On minimum algebraic connectivity of graphs whose complements are bicyclic

Lemma 2.1

[3] Let G be a simple graph. Then a(G) ≤ δ(G), where δ(G) = min{d_G(v), v ∈ V(G)}.

Lemma 2.2

[18] If Z_i for 1 ≤ i ≤ n is a non-increasing sequence, then, for any 1 ≤ i, j ≤ n, (Z_i – Z_j)² ≤ max{(Z_i – Z₁)², (Z_i – Z_n)²} ≤ (Z₁ – Z_n)².

Theorem 2.3

[21] Let n, m₁ and m₂ be any positive integers such that m₁ ≥ m₂ ≥ 1, n ≥ 11 and n = m₁ + m₂ + 5, then for any bicyclic graph with exactly two cycles G ∈ Ω_1,n,

where equalities hold if and only if G1′ (n – 6, 1)^c ≅ G1′ (m₁, m₂) ≅ G.

Lemma 3.1

Let m₁, m₂ and n be positive integers, such that m₁ ≥ m₂ ≥ 1, n ≥ 11 and m₁ + m₂ + 5 = n. Then a(G₁(n – 6, 1)^c) < … < a(G₁(m₁ + 1, m₂ – 1)^c) < a(G₁(m₁, m₂)^c).

Proof

Let G₁(m₁, m₂) be a graph with labeled vertices as shown in Figure 1 and Z be a unit Fiedler vector of G₁(m₁, m₂)^c. By Lemma 2.1 and LE-equation (3), a(G₁(m₁, m₂)^c) ≠ d_{G1(m1, m2)^c}(v) +1 for any v ∈ V(G₁(m₁, m₂)^c) and all the pendant vertices attached with same vertex have same values given by Z. Therefore, Z_i : = Z_vi for 1 ≤ i ≤ 7 and we have the following equations for a = a(G₁(m₁, m₂)^c),

Transform (A₁) into a matrix equation (M – aI)Z = 0, where Z = (Z₁, Z₂, Z₃, Z₄, Z₅, Z₆, Z₇)^T and

Let g(λ; m₁, m₂) := det(λI – M), then we have

g(λ; m₁, m₂) = λ (− 4 − m₁ − m₂ + λ) (− 3 − m₁ − m₂ + λ) (15 m₂ + 23 m₁ m₂ + 9 m12 m₂ + m13 m₂ + 8 m22 + 10 m₁ m22 + 2 m12m22 + m23 + m₁ m23 − 16 λ − 24 m₁ λ − 9 m12 λ − m13 λ − 32 m₂ λ − 28 m₁ m₂ λ − 5 m12 m₂ λ − 11 m22 λ − 5 m₁ m22 λ − m23 λ + 24 λ² + 18 m₁ λ² + 3 m12 λ² + 19 m₂ λ² + 7 m₁ m₂ λ² + 3 m22 λ² – 9 λ³ – 3 m₁ λ³ – 3 m₂ λ³ + λ⁴).

Since g(0; m₁, m₂) = 0 = g(a; m₁, m₂), thus a is the second smallest root of g(λ; m₁, m₂). Observe that

By Lemma 2.1, a = a(G₁(m₁, m₂)^c) < n. Then by m₁ ≥ m₂, a > 0 and g(a; m₁, m₂) = 0, we have g₁(a; m₁ + 1, m₂ – 1) = –a(2 + m₁ − m₂) (3 + m₁ + m₂ − a)² (4 + m₁ + m₂ − a) (n – a) < 0. This shows that a(G₁(m₁ + 1, m₂ – 1)^c) < a(G₁(m₁, m₂)^c). Similarly, we can prove a(G₁(m₁ + 2, m₂ – 2)^c) < a(G₁(m₁ + 1, m₂ – 1)^c). Consequently, a(G₁(n – 6, 1)^c) < … < a(G₁(m₁ + 1, m₂ – 1)^c) < a(G₁(m₁, m₂)^c).

Lemma 3.2

Let m₁, m₂ and n be positive integers, such that m₁ ≥ m₂ ≥ 1, n ≥ 11, m₁ + m₂ + 5 = n and 2 ≤ i ≤ 9. Then a(G₁(m₁, m₂)^c) ≤ a(G_i(m₁, m₂)^c), where equality holds if G₁(m₁, m₂) ≅ G_i(m₁, m₂).

