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Proceeding Paper

One-Node One-Edge Dimension-Balanced Hamiltonian Problem on Toroidal Mesh Graph †

by
Yancy Yu-Chen Chang
and
Justie Su-Tzu Juan
*
Department of Computer Science & Information Engineering, National Chi Nan University, Nantou 54561, Taiwan
*
Author to whom correspondence should be addressed.
Presented at the 2024 IEEE 7th International Conference on Knowledge Innovation and Invention, Nagoya, Japan, 16–18 August 2024.
Eng. Proc. 2025, 89(1), 17; https://doi.org/10.3390/engproc2025089017
Published: 23 February 2025

Abstract

Given a graph G = (V, E), the edge set can be partitioned into k dimensions, for a positive integer k. The set of all i-dimensional edges of G is a subset of E(G) denoted by Ei. A Hamiltonian cycle C on G contains all vertices on G. Let Ei(C) = E(C) ∩ Ei. For any 1 ≤ ik, C is called a dimension-balanced Hamiltonian cycle (DBH, for short) on G if ||Ei(C)| − |Ej(C)|| ≤ 1 for all 1 ≤ i < jk. The dimension-balanced cycle problem is generated with the 3-D scanning problem. Graph G is called p-node q-edge dimension-balanced Hamiltonian (p-node q-edge DBH) if it has a DBH after removing any p nodes and any q edges. G is called h-fault dimension-balanced Hamiltonian (h-fault DBH, for short) if it remains Hamiltonian after removing any h node and/or edges. The design for the network-on-chip (NoC) problem is important. One of the most famous NoC is the toroidal mesh graph Tm,n. The DBC problem on toroidal mesh graph Tm,n is appropriate for designing simple algorithms with low communication costs and avoiding congestion. Recently, the problem of a one-fault DBH on Tm,n has been studied. This paper solves the one-node one-edge DBH problem in the two-fault DBH problem on Tm,n.

1. Introduction

With the popularity of the Internet, the topology of its network allows for the improvement of the performance of the entire system. According to the connection structure of this network, graphs are used to display its topology. Vertices represent devices, and edges between two vertices represent communication devices. The path between the vertices in the graph shows data transmission and communication between the two. A Hamiltonian cycle connects all vertices and returns to the original vertex to ensure the smoothness of the network [1]. However, when vertices or edges fail, the Hamilton cycle is broken, and data cannot be transferred within it. This causes significant inconvenience. It is important to ensure that there are new routes in the network to provide data transfer between them. Therefore, it has been studied whether the Hamiltonian cycle exists when there are faulty vertices or edges in the network graph [2,3].
For a graph G = (V, E), the edge set E is partitioned into k dimensions {E1, E2, …, Ek} for a positive integer k. For any cycle C on G, the set of all i-dimensional edges of C, which is a subset of E(C), is denoted by Ei(C). If ||Ei(C)| − |Ej(C)|| ≤ 1 for 1 ≤ i < jk, C is called a dimension-balanced cycle or DBC for short. If C is a Hamiltonian cycle on graph G and C is also a DBC, then C is called a DBH of G. The dimension-balanced cycle problem is generated with the 3-D scanning problem [4]. Research results on the issues of DBCs and DBHs have been obtained in recent years [5,6,7].
A graph is called a toroidal mesh graph, denoted by Tm,n, where V(Tm,n) = {(x, y) | 0 ≤ xm − 1, 0 ≤ yn − 1} and the edge set E(Tm,n) = {(x1, y1)(x2, y2) | x1 = x2 and |y1y2| = (1 or n − 1), or |x1x2| = (1 or m − 1) and y1 = y2}. In other words, Tm,n is viewed as the Cartesian product of cycles Cm and Cn. In any toroidal mesh graph Tm,n, let E1 = {(x1, y1)(x2, y2) | |x1x2| = (1 or m − 1) and y1 = y2} and E2 = {(x1, y1)(x2, y2) | x1 = x2 and |y1y2| = (1 or n − 1)}. Figure 1 shows an example of T4,3. The design for the network-on-chip (NoC) problem has been an important issue recently. One of the most famous NoC problems is actually the toroidal mesh graph Tm,n.
Figure 1. T4,3.
Figure 1. T4,3.
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For the NoC design problem, how to choose a suitable routing path to prevent congestion and good performance is important. The DBC problem on toroidal mesh graph Tm,n is appropriate for designing simple algorithms with low communication costs, and it will avoid the situation of congestion. A graph G is called p-node q-edge dimension-balanced Hamiltonian (p-node q-edge DBH for short) if it has a DBH after removing any p nodes and any q edges. A graph G is called h-fault dimension-balanced Hamiltonian (h-fault DBH for short) if it remains Hamiltonian after removing any h node and/or edges [2]. Many researchers have studied the Hamiltonian problem on several topologies with faulty nodes or edges [3,8,9,10].
It has been proved that for any F with |F| ≤ 2, Tm,nF has a Hamiltonian cycle if m ≥ 3, n ≥ 3, and n is odd, where F is a subset of the union of the node set and the edge set of the graph Tm,n [2]. Recently, the problem of a one-fault DBH on Tm,n has been studied [11]. Therefore, this study investigated the two-fault DBH problem on Tm,n, on the one-node one-edge DBH problem.
The rest of this paper is organized as follows: Section 2 gives preliminaries, Section 3 shows the main results, and Section 4 is the conclusion.

