In previous chapter, we explained that the Hypercube Qn is n-diagnosable. In fact, the Crossed cube CQn, the M¨obius cube MQn, and the Twisted cube T Qn are all known as n-diagnosable but not (n + 1)-diagnosable. In this chapter, we will presented the concept of the strongly t-diagnosable system. Besides, we will also prove that the cube family with n-dimensional are all strongly n-diagnosable for n ≥ 4. Firstly, we take Q3 as an example to explained that why Q3 is not 4-diagnosable. The structure of Q3 as shown in Figure 3.1.
a 1
2
3
Figure 3.1: The structure of Q3.
Let S1 = {1, 2, 3} and S2 = {a, 1, 2, 3}, with | S1 |≤ 4 and | S2 |≤ 4. By lemma 1, S1 and S2 are indistinguishable-pair. Hence Q3 is not 4-diagnosable. For each of these cubes with n-dimension, we observe that for any two distinct sets of vertex S1 and S2,
| S1 |≤ n + 1 and | S2 |≤ n + 1, they are indistinguishable-pair implies that there exists some vertex v such that N(v) ⊂ S1T
S2. That is N(v) ⊂ S1 and N(v) ⊂ S2. We continue taking Q4 as an example, for each vertex v ∈ V (Q4) and each vertex set P ⊂ V (Q4), 0 ≤| P |≤ 4. Q4− P is connected if N(v) * P . It’s mean, the only component of Q4− P is itself. Let S1, S2 ⊂ V (Q4) be two distinct sets of vertex with | S1 |≤ 5, | S2 |≤ 5, and P = S1T
S2. We can get that the inequality | V (Q4) − P |= 24− | P |≥| S1 − P | + | S2−P | +1. Then there is at least one edge connecting S1∆S2 and V (Q4−(S1
SS2)). By
lemma 1, S1 and S2 are distinguishable-pair if for each v ∈ V (Q4), N(v) * P . Inversely, S1 and S2 are indistinguishable-pair, then there exists some vertex v ∈ V (Q4) such that N(v) ⊆ S1 and N(v) ⊆ S2. We observed the phenomenon and give a formally definition as follows:
Definition 9 A system G = (V, E) is strongly t-diagnosable if the following two condi-tions hold:
1. G is t-diagnosable, and
2. for any two distinct subsets S1, S2 ⊂ V with | S1 |≤ t + 1 and | S2 |≤ t + 1,
either (a) (S1, S2) is a distinguishable pair;
or (b) (S1, S2) is an indistinguishable pair and there exists a vertex v ∈ V
such that N(v) ⊆ F1 and N(v) ⊆ F2.
By lemma 5 and definition 9, we propose a sufficient condition for a system G to be strongly t-diagnosable as follows:
Proposition 1 Let G = (V, E) be the graph presentation of a system with | V |= n is strongly t-diagnosable if the following three conditions hold:
1. n ≥ 2(t + 1) + 1,
2. κ(G) ≥ t, and
3. for any vertex set P ⊂ V with | P |= t, G − P is disconnected implies that there exists a vertex v ∈ V such that N(v) ⊂ P .
Proof. To prove the proposition, we claim that condition (1) and (2) of definition 9 hold. Since condition (1) and (2), by lemma 5, G is t-diagnosable. For condition (2) of definition 9. Let S1 and S2 be an indistinguishable-pair, and P = S1T
S2, where S1 6= S2,
| S1 |≤ t+1 and | S2 |≤ t+1, then 0 ≤| P |≤ t. If G−P is connected, then there exists an edge between S14S2 and V − (S1S
S2). By lemma 1, S1 and S2 are distinguishable-pair.
This is a contradiction. Hence G − P is disconnected. By condition (2), κ(G) ≥ t and 0 ≤| P |≤ t. Therefore | P |= t. By condition (3), there exists a vertex v ∈ V such that N(v) ⊂ P . That is, N(v) ⊂ S1 and N(v) ⊂ S2. Hence condition(2) of definition 9 holds.