Proof

Using Lemma 2.1 and (3) (as in Lemma 3.1), we find the following polynomials of the graphs G_i(m₁, m₂)^c for 2 ≤ i ≤ 9.

g₂(a; m₁ + 1, m₂ – 1) = –a(n – a) (4 + m₁ + m₂ − a) (14 + 20 m₁ + 8 m12 + m13 + 10 m₂ + 6 m₁ m₂ + m12 m₂ – 2 m22 − m₁ m22 − m23 − 15a – 11 m₁ a – 2 m12 a – m₂a + 2 m22 a + 3a² + m₁ a² − m₂ a²),

g₃(a; m₁ + 1, m₂ – 1) = (m₁ + m₂ – a + 1) (− 23 – 25 m₁ − 9 m12 − m13 − 11 m₂ – 6 m₁ m₂ − m12 m₂ + 3 m22 + m₁ m22 + m23 − 50a – 97 m₁a – 57 m12 a – 13 m13 a – m14 a – 69 m₂a – 60 m₁ m₂a – 19 m12 m₂a – 2 m13 m₂a – 3 m22 a + m₁ m22 a + 7 m23 a + 2 m₁ m23 a + m24 a + 86a² + 88 m₁ a² + 29 m12 a² + 3 m13 a² + 32 m₂ a² + 18 m₁ m₂ a² + 3 m12 m₂ a² – 11 m22 a² – 3 m₁ m22 a² – 3 m23 a² – 30a³ – 19 m₁ a³ – 3 m12 a³ + m₂ a³ + 3 m22 a³ + 3a⁴ + m₁ a⁴ − m₂ a⁴),

g₄(a; m₁ + 1, m₂ – 1) = –a(4 + m₁ − m₂) (1 + m₁ + m₂ − a) (3 + m₁ + m₂ − a) (4 + m₁ + m₂ − a) (5 + m₁ + m₂ − a),

g₅(a, m₁ + 1, m₂ – 1) = –a(− 4 − m₁ − m₂ + a) (− 50 – 86 m₁ − 50 m12 − 12 m13 − m14 – 54 m₂ – 53 m₁ m₂ – 18 m12 m₂ – 2 m13 m₂ – 3 m22 + 6 m23 + 2 m₁ m23 + m24 + 71a + 77 m₁a + 27 m12 a + 3 m13 a + 30 m₂a + 18 m₁ m₂a + 3 m12 m₂a – 9 m22 a – 3 m₁ m22 a – 3 m23 a – 27a² – 18 m₁ a² – 3 m12 a² + 3 m22 a² + 3a³ + m₁ a³ − m₂ a³),

g₆(a, m₁ + 1, m₂ – 1) = (16 + 56 m₁ + 53 m12 + 18 m13 + 2 m14 + 56 m₂ + 113 m₁ m₂ + 67 m12 m₂ + 15 m13 m₂ + m14 m₂ + 53 m22 + 67 m₁ m22 + 26 m12m22 + 3 m13m22 + 18 m23 + 15 m₁ m23 + 3 m12m23 + 2 m24 + m₁ m24 – 171a – 353 m₁a – 260 m12 a – 90 m13 a – 15 m14 a – m15 a – 271 m₂a – 378 m₁ m₂a – 186 m12 m₂a – 39 m13 m₂a – 3 m14 m₂a – 105 m22 a – 88 m₁ m22 a – 24 m12m22 a – 2 m13m22 a + 8 m23 a + 9 m₁ m23 a + 2 m12m23 a + 9 m24 a + 3 m₁ m24 a + m25 a + 265a² + 394 m₁ a² + 210 m12 a² + 48 m13 a² + 4 m14 a² + 239 m₂ a² + 231 m₁ m₂ a² + 75 m12 m₂ a² + 8 m13 m₂ a² + 14 m22 a² + 3 m₁ m22 a² – 24 m23 a² – 8 m₁ m23 a² – 4 m24 a² – 152a³ – 158 m₁ a³ – 54 m12 a³ – 6 m13 a³ – 60 m₂ a³ – 37 m₁ m₂ a³ – 6 m12 m₂ a³ + 18 m22 a³ + 6 m₁ m22 a³ + 6 m23 a³ + 36a⁴ + 24 m₁ a⁴ + 4 m12 a⁴ – 4 m22 a⁴ – 3a⁵ – m₁ a⁵ + m₂ a⁵),

