Question: Let $n$ be a positive integer. A frog starts on the number line at $0$. Suppose it makes a finite sequence of hops, subject to two conditions: (i) The frog visits only points in $\{1, 2, \dots, 2^n-1\}$, each at most once. (ii) The length of each hop is in $\{2^0, 2^1, 2^2, \dots\}$. (The hops may be either direction, left or right.) Let $S$ be the sum of the (positive) lengths of all hops in the sequence. What is the maximum possible value of $S$?

We claim the answer is $\frac{4^n - 1}{3}$. We first prove the upper bound. Notice that the hop sizes must be in $\{2^0, 2^1, \dots, 2^{n - 1}\}$, since the frog must stay within the bounds of $\{0, 1, \dots, 2^n-1\}$ the whole time and the maximum distance between any two points in this set is $2^n-1 < 2^n$. Let $a_i$ be the number of hops of size $2^i$ the frog makes, for $0\le i\le n - 1$.

**Claim:** For any $k = 1, \dots, n$, we have $a_{n-1} + \dots + a_{n-k} \le 2^n - 2^{n-k}$.

**Proof:** Let $m = n-k$ and consider the positions modulo $2^m$. Call a jump *small* if its length is at most $2^{m-1}$, and *large* if it is at least $2^m$. A small jump changes the residue class of the frog modulo $2^m$, while a large jump does not. Within each fixed residue class modulo $2^m$, there are exactly $\frac{2^n}{2^m}$ points in $\{0, 1, \dots, 2^n-1\}$. Since each point is visited at most once, the frog can make at most $\frac{2^n}{2^m} - 1$ large jumps between points of this same residue class. Summing over all $2^m$ residue classes modulo $2^m$, the total number of large jumps is at most $2^m \left( \frac{2^n}{2^m} - 1 \right) = 2^n - 2^m$. This exactly states that $a_{n-1} + \dots + a_m \le 2^n - 2^m$, proving the claim.

Now, the total distance the frog travels is $S = a_0 + 2a_1 + 4a_2 + \dots + 2^{n-1} a_{n-1}$. We rewrite this using summation by parts. Let $A_k = a_{n-1} + \dots + a_{n-k}$ for $k = 1, \dots, n$, with $A_0 = 0$. Then $a_{n-k} = A_k - A_{k-1}$, and we can express $S$ as:
$$ S = A_n + A_{n-1} \cdot 2^0 + A_{n-2} \cdot 2^1 + \dots + A_1 \cdot 2^{n-2}. $$
Using the bounds $A_k \le 2^n - 2^{n-k}$, we get:
$$ S \le (2^n-2^0) + (2^n-2^1) + 2(2^n-2^2) + \dots + 2^{n-2}(2^n-2^{n-1}). $$
Evaluating this sum yields:
$$ S \le 2^n - 1 + \sum_{k=1}^{n-1} 2^{n-k-1}(2^n - 2^{n-k}) $$
$$ = 2^n - 1 + 2^{2n-1} \sum_{k=1}^{n-1} 2^{-k} - 2^{2n-1} \sum_{k=1}^{n-1} 4^{-k} $$
$$ = 2^n - 1 + 2^{2n-1} \left(1 - 2^{-(n-1)}\right) - 2^{2n-1} \frac{1/4 - 4^{-n}}{3/4} $$
$$ = 2^n - 1 + 2^{2n-1} - 2^n - \frac{2^{2n-1} - 2}{3} $$
$$ = \frac{3 \cdot 2^{2n-1} - 2^{2n-1} + 2 - 3}{3} = \frac{4^n - 1}{3}. $$

It remains to show that equality can hold. One can construct two families of paths such that there are $2^k$ hops of size $2^k$ for every $0\le k\le n - 1$, and the frog visits each point in $\{0, \dots, 2^n - 1\}$ exactly once, starting on $0$ and ending on $x$, for the two values $x\in\{1, 2^n - 1\}$. This achieves the theoretical maximum, completing the proof.

(Partial) 1. Observed that every hop length must be at most $2^{n-1}$ (since all visited points stay in $\{0,1,\dots,2^n-1\}$ and any hop of length $\ge 2^n$ is impossible). 2. Introduced counts $a_i$ of hops of length $2^i$ and derived a nontrivial global restriction on the numbers of “large” jumps using residues mod $2^m$ (e.g. proved for some $m$ that the number of hops with length $\ge 2^m$ is at most $2^n-2^m$). 3. Gave a construction idea that visits each point at most once and achieves total hop-sum $S=\sum_{k=0}^{n-1} 2^{2k}=\frac{4^n-1}{3}$ (or at least correctly constructed a family with $2^k$ hops of size $2^k$ for all $k$), even if the upper bound is not proved. (Almost) 1. Proved the key bound $a_{n-1}+\cdots+a_m\le 2^n-2^m$ for each $m$ (equivalently $A_k\le 2^n-2^{n-k}$), and set up the summation-by-parts expression for $S$ in terms of the $A_k$, but made a localized algebra/summation error when evaluating the final bound. 2. Proved the full upper bound $S\le \frac{4^n-1}{3}$ correctly, but the construction for equality is incomplete (e.g. only described informally, or verified the “visits each point exactly once” property for one endpoint but not the other). 3. Gave both the upper bound method and a matching construction, but left a minor gap in the residue-class argument (such as not justifying why large jumps preserve the residue mod $2^m$, or mishandling the count “$\frac{2^n}{2^m}-1$ per class”), while the rest of the solution is correct.


Question: Ana and Banana are playing a game. First Ana picks a word, which is defined to be a nonempty sequence of capital English letters. Then Banana picks a nonnegative integer $k$ and challenges Ana to supply a word with exactly $k$ subsequences which are equal to Ana's word. Ana wins if she is able to supply such a word, otherwise she loses. For example, if Ana picks the word “TST”, and Banana chooses $k = 4$, then Ana can supply the word “TSTST” which has $4$ subsequences which are equal to Ana's word. Which words can Ana pick so that she can win no matter what value of $k$ Banana chooses?

First we introduce some notation. Define a *block* of letters to be a maximal contiguous subsequence of consecutive identical letters. For example, the word $AABBBCAAA$ has four blocks, namely $AA$, $BBB$, $C$, and $AAA$. Throughout the solution, we fix the word $A$ that Ana picks, and introduce the following notation for its $m$ blocks:
$$A = A_1 A_2 \dots A_m = \underbrace{a_1 \dots a_1}_{x_1} \underbrace{a_2 \dots a_2}_{x_2} \dots \underbrace{a_m \dots a_m}_{x_m}.$$
A *rainbow* will be defined as a subsequence equal to Ana's initial word $A$ (meaning Ana seeks words with exactly $k$ rainbows). For brevity, let $A_i = \underbrace{a_i \dots a_i}_{x_i}$, so $A = A_1 \dots A_m$.

