Conditionally Diagnosable Systems
4.1 Conditional Diagnosability of Hypercube under the Comparison Model
In this section, we study the conditional diagnosability of hypercube under the comparison model. First, we give an example to show that the conditional diagnosability of the hypercube Qn is no greater than 3(n − 2) + 2, n ≥ 5. As shown in Figure 4.3, we take a cycle of length four in Qn. Let {v1, v2, v3, v4} be the four consecutive vertices on this cycle, and let F1 = N({v1, v3, v4})S{v1} and F2 = N({v1, v3, v4})S{v3}, then
|F1| = |F2| = 3(n−2)+2. It is straightforward to check that F1and F2are two conditional faulty sets, and F1 and F2 are indistinguishable by Theorem 3. Note that the hypercube Qnhas no cycle of length three and any two vertices have at most two common neighbors.
As we can see, |F1− F2| = |F2− F1| = 1 and |F1T F2| = 3(n − 2) + 1. Therefore, Qn is not conditionally (3(n − 2) + 2)-diagnosable and tc(Qn) ≤ 3(n − 2) + 1, n ≥ 3. Then, we shall show that Qn is conditionally t-diagnosable, where t = 3(n − 2) + 1.
Lemma 11 tc(Qn) ≤ 3(n − 2) + 1 under the comparison model, for n ≥ 3.
b
b b b
v1
v2
v3
v4
F1 F2
n-2 n-2n-2
n-2
S
Figure 4.3: An indistinguishable conditional-pair (F1, F2), where |F1| = |F2| = 3(n−2)+2.
Let F be a set of vertices F ⊂ V (Qn) and C be a connected component of Qn− F . We need some results on the cardinalities of F and V (C) under some restricted conditions.
The results are listed in Lemma 12 and 16. In Lemma 12, Lai et al. proved that deleting at most 2(n − 1) − 1 vertices from Qn, the incomplete hypercube Qn has one connected component containing at least 2n − |F | − 1 vertices. We expand this result further.
In Lemma 16, we show that deleting at most 3n − 6 vertices from Qn, the incomplete hypercube Qn has one connected component containing at least 2n− |F | − 2 vertices.
Lemma 12 [40] Let Qn be an n-dimensional hypercube, n ≥ 3, and let F be a set of vertices F ⊂ V (Qn) with n ≤ |F | ≤ 2(n − 1) − 1. Suppose that Qn− F is disconnected.
Then Qn− F has exactly two components, one is trivial and the other is nontrivial. The nontrivial component of Qn− F contains 2n− |F | − 1 vertices.
In order to prove Lemma 16, we need some preliminary results as follows.
Lemma 13 [46] Let Qnbe an n-dimensional hypercube. The connectivity of Qnis κ(Qn) = n.
Lemma 14 For any three vertices x, y, z in Q4, |N({x, y, z})| ≥ 7.
Proof.
A four-dimensional hypercube Q4 can be divided into two Q3’s, denoted by QL3 and QR3. Any two vertices in the Qn have at most two common neighbors. If these three vertices x, y, z all fall in QL3, then x, y, z have at least four neighboring vertices, all in QL3. Besides, x, y, z have three more neighboring vertices in QR3. Therefore, |N({x, y, z})| ≥ 4 + 3 = 7.
Suppose now x, y fall in QL3, z falls in QR3. Vertex x and y have at least four neighboring vertices, all in QL3. Vertex z will bring in at least three neighboring vertices in QR3.
Therefore, |N({x, y, z})| ≥ 4 + 3 = 7. 2
We are going to prove Lemma 16 by induction on n, and we need a base case to start with. As we observed, for n = 4, we found a counter example that the result of Lemma 16 does not hold. So we have to start with n = 5.
Lemma 15 Let Q5 be a five-dimensional hypercube, and let F be a set of vertices F ⊂ V (Q5) with |F | ≤ 3n − 6 = 9. Then Q5− F has a connected component containing at least 2n− |F | − 2 = 30 − |F | vertices.
Proof.
A five-dimensional hypercube Q5 can be divided into two Q4’s, denoted by QL4 and QR4. Let FL = FT V (QL4), 0 ≤ |FL| ≤ 9 and FR = FT V (QR4), 0 ≤ |FR| ≤ 9. Then
|F | = |FL| + |FR|. Without loss of generality, we may assume that |FL| ≥ |FR|. In the following proof, we consider three cases by the size of FR: 1) 0 ≤ |FR| ≤ 2, 2) |FR| = 3, and 3) |FR| = 4.
Case 1: 0 ≤ |FR| ≤ 2.
Since κ(Q4) = 4, QR4 − FR is connected and |V (QR4 − FR)| = 24 − |FR|. Let FR(L) ⊂ V (QL4) be the set of vertices which has neighboring vertices in FR. For each vertex v ∈ QL4−FL−FR(L), there is exactly one vertex v(R)in QR4−FR, such that (v, v(R)) ∈ E(Q5).