g₇(a, m₁ + 1, m₂ – 1) = –a(2 + m₁ − m₂) (− 5 − m₁ − m₂ + a) (− 4 − m₁ − m₂ + a) (− 3 − m₁ − m₂ + a) (− 1 − m₁ − m₂ + a),

g₈(a, m₁ + 1, m₂ – 1) = –a(1 + m₁ − m₂) (2 + m₁ + m₂ − a) (3 + m₁ + m₂ − a) (4 + m₁ + m₂ − a) (5 + m₁ + m₂ − a),

g₉(a, m₁ + 1, m₂ – 1) = (2 + m₁ + m₂ − a) (4 + m₁ + m₂ − a) (− 8 m₁ − 6 m12 − m13 + 8 m₂ − m12 m₂ + 6 m22 + m₁ m22 + m23 + 4a + 32 m₁a + 32 m12 a + 10 m13 a + m14 a – 30 m₂a + 10 m12 m₂a + 2 m13 m₂a – 32 m22 a – 10 m₁ m22 a – 10 m23 a – 2 m₁ m23 a – m24 a – a² – 31 m₁ a² – 20 m12 a² – 3 m13 a² + 31 m₂ a² – 3 m12 m₂ a² + 20 m22 a² + 3 m₁ m22 a² + 3 m23 a² + 10 m₁ a³ + 3 m12 a³ – 10 m₂ a³ – 3 m22 a³ − m₁ a⁴ + m₂ a⁴).

Now, consider g₁(a, m₁ + 1, m₂ – 1) – g₂(a, m₁ + 1, m₂ – 1) = a(1 + a – 2 m₂) (− 5 + a – m₁ − m₂) (− 4 + a – m₁ − m₂)² > 0.

Thus,

g₁(a, m₁ + 1, m₂ – 1) – g₃(a, m₁ + 1, m₂ – 1) = 23 – 333a + 266a² – 50a³ – 3a⁴ + a⁵ + 48 m₁ − 460a m₁ + 262a²m₁ − 36a³m₁ + 34 m12 − 222a m12 + 82a² m12 − 6a³ m12 + 10 m13 − 44a m13 + 8a² m13 + m14 – 3a m14 + 34 m₂ – 114a m₂ – 56a²m₂ + 54a³m₂ – 8a⁴m₂ + 42 m₁ m₂ – 112a m₁ m₂ – 18a²m₁ m₂ + 12a³m₁ m₂ + 16 m12 m₂ – 38a m12 m₂ + 2 m13 m₂ – 4a m13 m₂ + 8 m22 + 110a m22 − 100a² m22 + 18a³ m22 + 2 m₁ m22 + 56a m₁ m22 – 24a²m₁ m22 + 6a m12m22 − 4 m23 + 50a m23 − 16a² m23 – 2m₁ m23 + 12a m₁ m23 − m24 + 5a m24 < 0.

Thus,

g₁(a, m₁ + 1, m₂ – 1) – g₄(a, m₁ + 1, m₂ – 1) = 2a (1 + a – 2 m₂) (− 5 + a – m₁ − m₂) (− 4 + a – m₁ − m₂) (− 3 + a – m₁ − m₂) > 0.

Thus,

g₁(a, m₁ + 1, m₂ – 1) – g₅(a, m₁ + 1, m₂ – 1) = a(− 4 + a – m₁ − m₂) (40 – 7a – 5a² + a³ + 37 m₁ − 6am₁ − a²m₁ + 11 m12 − a m12 + m13 − 21 m₂ + 25am₂ – 5a²m₂ – 9m₁m₂ + 6am₁m₂ – m12 m₂ – 20 m22 + 7a m22 − 5 m₁ m22 − 3 m23 ) > 0.

Thus,

g₁(a, m₁ + 1, m₂ – 1) – g₆(a, m₁ + 1, m₂ – 1) = –16 – 189a + 137a² – 14a³ – 6a⁴ + a⁵ – 56 m₁ − 229am₁ + 139a²m₁ − 15a³m₁ − a⁴m₁ − 53 m12 − 107a m12 + 46a² m12 – 3a³ m12 − 18 m13 − 23a m13 + 5a² m13 − 2 m14 – 2a m14 – 56 m₂ + 49am₂ – 108a²m₂ + 53a³m₂ – 7a⁴m₂ – 113 m₁ m₂ + 46a m₁ m₂ – 51a²m₁ m₂ + 13a³m₁ m₂ – 67 m12 m₂ + 13a m12 m₂ – 6a² m12 m₂ – 15 m13 m₂ + a m13 m₂ − m14 m₂ – 53 m22 + 140a m22 − 90a² m22 + 15a³ m22 − 67m₁ m22 + 81a m₁ m22 − 24a²m₁ m22 − 26 m12m22 + 12a m12m22 − 3 m13m22 − 18 m23 + 45a m23 − 13a² m23 – 15 m₁ m23 + 13a m₁ m23 − 3 m12m23 − 2 m24 + 4a m24 − m₁ m24 > 0.

