Conditionally Diagnosable Systems
4.2 Conditional Diagnosability of BC Networks un- un-der the Comparison Model
An n-dimensional bijective connection network (BC network), denoted by Xn, is an n-regular graph with 2n vertices and n2n−1 edges. The set of all the n-dimensional BC networks is called the family of the n-dimensional BC networks, denoted by Ln. Xn and Ln may be recursively defined as below [24].
Definition 9 The one-dimensional BC graph X1 is a complete graph with two vertices.
The family of the one-dimensional BC graph is defined as L1 = {X1}. Let G be a graph.
G is an n-dimensional BC graph, denoted by Xn, if there exist V0, V1 ⊂ V (G) such that the following two conditions hold:
1. V (G) = V0S V1, V0 6= ∅, V1 6= ∅, V0T V1 = ∅; and
2. There exists an edge set M ⊂ E(G) such that M is a perfect matching between V0
and V1, G(V0) ∈ Ln−1 and G(V1) ∈ Ln−1.
Now, we use again Figure 4.3 to show that the conditional diagnosability of BC graph Xn is no greater than 3(n − 2) + 2, n ≥ 5. As shown in Figure 4.3, we take a cycle of length four in Xn and it is easy to check the two conditional faulty sets F1 and F2 are indistinguishable, where |F1| = |F2| = 3(n − 2) + 2. Therefore, Xn is not conditionally (3(n − 2) + 2)-diagnosable and tc(Xn) ≤ 3(n − 2) + 1, n ≥ 3. Next, we shall show that Xn is conditionally t-diagnosable, where t = 3(n − 2) + 1.
Lemma 19 tc(Xn) ≤ 3(n − 2) + 1 under the comparison model, for n ≥ 3.
Let F be a set of vertices F ⊂ V (Xn) and C be a connected component of Xn−F . We need some results on the cardinalities of F and V (C) under some restricted conditions.
The results are listed in Lemma 20 and 21. In Lemma 20, Zhu proved that deleting at most 2(n − 1) − 1 vertices from Xn, the incomplete BC graph Xn has one connected component containing at least 2n − |F | − 1 vertices. We expand this result further. In Lemma 20, we show that deleting at most 3n − 6 vertices from Xn, the incomplete BC graph Xn has one connected component containing at least 2n− |F | − 2 vertices.
Lemma 20 [57] ∀Xn ∈ Ln(n ≥ 3), let F be a set of vertices F ⊂ V (Xn) with n ≤
|F | ≤ 2(n − 1) − 1. Suppose that Xn− F is disconnected. Then Xn − F has exactly two components, one is trivial and the other is nontrivial. The nontrivial component of Xn− F contains 2n− |F | − 1 vertices.
The BC graph can be described as follows: Let Xndenote an n-dimensional BC graph.
X1 is a complete graph with two vertices labeled with 0 and 1, respectively. For n ≥ 2, each Xn consists of two Xn−1’s, denoted by Xn−1L and Xn−1R , with a perfect matching M between them. That is, M is a set of edges connecting the vertices of Xn−1L and the vertices of Xn−1R in a one-to-one manner. It is easy to see that there are 2n−1 edges between Xn−1L and Xn−1R . By using a simple induction, we can prove the following lemma.
Lemma 21 ∀Xn ∈ Ln(n ≥ 5), let F be a set of vertices F ⊂ V (Xn) with |F | ≤ 3n − 6.
Then Xn− F has a connected component containing at least 2n− |F | − 2 vertices.
Proof.
We prove the lemma by induction on n. For n = 5, it is straightforward to verify that the lemma holds. As the inductive hypothesis, we assume that the result is true for Xn−1, for |F | ≤ 3(n − 1) − 6, and for some n ≥ 6. Now we consider Xn, |F | ≤ 3n − 6. An n-dimensional BC graph Xn can be divided into two Xn−1’s, denoted by Xn−1L and Xn−1R . Let FL = FT V (Xn−1L ), 0 ≤ |FL| ≤ 3n − 6 and FR = FT V (Xn−1R ), 0 ≤ |FR| ≤ 3n − 6.
