ D and A ( { x q +1 , x q +2 , . . . , x q + n − } → x ) = ∅ . (23)Claim 5 and n ≤ p/ m ≤ n ≤ m + 2 ( m := ⌊ p/ ⌋ ).Suppose that n = m + 2. From q ≥ p − k −
1, (17) and od ( x ) ≤ m it follows that q = p − k − A ( x → { x q +1 , x q +2 , . . . , x p − k − n +1 } ) = ∅ . Together with q + n − ≥ p − k − n + 1 and (23) this implies that A ( x, { x q +1 , x q +2 , . . . , x p − k − n +1 } ) = ∅ . (24)Hence x = x p − k − n +2 (i.e., n = p − k + 1 and x p − = x p − k − n +1 ) and k ≥ m − ≥
4. From k ≥ x p − k − x ∈ D . By (24) and Lemma 1, we can assume that s = p − k − A ( C ( p − k − , p − → x p − ) = ∅ , (25)in particular, x p − k − x p − / ∈ D . Then, since d ( x p − ) ≥ p − x p − cannot be insertedinto the path x x . . . x p − k − , using Lemma 2(ii) and (25) we get that x p − x p − k ∈ D . From this itis easy to see that if x p − k x i ∈ D , i ∈ [2 , p − k − x i − x p − / ∈ D (otherwise, if i ≥
3, then C n = xx . . . x i − x p − x p − k x i x i +1 . . . x p − k − x and if i = 2, then C n = xx x p − x p − k x . . . x p − k − x ).Again using Lemma 1 and (24), we obtain A ( x p − k → C ( p − k + 2 , p − ∪ { x } ) = ∅ . Therefore x p − k dominates at least p/ − C (2 , p − k − x p − is notdominated at least by p/ − C (1 , p − k − k ≥ xx p − / ∈ D this implies that id ( x p − ) ≤ p/ −
2, a contradiction.9ow suppose that m ≤ n ≤ m + 1 and q = p − k −
1. From Claim 5 and n ≤ m + 1 we obtain that p − k − n − ≤ p − k + n − ≤ p − k − n + 1 . (26)From q = p − k −
1, (17) and (*) it follows that A ( { x p − k , x p − k +1 , . . . , x p − k + n − } → x ) = ∅ , and x p − k + n − x ∈ D (i.e., k ≤ n − d := f ( s + 1 , t − ≤ n −
2. If t ≥ p − k + 1,then C n = xx p − k − d . . . x p − k − d +1 · · · x s x t x t +1 · · · x p − k + n − x , a contradiction. Therefore t = p − k and s = p − k −
2. From our supposition that C n D it is not difficult to see that A ( x p − k → { x p − k − , x p − k − n , x p − k − n +1 } ) = ∅ . (27)If x p − k x p − k + i ∈ D , where i ∈ [2 , n − C n = x p − k x p − k + i x p − k + i +1 . . . x p − k + n − xx p − k − i +1 . . . x p − k by (17), a contradiction. So we may assume that A ( x p − k → { x p − k +2 , x p − k +3 , . . . , x p − k + n − } ) = ∅ . Together with (26) and (27) this implies that O ( x p − k ) ⊆ C ( p − k − n + 2 , p − k − ∪ { x p − k +1 } . Therefore n − ≥ p/ −
1. Then, since n ≤ m + 1, it follows that p = 2 m , n = m + 1, x p − k x p − k − n +2 ∈ D and x p − k − n +1 = x p − k + n − . Hence, by (23) it is obvious that there is a vertex x j ∈ C ( p − k − n +2 , p − k − x j x ∈ D . But then by (17), C n = x p − k x p − k − n +2 . . . x j xx j +1 . . . x p − k , a contradiction.Finally suppose that m ≤ n ≤ m + 1 and q ≥ p − k . Then p = 2 m and n = m + 1 by Claim 5. Using od ( x ) ≤ m and (17), we get that q = p − k , x a = x p − k − n +2 = x p − k + n − and A ( x, { x p − k +1 , x p − k +2 , . . . , x p − k + n − } ) = ∅ . Therefore a = 1 and x x ∈ D . Now using (14) with Lemma 1, we obtain A ( C ( p − k − , p − → x p − ) = A ( x p − k → C ( p − k + 2 , p − ∪ { x } ) = ∅ . Together with f ( p − k − , p −
3) = m − xx p − / ∈ D , x p − k x / ∈ D (by (14)) this implies that x p − k − x p − , x p − k x ∈ D and C n = x p − k x . . . x p − k − x p − x xx p − k , which is a contradiction and com-pletes the discussion of Subcase 2.1. Subcase 2.2. a + n − ≤ p − k − n .Put α := | I ( x ) ∩ C ( p − k − n + 2 , p − k − | , β := | I ( x ) ∩ C ( a, a + n − | , γ := | I ( x ) ∩ C ( a + n − , p − k − n + 1) | and B := C ( q + 2 , p − ∪ C (1 , a + n − ∪ C ( p − k − n + 2 , p − k − n + 1 + f ( p − k, q )) . It is clear that 1 ≤ α ≤ n − , ≤ β ≤ n − | B | = a + n + k − . (28)From Claims 1, 2 and 4(ii) it follows that A ( x q → B ) = ∅ and | B | ≤ m. (29)10or Subcase 2.2 first we will prove Claims 6-9. Claim 6. β ≤ k . Proof. If n − ≤ k , then β ≤ n − ≤ k by (28). So we may assume that n − ≥ k . Then from q ≥ p − k − x x / ∈ D . This means that a = 1 and a + n − n −
2. Then, since x → C ( p − k − n + 2 , q ) and | C ( p − k − n + 2 , q ) | ≥ k + 1 , from (*) it follows that A ( C (1 , n − k − → x ) = ∅ . Hence β ≤ | C (1 , n − \ C (1 , n − k − | ≤ k . Claim6 is proved.Note that from id ( x ) ≥ m − β ≤ k and (14) it follows that γ ≥ id ( x ) − ( α + β + a − ≥ id ( x ) − ( α + a + k − ≥ m − α − a − k. (30) Claim 7. If q = p − k −
1, then s = p − k − t = p − k and A ( x p − k → B \ { x p − k +1 } ) = ∅ . Proof.
Since q = p − k −
1, from the definitions of B and q it follows that B = C ( p − k + 1 , p − ∪ C (1 , a + n −
3) and xx p − k / ∈ D . Therefore x p − k + n − x ∈ D by (*). Together with (14) this implies that p − k + n − n − k − ≥ n ≥ k + 2 and x p − k + n − = x n − k − ). Therefore, if t = p − k , then C n = xx p − k − d . . . x s x t x t +1 . . . x n − k − x ( d := f ( s + 1 , t − t = p − k and thisimplies that s = p − k −
2. In particular, we also have A ( x p − k → C ( p − k + 2 , p − ∅ . If x p − k x a + i ∈ D , where i ∈ [1 , n − C n = x p − k x a + i x a + i +1 . . . x a + n − xx p − k − i . . . x p − k by (17),a contradiction. Therefore A ( x p − k → C ( a + 1 , a + n − ∅ . If x p − k x i ∈ D , where i ∈ [1 , a ], then by Lemma 6(ii), p − k = p − k = 1 and x p − k +1 = x ).Now from (17) and (*) we obtain that a = 1 and this completes the proof of Claim 7. Claim 8.
Either α ≥ γ ≥ Proof.
Suppose, on the contrary, that α = 1 and γ = 0. Then from (14) and Claim 6 we find that id ( x ) ≤ a + β ≤ a + k. Together with id ( x ) ≥ p/ − n ≥
5, (28) and (29) this implies that m ≥ | B | = a + n + k − ≥ id ( x ) + n − . Hence id ( x ) = m − β = k , n = 5, p = 2 m and | B | = m . Therefore x q → V ( D ) \ ( B ∪ { x q } ) by (29).Since x / ∈ B , we obtain x q x ∈ D , q = p − k − p − k − n = a + n − x p − k − n / ∈ B , x q x p − k − n ∈ D and C n = x q x p − k − n . . . x q ). Therefore A ( x p − k → B \ { x p − k +1 } ) = ∅ by Claim 7. Using this together with A ( x p − k → { x, x a + n − } ) = ∅ and | B \ { x p − k +1 }| ≥ m −
1, we getthat od ( x p − k ) ≤ m −
2, a contradiction. Claim 8 is proved.
Claim 9.
The vertex x q does not dominate at least γ + 1 vertices of C ( a + n − , p − k − n + 1). Proof.
First suppose that γ = 0. Then α ≥ i ∈ [1 , n −
3] suchthat x p − k − n +1+ i x ∈ D . If x q x p − k − n +1 ∈ D , then C n = x q x p − k − n +1 . . . x p − k − n + i +1 xx q − n + i +3 . . . x q , acontradiction. Therefore x q x p − k − n +1 / ∈ D , and for γ = 0 Claim 9 is true.Now suppose that γ ≥
1. Then, since x a + n − x / ∈ D , it follows that a + n − ≤ p − k − n . If x i x ∈ D for some i ∈ [ a + n − , p − k − n + 1], then A ( x q → { x i − , x i } ) = ∅ (for otherwise if x q x i ∈ D , then11 n = x q x i xx q − n +3 . . . x q and if x q x i − ∈ D , then C n = x q x i − x i xx q − n +4 . . . x q ). Therefore, x q does notdominate at least γ +1 vertices of C ( a + n − , p − k − n +1) since x a + n − x / ∈ D , and so Claim 9 is proved.Now we will complete the proof of Lemma 6 for Subcase 2.2.First suppose that q ≥ p − k . Then from Claims 1, 2, 4 and 9 it follows that x q does not dominate atleast a + n + k + γ − C . Since x q x / ∈ D , we see that m ≥ a + n + k + γ −
2. Fromthis and (30), we obtain m ≥ a + n + k + γ − ≥ m + n − α − . (31)Then, since α ≤ n −
2, it follows that α = n − C ( p − k − n + 2 , p − k − → x , m = a + n + k + γ − x q does not dominate exactly γ + 1 vertices from C ( a + n − , p − k − n + 1) (i.e., if γ ≥ x i ∈ C ( a + n − , p − k − n + 1), then x q x i / ∈ D if and only if x i x ∈ D or x i +1 x ∈ D ). From this,since k ≥ γ ≥ a ≥
1, by (31) we obtain n ≤ m −
1. Now using Claims 1, 2 and 4 we seethat A ( x q → C ( a + n − , p − k − n + 1)) = ∅ , because of | C ( p − k − n + 2 + f ( p − k, q ) , q | ≤ m − i ∈ [ a + n − , p − k − n + 1] such that x q x i , x i +2 x ∈ D . Thus we have a cycle C n = x q x i x i +1 x i +2 xx q − n +5 . . . x q , which is a contradiction.Now suppose that q = p − k −
1. Similarly as in Claim 9, one can show that x p − k does not dominateat least γ + 1 vertices from C ( a + n − , p − k − n + 1). By Claim 7, s = p − k − t = p − k , and A ( x p − k → B \ { x p − k +1 } ) = ∅ . Therefore, since x p − k x / ∈ D and | B | = a + n + k − x p − k does not dominate atleast a + n + k + γ − m ≥ a + n + k + γ − ≥ m + n − α − α ≥ n − ≥ x p − k x p − k − n +2 / ∈ D ( for otherwise we obtain a cycle C n ). Thus we see that x p − k doesnot dominate at least m + n − α − ≥ m vertices, since α ≤ n −
2. It follows that x p − k x p − k − ∈ D and D contains a cycle C n := xx p − k − n +2 . . . x p − k − x p − k x p − k − x , which is a contradiction and completes thediscussion of Subcase 2.2. Lemma 6 is proved. Main Result
We first introduce the following notations.