2. Preliminaries

In this section, we will first state some relevant theorems and definitions as a starting point for this study. For any vertex (i, j) in a toroidal mesh graph Tm,n, (i, j) is called black if i + j is even; otherwise, (i, j) is called white. If m and n are both even, every edge joins two vertices with different colors. This means Tm,n is a bipartite graph. Therefore, there is no odd cycle on Tm,n if both m and n are even, so Tm,nF has no Hamiltonian cycle when F is a set of one node. We have the following theorems.
Theorem 1
[11]. When m and n are both even, Tm,n is not one-node DBH.
This is because if there is no DBH on Tm,nF1 when F1 is a set of one node, then there is also no DBH on Tm,nF2 when F2 is a set of one node and one edge. The following theorem can be derived.
Theorem 2.
When m and n are both even, Tm,n is not two-fault DBH because Tm,n − F has no Hamiltonian cycle when F is a set of one node and one edge.
A vertex-transitive graph, or a node symmetric graph, is a graph such that every pair of vertices is equivalent under some element of its automorphism group [12]. It is not difficult to see that a toroidal mesh graph Tm,n is a vertex-transitive graph.

3. Main Results

This study explores the one-node one-edge DBH problem on Tm,n. Because of Theorem 2, we discuss this problem only for when m and n are not both even: (1) when one of m and n is odd and the other is even and (2) when m and n are both odd numbers. When one of m and n are is odd and the other is even, generality is decreased, assuming that m is an odd number and n is an even number. At first, Theorem 3 was used to analyze the situation when m = 3, and then Theorem 4 was used to analyze the situation when m > 3 based on the DBH on T5,4 with one faulty node and one faulty edge.
Theorem 3.
There is a DBH on T3,3 − F and T3,4 − F; however, when n > 4, there is no DBH on T3,n − F, where F is a set of any one node and one edge.
Proof. 
Figure 2 shows two DBHs on T3,3F and T3,4F, respectively. Since Tm,n is a vertex-transitive graph, the red vertex v is the faulty vertex (vF) without loss of generality. Due to the characteristics of vertex transitivity, the graphs of Figure 2 can be flipped and/or translated so that the faulty vertex is fixed at the original position. Six different extended DBHs are produced. Figure 3 shows those DBHs. Each edge does not exist on at least one DBH on T3,3F and T3,4F.
n > 4 is considered in Figure 4. When the fault set F = {(1, 0), (0, 0)(2, 0)}, for vertices (0, 0) and (2, 0), only these four edges drawn in Figure 3 are chosen. If DBH C exists in T3,nF, at most ⌈(3n − 1)/2⌉ − 4 horizontal edges can be selected in C. At least 2(n − 3) horizontal edges are needed in C to connect all vertices. This is because 2(n − 3) > ⌈(3n − 1)/2⌉ − 4 when n > 5. Therefore, it can be proven that when n > 4, there is no DBH on T3,nF. □
Figure 2. (a,b) Two DBHs on T3,3F; (c,d) two DBHs on T3,4F.
Figure 2. (a,b) Two DBHs on T3,3F; (c,d) two DBHs on T3,4F.
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Figure 3. (af) Six different extended DBHs of Figure 2c,d.
Figure 3. (af) Six different extended DBHs of Figure 2c,d.
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Figure 4. If F = {(1, 0), (0, 0)(2, 0)}, the edges must be used.
Figure 4. If F = {(1, 0), (0, 0)(2, 0)}, the edges must be used.
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Lemma 1.
For any T5,4 − F, there exists a DBH, where F is a set of any one node and one edge.
Proof. 
Since Tm,n is a vertex-transitive graph, without loss of generality, the faulty node is (2, 0) (the red node in Figure 5). Next, all possible cases where edges are faulty edges are considered. Figure 5a proves that when the faulty edges are edges not used in the graph, T5,4F exists DBH; the faulty edges that Figure 5a cannot solve are shown in Figure 5b. When a faulty edge occurs on the edges in Figure 5a, it is replaced with Figure 5b. □
Since the graph is vertex-transitive, we remove the faulty edge at the positions of unused edges through rotation, reflection, and translation. When m > 5 is odd and/or n > 4 is even, we perform two operations on Figure 5 to obtain a DBH of Tm,n − F recursively: (1) expanding one column to both the left and right of Tm−2,4 and (2) expanding two rows downward of Tm,n−2.
The following theorem is proved according to these two operations. For convenience, in the following figures, the red dashed lines represent the deleted edges, and the blue lines represent the added edges.
