A system G is strongly t-diagnosable if the following two conditions holds:

1. G is t-diagnosable, and

2. for any two distinct subsets F₁, F₂ ⊂ V (G) with |F_i| ≤ t + 1, i = 1, 2,

either (a) (F₁, F₂) is a distinguishable pair;

or (b) (F₁, F₂) is an indistinguishable pair and there exists a vertex v ∈ V

such that N (v)⊆ F₁ and N (v)⊆ F₂.

By Theorem 3 and Definition 6, we propose a sufficient condition for checking if a system G is strongly t-diagnosable as follows.

Lemma 5 A system G = (V, E) with |V | = n is strongly t-diagnosable if

1. n≥ 2t + 1

2. each node has order at least t

3. |T (G, U)| > p for each U ⊂ V (G) such that |U| = N − 2t + p and 0 ≤ p ≤ t − 1

4. for any two distinct subsets F₁, F₂ ⊂ V (G) with |F_i| ≤ t + 1, i = 1, 2,

either (a) (F₁, F₂) is a distinguishable pair;

or (b) (F₁, F₂) is an indistinguishable pair and there exists a vertex v ∈ V

such that N (v)⊆ F₁ and N (v)⊆ F₂.

Proof. With conditions 1, 2 and 3, by Theorem 3, G is t-diagnosable. Condition 4 is the same as condition 2 of Definition 6. So we complete the proof.

2

Theorem 6 A system G=(V,E) is strongly t-diagnosable if for each vertex set S ⊂ V with cardinality |S| = p, 0 ≤ p ≤ t, the following two conditions are satisfied

1. for 0≤ p ≤ t − 1, every component C of G − S |V_G−S(C; 3)| ≥ 2((t + 1) − p) + 1

2. for p = t, either every component C of G− S satisfies |VG−S(C; 3)| ≥ 3 or else G− S satisfies at least one trivial component.(Remark: 2((t + 1) − p) + 1 = 3 as p = t)

Proof. Assume S ⊂ V , |S| = p, 0 ≤ S ≤ t − 1, By condition 1, every component C of G− S satisfies |VG−S(C; 3)| ≥ 2((t + 1) − p) + 1 ≥ 2(t − p) + 1. By Theorem 4, G is t-diagnosable.

In order to prove that G is strongly t-diagnosable, we need to show that condition 2 of Definition 6 holds. Assume (F₁, F₂) be an indistinguishable pair, F₁ = F2, |F1| ≤ t + 1,

Proof. By definition 6, we want to prove the following two conditions: (i)G is (t + 1)-diagnosable (ii) for each indistinguishable pair (F₁, F₂), F_i ⊂ V , i = 1, 2, with |Fi| ≤ t+2, it implies that there exists a vertex v ∈ V such that N(v) ⊆ F₁ and N (v)⊆ F₂.

First, by Theorem 5, G is (t + 1)-diagnosable. The condition (i) holds. So we only need to prove condition (ii). Let (F₁, F₂) is an indistinguishable pair, F_i ⊂ V , i = 1, 2, with |F_i| ≤ t + 2. Let S = F₁∩ F₂,|S| = p, 0 ≤ p ≤ t + 1. If there exists a vertex v ∈ V , N (v)⊆ S. We finish the proof. Otherwise, N(v)
S for each vertex v ∈ V . We want to show that this is a contradiction. By Lemma 3, G− S is connected. The only component C of G−S is G−S itself. We divide this case into following two main cases: (1)0 ≤ p ≤ 3

Subcase 1.4: p = 3 graph with two faulty vertices x₂ and x₃. Then there exists at most 2t vertices such that the degree of these vertices is t− 1. The minimum number of degree greater than t in G₂ is 2t + 1− 2t = 1. |VG−S(C; 3)| ≥ t + (t − 2) + 1 = 2t − 1 ≥ 2((t + 2) − 3) + 1. By Lemma

2, F₁, F₂ is a distinguishable pair.