Notation . For any positive integer m , let H ( m, m ) denote the set of digraphs D on 2 m vertices suchthat V ( D ) = A ∪ B , h A i ≡ h B i ≡ K ∗ m , A ( B → A ) = ∅ and for every vertex x ∈ A (respectively, y ∈ B ) A ( x → B ) = ∅ (respectively, A ( A → y ) = ∅ ). Notation . For any integer m ≥
2, let H ( m, m − ,
1) denote the set of digraphs D on 2 m vertices suchthat V ( D ) = A ∪ B ∪ { a } , | A | = | B | + 1 = m , A ( h A i ) = ∅ , h B ∪ { a }i ⊆ K ∗ m , yz, zy ∈ D for each pairof vertices y ∈ A , z ∈ B and either I ( a ) = B and a → A or O ( a ) = B and A → a . Notation . For any integer m ≥ H (2 m ) as follows: V ( H (2 m )) = A ∪ B ∪ { x, y } , h A i ≡ h B i ≡ K ∗ m − , A ( A, B ) = ∅ , O ( x ) = { y } ∪ A , I ( x ) = O ( y ) = A ∪ B and I ( y ) = { x } ∪ B . H ′ (2 m ) is a digraph obtained from H (2 m ) by adding the arc yx . Theorem.
Let D be a digraph on p ≥
10 vertices with the minimum degree at least p − p/ − m := ⌊ p/ ⌋ ). Then D is pancyclic unless K ∗ m,m +1 ⊆ D ⊆ ( K m + K m +1 ) ∗ or p = 2 m and G ⊆ K ∗ m,m or else D ∈ H ( m, m ) ∪ H ( m, m − , ∪ { [( K m ∪ K m ) + K ] ∗ , H (2 m ) , H ′ (2 m ) } Proof.
Suppose that the theorem is false, in particular, for some n ∈ [3 , p ] the digraph D contains nocycle of length n . We recall that D is strong, hamiltonian and contains cycles of length 3, 4 and p − n ∈ [5 , p − C := C p − := x x . . . x p − x be an arbitrary cycle oflength p − D and let x be the vertex not containing in this cycle. From Lemmas 1 and 3 it followsthat d ( x ) = p − m − ≤ id ( x ) , od ( x ) ≤ m and for each i ∈ [1 , p − xx i ∈ D if and only if x i + n − x / ∈ D. ( ∗ )For the cycle C and for the vertex x we first prove the following claim: Claim 1.
There is a vertex x i , i ∈ [1 , p −
1] (to be definite, let x i := x p − ) and there are positive integers k and a with k + a ≤ p − A ( x, x p − ) = ∅ , A ( x, x ) = ∅ , x p − k − x, xx a ∈ D and A ( { x p − k , x p − k +1 , . . . , x p − } → x ) = A ( x → ( { x , x , . . . , x a } \ { x a } )) = ∅ . (32) Proof.
Using Lemma 5, we see that there is a vertex x i , i ∈ [1 , p − A ( x, x i ) = ∅ and A ( x, x i +1 ) = ∅ (say x i := x p − ). Then there are positive integers a and k such that xx a , x p − k − x ∈ D .We can choose a and k so that (32) holds. If k + a ≥ p −
1, then from od ( x ) = p − { x , x , . . . , x a } → x → { x a , x a +1 , . . . , x p − } . Hence, C n ⊂ D since n ∈ [5 , p − k + a ≤ p −
2. Claim 1 is proved.Claim 1 immediately implies that the conditions of Lemma 6 hold. Therefore n ≤ p − k − a + 2 andif p − k − ≤ i < j ≤ p − , then x i x j ∈ D if and only if j = i + 1 . (33)In particular, this implies that A ( C ( p − k − , p − → x p − ) = ∅ . (34)Consider the digraph ←− D , similarly to (34), one can show that A ( x p − → C (2 , a + 1)) = ∅ . (35) Case 1. n = p − k − a + 2.Then n = p − k − a + 2 ≤ p − k + a ≥
4. From (*) and (32) it follows that C ( p − k − a, p − k − → x → C ( a, a + k − . (36)Without loss of generality, we may assume that k ≥ ←− D ). Itfollows from statement (ii) of Lemma 6 that A ( x p − → C (1 , a )) = ∅ . (37)From (34) and (35), we see that x p − k − x p − / ∈ D and x p − x a +1 / ∈ D . Hence, since x p − cannot beinserted into the path x a +1 x a +2 . . . x p − k − , using (34), (35), d ( x p − ) ≥ p − C (1 , a ) → x p − → C ( p − k − , p −
2) and d ( x p − , C ( a + 1 , p − k − n − , (38)13n particular, x a x p − , x p − x p − k − ∈ D . Therefore, by Lemma 2, there is an l ∈ [ a + 1 , p − k −
2] suchthat A ( x p − , x l ) = ∅ and x l − x p − , x p − x l +1 ∈ D . Let l with these properties be the smallest. By(36) and (38), xx a +1 , x p − x p − ∈ D . Therefore, since xx a , x p − k − x ∈ D , if x l − x p − ∈ D , then C n = xx a . . . x l − x p − x p − x l +1 . . . x p − k − x , if x l x p − ∈ D , then C n = xx a +1 . . . x l x p − x p − x l +1 . . . x p − k − x and if x p − x l +1 ∈ D , then C n = xx a . . . x l − x p − x p − x l +1 . . . x p − k − x , a contradiction. So, we mayassume that A ( { x l − , x l } → x p − ) = A ( x p − → x l +1 ) = ∅ . (39)From Lemma 6(iii) it follows that A ( C ( p − k − , p − → x p − ) = ∅ . (40)Suppose that A ( x p − , x l ) = ∅ . Since x p − cannot be inserted into the path P := x a x a +1 . . . x p − k − and x p − x a / ∈ D by (37), x l − x p − / ∈ D by (39), using Lemma 2(iii), we see that d ( x p − , C ( a, l − ≤ l − a − . (41)Furthermore, using (39) and (40), similarly to (41), one can show that d ( x p − , C ( l + 1 , p − k − ≤ ( p − k − l − , if k = 2 ,p − k − l − , if k ≥ . Now by (37), (40), (41) and x p − x / ∈ D , d ( x p − ) = d ( x p − , C (1 , a − d ( x p − , C ( a, l − d ( x p − , C ( l + 1 , p − k − d ( x p − , C ( p − k, p − d ( x p − , x ) ≤ p − , a contradiction . Now suppose that A ( x p − , x l ) = ∅ . It follows from (39) that x p − x l ∈ D . Since x p − k − x , x p − x p − ∈ D and (36), we have if l ≥ a + 2, then C n = xx a +1 . . . x l − x p − x p − x l . . . x p − k − x , a contradiction. Hence l = a + 1. If x p − k − x ∈ D , then C n = xx a x p − x p − x l . . . x p − k − x , a contradiction. Thus we may assumethat x p − k − x / ∈ D . Then from (36) it follows that a = 1. Then k ≥
3. From (33) and (34), by Lemma2, we obtain d ( x p − , C (3 , p − k − p − k −
3. Using (34), (35), (38) and the minimality of l = a + 1it is easy to see that A ( C (2 , p − → x p − ) = ∅ . Hence id ( x p − ) ≤
2, a contradiction.
Case 2. n ≤ p − k − a + 1.Then from xx a ∈ D and (*) it follows that n ≤ p − k − a (i.e., a ≤ p − k − n ). Notation.
In the following let P := x a x a +1 . . . x p − k − , a := s , P := x s x s +1 . . . x s + n − and if i ≥
2, then P i − := x s i − x s i − +1 . . . x s i − + n − . For i ≥ s i − + n − ≤ p − k − n + 1 and there isan s i ∈ [ s i − + 2 , s i − + n −
3] such that xx s i ∈ D and xx s i − / ∈ D , then let P i := x s i x s i +1 . . . x s i + n − (the integers s i , i ≥ r − P i paths and let P r := x p − k − n +2 x p − k − n +3 . . . x p − k − . Since n ≤ p − k − a , wehave r ≥
2. If s r − + n − ≥ p − k − n +2, then we say that the path P is covered with paths P , P , . . . , P r .Note that each P i path has length n −
3. By the definition of s i , i ∈ [1 , r ] , and (*) we have xx s i , x s i + n − x ∈ D, x s i + n − x / ∈ D (42)and xx s i x s i +1 . . . x s i + n − x is a cycle of length n −
1, where s r := p − k − n + 2.We now divide the Case 2 into two subcases. 14 ubcase 2.1. The path P is covered with paths P , P , . . . , P r . Notation.
In the following , let A := C (1 , s ), A := C ( s + 1 , s + n − A r +1 := C ( p − k − , p − A i := C ( s i − + n − , s i + n −
4) if i ∈ [2 , r ].It is easy to see that C (1 , p −
2) = ∪ r +1 i =0 A i and A i ∩ A j = ∅ for each pair of distinct i, j ∈ [0 , r + 1].From n ≥ C n D it follows that if i ∈ [2 , r ], then | A i | = s i − s i − ≥ i ∈ { , , r + 1 } , then | A i | ≥ | A | = n − x p − cannot be inserted into no subpaths of P withvertices set A i for all i ∈ [1 , r ]. Therefore, using (34), (35) and Lemma 2, we obtain d ( x p − , A i ) ≤ ( | A i | , if i ∈ { , r }| A i | + 1 , if i ∈ [2 , r − . (43)First let us prove the following Claims 2-5. Claim 2. x p − x p − k − ∈ D . Proof.
