Some forbidden rectangular chessboards with an (a, b)-knight s move

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1 The 22 nd Annual Meeting in Mathematics (AMM 2017) Department of Mathematics, Faculty of Science Chiang Mai University, Chiang Mai, Thailand Some forbidden rectangular chessboards with an (a, b)-knight s move Krit Karudilok a,,, Sirirat Singhun a, and Ratinan Boonklurb b a Department of Mathematics, Faculty of Science, Ramkhamhaeng University, Bangkok 10240, Thailand b Department of Mathematics and Computer Science, Faculty of Science, Chulalongkorn University, Bangkok 10330, Thailand Abstract The m n chessboard is an array with squares arranged in m rows and n columns. An (a, b)-knight s move is a move from a square to another square by moving a knight passing a squares vertically or a squares horizontally and then passing b squares at 90 degrees angle. A closed (a, b)-knights tour is an (a, b)-knights move from a square to another square such that the knight lands on every square on m n chessboard once and returns to its starting square. In this paper, we obtain a new certain chessboard that admits no closed (a, b)-knight s tours depending on a and b. Moreover, the necessary condition that make the chessboard admits no closed (a, b)-knight s tours for any m n chessboard is obtained. Keywords: closed knight s tour, Hamiltonian cycle MSC: Primary 05C45; Secondary 05C38. 1 Introduction and Preliminaries The knight is a chess piece that can move one square vertically or one square horizontally and then two squares move at 90 degrees angle. In 1959, Leonhard Euler [3] found that on the 8 8 chessboard, an array with squares arranged in eight rows and eight columns, the knight can move from a square to another square such that it lands on every square once and returns to its starting square. This knight s move is called a closed knight s tour. The 8 8 chessboard is extended to the m n chessboard, an array with squares arranged in m rows and n columns. The knight s tour problem on the m n chessboard is converted to question about a certain graph. A graph G represented the m n chessboard is a graph with mn vertices where each square of board is replaced by a vertex and two vertices are joined by an edge if the knight can move from a square to another square. If G is a graph represented the m n chessboard, then the degree of vertex v, denoted by deg v, is the number of squares that the knight can move to. Then, the closed knight s tour is a Hamiltonian cycle in G. In 1991, Schwenk [2] obtained the sufficient and necessary conditions for the m n chessboard to admit a closed knight s tour. Corresponding author. Speaker. address: Kritkarudilok@hotmail.com (K. Karudilok), sin sirirat@ru.ac.th (S. Singhun), ratinan.b@chula.ac.th (R. Boonklurb). GRA-01-1

2 Theorem 1.1. [2] The m n chessboard with m n admits a closed knight s tour unless one or more of the following conditions hold: (i) m and n are both odd; or (ii) m = 1 or 2 or 4; or (iii) m = 3 and n = 4 or 6 or 8. In 2003, Chia et al. [1] continued to consider the m n chessboard while the knight s move was extended to a squares vertically or a squares horizontally and then b squares move at 90 degrees angle. This move is called an (a, b)-knight s move. Then, an (a, b)-knight s move and (b, a)-knight s move are the same. If the knight moves to each square of the chessboard only once with an (a, b)-knight s move and returns to the starting square, then this move is called a closed (a, b)-knight s tour. In the case that a = b, the knight can move from a square to another square with the same colour, black to black or white to white. Then, the m n chessboard admits no closed (a, b)-knight s tours. We shall assume that a < b. Some observations in [1] are obtained as follow. Theorem 1.2. [1] Suppose that the m n chessboard admits a closed (a, b)-knights tour, where a < b and m n. Then, (i) a + b is odd; and (ii) m or n is even; and (iii) m a + b; and (iv) n 2b. By Theorem 1.1 (ii), the 4 n chessboard admits no closed (1, 