This is the first in a series of three exercices de style to show some interesting aspects of the game of hats: a puzzle which was initially proposed by Lionel Levine, and arose from his work with Tobias Friedrich on
I borrow the introduction from :
Two players each have countably many hats put on their heads.
Each hat is either black or white with equal probability. Furthermore, the players are only able to see the hats on the other person’s head. Simultaneously each player points to a hat on their own head. They win if both players pick out a white hat.The question is what is the optimal strategy for this game? It should be noted that each individual player will pick a white hat with probability one half regardless of the strategy employed. The challenge then is to correlate their choices.
At a first glance, it may seem that there is no strategy with better than random ($1 \over 4$) chance of winning. This can be quickly discounted by the following simple strategy: each player looks for the first white hat on the partner’s head and chooses the corresponding hat on his or her own head.
Observe that if both players have a white hat first ($1 \over 4$ chance) then they win, if the two first hats are different ( $1 \over 2$ chance) then they lose, and if the two first hats are both black ($1 \over 4$ chance) then effectively they replay the game with the first hat removed. So, we see that the chance of winning is $x = {1 \over 4} + {1 \over 4} \cdot x$, so $x = {1 \over 3}$.
Finite case
Definitions and notations
Towers
Consider a tower of $n$ stacked hats ($n$ finite), each hat either black or white. We refer to $n$ as the tower’s height. By identifying each hat with a bit (value $0$ for black, $1$ for white), we can view the tower as the binary representation of a natural number, the first hat being the least significant bit.
Let $\mathbb{T}_n$ denote the set of all possible towers:
\begin{align}
\mathbb{T}_n \quad &\stackrel{\text{def}}{=} \quad \{0, 1\}^{[n]} \\
&= \quad [2^n], \nonumber
\end{align}
where $[n] = \{0, …, n-1\}$ is the set of natural numbers strictly smaller than $n$.
Let $t$ be a tower in $\mathbb{T}_n$.
As an element of $[2^n]$, $t$ is a plain natural number, which provide a convenient way to name it. For instance, $4 \in \mathbb{T}_3$ is the tower consisting of one white hat on top of two black hats.
$$
4 \quad = \quad
\begin{array}{c}
○\\
●\\
●
\end{array}
$$
Alternately, as an element of $\{0, 1\}^{[n]}$, $t$ is a map: $[n] \rightarrow \{0, 1\}$. Given a position $p$ in $[n]$, $t(p)$ tells us the color of the hat at position $p$ in the tower ($0$ for black or $1$ for white). $t(p)$ is also the value of the $p$th bit in the binary representation of $t$.
To avoid confusing $t(p)$ (map application) with $t \cdot (p)$ (multiplication of numbers), we write $\overline{(t)}(p)$ (overlined) for the map application. Also, we extend the domain of $\overline{(t)}$ to the set of integers:
\begin{equation}
\forall t \in \mathbb{T}_n, \forall p \in \mathbb{Z}, \quad
\overline{(t)}(p) \;\stackrel{\text{def}}{=}\; (t \gg p) \mathbin{\%} 2,
\end{equation}
(where $\gg$ is the bitwise right shift operator and $\mathbin{\%}$ the remainder operator).
Example
\begin{align*}
\overline{(4)}(-1) &= 0 \\
\overline{(4)}(0) &= 0 \\
\overline{(4)}(1) &= 0 \\
\overline{(4)}(2) &= 1 \\
\overline{(4)}(3) &= 0
\end{align*}
Strategies
A (deterministic) strategy $s$ for the game of $n$ hats is a way for each player, when presented with the tower of hats on top of their partner’s head, of choosing the position of a (hopefully white) hat on their own head. In other words: $s$ is a map from $\mathbb{T}_n$ to $\mathbb{Z}$.
Note that we allow a strategy to choose a position outside of the tower ($\ge n$ or $< 0$).
Let $\mathbb{S}_n = \mathbb{Z}^{\mathbb{T}_n}$ denote the set of all possible strategies for towers of $n$ hats:
\begin{equation}
\mathbb{S}_n \; \stackrel{\text{def}}{=} \; \mathbb{Z}^{\mathbb{T}_n}.
\end{equation}
A natural way to name a strategy is to use a $2^n$-uple for enumerating the choices it makes.
For instance if $s \in \mathbb{S}_2$ is such that:
$$
\left\{
\begin{array}{ll}
s\left(\begin{array}{c}
●\\
●
\end{array}\right) = s(0) = 0,\\
s\left(\begin{array}{c}
●\\
○
\end{array}\right) = s(1) = 0,\\
s\left(\begin{array}{c}
○\\
●
\end{array}\right) = s(2) = 1,\\
s\left(\begin{array}{c}
○\\
○
\end{array}\right) = s(3) = 0,
\end{array}
\right.
$$
then we write:
$$s = (0,0,1,0).$$
When there is no ambiguity (e.g.: all positions are single digit), we shorten the notation:
$$s = [0010].$$
And for clarity, when $n$ gets too big, we use hyphens to separate groups of digits, e.g.: write $[0100 \text{-} 2123 \text{-} 0100 \text{-} 2120]$ for $[0100212301002120] \in \mathbb{S}_4$.
Hit score
A configuration for the game of $n$ hats is a pair of towers $t$ and $u$ in $\mathbb{T}_n$: one for each player’s head. Given a strategy $s$ in $\mathbb{S}_n$, we denote by $\eta_{t,u}(s)$ the outcome of applying strategy $s$ in the configuration $(t, u)$: $1$ for hit (win), $0$ for miss (loss).
\begin{equation}
\eta_{t,u}(s) \; \stackrel{\text{def}}{=} \; \overline{(t)}\big(s(u)\big) \cdot \overline{(u)}\big(s(t)\big)
\end{equation}
We call hit score of $s$, denoted by $\eta(s)$, the sum of all such outcomes, i.e. the count of all game configurations where the team wins.:
\begin{equation}
\eta(s) \; \stackrel{\text{def}}{=} \; \sum_{t,u \in \mathbb{T}_n}\eta_{t,u}(s)
\end{equation}
Win rate
Given a strategy $s$ in $\mathbb{S}_n$, we denote by $\mu(s)$ the win rate (or measure) of $s$, i.e. the probability that a team applying the strategy wins.
\begin{equation}
\mu(s) \; \stackrel{\text{def}}{=} \; {1 \over{4^n}} \cdot \eta(s)
\end{equation}
Equivalence
We call two strategies $a$ and $b$ in $\mathbb{S}_n$ equivalent, when they agree on the winning configurations:
\begin{equation}
a \equiv b \; \stackrel{\text{def}}{\iff} \; \forall t,u \in \mathbb{T}_n, \eta_{t,u}(a) = \eta_{t,u}(b).