This completes the proof of the proposition. 2
Now, we propose a necessary and sufficient condition for a system to be strongly t-diagnosable as follows:
Lemma 8 Let G = (V, E) be the graph presentation of a system with | V |= n is strongly t-diagnosable if and only if
1. n ≥ 2(t + 1) + 1,
2. δ(G) ≥ t, and
3. for any indistinguishable-pair S1, S2 ⊂ V , S1 6= S2, with | S1 |≤ t+1 and | S2 |≤ t+1 it implies that there exists a vertex v ∈ V such that N(v) ⊂ S1 and N(v) ⊂ S2.
Proof.
To prove the necessity of condition (1), we show that the assumption n ≤ 2(t+1) leads to a contradiction. Assume n ≤ 2(t + 1). We can partition V into two disjoint vertex sets V1 and V2 with | V1 |≤ t + 1 and | V2 |≤ t + 1, where V = V1S
V2 and V1T
V2 = Ø. By lemma 1, V1 and V2 are indistinguishable-pair. Since G is strongly t-diagnosable. Then there exists some vertex v ∈ V such that N(v) ⊂ V1 and N(v) ⊂ V2. Hence V1
TV2 6= Ø.
That contradicts the assumption V1T
V2 = Ø.
To prove the necessity of condition (2), since G is strongly t-diagnosable. By definition 9, G is also t-diagnosable. By condition(2) of lemma 4, N(v) ≥ t for each vertex v ∈ V . Hence δ(G) ≥ t.
To prove the necessity of condition (3), that is the same as condition (2) of definition 9. This completes the proof for the necessity.
On the other hand, since condition (3) of this lemma and condition(2) of definition 9 are stated the same. We need only to prove that G is t-diagnosable. Assume not, then there exists an indistinguishable-pair S1, S2 ⊂ V , S1 6= S2, with | S1 |≤ t and | S2 |≤ t.
By condition (3), there exists a vertex v ∈ V such that N(v) ⊂ S1 and N(v) ⊂ S2. By condition (2), we know that | N(v) |≥ t. But, | S1 |≤ t and | S2 |≤ t. Hence S1 = S2 = N(v). This contradicts the S1 6= S2. We complete the proof of this lemma. 2
The lemma given above is a method for checking whether a system is strongly t-diagnosable. Now, we propose another necessary and sufficient condition. Let G = (V, E) be a strongly t-diagnosable system. If G is (t + 1)-diagnosable. By Theorem 1, for each vertex set P ⊂ V , | P |= p where 0 ≤ p ≤ t, each component C of G − P satisfies
| VC |≥ 2((t + 1) − p) + 1. Otherwise, G is t-diagnosable but not (t + 1)-diagnosable.
Then there exists an indistinguishable-pair(S1, S2), | S1 |≤ t + 1 and | S2 |≤ t + 1.
By condition(2) of Definition 9, there exists a vertex v ∈ V such that N(v) ⊂ S1 and N(v) ⊂ S2, where v /∈ S1S
1. for 0 ≤ p ≤ t − 1, each component C of G − P satisfies | VC |≥ 2((t + 1) − p) + 1, and
2. for p = t, either each component C of G − P satisfies | VC |≥ 3 or else G − P contains at least a trivial component.
Proof.
To prove the necessity of condition (1), assume that there exists a vertex set P ⊂ V with | P |= p, 0 ≤ p ≤ t−1, such that G−P has a component C with | VC |≤ 2((t+1)−p).
Since G is strongly t-diagnosable. By Definition 9, there exists a vertex v ∈ V such that N(v) ⊂ S1 and N(v) ⊂ S2. By lemma 4 | N(v) |≥ t. However, N(v) ⊂ S1
TS2 = P and
0 ≤ p ≤ t − 1, this is a contradiction.
To prove the necessity of condition (2), assume that there exists a component C of G − P with | VC |≤ 2. Then we have to prove that there is a trivial component in G − P . If | VC |= 1, we are done. Assume that | VC |= 2, we say VC = {v1, v2}. Let S1 = {v1}S
P and S2 = {v2}S
P . Then | S1 |=| S2 |= t + 1, and are indistinguishable-pair. Since G is strongly t-diagnosable. By definition 9, there exists a vertex v ∈ V such that N(v) ⊂ S1 and N(v) ⊂ S2. We have P = S1T
S2 and P = N(v). Therefore, {v} is a trivial component in G − P .