Thus,

g₁(a, m₁ + 1, m₂ – 1) – g₇(a, m₁ + 1, m₂ – 1) = 2a(− 5 + a – m₁ − m₂) (− 4 + a – m₁ − m₂) (− 3 + a – m₁ − m₂) (2 + m₁ − m₂) > 0.

Thus,

g₁(a, m₁ + 1, m₂ – 1) – g₈(a, m₁ + 1, m₂ – 1) = –a(− 4 + a – 2 m₁) (− 5 + a – m₁ − m₂) (− 4 + a – m₁ − m₂) (− 3 + a – m₁ − m₂) < 0.

Thus,

g₁(a, m₁ + 1, m₂ – 1) – g₉(a, m₁ + 1, m₂ – 1) = (− 4 + a – m₁ − m₂) (98a – 84a² + 23a³ – 2a⁴ – 16 m₁ + 199a m₁ – 178a²m₁ + 68a³m₁ − 13a⁴m₁ + a⁵m₁ − 20 m12 + 163a m12 – 131a² m12 + 39a³ m12 − 4a⁴ m12 − 8 m13 + 66a m13 – 39a² m13 + 6a³ m13 − m14 + 13a m14 – 4a² m14 + a m15 + 16 m₂ – 31am₂ + 86a²m₂ – 56a³m₂ + 13a⁴m₂ − a⁵m₂ + 46am₁ m₂ – 12a²m₁ m₂ – 8 m12 m₂ + 70a m12 m₂ – 39a² m12 m₂ + 6a³ m12 m₂ – 2 m13 m₂ + 26a m13 m₂ – 8a² m13 m₂ + 3a m14 m₂ + 20 m22 − 117a m22 + 119a² m22 − 39a³ m22 + 4a⁴ m22 + 8 m₁ m22 − 58am₁ m22 + 39a²m₁ m22 − 6a³m₁ m22 + 2a m13m22 + 8 m23 − 62a m23 + 39a² m23 − 6a³ m23 + 2 m₁ m23 − 26am₁ m23 + 8a²m₁ m23 − 2a m12m23 + m24 – 13a m24 + 4a² m24 – 3am₁ m24 – a m25 ) > 0.

Thus,

From (5) to (12), we have a(G₁(m₁ + 1, m₂ – 1)^c) ≤ a(G_i(m₁ + 1, m₂ – 1)^c). Consequently, for 2 ≤ i ≤ 9, a(G₁(m₁, m₂)^c) ≤ a(G_i(m₁, m₂)^c), where equality holds if G₁(m₁, m₂) ≅ G_i(m₁, m₂).

Lemma 3.3

Let m₁, m₂ and n be positive integers, such that m₁ ≥ m₂ ≥ 1, n ≥ 11, and m₁ + m₂ + 5 = n. Then a( G1′ (m₁, m₂)^c) < a(G₁(m₁, m₂)^c).

Proof

Let Z = {Z_vi} be a unit Fiedler vector of the graph G₁(m₁, m₂)^c for 1 ≤ i ≤ n. After deleting the edge v₄v₅ and adding v₄v₂ in G₁(m₁, m₂), we obtain G1′ (m₁, m₂) (see Figure 1 and Figure 3, for particular values of m₁ = 5 and m₂ = 1 see Figure 4). Now, by (2) and Lemma 2.2, we have

Figure 4

a( G1′ (5, 1)^c) < a(G₁(5, 1)^c).

Using (5), we obtain a(G₁(m₁, m₂)^c) = Z^TL(G₁(m₁, m₂)^c)Z = Z^T(nI – J)Z – Z^TL(G₁(m₁, m₂))Z > Z^T(nI – J)Z – Z^TL( G1′ (m₁, m₂))Z > a( G1′ (m₁, m₂)^c). Consequently, a( G1′ (m₁, m₂)^c) < a(G₁(m₁, m₂)^c).

