# Closed Knight’s Tours on (m,n,r)-Ringboards

^{1}

^{2}

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## Abstract

**:**

## 1. Introduction

**Theorem**

**1.**

**Theorem**

**2.**

**Definition**

**1.**

**Theorem**

**3.**

## 2. CKTs and OKTs on Some LBs and 7Bs

**Theorem**

**4.**

**Proof.**

- (i)
- For $1\le i\le t-1$, delete $(2,3)-(4,4)$ from ${P}_{1}$ and $(1,4)-(3,3)$ from ${P}_{2}$ of the ith CB($4\times 4$).
- (ii)
- For $1\le i\le t-1$, join $(2,3)$ and $(4,4)$ of the ith CB($4\times 4$) to $(1,1)$ and $(2,1)$ of the $(i+1)$th CB($4\times 4$), respectively
- (iii)
- For $1\le i\le t-1$, join $(1,4)$ and $(3,3)$ of the ith CB($4\times 4$) to $(3,1)$ and $(4,1)$ of the $(i+1)$th CB($4\times 4$), respectively.

- If m and n are odd integers, then $m\equiv 1$ or 3 (mod 4) and $n\equiv 1$ or 3 (mod 4).
- If m and n are even integers, then $m\equiv 0$ or 2 (mod 4) and $n\equiv 0$ or 2 (mod 4).
- If m and n are different parity, then $m\equiv 1$ or 3 (mod 4) and $n\equiv 0$ or 2 (mod 4); and $m\equiv 0$ or 2 (mod 4) and $n\equiv 1$ or 3 (mod 4).

- (i)
- if $t=0$, then delete $(1,1)-(2,3)$ and $(1,4)-(2,2)$ from the CKT of the LB$(a,b,4)$. If $t>0$, then further delete $(a-3,b)-(a-1,b-1)$ and $(a-2,b-1)-(a,b)$ from the CKT of the LB$(a,b,4)$.
- (ii)
- If $t=0$, then join $(4s,1)$, $(4s,2)$, $(4s,3)$ and $(4s,4)$ which are four end-points of two paths of the CB($4s\times 4$) to $(2,2)$, $(1,4)$, $(1,1)$ and $(2,3)$ of the LB$(a,b,4)$, respectively. If $t>0$, then further join $(1,1)$, $(2,1)$, $(3,1)$ and $(4,1)$ which are four end-points of two paths of the CB($4\times 4t$) to $(a-2,b-1)$, $(a,b)$, $(a-3,b)$ and $(a-1,b-1)$ of the LB$(a,b,4)$, respectively.

**Corollary**

**1.**

**Theorem**

**5.**

- (a)
- The LB$(m,n,3)$ contains an OKT from $(1,2)$ to $(1,3)$ if and only if (i) $m+n$ is odd and $m+n\ge 11$ or (ii) $m=5$ and $n=4$.
- (b)
- The LB$(m,n,3)$ contains an OKT from $(1,3)$ to $(2,2)$ if and only if (i) $m+n$ is even and $m+n\ge 12$ or (ii) $m=6$ and $n=4$.

**Proof.**

- (i)
- For $1\le i\le t-1$, delete $(1,3)-(3,4)$ from the OKT of the ith CB($3\times 4$);
- (ii)
- For $1\le i\le t-1$, join $(1,3)$ and $(3,4)$ of the ith CB($3\times 4$) to $(2,1)$ and $(1,1)$ of the $(i+1)$th CB($3\times 4$), respectively.

- Case 1: there exist nonnegative integers $s,t$ such that $m=4+4s$ and $n=7+4t$. We divided the LB$(m,n,3)$ into subboards, CB($4s\times 3$) (Figure 17) and LB$(4,7,3)$ (Figure 11a) if $s>0$ and $t=0$ and LB$(4,7,3)$ (Figure 11a) and CB($3\times 4t$) (Figure 16) if $s=0$ and $t>0$. Otherwise, we divide into three subboards, CB($4s\times 3$) (Figure 17), LB$(4,7,3)$ (Figure 11a) and CB($3\times 4t$) (Figure 16). Then, we construct the required OKT by the following two steps.
- (i)
- (ii)
- If $s>0$ and $t=0$, then join $(4s,1)$ and $(4s-1,3)$ of the CB($4s\times 3$) to $(1,3)$ and $(1,2)$ of the LB$(4,7,3)$, respectively. If $s=0$ and $t>0$, then join $(2,6)$ and $(4,7)$ of the LB$(4,7,3)$ to $(2,1)$ and $(1,1)$ of the CB($3\times 4t$) chessboard, respectively. Otherwise, join four pairs of vertices together.

- Case 2: there exist nonnegative integers $s,t$ such that $m=7+4s$ and $n=4+4t$. Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB$(4,7,3)$, $(2,6)$ and $(4,7)$ are replaced by LB$(7,4,3)$ (Figure 11b), $(5,3)$ and $(7,4)$, respectively.
- Case 3: there exist nonnegative integers $s,t$ such that $m=6+4s$ and $n=5+4t$. Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB$(4,7,3)$, $(2,6)$ and $(4,7)$ are replaced by LB$(6,5,3)$ (Figure 12a), $(4,4)$ and $(6,5)$, respectively.
- Case 4: there exist nonnegative integers $s,t$ such that $m=5+4s$ and $n=6+4t$. Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB$(4,7,3)$, $(2,6)$ and $(4,7)$ are replaced by LB$(5,6,3)$ (Figure 12b), $(3,5)$ and $(5,6)$, respectively.
- Case 5: there exist nonnegative integers $s,t$ such that $m=4+4s$ and $n=9+4t$. Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB$(4,7,3)$, $(2,6)$ and $(4,7)$ are replaced by LB$(4,9,3)$ (Figure 13a), $(2,8)$ and $(4,9)$, respectively.
- Case 6: there exist nonnegative integers $s,t$ such that $m=8+4s$ and $n=5+4t$. Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB$(4,7,3)$, $(2,6)$ and $(4,7)$ are replaced by LB$(8,5,3)$ (Figure 13b), $(6,4)$ and $(8,5)$, respectively.
- Case 7: there exist nonnegative integers $s,t$ such that $m=5+4s$ and $n=4+4t$. Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB$(4,7,3)$, $(2,6)$ and $(4,7)$ are replaced by LB$(5,4,3)$ (Figure 10), $(3,3)$ and $(5,4)$, respectively.
- Case 8: there exist nonnegative integers $s,t$ such that $m=6+4s$ and $n=7+4t$. Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB$(4,7,3)$, $(2,6)$ and $(4,7)$ are replaced by LB$(6,7,3)$ (Figure 14a), $(4,6)$ and $(6,7)$, respectively.
- Case 9: there exist nonnegative integers $s,t$ such that $m=7+4s$ and $n=6+4t$. Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB$(4,7,3)$, $(2,6)$ and $(4,7)$ are replaced by LB$(7,6,3)$ (Figure 14b), $(5,5)$ and $(7,6)$, respectively.