On the other hand, we claim that G is strongly t-diagnosable. We have to prove that G satisfies conditions (1) and (2) of definition 9. For condition (1) of definition 9, let P be a vertex set with | P |= p, 0 ≤ p ≤ t − 1. By condition (1), each component C of G − P satisfies | VC |≥ 2((t + 1) − p) + 1 ≥ 2(t − p) + 1. By Theorem 1, G is t-diagnosable.
For condition (2) of definition 9, let S1 and S2 be an indistinguishable-pair, S1 6= S2, with | S1 |≤ t + 1 and | S2 |≤ t + 1. Let P = S1T
S2, | P |= p, then 0 ≤ p ≤ t. Since S1 and S2 are indistinguishable-pair. Hence there is no edge between X = V − (S1S
S2) and S1∆S2. Therefore, S1∆S2 is disconnected from the other component in G − P . We observed that | S1∆S2 |≤ 2((t + 1) − p). By condition(1), p is not in the range from 0 to t − 1. So p = t and | S1∆S2 |≤ 2((t + 1) − p) = 2((t + 1) − t) = 2. By condition(2), G − P must have a trivial component {v}. Hence N(v) ⊂ P . Since G is t-diagnosable, by condition(2) of lemma 4, N(v) ≥ t. So P = N(v). Then N(v) ⊂ S1 and N(v) ⊂ S2. Therefore, G is strongly t-diagnosable.
Thus we complete the proof of this theorem. 2
Theorem 4 Let G1 = (V1, E1) and G2 = (V2, E2) be two t-diagnosable systems with the same number of vertices, where t ≥ 2. Then MCN G = G1
L
MG2 = (V, E) is strongly (t+1)-diagnosable, where V = V1S
V2 and E = E1S
and (2) S1 6= Ø and S2 6= Ø. We shall prove that: (i) | VC |≥ 2((t + 2) − p) + 1 for any component C of G − P as 0 ≤ p ≤ t, and (ii) for p = t + 1, either any component C of G − P satisfies | VC |≥ 3 or else G − P contains at least one trivial component.
Case 1: S1 = Ø or S2 = Ø
Without loss of generality, assume S1 = Ø and S2 = P . Since each vertex of V2 has an adjacent neighbor in V1. Hence G−P is connected. So, | VC |=| V −P |=| V1 | + | V2 | −p.
Since G1 and G2 are t-diagnosable. By lemma 4, | V1 |≥ 2t + 1 and | V2 |≥ 2t + 1. Hence
| VC |≥ 2(2t + 1) − p ≥ 2((t + 2) − p) + 1 for t ≥ 2 and 0 ≤ p ≤ t + 1.
Case 2: S1 6= Ø and S2 6= Ø
Since S1 6= Ø and S2 6= Ø. We know that 1 ≤ p1 ≤ t and 1 ≤ p2 ≤ t. In this case, we divide the case into two subcases: (2.a) 1 ≤ p1 ≤ t − 1 and 1 ≤ p2 ≤ t − 1, and (2.b) either p1 = t or p2 = t. In fact, for subcase (2.b),either p1 = t and p2 = 1, or, p2 = t and p1 = 1.
Subcase 2.a: 1 ≤ p1 ≤ t − 1 and 1 ≤ p2 ≤ t − 1
Let C1 be the component of G1 − S1. Since G1 is t-diagnosable. By theorem 1,
| VC1 |≥ 2(t − p1) + 1. We claim that there is at least one vertex in VC1 which is connected to V2 − S2. That is 2(t − p1) + 1 ≥ p2 + 1. Since p = p1 + p2, then 2(t − p1) + 1 = 2(t − p + p2) + 1 = 2p2+ 2(t − p) + 1. Suppose p ≤ t, then | VC1 |≥ 2(t − p1) + 1 ≥ p2+ 1.
Otherwise, p = t + 1. With p1 ≤ t − 1, then p2 ≥ 2 and 2p2 + 2(t − p) + 1 ≥ p2+ 1.
