A closed knight\u27s tour of a chessboard uses legal moves of the knight to visit every square exactly once and return to its starting position. In 1991 Schwenk completely classified the rectangular chessboards that admit a closed knight\u27s tour. For a rectangular chessboard that does not contain a closed knight\u27s tour, this paper determines the minimum number of squares that must be removed in order to admit a closed knight\u27s tour. Furthermore, constructions that generate a closed tour once appropriate squares are removed are provided

DeMaio, Joseph

Hippchen, Thomas

English

DigitalCommons@Kennesaw State University

Kennesaw State UniversityDigitalCommons@Kennesaw State UniversityFaculty Publications6-2009Closed Knight's Tours with Minimal SquareRemoval for All Rectangular BoardsJoseph DeMaioKennesaw State University, jdemaio@kennesaw.eduThomas HippchenNorth Metro Technical CollegeFollow this and additional works at: https://digitalcommons.kennesaw.edu/facpubsPart of the Applied Mathematics Commons, and the Mathematics CommonsThis Article is brought to you for free and open access by DigitalCommons@Kennesaw State University. It has been accepted for inclusion in FacultyPublications by an authorized administrator of DigitalCommons@Kennesaw State University. For more information, please contactdigitalcommons@kennesaw.edu.Recommended CitationDeMaio, J., & Hippchen, T. (2009). Closed knight's tours with minimal square removal for all rectangular boards. MathematicsMagazine, 82(3), 219-225. doi:10.4169/193009809X468869Integre Technical Publishing Co., Inc. Mathematics Magazine 82:3 March 9, 2009 7:40 a.m. notes.tex page 219VOL. 82, NO. 3, JUNE 2009 219with β0 < β1 < · · · < βn and bn = 0, has more than n roots. This contradiction to theinduction hypothesis concludes the proof.The Exponential Theorem generalizes the fundamental theorem of algebra to expo-nential functions the way the Generalized Polynomial Theorem does for generalizedpolynomials. A striking fact is that the two proofs follow the same path. Despite ap-pearances, the theorems are equivalent, as the following argument shows.Let f (t) = ∑nj=0 a jκ tj with 0 < κ0 < κ1 < · · · < κn , a j ∈ R, and an = 0. Let κ0 =ec0, κ1 = ec1, . . . , κn = ecn for some c0 < c1 < · · · < cn . By multiplying f (t) by tfor a suitable  > 1, we may assume that c0 > 0 to ascertain that 1 < c1/c0 < · · · <cn/c0. Thenf (t) = a0ec0t + a1ec1t + · · · + anecnt= a0ec0t + a1(ec0t)c1/c0 + · · · + an(ec0t)cn/c0= a0x + a1xc1/c0 + · · · + an xcn/c0 = g(x)with x = ec0t . By the Generalized Polynomial Theorem, with α0 = 1, α1 = c1/c0,. . . , αn = cn/c0, there exist at most k positive roots of the corresponding functiong(x). Certainly, when xi is such a root, ti := (ln xi )/c0 becomes a root of f (t), andvice-versa. This way, we have shown that the Generalized Polynomial Theorem im-plies the Exponential Theorem. The opposite implication comes from reversing theargument.REFERENCES1. V. N. Malozemov, Simultaneous Approximation of a Function and Its Derivatives (in Russian), Leningrad.Gos. Univ., Leningrad, 1973.2. Timo Tossavainen, The lost cousin of the Fundamental Theorem of Algebra, this MAGAZINE 80 (2007) 290–294.Closed Knight’s Tours with Minimal SquareRemoval for All Rectangular BoardsJOE DEMAIOKennesaw State UniversityKennesaw, GA 30144jdemaio@kennesaw.eduTHOMAS HIPPCHENKennesaw State UniversityKennesaw, GA 30144thippchen@gmail.comFinding a closed knight’s tour of a chessboard is a classic problem: Can a knight uselegal moves to visit every square on the board and return to its starting position? [1, 3]An open knight’s tour is a knight’s tour of every square that does not return to itsstarting position. While originally studied for the standard 8 × 8 board, the problem iseasily generalized to other rectangular boards. In 1991 Schwenk classified all rectan-gular