The Globally Bi- 3 ∗ -Connected Property of the Honeycomb
4.2 A Basic Algorithm
In this section, we present an algorithm. The purpose of this algorithm is to extend a 3∗-container C3∗(x, y) = {P1, P2, P3} of HReT(m, n) to a 3∗-container of HReT(m + 2, n).
Algorithm 1. For 0 ≤ i ≤ m − 1, let fi : V (HReT(m, n)) → V (HReT(m + 2, n)) be a function so assigned
fi(k, l) =
(k, l) if i ≥ k ≥ 0 (k + 2, l) otherwise.
For 0 ≤ i ≤ m−1 and 0 ≤ j, k ≤ n−1, let Qi(j, [j+k]n) denote the path(i, [j]n), (i, [j+
1]n), (i, [j + 2]n), ..., (i, [j + k]n) in HReT(m, n). Suppose that C3(x, y) is a 3-container of HReT(m, n) containing at least one edge joining vertices of column i to vertices of column [i + 1]m; i.e., ((i, j), ([i + 1]m, j)) in E(C3(x, y)) for some 0 ≤ j ≤ n − 1. Let 0≤ k0 < k1 < ... < kt≤ n − 1 be the indices such that ((i, kj), (i + 1, kj))∈ E(C3(x, y)).
We construct C3,i (x, y) as follows:
Let C3,i(x, y) be the image of C3(x, y)− {((i, kj), (i + 1, kj))| 0 ≤ kj ≤ n − 1} under fi. We set j = [j](t+1) and define Aj as
(i, [kj]n), ([i + 1]m+2, [kj]n), Q[i+1]m+2([kj]n, [kj− 1]n), ([i + 1]m+2, [kj − 1]n), ([i + 2]m+2, [kj − 1]n), Q−1[i+2]
m+2([kj]n, [kj− 1]n), ([i + 2]m+2, [kj]n), ([i + 3]m+2, [kj]n).
Obviously, Aj is a path joining (i, [kj]n) and (i + 3, [kj]n) for 0 ≤ j < t. It is easy to see that edges of C3,i(x, y) together with edges of Aj, with 0 ≤ j ≤ t form a 3-container C3,i (x, y) of HReT(m + 2, n). For example, a 3∗-container C3∗((0, 0), (2, 2)) of HReT(4, 12)− {(1, 7)} is shown in Figure 4.2(a). The corresponding C3,1 ((0, 0), (2, 2)) is shown in Figure 4.2(b). We have the following lemma.
Lemma 10. Suppose that C3(x, y) is a 3-container of HReT(m, n) containing at least one edge joining vertices of column i to vertices of column [i + 1]m. Then C3,i (x, y) forms a 3-container of HReT(m+2, n) containing at least one edge joining the vertices of column l
(a) (b)
Figure 4.2: Illustrations for Algorithm 1.
to the vertices of column [l +1]m for any l ∈ {i, [i+1]m, [i+2]m}. Moreover, C3∗,i(x, y) is a 3∗-container of HReT(m+2, n) if C3∗(x, y) is a 3∗-container of HReT(m, n). Furthermore, C3∗,i(x, y) is a 3∗-container of HReT(m + 2, n)− {fi(z)} if C3∗(x, y) is a 3∗-container of HReT(m, n)− {z}.
Lemma 11. Suppose that C3(x, y) is a 3-container of HReT(2, n) containing at least one edge in {((0, j), (1, j)) | j is odd} and at least one edge in {((0, j), (1, j)) | j is even}.
Then C3,i (x, y) with i∈ {0, 1} forms a 3-container of HReT(4, n) containing at least one edge joining the vertices of column l to the vertices of column l + 1 for any l ∈ {0, 1, 2, 3}.
Moreover, C3∗,i(x, y) is a 3∗-container of HReT(m + 2, n) if C3∗(x, y) is a 3∗-container of HReT(m, n). Furthermore, C3∗,i(x, y) is a 3∗-container of HReT(m + 2, n)− {fi(z)} if C3∗(x, y) is a 3∗-container of HReT(m, n)− {z}.
With Lemma 10 and Lemma 11, we say a 3-container C3(x, y) of HReT(2, n) is regular if C3(x, y) contains at least one edge in {((0, j), (1, j)) | j is odd} and at least one edge in {((0, j), (1, j)) | j is even}. Assume that m ≥ 4. We say a 3-container C3(x, y) of HReT(m, n) is regular if C3(x, y) contains at least one edge joining vertices in column i to vertices in column [i + 1]m for 0≤ i ≤ m − 1. We have the following lemma.
Lemma 12. Suppose that C3∗(x, y) is a regular 3∗-container for HReT(m, n). Then C3∗,i(x, y) is a regular 3∗-container for HReT(m + 2, n) for every 0 ≤ i < m. Moreover, suppose that C3∗(x, y) is a regular 3∗-container for HReT(m, n)− {z}. Then C3∗,i(x, y) is a regular 3∗-container for HReT(m + 2, n)− {fi(z)} for every 0 ≤ i < m.
4.3 The Globally Bi- 3
∗-Connected Property of Hon-eycomb Rectangular Torus HReT(2,n)
We first discuss the globally bi-3∗-connected property of the honeycomb rectangular torus HReT(m, n) for m = 2. Then we show the globally bi-3∗-connected properties of HReT(m, n) for m = 2 and general m in sections 4.4 and 4.5, respectively.
For h = {0, 1} and 0 ≤ j, k ≤ n−1, let Rh(j, [j +k]n) denote the path(h, [j]n), (h, [j + 1]n), ([h + 1]m, [j + 1]n), ([h + 1]m, [j + 2]n), (h, [j + 2]n), ..., ([h + 1]m, [j + k− 1]n), (h, [j + k− 1]n), (h, [j + k]n) in HReT(2, n).