Case 2: 4≤ p ≤ t + 1

Let U = G− F1− F2,|F1 F2| ≤ 2(t + 2 − p), |U| = |V (G)| − |F1∩ F2| ≥ 2(2t + 1) − (2(t + 2)− p) = 2t − 2 + p. Since G − S is connected, there exists (a, b) in E(G) such that a∈ F1 F2, b∈ U. Ui, 1≤ i ≤ k, be the connected components of subgraph U such that U =∪^k_i=1U_i. We assume |U_i| > 1. We can find the case such that the condition 1 of Theorem 1 is satisfied. Hence, (F₁, F₂) is a distinguishable pair. Otherwise|Ui| = 1, for all 1 ≤ i ≤ k. Hence, N_G−S(v)⊂ F₁  F₂, v ∈ U. Σ_v∈U|deg_G−S(v)| ≤ Σ_v∈F1F2|deg_G−S(v)|.

Σ_v∈U|degG−S(v)| ≥ ((2t − 2 + p) × t) − p × t = (2t − 2) × t. Σv∈F1F2|degG−S(v)| ≤ 2(t + 2− p) × t. Σ_v∈U|deg_G−S(v)| > Σ_v∈F1F2|deg_G−S(v)|, p ≥ 4. This is a contradiction.

2

Applying Theorem 7, we list the following corollary.

Corollary 1 The Hypercube Q_n, the Crossed cube CQ_n, the Twisted cube T Q_n, and the M¨obius cube M Q_n are all strongly n-diagnosable for n≥ 4.

In the following, we show that Q₃is not strongly 3-diagnosable. Let F₁ ={010, 100, 111}, F₂ = {001, 100, 111}, |F₁| = |F₂| = 3, S = F₁ ∩ F₂. Since N (v)
S, v ∈ V (Q₃) and (F₁, F₂) is a distinguishable pair. Hence, Q₃ is not strongly 3-diagnosable.

000 001

010 011

100 101

110 111

000

010 001

F₁ F₂

100

011

111

110 101

Q

Figure 3.1: An example of non-strongly 3-diagnosable system

Chapter 4 Conclusions

We observe that cube family are almost (n + 1)-diagnosable except the case that all the neighbors of some vertex are faulty simultaneously. In this thesis, We introduce a new concept, called a strongly t-diagnosable system under the comparison model. G₁, G₂ are two t-regular graph with the same number of vertices N , N ≥ 2t + 1, for t ≥ 3.

order_Gi(v) ≥ t for every node v in Gi and the connectivity κ(G_i) ≥ t for i = 1, 2. We prove that the MCN constructed from G₁and G₂ is strongly(t+1)-diagnosable. According to the result, we know that cube family with n-dimensional are all strongly n-diagnosable for n ≥ 4.

In the future work, we can try to solve the problem how large the maximum value of t such that cube family remains t-diagnosable under the condition that every fault-set F satisfies N (v)
F for each vertex v ∈ V . For example, {v1, v₂, v₃, v₄} is a subset Q2 of Q_n. Let F₁ ={v₂, v₄} ∪ N(v₁)∪ N(v₂)∪ N(v₃)− {v₁, v₃}, F₂ ={v₃, v₄} ∪ N(v₁)∪ N(v₂)∪ N (v₃)− {v1, v₂} (See Fig 4.1). |F1| = 3(n − 2) + 2, |F2| = 3(n − 2) + 2. Every vertex

has at most one good neighbor either F₁ or F₂ is faulty set. Because none of condition of Theorem 1 holds, (F₁, F₂) is an indistinguishable pair. There is an example to show that the conditional diagnosability of the Hypercube Q_n is no greater than 3(n− 2) + 2.

F₁ F₂

n-2 n-2 n-2

n-2 v₁

v₂ v₄ v₃

Figure 4.1: (F₁, F₂) is an indistinguishable pair of Q_n under every vertex has one good neighbor condition.

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