Two Sufficient Conditions for Hamilton and Dominating Cycles

Abstract

We prove that if 𝐺 is a 2-connect graph of size 𝑞 (the number of edges) and minimum degree 𝛿 with 𝛿≥2𝑞/3+𝜖/12−1/2, where 𝜖=11 when 𝛿=2 and 𝜖=31 when 𝛿≥3, then each longest cycle in 𝐺 is a dominating cycle. The exact analog of this theorem for Hamilton cycles follows easily from two known results according to Dirac and Nash-Williams: each graph with 𝛿≥𝑞+5/4−1/2 is hamiltonian. Both results are sharp in all respects.

1. Introduction

Only finite undirected graphs without loops or multiple edges are considered. We reserve 𝑛, 𝑞, 𝛿, and 𝜅 to denote the number of vertices (order), the number of edges (size), the minimum degree, and the connectivity of a graph, respectively. A graph 𝐺 is hamiltonian if 𝐺 contains a hamiltonian cycle, that is, a cycle of length 𝑛. Further, a cycle 𝐶 in 𝐺 is called a dominating cycle if the vertices in 𝐺⧵𝐶 are mutually nonadjacent. A good reference for any undefined terms is [1].

The following two well-known theorems provide two classic sufficient conditions for Hamilton and dominating cycles by linking the minimum degree 𝛿 and order 𝑛.

Theorem A (see [2]). Every graph with 𝛿≥(1/2)𝑛 is hamiltonian.

Theorem B (see [3]). If 𝐺 is a 2-connect graph with 𝛿≥(1/3)(𝑛+2), then each longest cycle in 𝐺 is a dominating cycle.

The exact analog of Theorem A that links the minimum degree 𝛿 and size 𝑞 easily follows from Theorem A and a particular result according to Nash-Williams [4] (see Theorem 1.1 below).

Theorem 1.1. Every graph is hamiltonian if 𝛿≥5𝑞+4−12. (1.1)

The hypothesis in Theorem 1.1 is equivalent to 𝑞≤𝛿²+𝛿−1 and cannot be relaxed to 𝑞≤𝛿²+𝛿 due to the graph 𝐾₁+2𝐾_𝛿 consisting of two copies of 𝐾_𝛿+1 and having exactly one vertex in common. Hence, Theorem 1.1 is best possible.

The main goal of this paper is to prove the exact analog of Theorem B for dominating cycles based on another similar relation between 𝛿 and 𝑞.

Theorem 1.2. Let 𝐺 be a 2-connect graph with 𝛿≥2𝑞3+𝜖−1122, (1.2) where 𝜖=11 when 𝛿=2 and 𝜖=31 when 𝛿≥3. Then each longest cycle in 𝐺 is a dominating cycle.

To show that Theorem 1.2 is sharp, suppose first that 𝛿=2, implying that the hypothesis in Theorem 1.2 is equivalent to 𝑞≤8. The graph 𝐾₁+2𝐾₂ shows that the connectivity condition 𝜅≥2 in Theorem 1.2 cannot be relaxed by replacing it with 𝜅≥1. The graph with vertex set {𝑣₁,𝑣₂,…,𝑣₈} and edge set 𝑣₁𝑣₂,𝑣₂𝑣₃,𝑣₃𝑣₄,𝑣₄𝑣₅,𝑣₅𝑣₆,𝑣₆𝑣₁,𝑣₁𝑣₇,𝑣₇𝑣₈,𝑣₈𝑣₄ (1.3) shows that the size bound 𝑞≤8 cannot be relaxed by replacing it with 𝑞≤9. Finally, the graph 𝐾₂+3𝐾₁ shows that the conclusion “each longest cycle in 𝐺 is a dominating cycle” cannot be strengthened by replacing it with “𝐺 is hamiltonian.” Analogously, we can use 𝐾₁+2𝐾_𝛿, 𝐾₂+3𝐾_𝛿−1, and 𝐾_𝛿+(𝛿+1)𝐾₁, respectively, to show that Theorem 1.2 is sharp when 𝛿≥3. So, Theorem 1.2 is best possible in all respects.

To prove Theorems 1.1 and 1.2, we need two known results, the first of which is belongs Nash-Williams [4].

Theorem C (see [4]). If 𝛿=(𝑛−1)/2, then either 𝐺 is hamiltonian or 𝐺=𝐾₁+2𝐾_𝛿, or 𝐺=𝐾_𝛿+1+𝐺_𝛿, where 𝐺_𝛿 denote an arbitrary graph on 𝛿 vertices.

The next theorem provides a lower bound for the length of a longest cycle in 2-connected graphs according to Dirac [2].

Theorem D (see [2]). Every 2-connected graph either has a hamiltonian cycle or has a cycle of length at least 2𝛿.

2. Notations and Preliminaries

The set of vertices of a graph 𝐺 is denoted by 𝑉(𝐺) and the set of edges by 𝐸(𝐺). For 𝑆, a subset of 𝑉(𝐺), we denote by 𝐺⧵𝑆 the maximum subgraph of 𝐺 with vertex set 𝑉(𝐺)⧵𝑆. We write 𝐺[𝑆] for the subgraph of 𝐺 induced by 𝑆. For a subgraph 𝐻 of 𝐺, we use 𝐺⧵𝐻 short for 𝐺⧵𝑉(𝐻). The neighborhood of a vertex 𝑥∈𝑉(𝐺) will be denoted by 𝑁(𝑥). Set 𝑑(𝑥)=|𝑁(𝑥)|. Furthermore, for a subgraph 𝐻 of 𝐺 and 𝑥∈𝑉(𝐺), we define 𝑁_𝐻(𝑥)=𝑁(𝑥)∩𝑉(𝐻) and 𝑑_𝐻(𝑥)=|𝑁_𝐻(𝑥)|.

A simple cycle (or just a cycle) 𝐶 of length 𝑡 is a sequence 𝑣₁𝑣₂⋯𝑣_𝑡𝑣₁ of distinct vertices 𝑣₁,…,𝑣_𝑡 with 𝑣_𝑖𝑣_𝑖+1∈𝐸(𝐺) for each 𝑖∈{1,…,𝑡}, where 𝑣_𝑡+1=𝑣₁. When 𝑡=2, the cycle 𝐶=𝑣₁𝑣₂𝑣₁ on two vertices 𝑣₁,𝑣₂ coincides with the edge 𝑣₁𝑣₂, and when 𝑡=1, the cycle 𝐶=𝑣₁ coincides with the vertex 𝑣₁. So, all vertices and edges in a graph can be considered as cycles of lengths 1 and 2, respectively.

Paths and cycles in a graph 𝐺 are considered as subgraphs of 𝐺. If 𝑄 is a path or a cycle, then the length of 𝑄, denoted by |𝑄|, is |𝐸(𝑄)|. We write 𝑄 with a given orientation by 𝑄^→. For 𝑥,𝑦∈𝑉(𝑄), we denote by 𝑥𝑄^→𝑦 the subpath of 𝑄 in the chosen direction from 𝑥 to 𝑦. For 𝑥∈𝑉(𝑄), we denote the ℎth successor and the ℎth predecessor of 𝑥 on 𝑄^→ by 𝑥^+ℎ and 𝑥^−ℎ, respectively. We abbreviate 𝑥⁺¹ and 𝑥⁻¹ by 𝑥⁺ and 𝑥⁻, respectively.

Special Definitions
Let 𝐺 be a graph, 𝐶 a longest cycle in 𝐺, and 𝑃=𝑥^→𝑃𝑦 a longest path in 𝐺⧵𝐶 of length 𝑝≥0. Let 𝜉₁,𝜉₂,…,𝜉_𝑠 be the elements of 𝑁_𝐶(𝑥)∪𝑁_𝐶(𝑦) occurring on 𝐶 in a consecutive order. Set 𝐼_𝑖=𝜉𝐶_𝑖^→𝜉_𝑖+1,𝐼^∗_𝑖=𝜉𝐶⁺_𝑖^→𝜉⁻_𝑖+1(𝑖=1,2,…,𝑠), (2.1) where 𝜉_𝑠+1=𝜉₁.(∗1) We call 𝐼₁,𝐼₂,…,𝐼_𝑠 elementary segments on 𝐶 created by 𝑁_𝐶(𝑥)∪𝑁_𝐶(𝑦).(∗2) We call a path 𝐿=𝑧^→𝐿𝑤 an intermediate path between two distinct elementary segments 𝐼_𝑎 and 𝐼_𝑏 if 𝑧∈𝑉𝐼^∗_𝑎,𝑤∈𝑉𝐼^∗_𝑏,𝑉(𝐿)∩𝑉(𝐶∪𝑃)={𝑧,𝑤}. (2.2)(∗3) The set of all intermediate paths between elementary segments 𝐼_𝑖1,𝐼_𝑖2,…,𝐼_𝑖𝑡 will be denoted by Υ(𝐼_𝑖1,𝐼_𝑖2,…,𝐼_𝑖𝑡).

Lemma 2.1. Let 𝐺 be a graph, 𝐶 a longest cycle in 𝐺, and 𝑃=𝑥^→𝑃𝑦 a longest path in 𝐺⧵𝐶 of length 𝑝≥1. If |𝑁_𝐶(𝑥)|≥2, |𝑁_𝐶(𝑦)|≥2 and 𝑁_𝐶(𝑥)≠𝑁_𝐶(𝑦), then |𝐶|≥3𝛿+max𝜎₁,𝜎₂−1≥3𝛿,𝑖𝑓2𝑝=1,max𝑝+8,4𝛿−2𝑝,𝑖𝑓𝑝≥2, (2.3) where 𝜎₁=|𝑁_𝐶(𝑥)⧵𝑁_𝐶(𝑦)| and 𝜎₂=|𝑁_𝐶(𝑦)⧵𝑁_𝐶(𝑥)|.

Lemma 2.2. Let 𝐺 be a graph, 𝐶 a longest cycle in 𝐺, and 𝑃=𝑥^→𝑃𝑦 a longest path in 𝐺⧵𝐶 of length 𝑝≥0. If 𝑁_𝐶(𝑥)=𝑁_𝐶(𝑦), |𝑁_𝐶(𝑥)|≥2 and 𝐼_𝑎, 𝐼_𝑏 are elementary segments induced by 𝑁_𝐶(𝑥)∪𝑁_𝐶(𝑦), then (a1) if 𝐿 is an intermediate path between 𝐼_𝑎 and 𝐼_𝑏, then 𝐼_𝑎+𝐼_𝑏≥2𝑝+2|𝐿|+4, (2.4) (a2) if Υ(𝐼_𝑎,𝐼_𝑏)⊆𝐸(𝐺) and |Υ(𝐼_𝑎,𝐼_𝑏)|=𝑖 for some 𝑖∈{1,2,3}, then 𝐼_𝑎+𝐼_𝑏≥2𝑝+𝑖+5. (2.5)

Lemma 2.3. Let 𝐺 be a graph, 𝑆 a cut set in 𝐺, and 𝐻 a connected component of 𝐺⧵𝑆 of order ℎ. Then 𝑞_𝐻≥ℎ(2𝛿−ℎ+1)2, (2.6) where 𝑞_𝐻=|{𝑥𝑦∈𝐸(𝐺)∶{𝑥,𝑦}∩𝑉(𝐻)≠∅}|.

