A Solution to the Checkerboard Problem

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1 A Solution to the Checkerboard Problem Futaba Okamoto Mathematics Department, University of Wisconsin La Crosse, La Crosse, WI 5460 Ebrahim Salehi Department of Mathematical Sciences, University of Nevada Las Vegas, Las Vegas, NV 8954 Ping Zhang Department of Mathematics, Western Michigan University, Kalamazoo, MI Abstract An m n checkerboard has m rows and n columns and its squares are alternately colored black and red. Two squares are said to be neighboring if they belong to the same row or to the same column and there is no square between them. A combinatorial problem called the Checkerboard Conjecture states that it is possible to place coins on some of the squares of an m n checkerboard (at most one coin per square) such that for every two squares of the same color the numbers of coins on neighboring squares are of the same parity, while for every two squares of different colors the numbers of coins on neighboring squares are of opposite parity. In this work, we show that the Checkerboard Conjecture is true for all m and n. AMS subject classification: 05B99. Keywords: Checkerboard conjecture, solution.. Introduction Suppose that the squares of an m n checkerboard (m rows and n columns), where mn, are alternately colored black and red. Figure (a) shows a 3 8 checkerboard where a shaded square represents a black square. Two squares are said to be neighboring if they belong to the same row or to the same column and there is no square between

2 4 Futaba Okamoto, Ebrahim Salehi, and Ping Zhang (a) (b) Figure : A 3 8 checkerboard and a coin placement on the checkerboard them. Thus every two neighboring squares are of different colors. A combinatorial problem was introduced in and the following conjecture was stated. The Checkerboard Conjecture. It is possible to place coins on some of the squares of an m n checkerboard (at most one coin per square) such that for every two squares of the same color the numbers of coins on neighboring squares are of the same parity, while for every two squares of different colors the numbers of coins on neighboring squares are of opposite parity. Figure (b) shows a placement of 6 coins on a 3 8 checkerboard such that the number of coins on neighboring squares of every black square is even and the number of coins on neighboring squares of every red square is odd. Thus for every two squares of different colors, the numbers of coins on neighboring squares are of opposite parity. Consequently, the Checkerboard Conjecture is true for a 3 8 checkerboard. Observe that all 6 coins on the 3 8 checkerboard of Figure (b) are placed only on black squares. Thus the number of coins on neighboring squares of every black square is 0, while the number of coins on neighboring squares of every red square is either or 3 as indicated in Figure (b). Indeed, for any m n checkerboard for which the Checkerboard Conjecture is true, there is always a solution in which all coins are placed only on squares of the same color. In this work, we show that the Checkerboard Conjecture is true for a checkerboard of any size. The Checkerboard Theorem. For every pair m, n of positive integers, it is possible to place coins on some of the squares of an m n checkerboard (at most one coin per square) such that for every two squares of the same color the numbers of coins on neighboring squares are of the same parity, while for every two squares of different colors the numbers of coins on neighboring squares are of opposite parity.. A Proof of the Checkerboard Theorem In order to verify the Checkerboard Theorem, we first introduce some additional definitions and notation and present some preliminary results in Section.. We then study certain extensions of coin placements in Section. and present a proof of the Checkerboard Theorem in Section.3.