Theorem 4.
When m ≥ 5 is an odd number and n ≥ 4 is an even number, there exists a DBH on Tm,n − F, where F is a set of any one node and one edge.
Proof. 
This theorem is proved by dividing it into two steps as mentioned above. Similarly to Lemma 1, two Hamiltonian cycles C1 and C2 in Tm,n − F, as in Figure 5a,b (say type (a) and (b) for short), are constructed.
For any odd m ≥ 5 and even n ≥ 4, since Tm,n is a vertex-transitive graph, the faulty node is (⌊m/2⌋, 0), without loss of generality. Lemma 1 shows that the theorem holds for m = 5 and n = 4. If m > 5 and n = 4, Step 1 proves that there exists a DBH on Tm,4 − F at first.
  • Step 1. One column is expanded to both the left and right of Tm−2,4: In this step, there exists a DBH on Tm,4 − F, which is proved by induction. Claim 1 is proved as follows.
  • Claim 1: |E1(C1)| = |E1(C2)| = ⌊(mn − 1)/2 ⌋ and |E2(C1)| = |E2(C2)| = ⌈(mn − 1)/2⌉.
By Lemma 1, two DBHs C1 and C2 exist with a type (a) and (b), respectively, on T5,4 − F. Figure 5a,b prove Claim 1. If m > 5, there exist two DBHs C1 and C2 on Tm−2,4 − F with type (a) and (b), respectively, satisfying Claim 1. We divide this step into two cases according to the number of (m mod 4) for discussion.
  • Case 1. (m mod 4) = 3: There are two DBHs on Tm,4 to be constructed. Therefore, this case is divided into two parts and discussed separately depending on whether the constructed DBH is type (a) or type (b).
  • Case 1.1. To construct a DBH with type (a): In the hypothesis, there is a DBH C1 on Tm−2,4 with type (a) satisfying Claim 1. First, C1 is put in the middle of Tm,4 with edges {(1, 0)(1, 3), (1, 3)(m − 2, 3), (m − 2, 0)(m − 2, 1), (m − 2, 1)(m − 2, 2)}, (m − 2, 2)(m − 2, 3)} in C1. Then, remove {(1, 3)(m − 2, 3), (m − 2, 1)(m − 2, 2)} from C1 and add {(m − 2, 1)(m − 1, 1), (m − 1, 1)(m − 1, 2), (m − 2, 2)(m − 1, 2), (m − 2, 3)(m − 1, 3), (m − 1, 0)(m − 1, 3), (0, 0)(m − 1, 0), (0, 0)(0, 1), (0, 1)(0, 2), (0, 2)(0, 3), (0, 3)(1, 3)} into C1. After calculation, the operation requires adding four first-dimensional edges and four second-dimensional edges. We removed one first-dimensional edge and one second-dimensional edge and added five first-dimensional edges and five second-dimensional edges. Therefore, the constructed Hamiltonian cycle is of type (a), satisfies Claim 1, and is a DBH. Figure 6 shows an example of T7,4 − F.
    Figure 6. The constructed DBH C1 on T7,4 − F.
    Figure 6. The constructed DBH C1 on T7,4 − F.
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  • Case 1.2. To construct a DBH with type (b): There is a DBH C2 on Tm−2,4 with type (b) and that satisfies Claim 1. First, C2 is put in the middle of Tm,4, with edge {(1, 1)(m − 2, 1)} in C2. Then, {(1, 0)(1, 3), (1, 1)(m − 2, 1)} is removed to add {(0, 0)(1, 0), (0, 1)(1, 1), (0, 1)(0, 2), (0, 2)(m − 1, 2), (0, 3)(1, 3), (0, 0)(0, 3), (m − 2, 1)(m − 1, 1), (m − 1, 0)(m − 1, 1), (m − 1, 0)(m − 1, 3), (m − 1, 2)( m − 1, 3)}. After calculation, this operation requires adding four first-dimensional edges and four second-dimensional edges. We removed one first-dimensional and one second-dimensional edges and added five first-dimensional edges and five second-dimensional edges. Therefore, the constructed Hamiltonian cycle becomes type (b), satisfies Claim 1, and is a DBH. Figure 7 shows an example of the constructed C2 on T7,4 − F.
    Figure 7. Constructed DBH C2 on T7,4 − F.
    Figure 7. Constructed DBH C2 on T7,4 − F.
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  • Case 2. (m mod 4) = 1: There are also two DBHs on Tm,4 need to be constructed. Therefore, this case is divided into two parts and discussed separately depending on whether the constructed DBH is type (a) or (b).
  • Case 2.1. To construct a DBH with type (a), a DBH C1 is used on Tm−2,4 with type (a) satisfying Claim 1. C1 is similar to Figure 6 in this subcase. First, C1 is put in the middle of Tm,4 with edges {(1, 0)(1, 1), (1, 1)(1, 2), (1, 2)(1, 3), (1, 0)(m − 2, 0), (m − 2, 0)(m − 2, 3)} in C1. Then, remove {(1, 0)(m − 2, 0), (1, 1)(1, 2)} and add {(0, 0)(1, 0), (0, 1)(1, 1), (0, 1)(0, 2), (0, 2)(1, 2), (0, 0)(0, 3), (0, 3)(m − 1, 3), (m − 2, 0)(m − 1, 0), (m − 1, 0)(m − 1, 1), (m − 1, 1)(m − 1, 2), (m − 1, 2)(m − 1, 3)} into C1. After calculation, this operation requires adding four first-dimensional edges and four second-dimensional edges. Similarly to case 1.1, we removed one first-dimensional edge and one second-dimensional edge and added five first-dimensional edges and five second-dimensional edges. Therefore, the constructed Hamiltonian cycle is of type (a), satisfies Claim 1, and is a DBH. Figure 8 shows an example of the constructed C1 on T9,4 − F.
