Research Article Knight s Tours on Rectangular Chessboards

Research ArticleKnightrsquos Tours on Rectangular ChessboardsUsing External Squares

Grady Bullington1 Linda Eroh1 Steven J Winters1 and Garry L Johns2

1University of Wisconsin Oshkosh Oshkosh WI 54901 USA2Saginaw Valley State University University Center MI 48710 USA

Correspondence should be addressed to Garry L Johns gljsvsuedu

Received 27 May 2014 Revised 12 November 2014 Accepted 17 November 2014 Published 9 December 2014

Academic Editor Stavros D Nikolopoulos

Copyright copy 2014 Grady Bullington et al This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

The classic puzzle of finding a closed knightrsquos tour on a chessboard consists of moving a knight from square to square in such a waythat it lands on every square once and returns to its starting pointThe 8times 8 chessboard can easily be extended to rectangular boardsand in 1991 A Schwenk characterized all rectangular boards that have a closed knightrsquos tour More recently Demaio and Hippcheninvestigated the impossible boards and determined the fewest number of squares that must be removed from a rectangular boardso that the remaining board has a closed knightrsquos tour In this paper we define an extended closed knightrsquos tour for a rectangularchessboard as a closed knightrsquos tour that includes all squares of the board and possibly additional squares beyond the boundariesof the board and answer the following question how many squares must be added to a rectangular chessboard so that the new boardhas a closed knightrsquos tour

1 Introduction

For many years individuals have been intrigued with thechessboard not just because of the intricacies and strategiesin the game of chess but also for the many puzzles that havebeen devised on the board In addition mathematicians seethese puzzles as special cases of major topics in graph theoryFor instance positioning queens on the board so that nonecan be captured illustrates independence choosing the fewestnumber of squares so that rooks on these squares can attackevery square is dominance andmoving a knight from squareto square in such a way that it lands on every square once andreturns to its starting point is an example of a Hamiltoniancycle One source that surveys the history and results of manychessboard puzzles as well as summarizing variations andextensions to other surfaces is [1] Specifically for the knightrsquostour problem if we label each square on an 8 times 8 chessboardwith an ordered pair using its row and column position andassign (1 1) to the upper left square then a knight beginningat (r c) can legally move to any of the eight squares (r plusmn 2c plusmn 1) or (r plusmn 1 c plusmn 2) unless (r c) is too close to the edgeof the board An open knightrsquos tour (sometimes called a path)

consists of a sequence of legal moves so that a knight lands oneach square exactly once See Figure 1 for some examples thatwill be used later where the tours begin at the square labeled119886 or 1 and end at squares 119897 or 20 A closed knightrsquos tour hasthe added condition that the knight must end its tour on theinitial square The 8 times 8 chessboard can easily be extendedto rectangular boards and in 1991 Schwenk characterized allrectangular boards that have a closed knightrsquos tour [2] Heused an induction proof with nine basis steps to constructclosed knightrsquos tours for all admissible boards to prove thefollowing

Schwenkrsquos Theorem An m times n chessboard with m le n has aclosed knightrsquos tour in all but the following cases

(a) 119898 = 1 for all 119899

(b) 119898 = 2 for all 119899

(c) 119898 = 3 and 119899 = 4 6 8 or is odd

(d) 119898 = 4 for all 119899

(e) 119898 gt 4 for all combinations of 119898 odd and 119899 odd

Hindawi Publishing CorporationJournal of Discrete MathematicsVolume 2014 Article ID 210892 9 pageshttpdxdoiorg1011552014210892

2 Journal of Discrete Mathematics

20 1 16 7 12

15 6 11 2 17

10 19 4 13 8

5 14 9 18 3

a d g j

l i b e

c f k h

Figure 1 The 3 times 4 and 4 times 5 rectangular boards with open knightrsquostours

More recently Demaio and Hippchen investigated theimpossible boards in [3] and determined the fewest numberof squares that must be removed from a rectangular boardso that the remaining board has a closed knightrsquos tour Ifwe define an extended closed knightrsquos tour for a rectangularchessboard as a closed knightrsquos tour that includes all squaresof the board and possibly additional squares beyond theboundaries of the board then our interest is to answer thefollowing how many squares must be added to a rectangularchessboard so that the new board has a closed knightrsquos tour Foran 119898 times 119899 board the minimum number called the extendedtour number is denoted as 120591(119898 119899) Every rectangular boardhas an extended knightrsquos tour and the extended tour numbersare summarized in our main result

Theorem 1 Every 119898 times 119899 chessboard with 0 lt 119898 le 119899 has anextended tour number of 0 except for the following

(a) 120591(1 119899) = 119899 + 2 for all 119899(b) 120591(2 119899) = 6 if n is even and 119899 = 4 otherwise 120591(2 119899) =4

(c) 120591(3 3) = 120591(3 5) = 3 120591(3 4) = 120591(3 6) = 120591(3 8) = 2and 120591(3 119899) = 1 if 119899 ge 7 and is odd

(d) 120591(4 119899) = 2 for all n(e) 120591(119898 119899) = 1 if119898 gt 4 with119898 and 119899 both being odd

Within the last twenty-five years much of the researchon knightrsquos tours have taken on the tone set by Schwenk [2]by being concerned about which boards allow knightrsquos toursKnuthrsquos work in [4] on leapers (ie generalized knights) andWatkinsrsquo book [1] have played no small part in the subjectrsquospopularity either A survey of the literature shows thatknightrsquos tours have been studied on a variety of shapes anddimensions (cf [5ndash7]) Generalized knightrsquos moves have beenstudied in [8ndash10] In [1] extensions of Schwenkrsquos Theoremwere given for four alternative surfaces the torus cylinderMobius strip and the Klein bottle Recently the authors in[11] considered the same four surfaces and determined thefewest number of squares that must be removed or added toinsure a knightrsquos tour

2 Proof of Main Result

We will assume that the board is 119898 times 119899 with 0 lt m le n theupper left corner of each board is black and the numbering

for the extended knightrsquos tour begins at that square Sinceevery closed knightrsquos tour alternates between black and whitesquares we have three useful observations about the numberof additional squares needed

(a) The total number of squares (original and additional)must be even and must consist of an equal number ofblack and white squares