Besides, |V (QL4−FL−FR(L))| ≥ 24−|FL|−|FR|. Hence Q5−F has a connected component that contains at least [24− |FR|] + [24− |FL| − |FR|] = 32 − |F | − |FR| ≥ 30 − |F | vertices.
Case 2: |FR| = 3.
Since κ(Q4) = 4, QR4−FRis connected and |V (QR4−FR)| = 24−|FR|. Let FR= {x, y, z}
and FR(L) = {x(L), y(L), z(L)} ⊂ V (QL4), where (x, x(L)), (y, y(L)), (z, z(L)) ∈ E(Q5). For each vertex v ∈ QL4 − FL− FR(L), there is exactly one vertex v(R) in QR4 − FR, such that (v, v(R)) ∈ E(Q5). If at least one of the three vertices x(L), y(L), z(L) belongs to FL, then
|V (QL4−FL−FR(L))| ≥ 24−|FL|−2. Hence Q5−F has a connected component that contains at least [24− |FR|] + [24− |FL| − 2] = 30 − |F | vertices; otherwise, |V (QL4 − FL− FR(L))| ≥ 24− |FL| − 3. Since |FL| ≤ 6, by Lemma 14, x(L), y(L), z(L) have at least one neighboring vertex in QL4 − FL− FR(L). Hence Q5 − F has a connected component that contains at least [24− |FR|] + [24− |FL| − 3] + 1 = 30 − |F | vertices.
Case 3: |FR| = 4.
Since |FR| = 4 and |FL| ≤ 5, by Lemma 12, QL4 − FL (QR4 − FR, respectively) has a connected component CL(CR, respectively) that contains at least 24−|FL|−1 (24−|FR|−1, respectively) vertices. Since |V (CL)| ≥ |FR| + 1, there exists a vertex u ∈ CLand a vertex v ∈ CRsuch that (u, v) ∈ E(Q5). Hence Q5− F has a connected component that contains at least [24 − |FL| − 1] + [24− |FR| − 1] = 30 − |F | vertices.
Consequently, the lemma holds. 2
We now prove Lemma 16.
Lemma 16 Let Qn be an n-dimensional hypercube, n ≥ 5, and let F be a set of vertices F ⊂ V (Qn) with |F | ≤ 3n − 6. Then Qn− F has a connected component containing at least 2n− |F | − 2 vertices.
Proof.
We prove the lemma by induction on n. By Lemma 15, the lemma holds for n = 5. As the inductive hypothesis, we assume that the result is true for Qn−1, for |F | ≤ 3(n−1)−6, and for some n ≥ 6. Now we consider Qn, |F | ≤ 3n − 6. An n-dimensional hypercube Qn
can be divided into two Qn−1’s, denoted by QLn−1 and QRn−1. Let FL= F T V (QLn−1), 0 ≤
|FL| ≤ 3n − 6 and FR = FT V (QRn−1), 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, QRn−1 − FR is connected and |V (QRn−1 − FR)| = 2n−1− |FR|. Let FR(L) ⊂ V (QLn−1) be the set of vertices which has neighboring vertices in FR. For each vertex v ∈ QLn−1 − FL − FR(L), there is exactly one vertex v(R) in QRn−1− FR, such that (v, v(R)) ∈ E(Qn). Besides, |V (QLn−1− FL− FR(L))| ≥ 2n−1− |FL| − |FR|. Hence Qn− 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, QLn−1− FL (QRn−1 − 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 QRn−1− FR is disconnected, 2.2) |V (CL)| = 2n−1− |FL| − 2 and QRn−1− 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 QRn−1− FR is disconnected.
This is an impossible case. Since κ(Qn−1) = n − 1, |FR| ≥ n − 1. By Lemma 12,
|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 QRn−1− FR is connected.
Since QRn−1−FR is connected, |V (QRn−1−FR)| = 2n−1−|FR|. Since |V (CL)| ≥ |FR|+1, there exists a vertex u ∈ CLand a vertex v ∈ CRsuch that (u, v) ∈ E(Qn). Hence Qn− 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(Qn). Hence Qn− 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 16, we have the following corollary.
Corollary 8 Let Qn be an n-dimensional hypercube, n ≥ 5, and let F be a set of vertices F ⊂ V (Qn) with |F | ≤ 3n − 6. Then Qn− F satisfies one of the following conditions:
1. Qn− F is connected.
2. Qn− F has two components, one of which is K1, and the other one has 2n− |F | − 1 vertices.
3. Qn− F has two components, one of which is K2, and the other one has 2n− |F | − 2 vertices.
4. Qn−F has three components, two of which are K1, and the third one has 2n−|F |−2 vertices
Let G(V, E) be a graph. A subset M of E(G) is called a matching in G if its elements are links and no two are adjacent in G; the two ends of an edge in M are said to be matched under M. A vertex cover of G is a subset K of V (G) such that every edge of G has at least one end in K. A subset I of V (G) is called an independent set of G if no two vertices of I are adjacent in G. As the description for Theorem 15, the maximum size of a matching in a bipartite graph is equal to the minimum size of a vertex cover. To prove the conditional diagnosability of the hypercube, we need the following classical result.