- (i)
- For $1\le i\le s$, delete $(2,2)-(4,3)$ from the OKT of the ith CB($4\times 3$);
- (ii)
- For $1\le i\le s-1$, join $(4,1)$ and $(4,3)$ of the ith CB($4\times 3$) to $(1,3)$ and $(2,2)$ of the $(i+1)$th CB($4\times 3$), respectively.

- Case 1: there exist nonnegative integers $s,t$ such that $m=5+4s$ and $n=7+4t$. We divide the LB$(m,n,3)$ into subboards, CB($4s\times 3$) (Figure 29) and LB$(5,7,3)$ (Figure 22a) if $s>0$ and $t=0$ and LB$(5,7,3)$ (Figure 22a) and CB($3\times 4t$) (Figure 16) if $s=0$ and $t>0$. Otherwise, we divide into three subboards, CB($4s\times 3$) (Figure 29), LB$(5,7,3)$ (Figure 22a) and CB($3\times 4t$) (Figure 16). Then, we construct the required OKT by the followings.
- (i)
- (ii)
- If $s>0$ and $t=0$, then join $(4s,1)$ and $(4s,3)$ of the CB($4s\times 3$) in Figure 29 to $(1,3)$ and $(2,2)$ of the LB$(5,7,3)$, respectively. If $s=0$ and $t>0$, then join $(3,6)$ and $(5,7)$ of the LB$(5,7,3)$ to $(2,1)$ and $(1,1)$ of the CB($3\times 4t$) in Figure 16, respectively. Otherwise, join four pairs of vertices together.

- Case 2: there exist nonnegative integers $s,t$ such that $m=7+4s$ and $n=5+4t$. Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB$(5,7,3)$, $(3,6)$ and $(5,7)$ are replaced by LB$(7,5,3)$ (Figure 22b), $(5,4)$ and $(7,5)$, respectively.
- Case 3: there exist nonnegative integers $s,t$ such that $m=7+4s$ and $n=7+4t$. Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB$(5,7,3)$, $(3,6)$ and $(5,7)$ are replaced by LB$(7,7,3)$ (Figure 23), $(5,6)$ and $(7,7)$, respectively.
- Case 4: there exist nonnegative integers $s,t$ such that $m=5+4s$ and $n=9+4t$. Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB$(5,7,3)$, $(3,6)$ and $(5,7)$ are replaced by LB$(5,9,3)$ (Figure 24a), $(3,8)$ and $(5,9)$, respectively.
- Case 5: there exist nonnegative integers $s,t$ such that $m=9+4s$ and $n=5+4t$. Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB$(5,7,3)$, $(3,6)$ and $(5,7)$ are replaced by LB$(9,5,3)$ (Figure 24b), $(7,4)$ and $(9,5)$, respectively.
- Case 6: there exist nonnegative integers $s,t$ such that $m=6+4s$ and $n=6+4t$. Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB$(5,7,3)$, $(3,6)$ and $(5,7)$ are replaced by LB$(6,6,3)$ (Figure 25), $(4,5)$ and $(6,6)$, respectively.
- Case 7: there exist nonnegative integers $s,t$ such that $m=8+4s$ and $n=6+4t$. Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB$(5,7,3)$, $(3,6)$ and $(5,7)$ are replaced by LB$(8,6,3)$ (Figure 26a), $(6,5)$ and $(8,6)$, respectively.
- Case 8: there exist nonnegative integers $s,t$ such that $m=4+4s$ and $n=10+4t$. Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB$(5,7,3)$, $(3,6)$ and $(5,7)$ are replaced by LB$(4,10,3)$ (Figure 26b), $(2,9)$ and $(4,10)$, respectively.
- Case 9: there exist nonnegative integers $s,t$ such that $m=6+4s$ and $n=4+4t$. Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB$(5,7,3)$, $(3,6)$ and $(5,7)$ are replaced by LB$(6,4,3)$ (Figure 21), $(4,3)$ and $(6,4)$, respectively.
- Case 10: there exist nonnegative integers $s,t$ such that $m=4+4s$ and $n=8+4t$. Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB$(5,7,3)$, $(3,6)$ and $(5,7)$ are replaced by LB$(4,8,3)$ (Figure 27a), $(2,7)$ and $(4,8)$, respectively.
- Case 11: there exist nonnegative integers $s,t$ such that $m=8+4s$ and $n=4+4t$. Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB$(5,7,3)$, $(3,6)$ and $(5,7)$ are replaced by LB$(8,4,3)$ (Figure 27b), $(6,3)$ and $(8,4)$, respectively.

**Corollary**

**2.**

- (a)
- The 7B$(m,n,3)$ contains an OKT from $(2,1)$ to $(3,1)$ if and only if (i) $m+n$ is odd and $m+n\ge 11$ or (ii) $m=4$ and $n=5$.
- (b)
- The 7B$(m,n,3)$ contains an OKT from $(3,1)$ to $(2,2)$ if and only if (i) $m+n$ is even and $m+n\ge 12$ or (ii) $m=4$ and $n=6$.

## 3. Existence of a Special OKT on CB($\mathit{m}\times \mathit{n}$)

**Theorem**

**6.**

- (a)
- Let $m\le 4$ and $n\ge m$. Then, a CB($m\times n$) contains an OKT from $(m,1)$ to $(2,n-1)$ if and only if $m=3$ and $n\ge 7$.
- (b)
- Let $n\ge m\ge 5$. Then, a CB($m\times n$) contains an OKT from $(m,1)$ to $(2,n-1)$ if and only if m and n are not both even.

**Proof.**

**Case 1:**$m=1$ and $n\ge 1$ or $m=2$ and $n\ge 2$ or $m=3$ and $n\in \{3,5,6\}$. A CB($m\times n$) contains no OKT by using Theorem 2, contradiction.

