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A226697 Central symmetric closed knight's tour on an 8 X 8 board, attributed to Euler. (n,m) position after (a(n,m)-1)th move. 1
37, 62, 43, 56, 35, 60, 41, 50, 44, 55, 36, 61, 42, 49, 34, 59, 63, 38, 53, 46, 57, 40, 51, 48, 54, 45, 64, 39, 52, 47, 58, 33, 1, 26, 15, 20, 7, 32, 13, 22, 16, 19, 8, 25, 14, 21, 6, 31, 27, 2, 17, 10, 29, 4, 23, 12, 18, 9, 28, 3, 24, 11, 30, 5 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
Each row length is 8, and there are 8 rows.
This is an 8 X 8 matrix with entries a(n,m) = a(8*(n-1) + m), n, m =1, 2, ..., 8, with the given length 64 sequence {a(n)}. The knight starts (move #0) at the square with position (n,m) = (5,1) because a(5,1) = 1. The square position (n,m) of the knight after move #k is found from a(n,m) = k + 1, for k = 0, 1, ..., 63. See A226698 for the inverse sequence:
A226698(a(n)-1) = n, n = 1, 2, ..., 64.
This is a Hamiltonian path which is re-entrant because with move #64 one can close the path at position (5,1) to obtain a Hamiltonian circuit.
In the Gardner reference this path is attributed to L. Euler (1759). The lower half of the board is covered first. The closed path has central symmetry. Each symmetric pair has an absolute difference of 32: |a(n,m) - a(9-n,9-m)| = 32, n,m = 1, 2, ..., 8.
For a figure with this central symmetric Hamiltonian cycle see also Diagram 4 on p. 24 of the Elkies-Stanley reference.
REFERENCES
Martin Gardner, Mathematical Magic Show, The Math. Assoc. of Am., Washington DC, 1989, Ch. 14, Knights of the Square Table,Fig. 86, p. 191. German Translation: Mathematische Hexereien, Ullstein, 1977, Abb. 86, S. 186.
LINKS
N. D. Elkies and R. P. Stanley, The mathematical knight, Math. Intelligencer, 25 (No. 1) (2003), 22-34.
EXAMPLE
The board as an 8 X 8 matrix a(n,m):
n\m 1 2 3 4 5 6 7 8
1: 37 62 43 56 35 60 41 50
2: 44 55 36 61 42 49 34 59
3: 63 38 53 46 57 40 51 48
4: 54 45 64 39 52 47 58 33
5: 1 26 15 20 7 32 13 22
6: 16 19 8 25 14 21 6 31
7: 27 2 17 10 29 4 23 12
8: 18 9 28 3 24 11 30 5
(5,1) marks the starting position of the knight because a(n,m) = 1 (move 0). Move 1 leads to square (7,2) because a(7,2) = 2, etc. Move 63 leads to a(4,3) = 64, the end of the re-entrant Hamiltonian path on the board.
Central symmetry: a(4,8) = 32 + 1 = 33, a(4,7) = 32 + 26 = 58, etc. such that the upper half can be found from the lower half.
CROSSREFS
Cf. A226698 (inverse).
Sequence in context: A103946 A107142 A158018 * A117475 A145486 A319673
KEYWORD
nonn,fini,full,tabf
AUTHOR
Wolfdieter Lang, Jun 25 2013
STATUS
approved

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