We prove two claims that resolve the problem.

**Claim 1:** If $x_i = 1$ for some $i$, then for any $k \ge 1$, the word
$$W = A_1 \dots A_{i-1} \underbrace{a_i \dots a_i}_{k} A_{i+1} \dots A_m$$
obtained by repeating the $i$th letter $k$ times has exactly $k$ rainbows.

*Proof:* Obviously there are at least $\binom{k}{k-1}=k$ rainbows, obtained by deleting $k-1$ choices of the letter $a_i$ in the repeated block. We show they are the only ones. Given a rainbow, consider the location of this singleton block in $W$. It cannot occur within the first $|A_1| + \dots + |A_{i-1}|$ letters, nor can it occur within the final $|A_{i+1}| + \dots + |A_m|$ letters. So it must appear in the $i$th block of $W$. That implies that all the other $a_i$'s in the $i$th block of $W$ must be deleted, as desired.

**Claim 2:** If $x_i \ge 2$ for all $i$, then no word $W$ has exactly two rainbows.

*Proof:* We prove that if there are two rainbows of $W$, then we can construct at least three rainbows. Let $W = w_1 \dots w_n$ and consider two rainbows of $W$. Since they are not the same, there must be a block $A_p$ of the rainbow, of length $\ell \ge 2$, which does not occupy the same locations in $W$. Assume the first rainbow uses $w_{i_1}, \dots, w_{i_\ell}$ for this block and the second rainbow uses $w_{j_1}, \dots, w_{j_\ell}$ for this block. Then among the letters $w_q$ for $\min(i_1, j_1) \le q \le \max(i_\ell, j_\ell)$, there must be at least $\ell+1$ copies of the letter $a_p$. Moreover, given a choice of $\ell$ copies of the letter $a_p$ in this range, one can complete the subsequence to a rainbow. So the number of rainbows is at least $\binom{\ell+1}{\ell} \ge \ell+1$. Since $\ell \ge 2$, this proves $W$ has at least three rainbows.

In summary, Ana wins if and only if $x_i = 1$ for some $i$, since she can duplicate the isolated letter $k$ times; but if $x_i \ge 2$ for all $i$ then Banana only needs to supply $k = 2$.

Ana wins if and only if at least one letter appears exactly once in her word.

(Partial) 1. Proved the sufficiency direction: if Ana’s word contains a letter that occurs exactly once (equivalently, some block has length 1), then for any k one can repeat that letter k times in the supplied word and obtain exactly k subsequences equal to Ana’s word. 2. Proved a usable obstruction when every letter occurs at least twice (e.g. every block has length ≥2): showed that Banana can choose a specific small k (typically k=2) that cannot be realized, by arguing that having at least two target subsequences forces the existence of a third. 3. Introduced a correct block/runs decomposition of Ana’s word and used it to relate different embeddings of the word as a subsequence to a lower bound on the number of subsequences. (Almost) 1. Correctly identified the winning criterion (“Ana wins iff at least one letter appears exactly once”), and proved the construction for all k, but did not fully justify uniqueness of the counted k subsequences (e.g. did not rule out extra subsequences using the repeated block in a different way). 2. Proved the necessity direction for the case “all blocks have length ≥2” by showing ‘2 subsequences ⇒ at least 3’, but left a localized gap in the combinatorial argument (e.g. why a shifted block forces at least ℓ+1 choices, or why any such choice extends to a full subsequence). 3. Solution establishes both directions with the right main ideas, but contains a minor counting or boundary-case slip (such as k=0 or k=1 not explicitly handled) that does not undermine the overall structure.


Question: In the coordinate plane are finitely many *walls*, which are disjoint line segments, none of which are parallel to either axis. A bulldozer starts at an arbitrary point and moves in the $+x$ direction. Every time it hits a wall, it turns at a right angle to its path, away from the wall, and continues moving. (Thus the bulldozer always moves parallel to the axes.) Prove that it is impossible for the bulldozer to hit both sides of every wall.

We say a wall $v$ is *above* another wall $w$ if some point on $v$ is directly above a point on $w$. (This relation is antisymmetric, as walls do not intersect.) The critical claim is as follows:

\begin{claim*}
There exists a lowest wall, i.e., a wall not above any other walls.
\end{claim*}

\begin{proof}
Assume not. Then we get a directed cycle of some length $n \ge 3$: it is possible to construct a series of points $P_i$ and $Q_i$ for $i = 1, \dots, n$ (indices modulo $n$), such that the point $Q_i$ is directly above $P_{i+1}$ for each $i$, the segment $\overline{Q_i P_{i+1}}$ does not intersect any wall in its interior, and finally each segment $\overline{P_i Q_i}$ is contained inside a wall. This gives us a broken line on $2n$ vertices which is not self-intersecting. Now consider the leftmost vertical segment $\overline{Q_i P_{i+1}}$ and the rightmost vertical segment $\overline{Q_j P_{j+1}}$. The broken line gives a path from $P_{i+1}$ to $Q_j$, as well as a path from $P_{j+1}$ to $Q_i$. These clearly must intersect, which is a contradiction.
\end{proof}

Thus, if the bulldozer eventually moves upwards indefinitely, it will never hit the bottom side of the lowest wall. Similarly, if the bulldozer eventually moves downwards indefinitely, it will never hit the upper side of the highest wall.

(Partial) 1. Introduced a correct invariant/partial order on walls (e.g. “v is above w if some point of v is vertically above some point of w”), and justified that this relation is antisymmetric because walls are disjoint. 2. Proved (or correctly reduced to proving) that among finitely many walls there exists a “lowest” wall (one not above any other), or dually a “highest” wall, and explained why such an extremal wall would obstruct hitting both of its sides. 3. Established a one-sided trapping statement: if the bulldozer ever starts moving vertically upward (resp. downward) without bound, then it can never hit the bottom side of a lowest wall (resp. the top side of a highest wall). (Almost) 1. Gave the full extremal-wall strategy and concluded the impossibility, but left a localized gap in proving existence of a lowest/highest wall (e.g. the “no directed cycle” argument is sketched but not rigorously justified). 2. Proved existence of a lowest (or highest) wall via the intended cycle/planarity contradiction, and correctly argued that one side of that wall is unhit, but did not fully justify the step that the bulldozer’s motion must eventually go upward indefinitely or downward indefinitely (missing a short case split). 3. Solution is almost complete, but made minor mistakes which are not negligible.