Then |F | = |FL| + |FR|. Without loss of generality, we may assume that |FL| ≥ |FR|.
In the following proof, we consider two cases by the size of FR: 1) 0 ≤ |FR| ≤ 2 and 2)
|FR| ≥ 3.
Case 1: 0 ≤ |FR| ≤ 2.
Since 0 ≤ |FR| ≤ 2, Xn−1R − FR is connected and |V (Xn−1R − FR)| = 2n−1− |FR|. Let FR(L) ⊂ V (Xn−1L ) be the set of vertices which has neighboring vertices in FR. For each vertex v ∈ Xn−1L − FL− FR(L), there is exactly one vertex v(R) in Xn−1R − FR, such that (v, v(R)) ∈ E(Xn). Besides, |V (Xn−1L − FL− FR(L))| ≥ 2n−1− |FL| − |FR|. Hence Xn− F has a connected component that contains at least [2n−1 − |FR|] + [2n−1− |FL| − |FR|] = 2n− |F | − |FR| ≥ 2n− |F | − 2 vertices.
Case 2: |FR| ≥ 3.
Since |FR| ≥ 3, 3 ≤ |FL| ≤ 3(n − 1) − 6 and 3 ≤ |FR| ≤ 3(n − 1) − 6. By the inductive hypothesis, Xn−1L − FL (Xn−1R − FR, respectively) has a connected component CL(CR, respectively) that contains at least 2n−1− |FL| − 2 (2n−1− |FR| − 2, respectively) vertices. Next, we divide the case into three subcases: 2.1) |V (CL)| = 2n−1− |FL| − 2 and Xn−1R − FR is disconnected, 2.2) |V (CL)| = 2n−1− |FL| − 2 and Xn−1R − FR is connected, and 2.3) |V (CL)| ≥ 2n−1− |FL| − 1 and |V (CR)| ≥ 2n−1− |FR| − 1.
Case 2.1: |V (CL)| = 2n−1− |FL| − 2 and Xn−1R − FR is disconnected.
This is an impossible case. Since κ(Xn−1) = n − 1, |FR| ≥ n − 1. By Lemma 20,
|FL| ≥ 2((n − 1) − 1). Then the total number of faulty vertices is at least (n − 1) + 2((n − 1) − 1) = 3n − 5 which is greater than 3n − 6, a contradiction.
Case 2.2: |V (CL)| = 2n−1− |FL| − 2 and Xn−1R − FR is connected.
Since Xn−1R −FRis connected, |V (Xn−1R −FR)| = 2n−1−|FR|. Since |V (CL)| ≥ |FR|+1, there exists a vertex u ∈ CLand a vertex v ∈ CR such that (u, v) ∈ E(Xn). Hence Xn−F has a connected component that contains at least [2n−1 − |FR|] + [2n−1 − |FL| − 2] = 2n− |F | − 2 vertices.
Case 2.3: |V (CL)| ≥ 2n−1− |FL| − 1 and |V (CR)| ≥ 2n−1− |FR| − 1.
Since |V (CL)| ≥ |FR| + 1, there exists a vertex u ∈ CL and a vertex v ∈ CR such that (u, v) ∈ E(Xn). Hence Xn− F has a connected component that contains at least [2n−1− |FL| − 1] + [2n−1− |FR| − 1] = 2n− |F | − 2 vertices.
This completes the proof of the lemma. 2
By Lemma 21, we have the following corollary.
Corollary 9 ∀Xn ∈ Ln(n ≥ 5), let F be a set of vertices F ⊂ V (Xn) with |F | ≤ 3n − 6.
Then Xn− F satisfies one of the following conditions:
1. Xn− F is connected.
2. Xn− F has two components, one of which is K1, and the other one has 2n− |F | − 1 vertices.
3. Xn− F has two components, one of which is K2, and the other one has 2n− |F | − 2 vertices.
4. Xn−F has three components, two of which are K1, and the third one has 2n−|F |−2 vertices
We are now ready to show that the conditional diagnosability of Xn is 3(n − 2) + 1 for n ≥ 5. Let F1, F2 ⊂ V (Xn) be two conditional faulty sets with |F1| ≤ 3(n − 2) + 1 and |F2| ≤ 3(n − 2) + 1, n ≥ 5. We shall show our result by proving that (F1, F2) is a distinguishable conditional-pair under the comparison model.
