The 22nd Annual Meeting in Mathematics (AMM 2017)
Department of Mathematics, Faculty of Science
Chiang Mai University, Chiang Mai, Thailand
Some forbidden rectangular chessboards
with an (a, b)-Knight’s move
Krit Karudiloka,†,‡ , Sirirat Singhuna , and Ratinan Boonklurbb
a
Department of Mathematics, Faculty of Science,
Ramkhamhaeng University, Bangkok 10240, Thailand
b
Department of Mathematics and Computer Science, Faculty of Science,
Chulalongkorn University, Bangkok 10330, Thailand
Abstract
The m × n chessboard is an array with squares arranged in m rows and n columns. An
(a, b)-knight’s move is a move from a square to another square by moving a knight passing a
squares vertically or a squares horizontally and then passing b squares at 90 degrees angle. A
closed (a, b)-knights tour is an (a, b)-knights move from a square to another square such that
the knight lands on every square on m×n chessboard once and returns to its starting square.
In this paper, we obtain a new certain chessboard that admits no closed (a, b)-knight’s tours
depending on a and b. Moreover, the necessary condition that make the chessboard admits
no closed (a, b)-knight’s tours for any m × n chessboard is obtained.
Keywords: closed knight’s tour, Hamiltonian cycle.
2010 MSC: Primary 05C45; Secondary 05C38.
1
Introduction and Preliminaries
The knight is a chess piece that can move one square vertically or one square horizontally and
then two squares move at 90 degrees angle. In 1959, Leonhard Euler [3] found that on the 8 × 8
chessboard, an array with squares arranged in eight rows and eight columns, the knight can
move from a square to another square such that it lands on every square once and returns to
its starting square. This knight’s move is called a closed knight’s tour. The 8 × 8 chessboard is
extended to the m × n chessboard, an array with squares arranged in m rows and n columns.
The knight’s tour problem on the m × n chessboard is converted to question about a certain
graph. A graph G represented the m × n chessboard is a graph with mn vertices where each
square of board is replaced by a vertex and two vertices are joined by an edge if the knight can
move from a square to another square. If G is a graph represented the m × n chessboard, then
the degree of vertex v, denoted by deg v, is the number of squares that the knight can move to.
Then, the closed knight’s tour is a Hamiltonian cycle in G. In 1991, Schwenk [2] obtained the
sufficient and necessary conditions for the m × n chessboard to admit a closed knight’s tour.
†
Corresponding author.
Speaker.
E-mail address: [email protected] (K. Karudilok), sin [email protected] (S. Singhun),
[email protected] (R. Boonklurb).
‡
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Theorem 1.1. [2] The m × n chessboard with m ≤ n admits a closed knight’s tour unless one
or more of the following conditions hold:
(i) m and n are both odd; or
(ii) m = 1 or 2 or 4; or
(iii) m = 3 and n = 4 or 6 or 8.
In 2003, Chia et al. [1] continued to consider the m × n chessboard while the knight’s move
was extended to a squares vertically or a squares horizontally and then b squares move at 90
degrees angle. This move is called an (a, b)-knight’s move. Then, an (a, b)-knight’s move and
(b, a)-knight’s move are the same. If the knight moves to each square of the chessboard only
once with an (a, b)-knight’s move and returns to the starting square, then this move is called a
closed (a, b)-knight’s tour. In the case that a = b, the knight can move from a square to another
square with the same colour, black to black or white to white. Then, the m × n chessboard
admits no closed (a, b)-knight’s tours. We shall assume that a < b. Some observations in [1] are
obtained as follow.
Theorem 1.2. [1] Suppose that the m × n chessboard admits a closed (a, b)-knights tour, where
a < b and m ≤ n. Then,
(i) a + b is odd; and
(ii) m or n is even; and
(iii) m ≥ a + b; and
(iv) n ≥ 2b.
