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Parberry proposed another algorihm to construct a closed knight’s tour on board of sizes n x n, or n X (n + 2) for even n > 6, and for board of size nx (n+ 1) for n > 6. The algorithm works by dividing the board into 4 quadrants, generate a closed knight’s tour for each quadrant, and finally remove 4 edges and add 4 new edges at the inside corners combine four smaller closed knight’s tours into a complete closed knight’s tour for the board. This kind of algorithm is called quadrisected. These two algorithms work for amxn board. Note that it is clear by parity argument that for mn is odd there is no closed knight’s tour on the m x n board. FIGURE 1. The structured knight’s tour.

Figure 1 Parberry proposed another algorihm to construct a closed knight’s tour on board of sizes n x n, or n X (n + 2) for even n > 6, and for board of size nx (n+ 1) for n > 6. The algorithm works by dividing the board into 4 quadrants, generate a closed knight’s tour for each quadrant, and finally remove 4 edges and add 4 new edges at the inside corners combine four smaller closed knight’s tours into a complete closed knight’s tour for the board. This kind of algorithm is called quadrisected. These two algorithms work for amxn board. Note that it is clear by parity argument that for mn is odd there is no closed knight’s tour on the m x n board. FIGURE 1. The structured knight’s tour.

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FIGURE 2. The removed edges A and B.
FIGURE 2. The removed edges A and B.
FIGURE 3. The added edges C and D.
FIGURE 3. The added edges C and D.
FIGURE 4. The structured knight’s tour on 6x6, 6x7, 6x8, and 6x9 boards.
FIGURE 4. The structured knight’s tour on 6x6, 6x7, 6x8, and 6x9 boards.
boards. board. Proof. For 6 < n < 11, the tours are given in Figures 6 and 7. Since n > 12, by Lemma 1 we get n = 6k, + 7ko + 8k3 + 9k4, for some ky, ko, k3,k4 € Z. Partition the original 8 xn board into (8 x 6k1), (8x 7ke, (8x 8k3), and (8 x 9k4) n each partition, we can construct a closed structured knight’s tour by horizontal merge operations. Compose all the obtained tours by horizontal merge operations into one larger closed structured knight’s tour on the original FIGURE 6. The structured knight’s tour on 8x6, 8x7, 8x8, and 8x9 boards.
boards. board. Proof. For 6 < n < 11, the tours are given in Figures 6 and 7. Since n > 12, by Lemma 1 we get n = 6k, + 7ko + 8k3 + 9k4, for some ky, ko, k3,k4 € Z. Partition the original 8 xn board into (8 x 6k1), (8x 7ke, (8x 8k3), and (8 x 9k4) n each partition, we can construct a closed structured knight’s tour by horizontal merge operations. Compose all the obtained tours by horizontal merge operations into one larger closed structured knight’s tour on the original FIGURE 6. The structured knight’s tour on 8x6, 8x7, 8x8, and 8x9 boards.
FIGURE 7. The structured knight’s tour on 8x10, and 8x11 boards.
FIGURE 7. The structured knight’s tour on 8x10, and 8x11 boards.
FIGURE 8. The structured knight’s tour on 10x6, 107, 10x8 and 10x9 boards. Proof. For 6 <n < 11, the tours are given in Figures 8 and 9. The construc- tion of a closed structured knight’s tour on the 10 x n board can be done in a similar way of the one in Lemma 2.
FIGURE 8. The structured knight’s tour on 10x6, 107, 10x8 and 10x9 boards. Proof. For 6 <n < 11, the tours are given in Figures 8 and 9. The construc- tion of a closed structured knight’s tour on the 10 x n board can be done in a similar way of the one in Lemma 2.
FIGURE 9. The structured knight’s tour on 10x10, and 10x11 boards.
FIGURE 9. The structured knight’s tour on 10x10, and 10x11 boards.
FIGURE 12. The deleted edges A,B,C,D,E,and F in a diamond board Proof. The diamond board of size n can be divided into four quadrants. | quadrant is exactly a step board of size m x m, where m = 4 is even. Fach If n > 24, then m > 12. By Lemma 5, we can obtain a closed structured knight’s tour on each step board of size m x m. In order to construct a struct knight’s tour on the original diamond board, delete edges A, B,C, D, E, ured and F shown in Figure 12 and add new six edges shown Figure 13. Thus, we get a closed structured knight’s tour on diamond board of size n.
FIGURE 12. The deleted edges A,B,C,D,E,and F in a diamond board Proof. The diamond board of size n can be divided into four quadrants. | quadrant is exactly a step board of size m x m, where m = 4 is even. Fach If n > 24, then m > 12. By Lemma 5, we can obtain a closed structured knight’s tour on each step board of size m x m. In order to construct a struct knight’s tour on the original diamond board, delete edges A, B,C, D, E, ured and F shown in Figure 12 and add new six edges shown Figure 13. Thus, we get a closed structured knight’s tour on diamond board of size n.
FIGURE 13. The added six edges in a diamond board
FIGURE 13. The added six edges in a diamond board
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Knight Tour Problem
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