    Figure 8. The constructed DBH C1 of T9,4 − F.
    Figure 8. The constructed DBH C1 of T9,4 − F.
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  • Case 2.2. To construct a DBH with type (b): There is a DBH C2 on Tm−2,4 with type (b) satisfying Claim 1. C2 is similar to Figure 7 in this subcase. First, C2 is put in the middle of Tm,4 with edge {(1, 2)(m − 2, 2)} in C2. Then, remove {(1, 0)(1, 3), (1, 2)(m − 2, 2)} and add {(0, 0)(1, 0), (0, 1)(0, 2), (0, 1)(m − 1, 1), (0, 2)(1, 2), (0, 3)(1, 3), (0, 0)(0, 3), (m − 2, 2)(m − 1, 2), (m − 1, 0)(m − 1, 1), (m − 1, 0)(m − 1, 3), (m − 1, 2)(m − 1, 3)}. Again, it can be determined that this operation requires adding four first-dimensional edges and four second-dimensional edges into C2 to keep it dimension-balanced. We removed one first-dimensional edge and one second-dimensional edge and added five first-dimensional edges and five second-dimensional edges. Therefore, the constructed Hamiltonian cycle is of type (b) satisfying Claim 1 and a DBH. Figure 9 shows an example of the constructed C2 on T9,4 − F.
    Figure 9. The constructed DBH C2 of T9,4 − F.
    Figure 9. The constructed DBH C2 of T9,4 − F.
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  • Step 2. Two rows downward of Tm,n−2.
By Step 1, there exist two DBHs C1 and C2 with types (a) and (b), respectively, on Tm,4 − F. |E1(C1)| = |E1(C2)| = 2m − 1, and |E2(C1)| = |E2(C2)| = 2m. Then, if n > 4, let k = (n − 4)/2. We divided this step into two cases according to the number of (m mod 4) for discussion. Ym is defined as the Cartesian product of Cm and P2, called a prism graph, where Pn is a path with n vertices. The vertex set of Ym is V(Ym) = {(x, y) | 0 ≤ xm − 1; 0 ≤ y ≤ 1}, and the edge set E(Ym) = {(x1, y1)(x2, y2) | x1 = x2 and |y1 − y2| = 1, or |x1 − x2| = (1 or m − 1) and y1 = y2}.
  • Case 1. (m mod 4) = 1: There are two DBHs on Tm,n to be constructed. Therefore, this case is divided into two parts and discussed separately depending on whether the constructed DBH is type (a) or (b).
  • Case 1.1. To construct a DBH with type (a): By Step 1, there is a DBH C1 on Tm,4 with type (a). Copy the last two rows of C1 to be a subgraph of Ym named Zm. Figure 10a,b show Z5 and Z9, respectively. Now, C1 is put in the upper part of Tm,n, and k Zms in the lower part of Tm,n. Then, remove {(0, 3)(m − 1, 3)} ∪ {(2i, 0)(2i, 3), (2i + 1, 3 + 2j)(2i + 2, 3 + 2j) | 0 ≤ i < (m − 1)/4, 0 ≤ j < k} ∪ {(2i + 1, 0)(2i + 1, 3) | (m + 3)/4 ≤ i < (m − 1)/2} and add {(2i, 0)(2i, n − 1), (2i, 3 + 2j)(2i, 4 + 2j), (2i, 3 + 2j)(2i + 1, 3 + 2j) | 0 ≤ i < (m − 1)/4, 0 ≤ j < k } ∪ {(2i + 1, 0)(2i + 1, n − 1), (2i + 1, 3 + 2j)(2i + 1, 4 + 2j) | (m + 3)/4 ≤ i < (m − 1)/2, 0 ≤ j < k} ∪ {((m − 1)/2, 3 + 2j)((m − 1)/2, 4 + 2j), (m − 1, 3 + 2j)(m − 1, 4 + 2j) | 0 ≤ j < k}. Figure 10c,d show examples for T5,6 − F, T9,6 − F. After calculation, the generated Hamiltonian cycle C satisfies |E1(C)| = |E1(C1)| + k(m − 1) = nm/2 − (k + 1) and |E2(C)| = |E1(C1)| + k(m + 1) = nm/2 + k. We added k or k + 1 first-dimensional edges and removed k or k + 1 second-dimensional edges from C to form a type (a) DBH of Tm,n − F. There are two sub-steps according to the value of n that can solve this situation.
  • Step 1.1.1. Let x = ⌊(n + 2)/8⌋ = ⌊(k + 3)/4⌋. Find x smallest i with {(0, i)(0, i + 1), (m − 1, i)(m − 1, i + 1)} ⊆ E(C). Then, remove (0, i)(0, i + 1), (m − 1, i)(m − 1, i + 1) and add (0, i)(m − 1, i), (0, i + 1)(m − 1, i + 1) to C. Figure 10e shows an example for T5,6 − F. This step adds 2⌊(k + 3)/4⌋ first-dimensional edge and removes 2⌊(k + 3)/4⌋ second-dimensional edges from C.
  • Step 1.1.2. Let x = ⌊(n − 2)/8⌋ = ⌊(k + 1)/4⌋. Find x smallest i with {(m − 2, i)(m − 2, i + 1), (m − 1, i)(m − 1, i + 1)} ⊆ E(C) and {(m − 2, i)(m − 1, i), (m − 2, i + 1)(m − 1, i + 1)} ⊄ E(C). Then, (m − 2, i)(m − 2, i + 1), (m − 1, i)(m − 1, i + 1) are removed, and (m − 2, i)(m − 1, i), (m − 2, i + 1)(m − 1, i + 1) are added into C. Figure 10f shows an example. This step adds 2⌊(k + 1)/4⌋ first-dimensional edges and removes 2⌊(k + 1)/4⌋ second-dimensional edges from C.
Note that 2⌊(k + 3)/4⌋ + 2⌊(k + 1)/4⌋ = k or k + 1. The created Hamiltonian cycle C is a DBH with type (a).
