Subtraction games. Chapter The Bachet game

Transcription

1 Chapter 7 Subtraction games 7.1 The Bachet game Beginning with a positive integer, two players alternately subtract a positive integer < d. The player who gets down to 0 is the winner. There is a set of winning positions in the form of a decreasing sequence of nonnegative integers, such that if you secure one of these positions, then your opponent cannot secure any of the winning positions, and no matter how he moves, you can always secure a (smaller) winning positions. By keeping track of these winning positions, you eventually secure 0 and win the game. In the present example (the Bachet game), the winning positions are precisely the multiples of d. Proof. If Player A occupies position kd, and his opponent subtract a < d, then A subtracts (d a) and occupies occupies position (k 1)d. The same strategy allows A to get to 0 through the multiples of d.

2 302 Subtraction games 7.2 The Sprague-Grundy sequence Let G be a two-person counter game in which two players alternately remove a positive amount of counters according to certain specified rules. The Sprague-Grundy sequence of G is the sequence (g(n)) of nonnegative integers defined recursively as follows. (1) g(n) = 0 for all n which have no legal move to another number. In particular, g(0) = 0. (2) Suppose from position n it is possible to move to any of positions m 1, m 2,..., m k, (all < n), then g(n) is the smallest nonnegative integer different from g(m 1 ), g(m 2 ),..., g(m k ). Theorem 7.1. The player who secures a position n with g(n) = 0 has a winning strategy. Example 7.1. The Bachet game: n g(n) , , 1, 0 3. d 1 (d 2),...,1, 0 d 1 d (d 1),...,1 0. More generally, g(kd + a) = a for integers k and a satisfying 0 a < d. The Sprague-Grundy sequence is the periodic sequence with period 0, 1,..., d 1. The winning positions are precisely the multiples of d. Example 7.2. The trivial counter game. Two players alternately subtract any positive amount. The only winning position is 0. The first player wins by removing all counters. In this case, g(n) = n for every n.

3 7.3 Subtraction of powers of Subtraction of powers of 2 n g(n) , , , 2, , 3, , 4, , 5, , 6, 4, , 7, 5, , 8, 6, 2 1 This suggests that the winning positions are the multiples of 3. Proof. If Player A occupies a multiple of 3, any move by Player B will results in a position 3k + 1 or 3k + 2. Player A can get to a smaller multiple of 3 by subtracting 1 or 2 accordingly. Exercise 1. What are the winning positions in the game of subtraction of powers of 3? Answer. Even numbers. 2. What are the winning positions in the game of subtraction of prime numbers or 1? Answer. Multiples of What are the winning positions in the game of subtraction of a proper divisor of the current number (allowing 1 but not the number itself)? Note that 1 is not a proper divisor of itself Answer. Odd numbers except 1. Solution. The clue is that all factors of an odd number are odd. Subtracting an odd leaves an even number. Hence the winning strategy is to leave an odd number so that you opponent will always leave you an even number. From this you get to an odd number by subtracting 1.

4 304 Subtraction games 7.4 Subtraction of square numbers Two players alternately subtract a positive square number. We calculate the Sprague-Grundy sequence. n g(n) , , , , , , 5, , 6, 1 0 The values of n 500 for which g(n) = 0 are as follows: 1 These are the winning positions Suppose we start with 74. Player A can subtract 9 to get to 65, or subtract 64 to get 10, which have value 0. In the latter case, B may move to 9, 6 or 1. A clearly wins if B moves to 9 or 1. But if B moves to 6, then A can move to 5 or 2 and win. Exercise 1. How would you win if the starting number is 200? or 500? 1 [Smith, p.68] incorrectly asserts that this sequence is periodic, with period 5.

