Which Rectangular Chessboards Have a Bishop s Tour?

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1 Which Rectangular Chessboards Have a Bishop s Tour? Gabriela R. Sanchis and Nicole Hundley Department of Mathematical Sciences Elizabethtown College Elizabethtown, PA November 27, Introduction Much has been written about the existence of knight s tours on a rectangular chessboard (see e.g. [2], [3], and [4]). The problem is the following: Given a chessboard of size m n, is it possible for a knight to traverse the chessboard using only legal knight moves (according to the rules of chess) so that each square on the board is visited exactly once? If so, we say that the chessboard has an open knight s tour. A closed knight s tour is one in which there is a legal move connecting the last square visited to the first. In [4], Allen J. Schwenk shows exactly which chessboards admit a closed knight s tour. The problem of determining which boards admit open knight s tours is similar. It would appear at first glance that chessboard tours using other chess pieces are not very interesting. It is easily seen that an open rook s tour exists on any rectangular board, and a closed rook s tour exists if either m or n is even. (A rook moves up and down or left to right, and we consider a square has been visited if the rook passes over it). For example, Figure 1 shows a closed rook s tour on a 7 8 board and an open rook s tour on a 7 9 board. This pattern can clearly be extended to any other size board. If both m and n are odd, then there are an odd number of squares, so any rook s tour must begin and end in the same color square. Since a rook cannot legally move to an adjacent square of the same color, there can 1

2 Figure 1: Rook s tours on 7 8 and 7 9 boards be no closed tour if m and n are both odd. Since kings and queens can move either diagonally, up and down, or left and right, constructing closed or open tours with these pieces is trivial. So the only piece left for consideration is the bishop. Since a bishop moves diagonally, it can only ever visit squares of one color. Thus, if it begins on a white square, it will never visit any of the black squares. So bishop s tours that visit every square are impossible. The best we can hope for is a bishop s tour that visits every black square, or one that visits every white square. The Bishop s Tour Problem Let us begin by defining the problem of interest to us. An m n chessboard is an array with square cells arranged in m rows and n columns. Without loss of generality we will assume that m n. We label the cell in the ith row and jth column (i, j), and we assume the standard coloring so that the cell labeled (i, j) is colored white if i + j is even, and it is colored black otherwise. So the cell labeled (1, 1) in the upper left corner is colored white. A legal bishop move is one from cell (i, j) to one of the 4 diagonally adjacent cells (i 1, j 1), (i + 1, j + 1), (i 1, j +1), (i+1, j 1) (some of these moves may not be possible if the cell (i, j) is too close to one of the edges of the board). We wish to find necessary and sufficient conditions on m and n that guarantee that a bishop can make successive legal moves and visit every black cell (or every white cell) exactly once. Clearly, if m = 1 then a bishop s tour visiting all the white cells is impossible if n 3, and one visiting all the black cells is impossible if n 4. Also, a bishop s tour on a 2 n board that visits all the cells of either color is always possible, as shown below in Figure 2. Now if m 3, a bishop s tour that visits all the white cells is impossible. To see this, let us first convert the 2

3 Figure 2: Bishops tours on 2 n boards problem into one about graphs. We let W (m, n) be a graph obtained by replacing each white cell of the board by a vertex and then joining two vertices by an edge if they are separated by a legal bishop s move. We let B(m, n) be the corresponding graph for black cells. A bishop can tour all the white cells if and only if W (m, n) contains a path containing all the vertices i.e. a Hamiltonian path. Suppose such a tour exists. If n is odd, then the corner cell labeled (1, n) is white. If m is odd then the corner cell labelled (m, 1) is white. And if both m and n are even then the corner cell labeled (m, n) is white. So there is at least one other white corner cell. Since there is only one legal move from a corner cell, any tour traversing all the white cells must begin at one of these corner cells and end at another corner cell. Our tour must then include two edges incident with each white cell on the board, except the two corner cells where the tour begins and ends. Without loss of generality, assume that the tour begins at vertex (1, 1). Now consider the vertices corresponding to cells labelled (1, 3) and (3, 1). Each of these vertices will have either one or two edges incident with it, and all these edges will have to be included in the Hamiltonian path. This is illustrated in Figure 3. The dotted edges may not exist if either of the cells (3, 1) or (1, 3) are corner cells Figure 3: A bishop s tour visiting all the white cells on an m n board with m 3 is impossible But then the path will contain three edges incident with the vertex corresponding to cell (2, 2), which can t be since a Hamiltonian path can contain at most 2 edges incident with each vertex. The argument above also shows that if m or n is even and both 3, then a bishop s tour on an m n board that visits all the black cells is impossible, since in this case the board contains two corner black cells. 3

4 We now pose the following question: On which m n boards (m n, m 3) is there a bishop s tour that visits each black cell exactly once? The answer to this question is given by the following theorem: Theorem 1 An m n chessboard with m n and m 3 admits a bishop s tour that visits every black cell exactly once if and only if both of the following two conditions hold: 1. m and n are both odd. 2. m = n or m = n 2. To prove Theorem 1, we first need to prove three lemmas. Lemma 1 There is no bishop s tour that visits every black cell on an 3 n board if n 7. Proof: Suppose such a tour existed. From the above discussion, we know that n is odd, say n = 2k + 1, k 3. In the graph B(3, n), there are k vertices in the first row corresponding to black cells, and k vertices in the third row. Each of these vertices has 2 edges incident with it. Since the tour must include two edges incident with each vertex except for the two vertices where the tour begins and ends, the tour must include both of the edges incident with at least 2k 2 of the 2k vertices in rows 1 and 3. There are k columns containing the 2k 2 vertices. Since k 3, we know that 2k 2 > k. So by the pidgeon-hole principle, at least two of the 2k 2 vertices must lie in the same column. But then the tour will contain the four edges incident with these two vertices, and these four edges form a cycle, as seen in Figure 4. This is clearly impossible, and so no such tour exists. Figure 4: A bishop s tour visiting all the black cells on a 3 n board is impossible when n 7 Lemma 2 Any bishop s tour that visits every black cell on an m n board (m n and m 5) must begin or end on one of the cells on the outer boundary of the board. 4