Suppose, on the contrary, that x p − x p − k − / ∈ D . From (34) and (35) it is easy to see that d ( x p − , A r +1 ) ≤ | A r +1 | and d ( x p − , A ) ≤ | A | + 1. Then, since p − P r +1 i =0 | A i | and p − ≤ d ( x p − ) = r +1 X i =0 d ( x p − , A i ) , from (43) it follows that there is an t ∈ { } ∪ [2 , r −
1] such that d ( x p − , A t ) = | A t | + 1.Suppose first that t ∈ [2 , r − x p − cannot be inserted into the subpath C [ x s t − + n − ,x s t + n − ], from d ( x p − , A t ) = | A t | + 1 and Lemma 2(ii) it follows that x s t + n − x p − and x p − x s t − + n − ∈ D . Therefore for each i ∈ [1 , t −
1] and for each l ∈ [ t, r ] it is easy to see that x s i + n − x p − / ∈ D and x p − x s l + n − / ∈ D. (44)Indeed, in the converse case, by (42) we have if x s i + n − x p − ∈ D , then C n = xx s i x s i +1 . . . x s i + n − x p − x s t − + n − x and if x p − x s l + n − ∈ D , then C n = xx s t x s t +1 . . . x s t + n − x p − x s l + n − x , a contradiction.Using (34), (35), (44) and Lemma 2, we obtain d ( x p − , A i ) ≤ | A i | − , if i ∈ { , r } , | A i | , if i ∈ [2 , r + 1] \ { r, t } , | A i | + 1 , if i ∈ { , t } . Therefore p − ≤ d ( x p − ) = r +1 X i =0 d ( x p − , A i ) ≤ r +1 X i =0 | A i | = p − , a contradiction.Now suppose that t = 0. Then from (35) and d ( x p − , A ) = | A | + 1 it follows that A → x p − . Hence x p − x s i +1 / ∈ D for each i ∈ [1 , r ] (otherwise by (42), C n = xx s x p − x s i +1 x s i +2 . . . x s i + n − x ). Now wedecompose the set C ( s + 1 , p − k −
2) into subsets B i , where B i := C ( s i + 1 , s i +1 ) if i ∈ [1 , r −
1] and B r := C ( s r + 1 , s r + n − x p − cannot be inserted into no subpaths of the path P withvertex set B i . Therefore, using (34), x p − x s i +1 / ∈ D , x p − x p − k − / ∈ D and Lemma 2, we obtain d ( x p − , B i ) ≤ ( | B i | , if i ∈ [1 , r − , | B i | − , if i = r. Hence it is not difficult to see that p − ≤ d ( x p − ) = r X i =1 d ( x p − , B i ) + d ( x p − , A ∪ A r +1 )) ≤ r X i =1 | B i | + | A ∪ A r +1 | ≤ p − , Claim 3. If i ∈ [1 , r ], then x s i + n − x p − / ∈ D . Proof.
Indeed, otherwise x s i + n − x p − ∈ D and C n = xx s i x s i +1 . . . x s i + n − x p − x p − k − x by (42) andClaim 2, a contradiction. Claim 4. x s x p − ∈ D . Proof.
Suppose, on the contrary, that x s x p − / ∈ D . Then from (35) it follows that d ( x p − , A ) ≤ | A | .Since x p − cannot be inserted into no subpaths of P with vertex set A i , i ∈ [1 , r ], using (34), (35), Claim3 and Lemma 2, we obtain d ( x p − , A i ) ≤ | A i | , if i ∈ { } ∪ [2 , r ] , | A i | − , if i = 1 , | A i | + 1 , if i = r + 1 . Hence p − ≤ d ( x p − ) = r +1 X i =0 d ( x p − , A i ) ≤ r +1 X i =0 | A i | = p − , a contradiction.Claim 4 is proved. Claim 5. If i ∈ [1 , r ], then x p − x s i +1 / ∈ D . Proof.
Indeed, otherwise x p − x s i +1 ∈ D and C n = xx s x p − x s i +1 . . . x s i + n − x by Claim 4 and (42), acontradiction.From (34), (35), Claim 3 and Lemma 2 it follows that d ( x p − , A i ) ≤ | A i | − , if i = 1 , | A i | + 1 , if i ∈ { , r + 1 } , | A i | , if i ∈ [2 , r ] . Therefore p − ≤ d ( x p − ) = r +1 X i =0 d ( x p − , A i ) ≤ r +1 X i =0 | A i | + 1 = p − . It follows that d ( x p − ) = p − d ( x p − , A i ) = | A i | − , if i = 1 , | A i | + 1 , if i ∈ { , r + 1 } , | A i | , if i ∈ [2 , r ] . (45)Using this together with (34), (35), Claim 3, the definitions of sets A i and Lemma 2 it is not difficult tosee that A → x p − → A r +1 ∪ { x s + n − , x s + n − , . . . , x s r + n − } . (46)Hence, by Claim 5, for all i ∈ [2 , r ] we have x s i +1 = x s i − + n − . We will now prove the following:
Claim 6. A ( x p − , { x s +1 , x s +3 , . . . , x p − k − } ) = ∅ ( p − k − s is odd) and { x s , x s +2 , . . . , x p − k − } → x p − → { x s +2 , x s +4 , . . . , x p − k − } . roof. Using the definition of s i , Claims 2-5, (46) and d ( x p − ) = p − n = 5 Claim 6 is true.Assume that n ≥
6. To prove Claim 6 for n ≥
6, it suffices to prove Claims 6.1-6.4.
Claim 6.1. If l ∈ [1 , r − i ∈ [ s l + 1 , s l + n −
4] and x i − x p − ∈ D , then x p − x i +2 / ∈ D . Proof.
Suppose, on the contrary, that l ∈ [1 , r − i ∈ [ s l + 1 , s l + n −
4] and x i − x p − , x p − x i +2 ∈ D .Without loss of generality, we may assume that i is as maximal as possible. If x s l + n − x ∈ D , then C n = xx s l . . . x i − x p − x i +2 . . . x s l + n − x , a contradiction. Therefore x s l + n − x / ∈ D , and it follows from (*)that xx s l +1 ∈ D . Hence x s l + n − x p − / ∈ D (otherwise by Claim 2, C n = xx s l +1 . . . x s l + n − x p − x p − k − x ).Since x p − cannot be inserted into the path x s l + n − x s l + n − . . . x s l +1 + n − , d ( x p − , A l +1 ) = | A l +1 | by(45), and x s l +1 + n − x p − / ∈ D by Claim 3, from Lemma 2 it follows that x p − x s l + n − ∈ D . Therefore byClaim 5 and (46), s l +1 ≤ s l + n − x s l +1 ∈ A l .Assume that i ≥ s l +1 −
1. Since xx s l +1 − / ∈ D (by the definition of s i ) and xx s l ∈ D , we see thatthere is an t ∈ [ s l + 1 , s l +1 −
2] such that xx t ∈ D and xx t +1 / ∈ D . Therefore x t + n − x ∈ D by (*), and C n = xx t x t +1 . . . x i − x p − x i +2 . . . x t + n − x , a contradiction.Now assume that i ≤ s l +1 −
2. Since x s l +1 ∈ A l , x p − → { x s l + n − , x s l + n − } , x p − x s l +1 +1 / ∈ D byClaim 5, and the path x s l +1 x s l +1 +1 . . . x s l + n − cannot be extended with x p − , there is an k ∈ [ s l +1 +2 , s l + n −
3] such that x p − → { x k , x k +1 , . . . , x s l + n − } and A ( x p − , x k − ) = ∅ . From the maximality of i it follows that x k − x p − / ∈ D . Let l = 1. Then, since the path x s +1 . . . x s + n − cannot be extendedwith x p − , by Lemma 2 we obtain d ( x p − , A ) = d ( x p − , C ( s + 1 , k − d ( x p − , C ( k, s + n − ≤ | A | − , which contradicts (45). Let now l ≥
2. Then, since x p − cannot be inserted into the path x s l − + n − x s l − + n − . . . x s l + n − , by Lemma 2(ii) we have d ( x p − , A l ) = d ( x p − , C ( s l − + n − , k − d ( x p − , C ( k, s l + n − ≤ s l − s l − − | A l | − , which also contradicts (45). Claim 6.1 is proved. Claim 6.2. If l ∈ [1 , r − A ( x p − , { x s l − + n − , x s l − + n , . . . , x s l + n − } ) = ∅ ( s l − s l − is even) and { x s l − + n − , x s l − + n − , . . . , x s l + n − } → x p − → { x s l − + n − , x s l − + n − , . . . , x s l + n − , x s l + n − } , where s + n − s + 2. Proof.
Note that for all i ∈ [ s , p − k − | A ( x i → x p − ) | + | A ( x p − → x i +1 ) | ≤ , (47)since x p − cannot be inserted into no subpath of P with vertex set A l , l ∈ [1 , r ].We prove Claim 6.2 by induction on l . Assume that l = 1. We first show the following statement: (a). For all i ∈ [ s , s + n −
4] the vertex x p − is adjacent at most with one vertex from { x i , x i +1 } . Proof of (a).
Suppose that Statement (a) is not true. Then for some i ∈ [ s , s + n −
4] the vertex x p − is adjacent with x i and x i +1 . Then by (47), we only need to consider the following three cases:(i) x p − → { x i , x i +1 } ; (ii) { x i , x i +1 } → x p − ; (iii) x p − x i , x i +1 x p − ∈ D .We show that all these cases cannot occur. (i) x p − → { x i , x i +1 } . Let i with these properties be the smallest. From Claims 4 and 5 we have x p − x s +1 / ∈ D and x s x p − ∈ D . Therefore i ≥ s + 4 by Claim 6.1. Now from (47), Claim 6.117nd the minimality of i it follows that A ( x p − , x i − ) = ∅ , x i − x p − / ∈ D . Hence, since the paths Q := x s +1 x s +2 . . . x i − and Q := x i x i +1 . . . x s + n − cannot be extended with x p − and x p − x s +1 / ∈ D (Claim 5), x s + n − x p − / ∈ D (Claim 3), using Lemma 2, we obtain d ( x p − , A ) = d ( x p − , Q ) + d ( x p − , Q ) ≤ | Q | − | Q | = | A | − , which contradicts (45).Similarly we obtain a contradiction for the cases (ii) and (iii). Statement (a) is proved.From statement (a) and Claims 3, 5 it follows that A ( x p − , { x s +1 , x s + n − } = ∅ . Therefore byLemma 4b and d ( x p − , A ) = | A | − n is odd and A ( x p − , { x s +1 , x s +3 , . . . , x s + n − } ) = ∅ , O ( x p − , A ) = I ( x p − , A ) = { x s +2 , x s +4 , . . . , x s + n − } . Hence Claim 6.2 is true for l = 1.Now assume that Claim 6.2 holds for l −
1, 2 ≤ l ≤ r −
1, and prove it for l . By (46), x p − x s l − + n − , x p − x s l + n − ∈ D , and by the inductive assumption, x s l − + n − x p − ∈ D . Therefore by Claim 6.1, x p − x s l − + n − / ∈ D . Now similarly to Statement (a) we can prove the following Statement (b): (b). For all i ∈ [ s l − + n − , s l + n − x p − is adjacent with at most one vertex from { x i , x i +1 } .From (46) and statement (b) it follows that A ( x p − , { x s l − + n − , x s l + n − } ) = ∅ . Now from Lemma4b, since d ( x p − , A l ) = | A l | by (45), it follows that for all l ∈ [2 , r − s l − s l − is even and A ( x p − , { x s l − + n − , x s l − + n , . . . , x s l + n − } ) = ∅ ,O ( x p − , A l ) = I ( x p − , A l ) = { x s l − + n − , x s l − + n − , . . . , x s l + n − } . Claim 6.2 for 2 ≤ l ≤ r − Claim 6.3. If i ∈ [ s r + 1 , s r + n −
5] and x i − x p − ∈ D , then x p − x i +2 / ∈ D . Proof.