\end{equation}
Note that two strategies $a$ and $b$ in $\mathbb{S}_n$ which differ only in the choice of $a(0) \neq b(0)$ are equivalent. This is because they make different choices only when one of the players is presented with tower $0$ (all black hats), which is always a losing situation for the team.
\begin{equation}
\big(\forall t \in \mathbb{T}_n \setminus \{0\}, \;\; a(t) = b(t)\big) \quad \implies \quad a \equiv b.
\end{equation}
Of course, equivalent strategies have the same hit score and win rate:
\begin{equation}
\forall a, b \in \mathbb{S}_n, \quad a \equiv b \implies \eta(a) = \eta(b).
\label{EquivRate}
\end{equation}
Examples of strategies
| Name | Tuple | Height | Hit score | Win rate |
|---|---|---|---|---|
| top | $\mathbf{T}_0 = [-1]$ | $0$ | $0$ | ${0 \over 1} = 0\%$ |
| top | $\mathbf{T}_1 = [00]$ | $1$ | $1$ | ${1 \over 4} = 25\%$ |
| lowest-white | $\mathbf{W}_2 = [1010]$ | $2$ | $5$ | ${5 \over 16} = 31.25\%$ |
| lowest-black | $\mathbf{B}_3 = [01020102]$ | $3$ | $21$ | ${21 \over 64} \approx 31.81\%$ |
Constant strategies
We call constant a strategy $s \in \mathbb{S}_n$ which always chooses the same position:
$$s \text{ constant } \;\stackrel{\text{def}}{\iff}\; \forall t, u \in \mathbb{T}_n, \; s(t) = s(u)$$
Observe:
- all constant strategies of height $0$ are equivalent,
- a constant strategy which chooses position $p \not\in [n]$ has hit score and win rate $0$,
- a constant strategy of height $>0$ which chooses position $p \in [n]$ has hit score $4^{n-1}$ and win rate of $1 \over 4$.
In particular, we denote by $\mathbf{T}_n$ (top $n$) the strategy in $\mathbb{S}_n$ which chooses position $n-1$ (i.e. the top position in the tower, when the tower is not empty).
\begin{equation}
\forall t \in \mathbb{T}_n, \quad \mathbf{T}_n(t) \stackrel{\text{def}}{=} n-1
\end{equation}
Lowest-white strategies
We denote by $\mathbf{W}_n$ (white n) the strategy in $\mathbb{S}_n$ which chooses the position of the lowest white hat, or $n-1$ if there is no white hat.
Example
$\mathbf{W}_2 = [1010]$ chooses position $1$ when the first hat is black, position $0$ otherwise.
| $\eta_{t_1,t_2}(\mathbf{W}_2)$ | $\mathbf{W}_2(t_2)$ | $1$ | $0$ | $1$ | $0$ |
| $\mathbf{W}_2(t_1)$ | $\downarrow\!t1 \quad t2\!\rightarrow$ | $0 = \begin{array}{c}●\\●\end{array}$ | $1 = \begin{array}{c}●\\○\end{array}$ | $2 = \begin{array}{c}○\\●\end{array}$ | $3 = \begin{array}{c}○\\○\end{array}$ |
| $1$ | $0 = \begin{array}{c}●\\●\end{array}$ | 0×0 | 0×0 | 0×1 | 0×1 |
| $0$ | $1 = \begin{array}{c}●\\○\end{array}$ | 0×0 | 1×1 | 0×0 | 1×1 |
| $1$ | $2 = \begin{array}{c}○\\●\end{array}$ | 1×0 | 0×0 | 1×1 | 0×1 |
| $0$ | $3 = \begin{array}{c}○\\○\end{array}$ | 1×0 | 1×1 | 1×0 | 1×1 |
$\mathbf{W}_2$ has hit score $5$ and win rate $5 \over 16$.
Lowest-black strategies
We denote by $\mathbf{B}_n$ (black n) the strategy in $\mathbb{S}_n$ which chooses the position of the lowest black hat, or $n-1$ if there is no black hat.
Example
$\mathbf{B}_3 = [01020102]$ chooses position $2$ when the lowest two hats are white, otherwise position $1$ when the lowest hat is white, otherwise position $0$.
$\mathbf{B}_3$ has hit score $21$ and win rate $21 \over 64$.
Operations on strategies
| Name | Example |
|---|---|
| embedding | $e_3([0100]) \;=\; [0100 \text{-} 0100]$ |
| left reset | $[0101] \rightarrow [-1] \;=\; [0101]$ |
| $[0101] \rightarrow [00] \;=\; [01 \text{-} 02 \text{-} 01 \text{-} 02]$ | |
| $[00] \rightarrow [0000] \;=\; [2000 \text{-} 2000]$ | |
| $[1010] \rightarrow [1010] \;=\; [3010 \text{-} 2010 \text{-} 3010 \text{-} 2010]$ | |
| right reset | $[0101] \leftarrow [-1] \;=\; [0101]$ |
| $[0101] \leftarrow [00] \;=\; [0102 \text{-} 0102]$ | |
| $[0000] \leftarrow [00] \;=\; [0002 \text{-} 0002]$ | |
| double reset | $DR([0010], [0010]) \;=\; [2012 \text{-} 2012 \text{-} 3012 \text{-} 2012]$ |
| left shift | $\uparrow\![0] \;=\; [0 \text{-} 0]$ |
| $\uparrow\![00] \;=\; [00 \text{-} 10]$ | |
| right shift | $[00]\!\uparrow \;=\; [01 \text{-} 00]$ |
| double shift | $[00]\!\Uparrow \;=\; [01 \text{-} 00 \text{–} 21 \text{-} 20]$ |
Embedding
| $e_1([-1]) = (-1, -1)$ |
| $e_3([1010]) = [1010 \text{-} 1010]$ |
Given a strategy $s$ in $\mathbb{S}_n$ and an integer $m > n$, we can use $s$ to play the game of $m$ hats. Simply: do not »looking » at the hats above position $n$ ($n$ included). This defines an embedding $e_m: \mathbb{S}_n \rightarrow \mathbb{S}_m$:
\begin{align*}
e_m: \mathbb{S}_n \rightarrow \mathbb{S}_m\\
s \mapsto e_m(s)
\end{align*}
\begin{equation}
\forall t \in \mathbb{T}_n, \;\; \big(e_m(s)\big)(t) \stackrel{\text{def}}{=} s(t \mathbin{\%} 2^n)
\end{equation}
(where $\mathbin{\%}$ is the remainder operator).