2)-knight s tours. Then, Chia et al. [1] also obtained some chessboard which admits no closed (a, b)-knight s tours. Theorem 1.3. [1] Suppose that m = a + b + 2t + 1, where 0 t a 1. Then, the m n chessboard admits no closed (a, b)-knights tours. Theorem 1.4. [1] Suppose that m = a(k + 2l), where 1 l k 2. Then, the m n chessboard admits no closed (a, ak)-knights tours, where a is odd and k is even. Theorem 1.5. [1] Suppose that m = 2(ak + l), where 1 k l a. Then, the m n chessboard admits no closed (a, a + 1)-knights tours. Theorem 1.6. [1] Suppose that m = 2a + 2t + 1, where 1 t a 1. Then, the m n chessboard admits no closed (a, a + 1)-knights tours. By Theorem 1.3, the certain chessboard that admits no closed (a, b)-knight s tours for any a < b depends on a and b. In this paper, we obtain a new certain chessboard that admits no closed (a, b)-knight s tours depending on a and b in Theorem 2.4. Moreover, the necessary condition that make the chessboard admitted no closed (a, b)-knight s tours for any m n chessboard is obtained in Theorem Main Results In this section, we present the certain chessboard that admits no closed (a, b)-knight s tours for any a and b. We consider two cases : a = 1 (in Lemma 2.1) and a > 1 (in Lemma 2.2 and 2.3). Next, the necessary condition that makes the chessboard admits no closed (a, b)-knight s tours for any m n chessboard is obtained. For convenience, to talk about the m n chessboard, we label the square (i, j), where i {1, 2, 3,..., m} and j {1, 2, 3,..., n}, counting from the upper left conner in matrix fashion. That is the square (1,1) is on the left upper side of the chessboard. These (i, j) labels also represent vertices in the graph representation of the m n chessboard. If the knight starts at the square or vertex (i, j) and moves with an (a, b)-knight s move, then the knight can move to one of eight squares or vertices : (i ± a, j ± b) or (i ± b, j ± a). Lemma 2.1. Suppose that a = 1, m = a + b and n = 2b. Then, the m n chessboard admits no closed (a, b)-knight s tours. GRA-01-2

3 Proof. Let a = 1. By Theorem 1.2, b is even. Then, the chessboard contains b + 1 rows and 2b columns. For i {1, 2, 3,..., b 2 }, deg (2i, b + 1) = 2 and the vertex (2i, b + 1) is adjacent to vertices (2i 1, 1) and (2i + 1, 1). Since the vertex (1, 1) is adjacent to (2, b + 1), let P 1 : (1, 1)(2, b + 1)(3, 1)(4, b + 1) (b, b + 1)(b + 1, 1) be a path from (1, 1) to (b + 1, 1). For i {1, 2, 3,..., b 2 }, deg (2i, b + 2) = 2 and the vertex (2i, b + 2) is adjacent to vertices (2i 1, 2) and (2i + 1, 2). Since the vertex (1, 2) is adjacent to (2, b + 2), let P 2 : (1, 2)(2, b + 2)(3, 2)(4, b + 2) (b, b + 2)(b + 1, 2) be a path from (1, 2) to (b + 1, 2). If the chessboard contains a closed (a, b)-knight s tour C, then P 1 and P 2 must be subpaths of C. Since the vertices (1, 1), (b, b + 2), (b + 1, 1) and (2, b + 2) are of degree two, the edges (1, 1)(b + 1, 2) and (1, 2)(b + 1, 1) must be edges in C. Thus, P 1 and P 2 together with such two edges form a cycle which does not contain all vertices in the chessboard as shown in Figure 1. This makes a contradiction. Figure 1: The forced cycle in the (b + 1) 2b chessboard. In the case that a > 1, let B be an m n chessboard where m = a + b and n = 2b. We partition the chessboard into three subboards, B 1, B 2 and B 3. The B 1 denote the subboard of B containing all four corners of a a chessboards. The top left corner of B 1 is located from the first row to the a th row and from the first column to the a th column of B. The top right corner of B 1 is located from the first row to the a th row and from the (2b a + 1) th column to the (2b) th column of B. The bottom left corner of B 1 is located from the (b + 1) th row to the (a + b) th row and from the first column to the a th column of B. The bottom right corner of B 1 is located from the (b + 1) th row to the (a + b) th row and from the (2b a + 1) th column to the (2b) th column of B. The B 2 denote the subboard of B located from the (a + 1) th row to the b th row and from the