3 A Solution to the Checkerboard Problem 43.. Definitions, Notation, and Preliminary Results We always assume that m and n are positive integers with m n and n unless otherwise noted. For an m n checkerboard C, let S = B R be the set of mn squares in C, where B and R are the sets of black and red squares, respectively, and let s i,j S be the square in the ith row and jth column for i m and j n. We may also assume that B ={s i,j S : i + j 0 (mod )}. We express a coin placement for C using a coin placement function f : S {0, } defined by f(s) = if and only if there is a coin placed on the square s. The corresponding neighbor sum of a square s, denoted by σ f (s) (or simply σ(s)), is the number of coins placed on the neighboring square of s. For simplicity, we further assume that σ(s)is expressed as one of 0 and modulo. We say that a coin placement f for a checkerboard C is a solution if either σ(s) = if and only if s R or σ(s) = if and only if s B. Therefore, the Checkerboard Conjecture is true for an m n checkerboard C if C has a solution. Let S = S S S n, where S j ={s i,j : i m} for j n. Hence, S j is the set of the m squares in the jth column. Furthermore, let S j = S S S j for j n. (Hence S = S and S n = S.) Let f : S {0, } be an arbitrary coin placement for the squares in S such that either f (s) = 0 for every s S R or f (s) = 0 for every s S B, say the former. Observe then that there exists a unique coin placement f : S {0, } such that (i) f (s) = 0 for every s S R, (ii) f restricted to S equals f, and (iii) σ(s) = for every s S R. After finding such a coin placement f, observe further that there exists a unique coin placement f 3 : S 3 {0, } such that (i) f 3 (s) = 0 for every s S 3 R, (ii) f 3 restricted to S equals f, and (iii) σ(s) = for every s S R. In general, for every integer j( j n ) suppose that f j : S j {0, } is a coin placement such that f j (s) = 0 for every s S j R and σ(s) = for every s S j R (if j ). Then there exists a unique coin placement f j+ : S j+ {0, } such that (i) f j+ (s) = 0 for every s S j+ R, (ii) f j+ restricted to S j equals f j, and (iii) σ(s) = for every s S j R.

4 44 Futaba Okamoto, Ebrahim Salehi, and Ping Zhang S5 S0 S6 S Figure : Trivial extensions for m = 4, 5 This yields the following lemma. Lemma.. Consider an m n checkerboard. For each coin placement f : S {0, } such that f (s) = 0 for every s S R, there exists a unique coin placement F : S {0, } such that (i) F(s) = 0 for every s S R, (ii) F restricted to S equals f, and (iii) σ(s) = for every s S n R. (Similarly, for each coin placement f : S {0, } such that f (s) = 0 for every s S B, there exists a unique coin placement F : S {0, } such that (i) F(s) = 0 for every s S B, (ii) F restricted to S equals f, and (iii) σ(s) = for every s S n B.) For the coin placements f and F for a checkerboard C described in Lemma., we say that F is the extension of f. We also say that F is an extension for C... Properties of Extensions of Coin Placements Consider an m n checkerboard, where n is sufficiently large. Let f : S {0, } be an arbitrary coin placement such that f (s) = 0 for every s S R and obtain the unique extension F of f. We will next show the following: If j 0 (mod m + ), then F(s) = 0 for every square s S j. (.) Definition.. For the trivial coin placement f : S {0, } such that f (s) = 0 for every s S, its unique extension is called the trivial extension and denoted by F. See Figure shows the trivial extensions for m = 4, 5 and note that both extensions satisfy (.). In fact, the trivial extension satisfies (.) for every m, which we state without a proof as follows.

5 A Solution to the Checkerboard Problem 45 S7 S4 (a) (b) Figure 3: A coin placement f and its extension F fora6 5 checkerboard. Observation.3. If F is the trivial extension for an m n checkerboard, then F (s) = 0 for every s S j, where j 0 (mod m + ). Before proving (.) for an arbitrary extension, let us look at another example. Consider a 6 5 checkerboard with a coin placement f : S {0, } placing two coins on black squares as shown in Figure 3(a). Figure 3(b) shows the extension F of f. Observe for every square s except those in the last column that σ(s) = if and only if s R. Furthermore, F(s) = 0 for every s S 7 S 4. Proposition.4. Let C be an m n checkerboard with m n. For an arbitrary coin placement f : S {0, } with f (s) = 0 for every s S R, let F : S {0, } be its extension. Then F(s) = 0 for every s S j where j 0 (mod m + ). Proof. Let f : S {0, } be given by { a i/ if i is odd f (s i, ) = 0 ifi is even for i m. Furthermore, for each integer j with 0 j m/ let A j = j i= a i if j m/ 0 if j = 0. We consider two cases according to the parity of m. Case. m is even. Define a coin placement f : S {0, } such that f(s) = 0 for every s R as follows. Suppose that i( i m) and j( j n) are positive