Question:
Find all integers $n \ge 2$ for which there exists a sequence of $2n$ pairwise distinct points $(P_1, \dots, P_n, Q_1, \dots, Q_n)$ in the plane satisfying the following four conditions:
\begin{enumerate}[label={(\roman*)}]
\item no three of the $2n$ points are collinear;
\item $P_iP_{i+1} \ge 1$ for all $i=1,2,\dots,n$, where $P_{n+1}=P_1$;
\item $Q_iQ_{i+1} \ge 1$ for all $i=1,2,\dots,n$, where $Q_{n+1}=Q_1$;
\item $P_i Q_j \le 1$ for all $i=1,2,\dots,n$ and $j=1,2,\dots,n$.
\end{enumerate}

The answer is even integers only.

\paragraph{Proof that even $n$ work.}
If we ignore the conditions that the points are pairwise distinct and form no collinear triples, we may take
\[
P_{2i-1}=(0.51, 0), \quad P_{2i}=(-0.51, 0), \quad Q_{2i-1}=(0, 0.51), \quad Q_{2i}=(0,-0.51)
\]
for $i = 1, 2, \dots, n/2$. The distances $P_iP_{i+1}$ and $Q_iQ_{i+1}$ (including $P_nP_1$ and $Q_nQ_1$) are $1.02 > 1$, while all distances $P_iQ_j$ are $0.51\sqrt{2} \approx 0.72 < 1$. Because these inequalities are strict, we may perturb each point by a sufficiently small amount to ensure that the strict distance inequalities still hold, while making the points pairwise distinct and ensuring no three points are collinear.

\paragraph{Proof that odd $n$ do not work.}
The main claim is the following.
\begin{claim*}
For $1\leq i\leq n$, points $Q_i$ and $Q_{i+1}$ must lie on opposite sides of line $P_1P_2$.
\end{claim*}
To isolate the geometry component of the problem, we rewrite the claim in the following contrapositive form, without referencing the points $Q_i$ and $Q_{i+1}$:
\begin{lemma*}
Suppose $P_1, P_2$ are points with $P_1P_2 \ge 1$, and $A, B$ are two points such that $\max(P_1A, P_1B, P_2A, P_2B) \le 1$. Moreover, assume $A$ and $B$ lie on the same side of line $P_1 P_2$, and no three of $\{P_1, P_2, A, B\}$ are collinear. Then $AB < 1$.
\end{lemma*}
\begin{proof}[Proof of Lemma]
Since $A$ and $B$ lie on the same side of $P_1P_2$, the convex hull of these four points is either a quadrilateral or a triangle.
\begin{itemize}
    \item If the convex hull is a quadrilateral, assume without loss of generality that the vertices are $P_1, P_2, A, B$ in order. Let $X$ denote the intersection of the diagonals $P_1A$ and $P_2B$. Since $P_1P_2 \ge 1$, we have
    \[
    1 + AB \leq P_1P_2 + AB < (P_1X + XP_2) + (AX + XB) = P_1A + P_2B \leq 2.
    \]
    Thus, $AB < 1$.
    \item Otherwise, the convex hull is a triangle. Since $A$ and $B$ are on the same side of $P_1P_2$, the segment $P_1P_2$ must be an edge of this convex hull. Assume without loss of generality that $B$ is in the interior of triangle $P_1P_2A$. Since $\angle P_1BA + \angle P_2BA = 360^\circ-\angle P_1BP_2 > 180^\circ$, at least one of $\angle P_1BA$ and $\angle P_2BA$ is obtuse. Assume without loss of generality the former angle is obtuse; then in $\triangle P_1BA$, the side opposite the obtuse angle is strictly the longest, so $AB < P_1A \leq 1$.
\end{itemize}
\end{proof}
Applying the lemma to $A = Q_i$ and $B = Q_{i+1}$, we are given $P_1P_2 \ge 1$ and $P_1Q_i, P_1Q_{i+1}, P_2Q_i, P_2Q_{i+1} \leq 1$. Since $Q_iQ_{i+1} \geq 1$, the points $Q_i$ and $Q_{i+1}$ cannot lie on the same side of $P_1P_2$. This proves the claim.

It follows from the claim that $Q_i$ is on the same side of line $P_1P_2$ as $Q_1$ if $i$ is odd, and on the opposite side if $i$ is even. Since $Q_1 = Q_{n+1}$, this alternation means the construction is not possible when $n$ is odd.

(Partial) 1. Gave a correct construction for infinitely many even n (e.g. using two alternating locations for the Pi and two alternating locations for the Qj with strict inequalities), with a brief explanation that small perturbations can enforce “pairwise distinct” and “no three collinear” while preserving all distance inequalities. 2. Proved the key geometric lemma/claim: if P1P2≥1 and A,B satisfy P1A,P1B,P2A,P2B≤1 and A,B lie on the same side of line P1P2 (with no three collinear), then AB<1; and correctly applied it to deduce that Qi and Qi+1 must lie on opposite sides of P1P2. 3. From the “opposite sides” claim, correctly derived the parity alternation of the Qi relative to line P1P2 and concluded that odd n is impossible (even if the even-n construction is missing). (Almost) 1. Correctly proved that odd n do not work via the opposite-sides lemma/claim and the resulting alternation argument, but the even-n existence part is missing or only sketched without a perturbation/distinctness justification. 2. Gave the even-n construction and the parity obstruction for odd n, but left a localized gap in the lemma proof (e.g. the convex-hull case split/diagonal inequality or the obtuse-angle argument is not fully justified). 3. Solution is almost complete, but made minor mistakes which are not negligible.


Question: There are $2022$ equally spaced points on a circular track $\gamma$ of circumference $2022$. The points are labeled $A_1$, $A_2$, \dots, $A_{2022}$ in some order, each label used once. Initially, Bunbun the Bunny begins at $A_1$. She hops along $\gamma$ from $A_1$ to $A_2$, then from $A_2$ to $A_3$, until she reaches $A_{2022}$, after which she hops back to $A_1$. When hopping from $P$ to $Q$, she always hops along the shorter of the two arcs $\arc{PQ}$ of $\gamma$; if $\overline{PQ}$ is a diameter of $\gamma$, she moves along either semicircle. Determine the maximal possible sum of the lengths of the $2022$ arcs which Bunbun traveled, over all possible labellings of the $2022$ points.

Replacing $2022$ with $2n$, the answer is $2n^2 - 2n + 2$. (When $n=1011$, the number is $2042222$.)