Figure 10. Examples for case 1.1 of Step 2 of Theorem 4. (a) Z5, (b)Z9, (c) a Hamiltonian cycle C on T5,6 − F, (d) a Hamiltonian cycle C on T9,6 − F, (e) the construct DBH of T5,6 − F, (f) the construct DBH of T5,10 − F.
Figure 10. Examples for case 1.1 of Step 2 of Theorem 4. (a) Z5, (b)Z9, (c) a Hamiltonian cycle C on T5,6 − F, (d) a Hamiltonian cycle C on T9,6 − F, (e) the construct DBH of T5,6 − F, (f) the construct DBH of T5,10 − F.
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  • Case 1.2. To construct a DBH with type (b): By Step 1, there is a DBH C2 on Tm,4 with type (b). The last two rows of C2 are copied to be a subgraph of Ym named Zm. Modify Zm by adding edges {(0, 2i + 1)(0, 2i + 2) | 0 ≤ i ≤ (m − 9)/4} and deleting edges {(0, (m − 5)/2)(0, (m − 3)/2} ∪ {(0, 2i)(0, 2i + 1) | (m + 5)/4 ≤ i ≤ (m − 2)/2}. Figure 10a,b show Z5 and Z9, respectively. Then, C2 is put in the upper part of Tm,n, and k Zms in the lower part of Tm,n. Then, for any 0 ≤ im − 1, if edge (0, i)(3, i) is in C2, this edge is removed, and edges {(0, i)(n − 1, i), (3 + 2j, i)(4 + 2j, i) | 0 ≤ j < k} are added to C2 and k Zms to form a new Hamiltonian cycle C of Tm,n − F. Figure 11c,d show an example of T5,6 − F and T9,6 − F. Similarly to case 1.1, after calculation, the generated Hamiltonian cycle C satisfies |E1(C)| = |E1(C1)| + k(m − 1) = nm/2 − (k + 1) and |E2(C)| = |E1(C1)| + k(m + 1) = nm/2 + k. We added k or k + 1 first-dimensional edges and removed k or k + 1 second-dimensional edges from C to form a type (b) DBH of Tm,n − F. There are two sub-steps according to the value of n to solve this situation. m = 5 is a special case and needs to be discussed separately.
  • Step 1.2.1. Let x = ⌊(n + 2)/8⌋ = ⌊(k + 3)/4⌋. For m = 5, find x smallest i with {(0, i)(0, i + 1), (m − 1, i)(m − 1, i + 1)} ⊆ E(C). Then, remove these two edges and add (0, i)(m − 1, i), (0, i + 1)(m − 1, i + 1) into C. Figure 11e shows an example. For m > 5, find x smallest i with {((m − 1)/2 + 2, i)((m − 1)/2 + 2, i + 1), ((m − 1)/2 + 3, i)((m − 1)/2 + 3, i + 1)} ⊆ E(C). Then, these two edges are removed, and ((m − 1)/2 + 2, i)((m − 1)/2 + 3, i), ((m − 1)/2 + 2, i + 1)((m − 1)/2 + 3, i + 1) are added to C. Figure 11f shows an example for T9,6 − F.
  • Step 1.2.2. Let x = ⌊(n − 2)/8⌋ = ⌊(k + 1)/4⌋. For m = 5, find the 2x edges (0, i)(0, i + 1), (1, i)(1, i + 1) such that (0, i)(1, i), (0, i + 1)(1, i + 1) ∉ E(C) (after performing Step 1.2.1). Then, (0, i)(0, i + 1), (1, i)(1, i + 1) are removed, and (0, i)(1, i), (0, i + 1)(1, i + 1) are added to C. Figure 11g shows an example for T5,10 − F. For m > 5, find the 2x edges ((m − 1)/2 + 3, i)((m − 1)/2 + 3, i + 1), ((m − 1)/2 + 4, i)((m − 1)/2 + 4, i + 1) such that ((m − 1)/2 + 3, i)((m − 1)/2 + 4, i), ((m − 1)/2 + 3, i + 1)((m − 1)/2 + 4, i + 1) ∉ E(C) and i > 0. Then, ((m − 1)/2 + 3, i)((m − 1)/2 + 3, i + 1), ((m − 1)/2 + 4, i)((m − 1)/2 + 4, i + 1) are removed, and ((m − 1)/2 + 3, i)((m − 1)/2 + 4, i), ((m − 1)/2 + 3, i + 1)((m − 1)/2 + 4, i + 1) are added to C. Figure 11f shows an example for T9,10 − F.
Note that 2⌊(k + 3)/4⌋ + 2⌊(k + 1)/4⌋ = k or k + 1, so the created Hamiltonian cycle C is a DBH with type (b).
Figure 11. Examples for case 1.2 of Step 2 of Theorem 4. (a) Z5, (b)Z9, (c) a Hamiltonian cycle C on T5,6 − F, (d) a Hamiltonian cycle C on T9,6 − F, (e) the construct DBH of T5,6 − F, (f) the construct DBH of T9,6 − F, (g) the construct DBH of T5,10 − F, (h) the construct DBH of T9,10 − F.
Figure 11. Examples for case 1.2 of Step 2 of Theorem 4. (a) Z5, (b)Z9, (c) a Hamiltonian cycle C on T5,6 − F, (d) a Hamiltonian cycle C on T9,6 − F, (e) the construct DBH of T5,6 − F, (f) the construct DBH of T9,6 − F, (g) the construct DBH of T5,10 − F, (h) the construct DBH of T9,10 − F.
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  • Case 2. (m mod 4) = 3: There are two DBHs on Tm,n constructed. Therefore, this case is divided into two parts and discussed separately depending on whether the constructed DBH is type (a) or type (b).