Lemma 2.4. Let 𝐺 be a 2-connect graph. If 𝛿≥(𝑛−2)/3, then either 𝑞≥3()9𝑤ℎ𝑒𝑛𝛿=2,𝛿−1)(𝛿+22𝑤ℎ𝑒𝑛𝛿≥3, (2.7) or each longest cycle in 𝐺 is a dominating cycle.

3. Proofs

Proof of Lemma 2.1. Put 𝐴₁=𝑁_𝐶(𝑥)⧵𝑁_𝐶𝑦,𝐴₂=𝑁_𝐶𝑦⧵𝑁_𝐶(𝑥),𝑀=𝑁_𝐶(𝑥)∩𝑁_𝐶𝑦. (3.1) By the hypothesis, 𝑁_𝐶(𝑥)≠𝑁_𝐶(𝑦), implying that max𝐴₁,𝐴₂≥1. (3.2) Let 𝜉₁,𝜉₂,…,𝜉_𝑠 be the elements of 𝑁_𝐶(𝑥)∪𝑁_𝐶(𝑦) occuring on 𝐶 in a consecutive order. Put 𝐼_𝑖=𝜉𝐶_𝑖^→𝜉_𝑖+1(𝑖=1,2,…,𝑠), where 𝜉_𝑠+1=𝜉₁. Clearly, 𝑠=|𝐴₁|+|𝐴₂|+|𝑀|. Since 𝐶 is extreme, |𝐼_𝑖|≥2(𝑖=1,2,…,𝑠). Next, if {𝜉_𝑖,𝜉_𝑖+1}∩𝑀≠∅ for some 𝑖∈{1,2,…,𝑠}, then |𝐼_𝑖|≥𝑝+2. Further, if either 𝜉_𝑖∈𝐴₁, 𝜉_𝑖+1∈𝐴₂ or 𝜉_𝑖∈𝐴₂, 𝜉_𝑖+1∈𝐴₁, then again |𝐼_𝑖|≥𝑝+2.
Case  1. (𝑝=1).
Case  1.1 (|𝐴_𝑖|≥1(𝑖=1,2)). It follows that among 𝐼₁,𝐼₂,…,𝐼_𝑠 there are |𝑀|+2 segments of length at least 𝑝+2. Observing also that each of the remaining 𝑠−(|𝑀|+2) segments has a length at least 2, we have |𝐶|≥𝑀|)(|𝑀|)𝑀|)𝐴𝑝+2(|+2+2𝑠−−2=3(|+2+2₁+𝐴₂𝐴−2=2₁+2𝐴₂+3|𝑀|+2. (3.3)
Since |𝐴₁|=𝑑(𝑥)−|𝑀|−1 and |𝐴₂|=𝑑(𝑦)−|𝑀|−1, |𝐶|≥2𝑑(𝑥)+2𝑑𝑦−|𝑀|−2≥3𝛿+𝑑(𝑥)−|𝑀|−2. (3.4) Recalling that 𝑑(𝑥)=|𝑀|+|𝐴₁|+1, we get |𝐶|≥3𝛿+𝐴₁−1=3𝛿+𝜎₁−1. (3.5) Analogously, |𝐶|≥3𝛿+𝜎₂−1. So, |𝐶|≥3𝛿+max𝜎₁,𝜎₂−1≥3𝛿. (3.6)Case  1.2 (either |𝐴₁|≥1, |𝐴₂|=0 or |𝐴₁|=0, |𝐴₂|≥1). Assume without loss of generality that |𝐴₁|≥1 and |𝐴₂|=0, that is, |𝑁_𝐶(𝑦)|=|𝑀|≥2 and 𝑠=|𝐴₁|+|𝑀|. Hence, among 𝐼₁,𝐼₂,…,𝐼_𝑠 there are |𝑀|+1 segments of length at least 𝑝+2=3. Taking into account that each of the remaining 𝑠−(|𝑀|+1) segments has a length at least 2 and |𝑀|+1=𝑑(𝑦), we get |𝐶|≥3(|𝑀|+1)+2(𝑠−|𝑀|−1)=3𝑑𝑦+2𝐴₁𝐴−1≥3𝛿+₁−1=3𝛿+max𝜎₁,𝜎₂−1≥3𝛿. (3.7)Case  2 (𝑝≥2). We first prove that |𝐶|≥2𝑝+8. Since |𝑁_𝐶(𝑥)|≥2 and |𝑁_𝐶(𝑦)|≥2, there are at least two segments among 𝐼₁,𝐼₂,…,𝐼_𝑠 of length at least 𝑝+2. If |𝑀|=0, then clearly 𝑠≥4 and |𝐶|≥2𝑝+2+2(𝑠−2)≥2𝑝+8. (3.8) Otherwise, since max{|𝐴₁|,|𝐴₂|}≥1, there are at least three elementary segments of length at least 𝑝+2, that is, |𝐶|≥3𝑝+2≥2𝑝+8. (3.9) So, in any case, |𝐶|≥2𝑝+8.
To prove that |𝐶|≥4𝛿−2𝑝, we distinguish two main cases.
Case  2.1 (|𝐴_𝑖|≥1(𝑖=1,2)). It follows that among 𝐼₁,𝐼₂,…,𝐼_𝑠 there are |𝑀|+2 segments of length at least 𝑝+2. Further, since each of the remaining 𝑠−(|𝑀|+2) segments has a length at least 2, we get |𝐶|≥𝑀|)(|𝑀|)=𝑝+2(|+2+2𝑠−−2𝑝−2|𝑀|+2𝑝+4|𝑀|+4+2𝐴₁+𝐴₂𝐴−2≥2₁+2𝐴₂+4|𝑀|+2𝑝. (3.10) Observing also that 𝐴₁+|𝑀|+𝑝≥𝑑(𝑥),𝐴₂+|𝑀|+𝑝≥𝑑𝑦, (3.11) we have 2𝐴₁+2𝐴₂+4|𝑀|+2𝑝≥2𝑑(𝑥)+2𝑑𝑦−2𝑝≥4𝛿−2𝑝, (3.12) implying that |𝐶|≥4𝛿−2𝑝.
Case  2.2 (either |𝐴₁|≥1, |𝐴₂|=0 or |𝐴₁|=0, |𝐴₂|≥1). Assume without loss of generality that |𝐴₁|≥1 and |𝐴₂|=0, that is, |𝑁_𝐶(𝑦)|=|𝑀|≥2 and 𝑠=|𝐴₁|+|𝑀|. It follows that among 𝐼₁,𝐼₂,…,𝐼_𝑠 there are |𝑀|+1 segments of length at least 𝑝+2. Observing also that |𝑀|+𝑝≥𝑑(𝑦)≥𝛿, that is, 2𝑝+4|𝑀|≥4𝛿−2𝑝, we get |𝐶|≥𝑝+2(|𝑀|+1)≥𝑝−2(|𝑀|−1)+2𝑝+4|𝑀|≥2𝑝+4|𝑀|≥4𝛿−2𝑝. (3.13)