\paragraph{Construction.} Let the $2n$ points along the track be labeled $0, 1, \dots, 2n-1$ in order. The maximum distance is achieved by the sequence of points $A_{2i-1} = i-1$ and $A_{2i} = n+i-1$ for $i=1, 2, \dots, n$. (For example, when $n=5$, the sequence of points is $0, 5, 1, 6, 2, 7, 3, 8, 4, 9$.) This corresponds to alternating between points in the two halves of the circle.

\paragraph{First proof of bound.} Let $d_i$ be the shorter distance from $A_{2i-1}$ to $A_{2i+1}$.

\begin{claim*}
The distance of the leg of the journey $A_{2i-1} \to A_{2i} \to A_{2i+1}$ is at most $2n - d_i$.
\end{claim*}

\begin{proof}
The point $A_{2i}$ lies on one of the two arcs from $A_{2i-1}$ to $A_{2i+1}$, which have lengths $d_i$ and $2n-d_i$. The sum of the shorter arc distances $A_{2i-1} \to A_{2i}$ and $A_{2i} \to A_{2i+1}$ is at most the length of the arc containing $A_{2i}$. Thus, we get a bound of $\max(d_i, 2n-d_i) = 2n - d_i$.
\end{proof}

That means the total distance is at most
\[
\sum_{i=1}^n \left( 2n - d_i \right) = 2n^2 - (d_1 + d_2 + \dots + d_n).
\]

\begin{claim*}
We have
\[
d_1 + d_2 + \dots + d_n \ge 2n-2.
\]
\end{claim*}

\begin{proof}
The left-hand side is the total length of the closed walk $A_1 \to A_3 \to \dots \to A_{2n-1} \to A_1$. Among the $n$ distinct points visited, two of them must be at a distance of at least $n-1$ apart. The closed walk is divided into two paths connecting these two points. If one path consists of a single step, its length is at least $n-1$, and the other path consists of $n-1$ steps which contribute at least $1$ each, giving a sum of at least $(n-1) + (n-1) \cdot 1 = 2n-2$. More generally, by the triangle inequality, each of the two paths connecting the two points has length at least $n-1$, so the total length of the walk is always at least $2(n-1) = 2n-2$.
\end{proof}

\paragraph{Second proof of bound.} Draw the $n$ diameters through the midpoints of the $2n$ arcs between adjacent points on the track.

\begin{claim*}[Interpretation of distances]
The distance between any two points equals the number of diameters crossed to travel between the points.
\end{claim*}

\begin{proof}
Clear, as each step of length $1$ along the shorter arc between two points crosses exactly one such diameter.
\end{proof}

With this in mind, call a diameter \emph{critical} if it is crossed by all $2n$ arcs.

\begin{claim*}
At most one diameter is critical.
\end{claim*}

\begin{proof}
Suppose there were two critical diameters; these divide the circle into four sectors, each containing at least one of the $2n$ points. Then all $2n$ arcs cross both diameters, and so every hop must travel between opposite sectors. But this means that the sequence of hops alternates between two opposite sectors, so the points in the other two sectors are never accessed --- contradiction.
\end{proof}

\begin{claim*}
Every diameter is crossed an even number of times.
\end{claim*}

\begin{proof}
Clear: crossing a diameter moves Bunbun from one semicircle to the opposite one, so the diameter needs to be crossed an even number of times for the closed loop to return to its origin.
\end{proof}

This immediately implies that the maximum possible total distance is achieved when one diameter is crossed all $2n$ times, and every other diameter is crossed $2n-2$ times, for a total distance of at most
\[
n \cdot (2n-2) + 2 = 2n^2 - 2n + 2.
\]

(Partial) 1. Gave (and correctly evaluated) a concrete labeling achieving total distance $2n^2-2n+2$ (e.g. alternating between opposite halves: $0,n,1,n+1,\dots,n-1,2n-1$). 2. Proved the key local bound for a 2-step leg: for $d_i$ the shorter distance from $A_{2i-1}$ to $A_{2i+1}$, showed \(\operatorname{dist}(A_{2i-1},A_{2i})+\operatorname{dist}(A_{2i},A_{2i+1})\le 2n-d_i\). 3. Introduced the diameter-crossing interpretation (distance equals number of crossed mid-diameters) and proved at least one of: “every diameter is crossed an even number of times” or “at most one diameter can be crossed by all $2n$ hops (critical diameter)”. (Almost) 1. Obtained the global upper bound $\sum (2n-d_i)=2n^2-\sum d_i$ and argued $\sum d_i\ge 2n-2$, but left a gap in justifying $\sum d_i\ge 2n-2$ (e.g. the choice of two odd-index points at distance $\ge n-1$ or the triangle-inequality argument for the two connecting paths). 2. Used the diameter-crossing method to derive the bound $2n^2-2n+2$, but did not fully justify one of the structural claims (e.g. why at most one diameter is critical, or why each diameter is crossed an even number of times in the closed tour). 3. Solution gives the correct maximum value and a matching construction, but has a minor indexing/boundary omission (e.g. handling the final hop $A_{2n}\to A_1$ or the diameter case) while the main method is correct.


Question:
Let $\alpha \ge 1$ be a real number.
Hephaestus and Poseidon play a turn‑based game on an infinite grid of unit squares.
Before the game starts, Poseidon chooses a finite number of cells to be flooded.
Hephaestus is building a levee, which is a subset of unit edges of the grid, called walls, forming a connected, non‑self‑intersecting path or loop.

The game then begins with Hephaestus moving first.
On each of Hephaestus's turns, he adds one or more walls to the levee, as long as the total length of the levee is at most $\alpha n$ after his $n$th turn.
On each of Poseidon's turns, every cell which is adjacent to an already flooded cell and with no wall between them becomes flooded as well.

Hephaestus wins if the levee forms a closed loop such that all flooded cells are contained in the interior of the loop — hence stopping the flood and saving the world.
For which $\alpha$ can Hephaestus guarantee victory in a finite number of turns no matter how Poseidon chooses the initial cells to flood?

We show that if $\alpha > 2$ then Hephaestus wins, but when $\alpha \le 2$ Hephaestus cannot contain even a single-cell flood initially.

\bigskip

**Strategy for $\alpha > 2$**:
Impose $\mathbb{Z}^2$ coordinates on the cells. Adding more flooded cells does not make our task easier, so let us assume that initially the cells $(x,y)$ with $|x|+|y| \le d$ are flooded for some $d \ge 2$; thus on Hephaestus's $k$-th turn, the water is contained in $|x|+|y| \le d+k-1$. Our goal is to contain the flood with a large rectangular levee.