first to the last column of B. The B 3 denote the subboard of B containing two parts. The first one is located from the first row of B to the a th row and the (a + 1) th column to the (2b a) th column of B. The second one is located from the (b + 1) th row to the last row of B and the (a + 1) th column to the (2b a) th column of B. The partition of the chessboard is shown in Figure 2. Then, we see that (i) if v belongs to B 1, then deg v = 2, (ii) if v belongs to B 2, then deg v = 2, and (iii) if v belongs to B 3, then deg v = 3 Lemma 2.2. Suppose that a > 1, b 2a 1, m = a+b and n = 2b. Then, the m n chessboard admits no closed (a, b)-knight s tours. Proof. Suppose that the m n chessboard to admit a closed (a, b)-knight s tour C of B. We consider the vertex (1, b + 1) that adjacent to (a + 1, 1), (b + 1, b a + 1) and (b + 1, a + b + 1). GRA-01-3

4 Figure 2: The partition of the m n chessboard into three subboards, B 1, B 2 and B 3. It is clear that the vertex (a + 1, 1) belongs to B 2. Since the b 2a 1, b a + 1 a. Then the vertex (b + 1, b a + 1) belongs to B 1. In order to show that (b + 1, b a + 1) belongs to B 1, it suffices to show that 2b (a + b + 1) < a. Since the 2b (a + b + 1) = b a 1 and b 2a 1, b a 1 a 2 < a. Then the vertex (b + 1, a + b + 1) belongs to B 1. Those three vertices are degree-2. Then it is impossible that three edges, (1, b + 1)(a + 1, 1), (1, b + 1)(b + 1, a) and (1, b + 1)(b + 1, a + b + 1) belong to C as shown Figure 3. Therefore, the m n chessboard admits no closed (a, b)-knight s tours C of B. Figure 3: Three edges that incident to (1, b + 1) belonging to C. Lemma 2.3. Suppose that a > 1, b > 2a 1, m = a+b and n = 2b. Then, the m n chessboard admits no closed (a, b)-knight s tours. Proof. Let B be an m n chessboard. Let k = b a, the floor of b a. Then, b = ka + r for some 0 r a 1. If k is even, let L 1 = {v i = (1, b ia) i {0, 2, 4,..., k}}, L 2 = {v i = (b + 1, b ia) i {1, 3, 5,..., k 1}}, R 1 = {u i = (1, b + ia) i {0, 2, 4,..., k}}, R 2 = {u i = (b + 1, b + ia) i {1, 3, 5,..., k 1}}, S e = {s i = (a + 1, 2b ia) i {0, 2, 4,..., k}, S o = {s i = (b a + 1, 2b ia) i {1, 3, 5,..., k 1}}, T e = {t i = (a + 1, ia) i {2, 4, 6,..., k}}, and T o = {t i = (b a + 1, ia) i {1, 3, 5,..., k 1}}. If k is odd, let L 1 = {v i = (1, b ia) i {0, 2, 4,..., k 1}}, L 2 = {v i = (b + 1, b ia) i {1, 3, 5,..., k}}, R 1 = {u i = (1, b + ia) i {0, 2, 4,..., k 1}}, GRA-01-4

5 R 2 = {u i = (b + 1, b + ia) i {1, 3, 5,..., k}}, S e = {s i = (a + 1, 2b ia) i {0, 2, 4,..., k 1}}, S o = {s i = (b a + 1, 2b ia) i {1, 3, 5,..., k}}, T e = {t i = (a + 1, ia) i {2, 4, 6,..., k 1}}, and T o = {t i = (b a + 1, ia) i {1, 3, 5,..., k}}. Then, we obtain the following observations. 1. u 0 = v For v L 1 L 2 R 1 R 2 {u k, v k }, v belongs to B 3 and deg v = The vertex v k belongs to B 1 and deg v k = 2. This can be seen by considering 2 cases. First, if k is even, then v k = (1, b ka) and b ka = r < a. Second, if k is odd, then v k = (b + 1, b ka), b + 1 > b = (a + b) a and b ka = r < a. 4. For v T e T o S e S o, v belongs to B 2 and deg v = 2. This can be seen that if v S e T e is located in (a + 1) th row, then vertex v belongs to B 2. Since b > 2a 1 and a > 1, b a + 1 > a and b a + 1 < b. If v S o T o is located in (b a + 1) th row, then vertex v belongs to B The vertex u k belongs to B 1 and deg u k = 2. This can be seen by considering 2 cases. First, if k is even, then u k = (1, b + ka) and ka = b r where 0 r a 1. Second, if k is odd, then u k = (b + 1, b + ka), b + 1 > b = (a + b) a and b + ka = b + (b r) = 2b r > 2b a. Here, we note that if a chessboard B admits a closed (a, b)-knight s tour C and v is a degree-2 vertex of B, then the two edges incident to v must belong to C. Suppose that there is a closed (a, b)-knight s tour C of B. By observation 4, the edges incident to v for all v T e T o S e S o {u k, v k } must belong to C. We shall find edges must belong to C by considering two cases of k. Case 1: k is even. Consider the left-hand side of B. Since for i {0, 1, 2,..., k}, deg