By Theorem 1.1 (ii), the 4 × n chessboard admits no closed (1, 2)-knight’s tours. Then, Chia
et al. [1] also obtained some chessboard which admits no closed (a, b)-knight’s tours.
Theorem 1.3. [1] Suppose that m = a + b + 2t + 1, where 0 ≤ t ≤ a − 1. Then, the m × n
chessboard admits no closed (a, b)-knights tours.
Theorem 1.4. [1] Suppose that m = a(k + 2l), where 1 ≤ l ≤ k2 . Then, the m × n chessboard
admits no closed (a, ak)-knights tours, where a is odd and k is even.
Theorem 1.5. [1] Suppose that m = 2(ak + l), where 1 ≤ k ≤ l ≤ a. Then, the m × n
chessboard admits no closed (a, a + 1)-knights tours.
Theorem 1.6. [1] Suppose that m = 2a + 2t + 1, where 1 ≤ t ≤ a − 1. Then, the m × n
chessboard admits no closed (a, a + 1)-knights tours.
By Theorem 1.3, the certain chessboard that admits no closed (a, b)-knight’s tours for any
a < b depends on a and b. In this paper, we obtain a new certain chessboard that admits
no closed (a, b)-knight’s tours depending on a and b in Theorem 2.4. Moreover, the necessary
condition that make the chessboard admitted no closed (a, b)-knight’s tours for any m × n
chessboard is obtained in Theorem 2.5.
2
Main Results
In this section, we present the certain chessboard that admits no closed (a, b)-knight’s tours for
any a and b. We consider two cases : a = 1 (in Lemma 2.1) and a > 1 (in Lemma 2.2 and 2.3).
Next, the necessary condition that makes the chessboard admits no closed (a, b)-knight’s tours
for any m × n chessboard is obtained. For convenience, to talk about the m × n chessboard,
we label the square (i, j), where i ∈ {1, 2, 3, . . . , m} and j ∈ {1, 2, 3, . . . , n}, counting from the
upper left conner in matrix fashion. That is the square (1,1) is on the left upper side of the
chessboard. These (i, j) labels also represent vertices in the graph representation of the m × n
chessboard. If the knight starts at the square or vertex (i, j) and moves with an (a, b)-knight’s
move, then the knight can move to one of eight squares or vertices : (i ± a, j ± b) or (i ± b, j ± a).
Lemma 2.1. Suppose that a = 1, m = a + b and n = 2b. Then, the m × n chessboard admits
no closed (a, b)-knight’s tours.
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Proof. Let a = 1. By Theorem 1.2, b is even. Then, the chessboard contains b + 1 rows and
2b columns. For i ∈ {1, 2, 3, . . . , 2b }, deg (2i, b + 1) = 2 and the vertex (2i, b + 1) is adjacent to
vertices (2i − 1, 1) and (2i + 1, 1). Since the vertex (1, 1) is adjacent to (2, b + 1), let
P1 : (1, 1)(2, b + 1)(3, 1)(4, b + 1) · · · (b, b + 1)(b + 1, 1)
be a path from (1, 1) to (b + 1, 1).
For i ∈ {1, 2, 3, . . . , 2b }, deg (2i, b + 2) = 2 and the vertex (2i, b + 2) is adjacent to vertices
(2i − 1, 2) and (2i + 1, 2). Since the vertex (1, 2) is adjacent to (2, b + 2), let
P2 : (1, 2)(2, b + 2)(3, 2)(4, b + 2) · · · (b, b + 2)(b + 1, 2)
be a path from (1, 2) to (b + 1, 2).
If the chessboard contains a closed (a, b)-knight’s tour C, then P1 and P2 must be subpaths
of C. Since the vertices (1, 1), (b, b + 2), (b + 1, 1) and (2, b + 2) are of degree two, the edges
(1, 1)(b + 1, 2) and (1, 2)(b + 1, 1) must be edges in C. Thus, P1 and P2 together with such two
edges form a cycle which does not contain all vertices in the chessboard as shown in Figure 1.