Theorem 4.1

Let n, m₁ and m₂ be any positive integers such that m₁ ≥ m₂ ≥ 1, n ≥ 11 and n = m₁ + m₂ + 5, then for any bicyclic graph with three cycles G ∈ Ω_2,n,

where 1 ≤ i ≤ 9.

Proof

Let G be a bicyclic graph with three cycles C₁(l₁), C₂(l₂) and C₃(l₃) with lengths l₁ ≥ 3, l₂ ≥ 3 and l₃ ≥ 4, respectively. The cycles C₁(l₁) and C₂(l₂) are inner cycles with at least one common edge and C₃(l₃) is an outer cycle such that l₃ = l₁ + l₂ – 2k, where k are common edges in C₁(l₁) and C₂(l₂). Let Z be a unit Fiedler vector of G^c. Then, we have a sequence {Z_vn} such that

For d_G(v₁, v_n) > 1, we can assume the path v₁Gv_n = v₁w₁ … w₂v_n. In the path v₁Gv_n, d_G(v₁, v_n) = 2 if w₁ = w₂. Add the edge v₁v_n and delete v₁w₁ or w₂v_n such that the resulting bicyclic graph G_α is not H(n, 2). Then by (2) and Lemma 2.2, we have

where G_α is a bicyclic graph with three cycles C₁( l1′ ), C₂( l2′ ) and C₃( l3′ ) having some trees attached with the vertices of one or both the cycles C₁( l1′ ) and C₂( l2′ ). The lengths l1′ , l2′ and l2′ may or may not different from l₁, l₂ and l₃ respectively. Most importantly, we note that d_Gα(v₁, v_n) = 1. If G_α ∉ {G_i : 1 ≤ i ≤ 9}, then we have the following three cases for G_α:

1. Both the vertices v₁ and v_n are cycle vertices. In this case, we discuss further four possibilities (1) both the vertices are on exactly one inner cycle, (2) one of v₁ and v_n is a common vertex of both the inner cycles, (3) both the vertices v₁ and v_n are common vertices of the inner cycles, and (4) each inner cycle contains exactly one of v₁ and v_n.

   1. We assume without loss of generality that both the vertices v₁ and v_n are on the cycle C₁( l1′ ). Since, for l1′ ≥ 4 and l2′ = 3 the cycles C₁( l1′ ) and C₂( l2′ ) have two common vertices, therefore, we can assume C₁( l1′ ) = v₁ v_n w₁ w₂ w₃ … w_iw_i+1 … w_mv₁, where m = l1′ – 2 and two vertices other than v₁ and v_n are also of C₂( l2′ ).

      1. Suppose that w_m–1 and w_m are common vertices of the inner cycles. If (Z_wm–1 – Z_v1)² ≥ (Z_wm–1 – Z_vn)², we delete the edge w_m–2w_m–1 and add w_m–1v₁ (as (b) is obtained from (a) in Figure 5). The resulting graph G_αα is a bicyclic graph with two inner cycles C₁( l1′ ) and C₂( l2′ ), and an outer cycle C₃( l3′ ) such that l1′ = 3 = l2′ , l3′ = 4, some trees are attached on v₁ in C₁( l1′ ) and some trees are attached on v_n which is non cycle. Thus, G_αα is a bicycle graph G₃(m₁, m₂) which is obtained when we identify B₁ by a vertex of degree 2 with end point say v₁ of an edge v₁v_n having some trees on v₁ and v_n (see Figure 1 with v₁v_n = v₂v₃).

         

         Figure 5

         If (Z_wm–1 – Z_v1)² < (Z_wm–1 – Z_vn)², we delete the edge w_m–2w_m–1 and add w_m–1v_n (as (c) is obtained from (a) in Figure 5). The resulting graph G_αα is a bicyclic graph with two inner cycles C₁( l1′ ) and C₂( l2′ ), and an outer cycle C₃( l3′ ) such that l1′ = 4, l2′ = 3, l3′ = 5 and some trees are attached on both v₁ and v_n in C₁( l1′ ). Thus, G_αα is a bicycle graph which is infect B₂ with some trees on the two adjacent vertices of degree 2 i.e G₆(m₁, m₂) (see Figure 2). If we proceed from the other side of the path, then for (Z_w2 – Z_v1)² ≥ (Z_w2 – Z_vn)², we delete the edge w₁w₂ and add the edge w₂v₁ otherwise, we delete w₁w₂ and add w₂v_n. Thus, the resulting graph G_αα is a bicyclic graph with three cycles such that the lengths of the inner cycles are l1′ – 2 and l2′ = 3 or l1′ – 1 and l2′ = 3. Now, repeat the process for the vertex w₃ and continue up to the vertex w_m–1. Thus, we obtain the same graphs G₃(m₁, m₂) and G₆(m₁, m₂).

