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

A (legal) knight’s move is the result of moving the knight two squares horizontally or vertically on the board and then turning and moving one square in the perpendicular direction. A closed knight’s tour is a knight’s move that visits every square on a given chessboard exactly once and returns to its start square. A closed knight’s tour and its variations are studied widely over the rectangular chessboard or a three-dimensional rectangular box. For m,n>2r, an (m,n,r)-ringboard or (m,n,r)-annulus-board is defined to be an m×n chessboard with the middle part missing and the rim contains r rows and r columns. In this paper, we obtain that a (m,n,r)-ringboard with m,n≥3 and m,n>2r has a closed knight’s tour if and only if (a) m=n=3 and r=1 or (b) m,n≥7 and r≥3. If a closed knight’s tour on an (m,n,r)-ringboard exists, then it has symmetries along two diagonals.


Introduction
The m × n chessboard or CB(m × n) is the generalization of the regular CB (8 × 8). It consists of m rows of n arrays of squares. Suppose the squares of the CB(m × n) are labeled by (i, j) in the matrix fashion. A legal knight's move is the result of a moving the knight two squares horizontally or vertically on the CB(m × n) and then turning and moving one square in the perpendicular direction. That is, if we start at (i, j), then the knight can move to one of eight squares: (i ± 2, j ± 1) or (i ± 1, j ± 2) (if exists).
A closed knight's tour (CKT) is a legal knight's move that visits every square on a given chessboard exactly once and returns to its start square. While, an open knight's tour (OKT) is a legal knight's move that visits every square on a given chessboard exactly once and the starting and terminating squares are different. Both CKT and OKT problems on a two-dimensional or three-dimensional chessboard are one of the interesting mathematical problems as you can see some of them listed in [1][2][3][4][5][6]. Not only the legal knight's move, but some researchers also extended it to be an (a, b)-knight's move which is the result of a moving the knight a squares horizontally or vertically on the CB(m × n) and then turning and moving b squares in the perpendicular direction. Several mathematical problems along this direction were considered, see for examples [7][8][9] and references therein for details.
In 1991, Schwenk [10] obtained necessary and sufficient conditions for the existence of a CKT for the CB (m × n) as follows.
For the CB(m × n) that contains no CKTs, DeMaio and Hippchen [11] and Bullington et al. [12] can provide the minimal number of squares to be removed or to be added in order for the obtained new board to have a CKT. In particular, for m = 3 or m and n are odd, Miller and Farnsworth [13] and Bi el al. [14] provided the exact position of a square to be removed from CB(m × n) so that the remaining board admits a CKT. However, for the case m = 4, the exact positions for two squares to be removed still open for researchers to explore.
In 2005, Chia and Ong [9] obtained necessary and sufficient conditions for the existence of an OKT for the CB(m × n) as follows.
In this article, we consider one of the variations of the CKT problem by considering the chessboard that the middle part is missing which is called (m, n, r)-ringboard or (m, n, r)-annulus board and we denote it by RB(m, n, r). Definition 1. Let m, n and r be integers such that m, n > 2r. An RB(m, n, r) is defined to be a CB(m × n) with the middle part missing and the rim containing exactly r rows and r columns.
In 1996, Wiitala [15] showed that the RB(m, m, 2) contains no CKT. However, the characterization of the general RB(m, n, r) has not been given. Thus, we try to establish the characterization like the one given by Schwenk [10]. Actually, the CKT problem on the RB(m, n, r) can be converted to a certain graph problem. If we regard each square of the RB(m, n, r) as a vertex, then a knight graph G(m, n, r) represented all legal knight's moves on RB(m, n, r) is a graph with 2r(m + n − 2r) vertices and two vertices (a, b) and (c, d) are joined by an edge whenever the knight can be moved from one square to another by a legal knight's move and this edge is denoted by (a, b) − (c, d). Then, a CKT (respectively, OKT) on the RB(m, n, r) is a Hamiltonian cycle (respectively, Hamiltonian path) in G(m, n, r). The following theorem is a necessary condition for the existence of a Hamiltonian path in a graph that we often use in this article. The goal of this article is to prove that for m, n ≥ 3 and m, n > 2r, the RB(m, n, r) admits a closed knight's tour if and only if (a) m = n = 3 and r = 1 or (b) m, n ≥ 7 and r ≥ 3. In order to reach our goal, we need to divide our RB(m, n, r) into small pieces depending on r. If r ≥ 5 is even, then RB(m, n, r) is divided into four smaller rectangular chessboard and we can use Theorem 1 to construct the CKT for RB(m, n, r) which will be elaborated in Case 3.1 of Theorem 8 in Section 4. However, if r ≥ 5 is odd and RB(m, n, r) is divided into four smaller rectangular chessboards, then there is a case that Theorem 1 cannot be used (Case 3.2 of Theroem 8). Thus, we need to construct our own CKT base on the existence of an OKT on some rectangular chessboards which will be constructed in Theorem 6 in Section 3. For small r, namely r ∈ {3, 4}, we need to divide RB(m, n, r) into two parts, namely an L-board and a 7-board of widths 3 or 4 which we denote by LB(r, c, 3), LB(r, c, 4), 7B(r, c, 3) and 7B(r, c, 4) depending on the numbers of row r and columns c (See Cases 1 and 2 of Theorem 8). For example, Figure 1 illustrates that RB (10,11,3) is divided into LB (10,8,3) and 7B (10,8,3) and RB (11,13,4) is divided into LB (11,9,4) and 7B (11,9,4).  (11,9,4) and 7B (11,9,4).
Therefore, to construct the CKT on the ringboard for this case, we prove the existence of a CKT on LB(m, n, 4) and 7B(m, n, 4) and the existence of some OKTs on LB(m, n, 3) and 7B(m, n, 3) are given in Theorems 4 and 5 in Section 2. For r = 2, we prove the extension of Wiitala's result in [15] which is the non-existence of the CKT on the RB(m, n, 2) in Theorem 7 in Section 4. Finally, the conclusion and discussion about our future research are in Section 5.
We note that Theorem 4 and Corollary 1 will be used in Case 2 of Theorem 8 in Section 4. Next, we construct two OKTs for the LB(m, n, 3) and two OKTs for the 7B(m, n, 3) for m, n ≥ 4.  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 ≥ 12 or (ii) m = 6 and n = 4.
Proof. Let m, n ≥ 4. (a) We assume that the LB(m, n, 3) contains an OKT from (1, 2) to (1, 3) and let m + n is even; or m = 5 and m + n < 11; or n = 4 and m + n < 11. If m + n is even, then the numbers of white squares and black squares are not the same. Thus, the two end-points of this OKT must have the same color. However, (1, 2) and (1, 3) are next to each other and have different colors, which is a contradiction.
Let m = 5 or n = 4 and m + n < 11. By the above argument m + n must be odd and since m, n ≥ 4, we have m + n = 9 which implies that m = 4 and n = 5.
If m + n is odd, then the numbers of white squares and black squares are the same. Thus, the two end-points of this OKT must have the different color. However, (1, 3) and (2, 2) have the same color, a contradiction. Next, we consider the cases that m = 4 and n = 4; or m = 4 and n = 6; or m = 5 and n = 5.
For m = 4 and n = 4, let G 1 be a knight graph of the LB(4, 4, 3). Consider Figure 18. By Theorem 3, we obtain a contradiction. For m = 4 and n = 6, let G 2 be a knight graph of the LB(4, 6, 3). Consider Figure 19. By Theorem 3, we obtain a contradiction.  On the other hand, let us assume that m + n is even and m + n ≥ 12; or m = 6 and n = 4. If m = 6 and n = 4, then the required OKT is presented in Figure 21.   Next, for the larger LB, we start by constructing an OKT on CB(4 × 3) from (1, 3) to (4, 1) and contains an edge (2, 2) − (4, 3) as shown in Figure 28. Then, we construct two paths on CB(4s × 3), where s ≥ 2. Let us connect s CB(4 × 3)'s in Figure 28 on the top of each other and do the following.
We note that Theorem 5(b) and Corollary 2(b) will be used in Case 1.1 of Theorem 8 in Section 4. While, Theorem 5(a) and Corollary 2(a) will be used in Case 1.2 of Theorem 8 in Section 4.