This makes a contradiction.
Figure 1: The forced cycle in the (b + 1) × 2b chessboard.
In the case that a > 1, let B be an m × n chessboard where m = a + b and n = 2b. We
partition the chessboard into three subboards, B1 , B2 and B3 . The B1 denote the subboard of
B containing all four corners of a × a chessboards. The top left corner of B1 is located from
the first row to the ath row and from the first column to the ath column of B. The top right
corner of B1 is located from the first row to the ath row and from the (2b − a + 1)th column to
the (2b)th column of B. The bottom left corner of B1 is located from the (b + 1)th row to the
(a + b)th row and from the first column to the ath column of B. The bottom right corner of
B1 is located from the (b + 1)th row to the (a + b)th row and from the (2b − a + 1)th column
to the (2b)th column of B. The B2 denote the subboard of B located from the (a + 1)th row
to the bth row and from the first to the last column of B. The B3 denote the subboard of B
containing two parts. The first one is located from the first row of B to the ath row and the
(a + 1)th column to the (2b − a)th column of B. The second one is located from the (b + 1)th
row to the last row of B and the (a + 1)th column to the (2b − a)th column of B. The partition
of the chessboard is shown in Figure 2. Then, we see that (i) if v belongs to B1 , then deg v = 2,
(ii) if v belongs to B2 , then deg v = 2, and (iii) if v belongs to B3 , then deg v = 3
Lemma 2.2. Suppose that a > 1, b ≤ 2a − 1, m = a + b and n = 2b. Then, the m × n chessboard
admits no closed (a, b)-knight’s tours.
Proof. Suppose that the m × n chessboard to admit a closed (a, b)-knight’s tour C of B. We
consider the vertex (1, b + 1) that adjacent to (a + 1, 1), (b + 1, b − a + 1) and (b + 1, a + b + 1).
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Figure 2: The partition of the m × n chessboard into three subboards, B1 , B2 and B3 .
It is clear that the vertex (a + 1, 1) belongs to B2 . Since the b ≤ 2a − 1, b − a + 1 ≤ a. Then the
vertex (b + 1, b − a + 1) belongs to B1 . In order to show that (b + 1, b − a + 1) belongs to B1 , it
suffices to show that 2b − (a + b + 1) < a. Since the 2b − (a + b + 1) = b − a − 1 and b ≤ 2a − 1,
b − a − 1 ≤ a − 2 < a. Then the vertex (b + 1, a + b + 1) belongs to B1 . Those three vertices
are degree-2. Then it is impossible that three edges, (1, b + 1)(a + 1, 1), (1, b + 1)(b + 1, a) and
(1, b + 1)(b + 1, a + b + 1) belong to C as shown Figure 3. Therefore, the m × n chessboard
admits no closed (a, b)-knight’s tours C of B.
Figure 3: Three edges that incident to (1, b + 1) belonging to C.
Lemma 2.3. Suppose that a > 1, b > 2a − 1, m = a + b and n = 2b. Then, the m × n chessboard
admits no closed (a, b)-knight’s tours.
Proof. Let B be an m × n chessboard. Let k = ⌊ ab ⌋, the floor of ab . Then, b = ka + r for some
0 ≤ r ≤ a − 1.
If k is even, let
L1 = {vi = (1, b − ia) | i ∈ {0, 2, 4, . . . , k}},
L2 = {vi = (b + 1, b − ia) | i ∈ {1, 3, 5, . . . , k − 1}},
R1 = {ui = (1, b + ia) | i ∈ {0, 2, 4, . . . , k}},
R2 = {ui = (b + 1, b + ia) | i ∈ {1, 3, 5, . . . , k − 1}},
Se = {si = (a + 1, 2b − ia) | i ∈ {0, 2, 4, . . . , k},
So = {si = (b − a + 1, 2b − ia) | i ∈ {1, 3, 5, . . . , k − 1}},
Te = {ti = (a + 1, ia) | i ∈ {2, 4, 6, . . . , k}}, and
To = {ti = (b − a + 1, ia) | i ∈ {1, 3, 5, . . . , k − 1}}.