Lemma 13. Let x and y be any two vertices of HReT(2, n) = (V0 ∪ V1, E) with x ∈ V0
and y ∈ V1. Then there exists a regular 3∗-container C3∗(x, y) of HReT(2, n). Hence HReT(2, n) is globally bi-3∗-connected.
Proof. Without loss of generality, we may assume that x = (0, 0) and y = (i, j). In order to prove this lemma, we will construct a regular 3∗-container C3∗(x, y) = {P1, P2, P3} in HReT(2, n). We have the following cases:
Case 1: i = 0 and j is odd. The corresponding paths are:
P1 = (0, 0), Q0(0, j), (0, j);
P2 = (0, j), R0(j, 0), (0, 0);
P3 = (0, 0), (1, 0), Q1(0, j), (1, j), (0, j).
Case 2: i = 1 and j is even.
Case 2.1: j = 0. The corresponding paths are:
P1 = (0, 0), Q0(0, n− 2), (0, n − 2), (1, n − 2), Q−11 (0, n− 2), (1, 0);
P2 = (0, 0), (1, 0);
P3 = (0, 0), (0, n − 1), (1, n − 1), (1, 0).
Case 2.2: j > 0. The corresponding paths are:
P1 = (0, 0), Q0(0, j), (0, j), (1, j);
P2 = (1, j), (1, j + 1), (0, j + 1), R0(j + 1, 0), (0, 0);
P3 = (0, 0), (1, 0), Q1(0, j), (1, j).
Hence HReT(2, n) is globally bi-3∗-connected. See Figure 4.3 for illustrations.
x
x x
y y
y
Case 1 Case 2.1 Case 2.2
Figure 4.3: Illustrations for Lemma 13.
Lemma 14. Let x, y, and z be any three different vertices of HReT(2, n) = (V0∪ V1, E) in V0. Then there exists a regular 3∗-container C3∗(x, y) of HReT(2, n) − {z}. Hence HReT(2, n) is hyper globally bi-3∗-connected.
Proof. Without loss of generality, we may assume that x = (0, 0), y = (i, j), and z = (k, l). In order to prove this lemma, we will construct a regular 3∗-container C3∗(x, y) = {P1, P2, P3} in HReT(2, n) − {z}. We have the following cases:
Case 1: i = 0. Then j is even.
Case 1.1: k = 0. Then l is even. By the symmetric property of HReT(2, n), we may assume that l < j. The corresponding paths are:
P1 = (0, j), Q0(j, 0), (0, 0);
P2 = (0, 0), R0(0, l− 1), (0, l − 1), (1, l − 1), (1, l), (1, l + 1), (0, l + 1), R0(l + 1, j), (0, j);
P3 = (0, j), (1, j), Q1(j, 0), (1, 0), (0, 0).
Case 1.2: k = 1. Then l is odd. By the symmetric property of HReT(2, n), we may assume that l < j. The corresponding paths are:
P1 = (0, j), Q0(j, 0), (0, 0);
P2 = (0, 0), R0(0, l), (0, l), R0(l, j), (0, j);
P3 = (0, j), (1, j), Q1(j, 0), (1, 0), (0, 0).
y y
y
x x
Case 1.1 Case 1.2 Case 2 z
z z
x
Figure 4.4: Illustrations for Lemma 14.
Case 2: i = 1. Then j is odd. k = 0. Then l is even. By the symmetric property of HReT(2, n), we may assume that l < j. The corresponding paths are:
P1 = (1, j), (0, j), Q0(j, 0), (0, 0);
P2 = (0, 0), R0(0, l− 1), (0, l − 1), (1, l − 1), (1, l), (1, l + 1), (0, l + 1), R0(l + 1, j− 1), (0, j − 1), (1, j − 1), (1, j);
P3 = (1, j), Q1(j, 0), (1, 0), (0, 0).
Hence HReT(2, n) is hyper globally bi-3∗-connected. See Figure 4.4 for illustrations.
4.4 The Globally Bi- 3
∗-Connected Property of Hon-eycomb Rectangular Torus HReT(4,n)
In this section, we need the following path patterns. For 0≤ i ≤ m−1 and 0 ≤ j, k ≤ n−1, we set
SiL(j) = ([i]m, [j]n), ([i− 1]m, [j]n), ([i− 1]m, [j + 1]n), ([i− 2]m, [j + 1]n), ([i− 2]m, [j + 2]n), ([i− 3]m, [j + 2]n), ([i− 3]m, [j + 3]n),
([i− 4]m, [j + 3]n), ([i− 4]m, [j + 2]n);
SiR(j) = ([i]m, [j]n), ([i + 1]m, [j]n), ([i + 1]m, [j + 1]n), ([i + 2]m, [j + 1]n),
) 2 , 4
0
(
Q R
0( 4 , 1 ) S
1L( 3 ) S
2L( 0 , 4 ) S
3R( 2 )
S2R(1,5)Figure 4.5: The path patterns Q0(4, 2), R0(4, 1), S1L(3), S2L(0, 4), S3R(2), and S2R(1, 5).
([i + 2]m, [j + 2]n), ([i + 3]m, [j + 2]n), ([i + 3]m, [j + 3]n), ([i + 4]m, [j + 3]n), ([i + 4]m, [j + 2]n);
SiL(j, k) = ([i]m, [j]n), S[i]Lm(j), ([i− 4]m, [j + 2]n), S[i−4]L m([j + 2]n), ([i− 8]m, [j + 4]n), ..., ([i− 2(k − j − 2)]m, [k− 2]n), S[i−2(k−j−2)]L m([k− 2]n), ([i− 2(k − j)]m, [k]n); and SiR(j, k) = ([i]m, [j]n), S[i]Rm(j), ([i + 4]m, [j + 2]n), S[i+4]R m([j + 2]n),
([i + 8]m, [j + 4]n), ..., ([i + 2(k− j − 2)]m, [k− 2]n), S[i+2(k−j−2)]R m([k− 2]n), ([i + 2(k− j)]m, [k]n).