Proof of Lemma 2.2. Let 𝜉₁,𝜉₂,…,𝜉_𝑠 be the elements of 𝑁_𝐶(𝑥) occuring on 𝐶 in a consecutive order. Put 𝐼_𝑖=𝜉𝐶_𝑖^→𝜉_𝑖+1(𝑖=1,2,…,𝑠), where 𝜉_𝑠+1=𝜉₁. To prove (𝑎1), let 𝐿=𝑧^→𝐿𝑤 be an intermediate path between elementary segments 𝐼_𝑎 and 𝐼_𝑏 with 𝑧∈𝑉(𝐼^∗_𝑎) and 𝑤∈𝑉(𝐼^∗_𝑏). Put 𝜉𝐶_𝑎^→𝑧=𝑑₁,𝑧𝐶^→𝜉_𝑎+1=𝑑₂,𝜉𝐶_𝑏^→𝑤=𝑑₃,𝑤𝐶^→𝜉_𝑏+1=𝑑₄,𝐶^′=𝜉_𝑎𝑥^→𝑃𝑦𝜉𝐶_𝑏^←𝑧^→𝐿𝑤𝐶^→𝜉_𝑎. (3.14) Clearly, 𝐶^′=|𝐶|−𝑑₁−𝑑₃+|𝐿|+|𝑃|+2. (3.15) Since 𝐶 is extreme, we have |𝐶|≥|𝐶^′|, implying that 𝑑₁+𝑑₃≥𝑝+|𝐿|+2. By a symmetric argument, 𝑑₂+𝑑₄≥𝑝+|𝐿|+2. Hence 𝐼_𝑎+𝐼_𝑏=⁴_𝑖=1𝑑_𝑖≥2𝑝+2|𝐿|+4. (3.16)
The proof of (𝑎1) is complete. To prove (𝑎2), let Υ(𝐼_𝑎,𝐼_𝑏)⊆𝐸(𝐺) and |Υ(𝐼_𝑎,𝐼_𝑏)|=𝑖 for some 𝑖∈{1,2,3}.
Case  1 (𝑖=1). It follows that Υ(𝐼_𝑎,𝐼_𝑏) consists of a unique intermediate edge 𝐿=𝑧𝑤. By (𝑎1), 𝐼_𝑎+𝐼_𝑏≥2𝑝+2|𝐿|+4=2𝑝+6. (3.17)Case 2 (𝑖=2). It follows that Υ(𝐼_𝑎,𝐼_𝑏) consists of two edges 𝑒₁, 𝑒₂. Put 𝑒₁=𝑧₁𝑤₁ and 𝑒₂=𝑧₂𝑤₂, where {𝑧₁,𝑧₂}⊆𝑉(𝐼^∗_𝑎) and {𝑤₁,𝑤₂}⊆𝑉(𝐼^∗_𝑏).
Case  2.1 (𝑧₁≠𝑧₂ and 𝑤₁≠𝑤₂). Assume without loss of generality that 𝑧₁ and 𝑧₂ occur in this order on 𝐼_𝑎.
Case  2.1.1. 𝑤₂ and 𝑤₁ occur in this order on 𝐼_𝑏.
Put 𝜉𝐶_𝑎^→𝑧₁=𝑑₁,𝑧𝐶₁^→𝑧₂=𝑑₂,𝑧𝐶₂^→𝜉_𝑎+1=𝑑₃,𝜉𝐶_𝑏^→𝑤₂=𝑑₄,𝑤𝐶₂^→𝑤₁=𝑑₅,𝑤𝐶₁^→𝜉_𝑏+1=𝑑₆,𝐶^′=𝜉𝐶_𝑎^→𝑧₁𝑤𝐶₁^←𝑤₂𝑧𝐶₂^→𝜉_𝑏𝑥^→𝑃𝑦𝜉^→_𝑏+1𝐶𝜉_𝑎. (3.18) Clearly, 𝐶^′=|𝐶|−𝑑₂−𝑑₄−𝑑₆+𝑒₁+𝑒₂+|𝑃|=|𝐶|+2−𝑑₂−𝑑₄−𝑑₆+𝑝+4. (3.19) Since 𝐶 is extreme, |𝐶|≥|𝐶^′|, implying that 𝑑₂+𝑑₄+𝑑₆≥𝑝+4. By a symmetric argument, 𝑑₁+𝑑₃+𝑑₅≥𝑝+4. Hence 𝐼_𝑎+𝐼_𝑏=⁶_𝑖=1𝑑_𝑖≥2𝑝+8. (3.20)Case  2.1.2. 𝑤₁ and 𝑤₂ occur in this order on 𝐼_𝑏.
Putting 𝐶^′=𝜉𝐶_𝑎^→𝑧₁𝑤𝐶₁^→𝑤₂𝑧𝐶₂^→𝜉_𝑏𝑥^→𝑃𝑦𝜉^→_𝑏+1𝐶𝜉_𝑎, (3.21) we can argue as in Case  2.1.1.
Case  2.2 (either 𝑧₁=𝑧₂, 𝑤₁≠𝑤₂ or 𝑧₁≠𝑧₂, 𝑤₁=𝑤₂). Assume without loss of generality that 𝑧₁≠𝑧₂, 𝑤₁=𝑤₂ and 𝑧₁,𝑧₂ occur in this order on 𝐼_𝑎. Put 𝜉𝐶_𝑎^→𝑧₁=𝑑₁,𝑧𝐶₁^→𝑧₂=𝑑₂,𝑧𝐶₂^→𝜉_𝑎+1=𝑑₃,𝜉𝐶_𝑏^→𝑤₁=𝑑₄,𝑤𝐶₁^→𝜉_𝑏+1=𝑑₅,𝐶^′=𝜉_𝑎𝑥^→𝑃𝑦𝜉𝐶_𝑏^←𝑧₁𝑤𝐶₁^→𝜉_𝑎,𝐶^′′=𝜉𝐶_𝑎^→𝑧₂𝑤𝐶₁^←𝜉_𝑎+1𝑥^→𝑃𝑦𝜉^→_𝑏+1𝐶𝜉_𝑎. (3.22)
Clearly, 𝐶^′=|𝐶|−𝑑₁−𝑑₄+𝑒₁+|𝑃|+2=|𝐶|−𝑑₁−𝑑₄+𝐶𝑝+3,_′′=|𝐶|−𝑑₃−𝑑₅+𝑒₂+|𝑃|+2=|𝐶|−𝑑₃−𝑑₅+𝑝+3. (3.23) Since 𝐶 is extreme, |𝐶|≥|𝐶^′| and |𝐶|≥|𝐶^′′|, implying that 𝑑₁+𝑑₄≥𝑝+3,𝑑₃+𝑑₅≥𝑝+3. (3.24) Hence, 𝐼_𝑎+𝐼_𝑏=⁵_𝑖=1𝑑_𝑖≥𝑑₁+𝑑₃+𝑑₄+𝑑₅+1≥2𝑝+7. (3.25)Case  3 (𝑖=3). It follows that Υ(𝐼_𝑎,𝐼_𝑏) consists of three edges 𝑒₁, 𝑒₂, 𝑒₃. Let 𝑒_𝑖=𝑧_𝑖𝑤_𝑖  (𝑖=1,2,3), where {𝑧₁,𝑧₂,𝑧₃}⊆𝑉(𝐼^∗_𝑎) and {𝑤₁,𝑤₂,𝑤₃}⊆𝑉(𝐼^∗_𝑏). If there are two independent edges among 𝑒₁, 𝑒₂, 𝑒₃, then we can argue as in Case  2.1. Otherwise, we can assume without loss of generality that 𝑤₁=𝑤₂=𝑤₃ and 𝑧₁, 𝑧₂, 𝑧₃ occur in this order on 𝐼_𝑎. Put 𝜉𝐶_𝑎^→𝑧₁=𝑑₁,𝑧𝐶₁^→𝑧₂=𝑑₂,𝑧𝐶₂^→𝑧₃=𝑑₃,𝑧𝐶₃^→𝜉_𝑎+1=𝑑₄,𝜉𝐶_𝑏^→𝑤₁=𝑑₅,𝑤𝐶₁^→𝜉_𝑏+1=𝑑₆,𝐶^′=𝜉_𝑎𝑥^→𝑃𝑦𝜉𝐶_𝑏^←𝑧₁𝑤𝐶₁^→𝜉_𝑎,𝐶^′′=𝜉𝐶_𝑎^→𝑧₃𝑤𝐶₁^←𝜉_𝑎+1𝑥^→𝑃𝑦𝜉^→_𝑏+1𝐶𝜉_𝑎. (3.26)
Clearly, 𝐶^′=|𝐶|−𝑑₁−𝑑₅+𝑒₁+𝐶𝑝+2,_′′=|𝐶|−𝑑₄−𝑑₆+𝑒₃+𝑝+2. (3.27) Since 𝐶 is extreme, we have |𝐶|≥|𝐶^′| and |𝐶|≥|𝐶^′′|, implying that 𝑑₁+𝑑₅≥𝑝+3,𝑑₄+𝑑₆≥𝑝+3. (3.28) Hence, 𝐼_𝑎+𝐼_𝑏=⁶_𝑖=1𝑑_𝑖≥𝑑₁+𝑑₄+𝑑₅+𝑑₆+2≥2𝑝+8. (3.29)

Proof of Lemma 2.3. Put 𝑉(𝐻)=𝑣₁,…,𝑣_ℎ,𝑁𝑣_𝑖∩𝑆=𝛽_𝑖(𝑖=1,…,ℎ). (3.30)
Observing that ℎ≥𝑑(𝑣_𝑖)−𝛽_𝑖+1≥𝛿−𝛽_𝑖+1 for each 𝑖∈{1,2,…,ℎ}, we have 𝛽_𝑖≥𝛿−ℎ+1  (𝑖=1,2,…,ℎ). Therefore, 𝑞_𝐻=𝑞(𝐻)+^ℎ_𝑖=1𝛽_𝑖=12^ℎ_𝑖=1𝑑_𝐻𝑣_𝑖+^ℎ_𝑖=1𝛽_𝑖,=12^ℎ_𝑖=1𝑑_𝐻𝑣_𝑖+𝛽_𝑖+12^ℎ_𝑖=1𝛽_𝑖=12^ℎ_𝑖=1𝑑𝑣_𝑖+12^ℎ_𝑖=1(),≥1𝛿−ℎ+121ℎ𝛿+2ℎ()=ℎ()𝛿−ℎ+12𝛿−ℎ+12. (3.31)