      2. Suppose that w_i and w_i+1 are common vertices of the inner cycles, where 2 ≤ i ≤ m – 2. If (Z_wi – Z_v1)² ≥ (Z_wi – Z_vn)², we delete w_i–1 w_i and add w_iv₁. Now, if (Z_wi+1 – Z_v1)² ≥ (Z_wi+1 – Z_vn)², we delete w_i+1w_i+2 and add w_i+1v₁, otherwise delete w_i+1w_i+2 and add w_i+1v_n. Thus, the resulting graphs are G₃(m₁, m₂) or G₆(m₁, m₂), respectively.

         If (Z_wi – Z_v1)² < (Z_wi – Z_vn)², we delete w_i–1w_i and add w_iv_n. Now, if (Z_wi+1 – Z_v1)² ≥ (Z_wi+1 – Z_vn)², we delete w_i+1w_i+2 and add w_i+1v₁, otherwise delete w_i+1w_i+2 and add w_i+1v_n. Thus, the resulting graphs are G₆(m₁, m₂) or G₃(m₁, m₂), respectively.

      3. Suppose that w₁ and w₂ are common vertices of the inner cycles, then we repeat (i) and the obtained graphs are same as there.

         If in (1(i)-(iii)), l2′ ≥ 4, then we can assume C₂( l2′ ) = u₁ u₂ … u_l w_iw_i+1, where w_i and w_i+1 are two common vertices of the cycles C₁( l1′ ) and C₂( l2′ ) for 1 ≤ i ≤ m – 1 and l = l2′ – 2. By the use of (1(i)-(iii)), we have C₁( l1′ ) with l1′ = 3, some trees attached on v₁ ∈ C₁( l1′ ) and some trees attached on v_n (pendant vertex) or with l1′ = 4 and some trees attached on v₁ and v_n (both are in C₁( l1′ )). Now, for C₂( l2′ ), if (Z_u1 – Z_v1)² ≥ (Z_u1 – Z_vn)², delete the edge u₁u₂ and add the edge u₁v₁, otherwise delete u₁u₂ and add u₁v_n. Thus, the resulting graphs are G₄(m₁, m₂) and G₅(m₁, m₂) or G₅(m₁, m₂) and G₇(m₁, m₂), (see Figure 1 and Figure 2) respectively. Moreover, l1′ = 3 = l2′ is not possible as both the vertices v₁ and v_n can not appear on only C₁( l1′ ).

   2. Without loss of generality suppose that v_n is on the cycle C₁( l1′ ) and v₁ is a common vertex of the inner cycles. Assume that l1′ ≥ 4, l2′ = 3 and C₁( l1′ ) = v₁ v_n w₁ w₂ w₃ … w_mv₁, where w_m is also a common vertex of the inner cycles and m = l1′ – 2.

      If (Z_wm–1 – Z_v1)² ≥ (Z_wm–1 – Z_vn)², we delete the edge w_m–2w_m–1 and add the edge w_m–1v₁. The resulting graph G_αα is a bicyclic graph with two inner cycles C₁( l1′ ) and C₂( l2′ ), and an outer cycle C₃( l3′ ) such that l1′ = 3 = l2′ , l3′ = 4, some trees are attached on v₁ which is a common vertex of both the inner cycles and some trees are attached on v_n in C₂( l2′ ). Thus, G_αα is a bicycle graph G₄(m₁, m₂) which is obtained when we identify B₁ by a vertex of degree 3 with an end point v₁ of an edge v₁v_n having some trees on v₁ and v_n (see Figure 1 with v₁v_n = v₂v₄).

