IRL ABooklet

Let $x$ be the number of heads and $y$ the number of tails at each point of the game. The pair $(x, y)$ completely describes the state of the game. Let $D(x, y) = x - y$. This quantity behaves as follows under each possible move:

a. $(x, y) \to (y, x)$ changes $D(x, y)$ to $-D(x, y)$.

b. $(x, y) \to (3x, 3y)$ changes $D(x, y)$ to $3D(x, y)$.

c. If $y \ge 4$, move $(x, y) \to (x + 4, y - 4)$ changes $D(x, y)$ to $D(x, y) + 8$.

d. If $y \ge 5$, move $(x, y) \to (x + 3, y - 5)$ changes $D(x, y)$ to $D(x, y) + 8$.

We will indicate composition of moves using the symbol $\circ$ so that, for example, (b)$\circ$(a) is the move (a) followed by move (b), which sends $(x, y)$ to $(3y, 3x)$, and $(b)^m$ is the $m$-fold composition of (b) sending $(x, y)$ to $(3^m x, 3^m y)$. We introduce the following moves:
$$(c') = (a) \circ (c) \circ (a): \text{ If } x \ge 4, \text{ it sends } (x, y) \text{ to } (x - 4, y + 4).$$
$$(d') = (a) \circ (d) \circ (a): \text{ If } x \ge 5, \text{ it sends } (x, y) \text{ to } (x - 3, y + 5).$$
We also define a move (e) which sends $(x, y)$ to $(x - 1, y - 1)$ if $x$ and $y$ are both positive and one of them is at least $4$:
$$(e) = \begin{cases} (d') \circ (c) & \text{if } x \ge 1, y \ge 4 \\ (d) \circ (c') & \text{if } x \ge 4, y \ge 1. \end{cases}$$
Finally, we define moves (f) and (f') as follows:
$$(f) = (c) \circ (e)^4: \text{ If } x \ge 4 \text{ and } y \ge 8, \text{ it sends } (x, y) \text{ to } (x, y - 8).$$
$$(f') = (c') \circ (e)^4: \text{ If } x \ge 8 \text{ and } y \ge 4, \text{ it sends } (x, y) \text{ to } (x - 8, y).$$
To prove that each pair $(2024, 8n)$ with $n \ge 0$ can be reached during the game, we first apply move (b) $k$ times, where $3^k \ge a$, and then move (e) a number of times to get
$$(1, 1) \to (3, 3) \to (9, 9) \to \dots \to (3^k, 3^k) \to (3^k - 1, 3^k - 1) \to \dots \to (a, a).$$
This shows that we can get any pair $(a, a)$ with $a \ge 3$. In particular, we can get any pair $(8n, 8n)$ with $n \ge 1$. If $8n \ge 2024 = 8 \cdot 253$ we apply move $(f')$ a number of times to reach $(2024, 8n)$ for any $n \ge 253$. To reach $(2024, 8n)$ for $0 \le n < 253$, we start with $(2024, 2024)$ and apply move (f) a suitable number of times. Hence, for any $n \ge 0$ we can reach $(2024, 8n)$ from $(1, 1)$ after a finite number of moves.