Proof of Lemma 2.4. Let 𝐶 be a longest cycle in 𝐺 and 𝑃=𝑥₁^→𝑃𝑥₂ a longest path in 𝐺⧵𝐶 of length 𝑝. If |𝑉(𝑃)|≤1, then 𝐶 is a dominating cycle and we are done. Let |𝑉(𝑃)|≥2, that is, 𝑝≥1. By the hypothesis, |𝐶|+𝑝+1≤𝑛≤3𝛿+2. Further, by Theorem D, |𝐶|≥2𝛿. From these inequalities, we get𝑛≤3𝛿+2,|𝐶|≤3𝛿−𝑝+1,1≤𝑝≤𝛿+1. (3.32)
Let 𝜉₁,𝜉₂,…,𝜉_𝑠 be the elements of 𝑁_𝐶(𝑥₁)∪𝑁_𝐶(𝑥₂) occuring on 𝐶 in a consecutive order. Put 𝐼_𝑖=𝜉𝐶_𝑖^→𝜉_𝑖+1,𝐼^∗_𝑖=𝜉𝐶⁺_𝑖^→𝜉⁻_𝑖+1(𝑖=1,2,…,𝑠), (3.33) where 𝜉_𝑠+1=𝜉₁.
Case  1 (𝛿=2). Let 𝑄 be a longest path in 𝐺 with 𝑄=𝜉𝑄^→𝜂 and 𝑉(𝑄)∩𝑉(𝐶)={𝜉,𝜂}. Since 𝐶 is extreme, we have |𝜉𝐶^→𝜂|≥|𝑄| and |𝜂𝐶^→𝜉|≥|𝑄|, implying that |𝐶|=𝜉𝐶^→𝜂+𝜂𝐶^→𝜉≥2|𝑄|. (3.34) Since 𝜅≥2 and 𝑝≥1, we have |𝑄|≥3. By (3.34), |𝐶|≥2|𝑄|≥6, implying that 𝑞≥|𝐶|+|𝑄|≥9.
Case  2. (𝛿≥3).
Case  2.1 (𝑝=1). By (3.32), |𝐶|≤3𝛿. (3.35)Case  2.1.1 (𝑁_𝐶(𝑥₁)≠𝑁_𝐶(𝑥₂)). It follows that max{𝜎₁,𝜎₂}≥1, where 𝜎₁=𝑁_𝐶𝑥₁⧵𝑁_𝐶𝑥₂,𝜎₂=𝑁_𝐶𝑥₂⧵𝑁_𝐶𝑥₁. (3.36) By Lemma 2.1, |𝐶|≥3𝛿. Recalling (3.35), we get |𝐶|=3𝛿. If max{𝜎₁,𝜎₂}≥2, then by Lemma 2.1, |𝐶|≥3𝛿+1, contradicting (3.35). Let max{𝜎₁,𝜎₂}=1. Clearly, 𝑠≥𝛿 and |𝐼_𝑖|≥3(𝑖=1,2,…,𝑠). Further, if 𝑠≥𝛿+1, then |𝐶|≥3𝑠≥3𝛿+3, again contradicting (3.35). Let 𝑠=𝛿, implying that |𝐼_𝑖|=3  (𝑖=1,2,…,𝑠). By Lemma 2.2, Υ(𝐼₁,𝐼₂,…,𝐼_𝑠)=∅. Let 𝐻₁,𝐻₂,…,𝐻_𝑠+1 be the connected components of 𝐺⧵{𝜉₁,𝜉₂,…,𝜉_𝑠} with 𝑉(𝐻_𝑖)=𝑉(𝐼^∗_𝑖)(𝑖=1,2,…,𝑠) and 𝑉(𝐻_𝑠+1)={𝑥₁,𝑥₂}. For each 𝑖∈{1,2,…,𝑠+1}, put ℎ_𝑖=𝑉𝐻_𝑖,𝑞_𝑖=𝑥𝑦∈𝐸(𝐺)∶𝑥,𝑦∩𝑉𝐻_𝑖≠∅. (3.37) Clearly, ℎ_𝑖=2(𝑖=1,2,…,𝑠+1). By Lemma 2.3, 𝑞_𝑖≥ℎ_𝑖2𝛿−ℎ_𝑖+12=2𝛿−1(𝑖=1,2,…,𝑠+1), (3.38) implying that 𝑞≥^𝑠+1_𝑖=1𝑞_𝑖≥()=()>3()𝑠+1)(2𝛿−1𝛿+1)(2𝛿−1𝛿−1)(𝛿+22. (3.39)Case  2.1.2 (𝑁_𝐶(𝑥₁)=𝑁_𝐶(𝑥₂)). Clearly, 𝑠≥𝛿−1. If 𝑠≥𝛿, then we can argue as in Case  2.1.1. Let 𝑠=𝛿−1. Further, if |𝐼_𝑖|+|𝐼_𝑗|≥10 for some distinct 𝑖,𝑗∈{1,2,…,𝑠}, then |𝐶|≥10+3(𝑠−2)=3𝛿+1, contradicting (3.35). Hence 𝐼_𝑖+𝐼_𝑗≤9foreachdistinct𝑖,𝑗∈{1,2,…,𝑠}. (3.40)Claim 1. Υ(𝐼₁,𝐼₂,…,𝐼_𝑠)⊆𝐸(𝐺) and(1)if max_𝑖|𝐼_𝑖|≤4 then |Υ(𝐼₁,𝐼₂,…,𝐼_𝑠)|≤3,(2)if max_𝑖|𝐼_𝑖|=5 then |Υ(𝐼₁,𝐼₂,…,𝐼_𝑠)|≤𝛿−1,(3)if max_𝑖|𝐼_𝑖|=6 then |Υ(𝐼₁,𝐼₂,…,𝐼_𝑠)|≤2(𝛿−2).
Proof. If Υ(𝐼₁,𝐼₂,…,𝐼_𝑠)=∅ then we are done. Otherwise, Υ(𝐼_𝑎,𝐼_𝑏)≠∅, for some distinct 𝑎,𝑏∈{1,2,…,𝑠}. By definition, there is an intermediate path 𝐿 between 𝐼_𝑎 and 𝐼_𝑏. If |𝐿|≥2, then by Lemma 2.2, 𝐼_𝑎+𝐼_𝑏≥2𝑝+2|𝐿|+4≥10, (3.41) contradicting (3.40). Otherwise, |𝐿|=1 and therefore, Υ(𝐼₁,𝐼₂,…,𝐼_𝑠)⊆𝐸(𝐺). By Lemma 2.2, |𝐼_𝑎|+|𝐼_𝑏|≥2𝑝+6=8. Combining this with (3.40), we have 8≤𝐼_𝑎+𝐼_𝑏≤9. (3.42) Furthermore, if |Υ(𝐼_𝑎,𝐼_𝑏)|≥3, then by Lemma 2.2, |𝐼_𝑎|+|𝐼_𝑏|≥2𝑝+8=10, contradicting (3.42). So, 1≤Υ𝐼_𝑖,𝐼_𝑗≤2foreachdistinct𝑖,𝑗∈{1,2,…,𝑠}. (3.43) Put 𝑟=|{𝑖∶|𝐼_𝑖|≥4}|. If 𝑟≥4, then |𝐶|≥16+3(𝑠−4)=3𝛿+1, contradicting (3.35). Further, if 𝑟=0, then by Lemma 2.2, Υ(𝐼₁,𝐼₂,…,𝐼_𝑠)=∅. Let 1≤𝑟≤3.
Case 𝑎1 (𝑟=3). It follows that |𝐼_𝑖|≥4(𝑖=𝑎,𝑏,𝑐) for some distinct 𝑎,𝑏,𝑐∈{1,2,…,𝑠} and |𝐼_𝑖|=3 for each 𝑖∈{1,2,…,𝑠}⧵{𝑎,𝑏,𝑐}. Recalling that 𝑠=𝛿−1 and |𝐶|=3𝛿, we have |𝐼_𝑎|=|𝐼_𝑏|=|𝐼_𝑐|=4, that is, max_𝑖|𝐼_𝑖|=4. By Lemma 2.2, |Υ(𝐼_𝑖,𝐼_𝑗)|≤1 for each distinct 𝑖,𝑗∈{𝑎,𝑏,𝑐}. Moreover, we have |Υ(𝐼_𝑖,𝐼_𝑗)|=0 if either 𝑖∉{𝑎,𝑏,𝑐} or 𝑗∉{𝑎,𝑏,𝑐}. So, Υ𝐼₁,𝐼₂,…,𝐼_𝑠=Υ𝐼_𝑎,𝐼_𝑏,𝐼_𝑐≤3. (3.44)Case 𝑎2 (𝑟=2). It follows that |𝐼_𝑎|≥4 and |𝐼_𝑏|≥4 for some distinct 𝑎,𝑏∈{1,2,…,𝑠} and |𝐼_𝑖|=3 for each 𝑖∈{1,2,…,𝑠}⧵{𝑎,𝑏}. By (3.42), we can assume without loss of generality that either |𝐼_𝑎|=|𝐼_𝑏|=4 or |𝐼_𝑎|=5, |𝐼_𝑏|=4.
Case 𝑎2.1 (|𝐼_𝑎|=|𝐼_𝑏|=4). It follows that max_𝑖|𝐼_𝑖|=4. By Lemma 2.2, |Υ(𝐼_𝑎,𝐼_𝑏)|≤1 and Υ(𝐼_𝑖,𝐼_𝑗)=∅ if {𝑖,𝑗}≠{𝑎,𝑏}, implying that |Υ(𝐼₁,𝐼₂,…,𝐼_𝑠)|=|Υ(𝐼_𝑎,𝐼_𝑏)|≤1.
Case 𝑎2.2 (|𝐼_𝑎|=5, |𝐼_𝑏|=4). It follows that max_𝑖|𝐼_𝑖|=5. By Lemma 2.2, we have |Υ(𝐼_𝑎,𝐼_𝑏)|≤2 and |Υ(𝐼_𝑎,𝐼_𝑖)|≤1 for each 𝑖∈{1,2,…,𝑠}⧵{𝑎,𝑏}. Furthermore, Υ(𝐼_𝑖,𝐼_𝑗)=∅ if 𝑎∉{𝑖,𝑗}. Thus, |Υ(𝐼₁,𝐼₂,…,𝐼_𝑠)|≤𝛿−1.
Case 𝑎3 (𝑟=1). It follows that |𝐼_𝑎|≥4 for some 𝑎∈{1,2,…,𝑠} and |𝐼_𝑖|=3 for each 𝑖∈{1,2,…,𝑠}⧵{𝑎}. By (3.42), 4≤|𝐼_𝑎|≤6.
Case 𝑎3.1 (|𝐼_𝑎|=4). It follows that max_𝑖|𝐼_𝑖|=4. By Lemma 2.2, Υ(𝐼_𝑎,𝐼_𝑖)=∅ for each 𝑖∈{1,2,…,𝑠}⧵{𝑎}, implying that |Υ(𝐼₁,𝐼₂,…,𝐼_𝑠)|=0.
Case 𝑎3.2 (|𝐼_𝑎|=5). It follows that max_𝑖|𝐼_𝑖|=5. By Lemma 2.2, |Υ(𝐼_𝑎,𝐼_𝑖)|≤1 for each 𝑖∈{1,2,…,𝑠}⧵{𝑎} and Υ(𝐼_𝑖,𝐼_𝑗)=∅ if 𝑎∉{𝑖,𝑗}, that is, |Υ(𝐼₁,𝐼₂,…,𝐼_𝑠)|≤𝛿−2.
Case 𝑎3.3 (|𝐼_𝑎|=6). It follows that max_𝑖|𝐼_𝑖|=6. By Lemma 2.2, |Υ(𝐼_𝑎,𝐼_𝑖)|≤2 for each 𝑖∈{1,2,…,𝑠}⧵{𝑎} and Υ(𝐼_𝑖,𝐼_𝑗)=∅ if 𝑎∉{𝑖,𝑗}, that is, |Υ(𝐼₁,𝐼₂,…,𝐼_𝑠)|≤2(𝛿−2). Claim 1 is proved.
Let 𝑒∈Υ(𝐼₁,𝐼₂,…,𝐼_𝑠) and let 𝑒=𝑧𝑤, where 𝑧∈𝑉(𝐼^∗_𝑎) and 𝑤∈𝑉(𝐼^∗_𝑏) for some distinct 𝑎,𝑏∈{1,2,…,𝑠}. Put 𝐺^′=𝐺⧵𝑒. Form a graph 𝐺^′′ in the following way. If 𝑑(𝑧)≥𝛿 and 𝑑(𝑤)≥𝛿 in 𝐺^′ then we take 𝐺^′′=𝐺^′. Next, suppose that 𝑑(𝑧)=𝛿−1 and 𝑑(𝑤)≥𝛿 in 𝐺^′. Put𝑈₁=𝜉₁,𝜉₂,…,𝜉_𝑠∪𝑉𝐼^∗_𝑎⧵{𝑧},𝑈₂=𝜉₁,𝜉₂,…,𝜉_𝑠∪𝑉𝐼^∗_𝑏⧵{𝑤}. (3.45) If 𝑈₁⊆𝑁(𝑧), then clearly 𝑑(𝑧)≥|𝑈₁|=𝛿 in 𝐺^′, contradicting the hypothesis. Otherwise, 𝑧𝑣∉𝐸(𝐺^′) for some 𝑣∈𝑈₁ and we take 𝐺^′′=𝐺^′+{𝑧𝑣}. Finally, if 𝑑(𝑧)=𝑑(𝑤)=𝛿−1, then as above, 𝑧𝑣∉𝐸(𝐺^′) and 𝑤𝑢∉𝐸(𝐺^′) for some 𝑣∈𝑈₁, 𝑢∈𝑈₂ and we take 𝐺^′′=𝐺^′+{𝑧𝑣,𝑤𝑢}. Clearly, 𝛿(𝐺^′′)=𝛿(𝐺) and 𝑞=𝑞(𝐺)≥𝑞(𝐺^′′)−1. This procedure may be repeated for all edges of Υ(𝐼₁,𝐼₂,…,𝐼_𝑠). The resulting graph 𝐺^∗ satisfies the following conditions: 𝛿𝐺^∗=𝛿(𝐺),𝑞(𝐺)≥𝑞𝐺^∗−Υ𝐼₁,𝐼₂,…,𝐼_𝑠. (3.46) In fact, 𝐺^∗=𝐺⧵Υ𝐼₁,𝐼₂,…,𝐼_𝑠+𝐸^∗, (3.47) where 𝐸^∗ consists of at most 2|Υ(𝐼₁,𝐼₂,…,𝐼_𝑠)| appropriate new edges such that 𝐺^∗⧵{𝜉₁,𝜉₂,…,𝜉_𝑠} is disconnected. Let 𝐻₁,𝐻₂,…,𝐻_𝑡 be the connected components of 𝐺^∗⧵{𝜉₁,𝜉₂,…,𝜉_𝑠} with 𝑉(𝐼^∗_𝑖)⊆𝑉(𝐻_𝑖)  (𝑖=1,2,…,𝑠) and 𝑉(𝐻_𝑠+1)={𝑥₁,𝑥₂}. For each 𝑖∈{1,2,…,𝑠+1}, put ℎ_𝑖=𝑉𝐻_𝑖,𝑞_𝑖=𝑥𝑦∈𝐸𝐺^∗∶𝑥,𝑦∩𝑉𝐻_𝑖≠∅. (3.48) Clearly, ℎ_𝑖≥2(𝑖=1,2,…,𝑠+1). If ℎ_𝑖≥6 for some 𝑖∈{1,2,…,𝑠}, then 𝑛≥^𝑠+1_𝑖=1ℎ_𝑖+𝜉₁,𝜉₂,…,𝜉_𝑠≥6+3𝑠=3𝛿+3, (3.49) contradicting (3.32). Otherwise, 2≤ℎ_𝑖≤5≤2𝛿−1(𝑖=1,2,…,𝑠+1). It follows that (ℎ_𝑖−2) (2𝛿−ℎ_𝑖−1)≥0 which is equivalent to ℎ_𝑖2𝛿−ℎ_𝑖+12≥2𝛿−1(𝑖=1,2,…,𝑠+1). (3.50)Case  2.1.2.1 (max_𝑖|𝐼_𝑖|≤4). By (3.50) and Lemma 2.3, 𝑞_𝑖(𝐺^∗)≥2𝛿−1(𝑖=1,2,…,𝑠+1). Hence 𝑞𝐺^∗≥^𝑠+1_𝑖=1𝑞_𝑖𝐺^∗≥(𝑠+1)(2𝛿−1)=𝛿(2𝛿−1). (3.51) Using (3.46) and Claim 1, we have 𝑞≥𝑞𝐺^∗()3()−3≥𝛿2𝛿−1−3≥𝛿−1)(𝛿+22. (3.52)Case  2.1.2.2 (max_𝑖|𝐼_𝑖|=5). Assume without loss of generality that max_𝑖|𝐼_𝑖|=|𝐼₁|=5, that is, 4≤ℎ₁≤5. By (3.50) and Lemma 2.3, 𝑞_𝑖(𝐺^∗)≥2𝛿−1(𝑖=2,…,𝑠+1) and 𝑞₁𝐺^∗≥ℎ₁2𝛿−ℎ₁+12≥2(2𝛿−3). (3.53) Hence 𝑞𝐺^∗≥𝑠(2𝛿−1)+2(2𝛿−3)=2𝛿²+𝛿−5. (3.54) By (3.46) and Claim 1, 𝑞≥𝑞𝐺^∗−(𝛿−1)≥2𝛿²3()−4>𝛿−1)(𝛿+22. (3.55)Case  2.1.2.3 (max_𝑖|𝐼_𝑖|=6). Assume without loss of generality that max_𝑖|𝐼_𝑖|=|𝐼₁|=6, that is, ℎ₁=5. By (3.50) and Lemma 2.3, 𝑞_𝑖(𝐺^∗)≥2𝛿−1(𝑖=2,…,𝑠+1) and 𝑞₁𝐺^∗≥ℎ₁2𝛿−ℎ₁+12=5(𝛿−2). (3.56) Hence 𝑞𝐺^∗≥𝑠(2𝛿−1)+5(𝛿−2)=2𝛿²+2𝛿−9. (3.57) By (3.46) and Claim 1, 𝑞≥𝑞𝐺^∗−2(𝛿−2)≥2𝛿²3()−5>𝛿−1)(𝛿+22. (3.58)Case  2.2 (𝑝≥2). According to (3.32), we can distinguish five main cases, namely, 2≤𝑝≤𝛿−3, 𝑝=𝛿−2, 𝑝=𝛿−1, 𝑝=𝛿, and 𝑝=𝛿+1.