Suppose that i ∈ [ s r + 1 , s r + n −
5] and x i − x p − , x p − x i +2 ∈ D . If xx s r − ∈ D , then C n = xx s r − x s r − . . . x i − x p − x i +2 . . . x p − k − x , a contradiction. Hence xx s r − / ∈ D and x s r + n − x ∈ D by (*). From Claims 3 and 6.1 it follows that i ≥ s r − + n −
2. Let d := f ( s r , s r − + n − x p − x s r − + n − ∈ D by (46), if d is even, then C n = xx s x s +1 . . . x s + d x p − x s r − + n − . . . x s r + n − x and if d is odd, then C n = xx s x s +1 . . . x s + d − x p − x s r + n − . . . x p − k − x by Claim 6.2 and (46), a con-tradiction. Claim 6.3 is proved. Claim 6.4. A ( x p − , { x s r − + n − , x s r − + n , . . . , x s r + n − } ) = ∅ ( s r − s r − is even) and { x s r − + n − , x s r − + n − , . . . , x s r + n − } → x p − → { x s r − + n − , x s r − + n − , . . . , x s r + n − , x s r + n − } . Proof.
From Claim 6.2 we have x s r − + n − x p − ∈ D . Hence, x p − x s r − + n − / ∈ D by Claim 6.1. Then,since d ( x p − , A r ) = | A r | by (45), x p − k − x p − / ∈ D by (34), using Lemma 2 we obtain x s r − + n − x p − ∈ D .Again using Lemma 2, Claim 6.3 and (45), similarly to Satament (a), we can prove the following statement: (c). For all i ∈ [ s r − + n − , s r + n − x p − is adjacent at most with one vertex from { x i , x i +1 } .By (46), x p − → { x s r − + n − , x s r + n − } . Therefore, A ( x p − , { x s r − + n − , x s r + n − } ) = ∅ by Statement(c). Together with (45) and Lemma 4b this implies that O ( x p − , A r ) = I ( x p − , A r ) = { x s r − + n − , x s r − + n − , . . . , x s r + n − } . d ( x p − ) = p − δ o ( x p − ) ≥ p/ − | k − s | ≤ . (48) Claim 7 . s := a = 1. Proof.
Suppose, on the contrary, that a ≥
2. Then x x ∈ D and from Claim 6 it follows that { x a + n − , x a + n − } → x p − . (49)From this it is easy to see that a ≥ a = 2, a + n − n − C n = x x . . . x n − x p − x ).If x x ∈ D or xx a +1 ∈ D , then C n = xx a x a +1 . . . x a + n − x p − x x x or C n = xx a +1 . . . x a + n − x p − x x by (49), a contradiction. So we may assume that a ≥ x x / ∈ D and xx a +1 / ∈ D . Together with(32) this implies that d ( x, C (1 , a + 1)) ≤ a + 1. By Claim 6, x p − x a +2 ∈ D , and it is easy to seethat the cycle C ′ := x p − x a +2 x a +3 . . . x p − x p − has length at least n −
1. Using Lemma 1 we ob-tain that d ( x, V ( C ′ )) = | V ( C ′ ) | , d ( x, C (1 , a + 1)) = a + 1 and x a +1 x ∈ D . Hence it is clear that x x ∈ D . From (48) and a ≥ k ≥
2. Then by (*), xx s r +1 ∈ D . If n ≥
6, then C n = xx s r +1 x s r +2 . . . x p − k − x p − x x x x by Claim 6, a contradiction. So, we may assume that n = 5.It is easy to see that a = 3 and x x ∈ D (if a ≥
4, then by (46) x x p − ∈ D and C = x p − x x x x x p − ).Since n = 5 and the path P is covered with paths P , P , . . . , P r it follows that x → { x , x , . . . , x p − k − } .It is not difficult to see that if x i x ∈ D for some i ∈ [5 , p − k − C = xx i x x x x or C = xx i − x i x x x respectively for odd i and for even i , a contradiction. Thus we may assume that A ( { x , x , . . . , x p − k − } → x ) = ∅ . Together with Lemma 6(ii), (35) and xx / ∈ D this implies that I ( x ) ⊆ { x , x , x } (i.e., id ( x ) ≤ Claim 8. k = 1. Proof.
Suppose that Claim 8 is false, that k ≥
2. Then from (48) and Claim 7 it is easy to see that k = 2. Note that x p − x ∈ D and x p − x / ∈ D . By Claim 6, x p − x ∈ D . Hence if x x p − ∈ D , then C n = xx x p − x p − x . . . x n − x by (42), a contradiction. Thus A ( x , x p − ) = ∅ . By (46), x p − x p − and x p − x n − ∈ D . Now from x p − x n − ∈ D and Claim 6 it follows that x n − x p − ∈ D . There-fore if x p − x ∈ D , then C n = x x . . . x n − x p − x p − x p − x , a contradiction. So we may assume that x p − x / ∈ D . Then, since x p − x p − / ∈ D by (33), and the path x x . . . x p − cannot be extended with x p − (otherwise some P i , i ∈ [1 , r ], path can be extended with x p − and C n ⊂ D ), from Lemma 2 it fol-lows that d ( x p − , { x , x , . . . , x p − } ) ≤ p −
6. Now from d ( x p − ) ≥ p − x p − x / ∈ D and A ( x p − , x ) = ∅ we obtain that x p − x p − and x p − x p − ∈ D . By Claim 6 we also have x n − x p − ∈ D . Therefore C n = xx x . . . x n − x p − x p − x p − x , a contradiction. Claim 8 is proved.Now we shall complete the discussion of Case 2.1.Using the Claims 6-8 it is not difficult to see that p = 2 m + 1 and n is odd (if n is even, then by Claim6, x n − x p − ∈ D and C n = x p − x x . . . x n − x p − ). Now we consider the cycle C ′ := xx x . . . x p − x of the length p −
1. It is easy to see that for this cycle C ′ we have the considered Subcase 1.2. Thenanalogously to Claim 6, we obtain { x , x , . . . , x p − } → x → { x , x , . . . , x p − } , A ( x, { x , x , . . . , x p − } ) = ∅ and A ( h{ x, x , x , . . . , x p − , x p − }i ) = ∅ , i.e. K ∗ m,m +1 ⊆ D ⊆ ( K m + K m +1 ) ∗ This contradicts the our initial supposition, and completes the discussion of Case 2.1.19 ase 2.2.
The path P cannot be covered with paths P , P , . . . , P r − , P r . Then n ≤ m + 1. Remark.
It is easy to see that in digraph ←− D the path x p − k − x p − k − . . . x s +1 x s also cannot be coveredwith corresponding paths. Therefore in the further we assume that id ( x ) = ⌊ p/ ⌋ := m .For convenience, in the following let q := s r − . From the maximality of q it follows that if x j ∈ P q and xx j ∈ D , then x → C ( q, j ). Since xx p − k − n +1 / ∈ D , xx q ∈ D and q + n − ≤ p − k − n + 1, there isan s ∈ [ q, p − k − n ] such that xx s +1 / ∈ D and x → C ( q, s ) . (50)Therefore x s + n − x ∈ D and A ( C ( q + n − , s + n − → x ) = ∅ by (*) . (51) Subcase 2.2.1. s ≥ q + n − f ( q, s ) ≥ n − A ( x s +1 → C ( s + 3 , s + n − ∅ , (52)(otherwise i ∈ [0 , n − x s +1 x s +3+ i ∈ D and C n = xx s − i x s − i +1 . . . x s +1 x s + i +3 . . . x s + n − x by (50)and (51)). If z ∈ C ( s + n − , p − ∪ C (1 , s − n + 3) and zx ∈ D , then x s +1 z / ∈ D (otherwise C n = xx s − n +4 x s − n +5 . . . x s +1 zx by (50)). From this and (51) it follows that x s +1 does not dominate atleast id ( x ) − id ( x, C ( s − n + 4 , q + n − C ( s + n − , p − ∪ C (1 , s − n + 3).Therefore, since x s +1 x / ∈ D , f ( s + 3 , s + n −
2) = n − id ( x ) − id ( x, C ( s − n + 4 , q + n − n − ≤ m. Hence n − ≤ m − id ( x ) + id ( x, C ( s − n + 4 , q + n − . Together with id ( x ) = m and id ( x, C ( s − n +4 , q + n − ≤ n − id ( x, C ( s − n +4 , q + n −
3) = n −
3. Hence s = q + n − C ( q + 1 , q + n − → x . Now from q ≥ x p − x / ∈ D we obtainthat there is an i ∈ [1 , q + 1] such that x i x ∈ D and x i − x / ∈ D . By our arguments above we have x s +1 x i − ∈ D and C n = xx q +2 . . . x s +1 x i − x i x , a contradiction. Subcase 2.2.2. s ≤ q + n − s and (*) immediately follows the following: Claim 9. (i) A ( x → C ( s + 1 , q + n − ∅ and (ii) C ( s + n − , q + 2 n − → x .Let h := f ( q, s −
1) and h := f ( s, q + n − ≤ h ≤ n − h ≥ h + h = n − Notation.
Let Y denote the set of vertices x i ∈ C ( q + n − , p − ∪ C (1 , q −
1) for which there is a vertex x j ∈ C ( s + n − , p − ∪ C (1 , s −
1) such that x j x ∈ D and the path x i x i +1 . . . x j has at most h +1 vertices.From Claim 13 and f ( q + n − , s + n −
2) = h it follows that C ( q + n − , s + n − ⊂ Y . Claim 10. A ( x q + n − → Y ) = ∅ and | Y | ≤ m . Proof.
Suppose that A ( x q + n − → Y ) = ∅ , that is x q + n − x i ∈ D for some x i ∈ Y . Then by thedefinition of Y there is a vertex x j ∈ C ( s + n − , p − ∪ C (1 , s −
1) such that x j x ∈ D and the path x i x i +1 . . . x j contains at most h +1 vertices. Therefore C n = x q + n − x i x i +1 . . . x j xx q + d − x q + d . . . x q + n − by (50), where d := |{ x i , x i +1 , . . . , x j }| , a contradiction. This proves that A ( x q + n − → Y ) = ∅ . Hence20t is clear that | Y | ≤ m . The claim is proved. Notation.
For each t ∈ [0 , h −
1] let R t denote the set of vertices x i / ∈ C ( q + n − , q + 2 n −
5) for which x i − t / ∈ C ( q + n − , s + n −
1) and xx i − t ∈ D , and let Z := ∪ h − t =0 R t . Claim 11. If t ∈ [0 , h − A ( R t → x q + n − ) = ∅ (i.e., A ( Z → x q + n − ) = ∅ ) and | Z | ≤ m . Proof.
Suppose that the claim is false, that is t ∈ [0 , h −
2] and there is a vertex x i ∈ R t such that x i x q + n − ∈ D . By the definition of R t we have xx i − t ∈ D and, by Claim 9(ii), x q +2 n − − t x ∈ D . There-fore C n = xx i − t . . . x i x q + n − . . . x q +2 n − − t x , a contradiction. The claim is proved. Claim 12. If x p − ∈ Y ∪ Z , then n < p − m . Proof.