boards that admit a closed knight’s tour [2]. He described every board that cannotadmit a closed knight’s tour and constructed closed knight’s tours for all other boards.Integre Technical Publishing Co., Inc. Mathematics Magazine 82:3 March 9, 2009 7:40 a.m. notes.tex page 220220 MATHEMATICS MAGAZINESCHWENK’S THEOREM. An m × n chessboard with m ≤ n has a closed knight’stour unless one or more of the following three conditions hold:(a) m and n are both odd;(b) m ∈ {1, 2, 4};(c) m = 3 and n ∈ {4, 6, 8}.How close to admitting a closed knight’s tour are those boards that satisfySchwenk’s conditions (a), (b), or (c)? Let us call these obstructed boards, sincethey admit no closed knight’s tour. The 3 × 3 board is obstructed; however, once thecenter square is removed a closed knight’s tour does indeed exist as seen in FIGURE 1.17435862Figure 1 A closed knight’s tour of the 3× 3 board with the center square removedLet the tour number, T (m, n), with m ≤ n be the minimum number of squareswhose removal from an m × n chessboard will allow a closed knight’s tour. Thus,T (3, 3) = 1. Note that unless m and n are the dimensions of an obstructed board,T (m, n) = 0; no squares need to be removed. Also note that removing T (m, n)squares randomly from an obstructed board does not guarantee the existence of aclosed knight’s tour. For instance, removing any square other than the center does notallow for a closed knight’s tour of the resulting 3 × 3 board. Furthermore T (1, n) andT (2, 2) are undefined since the knight cannot move from its starting position. AlsoT (2, n) = 2n − 2 for n ≥ 3 since a knight can move down a 2 × n board but cannotreturn to its starting position unless only one move has been made.Parity considerations restrict the number of squares we can remove from an ob-structed board if we hope to get a closed knight’s tour. Throughout this paper, when-ever we color the squares of a chessboard black and white, we will always begin witha black square in the upper left-hand corner. A legal move for a knight whose initialposition is a white square will always result in an ending position on a black squareand vice versa. Hence, any closed knight’s tour must visit an equal number of blacksquares and white squares. This quickly determines that an odd number of squaresmust be removed from a board where both m and n are odd and an even number ofsquares must be removed from all other boards.Thus, for an obstructed board, the smallest possible tour numbers are 1 and 2 re-spectively for boards with an odd or even number of squares. Recursive constructionsof closed knight’s tours will show that tour numbers are actually as small as possible,except in a few special cases. The constructions start with small boards, called baseboards, and build by tacking on boards with open tours.We compute all nonzero tour numbers by considering the three cases fromSchwenk’s Theorem: m = 3, m = 4, and the case where m and n are both odd.The case of m = 3 For odd n, two base boards are needed for n ≡ 1, 3 mod 4and one exceptional case exists for n = 5. For even n, three boards are examined forn = 4, 6, and 8.To construct a closed knight’s tour for the 3 × 7 board, start with the open 3 × 4 tourfrom FIGURE 2, which begins at a and ends at l. Next, take the tour for the 3 × 3 boardIntegre Technical Publishing Co., Inc. Mathematics Magazine 82:3 March 9, 2009 7:40 a.m. notes.tex page 221VOL. 82, NO. 3, JUNE 2009 221acljhegkbdfiFigure 2 An open knight’s tour of the 3 × 4 boardwith the center square removed as in FIGURE 1 and delete the 5–6 move. Connect theboards by creating the 5–a and 6–l moves (these correspond to legal knight moves) asshown in FIGURE 3. This is typical of our approach throughout the paper.17435862acljhegkbdfiFigure 3 A closed knight’s tour of the 3× 7 board after minimal square removalThis game can be played an infinite