Case  2.2.1 (2≤𝑝≤𝛿−3). It follows that |𝑁_𝐶(𝑥_𝑖)|≥𝛿−𝑝≥3(𝑖=1,2) and 𝛿≥5,𝛿−𝑝≥3. (3.59) If 𝑁_𝐶(𝑥₁)≠𝑁_𝐶(𝑥₂), then by (3.59) and Lemma 2.1, |𝐶|≥4𝛿−2𝑝≥3𝛿−𝑝+3, contradicting (3.32). Let 𝑁_𝐶(𝑥₁)=𝑁_𝐶(𝑥₂). Clearly, 𝑠≥|𝑁_𝐶(𝑥₁)|−(|𝑉(𝑃)|−1)≥𝛿−𝑝 and |𝐼_𝑖|≥𝑝+2(𝑖=1,2,…,𝑠). If 𝑠≥𝛿−𝑝+1, then |𝐶|≥𝑠𝑝+2≥𝛿−𝑝+1,=𝑝+2𝛿−𝑝−1𝑝−1+3𝛿−𝑝+1≥3𝛿−𝑝+3, (3.60) again contradicting (3.32). Let 𝑠=𝛿−𝑝. It means that 𝑥₁𝑥₂∈𝐸(𝐺), that is, 𝐺[𝑉(𝑃)] is hamiltonian. By symmetric arguments, 𝑁_𝐶(𝑦)=𝑁_𝐶(𝑥₁) for each 𝑦∈𝑉(𝑃). Assume that Υ(𝐼₁,𝐼₂,…,𝐼_𝑠)≠∅, that is, Υ(𝐼_𝑎,𝐼_𝑏)≠∅ for some elementary segments 𝐼_𝑎 and 𝐼_𝑏. By the definition, there is an intermediate path 𝐿 between 𝐼_𝑎 and 𝐼_𝑏. If |𝐿|≥2, then by Lemma 2.2𝐼_𝑎+𝐼_𝑏≥2𝑝+2|𝐿|+4≥2𝑝+8. (3.61) Hence |𝐶|=𝐼_𝑎+𝐼_𝑏+_{{}⧵{}𝑖∈1,…,𝑠𝑎,𝑏}𝐼_𝑖≥2𝑝+8+(𝑠−2),=𝑝+2𝛿−𝑝−2𝑝−1+3𝛿−𝑝+2≥3𝛿−𝑝+3, (3.62) contradicting (3.32). Thus, |𝐿|=1, that is, Υ(𝐼₁,𝐼₂,…,𝐼_𝑠)⊆𝐸(𝐺). By Lemma 2.2, 𝐼_𝑎+𝐼_𝑏≥2𝑝+2|𝐿|+4=2𝑝+6, (3.63) which yields |𝐶|=𝐼_𝑎+𝐼_𝑏+_{𝑖∈{1,…,𝑠}⧵{𝑎,𝑏}}𝐼_𝑖≥2𝑝+6+(𝑠−2)=()𝑝+2𝑠−2𝑝−2+4𝛿−2𝑝−2≥3𝛿−𝑝−2+𝛿−𝑝. (3.64) If 𝛿−𝑝≥4, then |𝐶|≥3𝛿−𝑝+2, contradicting (3.32). Let 𝛿−𝑝≤3. Recalling (3.59), we have 𝛿−𝑝=3, that is, 𝑝=𝛿−3 and 𝑠=𝛿−𝑝=3. Hence, |𝐶|≥𝑠(𝑝+2)=3(𝛿−1). On the other hand, by (3.32) and the fact that 𝑝≥2, we have |𝐶|≤3𝛿−𝑝+1≤3𝛿−1. Thus 3𝛿−3≤|𝐶|≤3𝛿−1. (3.65) Put 𝐺^′=𝐺⧵Υ(𝐼₁,𝐼₂,𝐼₃). As in Case  2.1.2, form a graph 𝐺^∗ by adding at most 2|Υ(𝐼₁,𝐼₂,𝐼₃)| new edges in 𝐺^′ such that 𝛿(𝐺^∗)=𝛿(𝐺) and 𝐺^∗⧵{𝜉₁,𝜉₂,𝜉₃} are disconnected. We denote 𝐺^∗=𝐺 immediately if Υ(𝐼₁,𝐼₂,𝐼₃)=∅. Hence 𝑞(𝐺)≥𝑞𝐺^∗−Υ𝐼₁,𝐼₂,𝐼₃. (3.66) Let 𝐻₁,𝐻₂,…,𝐻_𝑡 be the connected components of 𝐺^∗⧵{𝜉₁,𝜉₂,𝜉₃} with 𝑉(𝐼^∗_𝑖)⊆𝑉(𝐻_𝑖)(𝑖=1,2,3) and 𝑉(𝑃)⊆𝑉(𝐻₄). Since 𝑥₁𝑥₂∈𝐸(𝐺)  (i.e., 𝐺[𝑉(𝑃)] is hamiltonian) and 𝑃 is extreme, we have 𝑉(𝐻₄)=𝑉(𝑃). Using notation (3.48) for 𝐺^∗, we have ℎ_𝑖≥|𝐼_𝑖|−1≥𝑝+1=𝛿−2  (𝑖=1,2,3) and ℎ₄=𝛿−2. If ℎ_𝑖≥𝛿+1 for some 𝑖∈{1,2,3}, then 𝑛≥ℎ₁+ℎ₂+ℎ₃+ℎ₄+𝑠≥4𝛿−2. (3.67) By (3.59), 𝛿≥5, implying that 4𝛿−2≥3𝛿+3 and 𝑛≥3𝛿+3, contradicting (3.32). Let 𝛿−2≤ℎ_𝑖≤𝛿  (𝑖=1,2,3,4). It follows that ℎ_𝑖2𝛿−ℎ_𝑖+12≥(𝛿−2)(𝛿+3)2(𝑖=1,2,3,4). (3.68) By Lemma 2.3, 𝑞_𝑖(𝐺^∗)≥(𝛿−2)(𝛿+3)/2(𝑖=1,2,3,4), implying that 𝑞𝐺^∗≥⁴_𝑖=1𝑞_𝑖𝐺^∗≥2(𝛿−1)(𝛿+3). (3.69) If |Υ(𝐼₁,𝐼₂,𝐼₃)|≥4, then |Υ(𝐼_𝑎,𝐼_𝑏)|≥2 for some distinct 𝑎,𝑏∈{1,2,3}. By Lemma 2.2𝐼_𝑎+𝐼_𝑏≥2𝑝+7=2𝛿+1 (3.70) and hence |𝐶|≥3𝛿, contradicting (3.65). So, |Υ(𝐼₁,𝐼₂,𝐼₃)|≤3. By (3.66) and (3.69), 𝑞≥𝑞𝐺^∗()3()−3≥2𝛿−1)(𝛿+3−3≥𝛿−1)(𝛿+22. (3.71)Case  2.2.2 (𝑝=𝛿−2). It follows that |𝑁_𝐶(𝑥_𝑖)|≥𝛿−𝑝=2(𝑖=1,2). By (3.32), |𝐶|≤3𝛿+1−𝑝=2𝛿+3. (3.72) If 𝑁_𝐶(𝑥₁)≠𝑁_𝐶(𝑥₂), then by Lemma 2.1, |𝐶|≥4𝛿−2𝑝=2𝛿+4, contradicting (3.72). Let 𝑁_𝐶(𝑥₁)=𝑁_𝐶(𝑥₂). Further, if 𝑠≥3, then |𝐶|≥𝑠𝑝+2≥3𝛿=2𝛿+𝑝+2≥2𝛿+4, (3.73) again contradicting (3.72). Let 𝑠=2. It follows that 𝑥₁𝑥₂∈𝐸(𝐺), that is, 𝐺[𝑉(𝑃)] is hamiltonian. By symmetric arguments, 𝑁_𝐶(𝑦)=𝑁_𝐶(𝑥₁)={𝜉₁,𝜉₂} for each 𝑦∈𝑉(𝑃). Clearly, |𝐼_𝑖|≥𝑝+2=𝛿(𝑖=1,2).
Case  2.2.2.1 (Υ(𝐼₁,𝐼₂)=∅). It follows that 𝐺⧵{𝜉₁,𝜉₂} is disconnected. Let 𝐻₁,𝐻₂,…,𝐻_𝑡 be the connected components of 𝐺⧵{𝜉₁,𝜉₂} with 𝑉(𝐼^∗_𝑖)⊆𝑉(𝐻_𝑖)(𝑖=1,2) and 𝑉(𝑃)⊆𝑉(𝐻₃). Since 𝑃 is extreme and 𝐺[𝑉(𝑃)] is hamiltonian, we have 𝑉(𝐻₃)=𝑉(𝑃). By notation (3.48), ℎ_𝑖≥|𝐼_𝑖|−1≥𝛿−1(𝑖=1,2) and ℎ₃=𝛿−1. If ℎ_𝑖≥𝛿+3 for some 𝑖∈{1,2}, then 𝑛≥ℎ₁+ℎ₂+ℎ₃+𝜉₁,𝜉₂≥3𝛿+3, (3.74) contradicting (3.32). So, 𝛿−1≤ℎ_𝑖≤𝛿+2(𝑖=1,2,3). By Lemma 2.3, 𝑞_𝑖≥ℎ_𝑖2𝛿−ℎ_𝑖+12≥(𝛿−1)(𝛿+2)2(𝑖=1,2,3). (3.75) Hence, 𝑞≥³_𝑖=1𝑞_𝑖≥3(𝛿−1)(𝛿+2)2. (3.76)Case  2.2.2.2 (Υ(𝐼₁,𝐼₂)≠∅). By definition, there is an intermediate path 𝐿 between 𝐼₁ and 𝐼₂. If |𝐿|≥2, then by Lemma 2.2|𝐶|=𝐼₁+𝐼₂≥2𝑝+2|𝐿|+4≥2𝛿+4, (3.77) contradicting (3.72). Otherwise, Υ(𝐼₁,𝐼₂)⊆𝐸(𝐺). Further, if |Υ(𝐼₁,𝐼₂)|≥3, then by Lemma 2.2|𝐶|=𝐼₁+𝐼₂≥2𝑝+8=2𝛿+4, (3.78) again contradicting (3.72). Thus |Υ(𝐼₁,𝐼₂)|≤2.
Case  2.2.2.2.1 (|Υ(𝐼₁,𝐼₂)|=1). Put 𝐺^′=𝐺⧵Υ(𝐼₁,𝐼₂). As in Case  2.1.2, form a graph 𝐺^∗ by adding at most two new edges in 𝐺^′ such that 𝛿(𝐺^∗)=𝛿(𝐺), 𝐺^∗⧵{𝜉₁,𝜉₂} is disconnected and 𝑞(𝐺)≥𝑞(𝐺^∗)−1. Let 𝐻₁,𝐻₂,…,𝐻_𝑡 be the connected components of 𝐺^∗⧵{𝜉₁,𝜉₂} with 𝑉(𝐼^∗_𝑖)⊆𝑉(𝐻_𝑖)(𝑖=1,2) and 𝑉(𝑃)=𝑉(𝐻₃). Using notation (3.48) for 𝐺^∗, as in Case  2.2.2.1, we have 𝛿−1≤ℎ_𝑖≤𝛿+2(𝑖=1,2,3). By Lemma 2.2, |𝐼₁|+|𝐼₂|≥2𝑝+6=2𝛿+2. It means that max_𝑖|𝐼_𝑖|≥𝛿+1, that is, max_𝑖ℎ_𝑖≥𝛿. Assume without loss of generality that ℎ₁≥𝛿. By Lemma 2.3, 𝑞₁𝐺^∗≥ℎ₁2𝛿−ℎ₁+12≥𝛿(𝛿+1)2,𝑞_𝑖𝐺^∗≥ℎ_𝑖2𝛿−ℎ_𝑖+12≥(𝛿−1)(𝛿+2)2(𝑖=2,3), (3.79) implying that 𝑞𝐺^∗≥𝛿(𝛿+1)2+(𝛿−1)(𝛿+2). (3.80) Hence 𝑞≥𝑞𝐺^∗𝛿()−1≥𝛿+12+()3()𝛿−1)(𝛿+2−1≥𝛿−1)(𝛿+22. (3.81)Case  2.2.2.2.2 (|Υ(𝐼₁,𝐼₂)|=2). By Lemma 2.2, |𝐶|=𝐼₁+𝐼₂≥2𝑝+7=2𝛿+3. (3.82) Recalling (3.72), we get |𝐶|=2𝛿+3 and 𝑉(𝐺)=𝑉(𝑃∪𝐶). Put 𝐺^′=𝐺⧵Υ(𝐼₁,𝐼₂). As in Case  2.1.2, form a graph 𝐺^∗ by adding at most four new edges in 𝐺^′ such that 𝛿(𝐺^∗)=𝛿(𝐺), 𝐺^∗⧵{𝜉₁,𝜉₂} is disconnected and 𝑞(𝐺)≥𝑞(𝐺^∗)−2. Let 𝐻₁,𝐻₂, and 𝐻₃ be the connected components of 𝐺^∗⧵{𝜉₁,𝜉₂} with 𝑉(𝐻_𝑖)=𝑉(𝐼^∗_𝑖)(𝑖=1,2) and 𝑉(𝐻₃)=𝑉(𝑃). Using notation (3.48) for 𝐺^∗, we have as in Case  2.2.2.1, 𝛿−1≤ℎ_𝑖≤𝛿+2(𝑖=1,2,3). Since |𝐼_𝑖|≥𝛿(𝑖=1,2) and |𝐶|=|𝐼₁|+|𝐼₂|=2𝛿+3, we can assume without loss of generality that either |𝐼₁|=𝛿+2, |𝐼₂|=𝛿+1 or |𝐼₁|=𝛿+3,  |𝐼₂|=𝛿.
Case  2.2.2.2.2.1 (|𝐼₁|=𝛿+2, |𝐼₂|=𝛿+1). It follows that ℎ₁=𝛿+1, ℎ₂=𝛿 and ℎ₃=𝛿−1. By Lemma 2.3, 𝑞_𝑖𝐺^∗≥ℎ_𝑖2𝛿−ℎ_𝑖+12=𝛿(𝛿+1)2(),𝑞𝑖=1,2₃𝐺^∗≥ℎ₃2𝛿−ℎ₃+12=(𝛿−1)(𝛿+2)2. (3.83) Hence 𝑞≥³_𝑖=1𝑞_𝑖𝐺^∗()()−2≥𝛿𝛿+1−2+𝛿−1)(𝛿+22=3(𝛿−1)(𝛿+2)2. (3.84)
Case  2.2.2.2.2.2 (|𝐼₁|=𝛿+3, |𝐼₂|=𝛿). Let Υ(𝐼₁,𝐼₂)={𝑒₁,𝑒₂}, where 𝑒₁=𝑦₁𝑧₁,𝑒₂=𝑦₂𝑧₂,𝑦₁,𝑦₂⊆𝑉𝐼^∗₁,𝑧₁,𝑧₂⊆𝑉𝐼^∗₂. (3.85) If 𝑦₁≠𝑦₂ and 𝑧₁≠𝑧₂, then as in proof of Lemma 2.2 (Case  2.1), |𝐶|=𝐼₁+𝐼₂≥2𝑝+8=2(𝛿−2)+8=2𝛿+4, (3.86) contradicting (3.72). Let either 𝑦₁≠𝑦₂ and 𝑧₁=𝑧₂ or 𝑦₁=𝑦₂ and 𝑧₁≠𝑧₂.
Case  2.2.2.2.2.2.1 (𝑦₁≠𝑦₂ and 𝑧₁=𝑧₂). Assume without loss of generality that 𝑦₁,𝑦₂ occur on 𝐼₁ in this order. If 𝑦₂=𝑦⁺₁, then |𝐶|≥𝜉𝐶₁^→𝑦₁𝑧₁𝑦𝐶₂^→𝜉₂𝑥₂^←𝑃𝑥₁𝜉₁=2𝛿+4, (3.87) contradicting (3.72). Let 𝑦₂≠𝑦⁺₁, that is, |𝑦𝐶₁^→𝑦₂|≥2. Put 𝐶^′=𝜉𝐶₁^→𝑦₂𝑧𝐶₁^←𝜉₂𝑥₂^←𝑃𝑥₁𝜉₁,𝐶^′′=𝜉𝐶₁^←𝑧₁𝑦𝐶₁^→𝜉₂𝑥₂^←𝑃𝑥₁𝜉₁. (3.88) Clearly, |𝐶|≥𝐶^′=𝜉𝐶₁^→𝑦₁+𝑦𝐶₁^→𝑦₂+𝑒₂+𝜉𝐶₂^→𝑧₁+,|𝐶|≥𝐶𝑝+2_′′=𝜉𝐶₁^←𝑧₁+𝑒₁+𝑦𝐶₁^→𝑦₂+𝑦𝐶₂^→𝜉₂+𝑝+2. (3.89) By summing and observing that 𝜉𝐶₁^→𝑦₁+𝑦𝐶₁^→𝑦₂+𝑦𝐶₂^→𝜉₂+𝜉𝐶₂^→𝑧₁+𝑧𝐶₁^→𝜉₁=|𝐶|, (3.90) we get 2|𝐶|≥|𝐶|+𝑦𝐶₁^→𝑦₂+2𝑝+2+2≥|𝐶|+2𝛿+4. (3.91) Hence |𝐶|≥2𝛿+4, again contradicting (3.72).
Case  2.2.2.2.2.2.2 (𝑦₁=𝑦₂ and 𝑧₁≠𝑧₂). Assume without loss of generality that 𝑧₂,𝑧₁ occur on 𝐼₂ in this order.
Case  2.2.2.2.2.2.2.1 (𝛿≥6). If |𝜉𝐶₁^→𝑦₁|≥𝛿−1 and |𝑦𝐶₁^→𝜉₂|≥𝛿−1, then |𝐼₁|≥2𝛿−2≥𝛿+4, contradicting the hypothesis. Thus, we can assume without loss of generality that |𝜉𝐶₁^→𝑦₁|≤𝛿−2. If 𝑦⁻₁=𝜉₁, then |𝐶|≥𝜉𝐶₁^←𝑧₂𝑦𝐶₁^→𝜉₂𝑥₂^←𝑃𝑥₁𝜉₁≥2𝛿+5, (3.92) contradicting (3.72). Let 𝑦⁻₁≠𝜉₁, that is, 𝑦⁻₁∈𝑉(𝐼^∗₁). Since Υ(𝐼₁,𝐼₂)={𝑦₁𝑧₁,𝑦₁𝑧₂}, we have 𝑁(𝑦⁻₁)⊂𝑉(𝐼₁). If 𝑁(𝑦⁻₁)∩𝑉(𝑦𝐶⁺₁^→𝜉⁻₂)=∅, then |𝑁(𝑦⁻₁)|≤𝛿−1, a contradiction. Otherwise, 𝑦⁻₁𝑤∈𝐸(𝐺) for some 𝑤∈𝑉(𝑦𝐶⁺₁^→𝜉⁻₂). Put 𝑅=𝜉𝐶₁^→𝑦⁻₁𝑤𝐶^←𝑦₁,𝐶^′=𝜉₁^→𝑅𝑦₁𝑧𝐶₁^←𝜉₂𝑥₂^←𝑃𝑥₁𝜉₁,𝐶^′′=𝜉𝐶₁^←𝑧₂𝑦𝐶₁^→𝜉₂𝑥₂^←𝑃𝑥₁𝜉₁. (3.93) Clearly, |𝐶|≥𝐶^′=|𝑅|+𝑦₁𝑧₁+𝑧𝐶₁^←𝜉₂+,|𝐶|≥𝐶𝑝+2_′′=𝜉𝐶₁^←𝑧₁+𝑧𝐶₁^←𝑧₂+𝑦₁𝑧₂+𝑦𝐶₁^→𝜉₂+𝑝+2. (3.94) By summing and observing that |𝑅|≥|𝜉𝐶₁^→𝑦₁|+1, we get 2|𝐶|≥𝜉𝐶₁^→𝑦₁+𝑦𝐶₁^→𝜉₂+𝜉𝐶₂^→𝑧₁+𝑧𝐶₁^→𝜉₁+2𝑝+2+4=|𝐶|+2𝛿+4. (3.95) Hence |𝐶|≥2𝛿+4, contradicting (3.72).