6 46 Futaba Okamoto, Ebrahim Salehi, and Ping Zhang integers such that i + j is even. (Hence s i,j B.) Then A i+j + A i j + F (s i,j ) if j i and i + j m A (m+) i+j + A i j + F (s i,j ) if j i and i + j m + f(s j,i ) if i + j m f(s i,j ) = 0 if j = m + f(s (m+) i,j (m+) ) if m + j m + f(s i,j (m+) ) if j m + 3, where F is the trivial extension defined in Definition.. Note that f(s i,m+ ) = f(s (m+) i,m+ ) = 0 and so f(s i,j ) = 0 whenever j 0 (mod m + ). Also f(s i, ) = A i+ + A i + F (s i, ) = a i+ + 0 = c (s i, ) for i =, 3,...,m and so f restricted to S equals f. We now show that f is the extension of f. To do this, we need only verify that σ(s i,j ) = for every s i,j S n R. That is, we show that σ(s i,j) = for integers i and j with i m and j n such that i + j is odd. By symmetry, we may further suppose that either (i) j<i m or (ii) j = m +. Subcase.. j<i m. First suppose that j =. If i m, then σ(s i, ) = f(s i, ) + f(s i, ) + f(s i+, ) = A i+ + A i + F (s i, ) + A i + A i + F (s i, ) + A i+ + A i + F (s i+, ) = F (s i, ) + F (s i, ) + F (s i+, ) = =, while σ(s m, ) = f(s m, ) + f(s m, ) = A (m+) m+ + A m + F (s m, ) + A m + A m + F (s m, ) = F (s m, ) + F (s m, ) = + 0 =. Next suppose that i = m and 3 j m. Then σ(s m,j ) = f(s m,j ) + f(s m,j+ ) + f(s m,j ) = A (m+) m+j + A m j+ + F (s m,j ) + A (m+) m+j+ + A m j + F (s m,j+ ) + A (m+) m+j + A m j + F (s m,j ) = F (s m,j ) + F (s m,j+ ) + F (s m,j ) = + + =.

7 A Solution to the Checkerboard Problem 47 Hence, suppose next that j<i m. If i + j m, then σ(s i,j ) = f(s i,j ) + f(s i,j+ ) + f(s i,j ) + f(s i+,j ) = A i+j + A i j+ + F (s i,j ) + A i+j+ + A i j + F (s i,j+ ) + A i+j + A i j + F (s i,j ) + A i+j+ + A i j+ + F (s i+,j ) = F (s i,j ) + F (s i,j+ ) + F (s i,j ) + F (s i+,j ) =. For i + j = m +, σ(s i,j ) = f(s i,j ) + f(s i,j+ ) + f(s i,j ) + f(s i+,j ) = A m + A i j+ + F (s i,j ) + A (m+) m+ + A i j + F (s i,j+ ) + A m + A i j + F (s i,j ) + A (m+) m+ + A i j+ + F (s i+,j ) = F (s i,j ) + F (s i,j+ ) + F (s i,j ) + F (s i+,j ) =. Similarly, if i + j m + 3, then σ(s i,j ) = f(s i,j ) + f(s i,j+ ) + f(s i,j ) + f(s i+,j ) = A (m+) i+j + A i j+ + F (s i,j ) + A (m+) i+j+ + A i j + F (s i,j+ ) + A (m+) i+j + A i j + F (s i,j ) + A (m+) i+j+ + A i j+ + F (s i+,j ) = F (s i,j ) + F (s i,j+ ) + F (s i,j ) + F (s i+,j ) =. Subcase.. j = m +. Then i is even and i m. If i m, then Finally, σ(s i,m+ ) = f(s i,m ) + f(s i,m+ ) + f(s i,m+ ) + f(s i+,m+ ) = f(s m,i ) + f(s m i+, ) = A (m+) m+i + A m i + F (s m,i ) + A m i+ + A m i + F (s m i+, ) = F (s m,i ) + F (s m i+, ) = + 0 =. σ(s m,m+ ) = f(s m,m ) + f(s m,m+ ) + f(s m,m+ ) = f(s m,m ) + f(s, ) + 0 = A (m+) m + A 0 + F (s m,m ) + A + A 0 + F (s, ) = F (s m,m ) + F (s, ) = + 0 =.