See Figure 4.5 for illustrations.
Lemma 15. Let x and y be any two vertices of HReT(4, n) = (V0 ∪ V1, E) with x ∈ V0 and y ∈ V1. Then there exists a regular 3∗-container C3∗(x, y) of HReT(4, n). Hence HReT(4, n) is globally bi-3∗-connected.
Proof. Without loss of generality, we may assume that x = (0, 0) and y = (i, j). In order to prove this lemma, we will construct a regular 3∗-container C3∗(x, y) = {P1, P2, P3} in HReT(4, n). By the symmetric property of HReT(4, n), we may assume that i∈ {0, 1, 2}.
Hence we have the following cases:
Case 1: Suppose that i ∈ {0, 1}. By Lemma 13, there exists a regular 3∗-container C3∗((0, 0), (i, j)) of HReT(2, n). By Lemma 12, C3∗,1((0, 0), (i, j)) forms a 3∗-container of
Case 2.1
x x
y
y
Case 2.2
Figure 4.6: Illustrations for Lemma 15.
HReT(4, n).
Case 2: i = 2. Then j is odd.
Case 2.1: Suppose that j = 1. The corresponding paths are:
P1 = (0, 0), (0, n − 1), (1, n − 1), Q−11 (0, n− 1), (1, 0), (2, 0), (2, 1);
P2 = (0, 0), Q0(0, n− 2), (0, n − 2), (3, n − 2), Q−13 (1, n− 2), (3, 1), (2, 1);
P3 = (0, 0), (3, 0), (3, n − 1), (2, n − 1), Q−12 (1, n− 1), (2, 1).
Case 2.2: Suppose that j = 1. The corresponding paths are:
P1 = (0, 0), Q0(0, j− 1), (0, j − 1), (3, j − 1), (3, j), (2, j);
P2 = (0, 0), (3, 0), Q3(0, j− 2), (3, j − 2), (2, j − 2), Q−12 (0, j− 2), (2, 0), (1, 0), Q1(0, j− 1), (1, j − 1), (2, j − 1), (2, j);
P3 = (0, 0), (0, n − 1), SL−1(j + 3, n− 1), (0, j + 3), (0, j + 2), (1, j + 2), (1, j + 1), (1, j), (0, j), (0, j + 1), (3, j + 1), (3, j + 2), (2, j + 2), (2, j + 1), (2, j).
Hence HReT(4, n) is globally bi-3∗-connected. See Figure 4.6 for illustrations.
Lemma 16. Let x, y, and z be any three different vertices of HReT(4, 6) = (V0∪ V1, E) in V0. Then there exists a regular 3∗-container C3∗(x, y) of HReT(4, 6)− {z}. Hence HReT(4, 6) is hyper globally bi-3∗-connected.
Proof. Without loss of generality, we may assume that x = (0, 0), y = (i, j), and z = (k, l).
Hence HReT(4, 6) is hyper globally bi-3∗-connected.
Lemma 17. Assume that n ≥ 8. Let x, y, and z be any three different vertices of HReT(4, n) = (V0 ∪ V1, E) in V0. Then there exists a regular 3∗-container C3∗(x, y) of HReT(4, n)− {z}. Hence HReT(4, n) is hyper globally bi-3∗-connected.
Proof. Without loss of generality, we may assume that x = (0, 0), y = (i, j), and z = (k, l). In order to prove this lemma, we will construct a regular 3∗-container C3∗(x, y) = {P1, P2, P3} in HReT(4, n) − {z}. By the symmetric property of HReT(4, n), we may assume that i∈ {0, 1, 2}. We have the following cases:
Case 1: Suppose that i∈ {0, 1} and z ∈ {0, 1}. By Lemma 14, there exists a regular 3∗-container C3∗((0, 0), (i, j)) of HReT(2, n)− {(k, l)}. By Lemma 12, C3∗,1((0, 0), (i, j)) forms a 3∗-container of HReT(4, n)− {(k, l)}.
Case 2: i = 0 and k = 2. Then j and l are even. By the symmetric property, we have the following subcases.
Case 2.1: Suppose that j = 4 and l = 2. The corresponding paths are:
P1 = (0, 0), Q0(0, 4), (0, 4);
P2 = (0, 0), (0, n − 1), (0, n − 2), (3, n − 2), Q−13 (4, n− 2), (3, 4), (0, 4);
P3 = (0, 4), Q0(4, n− 3), (0, n − 3), (1, n − 3), Q−11 (0, n− 3), (1, 0), (1, n − 1), (1, n− 2), (2, n − 2), Q−12 (3, n− 2), (2, 3), (3, 3), (3, 2), (3, 1), (2, 1), (2, 0), (2, n− 1), (3, n − 1), (3, 0), (0, 0).
Case 2.2: Suppose that n− 4 > j ≥ 2 and l = j + 2. The corresponding paths are:
P1 = (0, 0), Q0(0, j), (0, j);
P2 = (0, 0), (3, 0), Q3(0, j), (3, j), (0, j);
P3 = (0, j), Q0(j, j + 4), (0, j + 4), (3, j + 4), (3, j + 5), (2, j + 5), (2, j + 4), (2, j + 3), (3, j + 3), (3, j + 2), (3, j + 1), (2, j + 1), Q−12 (0, j + 1), (2, 0), (1, 0), Q1(0, j + 5), (1, j + 5), (0, j + 5), (0, j + 6), (3, j + 6), (3, j + 7), (2, j + 7), (2, j + 6),
S2L(j + 6, n− 2), (2, n − 2), (1, n − 2), (1, n − 1), (0, n − 1), (0, 0).