(b) An odd number of squares are needed to connect twosquares of the same color

(c) An even number of squares are needed to connect twosquares of opposite color

In [2 3] where tours are restricted to squares on the originalboard no closed knightrsquos tour exists whenm= 1 or 2 howeverby allowing additional squares the extended tour numbersdo exist and are generally larger than those for larger boardsIn addition the arrangement of the added squares is notunique For instance consider the three variations for the 2 times2 board shown in Figure 2 Probably the second configurationis preferred over the first because the additional squares areall adjacent to the original board but the third arrangementwill be used in case (b) of the proof to show that 120591(2 2) = 6because of its simplicity symmetry and the ability to extendit to longer boards

The proof will be organized by considering the cases (a)ndash(e) from Schwenkrsquos Theorem in order

(a) The Case 119898 = 1 A special situation applies for the 1times 1 chessboard One solution is to move to an additionalsquare and simply return to the original square however ifbacktracking is to be avoided at least three additional squaresare needed (by observation (b)) One configuration that usesexactly three additional squares is in Figure 3with the original1 times 1 chessboard on square 1

For m = 1 and n ge 2 a useful variation of the rectangularboard is a graph where each square is represented by a vertex(colored white or black) and an edge joins two vertices ifa knight can move between the corresponding squares in asingle move Thus if m = 1 and n ge 2 the 1 times 119899 board canbe thought of as a graph with 119899 isolated vertices A cycleis formed by selecting a vertex one at a time either addingan odd number of new vertices after it if the next selectedvertex has the same color or adding an even number ofvertices after it if the next selected vertex has the oppositecolor and repeating this between the last selected vertex andthe first one Note that at least twice on the cycle the vertexcolor will change so 120591(1 n) ge (119899 minus 2) + 2 + 2 = n + 2for n ge 2 This bound would be achieved if a shortest cyclecould be formed by selecting the black vertices in successionand inserting one vertex between each pair inserting twovertices between the last black vertex and the firstwhite vertexselected inserting one vertex after each white vertex until allvertices are selected and then adding two vertices betweenthe last white vertex and the first black vertex Since there are119908 white squares and 119887 black squares then n = w + b and thisshortest cycle includes (119908 minus 1) + 2 + (119887 minus 1) + 2 = 119899 + 2 newvertices This is accomplished by the following process

Journal of Discrete Mathematics 3

3

5 2

1 4

6 9

10 7

8

9

10 7 2

1 8 5

6 3

4

4 2

1 6 9

5 8 3

10 7

Figure 2 Three arrangements for the six additional vertices needed for the 2 times 2 board

1 8 3 6

10 5

9 2 7 4

1 10 3 8 5

12 6

11 2 9 4 7

1

4 2

3

Figure 3 The 1 times 1 1 times 4 and 1 times 5 rectangles

(1) Place the 1 times 119899 board in the top row of a 3 times (n + 4)rectangle so that the upper left square of the board isin the (1 3) position and color that square black

(2) Begin moving (1 3) (3 4) (1 5) (1 119894) (3 119894 +1) (1 119894 + 2) (1 119898) wherem = n + 1 if 119899 is even orn + 2 if 119899 is odd

(3) If 119899 is even continue with (1 n + 1) (3 n + 2) (2 n +4) (1 n + 2)If 119899 is odd continue with (1 n + 2) (2 n + 4) (3 n +2) and (1 n + 1)

(4) Now proceed backwards from (1 j) (3 119895 minus 1) (1 119895 minus2) (1 4) (3 3) (2 1) (1 3) where j = n + 1 if 119899 isodd or n + 2 if 119899 is even

The forward and backward passes through row 3 do notoverlap because the first pass uses white squares and thereverse pass uses only black squares (These cycles are

illustrated in Figure 3 for 119899 = 4 and 5) So 120591(1 119899) = 119899 + 2for n ge 1

(b)The Case119898 = 2 Consider that 119899 is even first If 119899 = 2 thenthe corresponding graph consists of four isolated vertices andif n gt 4 is even then the corresponding graph consists of fourpaths with at least four additional vertices needed to connectthem If 119899 = 2 + 4k for some nonnegative integer k then thepaths are from (1 1) to (1 119899 minus 1) (2 1) to (2 119899 minus 1) (1 2) to (1n) and (2 2) to (2 n) Notice that two of the paths begin (onthe left) and end (on the right) on black squares while two ofthe paths begin and end on white squares On the other handif 119899 = 4k for some positive integer k ge 2 then the paths arefrom (1 1) to (2 119899 minus 1) (2 1) to (1 119899 minus 1) (1 2) to (2 n) and(2 2) to (1 n) In this case all four of the paths begin (on theleft) and end (on the right) on squares of opposite colors Let119875(11) 119875(12) 119875(21) and 119875(22) denote the four paths where thesubscript indicates the beginning vertex

For each of these values of 119899 it is possible to link thefour paths (or isolated vertices) together with exactly fouradditional vertices however then 119875(11) and 119875(22) will bejoined at both ends and 119875(12) and 119875(21) will be joined atboth ends creating two disjoint cycles instead of a singleknightrsquos tour Therefore to create a single cycle we will haveto connect a white square to a white square a black square toa black square a white square to a black square and a blacksquare to awhite square requiring at least 6 additional squaresby observations (b) and (c) Notice also that for 119899 at least 6joining the beginning of one path to the end of another wouldrequire at least 2 additional squares so we assume that wejoin the beginning of one path to the beginning of another(or the right end of one path to the right end of another)This can be done as follows Connect (1 1) to (2 2) usingthe additional square (3 0) connect (1 2) to (2 1) using theadditional square (0 0) connect (1 119899 minus 1) to (1 n) using (0 n+ 1) and (minus1 119899 minus 1) and connect (2 119899 minus 1) to (2 n) using (3 n+ 1) and (4 119899minus1)Thus only the six additional squares (3 0)(0 0) (0 n + 1) (minus1 119899minus1) (3 n + 1) and (4 119899minus1) are neededFor example for the 2 times 2 board the six external squares are(3 0) (0 0) (0 3) (minus1 1) (3 3) and (4 1) and a tour is (1 1)rarr (3 0) rarr (2 2) rarr (4 1) rarr (3 3) rarr (2 1) rarr (0 0) rarr(1 2) rarr (minus1 1) rarr (0 3) rarr (1 1) as shown in the third boardarrangement in Figure 2 Sample boards are shown for the 2


