Subfield: Combinatorics


Question: Players $A$ and $B$ play a paintful game on the real line. Player $A$ has a pot of paint with four units of black ink. A quantity $p$ of this ink suffices to blacken a (closed) real interval of length $p$. In every round, player $A$ picks some positive integer $m$ and provides $1 / 2^{m}$ units of ink from the pot. Player $B$ then picks an integer $k$ and blackens the interval from $k / 2^{m}$ to $(k+1) / 2^{m}$ (some parts of this interval may have been blackened before). The goal of player $A$ is to reach a situation where the pot is empty and the interval $[0,1]$ is not completely blackened.

Prove that there doesn't exist a strategy for player $A$ to win in a finite number of moves.

Solution:
We will present a strategy for player $B$ that guarantees that the interval $[0,1]$ is completely blackened, once the paint pot has become empty.

At the beginning of round $r$, let $x_{r}$ denote the largest real number for which the interval between 0 and $x_{r}$ has already been blackened; for completeness we define $x_{1}=0$. Let $m$ be the integer picked by player $A$ in this round; we define an integer $y_{r}$ by

$$
\frac{y_{r}}{2^{m}} \leqslant x_{r}<\frac{y_{r}+1}{2^{m}}
$$

Note that $I_{0}^{r}=\left[y_{r} / 2^{m},\left(y_{r}+1\right) / 2^{m}\right]$ is the leftmost interval that may be painted in round $r$ and that still contains some uncolored point.

Player $B$ now looks at the next interval $I_{1}^{r}=\left[\left(y_{r}+1\right) / 2^{m},\left(y_{r}+2\right) / 2^{m}\right]$. If $I_{1}^{r}$ still contains an uncolored point, then player $B$ blackens the interval $I_{1}^{r}$; otherwise he blackens the interval $I_{0}^{r}$. We make the convention that, at the beginning of the game, the interval $[1,2]$ is already blackened; thus, if $y_{r}+1=2^{m}$, then $B$ blackens $I_{0}^{r}$.

Our aim is to estimate the amount of ink used after each round. Firstly, we will prove by induction that, if before $r$ th round the segment $[0,1]$ is not completely colored, then, before this move,

(i) the amount of ink used for the segment $\left[0, x_{r}\right]$ is at most $3 x_{r}$; and

(ii) for every $m, B$ has blackened at most one interval of length $1 / 2^{m}$ to the right of $x_{r}$.

Obviously, these conditions are satisfied for $r=0$. Now assume that they were satisfied before the $r$ th move, and consider the situation after this move; let $m$ be the number $A$ has picked at this move.

If $B$ has blackened the interval $I_{1}^{r}$ at this move, then $x_{r+1}=x_{r}$, and $(i)$ holds by the induction hypothesis. Next, had $B$ blackened before the $r$ th move any interval of length $1 / 2^{m}$ to the right of $x_{r}$, this interval would necessarily coincide with $I_{1}^{r}$. By our strategy, this cannot happen. So, condition (ii) also remains valid.

Assume now that $B$ has blackened the interval $I_{0}^{r}$ at the $r$ th move, but the interval $[0,1]$ still contains uncolored parts (which means that $I_{1}^{r}$ is contained in $[0,1]$ ). Then condition (ii) clearly remains true, and we need to check $(i)$ only. In our case, the intervals $I_{0}^{r}$ and $I_{1}^{r}$ are completely colored after the $r$ th move, so $x_{r+1}$ either reaches the right endpoint of $I_{1}$ or moves even further to the right. So, $x_{r+1}=x_{r}+\alpha$ for some $\alpha>1 / 2^{m}$.

Next, any interval blackened by $B$ before the $r$ th move which intersects $\left(x_{r}, x_{r+1}\right)$ should be contained in $\left[x_{r}, x_{r+1}\right]$; by (ii), all such intervals have different lengths not exceeding $1 / 2^{m}$, so the total amount of ink used for them is less than $2 / 2^{m}$. Thus, the amount of ink used for the segment $\left[0, x_{r+1}\right]$ does not exceed the sum of $2 / 2^{m}, 3 x_{r}$ (used for $\left[0, x_{r}\right]$ ), and $1 / 2^{m}$ used for the



segment $I_{0}^{r}$. In total it gives at most $3\left(x_{r}+1 / 2^{m}\right)<3\left(x_{r}+\alpha\right)=3 x_{r+1}$. Thus condition $(i)$ is also verified in this case. The claim is proved.

Finally, we can perform the desired estimation. Consider any situation in the game, say after the $(r-1)$ st move; assume that the segment $[0,1]$ is not completely black. By $(i i)$, in the segment $\left[x_{r}, 1\right]$ player $B$ has colored several segments of different lengths; all these lengths are negative powers of 2 not exceeding $1-x_{r}$; thus the total amount of ink used for this interval is at most $2\left(1-x_{r}\right)$. Using $(i)$, we obtain that the total amount of ink used is at most $3 x_{r}+2\left(1-x_{r}\right)<3$. Thus the pot is not empty, and therefore $A$ never wins.