Win rate is invariant under embedding into $\mathbb{S}_m$:
\begin{equation}
\forall s \in \mathbb{S}_n, \forall m > n, \;\; \mu(s) = \mu(e_m(s))
\end{equation}
Left reset
| $[00] \rightarrow [-1] = [00]$ |
| $[00] \rightarrow [00] \rightarrow [-1] = [10 \text{-} 10]$ |
| $[00] \rightarrow [00] \rightarrow [00] \rightarrow [-1] = [2010 \text{-} 2010]$ |
Given two strategies $a \in \mathbb{S}_m$ and $b \in \mathbb{S}_n$, we call « $a$ left-reset by $b$ », denoted by $LR(a,b)$, the following strategy in $\mathbb{S}_{m+n}$: whenever the lowest $m$ hats in the tower are not all black, use strategy $a$; otherwise, use strategy $b$ on the remaining $n$ hats. Formally:
\begin{align}
LR(a,b): \; & \mathbb{T}_{m+n} \rightarrow [m+n] \nonumber \\
\nonumber \\
& t \stackrel{\text{def}}{\mapsto}
\left\{
\begin{array}{l}
a(t) \quad\text{if } t \not \equiv 0 \bmod 2^m\\
m + b(t \gg m) \quad\text{otherwise}
\end{array}
\right.
\end{align}
(where $\gg$ is the bitwise right shift operator).
Left reset is associative:
$$\forall a, b, c, \quad LR(a, LR(b, c)) \,=\, LR(LR(a, b), c)$$
so we make it into a binary operator:
\begin{equation}
b \rightarrow a \;\stackrel{\text{def}}{=}\; LR(a, b),
\end{equation}
and the following holds:
\begin{equation}
\forall a, b, c, \;\; (c \rightarrow b) \rightarrow a \,=\, c \rightarrow (b \rightarrow a) \,=\, c \rightarrow b \rightarrow a
\end{equation}
Right reset
| $[-1] \leftarrow [00] = [00]$ |
| $[-1] \leftarrow [00] \leftarrow [00] = [01 \text{-} 01]$ |
| $[-1] \leftarrow [00] \leftarrow [00] \leftarrow [00] = [0102 \text{-} 0102]$ |
Given two strategies $a \in \mathbb{S}_m$ and $b \in \mathbb{S}_n$, we call « $a$ right-reset by $b$ », denoted by $RR(a,b)$, the following strategy in $\mathbb{S}_{m+n}$: whenever the lowest $m$ hats in the tower are not all white, use strategy $a$; otherwise, use strategy $b$ on the remaining $n$ hats. Formally:
\begin{align}
RR(a,b): \; & \mathbb{T}_{m+n} \rightarrow [m+n] \nonumber \\
\nonumber \\
& t \stackrel{\text{def}}{\mapsto}
\left\{
\begin{array}{l}
a(t) \quad \text{if } t \not\equiv -1 \bmod 2^m\\
m + b(t \gg m) \quad \text{otherwise,}
\end{array}
\right.
\end{align}
where $\gg$ is the bitwise right shift operator.
Right reset is associative:
$$\forall a, b, c, \;\; RR(a, RR(b, c)) \,=\, RR(RR(a, b), c)$$
so we make it into a binary operator:
\begin{equation}
a \leftarrow b \;\stackrel{\text{def}}{=}\; RR(a, b),
\end{equation}
and the following holds:
\begin{equation}
\forall a, b, c, \;\; (a \leftarrow b) \leftarrow c \,=\, a \leftarrow (b \leftarrow c) \,=\, a \leftarrow b \leftarrow c
\end{equation}
Double reset
| $DR([00], [00]) = [1111]$ |
| $DR(DR([1010], [00]), [00]) = [30122013 \text{-} 30122013]$ |
| $DR([1010], DR([00], [00])) = [30133013 \text{-} 30133013]$ |
Given two strategies $a \in \mathbb{S}_m$ and $b \in \mathbb{S}_n$, we call « $a$ double-reset by $b$ », denoted by $DR(a, b)$, the following strategy in $\mathbb{S}_{m+n}$: whenever the lowest $m$ hats in the tower are not all black and not all white, use strategy $a$; otherwise, use strategy $b$ on the remaining $n$ hats. Formally:
\begin{align}
DR(a,b): \; & \mathbb{T}_{m+n} \rightarrow [m+n] \nonumber \\
\nonumber \\
& t \stackrel{\text{def}}{\mapsto}
\left\{
\begin{array}{l}
a(t) \quad \text{if } t \not\equiv 0 \text{ and } t \not\equiv -1 \bmod 2^m \\
m + b(t \gg m) \quad \text{otherwise,}
\end{array}
\right.
\end{align}
where $\gg$ is the bitwise right shift operator.
Double-reset is not associative.
Left shift
| $\uparrow\![0] = [0 \text{-} 0]$ |
| $\uparrow\![00] = [00 \text{-} 10]$ |
| $\uparrow\![0000] = [0000 \text{-} 2000]$ |
Given a strategie $s \in \mathbb{S}_n$, we call « $s$ left-shifted« , denoted by $\uparrow\!s$, the following strategy in $\mathbb{S}_{n+1}$:
\begin{equation}
\forall t \in \mathbb{T}_{n+1}, \quad (\uparrow\!s)(t) \stackrel{\text{def}}{=}
\left\{
\begin{array}{ll}
n & \text{if } t = 2^n\\
s(t \mathbin{\%} 2^n) & \text{otherwise,}
\end{array}
\right.
\end{equation}
where $\mathbin{\%}$ is the remainder operator.
Right shift
| $[0]\!\uparrow = [0 \text{-} 0]$ |
| $[00]\!\uparrow = [01 \text{-} 00]$ |
| $[0000]\!\uparrow = [0002 \text{-} 0000]$ |
Given a strategie $s \in \mathbb{S}_n$, we call « $s$ right-shifted« , denoted by $s\!\uparrow$, the following strategy in $\mathbb{S}_{n+1}$:
\begin{equation}
\forall t \in \mathbb{T}_{n+1}, \quad (s\!\uparrow)(t) \stackrel{\text{def}}{=}
\left\{
\begin{array}{ll}
n & \text{if } t = 2^n – 1\\
s(t \mathbin{\%} 2^n) & \text{otherwise,}
\end{array}
\right.