17

11 16

1 10 13 20 3 8 15 18

12 21 2 9 14 19 4 7

22 5

6

5

9 4

1 8 11 16 3 6

10 17 2 7 12 15

18 13

14

Figure 4 The 2 times 6 and 2 times 8 boards illustrate that 120591(2 119899) = 6 if 119899 ge 6 is even

9 5

1 8 11 16 3 6 13

10 17 2 7 12 15 4

18 14

6

1 4 11 8

10 7 2 5

3 12 9

7 11

1 6 9 12 3

8 13 2 5 10

14 4

Figure 5 120591(2 119899) = 4 if 119899 = 4 or is odd

1 6 3

7 4 9 12

11 2 5

8 10

18 12

17

1 4 13 8 11

14 9 6 3 16

5 2 15 10 7

Figure 6 120591(3 3) = 120591(3 5) = 3

times 6 and 2 times 8 rectangles in Figure 4 Thus 120591(2 119899) = 6 if n = 4and is even

If 119899 = 4 a knight at any square on the 2 times 4 board canmove to only one other square so the corresponding graphconsists of four components (all paths with one edge) To

form an extended closed knightrsquos tour at least four verticesmust be added to connect the pathsThis can be accomplishedas shown in Figure 5

Now consider when 119899 is odd Again the correspondinggraph consists of four paths and at least four vertices areneeded to connect them This can be done by using oneadditional vertex (or square) at each corner of the chessboardand applying the following procedure

(1) If 119899 = 1+4119896 for some positive integer k then the pathsare from (1 1) to (1 n) (2 1) to (2 n) (1 2) to (2 119899minus1)and (2 2) to (1 119899 minus 1)If 119899 = 3 + 4k for some nonnegative integer k then thepaths are from (1 1) to (2 n) (2 1) to (1 n) (1 2) to (1119899 minus 1) and (2 2) to (2 119899 minus 1)

(2) Insert the vertex (3 n + 1) between (1 n) and (2 119899minus1)Insert (0 0) between (1 2) and (2 1) Insert (0 n +1) between (2 n) and (1 119899 minus 1) Finally insert (3 0)between (2 2) and (1 1)


1 4 19 10 13 6 15

18 11 2 5 16 9

3 20 17 12 7 14

8

14

7

1 4 13

8 11 6

5 2 9

10

3

12

Figure 7 120591(3 4) = 120591(3 6) = 2

1 18 11 20 3 14

10 25 2 13 6 21

26 17 12 19 8 15 4

24 9 16 5 22 7

23

1 12 19 8 3

18 7 2 13 20

17 6 11 22 15 4 9

16 5 10 21 14

1 14 5 10

18 13 8 11 2 15

17 4 9 6

12 7 16 3

Figure 8 120591(4 4) = 120591(4 5) = 120591(4 6) = 2

The extended closed knightrsquos tours for the 2 times 5 and 2 times 7boards are in Figure 5 and 120591(2 119899) = 4 if 119899 = 4 or is odd

(c) The Case 119898 = 3 First let 119899 = 3 or 5 Since a 3 times 3 or3 times 5 board has an odd number of squares an odd numberof squares must be added to complete an extended cycle (byobservation (a)) For the 3 times 3 board if the knight is on thecenter square there are no squares towhich it canmoveThusan extra square is needed Similarly since the knight needs toreturn to the center square a second square must be addedTherefore 120591(3 3) ge 3 In fact exactly three additional verticesare needed as shown in Figure 6

Suppose 120591(3 5) = 1 Since square (1 1) is black (perour assumptions) then the added square must be white(for parity) and must be adjacent to two black squares inthe extended knightrsquos tour If we consider the cyclic graphassociated with the tour and remove the added vertex thena Hamiltonian path remains on the fifteen vertices with bothendpoints corresponding to black squares Since squares (2 1)and (2 5) are only adjacent to (1 3) and (3 3) theHamiltonianpath includes the cycle (2 1) (1 3) (2 5) (3 3) and (2 1)

which cannot happen on a path Therefore 120591(3 5) gt 2 and isin fact 3 as shown in Figure 6

Now let 119899 = 4 6 or 8 Based on observation (a) theextended tour numbers for the 3 times 4 3 times 6 and 3 times 8 boardsmust be at least 2 In fact 120591(3 4) = 120591(3 6) = 2 as shownin Figure 7 To show that 120591(3 8) = 2 we introduce a usefulconstruction for larger boards that consists of combining asmaller board with a rectangle having an open knightrsquos tourFor instance to obtain a 3 times 8 board attach the 3 times 4 boardfrom Figure 1 to the right side of the 3 times 4 board in Figure 7Begin the closed knightrsquos tour from square 1 to square 12 jumpfrom square 12 to square 119886 follow the open knightrsquos tour from119886 to l jump back from square 119897 to square 13 and resume theoriginal closed knightrsquos tour to square 1

Finally the proof that 120591(3 119899) = 1 if n ge 7 and is odd isincluded in subcases 3 and 4 in the proof for case (e) below

(d) The Case 119898 = 4 Since a 4 times 119899 board has an even numberof squares an even number of squares must be added tocomplete an extended cycle (by observation (a)) The 4 times4 4 times 5 and 4 times 6 boards are shown in Figure 8 Each of


a Ae

aA e

a A e

d D b d D b d D b

F fB

Ff B

F f B

C c E C c E C c E

Figure 9 Connections between 4 times 3 expansion boards for 4 times (n + 3k) boards with 119899 = 4 5 or 6

12

1 26 3 20 13 22 9 16 11

4 19 28 25 6 17 14 23 8

27 2 5 18 21 24 7 10 15

24

1 32 17 8 25 30 15

18 23 36 31 16 7 26

33 2 9 22 27 14 29

10 19 4 35 12 21 6

3 34 11 20 5 28 13

26

1 18 7 12 25

8 13 24 17 22

19 2 21 6 11

14 9 4 23 16

3 20 15 10 5

16

1 20 17 14 11 8 5

18 13 22 3 6 15 10

21 2 19 12 9 4 7

Figure 10 The 3 times 7 3 times 9 5 times 5 and 5 times 7 boards have extended tour numbers of 1

these ldquobaserdquo 4 times 119899 boards can be extended to a 4 times (n + 3k)board by adding 119896 copies of a 4 times 3 board on the right andlinking them as shown in Figure 9 For instance if k = 1then the extended closed knightrsquos tour is formed by alteringtwo moves Consider the 4 times 4 board and begin traversing atsquare 1 until you get to square 2 Now replace the move from2 to 3 with the following from square 2 move to a continueto f (using the lowercase letters) and then return to square 3Second replace the move between squares 15 and 16 with thefollowing from square 15 move to A continue to F (usingthe uppercase letters) and then return to square 16 Finallycomplete the tour by continuing to 1 in the original 4 times 4board Similarly for the 4 times 5 board use the tour 1 rarr 2 rarrsdot sdot sdot rarr 13 rarr 119886 rarr 119887 rarr sdot sdot sdot rarr 119891 rarr 14 rarr 15 rarr sdot sdot sdot rarr