number of times replacing the role of squares5 and 6 by g and h. Thus, T (3, n) = 1 for all n ≡ 3 mod 4. Note that the lower righthand corner of any board, when used, must contain the g–h move as there are only twolegal moves for a knight from that corner square.For the 3 × 5 board note that the corner squares have only two legal moves, whereone move is the center square. At most two of these four squares may be includedin a tour. Hence at least two of these squares must be removed. Furthermore, all fourof those corner squares are black and it will also be necessary to remove at least onewhite square. Thus, T (3, 5) ≥ 3. The existence of the tour in FIGURE 4 shows thatT (3, 5) = 3.111610837512429Figure 4 A closed knight’s tour of the 3× 5 board after minimal square removalBut the 3 × 5 board is the lone exception for all boards with an odd number ofsquares. FIGURE 5 shows that T (3, 9) = 1. As before, the open tour of FIGURE 2yields T (3, n) = 1 for all n ≡ 1 mod 4 where n = 5.1513101361820257524261921812162391114242217Figure 5 A closed knight’s tour of the 3× 9 board after minimal square removalIntegre Technical Publishing Co., Inc. Mathematics Magazine 82:3 March 9, 2009 7:40 a.m. notes.tex page 222222 MATHEMATICS MAGAZINEThe case of the 3 × 4 board is a very straightforward one. As shown in FIGURE 6,if two squares are removed, a knight’s tour exists. Thus, T (3, 4) ≤ 2. Since an evennumber of squares is required, T (3, 4) = 2. Furthermore, tacking on the open tour ofFIGURE 2 shows that T (3, 8) = 2.13689524107Figure 6 A closed knight’s tour of the 3× 4 board after minimal square removalThe 3 × 6 board is the only other obstructed one with m = 3. We will analyzeFIGURE 7 to show that T (3, 6) = 4.132798465101211161817131514Figure 7 A way to label the 3× 6 boardUsing all four corners immediately forces the paths 6–1–8–3–4 and 13–18–11–16–15. Since squares 8 and 11 can have no further connections, these paths are necessarilyextended to 7–6–1–8–3–4–9 and 12–13–18–11–16–15–10. However none of the fourremaining squares (2, 5, 14, and 17) can be included without closing one of these pathsbefore connecting to the other. No tour exists using all four corners that omits exactlytwo squares. Furthermore, including the 7–12 and 9–10 moves creates a tour whenomitting 4 squares and T (3, 6) ≤ 4.Next note that squares 5 and 17 cannot both be used without creating a closed cyclewith 10 and 12. Similarly, squares 2 and 14 cannot both be used without creating aclosed cycle with 7 and 9. When combined with the previous fact that no tour existsusing all four corners that omits exactly two squares we achieve T (3, 6) ≥ 3. SinceT (3, 6) is even, T (3, 6) = 4.The case of m = 4 Any board with m = 4 will have an even number of squares; thus,T (4, n) will always be even. The boards of FIGURE 8 show that T (4, 4) = T (4, 5) =T (4, 6) = 2.8114131471012925613121492751811161383417161510 21191638611221320721121051417151894Figure 8 Closed knight’s tours of the 4× 4, 4× 5, and 4× 6 boards after minimal squareremovalIntegre Technical Publishing Co., Inc. Mathematics Magazine 82:3 March 9, 2009 7:40 a.m. notes.tex page 223VOL. 82, NO. 3, JUNE 2009 223Next consider any tour on a 4 × k board that contains the a–b and c–d moves inthe lower right-hand corner as in the left-hand side of FIGURE 9. This 4 × k tour canbe extended to a 4 × (k + 3) tour by removing moves a–b and c–d and connecting toa 4 × 3 extension with the moves 1–c, 6–d, a–a1, and b–a6. Note that all three baseboards and the 4 × 3 extension contain the a–b and c–d moves in the lower right-handcorner as in FIGURE 9. This proves T (4, n) = 2 for all n ≥ 4.bcdabcdaa152a2a5a3a64136a4Figure 9 A closed knight’s tour of the 4 × (n + 3) board from a closed knight’s tour ofthe 4× n boardThe case of both m and n odd Much like the m = 3 and m = 4 cases, we useinduction with an