  • Case 2.1. To construct a DBH with type (a): By Step 1, there is a DBH C1 on Tm,n−2 with type (a). The last two rows of C1 are copied to be a subgraph of Ym named Zm. Figure 12a,b show that Z7 and Z11. C1 are put in the upper part of Tm,n, and k Zms in the lower part of Tm,n. Then, {(1 + 2i, 0)(1 + 2i, 3), (m − 1 − 2i, 0)(m − 1 − 2i, 3)| 0 ≤ i < (m − 3)/4} ∪ {(2i, 3 + 2j)(2i + 1, 3 + 2j)| 0 ≤ i < (m + 1)/4, 0 ≤ j < k} are removed, and {(1 + 2i, 0)(1 + 2i, n − 1), (m − 1 − 2i, 0)(m − 1 − 2i, n − 1), (1 + 2i, 3 + 2j)(2 + 2i, 3 + 2j), (1 + 2i, 3 + 2j)(1 + 2i, 4 + 2j), (m − 1 − 2i, 3 + 2j)(m − 1 − 2i, 4 + 2j)| 0 ≤ i < (m − 3)/4, 0 ≤ j < k } ∪ {((m − 1)/2, 3 + 2j)((m − 1)/2, 4 + 2j), (0, 3 + 2j)(0, 4 + 2j) | 0 ≤ j < k} are added. Figure 12c,d show T7,6 − F and T11,6 − F. After calculation, the generated Hamiltonian cycle C will satisfy |E1(C)| = |E1(C1)| + k(m − 1) = nm/2 − (k + 1) and |E2(C)| = |E2(C1)| + k(m + 1) = nm/2 + k. We added k or k + 1 first-dimensional edges and removed k or k + 1 second-dimensional edges from C to form a type (a) DBH of Tm,n − F. There are two sub-steps according to the value of n that can solve this situation.
  • Step 2.1.1. Let x = ⌊(n + 2)/8⌋ = ⌊(k + 3)/4⌋. First, {(0, 4i + 1)(0, 4i + 2), (m − 1, 4i + 1)(m − 1, 4i + 2) | 0 ≤ i < x} are removed from C, and then {(0, 4i + 1)(m − 1, 4i + 1), (0, 4i + 2)(m − 1, 4i + 2) | 0 ≤ i < x} are added to C. Figure 12e shows an example for T7,6 − F. In total, 2⌊(k + 3)/4⌋ first-dimensional edges are added to remove 2⌊(k + 3)/4⌋ second-dimensional edges from C.
  • Step 2.1.2. Let x = ⌊(n − 2)/8⌋ = ⌊(k + 1)/4⌋. Then, {(0, 4i + 3)(0, 4i + 4), (1, 4i + 3)(1, 4i + 4) | 0 ≤ i < x} are removed from C, and {(0, 4i + 3)(1, 4i + 3), (0, 4i + 4)(1, 4i + 4) | 0 ≤ i < x} are added to C. Figure 12f shows an example of T7,10 − F. This step adds 2⌊(k + 1)/4⌋ first-dimensional edges and removes 2⌊(k + 1)/4⌋ second-dimensional edges from C.
Note that 2⌊(k + 3)/4⌋ + 2⌊(k + 1)/4⌋ = k or k + 1, so the created Hamiltonian cycle C is a DBH with type (a).
Figure 12. Examples for case 2.1 of Step 2 of Theorem 4. (a) Z7, (b)Z11, (c) a Hamiltonian cycle C on T7,6 − F, (d) a Hamiltonian cycle C on T11,6 − F, (e) the construct DBH of T7,6 − F, (f) the construct DBH of T7,10 − F.
Figure 12. Examples for case 2.1 of Step 2 of Theorem 4. (a) Z7, (b)Z11, (c) a Hamiltonian cycle C on T7,6 − F, (d) a Hamiltonian cycle C on T11,6 − F, (e) the construct DBH of T7,6 − F, (f) the construct DBH of T7,10 − F.
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  • Case 2.2. To construct a DBH with type (b): By Step 1, there is a DBH C2 on Tm,4 with type (b). The last two rows of C2 are copied to be a subgraph of Ym named Zm. Zm is modified by adding edges {(0, 2i)(0, 2i + 1) | 0 ≤ i ≤ (m − 7)/4} and deleting edges {(0, (m − 5)/2)(0, (m − 3)/2} ∪ {(0, 2i)(0, 2i + 1) | (m + 5)/4 ≤ i ≤ (m − 3)/2}. Figure 13a,b show Z7 and Z11, respectively. Then, C2 is put in the upper part of Tm,n, and k Zms in the lower part of Tm,n. Then, for any 0 ≤ im − 1, if edge (0, i)(3, i) is in C2, then this edge is removed to add edges {(0, i)(n − 1, i), (3 + 2j, i)(4 + 2j, i) | 0 ≤ j < k} into C2 and k Zms to form a new Hamiltonian cycle C of Tm,n − F. Figure 13c,d show examples of T7,6 − F and T11,6 − F. Similarly to case 2.1, after calculation, the generated Hamiltonian cycle C satisfies |E1(C)| = |E1(C1)| + k(m − 1) = nm/2 − (k + 1) and |E2(C)| = |E1(C1)| + k(m + 1) = nm/2 + k. We added k or k + 1 first-dimensional edges and removed k or k + 1 second-dimensional edges from C to form a type (b) DBH of Tm,n − F. There are two sub-steps according to the value of n to solve this problem.
  • Step 2.2.1. Let x = ⌊(n + 2)/8⌋ = ⌊(k + 3)/4⌋. Find the x smallest i with {(0, i)(0, i + 1), (m − 1, i)(m − 1, i + 1)} ⊆ E(C). Then, these two edges are removed, and (0, i)(m − 1, i), (0, i + 1)(m − 1, i + 1) are added to C. Figure 13e shows an example of T7,6 − F.
  • Step 2.2.2. Let x = ⌊(n − 2)/8⌋ = ⌊(k + 1)/4⌋. Find the x edge pairs ((m − 1)/2 − 2, i)((m − 1)/2 − 2, i + 1), ((m − 1)/2 − 1, i)((m − 1)/2 − 1, i + 1) such that ((m − 1)/2 − 2, i)((m − 1)/2 − 1, i), ((m − 1)/2 − 2, i + 1)((m − 1)/2 − 1, i + 1) ∉ E(C), and i > 3. Then, ((m − 1)/2 − 2, i)((m − 1)/2 − 2, i + 1), ((m − 1)/2 − 1, i)((m − 1)/2 − 1, i + 1) are removed, and ((m − 1)/2 − 2, i)((m − 1)/2 − 1, i), ((m − 1)/2 − 2, i + 1)((m − 1)/2 − 1, i + 1) are added to C. Figure 13f shows an example of T7,10 − F.
Again, because of 2⌊(k + 3)/4⌋ + 2⌊(k + 1)/4⌋ = k or k + 1, so the created Hamiltonian cycle C is a DBH with type (b). □
Figure 13. Examples for case 2.2 of Step 2 of Theorem 4. (a) Z7, (b)Z11, (c) a Hamiltonian cycle C on T7,6 − F, (d) a Hamiltonian cycle C on T11,6 − F, (e) the construct DBH of T7,6 − F, (f) the construct DBH of T7,10 − F.
Figure 13. Examples for case 2.2 of Step 2 of Theorem 4. (a) Z7, (b)Z11, (c) a Hamiltonian cycle C on T7,6 − F, (d) a Hamiltonian cycle C on T11,6 − F, (e) the construct DBH of T7,6 − F, (f) the construct DBH of T7,10 − F.
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4. Conclusions