20 rarr 119860 rarr 119861 rarr sdot sdot sdot rarr 119865 rarr 21 rarr 22 rarr 1 and for the4 times 6 board use the tour 1 rarr 2 rarr sdot sdot sdot rarr 6 rarr 119886 rarr 119887 rarrsdot sdot sdot rarr 119891 rarr 7 rarr 8 rarr sdot sdot sdot rarr 21 rarr 119860 rarr 119861 rarr sdot sdot sdot rarr

119865 rarr 22 rarr sdot sdot sdot rarr 26 rarr 1 If an extended closed knightrsquostour exists with 119896 minus 1 copies of the 4 times 3 board then the kthcopy can be attached by linking the squares D E b and 119888 inthe (119896 minus 1)st copy with squares a f A and F respectively inthe kth copy Thus 120591(4 119899) = 2 for n ge 4

(e) The Case 119898 gt 4 with m and n Both Being Odd Figure 10shows thatby observation (a) the values for 120591(3 7) =120591(3 9) = 120591(5 5) = 120591(5 7) = 1 Finally to show that120591(119898 119899) = 1 if 119898 and 119899 are odd with 3 le m 7 le n weconsider four subcases whose general structures are shown in(1) Each subcase is formed by constructing an119898times119899 rectanglefrom the smaller rectangles in Figures 10 1 and 11 and thenldquoconnectingrdquo them according to the rules given in (2) Ingeneral the tours begin in the upper left corner move upand down the first column of boards and travel across therows of boards and back The external square is always abovethe small rectangular board in the upper left corner of the119898 times 119899 board For clarity note that in subcase 1 if one beginsthe knightrsquos tour with the upper left square labeled 1 uses thelabels for the squares of the boards as given in the figures andfollows the tour using the connecting rules then the numbersand letters will always proceed in ascending order howeverin subcases 3 and 4 when one uses rule 13 to connect thestring of 3 times 4 boards in the top row the second fourthsixth and so forth copies of the board will be traversed inreverse order That is the labels will be selected from 119897 downto 119886 instead of from 119886 to 119897 Similarly in subcases 2 and 4 the


3 28 1 18 9 22 7

14 17 4 25 6 19 10

27 2 15 12 21 8 23

16 13 26 5 24 11 20

12 17 8 3

7 2 13 18

16 11 4 9

1 6 19 14

20 15 10 5

A e E c

H b B f

h D d F

a G g C

36 1 32 15 20 3 24 7 26

31 14 19 2 33 10 27 4 23

18 35 12 29 16 21 6 25 8

13 30 17 34 11 28 9 22 5

Figure 11 The additional 5 times 4 4 times 4 4 times 7 and 4 times 9 boards needed for the four subcases

entire third fifth seventh and so forth row of boards will betraversed in reverse order

The four subcases for the general case when m and n are bothodd are as follows

Subcase 1 119898 = 1 mod 4 119899 = 1 mod 4

[[[[[[[[[[[[[

[

[5 times 5] 997888rarr1[5 times 4] 997888rarr

4sdot sdot sdot 997888rarr

4[5 times 4]

darr2



6[4 times 4]

darr3

darr3



6[4 times 4]

]]]]]]]]]]]]]

]

Subcase 2 119898 = 1 mod 4 119899 = 3 mod 4

[[[[[[[[[[[

[



4[5 times 4]

darr8



6[4 times 4]

darr11

darr11



6[4 times 4]

]]]]]]]]]]]

]


Subcase 3 119898 = 3 mod 4 119899 = 1 mod 4

[[[[[[[[[[[

[



13[3 times 4]

darr14



6[4 times 4]

darr17

darr17



6[4 times 4]

]]]]]]]]]]]

]

Subcase 4 119898 = 3 mod 4 119899 = 3 mod 4

[[[[[[[[[[[

[



13[3 times 4]

darr16



6[4 times 4]

darr11

darr11



6[4 times 4]

]]]]]]]]]]]

]

(1)

Connection rules for the four subcases are as follows

1 (2 5)ndash(4 1) 5 (2 4)ndash(1 1) 9 (2 6)ndash(1 1) 13 (3 4)ndash(1 1) 17 (3 1)ndash(1 2)(4 4)ndash(5 1) (4 5)ndash(2 1) (4 7)ndash(2 1) (1 3)ndash(2 1) (3 2)ndash(1 1)

(2 5)ndash(4 1) (4 6)ndash(3 1)(4 4)ndash(3 1) (2 7)ndash(4 1)

2 (5 3)ndash(1 1) 6 (2 3)ndash(1 1) 10 (2 8)ndash(1 1) 14 (3 3)ndash(1 1)(4 1)ndash(1 2) (4 4)ndash(2 1) (4 9)ndash(2 1) (2 1)ndash(1 2)




(2)

3 Conclusion

Every rectangular board has an extended knightrsquos tour and theextended tour numbers are summarized in Theorem 1 andthe constructions in the proof indicate where the extendedsquares should be

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The authors wish to thank the referees and the editorfor helpful suggestions which improved the mathematicalaccuracy and the presentation of this paper

References

[1] J J Watkins Across the Board The Mathematics of ChessboardProblems Princeton University Press Princeton NJ USA2004

[2] A J Schwenk ldquoWhich rectangular chessboards have a knightrsquostourrdquoMathematics Magazine vol 64 no 5 pp 325ndash332 1991

[3] JDemaio andTHippchen ldquoClosed knightrsquos tourswithminimalsquare removal for all rectangular boardsrdquoMathematics Maga-zine vol 82 pp 219ndash225 2009

