Explorations of Conway’s Sylver Coinage Game

Sylver Coinage is a game in which two players, A and B, alternately name positive integers that are not the sum of nonnegative multiples of previously named integers. The person who names 1 is the loser! This seemingly innocent looking game is the subject of one of John Conway's open problems with monetary rewards. One such open problem is: If player A names 16 to start, and both players play optimally thereafter, then who wins? In this talk, we will discuss a simplified version of the game in which a fixed positive integer n (greater than 2) is agreed upon in advance. Then A and B alternately name positive integers from the set {1,2,...,n} that are not linear combinations with positive coefficients of previously named numbers. As in the original game, the person who is forced to name 1 is the loser. We will investigate who wins under optimal play for given values of n and determine the Nim-values for the simplified game under certain conditions.

This talk was given by my undergraduate research students Joni Hazelman and Parker Montfort at the 2016 Nebraska Conference for Undergraduate Women in Mathematics at the University of Nebraska-Lincoln.

Transcript

1. Explorations of Conway’s Sylver Coinage Game Joni Hazelman, Parker Montfort, Robert Voinescu, & Ryan Wood Department of Mathematics & Statistics, Northern Arizona University Definition Sylver Coinage is a game in which two players, A and B, alternately name positive integers that are not the sum of nonnegative multiples of previously named integers. The person that is forced to name 1 loses. Example Here is an example game between A and B. A: (5) = {5, 10, 15, . . .} B: (5, 4) = {4, 5, 8, 9, 10, 12, →} A: (5, 4, 11) = {4, 5, 8, →} B: (5, 4, 11, 6) = {4, 5, 6, 8, →} A: (5, 4, 11, 6, 7) = {4, →} B: (5, 4, 11, 6, 7, 2) = {2, 4, →} A: (5, 4, 11, 6, 7, 2, 3) = {2, →} B: Forced to choose 1 and loses! Known Results 1. Choosing 2 or 3 are bad opening moves for A. In either case, B can choose the other to guarantee a win. Choosing 4, 6, 8, 9, and 12 are also bad opening moves for A. The sequences (2, 3), (4, 6), (6, 9), and (8, 12) are winning positions for B. 2. Hutching’s Theorem: If gcd(a, b) = 1 and {a, b} = {2, 3}, then (a, b) is winning position for A. 3. If p is prime with p ≥ 5, then (p) is a winning position for A. 4. If n is composite & not equal to 2a3b, then (n) is a winning position for B. Open Question (John Conway) If player A names 16, and both players play optimally thereafter, then who wins? Note that 16 is the smallest number not handled by the facts above. Simplified Sylver Coinage In Simplified Sylver Coinage, A and B alternately name positive integers from the set [n] := {2, . . . , n} that are not the sum of nonnegative multiples of previously named numbers among [n]. The player that eliminates the last remaining number is the winner. Example Suppose n = 10. Below is one possible sequence of moves. A: (4) = {4, 8} B: (4, 5) = {4, 5, 8, 9, 10} A: (4, 5, 6) = {4, 5, 6, 8, 9, 10} B: (4, 5, 6, 3) = {3, 4, 5, 6, 7, 8, 9, 10} A: (4, 5, 6, 3, 2) = [10], and so A wins! Terminology of Impartial Games 1. An impartial game is a finite set X of positions together with a starting position and a collection {Opt(P) ≤ x | P ∈ X}, where Opt(P) is the set of possible options for a position P. 2. Two players take turns replacing the current position P with one of the available options in Opt(P). The player that encounters the empty option loses. 3. The minimum excludant mex(N) of a set of ordinals N is the smallest ordinal not contained in the set. Note that mex(∅) = 0. 4. The nim-number nim(P) of a position P is the mex of the set of nim- numbers of the options of P: nim(P) = mex{nim(Q)|Q ∈ Opt(P)}. 5. Terminal positions have nim-number 0. 6. Player B has a winning strategy iff the nim-number for the game is 0. Example Game Trees If (x1, x2, . . . , xn ) is a sequence of moves, then the corresponding gap set is defined via G(x1, x2, . . . , xn ) = [n] \ (x1, x2, . . . , xn ). Game tree for n = 2 [2]={2} mex(0)=1 (A wins) G(2)=∅ mex(∅)=0 2 Game tree for n = 3 [3]={2,3} mex(1,1)=0 (B wins) G(2)={3} mex(0)=1 G(3)={2} mex(0)=1 G(2,3)=∅ mex(∅)=0 G(3,2)=∅ mex(∅)=0 3 2 3 2 Game tree for n = 4 [4]={2,3,4} mex(1,2,0)=3 (A wins) G(3)={2,4} mex(0,1)=2 G(2)={3} mex(0)=1 G(4)={2,3} mex(1,1)=0 G(3,4)={2} mex(0)=1 G(3,2)=∅ mex(∅)=0 G(2,3)=∅ mex(∅)=0 G(4,2)={3} mex(0)=1 G(4,3)={2} mex(0)=1 G(3,4,2)=∅ mex(∅)=0 G(4,2,3)=∅ mex(∅)=0 G(4,3,2)=∅ mex(∅)=0 4 2 3 3 3 2 4 2 2 3 2 Data Table n Winner nim Strategy 2 A 1 A chooses 2 3 B 0 If A chooses 2 (resp 3), then B chooses 3 (resp 2) 4 A 3 A chooses 4 then mimics B’s strategy on n = 3 5 B 0 B stalls until A is forced to choose 2 or 3 6 A 3 A chooses 6 then mimics B’s strategy on n = 5 7 B 0 B stalls until A is forced to choose 2 or 3 8 A 5 A chooses 8 then mimics B’s strategy on n = 7 9 B 0 B intelligently stalls until A is forced to choose 2 or 3 10 A 1 A chooses 10 then mimics B’s strategy on n = 9 11 B 0 Complicated . . . 12 A 4 A chooses 12 then mimics B’s strategy on n = 11 Proposition If B wins on [n], then A wins on [n + 1]. Proof A’s opening move is n + 1. Then mimic B’s strategy on [n]. Corollary We never have two consecutive B’s if we continue our table. Question Will we ever encounter two consecutive A’s if we continue our table? Conjecture A wins on [n] iff n is even. Potential strategy: A chooses n. Even if this is an appropriate strategy, there may be other winning strategies. One potential strategy to proving the conjecture is to show that if nim([n]) = 0, then nim([n + 2]) = 0. Knock Out In Knock Out (inspired by Simplified Sylver Coinage), A and B alternately name positive integers from the set [n], but this time we leave unselected numbers in play. The loser is the player that knocks out n as the sum of nonnegative multiples of previously named numbers among [n]. What happens in this game??? Example Suppose n = 6. Player A will lose immediately if they choose 2, 3, or 6. Suppose A chooses 4. Then B cannot choose 4 and again, 2, 3, or 6 are losing moves. If B is wise, they will choose 5. This forces A to choose 2, 3, or 6, and hence A loses. Email: Joni Hazelman [jbh227@nau.edu], Parker Monfort [plm68@nau.edu], Robert Voinescu [rv334@nau.edu], Ryan Wood [rpw54@nau.edu] Directed by Dana C. Ernst