**Case 2:**For $m=3$ and $n=4$, let ${G}_{1}$ be a knight graph of the CB($3\times 4$). We assume that ${G}_{1}$ contains a Hamiltonain path from $(3,1)$ to $(2,3)$. Consider ${G}_{1}^{\prime}={G}_{1}-\left\{(2,3)\right\}$. By assumption, ${G}_{1}^{\prime}$ has a Hamiltonian path. Let $S=\left\{\right(1,2),(3,2\left)\right\}$. Then, $\omega ({G}_{1}^{\prime}-S)=4>3=\left|S\right|+1$ as shown in Figure 30. By Theorem 3, we obtain a contradiction.

**Case 3:**For $m=4$ and n is odd such that $n\ge 5$. Let ${G}_{2}$ be a knight graph of the CB($4\times n$). We assume that ${G}_{2}$ contains a Hamiltonian path from $(m,1)$ to $(2,n-1)$. Consider ${G}_{2}^{\prime}={G}_{2}-\left\{(2,n-1)\right\}$. Let $S=\left\{\right(2,j),(3,l\left)\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\right|\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}j\phantom{\rule{4.pt}{0ex}}\mathrm{is}\phantom{\rule{4.pt}{0ex}}\mathrm{even},\phantom{\rule{4.pt}{0ex}}2\le j\le n-3,l\phantom{\rule{4.pt}{0ex}}\mathrm{is}\phantom{\rule{4.pt}{0ex}}\mathrm{odd}\phantom{\rule{4.pt}{0ex}}\mathrm{and}\phantom{\rule{4.pt}{0ex}}1\le l\le n\}$. Then, we can use mathematical induction to show that $\omega ({G}_{2}^{\prime}-S)=n+1>n=\left|S\right|+1$ as shown in Figure 31. By Theorem 3, we have a contradiction.

**Case 4:**For $m=4$ and n is even such that $n\ge 4$. Assume that CB($4\times n$) contains an OKT from $(4,1)$ to $(2,n-1)$. Since CB($4\times n$) contains the same numbers of black and white squares, this OKT must have end-points at two squares with different color. However, $4+1=5$ and $2+n-1=n+1$ are odd. Thus, $(m,1)$ and $(2,n-1)$ are two squares of the same color, contradiction.

**Case 1:**m and n are both odd such that $m,n\ge 5$. Let us construct OKTs from $(m,1)$ to $(2,n-1)$ containing the edge $(1,n)-(3,n-1)$ on some small size CB($m\times n$) where $m,n\in \{5,7\}$ as shown in Figure 34.

**Case 1.1:**$m,n\equiv 1$ (mod 4). There are integers s and t such that $0\le s\le t$, $t\ne 0$, $m=5+4s$ and $n=5+4t$. If $s=0$, then we divide the CB($5\times n$) into subboards, CB($5\times 5$) (Figure 34a) and t CB($5\times 4$)’s (Figure 35a). Then, we construct the required OKT by the followings.

- (i)
- We delete $(1,5)-(3,4)$ of the OKT on the CB($5\times 5$) and connect $(2,4),(1,5)$ and $(3,4)$ of the CB($5\times 5$) to $(1,1),(2,2)$ and $(4,1)$ of the 1st CB($5\times 4$), respectively.
- (ii)
- We delete $(1,4)-(3,3)$ of the second path of the ith CB($5\times 4$) for all $1\le i\le t-1$. Then, we connect $(2,3),(1,4)$ and $(3,3)$ of the ith CB($5\times 4$) to $(1,1),(2,2)$ and $(4,1)$ of the $(i+1)$th CB($5\times 4$).

- (i’)
- For each $1\le i\le s$, we divide the ith CB($4\times n$) into subboards, CB($4\times 5$) (Figure 36a) and CB($4\times 4t$) (Figure 6). Delete $(1,5)-(3,4)$ and $(2,4)-(4,5)$ of the OKT on CB($4\times 5$). Then, join $(2,4),(4,5),(1,5)$ and $(3,4)$ of the CB($4\times 5$) to $(1,1),(2,1),(3,1)$ and $(4,1)$ of the CB($4\times 4t$), respectively, to obtain an OKT on the CB($4\times n$) as shown in Figure 38.
- (ii’)
- Join $(5,1)$ of the OKT on CB($5\times n$) to $(1,3)$ of the 1st CB($4\times n$) shown in Figure 38.
- (iii’)

**Case 1.2:**$m\equiv 1$ (mod 4) and $n\equiv 3$ (mod 4). There are integers s and t such that $0\le s\le t$, $t\ne 0$, $m=5+4s$ and $n=7+4t$. If $s=0$, then we divide the CB($5\times n$) into subboards, CB($5\times 7$) (Figure 34b) and t CB($5\times 4$)’s (Figure 35a). Then, we construct the required OKT by (i) and (ii) in Case 1.1 but replace CB($5\times 5$) by CB($5\times 7$) (Figure 34b) and $(1,5)$, $(3,4)$ and $(2,4)$ by $(1,7)$, $(3,6)$ and $(2,6)$, respectively.

**Case 1.3:**$m\equiv 3$ (mod 4) and $n\equiv 1$ (mod 4). There are integers s and t such that $0\le s<t$, $m=7+4s$ and $n=5+4t$. If $s=0$, then we divide the CB($7\times n$) into subboards, CB($7\times 5$) (Figure 34c) and t CB($7\times 4$)’s (Figure 35c). Then, we construct the required OKT by (i) and (ii) in Case 1.1 but replace CB($5\times 5$) and CB($5\times 4$) by CB($7\times 5$) (Figure 34c) and CB($7\times 4$) (Figure 35c), respectively.

**Case 1.4:**$m\equiv 3$ (mod 4) and $n\equiv 3$ (mod 4). There are integers s and t such that $0\le s\le t$, $t\ne 0$, $m=7+4s$ and $n=7+4t$. If $s=0$, then we divide the CB($7\times n$) into subboards, CB($7\times 7$) (Figure 34d) and t CB($7\times 4$)’s (Figure 35c). Then, we construct the required OKT by (i) and (ii) in Case 1.1 but replace CB($5\times 5$) and CB($5\times 4$) by CB($7\times 7$) (Figure 34d) and CB($7\times 4$) (Figure 35c)) and $(1,5)$, $(3,4)$ and $(2,4)$ by $(1,7)$, $(3,6)$ and $(2,6)$, respectively.

**Case 2:**m is odd such that $m\ge 5$ and n is even such that $n\ge 6$. Let us construct OKTs from $(m,1)$ to $(2,n-1)$ containing the edge $(1,n)-(3,n-1)$ on some small size CB($m\times n$) where $m\in \{5,7\}$ and $n\in \{6,8\}$ as shown in Figure 43.