[4] D E Knuth ldquoLeaper graphsrdquoTheMathematical Gazette vol 78pp 274ndash297 1994

[5] A M Miller and D L Farnsworth ldquoKnightrsquos tours on cylin-drical and toroidal boards with one square removedrdquo ArsCombinatoria vol 108 pp 327ndash340 2013


[6] J Erde B Golenia and S Golenia ldquoThe closed knight tourproblem in higher dimensionsrdquoThe Electronic Journal of Com-binatorics vol 19 no 4 pp 9ndash17 2012

[7] J DeMaio and B Mathew ldquoWhich chessboards have a closedknightrsquos tour within the rectangular prismrdquo Electronic Journalof Combinatorics vol 18 no 1 pp 8ndash14 2011

[8] N Kamcev ldquoGeneralised Knightrsquos toursrdquo Electronic Journal ofCombinatorics vol 21 no 1 pp 1ndash31 2014

[9] S Bai X-F Yang G-B Zhu D-L Jiang and J HuangldquoGeneralized knightrsquos tour on 3D chessboardsrdquoDiscrete AppliedMathematics vol 158 no 16 pp 1727ndash1731 2010

[10] G L Chia and S-H Ong ldquoGeneralized knightrsquos tours onrectangular chessboardsrdquo Discrete Applied Mathematics vol150 no 1ndash3 pp 80ndash98 2005

[11] G Bullington L Eroh S JWinters andG L Johns ldquoVariationsfor the knights tour for rectangular chessboards on alternativesurfacesrdquo Congressus Numerantium vol 206 pp 199ndash214 2010

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20 1 16 7 12

15 6 11 2 17

10 19 4 13 8

5 14 9 18 3

a d g j

l i b e

c f k h

Figure 1 The 3 times 4 and 4 times 5 rectangular boards with open knightrsquostours

More recently Demaio and Hippchen investigated theimpossible boards in [3] and determined the fewest numberof squares that must be removed from a rectangular boardso that the remaining board has a closed knightrsquos tour Ifwe define an extended closed knightrsquos tour for a rectangularchessboard as a closed knightrsquos tour that includes all squaresof the board and possibly additional squares beyond theboundaries of the board then our interest is to answer thefollowing how many squares must be added to a rectangularchessboard so that the new board has a closed knightrsquos tour Foran 119898 times 119899 board the minimum number called the extendedtour number is denoted as 120591(119898 119899) Every rectangular boardhas an extended knightrsquos tour and the extended tour numbersare summarized in our main result

Theorem 1 Every 119898 times 119899 chessboard with 0 lt 119898 le 119899 has anextended tour number of 0 except for the following

(a) 120591(1 119899) = 119899 + 2 for all 119899(b) 120591(2 119899) = 6 if n is even and 119899 = 4 otherwise 120591(2 119899) =4

(c) 120591(3 3) = 120591(3 5) = 3 120591(3 4) = 120591(3 6) = 120591(3 8) = 2and 120591(3 119899) = 1 if 119899 ge 7 and is odd

(d) 120591(4 119899) = 2 for all n(e) 120591(119898 119899) = 1 if119898 gt 4 with119898 and 119899 both being odd

Within the last twenty-five years much of the researchon knightrsquos tours have taken on the tone set by Schwenk [2]by being concerned about which boards allow knightrsquos toursKnuthrsquos work in [4] on leapers (ie generalized knights) andWatkinsrsquo book [1] have played no small part in the subjectrsquospopularity either A survey of the literature shows thatknightrsquos tours have been studied on a variety of shapes anddimensions (cf [5ndash7]) Generalized knightrsquos moves have beenstudied in [8ndash10] In [1] extensions of Schwenkrsquos Theoremwere given for four alternative surfaces the torus cylinderMobius strip and the Klein bottle Recently the authors in[11] considered the same four surfaces and determined thefewest number of squares that must be removed or added toinsure a knightrsquos tour

2 Proof of Main Result

We will assume that the board is 119898 times 119899 with 0 lt m le n theupper left corner of each board is black and the numbering

for the extended knightrsquos tour begins at that square Sinceevery closed knightrsquos tour alternates between black and whitesquares we have three useful observations about the numberof additional squares needed

(a) The total number of squares (original and additional)must be even and must consist of an equal number ofblack and white squares

(b) An odd number of squares are needed to connect twosquares of the same color

(c) An even number of squares are needed to connect twosquares of opposite color

In [2 3] where tours are restricted to squares on the originalboard no closed knightrsquos tour exists whenm= 1 or 2 howeverby allowing additional squares the extended tour numbersdo exist and are generally larger than those for larger boardsIn addition the arrangement of the added squares is notunique For instance consider the three variations for the 2 times2 board shown in Figure 2 Probably the second configurationis preferred over the first because the additional squares areall adjacent to the original board but the third arrangementwill be used in case (b) of the proof to show that 120591(2 2) = 6because of its simplicity symmetry and the ability to extendit to longer boards

The proof will be organized by considering the cases (a)ndash(e) from Schwenkrsquos Theorem in order

(a) The Case 119898 = 1 A special situation applies for the 1times 1 chessboard One solution is to move to an additionalsquare and simply return to the original square however ifbacktracking is to be avoided at least three additional squaresare needed (by observation (b)) One configuration that usesexactly three additional squares is in Figure 3with the original1 times 1 chessboard on square 1

For m = 1 and n ge 2 a useful variation of the rectangularboard is a graph where each square is represented by a vertex(colored white or black) and an edge joins two vertices ifa knight can move between the corresponding squares in asingle move Thus if m = 1 and n ge 2 the 1 times 119899 board canbe thought of as a graph with 119899 isolated vertices A cycleis formed by selecting a vertex one at a time either addingan odd number of new vertices after it if the next selectedvertex has the same color or adding an even number ofvertices after it if the next selected vertex has the oppositecolor and repeating this between the last selected vertex andthe first one Note that at least twice on the cycle the vertexcolor will change so 120591(1 n) ge (119899 minus 2) + 2 + 2 = n + 2for n ge 2 This bound would be achieved if a shortest cyclecould be formed by selecting the black vertices in successionand inserting one vertex between each pair inserting twovertices between the last black vertex and the firstwhite vertexselected inserting one vertex after each white vertex until allvertices are selected and then adding two vertices betweenthe last white vertex and the first black vertex Since there are119908 white squares and 119887 black squares then n = w + b and thisshortest cycle includes (119908 minus 1) + 2 + (119887 minus 1) + 2 = 119899 + 2 newvertices This is accomplished by the following process