\end{equation}
where $\mathbin{\%}$ is the remainder operator.
Double shift
| $[00]\!\Uparrow = [01 \text{-} 00 \text{–} 21 \text{-} 20]$ |
| $[0000]\!\Uparrow = [0002 \text{-} 0000 \text{–} 3001 \text{-} 3000]$ |
Given a strategie $s \in \mathbb{S}_n$, we call « $s$ double-shifted« , denoted by $s\!\Uparrow$, the following strategy in $\mathbb{S}_{n+2}$:
\begin{equation}
\forall t \in \mathbb{T}_{n+2}, \quad (s\!\Uparrow)(t) \stackrel{\text{def}}{=}
\left\{
\begin{array}{ll}
n+1 & \text{if } t = 2^{n+1} + 2^n\\
\big(\uparrow\!(s\!\uparrow)\big)(t) & \text{otherwise.}
\end{array}
\right.
\end{equation}
Win rate computation
Brute-force computing the win rate would be expensive for large values of $n$. In the remainder of this section, we give simple formulae for calculating the win rate of certain strategies.
Formulae
| Operation | Condition | Win rate |
|---|---|---|
| left-reset | $a \in \mathbb{S}_n$, $b \in \mathbb{S}_m$ | $\mu(b \rightarrow a) = \mu(a) + {\mu(b) \over 4^n}$ |
| right-reset | $a \in [n]^{\mathbb{T}_n}$, $b \in [m]^{\mathbb{T}_m}$ | $\mu(a \leftarrow b) = \mu(a) + {\mu(b) \over 4^n}$ |
| double-reset | $a \in [n]^{\mathbb{T}_n}$, $b \in [m]^{\mathbb{T}_m}$ | $\mu(DR(a, b)) = \mu(a) + {{4 \mu(b) – 1} \over 4^n}$ |
| left-shift | $s \in {\mathbb{S}_n}$ | $\mu(\uparrow\!s) = \mu(s) + {1 \over 4^{n+1}}$ |
| right-shift | $s \in [n]^{\mathbb{T}_n}$ | $\mu(s\!\uparrow) = \mu(s) + {1 \over 4^{n+1}}$ |
| double-shift | $s \in [n]^{\mathbb{T}_n}$ | $\mu(s\!\Uparrow) = \mu(s) + {1 \over 4^{n+1}}+ {2 \over 4^{n+2}}$ |
Observations
Let $a$ be a strategy in $\mathbb{S}_n$. Recall that the hit score of $a$ is $\eta(a) = \sum_{t,u \in \mathbb{T}_n}\eta_{t,u}(a)$, where $\eta_{t,u}(a)$ denotes $\overline{(t)}(a(u)) \cdot \overline{(u)}(a(t))$.
Let $\sum\mathcal{M}$ denote the sum of all coefficients of matrix $\mathcal{M}$.
Then $\eta(a)$ can be written as:
\begin{align*}
\sum
\begin{pmatrix}
\big(\eta_{0,0}(a) & \eta_{0,1}(a) & \cdots & \eta_{0,2^n-2}(a) & \eta_{0,2^n-1}(a) \\
\big(\eta_{1,0}(a) & \eta_{1,1}(a) & \cdots & \eta_{1,2^n-2}(a) & \eta_{1,2^n-1}(a) \\
\vdots & \vdots & \ddots& \vdots& \vdots \\
\big(\eta_{2^n-2,0}(a) & \eta_{2^n-2,1}(a) & \cdots & \eta_{2^n-2,2^n-2}(a) & \eta_{2^n-2,2^n-1}(a) \\
\big(\eta_{2^n-1,0}(a) & \eta_{2^n-1,1}(a) & \cdots & \eta_{2^n-1,2^n-2}(a) & \eta_{2^n-1,2^n-1}(a)
\end{pmatrix}
\end{align*}
Which simplifies to:
\begin{equation}
\sum
\begin{pmatrix}
0 & 0 & \cdots & 0 & 0 \\
\cdot & \eta_{1,1}(a) & \cdots & \eta_{1,2^n-2}(a) & \eta_{1,2^n-1}(a) \\
\vdots & \vdots & \ddots& \vdots& \vdots \\
\cdot & \cdot & \cdots & \eta_{2^n-2,2^n-2}(a) & \eta_{2^n-2,2^n-1}(a) \\
\cdot & \cdot & \cdots & \cdot & \eta_{2^n-1,2^n-1}(a)
\end{pmatrix}
\end{equation}
Which when $a \in [n]^{\mathbb{T}_n}$ further simplifies to:
\begin{equation}
\!\!\!\!\!\!\!\!\!\!\!\!\sum
\begin{pmatrix}
0 & 0 & \cdots & 0 & 0 \\
\cdot & \eta_{1,1}(a) & \cdots & \eta_{1,2^n-2}(a) & \overline{(1)}\big(a(2^n-1)\big) \\
\vdots & \vdots & \ddots& \vdots& \vdots \\
\cdot & \cdot & \cdots & \eta_{2^n-2,2^n-2}(a) & \overline{(2^n-2)}\big(a(2^n-1)\big) \\
\cdot & \cdot & \cdots & \cdot & 1
\end{pmatrix}
\end{equation}
Note that:
- these matrices are symmetric
- the coefficients of the first line add up to $0$
- the coefficients of the last column add up to:
$\quad\quad
\left\{
\begin{array}{ll}
2^{n-1} & \text{when } a(2^m-1) \in [n]\\
0 & \text{otherwise}
\end{array}
\right.
$
Consequence: two strategies $a$ and $b$ in $[n]^{\mathbb{T}_n}$ which differ only in the choice of $a(2^n-1) \neq b(2^n-1)$ have the same hit score and win rate:
\begin{equation}
\big(\forall t \in \mathbb{T}_n \setminus \{2^n-1\}, \;\; a(t) = b(t)\big) \;\; \implies \;\; \eta(a) = \eta(b).