If k is odd, let
L1 = {vi = (1, b − ia) | i ∈ {0, 2, 4, . . . , k − 1}},
L2 = {vi = (b + 1, b − ia) | i ∈ {1, 3, 5, . . . , k}},
R1 = {ui = (1, b + ia) | i ∈ {0, 2, 4, . . . , k − 1}},
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R2 = {ui = (b + 1, b + ia) | i ∈ {1, 3, 5, . . . , k}},
Se = {si = (a + 1, 2b − ia) | i ∈ {0, 2, 4, . . . , k − 1}},
So = {si = (b − a + 1, 2b − ia) | i ∈ {1, 3, 5, . . . , k}},
Te = {ti = (a + 1, ia) | i ∈ {2, 4, 6, . . . , k − 1}}, and
To = {ti = (b − a + 1, ia) | i ∈ {1, 3, 5, . . . , k}}.
Then, we obtain the following observations.
1. u0 = v0 .
2. For v ∈ L1 ∪ L2 ∪ R1 ∪ R2 − {uk , vk }, v belongs to B3 and deg v = 3.
3. The vertex vk belongs to B1 and deg vk = 2. This can be seen by considering 2 cases.
First, if k is even, then vk = (1, b − ka) and b − ka = r < a. Second, if k is odd, then
vk = (b + 1, b − ka), b + 1 > b = (a + b) − a and b − ka = r < a.
4. For v ∈ Te ∪ To ∪ Se ∪ So , v belongs to B2 and deg v = 2. This can be seen that if
v ∈ Se ∪ Te is located in (a + 1)th row, then vertex v belongs to B2 . Since b > 2a − 1 and a > 1,
b − a + 1 > a and b − a + 1 < b. If v ∈ So ∪ To is located in (b − a + 1)th row, then vertex v
belongs to B2 .
5. The vertex uk belongs to B1 and deg uk = 2. This can be seen by considering 2 cases.
First, if k is even, then uk = (1, b + ka) and ka = b − r where 0 ≤ r ≤ a − 1. Second, if k is
odd, then uk = (b + 1, b + ka), b + 1 > b = (a + b) − a and b + ka = b + (b − r) = 2b − r > 2b − a.
Here, we note that if a chessboard B admits a closed (a, b)-knight’s tour C and v is a degree-2
vertex of B, then the two edges incident to v must belong to C.
Suppose that there is a closed (a, b)-knight’s tour C of B. By observation 4, the edges
incident to v for all v ∈ Te ∪ To ∪ Se ∪ So ∪ {uk , vk } must belong to C. We shall find edges must
belong to C by considering two cases of k.
Case 1: k is even. Consider the left-hand side of B. Since for i ∈ {0, 1, 2, . . . , k}, deg si =
2 and si is adjacent to vi , si vi belong to C. Since deg vk = 2 and vk is adjacent to vk−1 and
sk , vk vk−1 and vk sk belong to C. Then, vk−1 vk−2 does not belong to C because sk−1 vk−1 and
vk vk−1 belong to C. Next vk−2 vk−3 must belongs to C because sk−1 belongs to C but vk−1 vk−2
does not belong to C. By the same idea, for i ∈ {1, 3, 5, . . . , k − 1}, vi vi+1 and vi si belong to C
while vi vi−1 does not belong to C. Thus, v1 v0 does not belong to C.
For the right-hand side of B, the edges uk uk−1 and tk uk belong to C. Then, uk−1 uk−2 does
not belong to C as the same idea on the left-hand side. For i ∈ {1, 3, 5, . . . , k − 1}, ui ui+1 and
ui ti belong to C while ui−1 ui does not belong to C. Thus, u0 u1 must not belong to C. By
observation 1, u0 = v0 . Since deg v0 = 3 and v0 s0 is only one edge incident to v0 that belongs
to C, it is impossible. Therefore, the m × n chessboard admits no closed (a, b)-knight’s tours.