We pick large integers $N_1$ and $N_2$ such that
\[
  \alpha N_1 > 2N_1 + (2d + 3),\qquad
  \alpha (N_1+N_2) > 2N_2 + (6N_1 + 8d + 4).
\]
We construct the levee by defining a sequence of points and segments. The lengths of the segments to be added are as follows:
\begin{itemize}
  \item $X_1 Y_1$ is a horizontal wall of length $1$ placed to the north of the flood.
  \item $X_1 X_2$ and $Y_1 Y_2$ are horizontal extensions of length $N_1$ to the west and east, respectively.
  \item $X_4 X_3 X_2$ and $Y_4 Y_3 Y_2$ are broken lines of length $d+1$ each that turn southwards.
  \item $X_4 X_5$ and $Y_4 Y_5$ are vertical extensions of length $N_2$ to the south.
  \item $X_5 X_6 Y_6 Y_5$ is a broken line of length $4N_1 + 6d + 1$ that closes the loop at the south.
\end{itemize}

We follow the following plan:
\begin{itemize}
  \item \textbf{Turn $1$}: place wall $X_1 Y_1$. This cuts off the flood to the north.
  \item \textbf{Turns $2$ through $N_1+1$}: extend the levee to segment $X_2 Y_2$. This prevents further flooding to the north.
  \item \textbf{Turn $N_1+2$}: add in broken lines $X_4 X_3 X_2$ and $Y_4 Y_3 Y_2$ all at once. This cuts off the flood west and east.
  \item \textbf{Turns $N_1+3$ to $N_1+N_2+2$}: extend the levee along vertical segments $X_4 X_5$ and $Y_4 Y_5$. This prevents further flooding west and east.
  \item \textbf{Turn $N_1 + N_2 + 3$}: add in the broken line $X_5 X_6 Y_6 Y_5$ all at once and win.
\end{itemize}

\bigskip

**Proof for $\alpha \le 2$**:
Suppose Hephaestus contains the flood on his $(n+1)$-st turn. We prove that $\alpha > 2$ by showing that in fact at least $2n+4$ walls have been constructed.

Let $c_0, c_1, \dots, c_n$ be a path of cells such that $c_0$ is the initial cell flooded, and in general $c_i$ is flooded on Poseidon's $i$-th turn from $c_{i-1}$. The levee now forms a closed loop enclosing all $c_i$.

\begin{claim*}
  If $c_i$ and $c_j$ are adjacent then $|i-j|=1$.
\end{claim*}
\begin{proof}
  Assume $c_i$ and $c_j$ are adjacent but $|i-j|>1$. Then the two cells must be separated by a wall. But the levee forms a simple closed loop, so any wall of the levee separates its interior from its exterior. This would mean $c_i$ and $c_j$ are on opposite sides of the levee, contradicting that both are flooded and thus contained in the interior.
\end{proof}

Thus the cells $c_i$ actually form a simple path. We color \textit{green} any edge of the unit grid (wall or not) which is an edge of exactly one $c_i$ (i.e., the boundary of the polyomino). It is easy to see there are exactly $2n+4$ green edges.

Now, from the center of each cell $c_i$, shine a laser towards each green edge of $c_i$ (hence a total of $2n+4$ lasers are emitted).

\begin{claim*}
  No wall is hit by more than one laser.
\end{claim*}
\begin{proof}
  Assume for contradiction that a wall $w$ is hit by lasers from $c_i$ and $c_j$. Without loss of generality, assume that the laser is vertical, so $c_i$ and $c_j$ are in the same column.

  We consider two cases on the position of $w$:
  \begin{itemize}
    \item If $w$ is between $c_i$ and $c_j$, then the line segment connecting the centers of $c_i$ and $c_j$ intersects the levee exactly once (at $w$). However, both $c_i$ and $c_j$ lie in the interior of the levee. Any segment connecting two interior points of a simple closed loop must intersect the loop an even number of times, a contradiction.

    \item Suppose $w$ lies above both $c_i$ and $c_j$, and assume WLOG $i < j$. Since the laser from $c_i$ hits $w$ without hitting any walls in between, there is no levee at all between $c_i$ and $c_j$. Let $\rho \ge 1$ be the distance between the centers of $c_i$ and $c_j$. Since there are no walls between them, $c_j$ can be flooded in a straight line from $c_i$ within $\rho$ turns, and this straight line is the unique shortest possible path. Thus, this situation can only occur if $j = i+\rho$ and the cells $c_i, c_{i+1}, \dots, c_j$ form a contiguous column. But if they form a column, the top edge of $c_i$ is shared with $c_{i+1}$, meaning it is not a green edge, so no upward vertical laser can be emitted from $c_i$, a contradiction.
  \end{itemize}
  Since neither case is possible, the claim is proven.
\end{proof}

This implies the levee has at least $2n+4$ walls (the number of lasers) on Hephaestus's $(n+1)$-st turn. So $\alpha \ge \frac{2n+4}{n+1} > 2$.

(Partial) 1. For the winning direction, reduced to the case where the initially flooded region is a diamond/ball in Manhattan distance (e.g. all cells with |x|+|y|<=d), and correctly tracked that after k Poseidon moves the flood is contained in |x|+|y|<=d+k. 2. Gave a coherent construction idea for α>2: start with a short wall to block one side and then build a large rectangular (or similar) loop with long horizontal/vertical extensions, using the fact that Hephaestus may add many walls in one turn provided the total length constraint is met. 3. For the impossibility direction (α<=2), exhibited the flooded cells c0,c1,...,cn along a flooding chain and proved the key structural claim that if ci and cj are adjacent then |i−j|=1, hence the flooded cells form a simple path. (Almost) 1. Presented a complete strategy for α>2 based on building a large enclosing rectangle in stages (blocking north, extending left/right, turning down, extending down, then closing), with the correct inequalities on chosen parameters, but left a minor gap in checking that the flood cannot “leak around” an unfinished corner during the intermediate turns. 2. Proved α>2 is necessary by counting at least 2n+4 required walls from a length-n+1 containment time, including the green-edge/laser injectivity idea (no wall hit by two lasers), but had a localized flaw (e.g. parity/intersection argument, or handling of the ‘wall above both cells’ case) while the rest of the argument is correct. 3. Solution is almost complete, but made minor mistakes which are not negligible.


Question:
A \emph{snake of length $k$} is an animal which occupies an ordered $k$-tuple $(s_1, \dots, s_k)$ of cells in an $n \times n$ grid of square unit cells. These cells must be pairwise distinct, and $s_i$ and $s_{i+1}$ must share a side for $i=1,\dots,k-1$. If the snake is currently occupying $(s_1, \dots, s_k)$ and $s$ is an unoccupied cell sharing a side with $s_1$, the snake can \emph{move} to occupy $(s, s_1, \dots, s_{k-1})$ instead. The snake has \emph{turned around} if it occupied $(s_1, s_2, \dots, s_k)$ at the beginning, but after a finite number of moves occupies $(s_k, s_{k-1}, \dots, s_1)$ instead.

Determine whether there exists an integer $n > 1$ such that one can place some snake of length at least $0.9n^2$ in an $n \times n$ grid which can turn around.

The answer is yes. We can show that for any $c < 1$, there exists an $n$ such that a snake of length at least $c n^2$ can turn around in an $n \times n$ grid. In particular, this holds for $c = 0.9$.

Let $G$ be any undirected graph. Consider a snake of length $k$ lying within $G$, with each segment of the snake occupying one vertex, consecutive segments occupying adjacent vertices, and no two segments occupying the same vertex. One move of the snake consists of the snake's head advancing to an adjacent empty vertex and segment $i$ advancing to the vertex of segment $i - 1$ for $i = 2, 3, \dots, k$.

The proof proceeds in two stages. First, we construct a planar graph $G$ such that it is possible for a snake that occupies nearly all of $G$ to turn around inside $G$. Then, we construct a subgraph $H$ of the $n \times n$ grid graph such that $H$ is isomorphic to $G$ and $H$ occupies nearly all of the grid.

**Stage 1: Construction of the graph $G$**
Let $r$ and $\ell$ be positive integers. Start with $r$ disjoint *main* paths $p_1, p_2, \dots, p_r$, each of length at least $\ell$, with $p_i$ leading from $A_i$ to $B_i$ for $i = 1, 2, \dots, r$. Add to those $r$ *linking* paths, one leading from $B_i$ to $A_{i + 1}$ for each $i = 1, 2, \dots, r - 1$, and one leading from $B_r$ to $A_1$. Finally, add two families of *transit* paths: the first family contains one transit path joining $A_1$ to each of $A_2, A_3, \dots, A_r$, and the second family contains one path joining $B_r$ to each of $B_1, B_2, \dots, B_{r - 1}$. We require that all paths specified in the construction have no interior vertices in common, with the exception that transit paths in the same family may share vertices.

We claim that a snake of length $(r - 1)\ell$ can turn around inside $G$.

Let the concatenation $A_1 \to B_1 \to A_2 \to B_2 \to \dots \to A_r \to B_r \to A_1$ of all main and linking paths be the *great cycle*. We refer to this direction as the counterclockwise orientation of the great cycle, and the reverse direction as its clockwise orientation.

Place the snake so that its tail is at $A_1$ and its body extends counterclockwise along the great cycle. Then let the snake maneuver as follows. (We track only the snake's head, as its movement uniquely determines the movement of the complete body of the snake.)

- **Phase 1:** Advance counterclockwise along the great cycle to $B_{r - 1}$, take a detour along a transit path to $B_r$, and advance clockwise along the great cycle to $A_r$.
- **Phase $i$ (for $i = 2, 3, \dots, r - 1$):** Take a detour along a transit path to $A_1$, advance counterclockwise along the great cycle to $B_{r - i}$, take a detour along a transit path to $B_r$, and advance clockwise along the great cycle to $A_{r - i + 1}$.
- **Phase $r$:** Simply advance clockwise along the great cycle to $A_1$.

Throughout these phases, because the snake has length $(r - 1)\ell$ and each main path has length at least $\ell$, the required transit paths and segments of the great cycle are guaranteed to be unoccupied when the head enters them. By the end of phase $r$, the snake's tail is at $A_1$ and its body extends clockwise along the great cycle, meaning it has successfully reversed its orientation and turned around.

**Stage 2: Embedding $G$ into the grid**
Let $n$ be a sufficiently large positive integer. Consider an $n \times n$ grid $S$. Number the columns of $S$ from $1$ to $n$ from left to right, and its rows from $1$ to $n$ from bottom to top.

Let $a_1, a_2, \dots, a_{r + 1}$ be cells of $S$ such that all of them lie in column $2$, with $a_1$ in row $2$, $a_{r + 1}$ in row $n - 1$, and the cells approximately equally spaced. Let $b_1, b_2, \dots, b_r$ be cells of $S$ such that all of them lie in column $n - 2$, and $b_i$ lies in the same row as $a_{i + 1}$ for $i = 1, 2, \dots, r$.

Construct the subgraph $H \subset S$ as follows:
- For $i = 1, 2, \dots, r$, let the main path from $a_i$ to $b_i$ be a serpentine path that fills up the rectangle bounded by the rows and columns of $a_i$ and $b_i$ nearly completely. By doing so, every main path has a length of approximately $\frac{1}{r}n^2$.
- For $i = 1, 2, \dots, r - 1$, let the linking path that leads from $b_i$ to $a_{i + 1}$ lie inside the row shared by $b_i$ and $a_{i + 1}$.
- Let the linking path that leads from $b_r$ to $a_1$ lie inside row $n$, column $n$, and row $1$.
- Lastly, let the union of the first family of transit paths be column $1$, and let the union of the second family of transit paths be column $n - 1$, excluding their bottommost and topmost squares.

This grid subgraph $H$ is isomorphic to $G$. As shown in the first stage, a snake of length $k$ approximately equal to $\frac{r - 1}{r}n^2$ can turn around inside $H$, and therefore inside the $n \times n$ grid $S$.

When $r$ is fixed and $n$ tends to infinity, the ratio $\frac{k}{n^2}$ tends to $\frac{r - 1}{r}$. Furthermore, as $r$ tends to infinity, $\frac{r - 1}{r}$ tends to $1$. By choosing $r \ge 10$ and $n$ sufficiently large, we can achieve a snake of length at least $0.9n^2$ that can turn around, which completes the proof.

(Partial) 1. Reformulated the problem on an arbitrary graph, observing that it suffices to build a subgraph of the grid in which a long snake can reverse orientation. 2. Described a concrete “turnaround gadget”/scheme (e.g. a large cycle with detours/transit paths) and explained why the head can use a detour to start moving along the cycle in the opposite direction. 3. Gave a credible embedding idea into an n×n grid (e.g. using long serpentine paths to realize long disjoint paths and using one or two columns as shared transit corridors), with a correct density estimate in the limit (such as achieving length about (r−1)/r·n^2 for fixed r and large n). (Almost) 1. Constructed the graph G with main/linking/transit paths and gave the phase-based maneuver that reverses the snake, but left a small gap in the “unoccupied when needed” verification (e.g. an inequality involving ℓ, r, and how far the tail has moved). 2. Correctly embedded (or outlined embedding of) G into an n×n grid with serpentine main paths and transit columns, and obtained length ≥0.9n^2 for some choice of parameters, but omitted a localized check that the specified paths are internally vertex-disjoint / fit without unintended intersections at corners or endpoints. 3. Solution is almost complete, but made minor mistakes which are not negligible.


Question: Let $n$ be a positive integer and let $S \subseteq \{0,1\}^n$ be a set of binary strings of length $n$. Given an odd number $x_1, \dots, x_{2k+1} \in S$ of binary strings (not necessarily distinct), their *majority* is defined as the binary string $y \in \{0,1\}^n$ for which the $i$\textsuperscript{th} bit of $y$ is the most common bit among the $i$\textsuperscript{th} bits of $x_1$, \dots, $x_{2k+1}$. (For example, if $n=4$ the majority of $0000$, $0000$, $1101$, $1100$, $0101$ is $0100$.) Suppose that for some positive integer $k$, $S$ has the property $P_k$ that the majority of any $2k+1$ binary strings in $S$ (possibly with repetition) is also in $S$. Prove that $S$ has the same property $P_k$ for all positive integers $k$.

Let $M$ denote the majority function (of any length).

**First solution (induction).** We prove all $P_k$ are equivalent by induction on $n \ge 2$, with the base case $n = 2$ being easy to check by hand. (The case $n=1$ is also vacuous; however, the inductive step is not able to go from $n=1$ to $n=2$.)

For the inductive step, we proceed by contradiction; assume $S$ satisfies $P_{\ell}$, but not $P_{k}$, so there exist $x_1, \dots, x_{2k+1} \in S$ whose majority $y = M(x_1, \dots, x_{2k+1})$ is not in $S$. We contend that:

**Claim:** Let $y_i$ be the string which differs from $y$ only in the $i$\textsuperscript{th} bit. Then $y_i \in S$.

*Proof.* For a string $s \in S$ we let $\hat s$ denote the string $s$ with the $i$\textsuperscript{th} bit deleted (hence with $n-1$ bits). Now let
\[
T = \left\{ \hat s \mid s \in S \right\}.
\]
Since $S$ satisfies $P_\ell$, so does $T$; thus by the induction hypothesis on $n$, $T$ satisfies $P_{k}$. Consequently, $T \ni M(\hat{x}_1, \dots, \hat{x}_{2k+1}) = \hat y$. Thus there exists $s \in S$ such that $\hat s = \hat y$. This implies $s = y$ or $s = y_i$. But since we assumed $y \notin S$ it follows $y_i \in S$ instead.

Now take any $2\ell+1$ copies of the $y_i$, about equally often (i.e., the number of times any two $y_i$ are taken differs by at most $1$). We see the majority of these is $y$ itself, contradiction.

**Second solution (circuit construction).** Note that $P_k \implies P_1$ for any $k$, since
\[
M( \underbrace{a,\dots,a}_k, \underbrace{b,\dots,b}_k, c ) = M(a,b,c)
\]
for any $a$, $b$, $c$. We will now prove $P_1 + P_k \implies P_{k+1}$ for any $k$, which will prove the result. Actually, we will show that the majority of any $2k+3$ strings $x_1$, \dots, $x_{2k+3}$ can be expressed by $3$-majorities and $(2k+1)$-majorities. WLOG assume that $M(x_1, \dots, x_{2k+3}) = 0\dots0$, and let $\odot$ denote binary AND.

**Claim:** We have $M(x_1, x_2, M(x_3, \dots, x_{2k+3})) = x_1 \odot x_2$.

[Proof continues with circuit construction arguments]

(Partial) 1. Showed that for any k, property P_k implies P_1 (e.g. by padding M(a,b,c) as M(a,…,a,b,…,b,c) with k copies). 2. Set up the “delete one coordinate” reduction: defined \(\hat s\) by deleting the i-th bit, defined \(T=\{\hat s:s\in S\}\), and correctly observed that if S has P_k then T has P_k (majority commutes with coordinate deletion). 3. In a contradiction setup (S has P_\ell but fails P_k with majority y\notin S), proved the key claim that for each i the neighbor \(y_i\) obtained by flipping the i-th bit lies in S (using the reduction to T and induction on n, or any other valid argument producing all \(y_i\in S\)). (Almost) 1. Presented a complete contradiction argument via coordinate deletion and induction on n, and derived \(y_i\in S\) for all i, but did not fully justify the final multiset choice of \(2\ell+1\) copies of the \(y_i\) whose majority is exactly y (e.g. the “almost equally often” counting step is not checked). 2. Used the circuit/closure approach: proved P_k\(\Rightarrow\)P_1 and established (or nearly established) the recursive step P_1 + P_k \(\Rightarrow\) P_{k+1} by expressing a (2k+3)-majority using 3-majorities and (2k+1)-majorities, but left one of the circuit identities/boolean-case checks (or the final induction on k) with a small gap. 3. Solution is almost complete, but made minor mistakes which are not negligible.


Question: In a sports league, each team uses a set of at most $t$ signature colors. A set $S$ of teams is *color-identifiable* if one can assign each team in $S$ one of their signature colors, such that no team in $S$ is assigned *any* signature color of a different team in $S$. For all positive integers $n$ and $t$, determine the maximum integer $g(n,t)$ such that: In any sports league with exactly $n$ distinct colors present over all teams, one can always find a color-identifiable set of size at least $g(n,t)$.

The answer is $\lceil n/t \rceil$.

To see that $\lceil n/t \rceil$ is an upper bound, note that one can easily construct a sports league with exactly $\lceil n/t \rceil$ teams in total. (Since the total number of teams in the league is $\lceil n/t \rceil$, no subset of teams can be strictly larger than this.)

A quick warning: It is common to misread the definition of color-identifiable by ignoring the word "any". Here is an illustration. Suppose we have two teams, MIT and Harvard; the colors of MIT are red, gray, and black, and the colors of Harvard are red and white. (Thus $n=4$ and $t=3$.) The assignment of MIT to gray and Harvard to red is *not* acceptable because red is a signature color of MIT, even though it is not the one assigned to MIT.

We present two proofs of the lower bound.

**Approach by deleting teams (Gopal Goel):**
Initially, place all teams in a set $S$. Then we repeat the following algorithm: If there is a team all of whose signature colors are shared by some other team in $S$ already, then we delete that team. (If there is more than one such team, we pick one arbitrarily.) At the end of the process, all $n$ colors are still present at least once, so at least $\lceil n/t \rceil$ teams remain. Moreover, since the algorithm is no longer possible, the remaining set $S$ is already color-identifiable.

