The classic n-queens problem asks for placements of just n mutually non-attacking queens on an n × n board. By adding enough pawns, we can arrange to fill roughly one-quarter of the board with mutually non-attacking queens. How many pawns do we need? We discuss that question for square boards as well as rectangular m × n boards

Chatham, Doug

English

Morehead State University

Games and Puzzlesthe maximum queens problemwith pawnsDoug ChathamMorehead State University, Department of Mathematics and Physicsd.chatham@moreheadstate.eduAbstract: The classic n-queens problem asks for placements of just n mutuallynon-attacking queens on an n × n board. By adding enough pawns, we canarrange to fill roughly one-quarter of the board with mutually non-attackingqueens. How many pawns do we need? We discuss that question for squareboards as well as rectangular m× n boards.Keywords: chess, n-queens problem, combinatorics.The n-queens problemTake a standard chessboard and as many queens as possible. How many queenscan you put on the chessboard, with at most one queen in each square, so that noqueen “attacks” (i.e. is a vertical, horizontal, or diagonal move away from) anyother queen? Figure 1 shows a placement of eight mutuallynon-attacking queens. Finding such arrangements is the classic“8-queens problem”, first proposed by M. Bezzel in 1848. In 1869 this problemwas generalized to the “n-queens problem” of placing n mutually non-attackingqueens on an n × n board. The n-queens problem has solutions for N = 1and N > 4. Hundreds of papers have been written on this problem and itsvariations, including versions on other board shapes (including cylinders, toruses,and three-dimensional boards) and other piece types (including bishops, rooks,and fairy pieces) . We refer interested readers to the book Across the Board:The Mathematics of Chessboard Problems by J. J. Watkins [5], the n-queenssurvey article by J. Bell and B. Stevens [1], and the online n-queensbibliography maintained by W. Kosters [4].Recreational Mathematics Magazine, Number 6, pp. 95–102DOI 10.1515/rmm–2016–0010Bereitgestellt von  Morehead State University | Heruntergeladen  05.11.19 15:20   UTC96 the maximum queens problem with pawns8 0Z0Z0l0Z7 Z0ZqZ0Z06 0Z0Z0ZqZ5 l0Z0Z0Z04 0Z0Z0Z0l3 ZqZ0Z0Z02 0Z0ZqZ0Z1 Z0l0Z0Z0a b c d e f g hFigure 1: Eight mutually non-attacking queens on a standard chessboard.Peppering the problem with pawnsCan we do better? Can we put more than eight queens on a standardchessboard or more than n queens on an n × n chessboard? If we follow therules of the classic problem, the answer is clearly “no”: each queen attacks everyother square on its row (or column), so there is at most one queen per row (orcolumn) and thus at most n queens on an n× n board.We now change the rules in order to allow more queens on the board. Recallthat in chess, queens do not move through other pieces. If we put a pawn in asquare between two queens that are in the same row, column, or diagonal, thosequeens no longer attack each other. If we allow the placement of some pawns,how many mutually non-attacking queens can we place on the board? This isthe “maximum queens problem”, posed in 1998 by K. Zhao in her disseration [6].For n > 6, we can put n + 1 queens on the board with 1 pawn [3, Theorem 1].Figure 2 shows an example arrangement. Zhao proved that we need 3 pawns toput 6 queens on a 5× 5 board [6]. For each k > 0, we can place n+ k mutuallynon-attacking queens on an n× n board with k pawns, if n is sufficiently large[2, Theorem 11].Recreational Mathematics Magazine, Number 6, pp. 95–102DOI 10.1515/rmm–2016–0010Bereitgestellt von  Morehead State University | Heruntergeladen  05.11.19 15:20   UTCDoug Chatham 978 0Z0l0Z0Z7 l0ZpZ0Zq6 0Z0ZqZ0Z5 ZqZ0Z0Z04 0Z0Z0l0Z3 Z0l0Z0Z02 0Z0Z0ZqZ1 Z0ZqZ0Z0a b c d e f g hFigure 2: Nine mutually non-attacking queens with one pawn on a standardchessboard.How much further can we go? If we don’t care how many pawns we place onthe board, we can place n24 queens if n is even and(n+1)24 queens if n is odd.To see this, take an n × n board and divide it into two-column strips (witha one-column strip at the end if n is odd) and divide each strip into two-rowblock (with one-row blocks at the bottom if n is odd) as shown in Figure 3. Nowconsider the blocks. If we put two queens in any of these blocks, they will bein adjacent squares and will attack each other, regardless of how many pawnsare on the board. So, the maximm number of queens we can put on the boardequals the number of blocks, which is n24 queens if n is even and(n+1)24 queensif n is odd. To see that we can place that many queens, on the first, third, etc.rows place a queen in the first, third, etc. squares and then place pawns in everyempty square, as indicated in Figure 