Case 2.3: Suppose that n− 6 > j ≥ 2 and n − 4 > l > j + 2. The corresponding paths are:
P1 = (0, 0), Q0(0, j), (0, j);
P2 = (0, 0), (3, 0), Q3(0, j), (3, j), (0, j);
P3 = (1, j), (1, j + 1), (1, j + 2), (3, j + 2), (3, j + 1), (2, j + 1), Q−12 (0, j + 1), (2, 0), (1, 0), Q1(0, j + 2), (1, j + 2), (2, j + 2), (2, j + 3), (3, j + 3), (3, j + 4), (0, j + 4), (0, j + 3), S0R(j + 3, l− 3), (0, l − 3), (1, l − 3), (1, l − 2), (2, l − 2), (2, l − 1), (3, l− 1), (3, l), (3, l + 1), (2, l + 1), (2, l + 2), (2, l + 3), (3, l + 3), (3, l + 2), (0, l + 2), Q−10 (l− 1, l + 2), (0, l − 1), (1, l − 1), Q1(l− 1, l + 3), (1, l + 3), (0, l + 3), (0, l + 4), (3, l + 4), S2L(l + 4, n− 2), (2, n − 2), (1, n − 2), (1, n − 1), (0, n− 1), (0, 0).
Case 2.4: Suppose that n > 8 and j = l≥ 2. The corresponding paths are:
P1 = (0, 0), Q0(0, j), (0, j);
P2 = (0, 0), (3, 0), Q3(0, j− 1), (3, j − 1), (2, j − 1), Q−12 (0, j− 1), (2, 0), (1, 0), Q1(0, j + 1), (1, j + 1), (0, j + 1), (0, j);
P3 = (0, j), (3, j), (3, j + 1), (2, j + 1), (2, j + 2), (1, j + 2), (1, j + 3), (1, j + 4), (2, j + 4), (2, j + 3), (3, j + 3), (3, j + 2), (0, j + 2), (0, j + 3), (0, j + 4), (3, j + 4), (3, j + 5), (2, j + 5), (2, j + 6), (1, j + 6), (1, j + 5), S1L(j + 5, n− 5), (1, n − 5), (0, n − 5), (0, n− 4), (3, n − 4), (3, n − 3), (2, n − 3), (2, n − 2), (2, n − 1), (3, n − 1), (3, n − 2), (0, n− 2), (0, n − 3), (1, n − 3), (1, n − 2), (1, n − 1), (0, n − 1), (0, 0).
Case 2.5: Suppose that n = 8, j = 2, and l = 2. The corresponding paths are:
P1 = (0, 0), (0, 1), (0, 2);
P2 = (0, 2), (0, 3), (0, 4), (3, 4), Q3(4, 7), (3, 7), (2, 7), (2, 0), (2, 1), (3, 1), (3, 0), (0, 0);
P3 = (0, 2), (3, 2), (3, 3), (2, 3), Q2(3, 6), (2, 6), (1, 6), (1, 7), (1, 0), Q1(0, 5), (1, 5), (0, 5), (0, 6), (0, 7), (0, 0).
Case 2.6: Suppose that n = 8, j = 4, and l = 4. The corresponding paths are:
P1 = (0, 0), Q0(0, 4), (0, 4);
P2 = (0, 0), (0, 7), (1, 7), (1, 0), Q1(0, 6), (1, 6), (2, 6), (2, 5), (3, 5), (3, 4), (0, 4);
P3 = (0, 0), (3, 0), Q3(0, 3), (3, 3), (2, 3), Q−12 (0, 3), (2, 0), (2, 7), (3, 7), (3, 6), (0, 6), (0, 5), (0, 4).
Case 3: i = 1 and k = 2. Then j is odd and l is even. By the symmetric property, we have the following subcases.
Case 3.1: Suppose that n− 5 > j ≥ 1 and n − 4 > l > j + 2. The corresponding paths are:
P1 = (0, 0), Q0(0, j), (0, j), (1, j);
P2 = (0, 0), (3, 0), Q3(0, j), (3, j), (2, j), Q−12 (0, j), (2, 0), (1, 0), Q1(0, j), (1, j);
P3 = (1, j), (1, j + 1), (2, j + 1), (2, j + 2), (3, j + 2), (3, j + 1), S3L(j + 1, l− 2), (3, l − 2), (0, l− 2), (0, l − 1), (1, l − 1), (1, l), (1, l + 1), (1, l + 2), (2, l + 2), (2, l + 1), (3, l + 1), (3, l), (0, l), (0, l + 1), (0, l + 2), (3, l + 2), (3, l + 3), (2, l + 3), (2, l + 4), (1, l + 4), (1, l + 3), S1L(l + 3, n− 5), (1, n − 5), (0, n − 5), (0, n − 4), (3, n − 4), (3, n − 3), (2, n− 3), (2, n − 2), (2, n − 1), (3, n − 1), (3, n − 2), (0, n − 2), (0, n − 3), (1, n − 3), (1, n− 2), (1, n − 1), (0, n − 1), (0, 0).
Case 3.2: Suppose that n− 5 > j ≥ 1 and l = j + 1. The corresponding paths are:
P1 = (0, 0), Q0(0, j), (0, j), (1, j);
P2 = (0, 0), (3, 0), Q3(0, j), (3, j), (2, j), Q−12 (0, j), (2, 0), (1, 0), Q1(0, j), (1, j);
P3 = (1, j), Q1(j, j + 3), (1, j + 3), (2, j + 3), (2, j + 2), (3, j + 2), (3, j + 1), (0, j + 1), (0, j + 2), (0, j + 3), (3, j + 3), (3, j + 4), (2, j + 4), (2, j + 5), (1, j + 5), (1, j + 4), S1L(j + 4, n− 5), (1, n − 5), (0, n − 5), (0, n − 4), (3, n − 4), (3, n − 3), (2, n − 3), (2, n− 2), (2, n − 1), (3, n − 1), (3, n − 2), (0, n − 2), (0, n − 3), (1, n − 3), (1, n − 2), (1, n− 1), (0, n − 1), (0, 0).