\label{SameRate}
\end{equation}
Win rate after left reset
Let $a \in \mathbb{S}_n, b \in \mathbb{S}_m$:
\begin{equation}
\mu(b \rightarrow a) = \mu(a) + {\mu(b) \over 4^n}
\label{LeftResetRule}
\end{equation}
Proof:
\begin{align*}
& {\eta(b \rightarrow a) \over 4^m} \\
& \;\;=\;\; \sum
\begin{pmatrix}
\mu(b) & 0 & \cdots & 0 & 0 \\
\cdot & \eta_{1,1}(a) & \cdots & \eta_{1,2^n-2}(a) & \eta_{1,2^n-1}(a) \\
\vdots & \vdots & \ddots& \vdots& \vdots \\
\cdot & \cdot & \cdots & \eta_{2^n-2,2^n-2}(a) & \eta_{2^n-2,2^n-1}(a) \\
\cdot & \cdot & \cdots & \cdot & \eta_{2^n-1,2^n-1}(a)
\end{pmatrix} \\
& \;\;=\;\; \eta(a) + \mu(b) \\
& \;\;=\;\; 4^n \cdot \left(\mu(a) + {\mu(b) \over 4^n}\right) \quad \Box
\end{align*}
Win rate after right reset
This formula works for strategies which choose positions within the tower.
Let $a \in \mathbb{S}_n, b \in \mathbb{S}_m$:
\begin{equation}
\left\{
\begin{array}{l}
a \in [n]^{\mathbb{T}_n} \\
b \in [m]^{\mathbb{T}_m}
\end{array}
\right. \;\; \implies \;\; \mu(a \leftarrow b) = \mu(a) + {\mu(b) \over 4^n}
\label{RightResetRule}
\end{equation}
Proof:
\begin{align*}
& {\eta(a \leftarrow b) \over 4^m}
\\
& \;\;=\;\; \sum \\
& \begin{pmatrix}
0 & 0 & \cdots & 0 & {1 \over 4^m} \sum_{t,u\in\mathbb{T}_m} \big(\overline{(t)}(b(u)) \cdot \overline{(2^n-1)}(a(0))\big) \\
\cdot & \eta_{1,1}(a) & \cdots & \eta_{1,2^n-2}(a) & {1 \over 4^m} \sum_{t,u\in\mathbb{T}_m} \big(\overline{(t)}(b(u)) \cdot \overline{(2^n-1)}(a(1))\big) \\
\vdots & \vdots & \ddots& \vdots& \vdots \\
\cdot & \cdot & \cdots & \eta_{2^n-2,2^n-2}(a) & {1 \over 4^m} \sum_{t,u\in\mathbb{T}_m} \big(\overline{(t)}(b(u)) \cdot \overline{(2^n-1)}(a(2^n-2))\big) \\
\cdot & \cdot & \cdots & \cdot & \mu(b)
\end{pmatrix} \\
\\
& \;\;=\;\; \sum \\
& \begin{pmatrix}
0 & 0 & \cdots & 0 & {1 \over 4^m} \sum_{u\in\mathbb{T}_m}\sum_{t\in\mathbb{T}_m} \overline{(t)}(b(u)) \\
\cdot & \eta_{1,1}(a) & \cdots & \eta_{1,2^n-2}(a) & {1 \over 4^m} \sum_{u\in\mathbb{T}_m}\sum_{t\in\mathbb{T}_m} \overline{(t)}(b(u)) \\
\vdots & \vdots & \ddots& \vdots& \vdots \\
\cdot & \cdot & \cdots & \eta_{2^n-2,2^n-2}(a) & {1 \over 4^m} \sum_{u\in\mathbb{T}_m}\sum_{t\in\mathbb{T}_m} \overline{(t)}(b(u)) \\
\cdot & \cdot & \cdots & \cdot & \mu(b)
\end{pmatrix}
\end{align*}
Now since $0 \le b(u) < m, \quad \sum_{t\in\mathbb{T}_m} (\overline{(t)}(b(u)) = 2^{m-1}$.
Hence:
\begin{align*}
{\eta(a \leftarrow b) \over 4^m} \;\;
&= \;\; \sum
\begin{pmatrix}
0 & 0 & \cdots & 0 & {1 \over 2} \\
\cdot & \eta_{1,1}(a) & \cdots & \eta_{1,2^n-2}(a) & {1 \over 2} \\
\vdots & \vdots & \ddots& \vdots& \vdots \\
\cdot & \cdot & \cdots & \eta_{2^n-2,2^n-2}(a) & {1 \over 2} \\
\cdot & \cdot & \cdots & \cdot & \mu(b)
\end{pmatrix} \\
&= \;\; \eta(a) + \mu(b) \\
&= \;\; 4^n \cdot \left(\mu(a) + {\mu(b) \over 4^n}\right) \quad \Box
\end{align*}
Win rate after double reset
This formula works for strategies which choose positions within the tower.
Let $a \in \mathbb{S}_n, b \in \mathbb{S}_m$:
\begin{equation}
\left\{
\begin{array}{l}
a \in [n]^{\mathbb{T}_n} \\
b \in [m]^{\mathbb{T}_m}
\end{array}
\right. \;\; \implies \;\;
\mu(DR(a, b)) = \mu(a) + {{4 \mu(b) – 1} \over 4^n}
\label{DoubleResetRule}
\end{equation}
Proof:
\begin{align*}
& {\eta(DR(a, b)) \over 4^m} \\
\\
& \;\;=\;\; \sum \\
& \begin{pmatrix}
\mu(b) & 0 & \cdots & 0 & \mu(b) \\
\cdot & \eta_{1,1}(a) & \cdots & \eta_{1,2^n-2}(a) & {1 \over 4^m} \sum_{t,u\in\mathbb{T}_m} \big(\overline{(t)}(b(u)) \cdot \overline{(2^n-1)}(a(1))\big) \\
\vdots & \vdots & \ddots& \vdots& \vdots \\
\cdot & \cdot & \cdots & \eta_{2^n-2,2^n-2}(a) & {1 \over 4^m} \sum_{t,u\in\mathbb{T}_m} \big(\overline{(t)}(b(u)) \cdot \overline{(2^n-1)}(a(2^n-2))\big) \\
\cdot & \cdot & \cdots & \cdot & \mu(b)
\end{pmatrix} \\
\\
& \;\;=\;\; \sum \\
& \begin{pmatrix}
\mu(b) & 0 & \cdots & 0 & \mu(b) \\
\cdot & \eta_{1,1}(a) & \cdots & \eta_{1,2^n-2}(a) & {1 \over 2} \\
\vdots & \vdots & \ddots& \vdots& \vdots \\
\cdot & \cdot & \cdots & \eta_{2^n-2,2^n-2}(a) & {1 \over 2} \\
\cdot & \cdot & \cdots & \cdot & \mu(b)
\end{pmatrix} \\
\\
&= \;\; \eta(a) + 4 \mu(b) – 1 \\
\\
&= \;\; 4^n\left(\mu(a) + {{4 \mu(b) – 1} \over 4^n}\right) \quad \Box
\end{align*}
Win rate after left shift
Let $s \in \mathbb{S}_n$:
\begin{equation}
\mu(\uparrow\!s) = \mu(s) + {1 \over 4^{n+1}}
\label{LeftShiftRule}
\end{equation}
Proof:
Note that the strategy $(\uparrow\!s)$ is equivalent to $([00] \rightarrow s)$.