3.lplplpopoplplplpopoplplplFigure 3: A 5 × 5 chessboard divided into blocks, each of which can hold atmost one queen if no queens can attack other queens.Recreational Mathematics Magazine, Number 6, pp. 95–102DOI 10.1515/rmm–2016–0010Bereitgestellt von  Morehead State University | Heruntergeladen  05.11.19 15:20   UTC98 the maximum queens problem with pawnsCan we place the maximum number of queens with fewer pawns? If n is odd, theanswer is “no”. Consider Figure 3 again. When n is odd, thepartition produces a 1 × 1 block in the last row and column. To get themaximum number of queens, each block must have a queen, so the 1× 1 cornermust have a queen. This forces the placement of all the other queens. Thereis only one way to place the queens, and the squares without queens are eachbetween two queens and therefore require pawns. We now consider the casewhere n is even.Proposition 1. For n = 4k + 2 with k > 1, it is possible to place n24 queensand n24 − 3 pawns on an n× n board so that no queens attack each other.Proof Sketch: We present a pattern with n24 queens andn24 pawns. Label therows and columns 0, 1 . . . n− 1 as shown in Figure 4. In rows with labels of theform 4k + 1 (i.e. rows 1, 5, 9,etc.) in the squares whose row number exceedstheir column number, place queens in even-numbered columns and pawns inodd-numbered columns. In rows with labels of the form 4k + 3 (i.e. rows 3, 7,etc.) , in the squares whose row number exceeds their column number, placepawns in even-numbered columns and queens in odd numbered columns. Toobtain the rest of the pattern, reflect across the main diagonal and change thepiece type, so for each position (x, y), the piece in (y, x) is a queen if and onlyif the piece in (x, y) is a pawn.9 qoqoqoqoqZ8 Z0Z0Z0Z0ZP7 plplplpZ0L6 Z0Z0Z0ZQZP5 qoqoqZ0O0L4 Z0Z0ZPZQZP3 plpZ0L0O0L2 Z0ZQZPZQZP1 qZ0O0L0O0L0 ZPZQZPZQZP0 1 2 3 4 5 6 7 8 9Figure 4: “Corner” pattern.Recreational Mathematics Magazine, Number 6, pp. 95–102DOI 10.1515/rmm–2016–0010Bereitgestellt von  Morehead State University | Heruntergeladen  05.11.19 15:20   UTCDoug Chatham 99We can show that the board has n24 queens andn24 pawns and that none of thequeens attack each other. Finally, we note that we can remove the pawns inpositions (0, 1), (0, n− 1), and (n− 2, n− 1) and the queens will still not attackeach other. When n is a multiple of 4, we can do slightly better.Proposition 2. For n = 4k with k > 1, it is possible to place n24 queens andn24 − 4 pawns on an n× n board so that no queens attack each other.Proof Sketch: Again we present a pattern (as illustrated in Figure 5) thatproduces the desired results. Given a 4k × 4k board, label the rows andcolumns −2k . . . 2k−1. In the upper-right quadrant (i.e. the positions with bothcoordinates nonnegative), place queens and pawns in the “corner” pattern of theprevious Proposition, except do not remove any pawns like we didpreviously. Rotate the board a quarter-turn, relabel rows and columns, andplace queens and pawns in the new upper-right quadrant in the corner pattern.Repeat the previous sentence two more times.We now have a “dartboard” pattern, symmetric with respect to quarter-turnrotation, with n24 queens andn24 pawns. We can check that none of the queensattack any of the other queens. We conclude by noting that four pawns inthe outer ring formed by the first and last rows and the first and last columns(circled in Figure 5) can be removed. 0lplplpZo0Z0Z0ZqqZ0oqZ0oo0l0ZpZqqZpZ0l0oo0Zqo0ZqqZ0Z0Z0oZplplpl0Figure 5: “Dartboard” patternCan we do better? Are there patterns on even-order boards with the maximumnumber of queens but fewer pawns? Computer experiments for small boardsizes indicate that the answer is “no”, but we don’t have a general proof.Recreational Mathematics Magazine, Number 6, pp. 95–102DOI 10.1515/rmm–2016–0010Bereitgestellt von  Morehead State University | Heruntergeladen  05.11.19 15:20   UTC100 the maximum queens problem with pawnsStretching to RectanglesWe can extend the results to rectangular boards of size m × n where m is notnecessarily equal to n. Dividing the board into blocks as we did for squareboards, we get that the maximum number of non-attacking queens we canplace on such boards is dm2 edn2 e, where dxe is the smallest integer greaterthan or equal to x. For example, on a 4 × 7 board we can place at mostd 42ed72e = d2ed3.5e = (2)(4) = 8 queens, as illustrated in Figure 6. If both mand n are odd, there is only one way to place the queens and every other squarerequires a pawn.So we make at least one of the dimensions even.Proposition 3. If m or n is even, we can place dm2 edn2 e queens and dm2 edn2 e−2pawns on an m× n board so that no two queens attack each other.Proof sketch: Suppose without loss of generality that m is even. Label therows 0, 1, . . . ,m − 1 and the columns 