Case 3.3: Suppose that n− 5 > j ≥ 1 and l = n − 4. The corresponding paths are:
P1 = (0, 0), Q0(0, j), (0, j), (1, j);
P2 = (0, 0), (0, n − 1), (1, n − 1), (1, 0), Q1(0, j), (1, j);
P3 = (1, j), (1, j + 1), (2, j + 1), (2, j + 2), (3, j + 2), (3, j + 1), S3L(j + 1, n− 6), (0, n − 6), (0, n− 5), (1, n − 5), Q1(n− 5, n − 2), (1, n − 2), (2, n − 2), (2, n − 3), (3, n − 3), (3, n− 4), (0, n − 4), (0, n − 3), (0, n − 2), (3, n − 2), (3, n − 1), (2, n − 1), (2, 0), (2, 1), (3, 1), (3, 0), (0, 0).
Case 3.4: Suppose that j = n− 5 and l = n − 4. The corresponding paths are:
P1 = (0, 0), Q0(0, n− 5), (0, n − 5), (1, n − 5);
P2 = (0, 0), (0, n − 1), (1, n − 1), (1, 0), Q1(0, n− 5), (1, n − 5);
P3 = (1, n − 5), Q1(n− 5, n − 2), (1, n − 2), (2, n − 2), (2, n − 3), (3, n − 3), (3, n − 4), (0, n− 4), (0, n − 3), (0, n − 2), (3, n − 2), (3, n − 1), (2, n − 1), (2, 0), Q2(0, n− 5), (2, n− 5), (3, n − 5), Q−13 (0, n− 5), (3, 0), (0, 0).
Case 3.5: Suppose that n− 5 > j ≥ 1 and l = n − 2. The corresponding paths are:
P1 = (0, 0), Q0(0, j), (0, j), (1, j);
P2 = (0, 0), (3, 0), (3, n − 1), (2, n − 1), (2, 0), (1, 0), Q1(0, j), (1, j);
P3 = (1, j), (1, j + 1), (1, j + 2), (0, j + 2), (0, j + 1), (3, j + 1), Q−13 (1, j + 1), (3, 1), (2, 1), Q2(1, j + 2), (2, j + 2), (3, j + 2), (3, j + 3), (0, j + 3), (0, j + 4), (1, j + 4), (1, j + 3), S1R(j + 3, n− 6), (1, n − 6), (2, n − 6), (2, n − 5), (3, n − 5), (3, n − 4), (0, n − 4), (0, n− 3), (0, n − 2), (3, n − 2), (3, n − 3), (2, n − 3), (2, n − 4), (1, n − 4),
Q1(n− 4, n − 1), (1, n − 1), (0, n − 1), (0, 0).
Case 4: i = 2 and k = 0. Then j and l are even. By the symmetric property, we have the following subcases.
Case 4.1: Suppose that j = 0 are l > 0. The corresponding paths are:
P1 = (0, 0), (0, n − 1), (1, n − 1), (1, 0), (2, 0);
P2 = (0, 0), (0, 1), (1, 1), (1, 2), (2, 2), (2, 1), (2, 0);
P3 = (0, 0), (3, 0), (3, 1), (3, 2), (0, 2), (0, 3), (1, 3), (1, 4), (2, 4), (2, 3),
S2R(3, j− 1), (2, j − 1), (3, j − 1), (3, j), (3, j + 1), (2, j + 1), (2, j + 2), (1, j + 2), (1, j + 1), S1L(j + 1, n− 3), (1, n − 3), (0, n − 3), (0, n − 2), (3, n − 2), (3, n − 1), (2, n− 1), (2, 0).
Case 4.2: Suppose that l > j > 0. The corresponding paths are:
P1 = (0, 0), (0, 1), (1, 1), Q1(1, j), (1, j), (2, j);
P2 = (0, 0), (3, 0), (3, 1), (2, 1), Q2(1, j), (2, j);
P3 = (2, j), (2, j + 1), (2, j + 2), (1, j + 2), (1, j + 1), (0, j + 1), Q−10 (2, j + 1), (0, 2), (3, 2), Q3(2, j + 2), (3, j + 2), (0, j + 2), (0, j + 3), (1, j + 3), (1, j + 4), (2, j + 4), (2, j + 3), S2R(j + 3, l− 1), (2, l − 1), (3, l − 1), (3, l), (3, l + 1),
(2, l + 1), (2, l + 2), (1, l + 2), (1, l + 1), S1L(l + 1, n− 1), (1, n − 1), (0, n − 1), (0, 0).
Case 4.3: Suppose that j = l > 0. The corresponding paths are:
P1 = (0, 0), Q0(0, j− 1), (0, j − 1), (1, j − 1), Q−11 (0, j− 1), (1, 0), (2, 0), Q2(0, j), (2, j);
P2 = (0, 0), (3, 0), Q3(0, j + 1), (3, j + 1), (2, j + 1), (2, j);
P3 = (2, j), S2L(j, n− 1), (2, n − 2), (1, n − 2), (1, n − 1), (0, n − 1), (0, 0).
Case 5: i = 2 and k = 1. Then j is even and l is odd. By the symmetric property, we have the following subcases.