      If (Z_wm–1 – Z_v1)² < (Z_w2 – Z_vn)², we delete the edge w_m–2w_m–1 and add the edge w_m–1v_n. The resulting graph G_αα is a bicyclic graph with two inner cycles C₁( l1′ ) and C₂( l2′ ), and an outer cycle C₃( l3′ ) such that l1′ = 4, l2′ = 3, l3′ = 5, some trees are attached on v₁ which is a common vertex of both the inner cycles and some trees are attached on v_n in C₂( l2′ ). Thus, G_αα is a bicycle graph G₅(m₁, m₂) (see Figure 1) which is infect B₂ with some trees which are attached on two adjacent vertices of degree 2 and 3 in C₄ ⊆ B₂.

      If we proceed from the other side of the path, then for (Z_w2 – Z_v1)² ≥ (Z_w2 – Z_vn)², we delete the edge w₁w₂ and add the edge w₂v₁ otherwise, we delete the edge w₁w₂ and add the edge w₂v_n. Thus, the resulting graph H_αα is a bicyclic graph with three cycles such that the lengths of the inner cycles are l1′ – 2 and l2′ = 3 or l1′ – 1 and l2′ = 3. Now, repeat the process for the vertex w₃ and continue up to the vertex w_m–1. Thus, we obtain the same graphs G₄(m₁, m₂) and G₅(m₁, m₂).

      If in (2), l2′ ≥ 4, then we can assume C₂( l2′ ) = u₁ u₂ … u_l w_iv₁u₁, where w_i and v₁ are two common vertices of the cycles C₁( l1′ ) and C₂( l2′ ) for 1 ≤ i ≤ m and l = l2′ – 2. By the use of (2), we have C₁( l1′ ) with l1′ = 3, some trees attached on v₁ ∈ C₁( l1′ ) and some trees attached on v_n (pendant vertex) or with l1′ = 4 and some trees attached on v₁ and v_n (both are in C₁( l1′ )). Now, for C₂( l2′ ), if (Z_ul – Z_v1)² ≥ (Z_ul – Z_vn)², delete the edge u_l–1u_l and add the edge u_lv₁, otherwise delete the edge u_l–1u_l and add the edge u_lv_n. Thus, the resulting graphs G_αα are G₄(m₁, m₂) and G₅(m₁, m₂) or G₅(m₁, m₂) and G₇(m₁, m₂), respectively. Moreover, if l1′ = 3 = l2′ , then we obtain G₂(m₁, m₂).

   3. Suppose that v₁ and v_n both are common vertices of the inner cycles. Assume that l1′ ≥ 4, l2′ = 3 and C₁( l1′ ) = v₁ v_n w₁ w₂ w₃ … w_mv₁, where m = l1′ – 2.

      If (Z_wm–1 – Z_vn)² ≥ (Z_wm–1 – Z_v1)², we delete the edge w_m–2w_m–1 and add the edge w_m–1v_n. The resulting graph G_αα is a bicyclic graph with two inner cycles C₁( l1′ ) and C₂( l2′ ), and an outer cycle C₃( l3′ ) such that l1′ = 4, l2′ = 3, l3′ = 5, some trees are attached on v₁ and v_n which are common vertices of the inner cycles. Thus, G_αα is a bicycle graph G₈(m₁, m₂) (see Figure 2) which is obtained from B₂ by attaching some trees on both the vertices of degree 3.

      If (Z_wm–1 – Z_v1)² > (Z_wm–1 – Z_vn)², we delete the edge w_m–1w_m and add the edge w_m–1v₁. Then, check for w_m–2, if (Z_wm–2 – Z_vn)² ≥ (Z_wm–2 – Z_v1)², delete the edge w_m–3w_m–2 and add the edge w_m–2v_n. The resulting graph is H₈(m₁, m₂). If (Z_wm–2 – Z_v1)² > (Z_wm–2 – Z_vn)², delete w_m–2w_m–1 and add w_m–2v₁. Repeat this process until we reach on the vertex w₂. If (Z_w2 – Z_vn)² ≥ (Z_w2 – Z_v1)², delete the edge w₂w₁ and add the edge w₂v_n, otherwise delete w₂w₃ and add w₂v₁. The resulting graph is G₈(m₁, m₂).

      If in (3), l2′ ≥ 4, then we can assume C₂( l2′ ) = u₁ u₂… u_lv_nv₁u₁. By the use of (3), we have C₁( l1′ ) with l1′ = 4 and some trees attached on v₁ and v_n (both are common vertices). Now, again repeat (3) for C₂( l2′ ) and we obtain G₉(m₁, m₂). If l1′ = 3 = l2′ , then we obtain G₁(m₁, m₂).