5 7.5 More difficult games More difficult games 1. Subtraction of proper divisor of current number (not allowing 1 and the number itself). The winning positions within 500 are as follows Subtraction of primes (not allowing 1). The winning positions within 500 are as follows

6 306 Subtraction games

7 Chapter 8 The games of Euclid and Wythoff 8.1 The game of Euclid Two players alternately remove chips from two piles of a and b chips respectively. A move consists of removing a multiple of one pile from the other pile. The winner is the one who takes the last chip in one of the piles. Preliminary problem: Find a constant k such that for positive integers a and b satisfying b < a < 2b, (i) a < kb = b > k(a b), (ii) a > kb = b < k(a b). Analysis: If these conditions hold, then we have a = kb = b = k(a b). If a = k, then 1 = k ( a 1) = k(k 1). Such a number k b b must satisfy k(k 1) = 1; it is the golden ratio ϕ = 1 ( 5 + 1). 2 Proposition 8.1. Let ϕ := 1 2 ( 5 + 1). For any two real numbers a and b, (i) a < ϕb = b > ϕ(a b), (ii) a > ϕb = b < ϕ(a b). Proof. The golden ratio ϕ satisfies ϕ 1 = 1. ϕ (i) a < ϕb = a b < (ϕ 1)b = 1 b = ϕ(a b) < b. ϕ (ii) a > ϕb = a b > (ϕ 1)b = 1 b = ϕ(a b) > b. ϕ Theorem 8.2. In the game of Euclid (a, b), the first player has a winning strategy if and only if a > ϕb.

8 308 The games of Euclid and Wythoff Proof. The first player (A) clearly wins if a = kb for some integer k. Assume a > ϕb. Let q be the largest integer such that a > qb. If q = 1, the only move is (a, b) (b, a b). In this case, b < ϕ(a b) by Proposition 8.1(ii). If q 2, we consider the moves (i) (a, b) (b, a qb) and (ii) (a, b) (a (q 1)b, b) (Note that a (q 1)b > b). If b < ϕ(a qb), make move (i). Otherwise, b > ϕ(a qb). Make move (ii). In this case, a qb < 1 ϕ b = a (q 1)b < ( 1 ϕ + 1 ) b = ϕb. This means that A can make a move (a, b) (a, b ) with a > b such that a < ϕb. The second player B has no choice but only the move (a, b ) (a b, b ). By Proposition 8.1, b > ϕ(a b ). Assume a < ϕb. Then A has no choice except the move (a, b) (a b, b). Here, b > ϕ(a b). Winning strategy: Suppose a > ϕb. Let q be the largest integer such that a > qb. A chooses between (a, b ) = (b, a qb) or (a (q 1)b, b) for a < ϕb. Examples: (a,b) A moves to B moves to (50, 29) (29, 21) (21, 8) (8,5) (5,3) (3,2) (2,1) (1,0) wins (a,b) A moves to B moves to (50, 31) (31, 19) (19, 12) (12, 7) (7, 5) (5,3) (3,2) (2,1) (1,0) wins

9 8.2 Wythoff s game Wythoff s game Given two piles of chips, a player either removes an arbitrary positive amount of chips from any one pile, or an equal (positive) amount of chips from both piles. The player who makes the last move wins. We describe the position of the game by the amounts of chips in the two piles. If you can make (1, 2), then you will surely win no matter how your opponent moves. Now, to forbid your opponent to get to this position, you should occupy (3, 5). The sequence of winning positions: starting with (a 1, b 1 ) = (1, 2), construct (a n, b n ) by setting a n := min{c : c a i, b i, i < n}, b n :=a n + n. Here are the 18 smallest winning positions for Wythoff s game: Every positive integer appears in the two sequences. Proof. Suppose (for a contradiction) that not every positive integer appears in the two sequences. Let N be the smallest of such integers. There is a sufficiently large integer M such that all integers less than N are among a 1,...,a M and b 1,...,b M. Then by definition, a M+1 = N, a contradiction. 2. The sequence (a n ) is increasing. Proof. Suppose a n+1 a n. This means that a n+1 is the smallest integer not in the list a 1, a 2,...,a n 1, a n, b 1, b 2,...,b n 1, b n. In particular, a n+1 a n. This means that a n+1 < a n < a n + n = b n. Ignoring a n and b n from this list, a n+1 is the smallest integer not in a 1, a 2,...,a n 1, b 1, b 2,...,b n 1. This means, by definition, that a n = a n+1, a contradiction.