Figure 4: The bold edges represent edges that belong to C when k is even.
Case 2 : k is odd. Consider the left-hand side of B. Since for i ∈ {0, 1, 2, . . . , k}, deg si =
2 and si is adjacent to vi , si vi belong to C. Since deg vk = 2 and vk is adjacent to vk−1 and
sk , vk vk−1 and vk sk belong to C. Then, vk−1 vk−2 does not belong to C. By the same idea, for
i ∈ {0, 2, 4, . . . , k − 1}, vi vi+1 and vi si belong to C while vi vi−1 does not belong to C. Thus,
v1 v0 belong to C.
For the right-hand side of B, the edges uk uk−1 and tk uk belong to C. Then, uk−1 uk−2 does
not belong to C. For i ∈ {2, 4, 6, . . . , k − 1}, ui ui+1 and ui ti belong to C while ui−1 ui does not
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belong to C. That is, u1 u2 does not belong to C. Since deg u1 = 3 and u1 is adjacent to u0 , u2
and t1 , u0 u1 must belong to C. Since deg s0 = 2 and s0 is adjacent to u0 , u0 s0 must belong to
C. It is impossible that three edges, u0 v1 (= v0 v1 ), u0 u1 and u0 s0 (that incident to u0 ) belong
to C. Therefore, the m × n chessboard admits no closed (a, b)-knight’s tours.
Figure 5: The bold edges represent edges that belong to C when k is odd.
From Lemma 2.1, 2.2 and 2.3, we can conclude as follow.
Theorem 2.4. Suppose that m = a + b and n = 2b. Then, the m × n chessboard admits no
closed (a, b)-knight’s tours.
Next, we give a necessary condition that make the chessboard admits no closed (a, b)-knight’s
tours for any m × n chessboard.
Theorem 2.5. Let (a, b) ̸= 1. Then, there is no closed (a, b)-knight’s tours for any m × n
chessboard.
Proof. Assume that (a, b) = d, where d ̸= 1. Then, d|(a ± b) and d|(−a ± b). We label the
square (i, j) with the colour k ∈ {0, 1, 2, . . . , d − 1} if i + j ≡ k (mod d). That is, each square
must be labeled by one colour from {0, 1, 2, . . . , d − 1}. We claim that the knight can move from
a square to another square with the same colour.
Suppose that the knight starts at square (i, j) with colour k. Then, i + j ≡ k (mod d).
Then, the knight can move to the squares (i ± a, j ± b) or (i ± b, j ± a). We shall show that each
square is coloured by k.
If the knight move to the square (i ± a, j ± b), then (i ± a) + (j ± b) = (i + j) ± (a ± b) ≡ k
(mod d). Thus, the square is coloured by k.
If the knight move to the square (i ± b, j ± a), then (i ± b) + (j ± a) = (i + j) ± (a ± b) ≡ k
(mod d). Thus, the square is coloured by k.
Hence, the knight can move from a square to another square that is coloured by the same
colour. Thus, the knight can not move to all squares of the chessboard. Therefore, there is no
closed (a, b)-knight tour for any m × n chessboard.
Acknowledgment. The authors are grateful to the referees for their careful reading of the
manuscript and their useful comments.
References
[1] G.L. Chia and Siew-Hui Ong, Generalized knights tours on rectangular chessboards, Discrete
Applied Math. 150 (2005), 80–89.
[2] A.L. Schwenk, Which rectangular chessboards have a knight’s tour, Math. Magazine 64
(1991), 325–332.
[3] J.J. Watkins, Across the boards: The mathematics of the chessboard problems, Princeton
University, New Jersey, 2004.
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