Case  2.2.2.2.2.2.2.2 (𝛿=5). It follows that 𝐼₁=𝛿+3=8,𝐼₂=𝛿=5,|𝐶|=2𝛿+3=13. (3.96) If either |𝜉𝐶₁^→𝑦₁|≤𝛿−2=3 or |𝑦𝐶₁^→𝜉₂|≤𝛿−2=3, then we can argue as in Case  2.2.2.2.2.2.2.1. Otherwise, |𝜉𝐶₁^→𝑦₁|=|𝑦𝐶₁^→𝜉₂|=4. If |𝑧𝐶₁^←𝜉₂|≥4, then 𝜉𝐶₁^→𝑦₁𝑧𝐶₁^←𝜉₂𝑥₂^←𝑃𝑥₁𝜉₁≥14>|𝐶|, (3.97) a contradiction. Let |𝑧𝐶₁^←𝜉₂|≤3. Similarly, |𝜉𝐶₁^←𝑧₂|≤3, implying that 𝐼₂=𝜉₂𝜉⁺₂𝑧₂𝑧₁𝑧⁺₁𝜉₁. If 𝑧⁺₁𝑧₂∈𝐸(𝐺), then 𝜉𝐶₁^→𝑦₁𝑧₁𝑧⁺₁𝑧𝐶₂^←𝜉₂𝑥₂^←𝑃𝑥₁𝜉₁=14>|𝐶|, (3.98) a contradiction. So, 𝑁(𝑧⁺₁)⊆{𝜉₁,𝜉₂,𝑧₁,𝜉⁺₂}, again a contradiction, since |𝑁(𝑧⁺₁)|≥𝛿=5.
Case  2.2.2.2.2.2.2.3 (𝛿=4). It follows that 𝐼₁=𝛿+3=7,𝐼₂=𝛿=4,|𝐶|=2𝛿+3=11. (3.99) Since |𝐼₁|=7, we have either |𝜉𝐶₁^→𝑦₁|≥4 or |𝑦𝐶₁^→𝜉₂|≥4, say |𝜉𝐶₁^→𝑦₁|≥4. Put 𝐶^′=𝜉𝐶₁^→𝑦₁𝑧𝐶₁^←𝜉₂𝑥₂^←𝑃𝑥₁𝜉₁. (3.100) If |𝜉𝐶₁^→𝑦₁|≥5, then |𝐶^′|≥12>|𝐶|, a contradiction. This means that |𝜉𝐶₁^→𝑦₁|=4. If |𝑧𝐶₁^→𝜉₁|=1 then |𝐶^′|≥12>|𝐶|, a contradiction. Let |𝑧𝐶₁^→𝜉₁|≥2. Since |𝐼₂|=4, we have |𝑧𝐶₁^→𝜉₁|=2, that is, 𝐼₂=𝜉₂𝑧₂𝑧₁𝑧⁺₁𝜉₁. Further, if 𝑧⁺₁𝑧₂∈𝐸(𝐺), then 𝜉𝐶₁^→𝑦₁𝑧₁𝑧⁺₁𝑧₂𝜉₂𝑥₂^←𝑃𝑥₁𝜉₁=12>|𝐶|, (3.101) a contradiction. Let 𝑧⁺₁𝑧₂∉𝐸(𝐺). Since |Υ(𝐼₁,𝐼₂)|=2, we have 𝑁(𝑧⁺₁)⊆{𝜉₁,𝜉₂,𝑧₁}, contradicting the fact that |𝑁(𝑧⁺₁)|≥𝛿=4.
Case  2.2.3 (𝑝=𝛿−1). By (3.32), |𝐶|≤3𝛿+1−𝑝=2𝛿+2. (3.102) It follows that |𝑁_𝐶(𝑥_𝑖)|≥𝛿−𝑝=1(𝑖=1,2).
Case  2.2.3.1 (|𝑁_𝐶(𝑥_𝑖)|≥2(𝑖=1,2)). If 𝑁_𝐶(𝑥₁)≠𝑁_𝐶(𝑥₂), then by Lemma 2.1, |𝐶|≥2𝑝+8=2𝛿+6, contradicting (3.102). Let 𝑁_𝐶(𝑥₁)=𝑁_𝐶(𝑥₂). If 𝑠≥3, then |𝐶|≥𝑠𝑝+2≥3(𝛿+1)>2𝛿+2, (3.103) contradicting (3.102). Let 𝑠=2. It follows that |𝐶|≥𝑠(𝑝+2)≥2(𝛿+1). Recalling (3.102), we get |𝐶|=2𝛿+2,𝐼₁=𝐼₂=𝛿+1,𝑉(𝐺)=𝑉(𝐶∪𝑃). (3.104) Assume that 𝑦𝑧∈𝐸(𝐺) for some 𝑦∈𝑉(𝑃) and 𝑧∈𝑉(𝐶)⧵{𝜉₁,𝜉₂}. Besides, we can assume without loss of generality that 𝑧∈𝑉(𝐼^∗₁). Since 𝐶 is extreme, we have 𝜉𝐶₁^→𝑧≥𝑥₁^→𝑃𝑦+2,𝑧𝐶^→𝜉₂≥𝑦^→𝑃𝑥₂+2, (3.105) implying that 𝐼₁=𝜉𝐶₁^→𝑧+𝑧𝐶^→𝜉₂≥𝑥₁^→𝑃𝑦+𝑦^→𝑃𝑥₂+4=𝑝+4=𝛿+3, (3.106) a contradiction. So, 𝑁_𝐶(𝑦)⊆{𝜉₁,𝜉₂} for each 𝑦∈𝑉(𝑃). On the other hand, by Lemma 2.2, Υ(𝐼₁,𝐼₂)=∅ and hence 𝐺⧵{𝜉₁,𝜉₂} is disconnected. Let 𝐻₁,𝐻₂, and 𝐻₃ be the connected components of 𝐺⧵{𝜉₁,𝜉₂} with 𝑉(𝐻_𝑖)=𝑉(𝐼^∗_𝑖)(𝑖=1,2) and 𝑉(𝐻₃)=𝑉(𝑃). Using notation (3.48), we have ℎ_𝑖=𝛿(𝑖=1,2,3). By Lemma 2.3, 𝑞_𝑖≥ℎ_𝑖2𝛿−ℎ_𝑖+12=𝛿(𝛿+1)2(𝑖=1,2,3), (3.107) implying that 𝑞≥³_𝑖=1𝑞_𝑖≥3𝛿²+𝛿2>3(𝛿−1)(𝛿+2)2. (3.108)Case  2.2.3.2 (either |𝑁_𝐶(𝑥₁)|=1 or |𝑁_𝐶(𝑥₂)|=1). Assume without loss of generality that |𝑁_𝐶(𝑥₁)|=1. Put 𝑁_𝐶(𝑥₁)={𝑦₁}.
Case  2.2.3.2.1 (𝑁_𝐶(𝑥₂)≠𝑁_𝐶(𝑥₁)). It follows that 𝑥₂𝑦₂∈𝐸(𝐺) for some 𝑦₂∈𝑉(𝐶)⧵{𝑦₁} and we can argue as in Case  2.2.3.1.
Case  2.2.3.2.2 (𝑁_𝐶(𝑥₂)=𝑁_𝐶(𝑥₁)={𝑦₁}). It follows that 𝑁𝑥_𝑖=𝑉(𝑃)⧵𝑥_𝑖∪𝑦₁(𝑖=1,2). (3.109) Since 𝜅≥2, there is an edge 𝑧𝑤 such that 𝑧∈𝑉(𝑃) and 𝑤∈𝑉(𝐶)⧵{𝑦₁}. Since 𝑁_𝐶(𝑥₁)=𝑁_𝐶(𝑥₂)={𝑦₁}, we have 𝑧∉{𝑥₁,𝑥₂}. By (3.109), 𝑥₂𝑧⁻∈𝐸(𝐺). Then replacing 𝑃 with 𝑥₁^→𝑃𝑧⁻𝑥₂^←𝑃𝑧, we can argue as in Case  2.2.3.1.
Case  2.2.4 (𝑝=𝛿). By (3.32), |𝐶|≤3𝛿+1−𝑝=2𝛿+1. Let 𝑄=𝜉𝑄^→𝜂 be a longest path in 𝐺 with 𝑉(𝑄)∩𝑉(𝐶)={𝜉,𝜂}. If |𝑄|≥𝛿+1, then by (3.34), |𝐶|≥2|𝑄|≥2𝛿+2, a contradiction. Let |𝑄|≤𝛿. (3.110)Case  2.2.4.1 (𝑥₁𝑥₂∉𝐸(𝐺)). It follows that |𝑁_𝐶(𝑥_𝑖)|≥1  (𝑖=1,2). If |𝑁_𝐶(𝑥_𝑖)|≥2 for some 𝑖∈{1,2}, then clearly |𝑄|≥𝑝+2=𝛿+2, contradicting (3.110). Let |𝑁_𝐶(𝑥₁)|=|𝑁_𝐶(𝑥₂)|=1. Further, if 𝑁_𝐶(𝑥₁)≠𝑁_𝐶(𝑥₂), then again |𝑄|≥𝛿+2, contradicting (3.110). Let 𝑁_𝐶(𝑥₁)=𝑁_𝐶(𝑥₂)={𝑧₁} for some 𝑧₁∈𝑉(𝐶). Since 𝜅≥2, there is a path 𝐿=𝑦𝑧₂ connecting 𝑃 and 𝐶 such that 𝑦∈𝑉(𝑃) and 𝑧₂∈𝑉(𝐶)⧵{𝑧₁}. Clearly, 𝑦∉{𝑥₁,𝑥₂}. If 𝑥₂𝑦⁻∈𝐸(𝐺), then |𝑄|≥𝑧₁𝑥₁^→𝑃𝑦⁻𝑥₂^←𝑃𝑦𝑧₂=𝛿+2, (3.111) contradicting (3.110). Let 𝑥₂𝑦⁻∉𝐸(𝐺). Further, if 𝑦⁻≠𝑥₁, then recalling that 𝑥₂𝑥₁∉𝐸(𝐺) we have |𝑁_𝐶(𝑥₂)|≥2, a contradiction. Otherwise, 𝑦⁻=𝑥₁ and |𝑄|≥|𝑧₁𝑥₂^←𝑃𝑦𝑧₂|=𝛿+1, contradicting (3.110).
Case  2.2.4.2 (𝑥₁𝑥₂∈𝐸(𝐺)). Put 𝐶^′=𝑥₁^→𝑃𝑥₂𝑥₁. Since 𝜅≥2, there are two disjoint paths 𝐿₁,𝐿₂ connecting 𝐶^′ and 𝐶. Further, since 𝑃 is extreme, we have |𝐿₁|=|𝐿₂|=1. Let 𝐿₁=𝑦₁𝑧₁ and 𝐿₂=𝑦₂𝑧₂, where 𝑦₁,𝑦₂∈𝑉(𝐶^′) and 𝑧₁,𝑧₂∈𝑉(𝐶). Since 𝐶^′ is a hamiltonian cycle in 𝐺[𝑉(𝑃)], we can assume that 𝑃 is chosen such that 𝑥₁=𝑦₁. If |𝑥₁^→𝑃𝑦₂|≤2, then |𝑄|≥|𝑧₁𝑥₁𝑥₂^←𝑃𝑦₂𝑧₂|≥𝛿+1, contradicting (3.110). Let |𝑥₁^→𝑃𝑦₂|≥3. If 𝑥₂𝑣∈𝐸(𝐺) for some 𝑣∈{𝑦₂⁻¹,𝑦₂⁻²}, then |𝑄|≥𝑧₁𝑥₁^→𝑃𝑣𝑥₂^←𝑃𝑦₂𝑧₂≥𝛿+1, (3.112) contradicting (3.110). Otherwise, |𝑁_𝐶(𝑥₂)|≥2, implying that 𝑥₂𝑧₃∈𝐸(𝐺) for some 𝑧₃∈𝑉(𝐶)⧵{𝑧₁}. Then |𝑄|≥𝑧₁𝑥₁^→𝑃𝑥₂𝑧₃≥𝛿+2, (3.113) again contradicting (3.110).
Case  2.2.5 (𝑝=𝛿+1). By (3.32), |𝐶|≤3𝛿+1−𝑝=2𝛿. On the other hand, by Theorem D, |𝐶|≥2𝛿, implying that |𝐶|=2𝛿 and 𝑉(𝐺)=𝑉(𝐶∪𝑃). Let 𝑄=𝜉𝑄^→𝜂 be a longest path in 𝐺 with 𝑉(𝑄)∩𝑉(𝐶)={𝜉,𝜂}. If |𝑄|≥𝛿+1, then by (3.35), |𝐶|≥2|𝑄|≥2𝛿+2, a contradiction. Let |𝑄|≤𝛿. (3.114)Case  2.2.5.1 (𝑥₁𝑥₂∈𝐸(𝐺)). Put 𝐶^′=𝑥₁^→𝑃𝑥₂𝑥₁. Since 𝜅≥2, there are two disjoint edges 𝑧₁𝑤₁ and 𝑧₂𝑤₂ connecting 𝐶^′ and 𝐶 such that 𝑧₁,𝑧₂∈𝑉(𝐶^′) and 𝑤₁,𝑤₂∈𝑉(𝐶). Since 𝐶^′ is a hamiltonian cycle in 𝐺[𝑉(𝑃)], we can assume without loss of generality that 𝑃 is chosen such that 𝑧₁=𝑥₁. If |𝑥₁^→𝑃𝑧₂|≤3, then |𝑄|≥|𝑤₁𝑥₁𝑥₂^←𝑃𝑧₂𝑤₂|≥𝛿+1, contradicting (3.114). Let |𝑥₁^→𝑃𝑧₂|≥4. Further, if 𝑥₂𝑣∈𝐸(𝐺) for some 𝑣∈{𝑧₂⁻¹,𝑧₂⁻²,𝑧₂⁻³}, then |𝑄|≥𝑤₁𝑥₁^→𝑃𝑣𝑥₂^←𝑃𝑧₂𝑤₂≥𝛿+1, (3.115) contradicting (3.114). Now let 𝑥₂𝑣∉𝐸(𝐺) for each 𝑣∈{𝑧₂⁻¹,𝑧₂⁻²,𝑧₂⁻³}. It follows that |𝑁_𝐶(𝑥₂)|≥2, that is, 𝑥₂𝑤₃∈𝐸(𝐺) for some 𝑤₃∈𝑉(𝐶)⧵{𝑤₁}. But then |𝑄|≥|𝑤₁𝑥₁^→𝑃𝑥₂𝑤₃|=𝛿+3, contradicting (3.114).
Case  2.2.5.2 (𝑥₁𝑥₂∉𝐸(𝐺)). If 𝑑_𝑃(𝑥₁)+𝑑_𝑃(𝑥₂)≥|𝑉(𝑃)|=𝑝+1=𝛿+2 then by Theorem C, 𝐺[𝑉(𝑃)] is hamiltonian and we can argue as in Case  2.2.5.1. Otherwise, 𝑑_𝑃(𝑥₁)+𝑑_𝑃(𝑥₂)≤𝛿+1, implying that 𝑑_𝐶𝑥₁+𝑑_𝐶𝑥₂≥𝛿−1≥2. (3.116) Assume without loss of generality that 𝑑_𝐶(𝑥₁)≥𝑑_𝐶(𝑥₂).
Case  2.2.5.2.1 (𝑑_𝐶(𝑥₂)=0). It follows that 𝑁(𝑥₂)=𝑉(𝑃)⧵{𝑥₁,𝑥₂}. By (3.116), 𝑑_𝐶(𝑥₁)≥2. Put 𝐶^′=𝑥⁺₁^→𝑃𝑥₂𝑥⁺₁. Since 𝜅≥2, there is a path 𝐿=𝑧^→𝐿𝑤 connecting 𝐶^′ and 𝐶 such that 𝑧∈𝑉(𝐶^′)⧵{𝑥⁺₁} and 𝑤∈𝑉(𝐶). If 𝑥₁∈𝑉(𝐿), that is, 𝑥₁𝑧∈𝐸(𝐺), then 𝑥₁^→𝑃𝑧⁻𝑥₂^←𝑃𝑧𝑥₁ is a hamiltonian cycle in 𝐺[𝑉(𝑃)] and we can argue as in Case  2.2.5.1. Let 𝑥₁∉𝑉(𝐿). Since 𝑉(𝐺)=𝑉(𝐶∪𝑃), we have 𝐿=𝑧𝑤. Further, since 𝑑_𝐶(𝑥₁)≥2, we have 𝑥₁𝑤₁∈𝐸(𝐺) for some 𝑤₁∈𝑉(𝐶)⧵{𝑤}. Hence, |𝑄|≥𝑤₁𝑥₁^→𝑃𝑧⁻𝑥₂^←𝑃𝑧𝑤=𝛿+3, (3.117) contradicting (3.114).
Case  2.2.5.2.2 (𝑑_𝐶(𝑥₂)=1). Let 𝑁_𝐶(𝑥₂)={𝑤₁}. By (3.116), 𝑑_𝐶(𝑥₁)≥1. If either 𝑑_𝐶(𝑥₁)≥2 or 𝑁_𝐶(𝑥₁)≠𝑁_𝐶(𝑥₂), then 𝑥₁𝑤∈𝐸(𝐺) for some 𝑤∈𝑉(𝐶)⧵{𝑤₁} and therefore |𝑄|≥𝑤𝑥₁^→𝑃𝑥₂𝑤₁=𝛿+3, (3.118) contradicting (3.114). Otherwise, 𝑁_𝐶(𝑥₁)=𝑁_𝐶(𝑥₂)={𝑤₁}. Since 𝜅≥2, there is an edge 𝑧𝑤 such that 𝑧∈𝑉(𝑃) and 𝑤∈𝑉(𝐶)⧵{𝑤₁}. Clearly, 𝑧∉{𝑥₁,𝑥₂}. Further, we can argue as in Case  2.2.5.1.
Case  2.2.5.2.3 (𝑑_𝐶(𝑥₂)≥2). Since 𝑑_𝐶(𝑥₁)≥𝑑_𝐶(𝑥₂), we have 𝑑_𝐶(𝑥₁)≥2. Hence |𝑄|≥𝑝+2=𝛿+3, contradicting (3.114).