10 310 The games of Euclid and Wythoff 3. The sequence (b n ) is increasing. Proof. b n+1 = a n+1 + n + 1 > a n + n = b n. 4. Since (a n ) and (b n ) are increasing, a n n and b n n for every n. 5. a m b n for all integers m and n. Proof. If m n, then a m a n < b n. If m > n, then n m 1 and a m is the smallest integer not among In particular, a m b n. a 1,...,a m 1, b 1,...,b n,...,b m Each positive integer appears in the list exactly once. Corollary of (1), (2), (3), (5). 7. It is not possible to move from (a n, b n ) to (a m, b m ) for m < n. Proof. A move (a n, b n ) (a m, b m ) must subtract the same number from a n and b n. This is impossible since b n a n = n > m = b m a m. 8. Let a < b. Suppose (a, b) (a n, b n ) for any n. Then there is a move into (a n, b n ) for some n. Proof. Each of a and b appears exactly once in one of the sequences (a n, b n ). There are five possibilities for (a, b) (in each case, p < q): (i) (a p, a q ), (ii) (b p, b q ), (iii) (a p, b q ), (iv) (a q, b p ), (v) (b p, a q ). In each case, we move (a, b) (a p, b p ) except when in (i) a q < b p. In this case, let n = b a. Note that n = b a < a q a p < b p a p = p, and b n a n = b a = b b n = a a n. The move (a, b) (a n, b n ) is done by subtracting equal numbers from a and b.

11 8.3 Beatty s Theorem Beatty s Theorem Theorem 8.3 (Beatty). If α and β are positive irrational numbers satisfying = 1, then the sequences α β and α, 2α, 3α,... β, 2β, 3β,... form a partition of the sequence of positive integers. Proof. (1) If an integer q appears in both sequences, then there are integers h and k such that From these, Combining these, we have q < hα < q + 1, q < kβ < q + 1. h q + 1 < 1 α < h q, k q + 1 < 1 β < k q. h + k q + 1 < 1 < h + k, q and q < h + k < q + 1, an impossibility. This shows that an integer q can appear in at most one of the sequences. (2) Now suppose an integer q does not appear in any of these sequence. Then there are integers h and k such that (h 1)α < q < q + 1 < hα, (k 1)β < q < q + 1 < kα. From these, h q + 1 > 1 α > h 1, q k q + 1 > 1 β > k 1. q

12 312 The games of Euclid and Wythoff Combining these, we have h + k q + 1 > 1 > h + k 2, q and h + k > q + 1 > q > h + k 2, an impossibility. This shows that every integer q appears in at least one of the sequences. From (1) and (2) we conclude that every integer appears in exactly one of the sequences. We try to find irrational numbers α and β satisfying 1 α + 1 β = 1 such that the two sequences ( nα ) and ( nβ ) are (a n ) and (b n ) for the winning positions of the Wythoff game. Since b n a n = n, we require nβ nα = n. This is the case if we choose α and β such that β α = 1. Since 1 α + 1 β = 1, we have 1 α + 1 α+1 = 1 or α2 = α + 1. Therefore, α = ϕ = 1 2 ( 5 + 1), the golden ratio, and β = ϕ + 1. Here is a succinct description of the Wythoff sequence. Theorem 8.4. The winning positions of Wythoff s game are the pairs ( nϕ, nϕ + n). Exercise 1. Where are the Fibonacci numbers in the Wythoff sequence? 2. Where are the Lucas numbers in the Wythoff sequence?

13 Chapter 9 Extrapolation problems 9.1 What is f(n + 1) if f(k) = 2 k for k = 0, 1, 2..., n? Let f(x) be a polynomial of degree n such that f(0) = 1, f(1) = 2, f(2) = 4,..., f(n) = 2 n. What is f(n + 1)? It is tempting to answer with 2 n+1, but this assumes f(x) = 2 x, not a polynomial. 1 Here is an easier approach to the problem, by considering the successive differences. If f(x) is a polynomial of degree n, then f(x+1) f(x) is a polynomial of degree n 1. Suppose the values of f(x) are given at n + 1 consecutive integers 0, 1,..., n. Then we can easily find the values of f(x + 1) f(x) at 1, 2..., n. By repeating the process, we obtain the successive differences. 1, 2, 4, 8, 16,31,57,99 1, 2, 4, 8,15,26,42 1, 2, 4,7,11,16 1, 2,3,4,5 1,1,1,1 f(n + 1) = 2 n+1 1. What is f(n + 2)? How does the sequence 1, 4, 11, 26, 57,... 1 There is also the famous Lagrange interpolation formula to find such a polynomial.