3

5 2

1 4

6 9

10 7

8

9

10 7 2

1 8 5

6 3

4

4 2

1 6 9

5 8 3

10 7


1 8 3 6

10 5

9 2 7 4

1 10 3 8 5

12 6

11 2 9 4 7

1

4 2

3











17

11 16

1 10 13 20 3 8 15 18

12 21 2 9 14 19 4 7

22 5

6

5

9 4

1 8 11 16 3 6

10 17 2 7 12 15

18 13

14


9 5

1 8 11 16 3 6 13

10 17 2 7 12 15 4

18 14

6

1 4 11 8

10 7 2 5

3 12 9

7 11

1 6 9 12 3

8 13 2 5 10

14 4


1 6 3

7 4 9 12

11 2 5

8 10

18 12

17

1 4 13 8 11

14 9 6 3 16

5 2 15 10 7

Figure 6 120591(3 3) = 120591(3 5) = 3








1 4 19 10 13 6 15

18 11 2 5 16 9

3 20 17 12 7 14

8

14

7

1 4 13

8 11 6

5 2 9

10

3

12

Figure 7 120591(3 4) = 120591(3 6) = 2

1 18 11 20 3 14

10 25 2 13 6 21

26 17 12 19 8 15 4

24 9 16 5 22 7

23

1 12 19 8 3

18 7 2 13 20

17 6 11 22 15 4 9

16 5 10 21 14

1 14 5 10

18 13 8 11 2 15

17 4 9 6

12 7 16 3

Figure 8 120591(4 4) = 120591(4 5) = 120591(4 6) = 2









a Ae

aA e

a A e

d D b d D b d D b

F fB

Ff B

F f B

C c E C c E C c E


12

1 26 3 20 13 22 9 16 11

4 19 28 25 6 17 14 23 8

27 2 5 18 21 24 7 10 15

24

1 32 17 8 25 30 15

18 23 36 31 16 7 26

33 2 9 22 27 14 29

10 19 4 35 12 21 6

3 34 11 20 5 28 13

26

1 18 7 12 25

8 13 24 17 22

19 2 21 6 11

14 9 4 23 16

3 20 15 10 5

16

1 20 17 14 11 8 5

18 13 22 3 6 15 10

21 2 19 12 9 4 7







3 28 1 18 9 22 7

14 17 4 25 6 19 10

27 2 15 12 21 8 23

16 13 26 5 24 11 20

12 17 8 3

7 2 13 18

16 11 4 9

1 6 19 14

20 15 10 5

A e E c

H b B f

h D d F

a G g C

36 1 32 15 20 3 24 7 26

31 14 19 2 33 10 27 4 23

18 35 12 29 16 21 6 25 8

13 30 17 34 11 28 9 22 5




Subcase 1 119898 = 1 mod 4 119899 = 1 mod 4

[[[[[[[[[[[[[

[



4[5 times 4]

darr2



6[4 times 4]

darr3

darr3



6[4 times 4]

]]]]]]]]]]]]]

]

Subcase 2 119898 = 1 mod 4 119899 = 3 mod 4

[[[[[[[[[[[

[



4[5 times 4]

darr8



6[4 times 4]

darr11

darr11



6[4 times 4]

]]]]]]]]]]]

]


Subcase 3 119898 = 3 mod 4 119899 = 1 mod 4

[[[[[[[[[[[

[



13[3 times 4]

darr14



6[4 times 4]

darr17

darr17



6[4 times 4]

]]]]]]]]]]]

]

Subcase 4 119898 = 3 mod 4 119899 = 3 mod 4

[[[[[[[[[[[

[



13[3 times 4]

darr16



6[4 times 4]

darr11

darr11



6[4 times 4]

]]]]]]]]]]]

]

(1)








(2)

3 Conclusion




Acknowledgment


References




















Volume 2014




Journal of











Journal of


Function Spaces






Algebra










3

5 2

1 4

6 9

10 7

8

9

10 7 2

1 8 5

6 3

4

4 2

1 6 9

5 8 3

10 7


1 8 3 6

10 5

9 2 7 4

1 10 3 8 5

12 6

11 2 9 4 7

1

4 2

3











17

11 16

1 10 13 20 3 8 15 18

12 21 2 9 14 19 4 7

22 5

6

5

9 4

1 8 11 16 3 6

10 17 2 7 12 15

18 13

14


9 5

1 8 11 16 3 6 13

10 17 2 7 12 15 4

18 14

6

1 4 11 8

10 7 2 5

3 12 9

7 11

1 6 9 12 3

8 13 2 5 10

14 4


1 6 3

7 4 9 12

11 2 5

8 10

18 12

17

1 4 13 8 11

14 9 6 3 16

5 2 15 10 7

Figure 6 120591(3 3) = 120591(3 5) = 3








1 4 19 10 13 6 15

18 11 2 5 16 9

3 20 17 12 7 14

8

14

7

1 4 13

8 11 6

5 2 9

10

3

12

Figure 7 120591(3 4) = 120591(3 6) = 2

1 18 11 20 3 14

10 25 2 13 6 21

26 17 12 19 8 15 4

24 9 16 5 22 7

23

1 12 19 8 3

18 7 2 13 20

17 6 11 22 15 4 9

16 5 10 21 14

1 14 5 10

18 13 8 11 2 15

17 4 9 6

12 7 16 3

Figure 8 120591(4 4) = 120591(4 5) = 120591(4 6) = 2









a Ae

aA e

a A e

d D b d D b d D b

F fB

Ff B

F f B

C c E C c E C c E


12

1 26 3 20 13 22 9 16 11

4 19 28 25 6 17 14 23 8

27 2 5 18 21 24 7 10 15

24

1 32 17 8 25 30 15

18 23 36 31 16 7 26

33 2 9 22 27 14 29

10 19 4 35 12 21 6

3 34 11 20 5 28 13

26

1 18 7 12 25

8 13 24 17 22

19 2 21 6 11

14 9 4 23 16

3 20 15 10 5

16

1 20 17 14 11 8 5

18 13 22 3 6 15 10

21 2 19 12 9 4 7







3 28 1 18 9 22 7

14 17 4 25 6 19 10

27 2 15 12 21 8 23

16 13 26 5 24 11 20

12 17 8 3

7 2 13 18

16 11 4 9

1 6 19 14

20 15 10 5

A e E c

H b B f

h D d F

a G g C

36 1 32 15 20 3 24 7 26

31 14 19 2 33 10 27 4 23

18 35 12 29 16 21 6 25 8

13 30 17 34 11 28 9 22 5




Subcase 1 119898 = 1 mod 4 119899 = 1 mod 4

[[[[[[[[[[[[[

[



4[5 times 4]

darr2



6[4 times 4]

darr3

darr3



6[4 times 4]