Proposition 6 [52] Let G(V, E) be a bipartite graph. The set I ⊂ V (G) is a maximum independent set of G if and only if V − I is a minimum vertex cover of G.
The hypercube can be described as follows: Let Qn denote an n-dimensional hyper-cube. Q1 is a complete graph with two vertices labeled with 0 and 1, respectively. For
n ≥ 2, each Qn consists of two Qn−1’s, denoted by Q0n−1 and Q1n−1, with a perfect match-ing M between them. That is, M is a set of edges connectmatch-ing the vertices of Q0n−1 and the vertices of Q1n−1 in a one-to-one manner. It is easy to see that there are 2n−1 edges between Q0n−1 and Q1n−1. The hypercube is a bipartite graph with 2n vertices. Hence, we have the following Lemma.
Lemma 17 Let Qn be an n-dimensional hypercube. In hypercube Qn, the maximum size of a matching, the minimum size of a vertex cover and the maximum size of an independent set are all 2n−1.
We are now ready to show that the conditional diagnosability of Qn is 3(n − 2) + 1 for n ≥ 5. Let F1, F2 ⊂ V (Qn) 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 diagnosis model.
Lemma 18 Let Qn be an n-dimensional hypercube with n ≥ 5. For any two conditional faulty sets F1, F2 ⊂ V (Qn), 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 diagnosis 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 Qn, 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 Qn is n, Qn− S is connected, the subgraph CF1∆F2,S is the only component in Qn− S. Since the hypercube Qn 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 two in CF1∆F2,S. Hence, the number of vertices having degree three or more is at least 2n− |S| − 2 which is greater than 6n − 2|S| − 9, for n ≥ 5. By Theorem 4, (F1, F2) is a distinguishable 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. Since
|S| = n and |F1∆F2| ≤ 2(2n − 5), |V (U)| ≥ 2n− 2(2n − 5) − n = 2n− 5n + 10. By Lemma 17, the maximum size of a independent set is 2n−1 in Qn. Comparing the lower bound 2n− 5n + 10 and the upper bound 2n−1, we have 2n− 5n + 10 > 2n−1 for n ≥ 5, a contradiction.
Case 3: n + 1 ≤ |S| ≤ 3(n − 2).
By Corollary 8, there are four cases in Qn− S we need to consider. For case 1 of Corollary 8, Qn − 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 8, Qn − 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 Qn− 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 F1
and 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 Qn − S.
Let U = Qn − (F1S F2) − {x, y}. If no two vertices of V (U) are adjacent, then V (U) is an independent set and |V (U)| ≥ 2n − 6n + |S| + 8. By Lemma 17, the maximum size of a matching is 2n−1 − 1 in Qn − {x, y}. By Theorem 15 and Proposition 6, the maximum size of a independent set is 2n−1−1 in Qn−{x, y}. Comparing the lower bound 2n− 6n + |S| + 8 and the upper bound 2n−1− 1, we have 2n− 6n + |S| + 8 > 2n−1− 1
for n ≥ 5, n + 1 ≤ |S| ≤ 3(n − 2), a contradiction. Hence, there exist two vertices u and v in V (U) such that u is adjacent to v, then condition 1 of Theorem 3 is satisfied and therefore (F1, F2) is a distinguishable conditional-pair.
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 11, tc(Qn) ≤ 3(n − 2) + 1, and by Lemma 18, Qn is conditionally (3(n − 2) + 1)−diagnosable for n ≥ 5. Hence, tc(Qn) = 3(n − 2) + 1 for n ≥ 5. For Q3 and Q4, we observe that Q3 is not conditionally four-diagnosable and Q4 is not conditionally six-diagnosable, as shown in Figure 4.4. So, tc(Q3) ≤ 3 and tc(Q4) ≤ 5. Hence, the conditional diagnosabilities of Q3 and Q4 are both strictly less than 3(n − 2) + 1.
bb bb bb bb b b b b b b b b
0000 1111 0101 0011 1100 1010 1001 0110
0100 0010 0001 0111 1101 1011
F1 F2
Figure 4.4: Two indistinguishable conditional-pairs for Q3 and Q4.
For the three-dimensional hypercube Q3, Q3 is three-diagnosable and it is not condi-tionally 4-diagnosable. It follows from Lemma 9 that tc(Q3) = 3. For the four-dimensional hypercube Q4, we can use the similar technique used in proving Lemma 18 to prove that for any two conditional faulty sets F1, F2 ⊂ V (Q4), and F1 6= F2, with |F1| ≤ 5 and
|F2| ≤ 5, then (F1, F2) is a distinguishable conditional-pair under the comparison diagno-sis model. Hence, the conditional diagnosability of Q4 is 5. In summary, the conditional
diagnosability of Qn is stated as follows:
Theorem 20 Under the comparison model, the conditional diagnosability of Qn is 3(n − 2) + 1 for n ≥ 5, tc(Q3) = 3 and tc(Q4) = 5.