14 314 Extrapolation problems continue? The differences are 3, 7, 15, 31,.... From these we have f(n + 2) = 1 + (2 2 1) + (2 3 1) + + (2 n+1 1) = (2 1 1) + (2 2 1) + (2 3 1) + + (2 n+1 1) = ( n+1 ) (n + 1) = 2 n+2 2 (n + 1) = 2 n+2 n 3. Exercise 1. Prove that f( 1) = { 0 if n is odd, 1 if n is even. 2. Show that the values of f( 2) are for n = 2, 3, 4, 5, 6, 7,.... 2, 2, 3, 3, 4, 4,...

15 9.2 What is f(n + 1) if f(k) = 1 k+1 for k = 0, 1, 2..., n? What is f(n + 1) if f(k) = 1 k+1 for k = 0, 1, 2..., n? Suppose we have a polynomial f(x) of degree n with given values x n f(x) n+1 What are f(n + 1) and f(n + 2)? Solution. From the given data, we have (x + 1)f(x) 1 = 0 for x = 0, 1,..., n. Therefore, (x + 1)f(x) 1 is a polynomial of degree n + 1 with n + 1 roots 0, 1,..., n. (x + 1)f(x) 1 = cx(x 1)(x 2) (x n), for some constant c. With x = 1, we have Hence, 1 = c( 1)( 2)( 3) ( 1 n) c = ( 1)n (n + 1)!. (x + 1)f(x) = 1 + ( 1)n x(x 1) (x n). (n + 1)! (i) If n is odd, we have (n + 1)! (n + 2)f(n + 1) = 1 = 1 1 = 0 f(n + 1) = 0, (n + 1)! (n + 2)! + 1 (n + 3)f(n + 2) = 1 = (n + 1) f(n + 2) = n (n + 1)! n + 3. (ii) If n is even, we have (n + 1)! (n + 2)f(n + 1) = 1 + (n + 1)! (n + 3)f(n + 2) = 1 + = 2 f(n + 1) = 2 n + 2, (n + 2)! = n + 3 f(n + 2) = 1. (n + 1)! Summary { { 0, if n is odd, n+1, if n is odd, f(n+1) = 2, if n is even; f(n+2) = n+3 1, if n is even. n+2

16 316 Extrapolation problems Exercise 1. Let f(x) be a polynomial of degree n such that for k = 1, 2,..., n, f(k) = F k, the k-th Fibonacci number. Find the value of f(n + 1). { F n, if n is odd, Answer. f(n + 1) = L n, if n is even. 2. Find a quadratic polynomial f(x) such that f(1 n ) = 1 2n for every integer n. Note: 1 n is the integer whose decimal expansion consists of n digits each equal to 1; similarly for 1 2n. Answer. f(x) = x(9x + 2). 3. Find a cubic polynomial g(x) such that g(1 n ) = 1 3n for every n 1.

17 9.3 Why is e x not a rational function? Why is e x not a rational function? We show why the exponential function, and some other elementary functions, are not rational functions by studying the degrees of rational functions. A rational function is one of the form f(x) = P(x), where P(x) Q(x) and Q(x) are polynomials without common divisors (other than scalars from the field of coefficients). Theorem 9.1. Let f and g be rational functions. (a) If f + g 0, then deg(f + g) max(deg f, deg g). (b) deg(fg) = deg f + deg g. (c) If f 0, then deg f < deg f. (a) and (b) are well known. We prove (c). This depends on the simple fact that for a polynomial P(x), if the derivative P (x) is not identically zero, then deg P = deg P 1. Now let f(x) = P(x). Then f = P Q PQ = P PQ. Q(x) Q 2 Q Q 2 Now, deg P Q = deg P deg Q = (deg P 1) deg Q deg PQ Q 2 = deg P deg Q 1 = deg f 1, = deg P + (deg Q 1) 2 deg Q = deg P deg Q 1 = deg f 1. It follows from (a) that deg f deg f 1. This proves (c). Corollary 9.2. e x is not a rational function. Proof. Note that f(x) = e x satisfies the differential equation (x) = f(x). Suppose f(x) = e x is a rational function. Then a contradiction. deg f = deg f deg f 1,

18 318 Extrapolation problems Exercise 1. Make use of the fact (tan x) = sec 2 x to show that tan x is not a rational function. 2. Make use of the identity cos 2 x = 1 (1 + cos 2x) to show that cosx 2 is not a rational function. 3. Deduce that sin x is not a rational function. 4. Show that the logarithm function log x is not a rational function.