Hence $(\ref{EquivRate})$ they have the same hit score and win rate. Applying $(\ref{LeftResetRule})$ we get:
\begin{align*}
\mu(\uparrow\!s) \;\;=\;\; \mu([00] \rightarrow s) \;\;=\;\; \mu(s) + {\mu([00]) \over 4^n} \quad \Box
\end{align*}
Win rate after right shift
This formula works for a strategy which chooses positions within the tower.
Let $s \in \mathbb{S}_n$:
\begin{equation}
s \in [n]^{\mathbb{T}_n} \;\;
\implies \;\; \mu(s\!\uparrow) = \mu(s) + {1 \over 4^{n+1}}
\label{RightShiftRule}
\end{equation}
Proof:
Note that the strategies $(s\!\uparrow)$ and $(s \leftarrow [00])$ make the same choices except when presented with a tower with only white hats.
Hence $(\ref{SameRate})$ they have the same hit score and win rate. Applying $(\ref{RightResetRule})$ we get:
\begin{align*}
\mu(\uparrow\!s) \;\;=\;\; \mu(s \leftarrow [00]) \;\;=\;\; \mu(s) + {\mu([00]) \over 4^n} \quad \Box
\end{align*}
Win rate after double shift
This formula works for a strategy which chooses positions within the tower.
Let $s \in \mathbb{S}_n$:
\begin{equation}
s \in [n]^{\mathbb{T}_n} \;\;
\implies \;\; \mu(s\!\Uparrow) = \mu(s) + {1 \over 4^{n+1}}+ {2 \over 4^{n+2}}
\label{DoubleShiftRule}
\end{equation}
Proof:
Consider $s_1 = \uparrow\!(s\!\uparrow)$ and $s_2 = s\!\Uparrow$:
| $s_1 = \uparrow\!(s\!\uparrow)$ | $s_2 = s\!\Uparrow$ |
|---|---|
| $s(0), s(1), \!\cdot\cdot, s(2^n\text{-}2), \pmb{n},$ $s(0), s(1), \!\cdot\cdot, s(2^n\text{-}2), s(2^n\text{-}1),$ $\pmb{n}\!\pmb{+}\!\pmb{1}, s(1), \!\cdot\cdot, s(2^n\text{-}2), \pmb{n},$ $\pmb{s(0)}, s(1), \!\cdot\cdot, s(2^n\text{-}2), s(2^n\text{-}1)$ | $s(0), s(1), \!\cdot\cdot, s(2^n\text{-}2), \pmb{n},$ $s(0), s(1), \!\cdot\cdot, s(2^n\text{-}2), s(2^n\text{-}1),$ $\pmb{n}\!\pmb{+}\!\pmb{1}, s(1), \!\cdot\cdot, s(2^n\text{-}2), \pmb{n},$ $\pmb{n}\!\pmb{+}\!\pmb{1}, s(1), \!\cdot\cdot, s(2^n\text{-}2), s(2^n\text{-}1)$ |
The only difference between $s_1$ and $s_2$ is the position they choose for tower $k = 2^{n+1} + 2^n$:
$$s_2(k) = n + 1 \neq s(0) = s_1(k)$$
Let’s compare the hit scores. Since the $\eta_{t,u}$ are equal unless $t$ or $u$ equals $k$, we have:
\begin{align}
\eta(s_2) – \eta(s_1) \;\; & =\;\; \sum_{t,u \in \mathbb{T}_{n+2}} \big(\eta_{t,u}(s_2) – \eta_{t,u}(s_1)\big) \nonumber \\
& =\;\; 2 \!\!\sum_{t \in \mathbb{T}_{n+2}, t \neq k}\!\!
\big(\eta_{t,k}(s_2) – \eta_{t,k}(s_1)\big) \label{PartialDoubleShift}\\
& \;\;\;\;\;\; + \big(\eta_{k,k}(s_2)\big)^2. \nonumber
\end{align}
Now:
\begin{align}
& \sum_{t \in \mathbb{T}_{n+2}, t \neq k}
\big(\eta_{t,k}(s_2) – \eta_{t,k}(s_1)\big) \nonumber \\
& \quad = \!\!\sum_{t \in \mathbb{T}_{n+2}, t \neq k}\!\!
\big(\overline{(t)}(s_2(k))\cdot\overline{(k)}(s_2(t))
– \overline{(t)}(s_1(k))\cdot\overline{(k)}(s_1(t))\big) \nonumber
\end{align}
Because $k = 2^{n+1} + 2^n$ (all bits null but the two most significant), for the terms in this sum it holds that:
$$
\left\{
\begin{array}{l}
\overline{(k)}(s_2(t)) \neq 0 \implies s_2(t) \ge n
\implies t \in \{2^n-1, 2^{n+1}, 2^{n+1}-1\} \\
\overline{(k)}(s_1(t)) \neq 0 \implies s_1(t) \ge n
\implies t \in \{2^n-1, 2^{n+1}, 2^{n+1}-1\}
\end{array}
\right.