appropriate base case to analyze all boards with an odd number ofsquares for m ≥ 5. In all cases, T (m, n) = 1 for both m and n odd with 5 ≤ m ≤ n.Four base cases exist, one for each combination of m, n ≡ 1, 3 mod 4. The boards ofFIGURE 10 are used for m, n ≡ 1 mod 4 and m ≡ 1, n ≡ 3 mod 4 respectively. Form, n ≡ 3 mod 4 use the 3 × 7 board of FIGURE 3 and for m ≡ 3, n ≡ 1 mod 4 usethe 3 × 9 board of FIGURE 5. The open 3 × 4 tour of FIGURE 2 and the open 5 × 4tour of FIGURE 11 can be used to extend the base boards to any length n ≡ 1, 3 mod 4as demonstrated in FIGURE 3. For the 5 × 5 board delete the 10–11 move and createthe 1–10 and 11–20 moves. For the 5 × 7 board delete the 26–27 move and create the1–16 and 20–27 moves.15161122102361712154212472092131831419815122516232033222981143417321929303103118136272415262114728Figure 10 Base boards for m ≡ 1 mod 4We have constructed tours for all 3 × n and 5 × n boards for odd n ≥ 7. Next weneed to extend these boards down to an arbitrary odd m. To do so, rotate clockwise theopen tours of FIGURE 2 and FIGURE 11 to a 4 × 3 tour and a 4 × 5 tour and extend thebase 3 × n and 5 × n boards down to any depth m ≡ 1, 3 mod 4. For m, n ≡ 1 mod 4use the 5 × n board (created with FIGURE 10) delete the 14–15 move and create the1–14 and 15–20 moves with FIGURE 11 rotated clockwise. For m ≡ 1, n ≡ 3 mod 4use the 5 × n board (created with FIGURE 10), delete the 3–4 move and create the3–a and 4–l moves with FIGURE 2 rotated clockwise. For m ≡ 3, n ≡ 1 mod 4 usethe 3 × n board (created with FIGURE 5), delete 5–6 move and create the 1–6 and5–20 moves with FIGURE 11 rotated clockwise. For m, n ≡ 3 mod 4 use 3 × n boardIntegre Technical Publishing Co., Inc. Mathematics Magazine 82:3 March 9, 2009 7:40 a.m. notes.tex page 224224 MATHEMATICS MAGAZINE1251491831019413815611217201167Figure 11 An open tour of the 5× 4 board(created with FIGURE 3), delete the 7–8 move and create the 7–a and 8–l moves withFIGURE 2 rotated clockwise. This process provides us with a closed knight’s tour ofthe top and left side of the board in FIGURE 12.(3, 5) x nj x k. . .4 x (3, 5)Figure 12 Constructing a closed knight’s tour of the m × n board after minimal squareremoval for m,n ≡ 1 mod 2Now a j × k gap with j, k ≡ 0 mod 4 needs to be filled in to complete the m × nboard. Finally, we use the 4 × 4 board of FIGURE 13 to fill in the j × k gap using thesame technique of FIGURE 9.bcdabcda136a6a3a7a42574a2a5a1a88Figure 13 Filling in a j × k gap for j, k ≡ 0 mod 4Integre Technical Publishing Co., Inc. Mathematics Magazine 82:3 March 9, 2009 7:40 a.m. notes.tex page 225VOL. 82, NO. 3, JUNE 2009 225Conclusion In summary, the tour number for obstructed boards is as small as pos-sible (1 or 2) based on an odd or even number of squares with the few noted exceptionsas indicated below.For the m × n chessboard with m ≤ n, either board has a closed knight’s tour, sothat T (m, n) = 0, or else(a) T (m, n) = 1, where m and n are both odd except for m = 3 and n = 5;(b) T (4, n) = 2 for all n ≥ 4;(c) T (3, 4) = T (3, 8) = 2, T (3, 5) = 3, T (3, 6) = 4;(d) T (2, n) = 2n − 2 for n ≥ 3;(e) T (1, n) and T (2, 2) are undefined.Acknowledgment. We thank the anonymous referee whose suggestions significantly improved the clarity andquality of this article.REFERENCES1. George Jelliss, Knight’s Tour Notes, http://www.ktn.freeuk.com/sitemap.htm.2. A. J. Schwenk, Which rectangular chessboards have a knight’s tour? this MAGAZINE 64 (1991) 325–332.3. John J. Watkins, Across the Board: The Mathematics of Chessboard Problems, Princeton University Press,Princeton, NJ, 2004.

Closed Knight\u27s Tours with Minimal Square Removal for All Rectangular Boards

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Closed Knight\u27s Tours with Minimal Square Removal for All Rectangular Boards

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