Lemma 22 Let Xn be an n-dimensional BC graph with n ≥ 5. For any two conditional faulty sets F1, F2 ⊂ V (Xn), and F1 6= F2, with |F1| ≤ 3(n − 2) + 1 and |F2| ≤ 3(n − 2) + 1.
Then (F1, F2) is a distinguishable conditional-pair under the comparison model.
Proof.
We use Theorem 4 to prove this result. Let S = F1T F2, then 0 ≤ |S| ≤ 3(n − 2).
We will show that, deleting S from Xn, the subgraph CF1∆F2,S containing F1∆F2 has
”many” vertices having degree three or more. More precisely, we are going to prove that, in the subgraph CF1∆F2,S the number of vertices having degree three or more is at least 2[3(n − 2) + 1 − |S|] + 1 = 6n − 2|S| − 9. In the following proof, we consider three cases by the size of S: 1) 0 ≤ |S| ≤ n − 1, 2) |S| = n, and 3) n + 1 ≤ |S| ≤ 3(n − 2).
Case 1: 0 ≤ |S| ≤ n − 1.
Since the connectivity of Xn is n [24], Xn − S is connected, the subgraph CF1∆F2,S
is the only component in Xn− S. Since the BC graph Xn has no cycle of length three and any two vertices have at most two common neighbors, it is straightforward, though tedious, to check that the number of vertices which has degree two or one is at most 2 in CF1∆F2,S. Hence, the number of vertices having degree three or more is at least 2n−|S|−2 v in V (U) such that u is adjacent to v, then the condition 1 of Theorem 3 holds and therefore (F1, F2) is a distinguishable conditional-pair; otherwise V (U) is an independent set. Hence, NXn−S(v) ⊂ F1∆F2, ∀v ∈ U, and we have the following inequality
Case 3: n + 1 ≤ |S| ≤ 3(n − 2).
By Corollary 9, there are four cases in Xn− S we need to consider. For case 1 of Corollary 9, Xn− S is connected, the proof is exactly the same as that of Case 2, and hence the detail is omitted. For case 2 and 4 of Corollary 9, Xn − S has at least one trivial component {v} such that N(v) ⊂ F1 and N(v) ⊂ F2. Since F1 and F2 are two conditional faulty sets, the two cases are disregarded. Therefore, we only need to consider that Xn− S has two components, one of which is K2 and the other one has 2n− |S| − 2 vertices. Let (x, y) be the component with only one edge. Since N({x, y}) ⊆ S and F1and F2 do not contain all the neighbors of any vertex, vertex x and y cannot belong to F1∆F2. So the subgraph CF1∆F2,S is the other large connected component of Xn− S. Let U = Xn−(F1S F2) −{x, y}. If there exist two vertices u and v in V (U) such that u is adjacent to v, then the condition 1 of Theorem 3 holds and therefore (F1, F2) is a distinguishable conditional-pair; otherwise V (U) is an independent set. Hence, NXn−S(v) ⊂ F1∆F2,
∀v ∈ U, and we have the following inequality P
In Case 1, we prove that at least one of the conditions of Theorem 3 is satisfied in subgraph CF1∆F2,S. In Case 2 and 3, the condition 1 of Theorem 3 holds in subgraph CF1∆F2,S. Therefore, (F1, F2) is a distinguishable conditional-pair under the comparison
diagnosis model. 2
By Lemma 19, tc(Xn) ≤ 3(n − 2) + 1, and by Lemma 22, Xn is conditionally (3(n − 2) + 1)-diagnosable for n ≥ 5. We now have the following theorem.
Theorem 21 Under the comparison model, the conditional diagnosability of Xn is 3(n −
2) + 1 for n ≥ 5.
Since Qn, CQn, T Qn, MQn ∈ Ln, the following corollary holds.
Corollary 10 tc(Qn) = tc(CQn) = tc(T Qn) = tc(MQn) = 3(n − 2) + 1 under the com-parison model, for n ≥ 5.