$$
Hence:
\begin{align*}
& \sum_{t \in \mathbb{T}_{n+2}, t \neq k}
\big(\eta_{t,k}(s_2) – \eta_{t,k}(s_1)\big) \\
\\
& \quad = \;\;\; \overline{(2^n-1)}(s_2(k)) \,- \overline{(2^n-1)}(s_1(k)) \\
&\quad \;\;\; + \overline{(2^{n+1})}(s_2(k)) \,- \overline{(2^{n+1})}(s_1(k)) \\
&\quad \;\;\; + \overline{(2^{n+1}-1)}(s_2(k)) \,- \overline{(2^{n+1}-1)}(s_1(k)) \\
\\
& \quad = \;\;\; \overline{(2^n-1)}(n+1) \,- \overline{(2^n-1)}(s(0)) \\
&\quad \;\;\; + \overline{(2^{n+1})}(n+1) \,-\overline{(2^{n+1})}(s(0)) \\
&\quad \;\;\; + \overline{(2^{n+1}-1)}(n+1) \,- \overline{(2^{n+1}-1)}(s(0))
\end{align*}
Finally since $0 \le s(0) < n$:
\begin{align*}
\sum_{t \in \mathbb{T}_{n+2}, t \neq k}
& \big(\eta_{t,k}(s_2) – \eta_{t,k}(s_1)\big) \\
\\
= \quad & \quad\; 0 \quad-\quad 1 \\
& + 1 \quad-\quad 0 \\
& + 1 \quad-\quad 1 \\
\\
= \quad & \quad\; 0
\end{align*}
Going back to $(\ref{PartialDoubleShift})$, we conclude:
\begin{align*}
\eta(s_2) – \eta(s_1) \;\; & =\;\; \big(\eta_{k,k}(s_2)\big)^2 \;\;=\;\; 1 \\
\eta(s\!\Uparrow) \;\; & =\;\; \eta(\uparrow\!(s\!\uparrow)) + 1 \\
& =\;\; \big( 16 \cdot \eta(s) + 4 + 1\big) + 1 \\
& =\;\; 4^{n+2} \cdot \big(\mu(s) + {1 \over 4^{n+1}} + {2 \over 4^{n+2}}\big)
\quad \Box
\end{align*}
Efficient strategies
| Strategy | Height | Win rate | Optimal? |
|---|---|---|---|
| $\mathbf{T}_0 = [-1]$ | $0$ | $0$ | yes |
| $\mathbf{T}_1 = [00]$ | $1$ | ${1 \over 4}$ | yes |
| $\mathbf{W}_2 = [1010]$ | $2$ | ${5 \over 16}$ | yes |
| $\mathbf{B}_2 = [0101]$ | $2$ | ${5 \over 16}$ | yes |
| $\mathbf{C}_2 = [0010]$ | $2$ | ${5 \over 16}$ | yes |
| $\mathbf{C}_3 = \mathbf{R}_1 =$ $[01002120]$ | $3$ | ${21 \over 64}$ | yes |
| $\mathbf{C}_4 =$ $[01002123$ $01002120]$ | $4$ | ${89 \over 256}$ | yes |
| $\mathbf{C}_5 =$ $[01002123$ $01002120$ $41002123$ $41002120]$ | $5$ | ${358 \over 1024}$ | yes |
| $\mathbf{T}_n$ | $n$ | ${1 \over 4}$ | no for $n>1$ |
| $\mathbf{W}_n$ | $n$ | ${1 \over 3}\big(1 – {1 \over 4^n}\big)$ | no for $n>2$ |
| $\mathbf{B}_n$ | $n$ | ${1 \over 3}\big(1 – {1 \over 4^n}\big)$ | no for $n>2$ |
| $\mathbf{C}_{2n+1}$ | $2n+1$ | ${7 \over 20} – {1 \over 10 (16)^n}$ | yes? |
| $\mathbf{C}_{2n+2}$ | $2n+2$ | ${7 \over 20} – {3 \over 80 (16)^n}$ | yes? |
| $\mathbf{R}_n$ | $3n$ | ${7 \over 20} – {1 \over {10 (16)^n}}$ | no for $n>1$ |
Top strategies
The top strategies are obtained by repeatedly double-resetting the strategy $[00]$:
\begin{align}
\mathbf{T}_0 \;
&\stackrel{\text{def}}{=} \; [-1] \\
\forall n \in \mathbb{N}, \quad \mathbf{T}_{n+1} \;
&\stackrel{\text{def}}{=} \; DR([00], \mathbf{T}_n) \nonumber
\end{align}
\begin{align}
\mu(\mathbf{T}_0) &= 0\\
\forall n > 0, \quad \mu(\mathbf{T}_n) &= {1 \over 4}\\
\end{align}
Lowest-white strategies
The lowest-white strategies are obtained by repeatedly left-resetting the strategy $[00]$:
\begin{align}
\mathbf{W}_0 \;
&\stackrel{\text{def}}{=} \; [-1] \\
\forall n \in \mathbb{N}, \quad \mathbf{W}_{n+1} \;
&\stackrel{\text{def}}{=} \; \mathbf{W}_n \rightarrow [00] \nonumber
\end{align}
Applying $(\ref{LeftResetRule})$ we get:
\begin{equation}
\mu(\mathbf{W}_n) = {1 \over 3} \cdot \big(1 – {1 \over 4^n}\big).
\end{equation}
As the height of the towers goes to infinity, the win rate of the lowest-white strategies converges to $1 \over 3$:
\begin{equation}
\lim_{n \to \infty}\mu(\mathbf{W}_n) = {1 \over 3} \approx 33.33\%.
\end{equation}
Lowest-black strategies
The lowest-black strategies are obtained by repeatedly right-resetting the strategy $[00]$:
\begin{align}
\mathbf{B}_0 \;
&\stackrel{\text{def}}{=} \; [-1] \\
\forall n \in \mathbb{N}, \quad \mathbf{B}_{n+1} \;
&\stackrel{\text{def}}{=} \; [00] \leftarrow \mathbf{B}_n \nonumber
\end{align}
Applying $(\ref{RightResetRule})$ we get:
\begin{equation}
\mu(\mathbf{B}_n) = {1 \over 3} \cdot \big(1 – {1 \over 4^n}\big).
\end{equation}
As the height of the towers goes to infinity, the win rate of the lowest-black strategies converges to $1 \over 3$:
\begin{equation}
\lim_{n \to \infty}\mu(\mathbf{B}_n) = {1 \over 3} \approx 33.33\%.