**Case 2.1:**$m\equiv 1$ (mod 4) and $n\equiv 0$ (mod 4). There are integers s and t such that $0\le s\le t$, $t\ne 0$, $m=5+4s$ and $n=8+4t$. If $s=0$, then we divide the CB($5\times n$) into subboards, CB($5\times 8$) (Figure 43b) and t CB($5\times 4$)’s (Figure 35a). Then, we construct the required OKT by (i) and (ii) in Case 1.1 but replace CB($5\times 5$) by CB($5\times 8$) (Figure 43b) and $(1,5)$, $(3,4)$ and $(2,4)$ by $(1,8)$, $(3,7)$ and $(2,7)$, respectively.

**Case 2.2:**$m\equiv 1$ (mod 4) and $n\equiv 2$ (mod 4). There are integers s and t such that $0\le s\le t$, $t\ne 0$, $m=5+4s$ and $n=6+4t$. If $s=0$, then we divide the CB($5\times n$) into subboards, CB($5\times 6$) (Figure 43a) and t CB($5\times 4$)’s (Figure 35a). Then, we construct the required OKT by (i) and (ii) in Case 1.1 but replace CB($5\times 5$) by CB($5\times 6$) (Figure 43a) and $(1,5)$, $(3,4)$ and $(2,4)$ by $(1,6)$, $(3,5)$ and $(2,5)$, respectively.

**Case 2.3:**$m\equiv 3$ (mod 4) and $n\equiv 0$ (mod 4). There are integers s and t such that $0\le s\le t$, $t\ne 0$, $m=7+4s$ and $n=8+4t$. If $s=0$, then we divide CB($7\times n$) into subboards, CB($7\times 8$) (Figure 43d) and t CB($7\times 4$)’s (Figure 35c). Then, we construct the required OKT by (i) and (ii) in Case 1.1 but replace CB($5\times 5$) and CB($5\times 4$) by CB($7\times 8$) (Figure 43d) and CB($7\times 4$) (Figure 35c) and $(1,5)$, $(3,4)$ and $(2,4)$ by $(1,8)$, $(3,7)$ and $(2,7)$, respectively.

**Case 2.4:**$m\equiv 3$ (mod 4) and $n\equiv 2$ (mod 4). There are integers s and t such that $0\le s<t$, $m=7+4s$ and $n=6+4t$. If $s=0$, then we divide the CB($7\times n$) into subboards, CB($7\times 6$) (Figure 43c) and t CB($7\times 4$)’s (Figure 35c). Then, we construct the required OKT by (i) and (ii) in Case 1.1. but replace CB($5\times 5$) and CB($5\times 4$) by CB($7\times 6$) (Figure 43c) and CB($7\times 4$) (Figure 35c) and $(1,5)$, $(3,4)$ and $(2,4)$ by $(1,6)$, $(3,5)$ and $(2,5)$, respectively.

**Case 3:**m is even such that $m\ge 6$ and n is odd such that $n\ge 5$. Let us construct OKTs from $(m,1)$ to $(2,n-1)$ containing the edge $(1,n)-(3,n-1)$ on some small size CB($m\times n$) where $m\in \{6,8\}$ and $n\in \{5,7\}$ as shown in Figure 50.

**Case 3.1:**$m\equiv 0$ (mod 4) and $n\equiv 1$ (mod 4). There are integers s and t such that $0\le s<t$, $m=8+4s$ and $n=5+4t$. If $s=0$, then we divide the CB($8\times n$) into subboards, CB($8\times 5$) (Figure 50c) and t CB($8\times 4$) (Figure 35d). Then, we construct the required OKT by (i) and (ii) in Case 1.1 replace CB($5\times 5$) and CB($5\times 4$) by CB($8\times 5$) (Figure 50c) and CB($8\times 4$) (Figure 35d), respectively.

**Case 3.2:**$m\equiv 0$ (mod 4) and $n\equiv 3$ (mod 4). There are integers s and t such that $0\le s<t$, $t\ne 0$, $m=8+4s$ and $n=7+4t$. If $s=0$, then we divide the CB($8\times n$) into subboards, CB($8\times 7$) (Figure 50d) and t CB($8\times 4$)’s (Figure 35d). Then, we construct the required OKT by (i) and (ii) in Case 1.1 but replace CB($5\times 5$) and CB($5\times 4$) by CB($8\times 7$) (Figure 50d) and CB($8\times 4$) (Figure 35d) and $(1,5)$, $(3,4)$ and $(2,4)$ by $(1,7)$, $(3,6)$ and $(2,6)$, respectively.

**Case 3.3:**$m\equiv 2$ (mod 4) and $n\equiv 1$ (mod 4). There are integers s and t such that $0\le s<t$, $t\ne 0$, $m=6+4s$ and $n=5+4t$. If $s=0$, then we divide the CB($6\times n$) into subboards, CB($6\times 5$) (Figure 50a) and t CB($6\times 4$) (Figure 35b). Then, we construct the required OKT by (i) and (ii) in Case 1.1 but replace CB($5\times 5$) and CB($5\times 4$) by CB($6\times 5$) (Figure 50a) and CB($6\times 4$) (Figure 35b), respectively.

**Case 3.4:**$m\equiv 2$ (mod 4) and $n\equiv 3$ (mod 4). There are integers s and t such that $0\le s\le t$, $t\ne 0$, $m=6+4s$ and $n=7+4t$. If $s=0$, then we divide the CB($6\times n$) into subboards, CB($6\times 7$) (Figure 50b) and t CB($6\times 4$) (Figure 35b). Then, we construct the required OKT by (i) and (ii) in Case 1.1 but replace CB($5\times 5$) and CB($5\times 4$) by CB($6\times 7$) (Figure 50b) and CB($6\times 4$) (Figure 35b) and $(1,5)$, $(3,4)$ and $(2,4)$ by $(1,7)$, $(3,6)$ and $(2,6)$, respectively.

