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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.

A knight’s tour on a chessboard is a path, consisting of at least two moves, in which the knight chess piece visits each square on the chessboard exactly once. In a closed knight’s tour, the knight returns to the square on which it started. In an open knight’s tour, the knight does not return to its starting square. For the purpose of this paper, any mention of knight’s tours will refer to closed knight’s tours, unless specified otherwise. In terms of graph theory, finding a knight’s tour on an chess-board is equivalent to finding a Hamiltonian cycle made up of knight’s moves on an graph, where the squares are the nodes and the moves are the edges.

Euler worked on knight’s tours. He constructed an 8 × 8 closed tour by joining open tours on two 4 × 8 boards [1, pp. 6-7]. The technique of combining two tours to create another tour is commonly used today for constructing knight’s tours. In 1991, Allen Schwenk settled the question of which chessboards contain knight’s tours and which do not [

Subsequently, interest shifted to the existence of knight’s tours on chess-boards with squares removed. DeMaio and Hippchen [

We determine which chessboards have a closed knight’s tour, when each square is removed in turn.

We color chessboards with alternating black and white squares and the upper-left corner square black, so that where matrix notation is used squares (i, j) with i + j odd are white.

In this section, we show that all 3 × n boards with one square removed contain a knight’s tour, except for those listed in Lemma 1.

Lemma 1. A knight’s tour does not exist on a 3 × n chessboard with one square removed if:

n is eventhe removed square is (i, j) with i + j oddn = 1n = 3 when any square other than the center square is removedn = 5n = 7 when any square other than square (2, 2) or (2, 6) is removedn = 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.

Proof. Because each knight’s move alternates colors, for a tour to exist, there must be an equal number of black and white squares and an even number of total squares. This implies that for a knight’s tour to exist on an m × n chessboard with one square removed, both m and n must be odd and the removed square must be black. So, on a 3 × n board with one square removed, a tour cannot exist if n is even or the removed square is (i, j) with i + j odd. The 3 × 1 board does not contain a knight’s tour because the knight is unable to move in the horizontal direction.

Knight’s moves are often forced on 3 × n boards.

A knight’s tour cannot exist if a non-Hamiltonian cycle is forced on a chessboard because a non-Hamiltonian cycle prevents the knight from visiting each square.

is removed, no tour can exist on the 3 × 7 board.

square (2, 4) must connect to squares (1, 2) and (3, 2), thus producing a cycle.

Figures 7 and 8 show the six squares that can never be removed from 3 × n boards, when n is odd and greater than or equal to 11. As seen in

Lemma 2. A knight’s tour exists on a 3 × n chessboard with one square removed if:

n = 3 when the center square is removedn = 7 when square (2, 2) or (2, 6) is removedn = 9 when square (1, 1), (1, 5), (3, 1), (3, 5), (1, 9), or (3, 9) is removed, or

and odd when any square (i, j) with i + j even other than (1, 3), (2, 4) (3, 3), (1, n – 2), (2, n – 3), or (3, n – 2) is removed.

Proof. The forced cycle on the 3 × 3 board, shown in

and so forth. “Hole” indicates the removed square. Due to the symmetry of the 3 × 7 board, showing that the removal of square (2, 2) permits a tour also shows that the removal of square (2, 6) permits a tour. In the same way, Tables 2 and 3 show that the removal of the corner squares and squares (1, 5) and (3, 5) permit a knight’s tour on the 3 × 9 board.

Tables 4-7 show that all black squares can be removed from the 3 × 11 board, except for squares (1, 3), (2, 4), (3, 3), (1, 9), (2, 8), and (3, 9). These 3 × 11 tours can be extended into 3 × (11 + 4) tours using the 3 × 4 extender board shown in

Two 3 × 4 extender boards can be combined to make a 3 × 8 extender board. Therefore, any 3 × 11 tour can be extended into a 3 × (11 + 4 k) tour, k = 1, 2, 3,···. This proves that all black squares can be removed from 3 × n boards, n ≥ 11, n ≡ 3 mod 4, n odd, except for the six squares that were specified in Lemma 1.

Specifically, all black squares can be removed from 3 × n boards, n ≥ 13, n ≡ 1 mod 4, n odd, except for the six squares that were specified in Lemma 1. Tables 9-13 show with examples that all black squares can be removed from a 3 × 13 board, except for squares (1, 3), (2, 4), (3, 3), (1, 11), (2, 10), and (3, 11). The 3 × 4

extender board can also expand these 3 × 13 tours into 3 × (13 + 4 k) tours, k = 1, 2, 3,··· .

Combining Lemmas 1 and 2 gives Theorem 1.

Theorem 1. A knight’s tour exists on all 3 × n chessboards with one square removed except for the boards listed in Lemma 1.