\end{equation}
Carter strategies
| $\mathbf{C}$ | Win rate | Tuple |
|---|---|---|
| $\mathbf{C}_1$ | $1 \over 4$ | $[00]$ |
| $\mathbf{C}_2$ | $5 \over 16$ | $[0100]$ |
| $\mathbf{C}_3$ | $22 \over 64$ | $[01002120]$ |
| $\mathbf{C}_4$ | $89 \over 256$ | $[0100212\text{-}01002120]$ |
| $\mathbf{C}_5$ | $358 \over 1024$ | $[01002123 \text{-} 01002120 \text{-} 41002123 \text{-} 41002120]$ |
| $\mathbf{C}_6$ | $1433 \over 4096$ | $[01002123 \text{-} 01002120 \text{-} 41002123 \text{-} 41002125$ $01002123 \text{-} 01002120 \text{-} 41002123 \text{-} 41002120]$ |
| $\mathbf{C}_7$ | $5734 \over 16384$ | $[01002123 \text{-} 01002120 \text{-} 41002123 \text{-} 41002125$ $01002123 \text{-} 01002120 \text{-} 41002123 \text{-} 41002120$ $61002123 \text{-} 01002120 \text{-} 41002123 \text{-} 41002125$ $61002123 \text{-} 01002120 \text{-} 41002123 \text{-} 41002120]$ |
Carter strategies are defined by repeatedly shifting and double-shifting the strategy $[00]$:
\begin{align}
\mathbf{C}_1 \quad &\stackrel{\text{def}}{=} \quad [00], \nonumber \\
\mathbf{C}_{2n+2} \quad &\stackrel{\text{def}}{=} \quad \mathbf{C}_{2n+1}\!\uparrow, \\
\mathbf{C}_{2n+3} \quad &\stackrel{\text{def}}{=} \quad \mathbf{C}_{2n+1}\!\Uparrow. \nonumber
\end{align}
Applying $(\ref{LeftShiftRule})$ & $(\ref{DoubleShiftRule})$ we get:
\begin{align}
\mu(\mathbf{C}_{2n+1}) \quad &= \quad {7 \over 20} – {1 \over 10 \cdot (16)^n}, \\
\mu(\mathbf{C}_{2n+2}) \quad &= \quad {7 \over 20} – {3 \over 80 \cdot (16)^n}.
\end{align}
As the height of the towers goes to infinity, the win rate of the Carter strategies converges to $7 \over 20$:
\begin{equation}
\lim_{n \to \infty}\mu(\mathbf{C}_n) = {7 \over 20} = 35\%.
\end{equation}
Reyes strategies
| $\mathbf{R}$ | Win rate | Tuple |
|---|---|---|
| $\mathbf{R}_1$ | $22 \over 64$ | $[21002122]$ |
| $\mathbf{R}_2$ | $358 \over 1024$ | $[51002125 \text{-} 41002124 \text{-} 31002123 \text{-} 31002123$ $51002125 \text{-} 41002124 \text{-} 51002125 \text{-} 51002125]$ |
| $\mathbf{R}_3$ | $5734 \over 16384$ | $[61002126 \text{-} 41002124 \text{-} 31002123 \text{-} 31002123$ $51002125 \text{-} 41002124 \text{-} 51002125 \text{-} 61002126$ $71002127 \text{-} 41002124 \text{-} 31002123 \text{-} 31002123$ $51002125 \text{-} 41002124 \text{-} 51002125 \text{-} 71002127$ $61002126 \text{-} 41002124 \text{-} 31002123 \text{-} 31002123$ $51002125 \text{-} 41002124 \text{-} 51002125 \text{-} 61002126$ $61002126 \text{-} 41002124 \text{-} 31002123 \text{-} 31002123$ $51002125 \text{-} 41002124 \text{-} 51002125 \text{-} 61002126$ $81002128 \text{-} 41002124 \text{-} 31002123 \text{-} 31002123$ $51002125 \text{-} 41002124 \text{-} 51002125 \text{-} 81002128$ $71002127 \text{-} 41002124 \text{-} 31002123 \text{-} 31002123$ $51002125 \text{-} 41002124 \text{-} 51002125 \text{-} 71002127$ $81002128 \text{-} 41002124 \text{-} 31002123 \text{-} 31002123$ $51002125 \text{-} 41002124 \text{-} 51002125 \text{-} 81002128$ $61002126 \text{-} 41002124 \text{-} 31002123 \text{-} 31002123$ $51002125 \text{-} 41002124 \text{-} 51002125 \text{-} 61002126]$ |
Reyes strategies are defined by repeatedly double-resetting the Carter strategy $\mathbf{C}_3$:
\begin{align}
\mathbf{R}_1 \;
&\stackrel{\text{def}}{=} \; \mathbf{C}_3 \\
\forall n > 0, \quad \mathbf{R}_{n+1} \;
&\stackrel{\text{def}}{=} \; DR(\mathbf{C}_3, \mathbf{R}_n). \nonumber
\end{align}
Applying $(\ref{DoubleResetRule})$ we get:
\begin{equation}
\mu(\mathbf{R}_n) = {7 \over 20} – {1 \over {10 \cdot (16)^n}}.
\end{equation}
As the height of the towers goes to infinity, the win rate of the Reyes strategies converges to $7 \over 20$:
\begin{equation}
\lim_{n \to \infty}\mu(\mathbf{R}_n) = {7 \over 20} = 35\%.
\end{equation}
Recap and Whereto?
I came across the game of hats through Numberphile’s . Then Google pointed to the interesting article by Stan Wagon .
Wagon provides a state of the art as of 2014:
- Gives a list of best known symmetric strategies due to Carter and Reyes, up to $n = 8$ (hence the christening of the $\mathbf{C}_n$ and $\mathbf{R}_n$ strategies in this article). Carter and Reyes strategies are proven optimal by brute force for $n < 5$. For $n = 5$ Wagon mentions an indirect proof by Jim Roche and him, further confirmed by a different proof by Rob Pratt. For $n > 5$ we don’t know.
- Mentions a proof by Walter Stromquist that $15\over40$ is an upper bound for the win rate of any symmetric strategy.
- Tells about asymmetric strategies (it is believed that they do not lead to better the win rates than the symmetric strategies).
- Tells about playing the game with more than two players.
- Shows how to leverage the axiom of choice to define a strategy for the game with countably infinitely many hats.
Wagon’s state of the art doesn’t give the explicit value of the Carter strategies beyond $\mathbf{C}_8$, nor does it give a procedural way to construct them. It even asks the question: « What about n = 9 or larger? ». By contrast this article provides an explicit construction of $\mathbf{C}_n$ for all $n > 0$.
In the next article I plan to describe a « hat calculator » which implements the operations defined in this article and lets you play with strategies. It also implements an efficient brute-force algorithm to look for optimal strategies (other than Carter’s).
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