   4. Suppose that v₁ is on C₁( l1′ ) and v_n is on C₂( l2′ ), where l1′ , l2′ ≥ 3. We note that d_Gα(v₁, v_n) ≥ 2, which is not possible.

2. One of v₁, v_n is a cycle vertex.

   We assume that v₁ is a cycle vertex and v_n is non cycle vertex without loss of generality. In this case, for v₁, we have three possibilities (1) v₁ is on C₁( l1′ ), (2) v₁ is a common vertex of both the inner cycles and (3) v₁ is on C₂( l2′ ).

   1. If v₁ is only on C₁( l1′ ), then for l1′ ≥ 4 and l2′ = 3, we have C₁( l1′ ) = v₁w₁ w₂ w₃ … w_iw_i+1 … w_mv₁. Assume w_i and w_i+1 are common vertices of the inner cycles, where 1 ≤ i ≤ m – 1. Now, we repeat (Case a (1)) and obtain G₃(m₁, m₂) or G₆(m₁, m₂). If l2′ ≥ 4 then again by (Case a (1)), the resulting graphs are G₄(m₁, m₂) and G₅(m₁, m₂) or G₅(m₁, m₂) and G₇(m₁, m₂). Moreover, if l1′ = 3 = l2′ , then the resulting graph is G₃(m₁, m₂).

   2. If v₁ is a common vertex of the inner cycles. Assume that w_m is an other common vertex such that l1′ ≥ 4, l2′ = 3, C₁( l1′ ) = v₁ v_n w₁ w₂ w₃ … w_mv₁ and m = l1′ – 1. Now, by (Case a (2)), we obtain G₄(m₁, m₂) or G₅(m₁, m₂). If l2′ ≥ 4 then again by (Case a (2)), the resulting graphs are G₄(m₁, m₂) and G₅(m₁, m₂) or G₅(m₁, m₂) and G₇(m₁, m₂). Moreover, if l1′ = 3 = l2′ , then the resulting graph is G₄(m₁, m₂). (3) If v₁ is only on C₂( l2′ ), then follow (Case b (1)).

3. Both v₁ and v_n are non cycle vertices.

   Suppose that u and v are common vertices of the inner cycles such that C₁( l1′ ) = uvw₁w₂w₃ … w_m and C₂( l2′ ) = uvu₁u₂u₃ … u_l, where l1′ ≥ 4, m = l1′ – 2 and l = l2′ – 2. Assume that there is a path P containing the vertices v₁ and v_n has one end point either u or v. If v is on P and w₁ is adjacent to v in C₁( l1′ such that (Z_w1 – Z_v1)² ≥ (Z_w1 – Z_vn)², then we delete w₁v and add w₁v₁, otherwise delete w₁v and add w₁v_n. Then the resulting bicyclic graph G_α,α is in Case a or Case b.

   Then by equation (2) and Lemma (3), we have

   ZTL(Gα)Z=∑vivj∈E(Gα)(Zvi−Zvj)2≤∑vivj∈E(Gαα)(Zvi−Zvj)2=ZTL(Gαα)Z, (14)

   If G_αα ∉ {G_i : 1 ≤ i ≤ 9} and there exists a pendant vertex v, whose neighbor a is neither v₁ nor v_n, satisfying (Z_v – Z_v1)² ≥ (Z_v – Z_vn)², then delete av and add vv₁; otherwise delete av and add vv_n. Repeat this rearranging until the resulting graph

   Gααα∈{Gi:1≤i≤9}.

   Then by equation (2) and Lemma (2.2), we have

   ZTL(Gαα)Z=∑vivj∈E(Gαα)(Zvi−Zvj)2≤∑vivj∈E(Gααα)(Zvi−Zvj)2=ZTL(Gααα)Z (15)

   By (13) – (15), we have

   a(Gc)=ZTL(Gc)Z=ZT(nI−J)Z−ZTL(G)Z≥ZT(nI−J)Z−ZTL(Gα)Z≥ZT(nI−J)Z−ZTL(Gαα)Z≥ZT(nI−J)Z−ZTL(Gααα)Z=ZTL(Gαααc)Z≥a(Gααα).

   Hence we have a(G^c) ≥ a( Gαααc ). Consequently, a(G_i(m₁, m₂)^c) ≤ a(G^c) and equality holds if and only if G_i(m₁, m₂) ≅ G, where G ∈ Ω_2,n and 1 ≤ i ≤ 9.