Existence of a Special OKT on CB(m × n)
The following theorem gives necessary and sufficient conditions on the existence of a special OKT on CB(m × n) from (m, 1) to (2, n − 1). This OKT will be used to prove our main result for r ≥ 5 when r is odd (Case 3.2 of Theorem 8 in Section 4).
Let n ≥ m ≥ 5. Then, a CB(m × n) contains an OKT from (m, 1) to (2, n − 1) if and only if m and n are not both even.
Case 2: For m = 3 and n = 4, let G 1 be a knight graph of the CB(3 × 4). We assume that G 1 contains a Hamiltonain path from (3, 1) to (2,3). Figure 30. By Theorem 3, we obtain a contradiction. Case 3: For m = 4 and n is odd such that n ≥ 5. Let G 2 be a knight graph of the CB(4 × n). We assume that G 2 contains a Hamiltonian path from (m, 1) to (2, n − 1). Consider (3, l) | j is even, 2 ≤ j ≤ n − 3, l is odd and 1 ≤ l ≤ n}. Then, we can use mathematical induction to show that ω(G 2 − S) = n + 1 > n = |S| + 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 ≥ 4. Assume that CB(4 × n) contains an OKT from (4, 1) to (2, n − 1). Since CB(4 × 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.
On the other hand, let us assume that m and n are not both even such that n ≥ m ≥ 5. Then, we consider three cases as follows.

Main Theorem
To characterize the RB(m, n, r) according to the existence of its CKT, let us first consider the case when r = 2. It is known from Wiitala [15] that RB(m, m, 2) admits no CKTs. The following theorem can be regarded as an extended result of Wiitala [15]. Recall that G(m, n, r) is the knight graph of the RB(m, n, r).

Theorem 7.
There are no CKT on RB(m, n, 2) for all n > m ≥ 5.
Proof. Let m and n be integers such that n > m ≥ 5. Then, there exist positive integers k and l and r, q ∈ {1, 2, 3, 4} such that m = 4k + r and n = 4l + q. Assume that there exists a CKT H on RB(m, n, 2) which is a Hamiltonian cycle on G(m, n, 2).

Conclusions and Discussion
In this paper, we have obtained necessary and sufficient conditions for the existence of a CKT for the RB(m, n, r). In every case of Theorem 8, it can be seen that the CKTs are constructed by smaller board-pieces that have diagonal or horizontal or vertical symmetries. As a consequence, to obtain our main result, we have to study the existence of a CKT on LB(m, n, 3) and LB(m, n, 4). In the future, an interesting study is to find necessary and sufficient conditions for the existence of a CKT for the general L-board, namely LB(m, n, l, u), which is the L-board consisting of m rows n with the lower leg of width l and the upper leg of width u, see Figure 76.

Conflicts of Interest:
The authors declare no conflict of interest.

Abbreviations
The following abbreviations are used in this manuscript: