Subfield: Combinatorics


Question: Five identical empty buckets of 2-liter capacity stand at the vertices of a regular pentagon. Cinderella and her wicked Stepmother go through a sequence of rounds: At the beginning of every round, the Stepmother takes one liter of water from the nearby river and distributes it arbitrarily over the five buckets. Then Cinderella chooses a pair of neighboring buckets, empties them into the river, and puts them back. Then the next round begins. The Stepmother's goal is to make one of these buckets overflow. Cinderella's goal is to prevent this. Can the wicked Stepmother enforce a bucket overflow?

Solution:
No, the Stepmother cannot enforce a bucket overflow and Cinderella can keep playing forever. Throughout we denote the five buckets by $B_{0}, B_{1}, B_{2}, B_{3}$, and $B_{4}$, where $B_{k}$ is adjacent to bucket $B_{k-1}$ and $B_{k+1}(k=0,1,2,3,4)$ and all indices are taken modulo 5 . Cinderella enforces that the following three conditions are satisfied at the beginning of every round:

(1) Two adjacent buckets (say $B_{1}$ and $B_{2}$ ) are empty.

(2) The two buckets standing next to these adjacent buckets (here $B_{0}$ and $B_{3}$ ) have total contents at most 1.

(3) The remaining bucket (here $B_{4}$ ) has contents at most 1 .

These conditions clearly hold at the beginning of the first round, when all buckets are empty.

Assume that Cinderella manages to maintain them until the beginning of the $r$-th round $(r \geq 1)$. Denote by $x_{k}(k=0,1,2,3,4)$ the contents of bucket $B_{k}$ at the beginning of this round and by $y_{k}$ the corresponding contents after the Stepmother has distributed her liter of water in this round.

By the conditions, we can assume $x_{1}=x_{2}=0, x_{0}+x_{3} \leq 1$ and $x_{4} \leq 1$. Then, since the Stepmother adds one liter, we conclude $y_{0}+y_{1}+y_{2}+y_{3} \leq 2$. This inequality implies $y_{0}+y_{2} \leq 1$ or $y_{1}+y_{3} \leq 1$. For reasons of symmetry, we only consider the second case.

Then Cinderella empties buckets $B_{0}$ and $B_{4}$.

At the beginning of the next round $B_{0}$ and $B_{4}$ are empty (condition (1) is fulfilled), due to $y_{1}+y_{3} \leq 1$ condition (2) is fulfilled and finally since $x_{2}=0$ we also must have $y_{2} \leq 1$ (condition (3) is fulfilled).

Therefore, Cinderella can indeed manage to maintain the three conditions (1)-(3) also at the beginning of the $(r+1)$-th round. By induction, she thus manages to maintain them at the beginning of every round. In particular she manages to keep the contents of every single bucket at most 1 liter. Therefore, the buckets of 2-liter capacity will never overflow.
We prove that Cinderella can maintain the following two conditions and hence she can prevent the buckets from overflow:

$\left(1^{\prime}\right)$ Every two non-adjacent buckets contain a total of at most 1.

$\left(2^{\prime}\right)$ The total contents of all five buckets is at most $\frac{3}{2}$.

We use the same notations as in the first solution. The two conditions again clearly hold at the beginning. Assume that Cinderella maintained these two conditions until the beginning of the $r$-th round. A pair of non-neighboring buckets $\left(B_{i}, B_{i+2}\right), i=0,1,2,3,4$ is called critical



if $y_{i}+y_{i+2}>1$. By condition $\left(2^{\prime}\right)$, after the Stepmother has distributed her water we have $y_{0}+y_{1}+y_{2}+y_{3}+y_{4} \leq \frac{5}{2}$. Therefore,

$$
\left(y_{0}+y_{2}\right)+\left(y_{1}+y_{3}\right)+\left(y_{2}+y_{4}\right)+\left(y_{3}+y_{0}\right)+\left(y_{4}+y_{1}\right)=2\left(y_{0}+y_{1}+y_{2}+y_{3}+y_{4}\right) \leq 5
$$

and hence there is a pair of non-neighboring buckets which is not critical, say $\left(B_{0}, B_{2}\right)$. Now, if both of the pairs $\left(B_{3}, B_{0}\right)$ and $\left(B_{2}, B_{4}\right)$ are critical, we must have $y_{1}<\frac{1}{2}$ and Cinderella can empty the buckets $B_{3}$ and $B_{4}$. This clearly leaves no critical pair of buckets and the total contents of all the buckets is then $y_{1}+\left(y_{0}+y_{2}\right) \leq \frac{3}{2}$. Therefore, conditions $\left(1^{\prime}\right)$ and $\left(2^{\prime}\right)$ are fulfilled.

Now suppose that without loss of generality the pair $\left(B_{3}, B_{0}\right)$ is not critical. If in this case $y_{0} \leq \frac{1}{2}$, then one of the inequalities $y_{0}+y_{1}+y_{2} \leq \frac{3}{2}$ and $y_{0}+y_{3}+y_{4} \leq \frac{3}{2}$ must hold. But then Cinderella can empty $B_{3}$ and $B_{4}$ or $B_{1}$ and $B_{2}$, respectively and clearly fulfill the conditions.

Finally consider the case $y_{0}>\frac{1}{2}$. By $y_{0}+y_{1}+y_{2}+y_{3}+y_{4} \leq \frac{5}{2}$, at least one of the pairs $\left(B_{1}, B_{3}\right)$ and $\left(B_{2}, B_{4}\right)$ is not critical. Without loss of generality let this be the pair $\left(B_{1}, B_{3}\right)$. Since the pair $\left(B_{3}, B_{0}\right)$ is not critical and $y_{0}>\frac{1}{2}$, we must have $y_{3} \leq \frac{1}{2}$. But then, as before, Cinderella can maintain the two conditions at the beginning of the next round by either emptying $B_{1}$ and $B_{2}$ or $B_{4}$ and $B_{0}$.