## 4. Main Theorem

**Theorem**

**7.**

**Proof.**

**Case 1:**$k<l$ and $r=q=1$. Since all vertice in $\left\{\right(1,4i+1),(m,4i+1\left)\right|0\le i\le l\}$ and $\left\{\right(4i+1,1),(4i+1,n\left)\right|0\le i\le k\}$ have only 2 incident edges and we collect all incident edges from these two sets, it happens to form a cycle $(1,1)$, $(2,3)$, $(1,5)$, $(2,7)$, … , $(2,n-2)$, $(1,n)$, $(3,n-1)$, $(5,n)$, $(7,n-1)$, … , $(m-2,n-1)$, $(m,n)$, $(m-1,n-2)$, $(m,n-4)$, $(m-1,n-6)$, … , $(m-1,3)$, $(m,1)$, $(m-2,2)$, $(m-4,1)$, $(m-6,2)$, … , $(3,2)$, $(1,1)$, see Figure 55 for a cycle on $G(13,17,2)$. This is a contradiction since this cycle does not contain all vertices of $G(m,n,2)$.

**Case 2:**$k<l$ and $r=q=2$. We obtain a contradiction similar to Case 1 by considering $\left\{\right(1,4i+1),(m-1,4i+1\left)\right|0\le i\le l\}$ and $\left\{\right(4i+1,1),(4i+1,n-1\left)\right|0\le i\le k\}$ instead, see Figure 56a for a cycle on $G(14,18,2)$.

**Case 3:**$k<l$ and $r=q=3$. We obtain a contradiction similar to Case 1 by considering $\left\{\right(1,4i+4),(m,4i+4\left)\right|0\le i\le l-1\}$ and $\left\{\right(4i+4,1),(4i+4,n\left)\right|0\le i\le k-1\}$ instead, see Figure 56b for a cycle on $G(15,19,2)$.

**Case 4:**$k<l$ and $r=q=4$. We obtain a contradiction similar to Case 1 by considering $\left\{\right(1,4i+1),(m,4i+4\left)\right|0\le i\le l\}$ and $\left\{\right(4i+1,1),(4i+4,n\left)\right|0\le i\le k\}$ instead, see Figure 56c for a cycle on $G(16,20,2)$.

**Case 5:**$k\le l$, $r=1$ and $q=2$. We obtain a contradiction similar to Case 1 by considering $\left\{\right(1,4i+1),(m,4i+1\left)\right|0\le i\le l\}$ and $\left\{\right(4i+1,1),(4i+1,n-1\left)\right|0\le i\le k\}$ instead, see Figure 56d for a cycle on $G(13,14,2)$.

**Case 6:**$k\le l$, $r=1$ and $q=3$. We obtain a contradiction similar to Case 1 by considering $\left\{\right(1,4i+2),(m,4i+2\left)\right|0\le i\le l\}$ and $\left\{\right(4i+1,2),(4i+1,n-1\left)\right|0\le i\le k\}$ instead, see Figure 56e for a cycle on $G(13,15,2)$.

**Case 7:**$k\le l$, $r=1$ and $q=4$.

**Case 7.1:**$k=1$. Since $(1,n)$ and $(5,n)$ have only 2 incident edges on the $G(5,n,2)$, $(1,n)-(3,n-1)$ and $(3,n-1)-(5,n)$ must be in H and it forces that $(1,n-2)-(3,n-1)$ and $(3,n-1)-(5,n-2)$ must not be in H. Then, it also forces that $(2,n-4)-(1,n-2)$, $(1,n-2)-(2,n)$, $(4,n-4)-(5,n-2)$ and $(5,n-2)-(4,n)$ must be in H. Next, since all vertice in $\left\{\right(1,4i+1),(5,4i+1\left)\right|0\le i\le l\}$, $\left\{\right(1,4i+2),(5,4i+2\left)\right|0\le i\le l-1\}$ and $\left\{\right(2,n),(4,n\left)\right\}$ have only 2 incident edges. Collect $(2,n-4)-(1,n-2)$, $(1,n-2)-(2,n)$, $(4,n-4)-(5,n-2)$ and $(5,n-2)-(4,n)$ which must be in H together with all incident edges from these three sets, it happen to form a cycle $(1,1)$, $(2,3)$, $(1,5)$, $(2,7)$, … , $(1,n-3)$, $(2,n-1)$, $(4,n)$, $(5,n-2)$, $(4,n-4)$, $(5,n-6)$, … , $(4,4)$, $(5,2)$, $(3,1)$, $(1,2)$, $(2,4)$, $(1,6)$, … , $(1,n-6)$, $(2,n-4)$, $(1,n-2)$, $(2,n)$, $(4,n-1)$, $(5,n-3)$, … , $(5,5)$, $(4,3)$, $(5,1)$, $(3,2)$, $(1,1)$, see Figure 57 for a cycle on $G(5,12,2)$. This is a contradiction since this cycle does not contain all vertices of $G(5,n,2)$.

**Case 7.2:**$k\ge 2$. We obtain a contradiction similar to Case 1 by considering $\left\{\right(1,4i+1),(1,4i+2),(m,4i+1),(m,4i+2\left)\right|0\le i\le l-1\}$, $\left\{\right(1,n-3),(2,n-4),(m-1,n-4),(m,n-3\left)\right\}$ and $\left\{\right(4i+1,1),(4i+1,2),(4i+2,n),(4i+4,n\left)\right|0\le i\le k-1\}$ instead, see Figure 58 for a cycle on $G(13,16,2)$.

**Case 8:**$k<l$, $r=2$ and $q=1$. We obtain a contradiction similar to Case 1 by considering $\left\{\right(1,4i+1),(m-1,4i+1\left)\right|0\le i\le l\}$ and $\left\{\right(4i+1,1),(4i+1,n\left)\right|0\le i\le k\}$ instead, see Figure 59a for a cycle on $G(14,17,2)$.

**Case 9:**$k\le l$, $r=2$ and $q=3$. We obtain a contradiction similar to Case 1 by considering $\left\{\right(1,4i+2),(m-1,4i+2\left)\right|0\le i\le l\}$ and $\left\{\right(4i+1,2),(4i+1,n-1\left)\right|0\le i\le k\}$ instead, see Figure 59b for a cycle on $G(14,15,2)$.