Hence | VC1 |≥ 2(t − p1) + 1 ≥ p2+ 1. The claim is completed. Let C2 be the component
of G2 − S2. Since G2 is t-diagnosable. By theorem 1, | VC2 |≥ 2(t − p2) + 1. We let C vertex v1 belongs to some component C1 of G1−S1, then the component C containing the two vertex sets VC1 and VC2 has at least four vertices. Otherwise, N(VC2, V1) ⊂ S1. With
| S1 |= p1− 1, | N(VC2, V1) |= 1. That is, | VC2 |= 1. Hence, C2 is a trivial component.
Thus we complete the proof of this theorem. 2
We will give an example to explain why the above result is not true when t = 1.
As shown in figure3.2(a), let G1 and G2 are 1-diagnosable systems with vertex sets {v1, v2, v3, v4, v5} and {u1, u2, u3, u4, u5}, respectively. Let G = G1
L
M G2 be a Matching Composition Network constructed by adding a perfect matching between G1 and G2. By lemma 5, G is 2-diagnosable. See Figure3.2(b), let F1 = {v1, v2, u2} and F2 = {u1, u2, v2}.
By lemma 1, F1 and F2 are indistinguishable-pair but there doesn’t exist any vertex v ∈ V1
SV2 such that N(v) ⊂ F1 and N(v) ⊂ F2. Hence G is not strongly 2-diagnosable.
v1
Figure 3.2: An example of non-strongly (t+1)-diagnosable as t=1.
According to theorem 4, we know that all systems of the cube family are strongly (t + 1)-diagnosable because their subcubes are t-diagnosable for t ≥ 3. The Hypercube Qn, the Crossed cube CQn, the Twisted cube T Qn, and the M¨obius cube MQnare famous parts in the cube family. Hence we hold the following corollary.
Corollary 1 The Hypercube Qn, the Crossed cube CQn, the Twisted cube T Qn, and the M¨obius cube MQn are all strongly n-diagnosable for n ≥ 4.
For n = 2, these cubes are all a cycle of length four. They are 1-diagnosable but not 2-diagnosable. For n = 3, these cubes are all 3-connected, by lemma 5, they are 3-diagnosable. We now show some examples which are not strongly t-diagnosable. Let us take the 3-dimensional Hypercube Q3 as an example. it is 3-diagnosable but not strongly 3-diagnosable from the fact that | V (Q3) |= 8 ≤ 2(t + 1) + 1 as t = 3, which contradicts the condition (1) of lemma 8. Let Cln be a cycle of length n, n ≥ 7. It is not difficult to
verify that Cln is 2-diagnosable, but it is not strongly 2-diagnosable. Another nontrivial example is shown in figure 3.3. This graph is 2-connected and 3-regular. We can use theorem 1 to verify that it is 3-diagnosable. As shown in figure 3.3, S1 = {1, 2, 5, 6} and S2 = {3, 4, 5, 6}. (S1, S2) is an indistinguishable-pair, but there does not exist any vertex v ∈ V (G) such that N(v) ⊂ S1 and N(v) ⊂ S2. By definition 9, this graph is not strongly 3-diagnosable.
2 1 5
6 3 4
7
8 12
11
10 9
S1 S S2
Figure 3.3: An example of non-strongly 3-diagnosable system.
Chapter 4 Conclusion
The fault diagnosis is a popular issue for interconnection network. There are some open problems which we can discuss. In recent years, researchers have considered a large number of strategies for self-diagnosis in interconnection network. The PMC model, first proposed by Preparata et al.[21], is used widely for fault diagnosis of interconnection network. In this thesis, we study the properties of fault diagnosis of the cube family. We also propose the concepts of strongly t-diagnosable systems under the PMC model. We show that the cube family with n-dimensional are all strongly n-diagnosable, where n ≥ 4.
The cube family include Hypercube, Crossed cube, Twisted cube and M¨obius cube et al.
There are many models which we can research except the PMC model. The Comparison model [16] that is another well-known fault diagnosis model. Hence, it is also interesting to investigate the issues of strongly t-diagnosable of a system under the Comparison model.
Besides, there are two attractive problems which are worth researching. Firstly, we want to know whether the recursive interconnection networks are all strongly t-diagnosable system. Secondly, what is the diagnosability of interconnection network when we allow
one good neighbor condition?
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