5 Proof: In the graph B(m, n), each vertex corresponding to a black cell on the outer boundary of the board has exactly two edges incident with it. If the bishop s tour contains all of these edges, it will contain a cycle that, for example, does not include vertex (4, 3): (4,3) Figure 5: A bishop s tour that includes all edges on the outer boundary of the chessboard is impossible This is impossible, so at least one of the edges incident with a vertex on the outer boundary of the graph is not contained in the given tour. But this means that the tour begins or ends at this vertex. Lemma 3 A bishop s tour that visits all the black cells exists on a m n board where m 1 and n 1, if and only if one also exists on an (m + 4) (n + 4) board. Proof: Note that given an m n board, we can construct an (m + 4) (n + 4) board by adding two rows and two columns along each edge of the board. This is illustrated in Figure 6(a), where a 7 9 board, shown in light shading, has been extended to a board. Suppose first that a tour exists on the m n board. By the above lemma, the tour must begin or end on one of the cells on the outer boundary of the board, corresponding to some vertex C as illustrated in Figure 6(a). Then we can easily extend the tour to cover the entire (m + 4) (n + 4) board as indicated in the figure. Now suppose a tour existed on an (m + 4) (n + 4) board. Let V 1 be the set of vertices corresponding to black cells (i, j) where i = 1 or j = 1 or i = m + 4 or j = n + 4. Let V 2 be the set of vertices not in V 1 corresponding to black cells (i, j) where i = 2 or j = 2 or i = m + 3 or j = n + 3. And let V 3 be the set of vertices in B(m + 4, n + 4) not in V 1 or V 2. Let E 1 be the set of edges that are incident with a vertex in V 1. The tour on the (m + 4) (n + 4) board must include two edges incident with each vertex except for the two 5

6 B C... A B F.... A (a) E D... C (b) Figure 6: Constructing a tour on an m n board given a tour on an (m + 4) (n + 4) board vertices at which the tour begins and ends. So the tour must include all the edges in E 1 except possibly one or two of these. It cannot include all the edges of E 1, since then it would include a cycle that does not visit every cell. So either exactly one or exactly two of the edges in E 1 is not contained in the tour. We will take each case separately. Suppose exactly one edge of E 1 is not contained in our tour. This edge connects a vertex in V 1, call it A, to a vertex in V 2, call it B. The situation is illustrated in Figure 6(a). Then the bishop following this tour must begin at vertex A, make its way around the board visiting all the vertices in V 1 and V 2 until it reaches vertex B. Since it cannot return to A, it must then visit a vertex in V 3, say C. By truncating the beginning part of the tour from A to C, we obtain a tour on the m n board whose cells corresponds to vertices in V 3. Suppose now that two edges of E 1 are not contained in our tour. The first edge connects some vertex A in V 1 to some vertex B in V 2. The second edge connects some vertex C in V 1 to some vertex D in V 2. The situation is illustrated on the board in Figure 6(b). This tour then must begin at A, and then end at C. If the bishop begins its tour at A, it must then visit vertices in V 1 and V 2 until it reaches vertex D. Since it cannot visit vertex C next, it must visit a vertex E in V 3 next. Eventually, the bishop visits vertex B, then makes its way around the board to C. If we truncate the beginning of the tour from vertex A to vertex D, and also truncate the end of the tour from vertex B to vertex C, what remains will be a tour on the m n board whose cells correspond to vertices in V 3. We now proceed with the proof of Theorem 1. 6

7 Proof of Theorem 1: Suppose a bishop s tour that visits all the black cells exists on an m n board where m n and m 3. We have already seen that m and n must both be odd. Now suppose condition (2) of the theorem is not satisfied, i.e. n m + 4. By Lemma 1, we know that m 3. Since m is odd, either m = 4k + 1 or m = 4k + 3 for some natural number k. If m = 4k + 1, then by applying Lemma 3 k times, there must be a bishop s tour on a 1 (n 4k) board. (Note that n 4k = n m + 1 5). This is clearly impossible. If on the other hand m = 4k + 3, then again applying Lemma 3 k times, there must be a bishop s tour that visits all the black cells on a 3 (n 4k) board. In this case n 4k = n m + 3 7, so this contradicts Lemma 1. Hence condition 2 must be satisfied. It remains to show that if m and n satisfy conditions (1) and (2), then a bishop s tour that visits all the black cells exists. The proof is by induction. By Lemma 3, if a tour exists on an m n board, then one exists on an (m + 4) (n + 4) board. The result will follow if we can construct tours on boards of sizes 3 3, 3 5, 5 5, and 5 7. These are shown in Figure 7. Figure 7: Bishop s tours on the four base boards This completes the proof of Theorem 1. References [1] Bonnie Averbach and Orin Chein. Problem solving through recreational mathematics, Dover Publications, [2] W.W. Rouse Ball and H.S.M. Coxeter. Mathematical Recreations and Essays, University of Toronto Press, [3] Miodrag Petkovic. Mathematics and Chess, Dover Publications, [4] Allen J. Schwenk. Which Rectangular Chessboards Have a Knight s Tour? The Mathematics Magazine, 64 (1991) [5] R.J. Wilson and J.J. Watkins, Graphs, an Introductory Approach, John Wiley and Sons, Inc., New York,

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

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