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# 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. January 31, 2016

## 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) = , 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}
mex(0)=1 (A wins)
G(2)=∅
mex(∅)=0
2
Game tree for n = 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
={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 [[email protected]], Parker Monfort [[email protected]], Robert Voinescu [rv3[email protected]], Ryan Wood [[email protected]] Directed by Dana C. Ernst