Proof of Theorem 1.1. Let 𝐺 be a graph satisfying the hypothesis of Theorem 1.1, which is equivalent to 𝑞≤𝛿²+𝛿−1. (3.119) Since 1𝑞=2_{𝑢∈𝑉(𝐺)}𝑑(𝑢)≥𝛿𝑛2, (3.120) we have 𝛿𝑛/2≤𝛿²+𝛿−1 which is equivalent to 𝛿≥𝑛−12−12+1𝛿. (3.121) If 𝑛 is even, that is, 𝑛=2𝑡 for some integer 𝑡, then 𝛿≥2𝑡−12−12+1𝛿1=𝑡−1+𝛿, (3.122) implying that 𝛿≥𝑡=𝑛/2. By Theorem A, 𝐺 is hamiltonian. Let 𝑛 is odd, that is, 𝑛=2𝑡+1 for some integer 𝑡. Then 𝛿≥𝑡−1/2+1/𝛿 implying that 𝛿≥𝑡≥(𝑛−1)/2. Recalling that 𝐺 is hamiltonian when 𝛿>(𝑛−1)/2, we can assume that 𝛿=(𝑛−1)/2. By Theorem C, either 𝐺 is hamiltonian or containers at least 𝛿²+𝛿 edges, contradicting (3.119). Theorem 1.1 is proved.