8 48 Futaba Okamoto, Ebrahim Salehi, and Ping Zhang Therefore, f = F by Lemma.. Case. m is odd. Define a coin placement f : S {0, } such that f(s)= 0 for every s R as follows. Suppose that i( i m) and j( j n) are positive integers such that i + j is even. Then f(s i,j ) = A i+j + A i j + F (s i,j ) if j i and i + j m + A (m+) i+j + A i j + F (s i,j ) if j i and i + j m + 3 f(s j,i ) if i + j m 0 ifj {m +, m + } A (m+) i+j + A j i + F (s i,j ) if m + j m + i and j i m + A (m+) i+j + A (m+) j i + F (s i,j ) if m + j m + i and j i m + 3 f(s (m+) j,(m+) i ) if m + 4 i j m + f(s i,j (m+) ) if j m + 3. Note that f(s i,j ) = 0 whenever j 0 (mod m + ). Also f(s i, ) = A i+ + A i + F (s i, ) = a i+ + 0 = f (s i, ) for i =, 3,...,mand so f restricted to S equals f. One can also verify that f = F by an argument similar to the one used in Case. Proposition.4 yields the following useful result. Corollary.5. Let F be an extension of an m n checkerboard such that F(s) = 0 for every s R. Ifl is a positive integer such that {s,l,s m,l } B and F(s) = 0 for every s S l, then F(s) = 0 for every s S j, where j l(mod m + )..3. New Solutions from Old and the Main Result In this section we study how one can obtain a solution for a checkerboard of certain size from a solution for another checkerboard of different size, which will lead to a proof of the Checkerboard Theorem. Figure 4(a) shows a solution for a 3 checkerboard such that σ(s) = ifand only if s R. Extending this coin placement, we are able to obtain a solution for a 8 checkerboard as well as a solution for a 3 6 checkerboard, both of which has the property that σ(s) = if and only if s R, as shown in Figure 4(b). Therefore, the solution for the 3 checkerboard in Figure 4(a) can be extended to solutions for 8 and 3 6 checkerboards. On the other hand, we may also say that the solution for the 8 checkerboard shown in Figure 4(b) can be reduced to a solution for a 3 checkerboard.

9 A Solution to the Checkerboard Problem 49 (a) (b) Figure 4: Examples of extension and reduction of solutions Recall that if f is a solution for an m n checkerboard, then σ f (s) = if and only if s R (or for every s B); while if F is an extension for an m n checkerboard, then σ F (s) = for every s R (for every s B) except possibly for those in S n R (in S n B). The following observation describes how an extension of an m n checkerboard induces a solution for an m l checkerboard for some special values of l<n. Observation.6. If F is an extension for an m n checkerboard satisfying that F(s) = 0 for every s S l+, then an m l checkerboard has a solution. As a consequence of Proposition.4 and Observation.6, we obtain a result for m m checkerboards. Corollary.7. For every integer m, an m m checkerboard C has a solution such that σ(s) = if and only if s R. The following result will be useful to us. Proposition.8. If n l(mod m + ), then an m l checkerboard has a solution with σ(s) = for every s R if and only if an m n checkerboard has a solution with σ(s) = for every s R. Proof. It suffices to show that an m l checkerboard has a solution with σ(s) = for every s R if and only if an m (l + m + ) checkerboard has a solution with σ(s) = for every s R. First consider an m l checkerboard with a solution f such that σ(s) = ifand only if s R. Let F R and F L be the coin placement obtained by extending f to the right and to the left, respectively. If either m is even or l is odd, then {s,l+,s m,l+ } B and F R (s) = 0 for every s S l+, implying that F R (s) = 0 for every s S l+m+ by Corollary.5. Then the result follows by Observation.6. For the converse, suppose that C is an m (l + m + ) checkerboard having a solution f such that σ(s) = if and only if s R. Then f(s) = 0 for every s S m+ by Proposition.4. Therefore, f restricted to the last l columns of C is a solution for an m l checkerboard with the desired property. Before finally proving the main theorem, we show that the Checkerboard Conjecture holds for checkerboards of special size.