]]]]]]]]]]]]]

]

Subcase 2 119898 = 1 mod 4 119899 = 3 mod 4

[[[[[[[[[[[

[



4[5 times 4]

darr8



6[4 times 4]

darr11

darr11



6[4 times 4]

]]]]]]]]]]]

]


Subcase 3 119898 = 3 mod 4 119899 = 1 mod 4

[[[[[[[[[[[

[



13[3 times 4]

darr14



6[4 times 4]

darr17

darr17



6[4 times 4]

]]]]]]]]]]]

]

Subcase 4 119898 = 3 mod 4 119899 = 3 mod 4

[[[[[[[[[[[

[



13[3 times 4]

darr16



6[4 times 4]

darr11

darr11



6[4 times 4]

]]]]]]]]]]]

]

(1)








(2)

3 Conclusion




Acknowledgment


References




















Volume 2014




Journal of











Journal of


Function Spaces






Algebra










17

11 16

1 10 13 20 3 8 15 18

12 21 2 9 14 19 4 7

22 5

6

5

9 4

1 8 11 16 3 6

10 17 2 7 12 15

18 13

14


9 5

1 8 11 16 3 6 13

10 17 2 7 12 15 4

18 14

6

1 4 11 8

10 7 2 5

3 12 9

7 11

1 6 9 12 3

8 13 2 5 10

14 4


1 6 3

7 4 9 12

11 2 5

8 10

18 12

17

1 4 13 8 11

14 9 6 3 16

5 2 15 10 7

Figure 6 120591(3 3) = 120591(3 5) = 3








1 4 19 10 13 6 15

18 11 2 5 16 9

3 20 17 12 7 14

8

14

7

1 4 13

8 11 6

5 2 9

10

3

12

Figure 7 120591(3 4) = 120591(3 6) = 2

1 18 11 20 3 14

10 25 2 13 6 21

26 17 12 19 8 15 4

24 9 16 5 22 7

23

1 12 19 8 3

18 7 2 13 20

17 6 11 22 15 4 9

16 5 10 21 14

1 14 5 10

18 13 8 11 2 15

17 4 9 6

12 7 16 3

Figure 8 120591(4 4) = 120591(4 5) = 120591(4 6) = 2









a Ae

aA e

a A e

d D b d D b d D b

F fB

Ff B

F f B

C c E C c E C c E


12

1 26 3 20 13 22 9 16 11

4 19 28 25 6 17 14 23 8

27 2 5 18 21 24 7 10 15

24

1 32 17 8 25 30 15

18 23 36 31 16 7 26

33 2 9 22 27 14 29

10 19 4 35 12 21 6

3 34 11 20 5 28 13

26

1 18 7 12 25

8 13 24 17 22

19 2 21 6 11

14 9 4 23 16

3 20 15 10 5

16

1 20 17 14 11 8 5

18 13 22 3 6 15 10

21 2 19 12 9 4 7







3 28 1 18 9 22 7

14 17 4 25 6 19 10

27 2 15 12 21 8 23

16 13 26 5 24 11 20

12 17 8 3

7 2 13 18

16 11 4 9

1 6 19 14

20 15 10 5

A e E c

H b B f

h D d F

a G g C

36 1 32 15 20 3 24 7 26

31 14 19 2 33 10 27 4 23

18 35 12 29 16 21 6 25 8

13 30 17 34 11 28 9 22 5




Subcase 1 119898 = 1 mod 4 119899 = 1 mod 4

[[[[[[[[[[[[[

[



4[5 times 4]

darr2



6[4 times 4]

darr3

darr3



6[4 times 4]

]]]]]]]]]]]]]

]

Subcase 2 119898 = 1 mod 4 119899 = 3 mod 4

[[[[[[[[[[[

[



4[5 times 4]

darr8



6[4 times 4]

darr11

darr11



6[4 times 4]

]]]]]]]]]]]

]


Subcase 3 119898 = 3 mod 4 119899 = 1 mod 4

[[[[[[[[[[[

[



13[3 times 4]

darr14



6[4 times 4]

darr17

darr17



6[4 times 4]

]]]]]]]]]]]

]

Subcase 4 119898 = 3 mod 4 119899 = 3 mod 4

[[[[[[[[[[[

[



13[3 times 4]

darr16



6[4 times 4]

darr11

darr11



6[4 times 4]

]]]]]]]]]]]

]

(1)








(2)

3 Conclusion




Acknowledgment


References




















Volume 2014




Journal of











Journal of


Function Spaces






Algebra










1 4 19 10 13 6 15

18 11 2 5 16 9

3 20 17 12 7 14

8

14

7

1 4 13

8 11 6

5 2 9

10

3

12

Figure 7 120591(3 4) = 120591(3 6) = 2

1 18 11 20 3 14

10 25 2 13 6 21

26 17 12 19 8 15 4

24 9 16 5 22 7

23

1 12 19 8 3

18 7 2 13 20

17 6 11 22 15 4 9

16 5 10 21 14

1 14 5 10

18 13 8 11 2 15

17 4 9 6

12 7 16 3

Figure 8 120591(4 4) = 120591(4 5) = 120591(4 6) = 2









a Ae

aA e

a A e

d D b d D b d D b

F fB

Ff B

F f B

C c E C c E C c E


12

1 26 3 20 13 22 9 16 11

4 19 28 25 6 17 14 23 8

27 2 5 18 21 24 7 10 15

24

1 32 17 8 25 30 15

18 23 36 31 16 7 26

33 2 9 22 27 14 29

10 19 4 35 12 21 6

3 34 11 20 5 28 13

26

1 18 7 12 25

8 13 24 17 22

19 2 21 6 11

14 9 4 23 16

3 20 15 10 5

16

1 20 17 14 11 8 5

18 13 22 3 6 15 10

21 2 19 12 9 4 7







3 28 1 18 9 22 7

14 17 4 25 6 19 10

27 2 15 12 21 8 23

16 13 26 5 24 11 20

12 17 8 3

7 2 13 18

16 11 4 9

1 6 19 14

20 15 10 5

A e E c

H b B f

h D d F

a G g C

36 1 32 15 20 3 24 7 26

31 14 19 2 33 10 27 4 23

18 35 12 29 16 21 6 25 8

13 30 17 34 11 28 9 22 5




Subcase 1 119898 = 1 mod 4 119899 = 1 mod 4

[[[[[[[[[[[[[

[



4[5 times 4]

darr2



6[4 times 4]

darr3

darr3



6[4 times 4]

