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 [[email protected]], Parker Monfort [[email protected]], Robert Voinescu [rv3[email protected]], Ryan Wood [[email protected]] Directed by Dana C. Ernst