Closed Knight’s Tours on (m,n,r)-Ringboards
Abstract
:1. Introduction
2. CKTs and OKTs on Some LBs and 7Bs
- (i)
- For , delete from and from of the ith CB().
- (ii)
- For , join and of the ith CB() to and of the th CB(), respectively
- (iii)
- For , join and of the ith CB() to and of the th CB(), respectively.
- If m and n are odd integers, then or 3 (mod 4) and or 3 (mod 4).
- If m and n are even integers, then or 2 (mod 4) and or 2 (mod 4).
- If m and n are different parity, then or 3 (mod 4) and or 2 (mod 4); and or 2 (mod 4) and or 3 (mod 4).
- (i)
- if , then delete and from the CKT of the LB. If , then further delete and from the CKT of the LB.
- (ii)
- If , then join , , and which are four end-points of two paths of the CB() to , , and of the LB, respectively. If , then further join , , and which are four end-points of two paths of the CB() to , , and of the LB, respectively.
- (a)
- The LB contains an OKT from to if and only if (i) is odd and or (ii) and .
- (b)
- The LB contains an OKT from to if and only if (i) is even and or (ii) and .
- (i)
- For , delete from the OKT of the ith CB();
- (ii)
- For , join and of the ith CB() to and of the th CB(), respectively.
- Case 1: there exist nonnegative integers such that and . We divided the LB into subboards, CB() (Figure 17) and LB (Figure 11a) if and and LB (Figure 11a) and CB() (Figure 16) if and . Otherwise, we divide into three subboards, CB() (Figure 17), LB (Figure 11a) and CB() (Figure 16). Then, we construct the required OKT by the following two steps.
- (i)
- (ii)
- If and , then join and of the CB() to and of the LB, respectively. If and , then join and of the LB to and of the CB() chessboard, respectively. Otherwise, join four pairs of vertices together.
- Case 2: there exist nonnegative integers such that and . Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB, and are replaced by LB (Figure 11b), and , respectively.
- Case 3: there exist nonnegative integers such that and . Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB, and are replaced by LB (Figure 12a), and , respectively.
- Case 4: there exist nonnegative integers such that and . Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB, and are replaced by LB (Figure 12b), and , respectively.
- Case 5: there exist nonnegative integers such that and . Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB, and are replaced by LB (Figure 13a), and , respectively.
- Case 6: there exist nonnegative integers such that and . Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB, and are replaced by LB (Figure 13b), and , respectively.
- Case 7: there exist nonnegative integers such that and . Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB, and are replaced by LB (Figure 10), and , respectively.
- Case 8: there exist nonnegative integers such that and . Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB, and are replaced by LB (Figure 14a), and , respectively.
- Case 9: there exist nonnegative integers such that and . Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB, and are replaced by LB (Figure 14b), and , respectively.
- (i)
- For , delete from the OKT of the ith CB();
- (ii)
- For , join and of the ith CB() to and of the th CB(), respectively.
- Case 1: there exist nonnegative integers such that and . We divide the LB into subboards, CB() (Figure 29) and LB (Figure 22a) if and and LB (Figure 22a) and CB() (Figure 16) if and . Otherwise, we divide into three subboards, CB() (Figure 29), LB (Figure 22a) and CB() (Figure 16). Then, we construct the required OKT by the followings.
- Case 2: there exist nonnegative integers such that and . Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB, and are replaced by LB (Figure 22b), and , respectively.
- Case 3: there exist nonnegative integers such that and . Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB, and are replaced by LB (Figure 23), and , respectively.
- Case 4: there exist nonnegative integers such that and . Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB, and are replaced by LB (Figure 24a), and , respectively.
- Case 5: there exist nonnegative integers such that and . Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB, and are replaced by LB (Figure 24b), and , respectively.
- Case 6: there exist nonnegative integers such that and . Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB, and are replaced by LB (Figure 25), and , respectively.
- Case 7: there exist nonnegative integers such that and . Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB, and are replaced by LB (Figure 26a), and , respectively.
- Case 8: there exist nonnegative integers such that and . Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB, and are replaced by LB (Figure 26b), and , respectively.
- Case 9: there exist nonnegative integers such that and . Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB, and are replaced by LB (Figure 21), and , respectively.
- Case 10: there exist nonnegative integers such that and . Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB, and are replaced by LB (Figure 27a), and , respectively.
- Case 11: there exist nonnegative integers such that and . Then, we construct the required OKT by using the same procedure as we did in Case 1 but LB, and are replaced by LB (Figure 27b), and , respectively.
- (a)
- The 7B contains an OKT from to if and only if (i) is odd and or (ii) and .
- (b)
- The 7B contains an OKT from to if and only if (i) is even and or (ii) and .
3. Existence of a Special OKT on CB()
- (a)
- Let and . Then, a CB() contains an OKT from to if and only if and .
- (b)
- Let . Then, a CB() contains an OKT from to if and only if m and n are not both even.
- (i)
- We delete of the OKT on the CB() and connect and of the CB() to and of the 1st CB(), respectively.
- (ii)
- We delete of the second path of the ith CB() for all . Then, we connect and of the ith CB() to and of the th CB().
- (i’)
- (ii’)
- Join of the OKT on CB() to of the 1st CB() shown in Figure 38.
- (iii’)
4. Main Theorem
5. Conclusions and Discussion
Author Contributions
Funding
Conflicts of Interest
Abbreviations
CKT | Closed Knight’s Tour |
OKT | Open Knight’s Tour |
CB | chessboard |
LB | L-board of size |
7B | 7-board of size |
RB | -ringboard |
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Srichote, W.; Boonklurb, R.; Singhun, S. Closed Knight’s Tours on (m,n,r)-Ringboards. Symmetry 2020, 12, 1217. https://doi.org/10.3390/sym12081217
Srichote W, Boonklurb R, Singhun S. Closed Knight’s Tours on (m,n,r)-Ringboards. Symmetry. 2020; 12(8):1217. https://doi.org/10.3390/sym12081217
Chicago/Turabian StyleSrichote, Wasupol, Ratinan Boonklurb, and Sirirat Singhun. 2020. "Closed Knight’s Tours on (m,n,r)-Ringboards" Symmetry 12, no. 8: 1217. https://doi.org/10.3390/sym12081217
APA StyleSrichote, W., Boonklurb, R., & Singhun, S. (2020). Closed Knight’s Tours on (m,n,r)-Ringboards. Symmetry, 12(8), 1217. https://doi.org/10.3390/sym12081217