Case 5.1: Suppose that j = 0 and l = 1. The corresponding paths are:
P1 = (0, 0), (0, n − 1), (1, n − 1), (1, 0), (2, 0);
P2 = (0, 0), (0, 1), (0, 2), (3, 2), (3, 1), (2, 1), (2, 0);
P3 = (0, 0), (3, 0), (3, n − 1), (3, n − 2), (0, n − 2), Q−10 (3, n− 2), (0, 3), (1, 3), (1, 2), (2, 2), (2, 3), (3, 3), Q3(3, n− 3), (3, n − 3), (2, n − 3), Q−12 (4, n− 3), (2, 4), (1, 4), Q1(4, n− 2), (1, n − 2), (2, n − 2), (2, n − 1), (2, 0).
Case 5.2: Suppose that j = 0 and n− 1 > l > 1. The corresponding paths are:
P1 = (0, 0), (0, n − 1), (1, n − 1), (1, 0), (2, 0);
P2 = (0, 0), (3, 0), (3, 1), (2, 1), (2, 0);
P3 = (0, 0), (0, 1), (1, 1), (1, 2), (2, 2), (2, 3), (3, 3), (3, 2), S3L(2, j− 3), (3, j − 3), (0, j − 3), (0, j− 2), (1, j − 2), (1, j − 1), (2, j − 1), (2, j), (2, j + 1), (1, j + 1), (1, j + 2), (1, j + 3), (2, j + 3), (2, j + 2), (3, j + 2), Q−13 (j− 1, j + 2), (3, j − 1), (0, j − 1), Q0(j− 1, j + 3), (0, j + 3), (3, j + 3), (3, j + 4), (2, j + 4), (2, j + 5), (1, j + 5), (1, j + 4), S1L(j + 4, n− 3), (1, n − 3), (0, n − 3), (0, n − 2), (3, n − 2), (3, n − 1), (2, n− 1), (2, 0).
Case 5.3: Suppose that n− 1 > l > j + 2 and j > 0. The corresponding paths are:
P1 = (0, 0), (0, 1), (1, 1), Q1(1, j), (1, j), (2, j);
P2 = (0, 0), (3, 0), (3, 1), (2, 1), Q2(1, j), (2, j);
P3 = (2, j), (2, j + 1), (3, j + 1), (3, j), S3L(j, l− 3), (3, l − 3), (0, l − 3), (0, l − 2), (1, l − 2), (1, l− 1), (2, l − 1), (2, l), (2, l + 1), (1, l + 1), (1, l + 2), (1, l + 3), (2, l + 3), (2, l + 2), (3, l + 2), (3, l + 1), (3, l), (3, l− 1), (0, l − 1), Q0(l− 1, l + 3), (0, l + 3), (3, l + 3), (3, l + 4), (2, l + 4), (2, l + 5), (1, l + 5), (1, l + 4), S1L(l + 4, n− 1), (1, n − 1), (0, n− 1), (0, 0).
Case 5.4: Suppose that n− 2 > j and l = n − 1. The corresponding paths are:
P1 = (0, 0), Q0(0, j + 1), (0, j + 1), (1, j + 1), (1, j + 2), (2, j + 2), (2, j + 1), (2, j);
P2 = (0, 0), (3, 0), (3, n − 1), (2, n − 1), (2, 0), (1, 0), Q1(0, j), (1, j), (2, j);
P3 = (2, j), Q−12 (1, j), (2, 1), (3, 1), Q3(1, j + 2), (3, j + 2), (0, j + 2), (0, j + 3), (1, j + 3), (1, j + 4), (2, j + 4), (2, j + 3), S2R(j + 3, n− 3), (2, n − 3), (3, n − 3), (3, n − 2), (0, n− 2), (0, n − 1), (0, 0).
Case 5.5: Suppose that j = n− 2 and l = n − 1. The corresponding paths are:
P1 = (0, 0), (0, n − 1), (0, n − 2), (0, n − 3), (1, n − 3), (1, n − 2), (2, n − 2);
P2 = (0, 0), Q0(0, n− 4), (3, n − 4), Q3(n− 4, n − 1), (2, n − 1), (2, n − 2);
P3 = (0, 0), (3, 0), Q3(0, n− 5), (3, n − 5), (2, n − 5), Q−12 (0, n− 5), (2, 0), Q1(0, n− 4), (1, n− 4), (2, n − 4), (2, n − 3), (2, n − 2).
Hence HReT(4, n) is hyper globally bi-3∗-connected for n ≥ 8. See Figure 4.7 for illustrations.
Case 2.1 Case 2.2 Case 2.3 Case 2.4 Case 2.5 Case 2.6 Case 3.1
Case 3.2 Case 3.3 Case 3.4 Case 3.5 Case 4.1 Case 4.2
Case 4.3 Case 5.1 Case 5.2 Case 5.3 Case 5.4 Case 5.5
x x x x
x x
x
x x x x x x
x x x x x x
y
y
y
y y
y
y
y
y y
y
y y
y
y y
y y
y z
z
z
z z
z z
z
z z
z
z z
z
z
z z
z z
Figure 4.7: Illustrations for Lemma 17.
4.5 The Globally Bi- 3
∗-Connected Property of Hon-eycomb Rectangular Torus HReT(m,n)
Lemma 18. Assume that m and n are positive even integers with m, n ≥ 4. Let x and y be any two vertices of HReT(m, n) = (V0∪ V1, E) with x∈ V0 and y ∈ V1. Then there exists a regular 3∗-container C3∗(x, y) of HReT(m, n).
Proof. Without loss of generality, we may assume that x = (0, 0) and y = (i, j). In order to prove this lemma, we will construct a regular 3∗-container C3∗(x, y) = {P1, P2, P3} in HReT(m, n). We prove the lemma by induction on m. With Lemma 15, our theorem holds for m = 4. Now, we consider the case that m≥ 6.
