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Knight’s Tours on 3 x <i>n</i> Chessboards with a Single Square Removed
School of Mathematical Sciences, Rochester Institute of Technology, Rochester, USA
School of Mathematical Sciences, Rochester Institute of Technology, Rochester, USA
- 1 School of Mathematical Sciences, Rochester Institute of Technology, Rochester, USA
- 2 School of Mathematical Sciences, Rochester Institute of Technology, Rochester, USA
Open Journal of Discrete Mathematics·Volume 03 (2013)·Pages 56–59·Published 29 January 2013·DOI10.4236/ojdm.2013.31012
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Abstract
The following theorem is proved : A knight ’ s tour exists on all 3 x n chessboards with one square removed unless : n is even, the removed square is ( i , j ) with i + j odd, n = 3 when any square other than the center square is removed, n = 5, n = 7 when any square other than square (2, 2) or (2, 6) is removed, n = 9 when square (1, 3), (3, 3), (1, 7), (3, 7), (2, 4), (2, 6), (2, 2), or (2, 8) is removed, or when square (1, 3), (2, 4), (3, 3), (1, n – 2), (2, n – 3), or (3, n – 2) is removed.
KeywordsKnight’s TourHamiltonian CycleForced EdgeExtender Board
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