**Case 10:**$k\le l$, $r=2$ and $q=4$.

**Case 10.1:**$k=1$ and $l=1$. Since $(2,8)$ and $(6,8)$ have only 2 incident edges on the $G(6,8,2)$, $(2,8)-(4,7)$ and $(4,7)-(6,8)$ must be in H and it forces that $(5,5)-(4,7)$ must not be in H. Then, it also forces that $(6,3)-(5,5)$ and $(5,5)-(6,7)$ must be in H. Next, since all vertice in $\left\{\right(1,1),(1,5),(2,7),(6,7),(5,1\left)\right\}$ have only 2 incident edges. Collect $(6,3)-(5,5)$ and $(5,5)-(6,7)$ which must be in H and together with all incident edges from the set $\left\{\right(1,1),(1,5),(2,7),(6,7),(5,1\left)\right\}$, it happens to form a cycle $(1,1)$, $(2,3)$, $(1,5)$, $(2,7)$, $(4,8)$, $(6,7)$, $(5,5)$, $(6,3)$, $(5,1)$, $(3,2)$, $(1,1)$, see Figure 60. This is a contradiction since this cycle does not contain all vertices of $G(6,8,2)$.

**Case 10.2:**$k\ge 1$ and $l\ge 2$. We obtain a contradiction similar to Case 1 by considering $\left\{\right(1,4i+1\left)\right|0\le i\le l\}$, $\left\{\right(4i+2,n-1),(4i+1,1\left)\right|0\le i\le k\}$, $\left\{\right(m-1,5\left)\right\}$ and $\left\{\right(m,4i+3\left)\right|1\le i\le l\}$ instead, see Figure 61 for a cycle on $G(14,16,2)$.

**Case 11:**$k<l$, $r=3$ and $q=1$. We obtain a contradiction similar to Case 1 by considering $\left\{\right(2,4i+1),(m-1,4i+1\left)\right|0\le i\le l\}$ and $\left\{\right(4i+2,1),(4i+2,n\left)\right|0\le i\le k\}$ instead, see Figure 62a for a cycle on $G(15,17,2)$.

**Case 12:**$k<l$, $r=3$ and $q=2$. We obtain a contradiction similar to Case 1 by considering $\left\{\right(2,4i+1),(m-1,4i+1\left)\right|0\le i\le l\}$ and $\left\{\right(4i+2,1),(4i+2,n-1\left)\right|0\le i\le k\}$ instead, see Figure 62b for a cycle on $G(15,18,2)$.

**Case 13:**$k\le l$, $r=3$ and $q=4$.

**Case 13.1:**$k=1$ and $l=1$. Since $(1,1)$, $(1,5)$, $(7,1)$ and $(7,5)$ have only 2 incident edges on $G(7,8,2)$, $(1,1)-(2,3)$, $(2,3)-(1,5)$, $(7,1)-(6,3)$ and $(6,3)-(7,5)$ must be in H and it forces that $(3,1)-(2,3)$ and $(5,1)-(6,3)$ must not be in H. Then, it also forces that $(1,2)-(3,1)$, $(3,1)-(5,2)$, $(3,2)-(5,1)$ and $(5,1)-(7,2)$ must be in H. Thus, $(3,2)$ and $(5,2)$ already have two incident edges on H and it forces again that $(3,2)-(2,4)$ and $(5,2)-(6,4)$ must not be in H. Next, since all vertice in $\left\{\right(1,1),(1,2),(1,5),(2,8),(4,8),(6,8),(7,5),(7,2),(7,1\left)\right\}$ have only 2 incident edges. Collect $(2,4)-(1,6)$, $(3,1)-(5,2)$, $(3,2)-(5,1)$ and $(6,4)-(7,6)$ which must be in H together with all incident edges from $\left\{\right(1,1),(1,2),(1,5),(2,8),(4,8),(6,8),(7,5),(7,2),(7,1\left)\right\}$, it happens to form a cycle $(1,1)$, $(2,3)$, $(1,5)$, $(2,7)$, $(4,8)$, $(6,7)$, $(7,5)$, $(6,3)$, $(7,1)$, $(5,2)$, $(3,1)$, $(1,2)$, $(2,4)$, $(1,6)$, $(2,8)$, $(4,7)$, $(6,8)$, $(7,6)$, $(6,4)$, $(7,2)$, $(5,1)$, $(3,2)$, $(1,1)$, see Figure 63. This is a contradiction since this cycle does not contain all vertices of $G(7,8,2)$.

**Case 13.2:**$k=1$ and $l\ge 2$. Since $(1,1)$, $(1,5)$, $(7,1)$ and $(7,5)$ have only 2 incident edges on $G(7,n,2)$, $(1,1)-(2,3)$, $(2,3)-(1,5)$, $(7,1)-(6,3)$ and $(6,3)-(7,5)$ must be in H and it forces that $(3,1)-(2,3)$ and $(5,1)-(6,3)$ must not be in H. Then, it also forces that $(1,2)-(3,1)$, $(3,1)-(5,2)$, $(3,2)-(5,1)$ and $(5,1)-(7,2)$ must be in H. Next, since all vertice in $\left\{\right(1,4i+1),(7,4i+1\left)\right|0\le i\le l\}$, $\left\{\right(1,4i+2),(7,4i+2\left)\right|0\le i\le l-1\}$ and $\left\{\right(2,n-4),(6,n-4),(2,n),(4,n),(6,n\left)\right\}$ have only 2 incident edges. Collect $(3,1)-(5,2)$ and $(3,2)-(5,1)$ which must be in H together with all incident edges from these two sets, it happen to form a cycle $(1,1)$, $(2,3)$, $(1,5)$, …, $(2,n-5)$, $(1,n-3)$, $(2,n-1)$, $(4,n)$, $(6,n-1)$, $(7,n-3)$, …, $(7,5)$, $(6,3)$, $(7,1)$, $(5,2)$, $(3,1)$, $(1,2)$, $(2,4)$, …, $(1,n-6)$, $(2,n-4)$, $(1,n-2)$, $(2,n)$, $(4,n-1)$, $(6,n)$, $(7,n-2)$, $(6,n-4)$, $(7,n-6)$, …, $(6,4)$, $(7,2)$, $(5,1)$, $(3,2)$, $(1,1)$, see Figure 64 for a cycle on $G(7,16,2)$. This is a contradiction since this cycle does not contain all vertices of $G(7,4l+4,2)$.

**Case 13.3:**$k\ge 2$. We obtain a contradiction similar to Case 1 by considering $\left\{\right(1,4i+1),(m,4i+1\left)\right|0\le i\le l\}$, $\left\{\right(1,4i+2),(m,4i+2\left)\right|0\le i\le l-1\}$, $\left\{\right(2,n-4),(m-1,n-4\left)\right\}$, $\left\{\right(4i+2,n\left)\right|0\le i\le k\}$, $\left\{\right(4i+4,n\left)\right|0\le i\le k-1\}$, $\left\{\right(4i+1,1),(4i+1,2\left)\right|0\le i\le k-2\}$ and $\left\{\right(4i+3,1),(4i+3,2\left)\right|1\le i\le k\}$ instead, see Figure 65 for a cycle on $G(15,16,2)$.