Proof of Theorem 1.2. Let 𝐺 be a 2-connected graph. The hypothesis of Theorem 1.2 is equivalent to 𝑞≤3()8when𝛿=2,𝛿−1)(𝛿+2−12when𝛿≥3. (3.123)Case  1 (𝛿=2 and 𝑞≤8). Let 𝐶 be a longest cycle in 𝐺 and 𝑃=𝑥₁^→𝑃𝑥₂ a longest path in 𝐺⧵𝐶 of length 𝑝. If 𝑝=0, then 𝐶 is a dominating cycle and we are done. Let 𝑝≥1. Since 𝜅≥2, there is a path 𝑄=𝜉𝑄^→𝜂 such that 𝑉(𝑄)∩𝑉(𝐶)={𝜉,𝜂} and |𝑄|≥3. Further, since 𝐶 is extreme, we have |𝐶|=|𝜉𝐶^→𝜂|+|𝜂𝐶^→𝜉|≥2|𝑄|≥6 and therefore, 𝑞≥|𝐶|+|𝑄|≥9, contradicting the hypothesis.
Case  2 (𝛿≥3 and 𝑞≤(3(𝛿−1)(𝛿+2)−1)/2). Since 1𝑞=2_{(𝐺)𝑢∈𝑉}𝑑(𝑢)≥𝛿𝑛2, (3.124) we have 𝛿𝑛/2≤(3(𝛿−1)(𝛿+2)−1)/2, which is equivalent to 𝛿≥𝑛−23−13+7.3𝛿 (3.125) If 𝑛=3𝑡 for some integer 𝑡, then 𝛿≥3𝑡−23−13+773𝛿=𝑡−1+,3𝛿 (3.126) implying that 𝛿≥𝑡=𝑛/3>(𝑛−2)/3. Next, if 𝑛=3𝑡+1 for some integer 𝑡, then 𝛿≥3𝑡−13−13+723𝛿=𝑡−3+7,3𝛿 (3.127) implying that 𝛿≥𝑡=(𝑛−1)/3>(𝑛−2)/3. Finally, if 𝑛=3𝑡+2 for some integer 𝑡, then 𝛿≥3𝑡3−13+713𝛿=𝑡−3+7,3𝛿 (3.128) implying that 𝛿≥𝑡=(𝑛−2)/3. So, 𝛿≥(𝑛−2)/3, in any case. By Lemma 2.4, each longest cycle in 𝐺 is a dominating cycle.

Acknowledgment

The author would like to thank the anonymous referee for careful reading of the paper and valuable suggestions.