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Fig. 8. A knight’s tour on a 27 x 27 board that is missing one square Fig. 7. Structured knight’s tours for (from left to right) 5 x 5, 7 x 7,9 x 9, and 11 x 11 boards that are missing the upper left corner square.

Figure 8 A knight’s tour on a 27 x 27 board that is missing one square Fig. 7. Structured knight’s tours for (from left to right) 5 x 5, 7 x 7,9 x 9, and 11 x 11 boards that are missing the upper left corner square.

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EAN LCALININIIAL YY CEs BAR haere ewes Ae ARS. weet OF et Now suppose that n>12 is even and that structured knight’s tours exist on m x m and m x (m+ 2) boards for all 6 < m <n. Divide the n x n board into four quadrants as evenly as possible. More precisely, each side of length n = 4k for some k < WN is divided into two parts of length 2k, and each side of length 44 + 2 for some k ¢ N is divided into a part of length 2k and a part of length 2(& + 1). In the construction of an n x n board in which n = 4k for some k < N, the four quadrants are each 2k x 2k. Alternatively, if n = 4k + 2 for some k € N, then the four quadrants are either 2k x 2k, 2k x 2(k+1), 2(k +1) x 2k, or 2(k +1) x 2(4 + 1). In the construction Fig. 1. Required moves for a structured knight’s tour.
EAN LCALININIIAL YY CEs BAR haere ewes Ae ARS. weet OF et Now suppose that n>12 is even and that structured knight’s tours exist on m x m and m x (m+ 2) boards for all 6 < m <n. Divide the n x n board into four quadrants as evenly as possible. More precisely, each side of length n = 4k for some k < WN is divided into two parts of length 2k, and each side of length 44 + 2 for some k ¢ N is divided into a part of length 2k and a part of length 2(& + 1). In the construction of an n x n board in which n = 4k for some k < N, the four quadrants are each 2k x 2k. Alternatively, if n = 4k + 2 for some k € N, then the four quadrants are either 2k x 2k, 2k x 2(k+1), 2(k +1) x 2k, or 2(k +1) x 2(4 + 1). In the construction Fig. 1. Required moves for a structured knight’s tour.
Fig. 2. Structured knight’s tours for (in row-major order) 6 x 6, 6 x 8, 8 x 8, 8 x 10, 10 x 10, and 10 x 12 boards.
Fig. 2. Structured knight’s tours for (in row-major order) 6 x 6, 6 x 8, 8 x 8, 8 x 10, 10 x 10, and 10 x 12 boards.
I. Parberry! Discrete Applied Mathematics 73 (1997) 251-260 Fig. 3. How to combine four structured knight’s tours into one: (a) the moves at the inside corners, (b) the edges 4,B,C,D to be deleted, and (c) the replacement edges E, F,G,H.
I. Parberry! Discrete Applied Mathematics 73 (1997) 251-260 Fig. 3. How to combine four structured knight’s tours into one: (a) the moves at the inside corners, (b) the edges 4,B,C,D to be deleted, and (c) the replacement edges E, F,G,H.
Fig. 4. A 16 x 16 knight’s tour constructed from the 8 x 8 knight’s tour in Fig. 2 using the technique of Theorem 2.1. The algorithm of Theorem 2.1 is particularly easy to implement, and can be used to sonstruct knight’s tours of size up to 1000 x 1000 in under 11s on a SUN SPARC 2. The construction described in the proof of Theorem 2.1 can also be used to obtain some knight’s tours with interesting properties. It should be clear that we have already solved the quadrisected knight’s tour problem. The only even values of for which we lave not constructed quadrisected knight’s tours are n = 8,10. A quadrisected knight’s our is known for an 8 x 8 board (see Dudeney [7—9]) leaving n = 10 open. Another minor modification to our algorithm also delivers symmetric knight’s tours.
Fig. 4. A 16 x 16 knight’s tour constructed from the 8 x 8 knight’s tour in Fig. 2 using the technique of Theorem 2.1. The algorithm of Theorem 2.1 is particularly easy to implement, and can be used to sonstruct knight’s tours of size up to 1000 x 1000 in under 11s on a SUN SPARC 2. The construction described in the proof of Theorem 2.1 can also be used to obtain some knight’s tours with interesting properties. It should be clear that we have already solved the quadrisected knight’s tour problem. The only even values of for which we lave not constructed quadrisected knight’s tours are n = 8,10. A quadrisected knight’s our is known for an 8 x 8 board (see Dudeney [7—9]) leaving n = 10 open. Another minor modification to our algorithm also delivers symmetric knight’s tours.
Fig. 5. 16 x 16, and 20 x 20 knight’s tours that are symmetric under a 180° rotation, constructed using the technique of Theorem 2.2. Fig. 6. Structured knight’s tours for (in row-major order) 6 x 7, 7 x 8, 8x9, 9x10, 10x 11, and 11 x 12 boards.
Fig. 5. 16 x 16, and 20 x 20 knight’s tours that are symmetric under a 180° rotation, constructed using the technique of Theorem 2.2. Fig. 6. Structured knight’s tours for (in row-major order) 6 x 7, 7 x 8, 8x9, 9x10, 10x 11, and 11 x 12 boards.
I. Parberry! Discrete Applied Mathematics 73 (1997) 251-260 Fig. 9. How to combine four knight’s tours into one: (a) the moves at the inside corners (the holes are shaded), (b) the edges A,B,C, D to be deleted, and (c) the replacement edges.
I. Parberry! Discrete Applied Mathematics 73 (1997) 251-260 Fig. 9. How to combine four knight’s tours into one: (a) the moves at the inside corners (the holes are shaded), (b) the edges A,B,C, D to be deleted, and (c) the replacement edges.
Fig. 10. 10 x 10, 14 x 14, and 18 x 18 knight’s tours that are symmetric under a 90 degree rotation, constructed using the technique of Theorem 2.5.
Fig. 10. 10 x 10, 14 x 14, and 18 x 18 knight’s tours that are symmetric under a 90 degree rotation, constructed using the technique of Theorem 2.5.
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Applied MathematicsDiscrete Applied Mathematics
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