Suppose that i < m− 2. By induction, there exists a regular 3∗-container C3∗(x, y) = {P1, P2, P3} in HReT(m − 2, n). By Lemma 12, C3∗,m−3((0, 0), (i, j)) forms a 3∗-container of HReT(m, n). Suppose that i≥ m − 2. By induction, there exists a regular C3∗(x, (i− 2, j)) = {P1, P2, P3} in HReT(m − 2, n). By Lemma 12, C3∗,1((0, 0), (i, j)) forms a 3∗ -container of HReT(m, n).
Lemma 19. Assume that m and n are positive even integers with m≥ 4 and n ≥ 6. Let x, y, and z be any three different vertices of HReT(m, n) = (V0∪ V1, E) in V0. Then there exists a regular 3∗-container C3∗(x, y) of HReT(m, n)− {z}.
Proof. Without loss of generality, we may assume that x = (0, 0), y = (i, j), and z = (k, l). In order to prove this lemma, we will construct a regular 3∗-container C3∗(x, y) = {P1, P2, P3} in HReT(m, n)−{z}. We prove the lemma by induction on m. With Lemmas 16 and 17, our theorem holds for m = 4. Now, we consider the case that m ≥ 6.
Suppose that i < m−2 and k < m−2. By induction, there exists a regular 3∗-container C3∗(x, y) = {P1, P2, P3} in HReT(m − 2, n) − {z}. By Lemma 12, C3∗,m−3((0, 0), (i, j)) forms a 3∗-container of HReT(m, n)− {z}. Suppose that i < m − 2 and k ≥ m − 2. By induction, there exists a regular 3∗-container C3∗(x, y) ={P1, P2, P3} in HReT(m−2, n)−
(k − 2, l). By Lemma 12, C3∗,i((0, 0), (i, j)) forms a 3∗-container of HReT(m, n)− {z}.
Suppose that i ≥ m − 2 and k < m − 2. By induction, there exists a regular 3∗ -container C3∗(x, (i − 2, j)) = {P1, P2, P3} in HReT(m − 2, n) − {z}. By Lemma 12, C3∗,k((0, 0), (i, j)) forms a 3∗-container of HReT(m, n)− {z}. Suppose that i ≥ m − 2 and k ≥ m−2. By induction, there exists a regular 3∗-container C3∗(x, (i−2, j)) = {P1, P2, P3} in HReT(m− 2, n) − (k − 2, l). By Lemma 12, C3∗,1((0, 0), (i, j)) forms a 3∗-container of HReT(m, n)− {z}.
Theorem 13. Assume that m and n are positive even integers with n ≥ 4. Then
x y z
Figure 4.8: Illustration for Theorem 13.
HReT(m, n) is strongly globally bi-3∗-connected. Moreover, HReT(m, n) is hyper glob-ally bi-3∗-connected if and only if n≥ 6 or m = 2.
Proof. With Lemmas 13 and 18, HReT(m, n) is globally bi-3∗-connected if m, n are even integers with n ≥ 4.
By Lemmas 14 and 19, HReT(m, n) is hyper globally bi-3∗-connected if m, n are even integers with n ≥ 6 or m = 2.
Now we consider the case HReT(m, 4) with m is an even integer and m≥ 4. We first prove that such HReT(m, 4) is not hyper globally bi-3∗-connected.
To prove this fact, let x = (1, 1), y = (1, 3) and z = (0, 2). Suppose that there exists a 3∗-container C3∗(x, y) ={P1, P2, P3} of HReT(m, 4)−{z}. Since degHReT (m,4)−z(v) = 2 for v ∈ {(0, 1), (0, 3), (3, 2)}, (1, 1), (1, 2), (1, 3) and (1, 1), (0, 1), (0, 0), (0, 3), (1, 3) are two paths in C3∗(x, y). Without loss of generality, we assmue that P1 = (1, 1), (1, 2), (1, 3)
and P2 = (1, 1), (0, 1), (0, 0), (0, 3), (1, 3). Since degHReT (m,4)−z((1, 1)) = degHReT (m,4)−z
((1, 3)) = 3, ((1, 3), (1, 0)) and ((1, 0), (1, 1)) are edges in P3. Thus P3 = (1, 1), (1, 0), (1, 3). Obviously, {P1∪ P2∪ P3} does not span HReT(m, 4) − {z}. See Figure 4.8 for an illustration. Hence HReT(m, 4) is not hyper globally bi-3∗-connected.
Although any HReT(m, 4) with m is an even integer and m≥ 4 is not hyper globally bi-3∗-connected, we will prove that such HReT(m, 4) is strongly globally bi-3∗-connected by induction.
We first prove that HReT(4,4) is strongly bi-3∗-connected. Let x and y be any two different vertices in the same partite set of HReT(4,4). Without loss of generality, we may
assume that x and y are vertices in V0 and x = (0, 0). We need to find a vertex z in V0− {x, y} such that there exists a 3∗-container C3∗(x, y) = {P1, P2, P3} of HReT(4,4)−{z}.
The corresponding vertex z and 3∗-container C3∗(x, y) are listed below.