10 50 Futaba Okamoto, Ebrahim Salehi, and Ping Zhang Lemma.9. If n,m,m ± (mod m + ), then an m n checkerboard has a solution such that σ(s) = for every s R. Proof. By Corollary.7 and Proposition.8 it suffices to show that an m k checkerboard has a solution such that σ(s) = if and only if s R for k {,m± }. Fora m checkerboard, the coin placement f defined by { j 3 ifm 0 (mod 4) f(s,j ) = if and only if j otherwise is a solution. For an m (m + ) checkerboard, the trivial extension F is a solution if m is even; otherwise, the extension of the coin placement f : S {0, } defined by f (s) = 0 if and only if s S R is a solution. This also implies that an m (m ) checkerboard has the desired property. Theorem.0. The Checkerboard Theorem An m n checkerboard has a solution for every pair m, n of positive integers with mn. Proof. We in fact show that every checkerboard has a solution such that σ(s) = for every s R. Assume, to the contrary, that there exists an m n checkerboard C(m n) with no such solution. Therefore, n m + 3 by Lemma.9. We may assume that C is a smallest such checkerboard, that is, every checkerboard having less than mn squares has a solution with σ(s) = if and only if s R. In particular, an m (n m ) checkerboard has the desired property. However, Proposition.8 implies that so does C, which is a contradiction. To actually find a solution for a given m n checkerboard C, let m = l, n = l 0, where then l l 0, and C = C 0. Let l be the integer with l l + and l 0 l (mod l + ). If l {,l,l ± }, then construct a solution for C 0 using Lemma.9. Otherwise, we consider an l l checkerboard C. Let l 3 be the integer with l 3 l + and l+ l 3 (mod l +). Ifl 3 {,l,l ±}, then construct a solution for C using Lemma.9. Otherwise, consider an l 3 l checkerboard C, and so on. We continue this process until we obtain an l k+ l k checkerboard C k (k 0) for which there exists a solution such that σ(s) = if and only if s R. Then by extending C k upward and rotating by 90, we obtain the l k l k checkerboard C k with a solution such that σ(s) = if and only if s R. In general, suppose that we have a solution for the l i+ l i checkerboard C i such that σ(s) = if and only if s R for some i( i k). Reposition C i, if necessary, so that {s,,s,li } / R. Then extending C i upward, we obtain a solution for an l i l i checkerboard, which can be rotated to result in a solution for the l i l i checkerboard C i with the desired property. We illustrate this process described above with the following two examples. Example.. Consider an 8 3 checkerboard. Since 3 4 (mod 8 + ) (4 / {, 8, 8 ± }) and 8 3 (mod 4 + ) (3 {, 4, 4 ± }), first obtain a solution for a 3 4 checkerboard C as described in Lemma.9. Extending this solution results in

11 A Solution to the Checkerboard Problem 5 Figure 5: Constructing solutions for 4 8 and 8 3 checkerboards a solution for a 8 4 checkerboard C, which can be extended again and we obtain a solution for an 8 3 checkerboard C 0 with a solution. (See Figure 5.) Example.. Consider an 8 checkerboard. Since 8 6 (mod + ) (6 / {,, ± }), 4 (mod 6 + ) (4 / {, 6, 6 ± }), and 6 (mod 4 + ) ( {, 4, 4±}), after obtaining a solution for a 4 checkerboard, one can recursively construct solutions for 4 6, 6, and 8 checkerboards, as shown in Figure 6. Figure 6: Constructing solutions for 4 6, 6, and 8 checkerboards We conclude this paper with related open questions. Problem.3. For a given checkerboard, how many solutions (up to symmetry) are there? Which checkerboard have unique solutions? Problem.4. For a given checkerboard, what is the minimum number of coins necessary to construct a solution? Also, what is the maximum number of coins that can be used in a solution? Acknowledgments We are grateful to Professor Gary Chartrand for suggesting the problem to us and kindly providing useful information on this topic.

12 5 Futaba Okamoto, Ebrahim Salehi, and Ping Zhang References G. Chartrand, L. Lesniak and P. Zhang, Graphs & Digraphs: Fifth Edition, Chapman & Hall/CRC, Boca Raton, FL, 00. F. Okamoto, E. Salehi, and P. Zhang, A checkerboard problem and modular colorings of graphs. Bull. Inst. Combin. Appl., 58, pp. 9 47, 00.