]]]]]]]]]]]]]

]

Subcase 2 119898 = 1 mod 4 119899 = 3 mod 4

[[[[[[[[[[[

[



4[5 times 4]

darr8



6[4 times 4]

darr11

darr11



6[4 times 4]

]]]]]]]]]]]

]


Subcase 3 119898 = 3 mod 4 119899 = 1 mod 4

[[[[[[[[[[[

[



13[3 times 4]

darr14



6[4 times 4]

darr17

darr17



6[4 times 4]

]]]]]]]]]]]

]

Subcase 4 119898 = 3 mod 4 119899 = 3 mod 4

[[[[[[[[[[[

[



13[3 times 4]

darr16



6[4 times 4]

darr11

darr11



6[4 times 4]

]]]]]]]]]]]

]

(1)








(2)

3 Conclusion




Acknowledgment


References




















Volume 2014




Journal of











Journal of


Function Spaces






Algebra










a Ae

aA e

a A e

d D b d D b d D b

F fB

Ff B

F f B

C c E C c E C c E


12

1 26 3 20 13 22 9 16 11

4 19 28 25 6 17 14 23 8

27 2 5 18 21 24 7 10 15

24

1 32 17 8 25 30 15

18 23 36 31 16 7 26

33 2 9 22 27 14 29

10 19 4 35 12 21 6

3 34 11 20 5 28 13

26

1 18 7 12 25

8 13 24 17 22

19 2 21 6 11

14 9 4 23 16

3 20 15 10 5

16

1 20 17 14 11 8 5

18 13 22 3 6 15 10

21 2 19 12 9 4 7







3 28 1 18 9 22 7

14 17 4 25 6 19 10

27 2 15 12 21 8 23

16 13 26 5 24 11 20

12 17 8 3

7 2 13 18

16 11 4 9

1 6 19 14

20 15 10 5

A e E c

H b B f

h D d F

a G g C

36 1 32 15 20 3 24 7 26

31 14 19 2 33 10 27 4 23

18 35 12 29 16 21 6 25 8

13 30 17 34 11 28 9 22 5




Subcase 1 119898 = 1 mod 4 119899 = 1 mod 4

[[[[[[[[[[[[[

[



4[5 times 4]

darr2



6[4 times 4]

darr3

darr3



6[4 times 4]

]]]]]]]]]]]]]

]

Subcase 2 119898 = 1 mod 4 119899 = 3 mod 4

[[[[[[[[[[[

[



4[5 times 4]

darr8



6[4 times 4]

darr11

darr11



6[4 times 4]

]]]]]]]]]]]

]


Subcase 3 119898 = 3 mod 4 119899 = 1 mod 4

[[[[[[[[[[[

[



13[3 times 4]

darr14



6[4 times 4]

darr17

darr17



6[4 times 4]

]]]]]]]]]]]

]

Subcase 4 119898 = 3 mod 4 119899 = 3 mod 4

[[[[[[[[[[[

[



13[3 times 4]

darr16



6[4 times 4]

darr11

darr11



6[4 times 4]

]]]]]]]]]]]

]

(1)








(2)

3 Conclusion




Acknowledgment


References




















Volume 2014




Journal of











Journal of


Function Spaces






Algebra










3 28 1 18 9 22 7

14 17 4 25 6 19 10

27 2 15 12 21 8 23

16 13 26 5 24 11 20

12 17 8 3

7 2 13 18

16 11 4 9

1 6 19 14

20 15 10 5

A e E c

H b B f

h D d F

a G g C

36 1 32 15 20 3 24 7 26

31 14 19 2 33 10 27 4 23

18 35 12 29 16 21 6 25 8

13 30 17 34 11 28 9 22 5




Subcase 1 119898 = 1 mod 4 119899 = 1 mod 4

[[[[[[[[[[[[[

[



4[5 times 4]

darr2



6[4 times 4]

darr3

darr3



6[4 times 4]

]]]]]]]]]]]]]

]

Subcase 2 119898 = 1 mod 4 119899 = 3 mod 4

[[[[[[[[[[[

[



4[5 times 4]

darr8



6[4 times 4]

darr11

darr11



6[4 times 4]

]]]]]]]]]]]

]


Subcase 3 119898 = 3 mod 4 119899 = 1 mod 4

[[[[[[[[[[[

[



13[3 times 4]

darr14



6[4 times 4]

darr17

darr17



6[4 times 4]

]]]]]]]]]]]

]

Subcase 4 119898 = 3 mod 4 119899 = 3 mod 4

[[[[[[[[[[[

[



13[3 times 4]

darr16



6[4 times 4]

darr11

darr11



6[4 times 4]

]]]]]]]]]]]

]

(1)








(2)

3 Conclusion




Acknowledgment


References




















Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Subcase 3 119898 = 3 mod 4 119899 = 1 mod 4

[[[[[[[[[[[

[



13[3 times 4]

darr14



6[4 times 4]

darr17

darr17



6[4 times 4]

]]]]]]]]]]]

]

Subcase 4 119898 = 3 mod 4 119899 = 3 mod 4

[[[[[[[[[[[

[



13[3 times 4]

darr16



6[4 times 4]

darr11

darr11



6[4 times 4]

]]]]]]]]]]]

]

(1)








(2)

3 Conclusion




Acknowledgment


References




















Volume 2014




Journal of











Journal of


Function Spaces






Algebra























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









Documents

Research Article Knight s Tours on Rectangular Chessboards