**Case 14:**$k<l$, $r=4$ and $q=1$.

**Case 14.1:**$k=1$. Since $(1,1)$, $(1,5)$, $(1,n-4)$ and $(1,n)$ have only 2 incident edges on $G(8,4l+1,2)$, $(1,1)-(2,3)$, $(2,3)-(1,5)$, $(1,n-4)-(2,n-2)$ and $(2,n-2)-(1,n)$ must be in H and it forces that $(4,2)-(2,3)$ and $(4,n-1)-(2,n-2)$ must not be in H. Then, it also forces that $(4,2)-(6,1)$ and $(4,n-1)-(6,n)$ must be in H. Next, since all vertice in $\left\{\right(1,4i+1),(2,4i+1\left)\right|0\le i\le l\}$, $\left\{\right(5,1),(5,n\left)\right\}$ and $\left\{\right(8,4i+2),(8,4i+4\left)\right|0\le i\le l-1\}$ have only 2 incident edges. Collect $(4,2)-(6,1)$ and $(4,n-1)-(6,n)$ which must be in H together with all incident edges from these three sets, it happens to form a cycle $(1,1)$, $(2,3)$, $(1,5)$, …, $(1,n-4)$, $(2,n-2)$, $(1,n)$, $(3,n-1)$, $(5,n)$, $(7,n-1)$, $(8,n-3)$, …, $(7,4)$, $(8,2)$, $(6,1)$, $(4,2)$, $(2,1)$, $(1,3)$, $(2,5)$, …, $(2,n-4)$, $(1,n-2)$, $(2,n)$, $(4,n-1)$, $(6,n)$, $(8,n-1)$, $(7,n-3)$, …, $(8,4)$, $(7,2)$, $(5,1)$, $(3,2)$, $(1,1)$, see Figure 66 for a cycle on $G(8,13,2)$. This is a contradiction since this cycle does not contain all vertices of $G(8,4l+1,2)$.

**Case 14.2:**$k\ge 2$. We obtain a contradiction similar to Case 1 by considering $\left\{\right(1,4i+1),(2,4i+1\left)\right|0\le i\le l\}$, $\left\{\right(4i+1,1),(4i+1,n\left)\right|0\le i\le k\}$, $\left\{\right(4i+2,1),(4i+2,n\left)\right|0\le i\le k-1\}$, $\left\{\right(m-4,2),(m-4,n-1\left)\right\}$ and $\left\{\right(m,4i+2),(m,4i+4\left)\right|0\le i\le l-1\}$ instead, see Figure 67 for a cycle on $G(16,17,2)$.

**Case 15:**$k<l$, $r=4$ and $q=2$.

**Case 15.1:**$k=1$ and $l\ge 2$. Since $(1,n-4)$ and $(1,n)$ have only 2 incident edges on $G(8,4l+2,2)$, $(1,n-4)-(2,n-2)$ and $(2,n-2)-(1,n)$ must be in H and it forces that $(2,n-2)-(3,n)$ must not be in H. Then, it also forces that $(1,n-1)-(3,n)$ and $(3,n)-(5,n-1)$ must be in H. Next, since all vertice in $\left\{\right(1,4i+1),(7,4i+2\left)\right|0\le i\le l\}$ and $\left\{\right(5,1\left)\right\}$ have only 2 incident edges. Collect $(1,n-1)-(3,n)$ and $(3,n)-(5,n-1)$ which must be in H together with all incident edges from these two sets, it happens to form a cycle $(1,1)$, $(2,3)$, $(1,5)$, …, $(1,n-5)$, $(2,n-3)$, $(1,n-1)$, $(3,n)$, $(5,n-1)$, $(7,n)$, $(8,n-2)$, $(7,n-4)$, …, $(8,4)$, $(7,2)$, $(5,1)$, $(3,2)$, $(1,1)$, see Figure 68 for a cycle on $G(8,14,2)$. This is a contradiction since this cycle does not contain all vertices of $G(8,4l+2,2)$.

**Case 15.2:**$k\ge 2$. We obtain a contradiction similar to Case 1 by considering $\left\{\right(1,4i+1),(m-1,4i+2\left)\right|0\le i\le l\}$, $\left\{\right(4i+1,1\left)\right|0\le i\le k\}$, $\left\{\right(m-5,n\left)\right\}$ and $\left\{\right(4i+1,n-1\left)\right|1\le i\le k-1\}$ instead, see Figure 69 for a cycle on $G(16,18,2)$.

**Case 16:**$k<l$, $r=4$ and $q=3$. We obtain a contradiction similar to Case 1 by considering $\left\{\right(1,4i+1),(2,4i+1\left)\right|0\le i\le l-1\}$, $\left\{\right(1,n-4),(2,n-4\left)\right\}$, $\left\{\right(4i+1,1),(4i+1,n\left)\right|0\le i\le k\}$, $\left\{\right(4i+2,1),(4i+2,n\left)\right|0\le i\le k-1\}$, $\left\{\right(m-4,2),(m-4,n-1\left)\right\}$, $\left\{\right(m,4i+2\left)\right|0\le i\le l\}$ and $\left\{\right(m,4i\left)\right|1\le i\le l\}$ instead, see Figure 70 for a cycle on $G(16,19,2)$.

**Theorem**

**8.**

**Proof.**

**Case 1:**$r=3$.

**Case 1.1:**m is odd and n is even, or m is even and n is odd. We partition the RB$(m,n,3)$ into LB$(m,n-3,3)$ and 7B$(m,n-3,3)$, see Figure 71a for RB$(10,11,3)$. Since $m+n-3$ is even and $m+n-3\ge 12$, by Theorem 5(b), the LB$(m,n-3,3)$ contains an OKT from $(1,3)$ to $(2,2)$ and by Corollary 2(b), the 7B$(m,n-3,3)$ contains an OKT from $(3,1)$ to $(2,2)$. By joining $(1,3)$ and $(2,2)$ of LB$(m,n-3,3)$ to $(2,2)$ and $(3,1)$ of 7B$(m,n-3,3)$, respectively, we obtain a CKT on RB$(m,n,3)$ as shown in Figure 71b for the RB$(10,11,3)$.