0, 1, . . . , n − 1. In columns with labelsof the form 4k, place queens in the even-numbered rows and pawns in theodd-numbered rows. In columns with labels of the form4k + 2, place pawnsin the even-numbered rows and queens in the odd-numbered rows. We get astriped pattern as illustrated in Figure 6. We can check that none of the queensattack each other and that there are the right number of queens. We concludeby eliminating a pawn in the first and last non-empty columns, as indicated inFigure 6.3 pZqZpZq2 l0o0l0o1 pZqZpZq0 l0o0l0o0 1 2 3 4 5 6Figure 6: “Stripe” patternIf m or n is 2, that is the best possible result: A 2-row board without pawnscan hold at most 2 queens, one per row. Each pawn increases the number ofavailable “rows” by at most 1. So the total number of queens on such a boardis at most 2 more than the number of pawns. The number of pawns needed toget dm2 edn2 e queens on a 2-row (or 2-column) board is dm2 edn2 e − 2.If both m and n are even and larger than 2, we can do better, using portions ofthe “corner” pattern.Recreational Mathematics Magazine, Number 6, pp. 95–102DOI 10.1515/rmm–2016–0010Bereitgestellt von  Morehead State University | Heruntergeladen  05.11.19 15:20   UTCDoug Chatham 101Proposition 4. If m > 2 and n > 2 are even, then we can place dm2 edn2 e queensand dm2 edn2 e − 3 pawns on an m × n board so that no two queens attack eachother.Proof Sketch: Suppose without loss of generality that m < n. Take an n × nboard and place queens and pawns according to the procedure for thecorner pattern, without the final pawn removal. If m is a multiple of 4, removerows m, . . . , n − 1, and note that the pawns at positions (0, 1), (m − 1, 0) andeither (0, n − 1) or (m − 1, n − 1) can be removed. If m is not a multiple of 4,remove columns m, . . . , n − 1, note that pawns at (0, 1), (0,m − 1), and either(n−1,m−1) or (n−1, 0) can be removed, and then transpose rows and columns.Can we do better? The results of computer experimentation with low values ofm and n indicate that the answer is “no”, but, again, we have no general proof.Other DirectionsThis paper leaves many questions open for future research. Here are a few:1. A common thing to do with the classic n-queens problem is to count thenumber of solutions for particular values of n. Except when m and nare both odd, we haven’t done that with the maximum queens problem.Given the maximum number of queens and the minimumnecessary number of pawns, how many ways can those pieces be arrangedon an m× n board (with m or n even) so that none of the queens attackeach other?2. Clearly, with enough pawns, we can place q mutually non-attacking queensfor 0 6 q 6 dm2 edn2 e. How many pawns are needed?3. What results can we get on other types of board, such as cylindricalboards, Mobius strips, and toruses? Observe that themaximum number of queens is the same as the maximum number of kingsyou can put on the board so that no kings are on adjacent squares: Ifno two pieces are on adjacent squares, we can separate them with pawns.The maximum number of nonadjacent kings is referred to as the “kingsindependence number” and it is known for many board types [5]. Forexample, the kings independence number on an m × n torus ismin{bmbn2 c2 c, bnbm2 c2 c} [5, Theorem 11.1] (where bxc is the largestinteger less than or equal to x) and that number is also the maximumnumber of non-attacking queens we can put on that board withsufficiently many pawns. So, the interesting question is “How many pawnsdo we need?”Recreational Mathematics Magazine, Number 6, pp. 95–102DOI 10.1515/rmm–2016–0010Bereitgestellt von  Morehead State University | Heruntergeladen  05.11.19 15:20   UTC102 the maximum queens problem with pawnsReferences[1] Bell, J., Stevens, B. “A survey of known results and research areas forn-queens”, Discrete Math, 309, 1–31, 2009.[2] Chatham, R. D., Doyle, M., Fricke, G. H., Reitmann, J., Skaggs, R.D., Wolff, M. “Independence and domination separation on chessboardgraphs”, J. Combin. Math. Combin. Comput, 68, 3–17, 2009.[3] Chatham, R. D., Fricke, G. H., Skaggs, R. D. “The queens separationproblem”, Util. Math, 69, 129–141, 2006.[4] Kosters, W. A. n-Queens bibliography, 2016.http://www.liacs.nl/home/kosters/nqueens/[5] Watkins, J. J. Across the Board: The Mathematics of Chessboard Problems,Princeton University Press, 2004.[6] Zhao, K. The Combinatorics of Chessboards, Ph.D. dissertation, CityUniversity of New York, 1998.Recreational Mathematics Magazine, Number 6, pp. 95–102DOI 10.1515/rmm–2016–0010Bereitgestellt von  Morehead State University | Heruntergeladen  05.11.19 15:20   UTC

The maximum queen’s problem with pawns

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The maximum queen’s problem with pawns

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