Suppose, on the contrary, that x p − ∈ Y ∪ Z and n ≥ p − m . Therefore m + 1 ≥ n ≥ m .We first prove that q = s (i.e., s = s r − and r = 2). Assume that q = s . By the definition of s i wehave s i − ≤ s i −
2. Hence 1 ≤ s ≤ q − q ≥
3. Then p − ≥ n + k + q − q + n − ≤ p − k − n +1.From this and m + 1 ≥ n ≥ m it is not difficult to see that n = m . Since n ≥ p − m , we obtain p = 2 m , k = s = 1, q = 3 and q + n − p − k − n + 2 = m + 1. From C n D it is easy to see that x p − x m / ∈ D and x m − x p − / ∈ D . It follows from (34) and (35) that x p − x p − / ∈ D and x p − x / ∈ D . Then, sincethe paths x x . . . x m − and P r = x m +1 x m +2 . . . x p − cannot be extended with x p − , using Lemma 2,(34), (35) and d ( x p − ) ≥ m − x p − x m +1 , x p − x p − , x m x p − , x x p − ∈ D . Therefore if x p − x ∈ D , then C n = xx x p − x m +1 . . . x p − x and if x p − x / ∈ D , then xx m − ∈ D (i.e., s = m −
1) by(*), xx ∈ D by (50), and C n = xx x . . . x m x p − x p − x , a contradiction. This proves that q = s .Let n = m + 1 and p = 2 m . Then k = s = 1 and n − p − k − n + 1 = m −
1. It is easy to seethat x m x p − / ∈ D . If x p − ∈ Y , then x m − x p − / ∈ D by Claim 10, if x p − ∈ Z , then x p − x m / ∈ D byClaim 11, and A ( x m , x p − ) = ∅ . Therefore, since the paths P and P r cannot be extended with x p − and x p − x / ∈ D by (34), x p − x p − / ∈ D by (35), using Lemma 2 we obtain that x x p − , x p − x m +1 ∈ D and C n = xx x p − x m +1 . . . x p − x , a contradiction.Let now n = m + 1 and p = 2 m + 1. It is easy to see that A ( x m , x p − ) = ∅ , 1 ≤ k, s ≤
2. By (34) and (35), x p − x p − / ∈ D and x p − x / ∈ D . Therefore, since d ( x p − ) ≥ p − x x . . . x m − and x m +1 x m +2 . . . x p − cannot be extended with x p − , using Lemma 2 we get that x x p − , x m − x p − , x p − x m +1 ∈ D . If s = 2, then x x, xx ∈ D and C n = x m − x p − x xx . . . x m − x ,a contradiction. So we may assume that s = 1. Then xx ∈ D , and since s ≤ q + n − m − xx m − / ∈ D . Hence x p − x ∈ D by (*), and C n = xx x p − x m +1 . . . x p − x , a contradiction.Let finally n = m . From n ≥ p − m it follows that p = 2 m , 1 ≤ k ≤ ≤ s = q ≤
3. Since C n D , it is easy to see that x m − x p − / ∈ D and x p − x m / ∈ D. (53)Now we shall consider the cases k = 3, k = 2 and k = 1 separately. Case. k = 3. Then q = s = 1 ( x q + n − = x m − ) and x p − x p − / ∈ D by (34). Hence, since the paths x x . . . x m − and x m − . . . x p − cannot be extended with x p − , from Lemma 2, (53) and (34) it followsthat x m − x p − , x p − x m − ∈ D , which contradicts Claim 10 or 11. Case. k = 2. Then p − k − n + 2 = m and s ≤
2. Let s = 2. Then, since x p − x p − / ∈ D by (34), x p − x / ∈ D by (35), and the paths x x . . . x m − and x m x m +1 . . . x p − cannot be extended with x p − ,using (53) and Lemma 2(iii), we get that d ( x p − ) ≤ m −
2, a contradiction. If s = 1, then again using(34), (53) and Lemma 2, we obtain x m − x p − , x p − x m − ∈ D , which also contradicts Claim 10 or 11.21 ase. k = 1. Then p − k − n + 2 = m + 1 and 1 ≤ s ≤ s = 3, then xx ∈ D and, since the paths x x . . . x m and x m +1 x m +2 . . . x p − cannot be extendedwith x p − , using (34), (35), (53) and Lemma 2, we obtain x p − x p − , x m x p − , x x p − , x p − x m +1 ∈ D .Therefore, if xx ∈ D , then C n = xx . . . x m x p − x p − x and if xx / ∈ D , then by (*) and the definition of s we have x p − x ∈ D and C n = xx x p − x m +1 . . . x p − x , a contradiction.If s = 2, then xx ∈ D . Using (35), (53) and Lemma 2, we obtain x x p − , x p − x m +1 ∈ D .From s ≤ q + n − xx m − / ∈ D . Then by (*), x p − x ∈ D ( p − m −
3) and C n = xx x p − x m +1 . . . x p − x , a contradiction.Now assume that s = 1. If x p − ∈ Y , then x m − x p − / ∈ D by Claim 10, and if x p − ∈ Z , then x p − x m − / ∈ D by Claim 11. Since the paths x x . . . x m − and x m +1 x m +2 . . . x p − cannot be extendedwith x p − and d ( x p − ) ≥ p −
1, using (34), (35), (53) and Lemma 2, we obtain x p − x m +1 , x x p − , x m x p − ∈ D . Therefore, if xx m − / ∈ D , then by (*), x p − x ∈ D and C n = xx x p − x m +1 . . . x p − x and if xx m − ∈ D , then C n = xx m − x m x p − x m +1 . . . x p − x since x p − x ∈ D , a contradiction. This completesthe proof of Claim 12. Claim 13.
Let x i ∈ C ( q + n − , p − ∪ C (1 , q −
1) and xx i ∈ D . Then(i) x i + n − x q + n − / ∈ D and (ii) if x q + n − x q + d ∈ D , where d ∈ [0 , n − x i + d x q + n − / ∈ D . Proof.
Assume that the claim is not true. Then (i) x i + n − x q + n − ∈ D and C n = xx i x i +1 . . . x i + n − x q + n − x ; (ii) x i + d x q + n − ∈ D and C n = xx i x i +1 . . . x i + d x q + n − x q + d . . . x q + n − x , a contradiction. Claim 14. If xx q + n − / ∈ D (i.e., s ≤ q + n − A ( C ( q, s ) → x q + n − ) = ∅ . Proof.
By (*), x q +2 n − x ∈ D . If the claim is not true, then x i x q + n − ∈ D , where x i ∈ C ( q, s ), and xx i ∈ D by (50). Hence C n = xx i x q + n − x q + n − . . . x q +2 n − x , a contradiction. Notation.
For all j ∈ [1 , n −
2] let H j denote the set of vertices x i / ∈ { x q + n − , x q + n − } for which x i + j − / ∈ C ( q + n − , q + 2 n −
5) and x i + j − x ∈ D . Claim 15. If xx q + n − ∈ D and x q +2 n − − j x q + n − ∈ D , where j ∈ [1 , n − A ( x q + n − → H j ) = ∅ . Proof.
If the claim is not true, then x q + n − x i ∈ D , where x i ∈ H j , and C n = xx q + n − . . . x q +2 n − − j x q + n − x i . . . x i + j − x , a contradiction.Further, let α := | O ( x ) ∩ C ( q + n − , s + n − | and α := | O ( x ) ∩ C ( s + n, q + 2 n − | .Note that α ≤ h + 2 and α ≤ h − Claim 16. s ≥ q + 1. Proof.
Suppose, on the contrary, that s = q . Then h = n − ≥ α ≤
2. By Claim 9(ii), C ( q + n − , q + 2 n − → x, i.e., | I ( x ) ∩ { x q + n − , x q + n − , . . . , x q +2 n − }| = n − . (54)Note that if x i / ∈ C ( q + n − , q + 2 n −
5) and x i x ∈ D , then x i ∈ H and x i − ∈ H . Therefore | H | = | H | = m − n + 3 by id ( x ) = m and (54). From id ( x ) = m it follows that there is a vertex x j / ∈ C ( q + n − , q + 2 n −
4) such that x j x / ∈ D . From this we obtain that the set ∪ n − i =2 H i (respectively, ∪ n − i =1 ,i =2 H i ) contains at least n − H (respectively, H ). Now it is not difficultto show the following inequalities:a) . (cid:12)(cid:12) n − [ j =1 H j (cid:12)(cid:12) ≥ m and b) . if i ∈ [1 , n − , then (cid:12)(cid:12) n − [ j =1 ,j = i H j (cid:12)(cid:12) ≥ m − xx q + n − ∈ D, xx q + n − / ∈ D, then (cid:12)(cid:12) n − [ j =1 H j (cid:12)(cid:12) ≥ m. (56)From xx q − / ∈ D , s = q , Claim 9(i) and p ≥
10 it follows that O ( x ) ∩ { x q + n , x q + n +1 , . . . , x q − } 6 = ∅ . By Claim 9(i) and the definitions of α , α we have od ( x, C ( q + 2 n − , p − ∪ C (1 , q )) = od ( x ) − α − α . Therefore | R | = od ( x ) − α − α and | R | ≥ od ( x ) − α − α . Note that { x q +1 , x q +2 , . . . , x q + n − } ⊂ n − [ t =1 R t . Now it is not difficult to see that for each j ∈ [0 ,
1] the following inequality holds (cid:12)(cid:12) n − j [ t = j R t \ { x q +2 n − j } (cid:12)(cid:12) ≥ od ( x ) − α − α + n − . (57)We now distinguish several cases. Case 1. xx q + n − / ∈ D . Then α ≤ α + α ≤ n −
3. From Claim 11 and (57) it follows that m ≥ (cid:12)(cid:12) n − [ t =0 R t (cid:12)(cid:12) + 1 ≥ od ( x ) − α − α + n − . Now it is easy to see that α + α = n −
3, (in particular, xx q + n − ∈ D ), od ( x ) = m − p = 2 m andby Claim 11, id ( x q + n − ) ≤ m − C ( q + n − , q + 2 n − → x q + n − . Therefore Claim 15 holds.On the other hand, by (55b) we have | ∪ n − i =1 H i | ≥ m −
1. Then, since x q + n − x / ∈ D , we obtain that od ( x q + n − ) ≤ m − d ( x q + n − ) ≤ m −
2, a contradiction.
Case 2. xx q + n − ∈ D . Note that, by (*), x q +2 n − x / ∈ D and similarly to Claim 15 , one can show that A ( x q + n − → H n − ) = ∅ . (58)For each t ∈ [1 , n −
3] it is easy to see that A ( R t \ { x q +2 n − } → x q + n − ) = ∅ . (59)Indeed, if x i x q + n − ∈ D for some x i ∈ R t \ { x q +2 n − } , then x q +2 n − − t x ∈ D by (54), and C n = xx i − t x i − t +1 . . . x i x q + n − . . . x q +2 n − − t x , a contradiction. Case 2.1. xx q + n − / ∈ D . Then α = 1. From (57) and (59) we have m − ≥ (cid:12)(cid:12) n − [ t =1 R t \ { x q +2 n − } (cid:12)(cid:12) ≥ od ( x ) − α − α + n − . It follows that α = n − x → C ( q + n, q + 2 n − od ( x ) = m − p = 2 m , id ( x q + n − ) ≤ m − C ( q + n, q + 2 n − → x q + n − . This implies that A ( x q + n − → H j ) = ∅ for each j ∈ [1 , n −
3] (otherwiseif x i ∈ H j and x q + n − x i ∈ D , then C n = xx q + n . . . x q +2 n − − j x q + n − x i x i +1 . . . x i + j − x ). Hence, using2356), we get od ( x q + n − ) ≤ m −
1. So d ( x q + n − ) ≤ m −
2, a contradiction.