y z C3∗(x, y)
(0, 2) (1, 3) (0, 0), (0, 1), (0, 2)
(0, 0), (0, 3), (0, 2)
(0, 0), (3, 0), (3, 1), (2, 1), (2, 0), (1, 0), (1, 1), (1, 2), (2, 2), (2, 3), (3, 3), (3, 2), (0, 2)
(1, 1) (1, 3) (0, 0), (0, 1), (1, 1)
(0, 0), (3, 0), (3, 1), (2, 1), (2, 0), (1, 0), (1, 1),
(0, 0), (0, 3), (0, 2), (3, 2), (3, 3), (2, 3), (2, 2), (1, 2), (1, 1)
(1, 3) (0, 2) (0, 0), (0, 3), (1, 3)
(0, 0), (0, 1), (1, 1), (1, 2), (1, 3)
(0, 0), (3, 0), Q3(0, 3), (3, 3), (2, 3), Q−12 (0, 3), (2, 0), (1, 0), (1, 3)
(2, 0) (0, 2) (0, 0), (0, 3), (1, 3), (1, 0), (2, 0)
(0, 0), (3, 0), (3, 3), (3, 2), (3, 1), (2, 1), (2, 0)
(0, 0), (0, 1), (1, 1), (1, 2), (2, 2), (2, 3), (2, 0)
(2, 2) (0, 2) (0, 0), (3, 0), Q3(0, 3), (3, 3), (2, 3), (2, 2)
(0, 0), (0, 3), (1, 3), (1, 0), (2, 0), (2, 1), (2, 2)
(0, 0), (0, 1), (1, 1), (1, 2), (2, 2)
(3, 1) (0, 2) (0, 0), (3, 0), (3, 1)
(0, 0), (0, 3), (1, 3), (1, 0), (2, 0), (2, 1), (3, 1)
(0, 0), (0, 1), (1, 1), (1, 2), (2, 2), (2, 3), (3, 3), (3, 2), (3, 1)
(3, 3) (0, 2) (0, 0), (3, 0), (3, 3)
(0, 0), (0, 3), (1, 3), (1, 0), (2, 0), (2, 1), (3, 1), (3, 2), (3, 3)
(0, 0), (0, 1), (1, 1), (1, 2), (2, 2), (2, 3), (3, 3)
Obviously, all these 3∗-containers of HReT(4,4)−{z} are regular.
Now we consider the case HReT(m, 4) with m > 4. Without loss of generality, we may assmue that x = (0, 0), y = (i, j), and z = (k, l). Suppose that i < m − 2 and k < m− 2. By induction, there exists a regular 3∗-container C3∗(x, y) = {P1, P2, P3} in HReT(m− 2, 4) − {z}. By Lemma 12, C3∗,m−3((0, 0), (i, j)) forms a 3∗-container of HReT(m, 4) − {z}. Suppose that i < m − 2 and k ≥ m − 2. By induction, there exists a regular 3∗-container C3∗(x, y) = {P1, P2, P3} in HReT(m − 2, 4) − (k − 2, l). By Lemma 12, C3∗,i((0, 0), (i, j)) forms a 3∗-container of HReT(m, 4) − {z}. Suppose that i ≥ m − 2 and k < m − 2. By induction, there exists a regular C3∗(x, (i − 2, j)) = {P1, P2, P3} in HReT(m − 2, 4) − {z}. By Lemma 12, C3∗,k((0, 0), (i, j)) forms a 3∗ -container of HReT(m, 4)−{z}. Suppose that i ≥ m−2 and k ≥ m−2. By induction, there exists a regular 3∗-container C3∗(x, (i−2, j)) = {P1, P2, P3} in HReT(m−2, 4)−(k −2, l).
By Lemma 12, C3∗,1((0, 0), (i, j)) forms a 3∗-container of HReT(m, 4)− {z}.
Thus the theorem is proved.
Chapter 5 Conclusion
There are a lot of studies on hamiltonian graphs. In this thesis, we are interested in some specific types of hamiltonian graphs. We introduce the concept of mutually independent hamiltonicity first. The concept of mutually independent hamiltonian arises from the following applications. If there are k pieces of data needed to be sent from u to v, and the data needed to be processed at every vertex, then we want mutually independent hamiltonian paths so that there will be no waiting time at a processor. Thus the mutually independent hamiltonian property is useful for communication algorithms. In chapter 2, we are interested in two families of graphs. The first family of graphs are those graphs with
¯
e ≤ n − 4 and n ≥ 4. It was proved [37] that such graphs are hamiltonian connected. In Theorem 6, we strengthen this classical result by proving that there are at least n− 2 − ¯e mutually independent hamiltonian paths between every pair of distinct vertices of G.
The second family of graphs are those graphs with the sum of the degree of any two non-adjacent vertices being at least n + 1. Assume that G is a graph with the sum of any two non-adjacent vertices being at least n + 2. Let u and v be any two distinct vertices of G. In Theorem 7, we show that there are degG(u) + degG(v)− n mutually independent hamiltonian paths between u and v if (u, v)∈ E(G), and there are degG(u)+degG(v)−n+2 mutually independent hamiltonian paths between u and v if otherwise.
In chapter 3, we proposed a new concept called panpositionable hamiltonicity. We showed that the arrangement graph An,k is panpositionable hamiltonian if k ≥ 1 and n−k ≥ 2 in Theorem 10. By applying this result, we can prove that An,k is panconnected and pancyclic if k ≥ 1 and n − k ≥ 2. We also explained some relationship between the panpositionable hamiltonian property and the panconnected property by giving an example to show that a panconnected graph G is not necessarily panpositionable hamil-tonian. Therefore, the panpositionable hamiltonian property is a stronger property for an interconnection network.
The honeycomb networks have been proposed as attractive alternatives to mesh and torus interconnection networks for computer architectures, interconnection topologies, parallel processes and distributed systems. In particular, the honeycomb rectangular torus HReT(m, n) is a well-structured 3-connected cubic network. In chapter 4, we study the globally bi-3∗-connected property of the honeycomb rectangular torus HReT(m, n).
We have proved that any HReT(m, n) is strongly globally bi-3∗-connected. We also proved that HReT(m, n) is hyper globally bi-3∗-connected if and only if n≥ 6 or m = 2.
Future work will be directed to explore the mutually independent hamiltonicity and the panpositionable hamiltonicity of other interconnection networks. Moreover, we will try to find the globally 3∗-connected property of other cubic interconnection networks. It would be interesting to study some relationship between these specific properties, such as panpositionable hamiltonicity, panconnectivity and pancyclicity, and the other criteria for measuring the performance of a network.
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