We discussed whether Tm,n is a two-fault DBH or not, focusing on the one-node one-edge DBH problem. For any m and n, the conclusions of this study are summarized in Table 1. When both m and n are even, Tm,n is not a one-node one-edge DBH; and when one of m and n is odd (>3) and the other is even, Tm,n is a one-node one-edge DBH. It is proven that T3,n is not the one-node one-edge DBH for any n ≥ 5. In the subsequent study where m and n are both odd, it is also proven that T5,5 is not the one-node one-edge DBH. Due to space limitations, this part of the proof is omitted in this article. Finally, we provide a conjecture that when both m and n are odd and at least one is >5, Tm,n is not the one-node one-edge DBH.
Conjecture 1.
There is no DBH on Tm,n − F when m and n ≥ 5, where F is a set of any one node and one edge.
Table 1. Summary of one-node one-edge DBHs on Tm,n.
Table 1. Summary of one-node one-edge DBHs on Tm,n.
m = 3m Is Odd (>3)m = 4m Is Even (>4)
n = 3Yes
(Thm 3)
No
(Thm 3)
Yes
(Thm 3)
No
(Thm 3)
n is odd
(>3)
No
(Thm 3)
No
(Conjecture 1)
Yes
(Thm 4)
n = 4Yes
(Thm 3)
Yes
(Thm 4)
No
(Thm 1)
n is even
(>4)
No
(Thm 3)