s i = 2 and s i is adjacent to v i, s i v i belong to C. Since deg v k = 2 and v k is adjacent to v k 1 and s k, v k v k 1 and v k s k belong to C. Then, v k 1 v k 2 does not belong to C because s k 1 v k 1 and v k v k 1 belong to C. Next v k 2 v k 3 must belongs to C because s k 1 belongs to C but v k 1 v k 2 does not belong to C. By the same idea, for i {1, 3, 5,..., k 1}, v i v i+1 and v i s i belong to C while v i v i 1 does not belong to C. Thus, v 1 v 0 does not belong to C. For the right-hand side of B, the edges u k u k 1 and t k u k belong to C. Then, u k 1 u k 2 does not belong to C as the same idea on the left-hand side. For i {1, 3, 5,..., k 1}, u i u i+1 and u i t i belong to C while u i 1 u i does not belong to C. Thus, u 0 u 1 must not belong to C. By observation 1, u 0 = v 0. Since deg v 0 = 3 and v 0 s 0 is only one edge incident to v 0 that belongs to C, it is impossible. Therefore, the m n chessboard admits no closed (a, b)-knight s tours. Figure 4: The bold edges represent edges that belong to C when k is even. Case 2 : k is odd. Consider the left-hand side of B. Since for i {0, 1, 2,..., k}, deg s i = 2 and s i is adjacent to v i, s i v i belong to C. Since deg v k = 2 and v k is adjacent to v k 1 and s k, v k v k 1 and v k s k belong to C. Then, v k 1 v k 2 does not belong to C. By the same idea, for i {0, 2, 4,..., k 1}, v i v i+1 and v i s i belong to C while v i v i 1 does not belong to C. Thus, v 1 v 0 belong to C. For the right-hand side of B, the edges u k u k 1 and t k u k belong to C. Then, u k 1 u k 2 does not belong to C. For i {2, 4, 6,..., k 1}, u i u i+1 and u i t i belong to C while u i 1 u i does not GRA-01-5

6 belong to C. That is, u 1 u 2 does not belong to C. Since deg u 1 = 3 and u 1 is adjacent to u 0, u 2 and t 1, u 0 u 1 must belong to C. Since deg s 0 = 2 and s 0 is adjacent to u 0, u 0 s 0 must belong to C. It is impossible that three edges, u 0 v 1 (= v 0 v 1 ), u 0 u 1 and u 0 s 0 (that incident to u 0 ) belong to C. Therefore, the m n chessboard admits no closed (a, b)-knight s tours. Figure 5: The bold edges represent edges that belong to C when k is odd. From Lemma 2.1, 2.2 and 2.3, we can conclude as follow. Theorem 2.4. Suppose that m = a + b and n = 2b. Then, the m n chessboard admits no closed (a, b)-knight s tours. Next, we give a necessary condition that make the chessboard admits no closed (a, b)-knight s tours for any m n chessboard. Theorem 2.5. Let (a, b) 1. Then, there is no closed (a, b)-knight s tours for any m n chessboard. Proof. Assume that (a, b) = d, where d 1. Then, d (a ± b) and d ( a ± b). We label the square (i, j) with the colour k {0, 1, 2,..., d 1} if i + j k (mod d). That is, each square must be labeled by one colour from {0, 1, 2,..., d 1}. We claim that the knight can move from a square to another square with the same colour. Suppose that the knight starts at square (i, j) with colour k. Then, i + j k (mod d). Then, the knight can move to the squares (i ± a, j ± b) or (i ± b, j ± a). We shall show that each square is coloured by k. If the knight move to the square (i ± a, j ± b), then (i ± a) + (j ± b) = (i + j) ± (a ± b) k (mod d). Thus, the square is coloured by k. If the knight move to the square (i ± b, j ± a), then (i ± b) + (j ± a) = (i + j) ± (a ± b) k (mod d). Thus, the square is coloured by k. Hence, the knight can move from a square to another square that is coloured by the same colour. Thus, the knight can not move to all squares of the chessboard. Therefore, there is no closed (a, b)-knight tour for any m n chessboard. Acknowledgment. The authors are grateful to the referees for their careful reading of the manuscript and their useful comments. References [1] G.L. Chia and Siew-Hui Ong, Generalized knights tours on rectangular chessboards, Discrete Applied Math. 150 (2005), [2] A.L. Schwenk, Which rectangular chessboards have a knight s tour, Math. Magazine 64 (1991), [3] J.J. Watkins, Across the boards: The mathematics of the chessboard problems, Princeton University, New Jersey, GRA-01-6