Case 2.2. xx q + n − ∈ D . Then α = 2.Suppose that α = n − x → C ( q + n − , q + 2 n − A ( { x q +2 n − , . . . , x q +3 n − } → x ) = ∅ by (*). For each j ∈ [1 , n −
2] it is easy to see that A ( x q +2 n − → H j ) = ∅ (otherwise if x i ∈ H j and x q +2 n − x i ∈ D , then C n = xx q + n + j − . . . x q +2 n − x i x i +1 . . . x i + j − x ). From this and (55a) it followsthat x q +2 n − x q + n − ∈ D . Then it is easy to see that A ( x q + n − → H j ) = ∅ for each j ∈ [1 , n − x q + n − x / ∈ D it follows that od ( x q + n − ) ≤ p/ −
2, a contradiction.Now suppose that α ≤ n −
5. If C ( q + n − , q + 2 n − → x q + n − , then, since x q + n − x / ∈ D , from(55a), Claim 15 and (58) it follows that od ( x q + n − ) ≤ p/ −
2, a contradiction. So we may assume thatthere is an l ∈ [1 , n −
3] such that x q +2 n − − l x q + n − / ∈ D . Hence, using Claim 11 and inequality (57)(when j = 0), we obtain α = n − od ( x ) = m − p = 2 m , id ( x q + n − ) ≤ m − C ( q + n − , q + 2 n − \ { x q +2 n − − l } → x q + n − . Therefore, A ( x q + n − → H j ) = ∅ for each j ∈ [1 , n − \ { l } by Claim 15. Hence from (55b), (58) and x q + n − x / ∈ D it follows that od ( x q + n − ) ≤ m −
1. Thus we have d ( x q + n − ) ≤ m −
2, a contradiction.Claim 16 is proved.
Notation.
Let x l / ∈ C ( q, q + n −
3) be a vertex such that x l x ∈ D and the path x l x l +1 . . . x q is as shortas possible, and let β := | I ( x ) ∩ C ( q, s − | , β := | I ( x ) ∩ C ( s, q + n − | and b + 1 := |{ x l , x l +1 , . . . , x q − }| . Claim 17. | Y | ≥ m − β + h . Proof.
Using (51) it is easy to see that | I ( x ) ∩ { x s + n − , x s + n , . . . , x l }| = m − β − β ,C ( q + n − , s + n − ∪ ( I ( x ) ∩ { x s + n − , x s + n , . . . , x l } ) ⊂ Y. (60)If b ≥ β , then Y contains at least β vertices from the set { x l , x l +1 , x l +2 , . . . , x q − } \ { x l } and | Y | ≥ | I ( x ) ∩ { x s + n − , x s + n , . . . , x l }| + h + β ≥ m − β + h . Therefore Claim 17 holds for b ≥ β . So we may assume that b ≤ β −
1. It is clear that β ≥ b + 1 ≥ { x l , x l +1 , x l +2 , . . . , x q − } ⊂ Y. (61)Suppose that |{ x s + n − , x s + n , . . . , x l }| ≤ id ( x ) − β − b − id ( x ) − ( β + β ) + β − b − . Then from β − b ≤ h , (60), (61) and the definition of Y it follows that Y = { x q + n − , x q + n , . . . , x q − } , | Y | = p − n and x p − ∈ Y . By Claim 12, n < p − m (i.e., m < p − n ). On the other hand, | Y | ≤ m byClaim 10, and hence p − n ≤ m . This contradicts that m < p − n .Now suppose that |{ x s + n − , x s + n , . . . , x l }| ≥ id ( x ) − ( β + β ) + β − b. Then, since β − b ≤ h , at least β − b vertices from Y ∩ { x s + n − , x s + n , . . . , x l } are not dominate thevertex x . Therefore, by (60) and (61), | Y | ≥ id ( x ) − ( β + β ) + ( β − b ) + b + h ≥ m − β + h , Claim 18. h ≤ h − Proof.
Suppose, on the contrary, that h ≥ h −
1. Note that h ≥ h ≥ s ≥ q + 1.Let F := O ( x ) ∩ { x q + n − , x q + n − , . . . , x q − } and if j ≥
1, then let F j := { x i /x i − ∈ F j − } .Using Claim 9(i) we see that for all j ≥ | F j | = od ( x ) − h − . (62)We now show that xx q + n − ∈ D (i.e., s = q + n −
4) or x q + n − x / ∈ D. (63)Suppose, on the contrary, that is xx q + n − / ∈ D and x q + n − x ∈ D . Then h ≥ h ≥ n ≥ A ( x → { x q − , x q − } ) = ∅ . Therefore F i ∩ C ( q, q + n −
3) = ∅ , i ∈ [0 , i, j ∈ [0 ,
2] the following holds | F i ∪ F j | ≥ od ( x ) − h . (64)From β ≤ h ≤ h +1 and Claim 17 it follows that | Y | ≥ m −
1. Then by Claim 10, x q + n − → { x q + i , x q + j } for some distinct i, j ∈ [0 , A ( F i ∪ F j → x q + n − ) = ∅ by Claim 13. Hence, using (64), xx q + n − / ∈ D and Claim 14, we see that id ( x q + n − ) ≤ p/ −
2, a contradiction. So (63) is proved.Let xx q + n − / ∈ D (i.e., s ≤ q + n − x q + n − x / ∈ D , A ( x, x q + n − ) = ∅ and β ≤ h −
1. Recall that | Y | ≤ m (Claim 10) and | Y | ≥ m − β + h (Claim 17). Hence h ≤ β .Therefore h − ≤ h ≤ β ≤ h − h = β = h − C ( s, q + n − → x . It follows fromClaims 10 and 17 that | Y | = m and x q + n − → C ( q, q + n − A ( F → x q + n − ) = ∅ , then C n = xx i x i +1 x q + n − x q x q +1 . . . x q + n − x , a contradiction. So we may assume that A ( F → x q + n − ) = ∅ .Since xx q + n − / ∈ D , using (62) and Claim 14, we see that id ( x q + n − ) ≤ p − od ( x ) − ≤ m −
1. Hence p = 2 m . On the other hand, from | Y | = m and Claim 10 it follows that od ( x q + n − ) ≤ m −
1. Therefore d ( x q + n − ) ≤ m −
2, a contradiction.Let now xx q + n − ∈ D (i.e., s = q + n − h = 2 and β ≤ n ≥
6. Then from h = 2 ≥ β ≥ h ≥ β = 2 (i.e., x q + n − x ∈ D ) and n = 6. By Claims 10 and 17, | Y | = m and x q + n − → C ( q, q + n − A ( x → { x q − , x q − } ) = ∅ and (62), we have | ∪ i =0 F i | ≥ od ( x ) − h + 2. It follows from Claim 13 that A ( ∪ i =0 F i → x q + n − ) = ∅ . Together with xx q + n − / ∈ D this implies that id ( x q + n − ) ≤ p − − od ( x ) ≤ m −
1. Thus p = 2 m . On the other hand, from | Y | = m and Claim 10 we have id ( x q + n − ) ≤ m − d ( x q + n − ) ≤ m −
2, which is a contradiction.So suppose next that n = 5. Note that x q + n − = x q +2 .Let x q +1 x / ∈ D . If x i ∈ I ( x ) \ { x q +2 } , then x q +1 x i − / ∈ D ( otherwise C = xx q +1 x i − x i − x i x ) andif x i ∈ I ( x ) \ { x q , x q +2 } , then x q +1 x i − / ∈ D ( otherwise C = xx q x q +1 x i − x i x ). Since x q +1 x / ∈ D , id ( x ) = m and the number of such vertices x i − and x i − at least m , we obtain od ( x q +1 ) ≤ p − m −
2, acontradiction.Let now x q +1 x ∈ D . Then β = 2 and A ( x → { x q − , x q − } ) = ∅ by (*). By (62), | [ i =0 F i | ≥ od ( x ) . (65)Assume that x q +2 → { x q , x q +1 } . From Claim 13 it follows that A ( ∪ i =0 F i → x q +2 ) = ∅ . Together with(65) and xx q +2 / ∈ D this implies that x q x q +2 ∈ D . It is not difficult to see that A ( x q +3 → I ( x ) \{ x q +2 } ) =25 . This and x q +3 x / ∈ D imply that od ( x q +3 ) ≤ p − id ( x ) − x q +3 x q +2 ∈ D . Since x q +2 → { x q , x q +1 } ,it follows that A ( O ( x ) ∪ { x } → x q +3 ) = ∅ . Hence id ( x q +3 ) ≤ p − − od ( x ). Together with the fact that od ( x q +3 ) ≤ p − id ( x ) − d ( x q +3 ) ≤ p −
2, a contradiction.Now assume that | A ( x q +2 → { x q , x q +1 } ) | ≤
1. Using Claims 10 and 17, we obtain that | Y | = m − x q +2 x q + j ∈ D for some j ∈ [0 , x q x / ∈ D . Then xx q − ∈ D by(*), and | B := I ( x ) ∩{ x q +3 , x q +4 , . . . , x q − }| = m − . From this and | Y | = m − B = { x q +5 , x q +6 , . . . , x q + m +2 } .Hence x q − x / ∈ D and x q +2 x q − ∈ D by Claim 10, x q − / ∈ Y . Now, since xx q − ∈ D , we obtain A ( x q − → I ( x ) ∪ { x } ) = ∅ . Thus od ( x q − ) ≤ p − m −
2, which is a contradiction. Let now x q x ∈ D .Then A ( x → { x q − , x q − , x q − } ) = ∅ by (*). It is not difficult to see that A ( F ∪ F ∪ { x, x q +2 , x q +3 } → x q − } ) = ∅ . Therefore, since | F ∪ F | ≥ od ( x ) −
1, it follows that xx q +3 ∈ D , | F ∪ F | = od ( x ) − F = { x q +3 , x q +4 , . . . , x q + od ( x ) } and { x q , x q +1 } → x q − . Hence, if x i / ∈ { x q , x q +1 } and xx i ∈ D , then A ( { x i + j , x i +2 } → x q +2 ) = ∅ . Since A ( { x, x q − } → x q +2 ) = ∅ , we conclude that id ( x q +2 ) ≤ p − od ( x ) − A ( x q +2 → Y ∪ { x q +1 − j } ) = ∅ . Because of this and | Y | = m − od ( x q +2 ) ≤ p − m −
1. Therefore d ( x q +2 ) ≤ p −
2, a contradiction. Claim 18 is proved.Note that Claims 16 and 18 imply that n ≥ Claim 19. (i) L := A ( x q + n − → C ( q + n, q + 2 n − ∅ and(ii) | A ( x q +2 n − → x q + n − ) | + | A ( x → x q + n − ) | ≤ Proof.
Note that s ≥ q + 1 by Claim 16.(i) Suppose that L = ∅ , and let x q + n − x i ∈ L . By (50) and Claim 9(ii), we have if x i ∈ C ( q + n, s + n − C n = xx s − f ( q + n − ,i − . . . x q + n − x i . . . x s + n − x and if x i ∈ C ( s + n − , q + 2 n − C n = xx q +1 . . . x q + n − x i x , a contradiction.(ii) Suppose that x q +2 n − x q + n − and xx q + n − ∈ D . Then for j = 1 Claim 15 holds (i.e., A ( x q + n − → H ) = ∅ ). From this, Claim 19(i) and (51) it follows that A ( x q + n − → C ( q + n, q + 2 n − ∪ I ( x )) = ∅ .Therefore, since x q + n − x / ∈ D and | I ( x ) | = m , we see that od ( x q + n − ) ≤ p/ −
2, a contradiction. Claim19 is proved.
Claim 20. If x q +2 n − x / ∈ D and i ∈ [0 , | A ( x q +2 n − i → x q + n − ) | + | A ( x → x q + n − i ) | ≤ . Proof.
Assume that the claim is false, that x q +2 n − x / ∈ D , i ∈ [0 ,
1] and x q +2 n − i x q + n − , xx q + n − i ∈ D . It is easy to see that if x j / ∈ A ( q + n − , q + 2 n −
4) and x j x ∈ D , then x q + n − x j − / ∈ D (oth-erwise C n = xx q + n − i . . . x q +2 n − i x q + n − x j − x j x ). Together with A ( { x q + n − , x q + n − } → x ) = ∅ (by s ≥ q + 1 and (*)) and Claim 19(i) this implies that the vertex x q + n − does not dominate at least id ( x ) + 1 = m + 1 vertices. Thus od ( x ) ≤ p/ −
2, a contradiction. Claim 20 is proved.
Claim 21. | Z | ≥ od ( x ) − α + h − Proof.
Let B := { x q +2 n − , x q +2 n − , . . . , x s − , x s } . Note that | B | ≥ od ( x ) − α − α + 1 and C ( s +1 , q + n − ⊆ Z by xx q − / ∈ D and by Claim 9(i). Hence for α = 0 Claim 21 is true. Assume that α ≥
1. If | B | ≥ od ( x ) − α , then at least α vertices of B are not in O ( x ), and the set Z contains atleast od ( x ) − α vertices from B since α ≤ h −
2. From this and C ( s + 1 , q + n − ⊂ Z , we obtain | Z | ≥ od ( x ) − α + h −
2, and the claim is holds for this case. Assume that | B | ≤ od ( x ) − α − . It is notdifficult to see that Z = C \ { x q + n − , x q + n − , . . . , x q +2 n − } . Therefore, by Claim 11, m ≥ | Z | = p − n and x p − ∈ Z . Hence, by Claim 12, n < p − m (i.e., m < p − n ), a contradiction. Claim 21 is proved. Claim 22. If L := A ( { x q +2 n − , x q +2 n − } → x ) = ∅ , then x q +2 n − x q + n − / ∈ D . Proof.
Assume that L = ∅ and x q +2 n − x q + n − ∈ D . Note that xx q + n − ∈ D by(*). Hence A ( x q + n − → H ) = ∅ by Claim 15, and A ( x q + n − → C ( q + n, q + 2 n − ∅ by Claim 19(i).26hen, since x q + n − x / ∈ D , (51) and | H | ≥ id ( x ) − n + 4, we obtain that x q + n − does not dominate atleast m + 1 vertices, a contradiction. Claim 22 is proved. Claim 23. | A ( x → { x q + n − , x q + n − } ) | ≤ Proof.
Suppose, on the contrary, that is x → { x q + n − , x q + n − } . Then L := A ( { x q +2 n − , x q +2 n − }→ x ) = ∅ by (*). Therefore from Claims 19(ii), 20 ( when γ = 0) and 22 it follows that L := A ( { x q +2 n − , x q +2 n − , x q +2 n − } → x q + n − ) = ∅ . Hence | Z | ≤ m − od ( x ) − α + h − ≤ | Z | ≤ m − od ( x ) = m − h ≤ α ≤ h + 2 ≤ h . Clearly, α = h = h + 2 ≥ | Z | = m − α = h + 2 means that x → C ( q + n − , s + n −
1) and hence by (*) we have L = A ( { x q +2 n − , x q +2 n − , x q +2 n − } → x ) = ∅ . Let E denote the set of vertices x j / ∈ C ( q + n − , q + 2 n −
5) for which x j +3 x ∈ D . Then | E | ≥ m − h + 1 by L = ∅ . Together with x q + n − x / ∈ D , L = ∅ and Claim 19(i) this impliesthat x q + n − x j ∈ D for some x j ∈ E . If n = 6, then C n = xx q + n − x j x j +1 x j +2 x j +3 x , a contradiction.Assume that n ≥
7. From | Z | = m − L = ∅ and Claim 11 it follows that x q +2 n − x q + n − ∈ D and C n = xx q + n − . . . x q +2 n − x q + n − x j x j +1 x j +2 x j +3 x , a contradiction. Claim 23 is proved. Claim 24. (i) xx q + n − / ∈ D , x q +2 n − x ∈ D and (ii) xx q + n − ∈ D , x q +2 n − x / ∈ D . Proof. (i) Suppose that xx q + n − ∈ D . Then xx q + n − / ∈ D by Claim 23 and x q +2 n − x q + n − / ∈ D byClaim 19(ii). Together with Claims 11, 18, 21 and α ≤ h + 1 ≤ h − | Z | = m − od ( x ) = m − p = 2 m , h = h − α = h + 1, id ( x q + n − ) ≤ m − x q +2 n − x q + n − ∈ D . Since xx q + n − / ∈ D and xx q + n − ∈ D , by (*) we have x q +2 n − x ∈ D and x q +2 n − x / ∈ D .Let E denote the set of vertices x j / ∈ C ( q + n − , q + 2 n −
4) for which x j +1 x ∈ D . Since | I ( x ) ∩ C ( q + n − , q + 2 n − | = h , (66)it is easy to see that | E | ≥ m − h . If x q + n − x i ∈ D , where x i ∈ E , then C n = xx q + n − . . . x q +2 n − x q + n − x i x i +1 x , a contradiction. So we can assume that A ( x q + n − → E ) = ∅ . From this, Claim 19(i)and x q + n − x / ∈ D it follows that od ( x q + n − ) ≤ m −
1. Therefore d ( x q + n − ) ≤ m −
2, a contradiction.This proves that xx q + n − / ∈ D , and hence x q +2 n − x ∈ D by (*).(ii) Suppose that xx q + n − / ∈ D . Then | Z | ≤ m − xx q + n − / ∈ D by Claims 24(i), and α ≤ h ≤ h − od ( x ) = m − p = 2 m , α = h and | Z | = m −
1. It follows from Claim 11 that id ( x q + n − ) ≤ m − x q +2 n − x q + n − ∈ D .Then, since α = h and A ( x → { x q + n − , x q + n − } ) = ∅ , we have xx q + n ∈ D .Let E ′ denote the set of vertices x j / ∈ C ( q + n − , q + 2 n −
5) for which x j +1 x ∈ D . It is easy to seethat | E ′ | = m − h by (66), and A ( x q + n − → E ′ ) = ∅ (otherwise if x q + n − x j ∈ D , where x j ∈ E ′ , then C n = xx q + n . . . x q +2 n − x q + n − x j x j +1 x ). This together with Claim 19(i) and x q + n − x / ∈ D implies that od ( x q + n − ) ≤ m −
1. Therefore d ( x q + n − ) ≤ m −
2, a contradiction. This shows that xx q + n − ∈ D ,and hence x q +2 n − x / ∈ D by (*). Claim 24 is proved. Claim 25. L := A ( x q + n − → { x q , x q +1 } ) = ∅ . Proof.
Suppose, on the contrary, that L = ∅ . It is easy to see that if x q + n − x q ∈ D , then C n = x q + n − x q x q +1 . . . x q + n − and if x q + n − x q +1 ∈ D , then C n = x q + n − x q +1 . . . x q + n − xx q + n − x q + n − since xx q + n − ∈ D by Claim 24(ii), a contradiction.27 laim 26. If s ≥ q + 2, then L := A ( x q + n − → C ( s + n − , q + 2 n −
5) = ∅ . Proof. If x q + n − x i ∈ L , then by Claim 9(ii), C n = xx q +2 x q +3 . . . x q + n − x i x , a contradiction. Claim 27. xx q + n / ∈ D . Proof.
Suppose, on the contrary, that xx q + n ∈ D . Note that x q +2 n − x / ∈ D (Claim 24(ii)), s ≥ q + 1(Claim 16) and x q +2 n − x q + n − / ∈ D (Claim 20, when γ = 1). Therefore from Claim 11 it follows that | Z | ≤ m −
1. Using Claim 9(ii) and (51), we obtain | I ( x ) ∩ C ( q + n − , q + 2 n − | = h − | H | ≥ m − h + 1 . (67)Now we distinguish two cases. Case 1. x q +2 n − x q + n − ∈ D . It is easy to see that L := A ( x q + n − → H ) = ∅ (otherwise x q + n − x i ∈ L and C n = xx q + n . . . x q +2 n − x q + n − x q + n − x i x i +1 x ). If s ≥ q + 2, then from x q + n − x / ∈ D , (67), L = ∅ and Claim 26 it follows that od ( x q + n − ) ≤ p/ −
2, a contradiction. So we may assume that s = q + 1.Then α = 2 and from h ≤ h − m − ≥ | Z | ≥ od ( x ) − α + h − ≤ h ≤ ≤ n ≤
7. If x q + n − x q + n +1 ∈ D , then by Claim 24(i), C n = xx q + n − x q + n − x q + n +1 . . . x q +2 n − x , a contradiction. Hence x q + n − x q + n +1 / ∈ D . If h = 4, then | Z | = m −
1, and using Claim 11, we obtain x q + n x q + n − ∈ D since x q +2 n − x q + n − / ∈ D . Therefore when h = 4,then x q + n − x q + n +2 / ∈ D (otherwise by Claim 24(i), C n = xx q + n x q + n − x q + n − x q + n +2 . . . x q +2 n − x ).Thus, since n = 6 or 7, we have L := A ( x q + n − → C ( q + n + 1 , q + 2 n − ∅ . From L = ∅ , (67) and x q + n − x / ∈ D it follows that for all x i / ∈ C ( q + n − , q + 2 n − x q + n − x i ∈ D if and only if x i +1 x / ∈ D. (68)Together with Claim 25 this implies that { x q +1 x q +2 } → x , in particular, x q + n − x ∈ D since 6 ≤ n ≤ x q + n − x q − ∈ D , then C n = x q + n − x q − x q . . . x q + n − xx q + n − x q + n − . So we may assume that x q + n − x q − / ∈ D . Hence x q x ∈ D by (68). Continuing in this manner, we obtain { x q +2 n − , x q +2 n − , . . . ,x q +2 } → x , which is a contradiction. Case 2. x q +2 n − x q + n − / ∈ D . Then from x q +2 n − x q + n − / ∈ D and Claim 11 it follows that | Z | ≤ m − h = h + 2, α = h + 1, od ( x ) = m − p = 2 m , C ( q + n − , q + 2 n − → x q + n − (69)and if x i / ∈ C ( q + n − , q + 2 n − x i x q + n − ∈ D if and only if x i / ∈ Z .We now show that s = q + 1. Suppose that s ≥ q + 2. Then n ≥
8, since h = h + 2 ≥
4. If x i x ∈ D , x i / ∈ C ( q + n − , q + 2 n −
3) and x q + n − x i − ∈ D or x q + n − x i − ∈ D then by (69), C n = xx q + n . . . x q +2 n − x q + n − x q + n − x i − x i − x i x or C n = xx q + n . . . x q +2 n − x q + n − x q + n − x i − x i − x i − x i x ,a contradiction. So we may assume that if x i x ∈ D and x i / ∈ C ( q + n − , q + 2 n − L := A ( x q + n − → { x i − , x i − } ) = ∅ . Since | I ( x ) ∩ { x q +2 n − , x q +2 n − , x q +2 n − , . . . , x q + n − }| = id ( x ) − h + 1 , (70)we see that the number of such x i − and x i − vertices at least id ( x ) − h + 1. Therefore from L = ∅ , x q + n − x / ∈ D and Claim 26 it follows that od ( x q + n − ) ≤ m −
2, a contradiction. The equality s = q + 1is proved. 28hen h = 1 since s = q + 1. From h = h + 2 it follows that h = 3 and n = 6. Note that x q + n − x q + n − ∈ D by (69), and A ( x q + n − → { x q + n , x q + n +1 } ) = ∅ ( x q + n +1 = x q +2 n − ) by Claim 19(i).From this, x q + n − x / ∈ D and (70) it follows that there is a vertex x l / ∈ C ( q + n − , q + 2 n −
5) such that x l x , x q + n − x l − ∈ D ( respectively, x j / ∈ C ( q + n − , q + 2 n −
5) such that x j x , x q + n − x j ∈ D ). It iseasy to see that A ( { x q + n , x q + n +1 } → x q + n − ) = ∅ , (71)(otherwise if x q + n +1 x q + n − ∈ D , then C n = xx q + n x q + n +1 x q + n − x q + n − x j x and if x q + n x q + n − ∈ D ,then C n = xx q + n x q + n − x q + n − x l − x l x ). On the other hand, if xx i ∈ D and x i + t / ∈ C ( q + n − , q +2 n − t ∈ [1 , x i + t x q + n − / ∈ D , since in the converse case, C n = xx i . . . x i + t x q + n − . . . x q +2 n − − t x ,a contradiction. Since | O ( x ) ∩ { x q +2 n − , x q +2 n − , . . . , x q +1 }| ≥ od ( x ) − xx q − / ∈ D it follows that the number of such x i + t , t ∈ [1 , od ( x ) −
1. Therefore,by (71) and xx q + n − / ∈ D we obtain id ( x q + n − ) ≤ m −
2, a contradiction. Claim 27 is proved.
Claim 28. α = h . Proof.
By Claims 24(i) and 27, L := A ( x → { x q + n − , x q + n } ) = ∅ . Suppose that Claim 28 is false (i.e., α = h ). Note that s ≥ q + 1 (in particular, h ≥
1) by Claim 16, and α ≤ h − L = ∅ . FromClaims 11, 21 and 18 it follows that α = h − h = h + 2, od ( x ) = m − p = 2 m and | Z | = m .Using | Z | = m and Claim 11, we obtain x i x q + n − ∈ D if and only if x i / ∈ Z , in particular, C ( q + n − , q + 2 n − → x q + n − . (72)By Claim 24(ii), xx q + n − ∈ D . Hence 1 ≤ α = h − h ≥ s ≥ q + 2).Suppose first that h ≥
3. In this case it is easy to see that L := A ( x q + n → C ( s + n − , q + 2 n − ∅ , (otherwise if x q + n x i ∈ L , then by Claim 9(ii), C n = xx q +3 x q +4 . . . x q + n x i x ). Using the fact that L = ∅ and α = h − ≥ xx q + n + j ∈ D for some j ∈ [1 , x i / ∈ C ( q + n − , q + 2 n −
5) and x i x ∈ D , then x q + n x i +1 − j / ∈ D (otherwise by (72) and xx q + n + j ∈ D we have C n = xx q + n + j . . . x q + n − x q + n − x q + n − x q + n x i +1 − j x i x ). Together with | I ( x ) ∩ { x q +2 n − , x q +2 n − , . . . , x q + n − }| ≥ m − h + 1 ,x q + n x / ∈ D and L = ∅ this implies that x q + n does not dominate at least m + 1 vertices, which is acontradiction.So suppose next that h = 2. Then from α = h − h − n = 8. From xx q + n − ∈ D , L = ∅ and α = 1 we have A ( x → C ( q + n − , q + n + 1)) = ∅ .For each l ∈ [1 ,
3] by R ′ l we denote the set of vertices x i + l / ∈ C ( q + n − , q + 2 n −
4) for which xx i ∈ D .Using Claim 9(i) and the definition of α and α , we obtain | [ l =1 R ′ l | ≥ od ( x ) − α − α + 2 . (73)It is easy to see that for all l ∈ [1 , L := A ( R ′ l → x q + n − ) = ∅ ) , (otherwise if x i + l x q + n − ∈ L , then by Claim 9(ii), C n = xx i . . . x i + l x q + n − . . . x q +2 n − − l x ). It followsfrom L = ∅ , α = 1, (73) and xx q + n − / ∈ D that | ∪ l =1 R ′ l | ≤ m − α ≥
1. From this, (73) and29 x q − / ∈ D it is not difficult to see that α = 2 (i.e., x → { x q +2 n − , x q +2 n − } ), x q = x and x q +2 n − = x p − (otherwise we obtain that | ∪ l =1 R ′ l | ≥ od ( x ) + 1, which is a contradiction). Therefore p = 2 n − p = 14, n = m + 1) and ∪ l =1 R ′ l = { x , x , . . . , x n − } . Then, since L = ∅ and | ∪ l =1 R ′ l | = m − id ( x q + n − ) = n − C ( q + n, p − → x q + n − . On the other hand, since C n D , it is easy to seethat A ( x q + n − → { x q + n +1 , . . . , x p − , x , x , x , x } ) = ∅ ( q + n + 1 = p − . This means that od ( x q + n − ) ≤ m −
1, and hence d ( x q + n − ) ≤ m −
2, a contradiction. Claim 28 isproved.
Claim 29. s = q + 1. Proof.
Suppose, on the contrary, that s = q + 1. Then by Claims 16, 24, 27 and 28 we have s ≥ q + 2, α = h and x → C ( q + n + 1 , s + n − ∪ { x q + n − } . (74)If x s + n − x i ∈ D , where x i ∈ C ( s + n − , q + 2 n − C n = xx s x s +1 . . . x s + n − x i x . So we may assume that A ( x s + n − → C ( s + n − , q + 2 n − ∅ . Then, since | H | = m − h + 1, | C ( s + n − , q + 2 n − | = h − x s + n − x / ∈ D it is easyto see that L := A ( x s + n − → H ) = ∅ . If x q +2 n − x q + n − ∈ D , then by (74), C n = xx s + n − . . . x q +2 n − x q + n − . . . x s + n − x i x i +1 x , where x s + n − x i ∈ L . So we may assume that x q +2 n − x q + n − / ∈ D . Since α = h ≤ h − x q +2 n − / ∈ Z ,from Claims 21 and 11 it follows that | Z | = m − od ( x ) = m − p = 2 m , h = h + 2 and C ( q + n − , q + 2 n − → x q + n − , (75)If s ≥ q + 3, then s + n − ≥ q + n + 1 and C n = xx s + n − . . . x q +2 n − x q + n − . . . x s + n − x i x i +1 x , where x s + n − x i ∈ L by (74) and (75), a contradiction. Thus we may assume that s = q + 2. Therefore h = 2, h = 4 and n = 8. From A ( x → { x q + n − , x q + n } ) = ∅ (by Claims 27 and 24(i)) and (*) it follows that { x q +2 n − , x q +2 n − } → x . Together with n = 8 this implies that for each i ∈ [0 , x q + n x q +2 n − − i / ∈ D, (76)(otherwise, since xx q + n − ∈ D , we have C n = xx q + n − x q + n − x q + n x q +2 n − − i . . . x q +2 n − − i x ). Moreover,it is easy to see that L := A ( x q + n → H l ) = ∅ , where l ∈ [2 ,
3] (otherwise if x q + n x i ∈ L , then by (74)and (75), C n = xx q + n +1 . . . x q +2 n − − l x q + n − x q + n − x q + n x i . . . x i + l − x ). Since xx q + n +1 ∈ D , by (*) wehave x q +2 n − x / ∈ D and it is not difficult to see that | H ∪ H | ≥ m −
2. From this, (76) and x q + n x / ∈ D it follows that od ( x q + n ) ≤ m − s = q + 1, h − ≥ h = α = 1 and 3 ≤ h ≤ ≤ n ≤ B := { x q +2 n − , x q +2 n − , . . . , x q − } and b := od ( x, B ). It follows from α = 1 and Claim 9(i) that b = od ( x ) − α − . (77)Let E denote the set of vertices x i + l / ∈ C ( q + n − , q + 2 n − l ∈ [1 , n − xx i ∈ D .It is easy to see that L := A ( E → x q + n − ) = ∅ (otherwise x i + l x q + n − ∈ L and C n = xx i x i +1 . . .x i + l x q + n − . . . x q +2 n − − l x by Claim 9(ii)). From this and L = A ( x → { x q + n − , x q + n } ) = ∅ (by Claims304(i) and 27), we obtain | E | ≤ m −
1. Remark that from x → { x q , x q +1 } and xx q − / ∈ D it followsthat { x q +1 , x q +2 , . . . , x q + n − } ⊂ E . Hence if b ≥
1, then the set E contains at least b + 1 vertices from B ∪ { x q } .We now show that α = n −
5, i.e. x → { x q + n +1 , . . . , x q +2 n − } . (78)Assume that α = n −
5. Then from L = ∅ we have α ≤ n −
6. Therefore b ≥ od ( x ) − n + 3 ≥ n ≤ m + 1. It follows immediately from the remark above that the set E contains at least b + n − ≥ m vertices. This contrary to | E | ≤ m − α = n − α = n − b = od ( x ) − n + 2. It is clear that x q +2 n − x q + n − / ∈ D , sinceotherwise, by (78) and Claim 9(ii), C n = x q +2 n − x q + n − x q + n xx q + n +1 . . . x q +2 n − . From x q +2 n − / ∈ C ( q + 1 , q + n −
3) it is easy to see that (in case b = 0 and in case b ≥
1) the set E contains at least b + 1vertices from the set B ∪{ x q } . Thus we have m − ≥ | E | ≥ b + n − od ( x ). Hence od ( x ) = | E | = m − p = 2 m , id ( x q + n − ) = m − L = ∅ , and { x q + n , x q + n +1 , . . . , x q +2 n − } → x q + n − . Therefore for all l ∈ [1 , n −
4] if x i x ∈ D and x i − l / ∈ C ( q + n − , q + 2 n − x q + n − x i − l / ∈ D , since otherwise, by(78), C n = xx q + n +1 . . . x q +2 n − − l x q + n − x i − l . . . x i x , a contradiction. It is not difficult to see that thenumber of such x i − l vertices at least m −
1. Therefore, since x q + n − x / ∈ D , we get od ( x q + n − ) ≤ m − d ( x q + n − ) ≤ m −
2, a contradiction. This completes the proof of the theorem.