**Case 1.2:**m and n are odd or even. We partition the RB$(m,n,3)$ into LB$(m,n-3,3)$ and 7B$(m,n-3,3)$, see Figure 72a for RB$(11,13,3)$. Since $m+n-3$ is odd and $m+n-3\ge 11$, by Theorem 5(a), the LB$(m,n-3,3)$ contains an OKT from $(1,2)$ to $(1,3)$ and by Corollary 2(a), the 7B$(m,n-3,3)$ contains an OKT from $(2,1)$ to $(3,1)$. By joining $(1,2)$ and $(1,3)$ of LB$(m,n-3,3)$ to $(2,1)$ and $(3,1)$ of 7B$(m,n-3,3)$, respectively, we obtain a CKT on RB$(m,n,3)$ as shown in Figure 72b for the RB$(11,13,3)$.

**Case 2:**for $r=4$. We partition the RB$(m,n,4)$ into LB$(m,n-4,4)$ and 7B$(m,n-4,4)$, see Figure 73a for RB$(11,13,4)$. By Theorem 4 and Corollary 1, the LB$(m,n-4,4)$ has a CKT that contains an edge $(1,4)-(3,3)$ and 7B$(m,n-4,4)$ has a CKT that contains an edge $(4,1)-(2,2)$. By deleting $(1,4)-(3,3)$ of LB$(m,n-4,4)$ and $(4,1)-(2,2)$ of 7B$(m,n-4,4)$ and joining $(1,4)$ and $(3,3)$ of LB$(m,n-4,4)$ to $(2,2)$ and $(4,1)$ of 7B$(m,n-4,4)$, respectively, we obtain a CKT on RB$(m,n,4)$, as show in Figure 73b for RB$(11,13,4)$.

**Case 3:**$r\ge 5$.

**Case 3.1:**r is even. We partition the RB$(m,n,r)$ into two CB($r\times (n-r)$) and two CB($(m-r)\times r$), see Figure 74a for RB$(13,14,6)$. There are three steps to obtain a CKT which has some edges on each partitioned board. First, we consider a CB($r\times (m-r)$). By Theorem 1, it contains a CKT having edges $(1,m-r-1)-(3,n-r)$ and $(r,2)-(r-1,4)$. Rotate CB($r\times (m-r)$) 90 degrees clockwise, we obtain a CKT on CB($(m-r)\times r$) of the upper right-hand side having edges $(m-r-1,r)-(m-r,r-2)$ and $(2,1)-(4,2)$. Next, rotate CB($r\times (m-r)$) 90 degrees counterclockwise, we obtain a CKT on CB($(m-r)\times r$) of the lower left-hand side having edge $(m-r-3,r-1)-(m-r-1,r)$. Finally, we consider a CB($r\times (n-r)$) on the upper left-hand side. By Theorem 1, it contains a CKT having edges $(1,n-r-1)-(3,n-r)$ and $(r,2)-(r-1,4)$. Rotate CB($r\times (n-r)$) 180 degrees clockwise, we obtain a CKT on CB($r\times (n-r)$) of the lower right-hand side having edges $(r-2,1)-(r,2)$ and $(1,n-r-1)-(3,n-r-3)$.

**Case 3.2:**r is odd. We partition the RB$(m,n,r)$ into two CB($r\times (n-r)$) and two CB($(m-r)\times r$), see Figure 75a for RB$(12,13,5)$. There are three steps to obtain an OKT having two end-points on each partitioned board. First, we consider a CB($r\times (m-r)$). By Theorem 6(b), it contains an OKT from $(r,1)$ to $(2,m-r-1)$. Rotate CB($r\times (m-r)$) 90 degrees clockwise, we obtain an OKT from $(1,1)$ to $(m-r-1,r-1)$ on CB($(m-r)\times r$) of the upper right-hand side. Next, rotate CB($r\times (m-r)$) 90 degrees counterclockwise, we obtain an OKT from $(m-r,r)$ to $(2,2)$ on CB($(m-r)\times r$) of the lower left-hand side. Finally, we consider a CB($r\times (n-r)$) on the upper left-hand side. By Theorem 6, it contains an OKT from $(r,1)$ to $(2,n-r-1)$. Rotate CB($r\times (n-r)$) 180 degrees clockwise, we obtain an OKT from $(1,n-r)$ and $(r-1,2)$ on CB($r\times (n-r)$) of the lower right-hand side.

## 5. Conclusions and Discussion

## Author Contributions

## Funding

## Conflicts of Interest

## Abbreviations

CKT | Closed Knight’s Tour |

OKT | Open Knight’s Tour |

CB$(m\times n)$ | $m\times n$ chessboard |

LB$(m,n,r)$ | L-board of size $(m,n,r)$ |

7B$(m,n,r)$ | 7-board of size $(m,n,r)$ |

RB$(m,n,r)$ | $(m,n,r)$-ringboard |

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**Figure 4.**CKTs for LB$(6,5,4)$, LB$(5,6,4)$, LB$(7,6,4)$, LB$(6,7,4)$, LB$(8,5,4)$, LB$(5,8,4)$, LB$(8,7,4)$ and LB$(7,8,4)$.

**Figure 43.**OKTs from $(m,1)$ to $(2,n-1)$ on the CB($m\times n$) where $m\in \{5,7\}$ and $n\in \{6,8\}$.

**Figure 50.**OKTs from $(m,1)$ to $(2,n-1)$ on the CB($m\times n$) where $m\in \{6,8\}$ and $n\in \{5,7\}$.

**Figure 56.**Cycles on $G(14,18,2)$, $G(15,19,2)$, $G(16,20,2)$, $G(13,14,2)$ and $G(13,15,2)$, respectively.

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**MDPI and ACS Style**

Srichote, W.; Boonklurb, R.; Singhun, S.
Closed Knight’s Tours on (*m*,*n*,*r*)-Ringboards. *Symmetry* **2020**, *12*, 1217.
https://doi.org/10.3390/sym12081217

**AMA Style**

Srichote W, Boonklurb R, Singhun S.
Closed Knight’s Tours on (*m*,*n*,*r*)-Ringboards. *Symmetry*. 2020; 12(8):1217.
https://doi.org/10.3390/sym12081217

**Chicago/Turabian Style**

Srichote, Wasupol, Ratinan Boonklurb, and Sirirat Singhun.
2020. "Closed Knight’s Tours on (*m*,*n*,*r*)-Ringboards" *Symmetry* 12, no. 8: 1217.
https://doi.org/10.3390/sym12081217