Author Contributions

Conceptualization, J.S.-T.J.; methodology, J.S.-T.J. and Y.Y.-C.C.; validation, J.S.-T.J.; formal analysis, J.S.-T.J. and Y.Y.-C.C.; investigation, J.S.-T.J.; data curation, Y.Y.-C.C.; writing—original draft preparation, Y.Y.-C.C.; writing—review and editing, J.S.-T.J.; visualization, Y.Y.-C.C.; supervision, J.S.-T.J.; project administration, J.S.-T.J.; funding acquisition, J.S.-T.J. All authors have read and agreed to the published version of the manuscript.

Funding

This work was partially supported by the National Science and Technology Council of Taiwan R.O.C. under grant NSTC 112-2115-M-260 -001 -MY2.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Conflicts of Interest

The authors declare no conflicts of interest.

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Figure 5. (a) A DBH of T5,4F; (b) another DBH of T5,4F.
Figure 5. (a) A DBH of T5,4F; (b) another DBH of T5,4F.
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Chang, Y.Y.-C.; Juan, J.S.-T. One-Node One-Edge Dimension-Balanced Hamiltonian Problem on Toroidal Mesh Graph. Eng. Proc. 2025, 89, 17. https://doi.org/10.3390/engproc2025089017

AMA Style

Chang YY-C, Juan JS-T. One-Node One-Edge Dimension-Balanced Hamiltonian Problem on Toroidal Mesh Graph. Engineering Proceedings. 2025; 89(1):17. https://doi.org/10.3390/engproc2025089017

Chicago/Turabian Style

Chang, Yancy Yu-Chen, and Justie Su-Tzu Juan. 2025. "One-Node One-Edge Dimension-Balanced Hamiltonian Problem on Toroidal Mesh Graph" Engineering Proceedings 89, no. 1: 17. https://doi.org/10.3390/engproc2025089017

APA Style

Chang, Y. Y.-C., & Juan, J. S.-T. (2025). One-Node One-Edge Dimension-Balanced Hamiltonian Problem on Toroidal Mesh Graph. Engineering Proceedings, 89(1), 17. https://doi.org/10.3390/engproc2025089017

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