It might seem counter-intuitive that we are *deleting* teams from the full set when the original problem is trying to get a large set $S$. This is less strange when one thinks of it instead as "safely deleting useless teams". Basically, if one deletes such a team, the problem statement implies that the task must still be possible, since $g(n,t)$ does not depend on the number of teams: $n$ is the number of colors present, and deleting a useless team does not change this. It turns out that this optimization is already enough to solve the problem.

**Approach by adding colors:**
For a constructive algorithmic approach, the idea is to greedily pick by color (rather than by team), taking at each step the least used color. Select the color $C_1$ with the *fewest* teams using it, and a team $T_1$ using it. Then delete all colors $T_1$ uses, and all teams which use $C_1$. Note that by the problem condition, this deletes at most $t$ colors total. Any remaining color $C$ still has at least one user. Indeed, if not, then $C$ had the same set of teams as $C_1$ did (by the minimality of $C$), but then it should have been deleted as a color of $T_1$. Now repeat this algorithm with $C_2$ and $T_2$, and so on. This operation uses at most $t$ colors each time, so we select at least $\lceil n/t \rceil$ colors, corresponding to at least $\lceil n/t \rceil$ teams.

A greedy approach by team *does not work*. For example, suppose we try to "grab teams until no more can be added". As before, assume our league has teams MIT and Harvard; the colors of MIT are red, gray, and black, and the colors of Harvard are red and white. (Thus $n=4$ and $t=3$.) If we start by selecting MIT and red, then it is impossible to select any more teams; but $g(n,t) = 2$.

(Partial) 1. Gave a correct extremal construction showing g(n,t) \le \lceil n/t\rceil (e.g. a league with exactly \lceil n/t\rceil teams using all n colors, each team using at most t colors). 2. Made the key observation that in any color-identifiable set S, each chosen team must have an assigned color that appears in no other team of S (equivalently, the assigned colors are pairwise disjoint from all other teams’ color-sets). 3. Presented a valid deletion or greedy framework (e.g. iteratively delete a team all of whose colors are shared with other remaining teams, or greedily select a least-used color and remove related colors/teams), even if the proof that it produces at least \lceil n/t\rceil teams is incomplete. (Almost) 1. Proved the lower bound \ge \lceil n/t\rceil using the deletion algorithm, but did not fully justify one local step (e.g. why the process stops with a color-identifiable set, or why all n colors remain present so the remaining number of teams is at least \lceil n/t\rceil). 2. Proved the lower bound via the color-greedy algorithm and correctly argued ‘at most t colors are removed per step’, but left a gap in the minimality argument that every remaining color still has a user, or did not clearly deduce that the selected teams can be assigned distinct identifying colors. 3. Established both bounds and stated g(n,t)=\lceil n/t\rceil, but made a minor definitional slip (typically mis-handling the word “any” in ‘no team is assigned any signature color of a different team’) in an example or sub-argument while the main proof structure is otherwise correct.


Question: You are cheating at a trivia contest. For each question, you can peek at each of the $n > 1$ other contestant's guesses before writing your own. For each question, after all guesses are submitted, the emcee announces the correct answer. A correct guess is worth $0$ points. An incorrect guess is worth $-2$ points for other contestants, but only $-1$ point for you, because you hacked the scoring system. After announcing the correct answer, the emcee proceeds to read out the next question. Show that if you are leading by $2^{n-1}$ points at any time, then you can surely win first place.

We will prove the result with $2^{n-1}$ replaced by the stronger bound $2^{n-2}+1$. We first make the following reductions. First, change the weights to be $+1$, $-1$, $0$ respectively for a correct answer, an incorrect answer by another contestant, and an incorrect answer by you (rather than $0$, $-2$, $-1$); this has no effect on the relative score differences. Without loss of generality, assume that all contestants except you initially have a score of zero, and that your score exceeds $2^{n-2}$. We may ignore rounds in which all answers are the same. Finally, we can ignore rounds in which you get the correct answer, since that leaves you at least as well off relative to the other contestants as before. In other words, we will assume your score is always fixed, but you can pick any group of people with the same answer and ensure they lose $1$ point, while the group of people with the correct answer gains $1$ point.

The key observation is the following. Consider two rounds $R_1$ and $R_2$ such that: in round $R_1$, some set $S$ of contestants gains a point; in round $R_2$, the set $S$ of contestants all have the same answer. Then, if we copy the answers of contestants in $S$ during $R_2$ (which is an incorrect answer, as we ignore rounds where we answer correctly), they lose a point in $R_2$, which cancels out the point they gained in $R_1$. For any contestant not in $S$, they lost a point in $R_1$ and gain at most one point in $R_2$, so their net score change across the two rounds is at most $0$. In other words, the combined effect of $R_1$ and $R_2$ does not increase any contestant's score, so we can ignore $R_1$ and $R_2$ forever.

We thus consider the following strategy. We keep a list $\mathcal{L}$ of subsets of $\{1, \dots, n\}$, initially empty. On each round, we apply the following rules:
Suppose there exists a set $S$ of people with the same answer such that $S \in \mathcal{L}$. (If multiple such sets exist, choose one arbitrarily.) Then, copy the answer of $S$, causing them to lose a point. Delete $S$ from $\mathcal{L}$. (Importantly, we do not add any new sets to $\mathcal{L}$ in this case.)

Otherwise, copy the answer of any set $T$ of contestants, selecting $T$ such that $|T| \ge n/2$ if possible. Let $S$ be the set of contestants who answer correctly (if any), and add $S$ to the list $\mathcal{L}$. Note that $|S| \le n/2$; if a set of size at least $n/2$ exists, we chose $|T| \ge n/2$, and since $S$ is disjoint from $T$ (as we assume we did not copy the correct answer), $|S| \le n - |T| \le n/2$. If no set of size at least $n/2$ exists, then $|S| < n/2$ holds automatically.

By construction, $\mathcal{L}$ has no duplicate sets. So the score of any contestant $c$ is bounded above by the number of times that $c$ appears among sets in $\mathcal{L}$. The number of sets in $\mathcal{L}$ containing $c$ is at most the number of subsets of $\{1, \dots, n\}$ containing $c$ of size at most $n/2$. This is equivalent to the number of subsets of an $(n-1)$-element set of size at most $(n-2)/2$, which is at most $\frac{1}{2} \cdot 2^{n-1} = 2^{n-2}$. So, if you lead by $2^{n-2}+1$, then you ensure victory. This completes the proof.

(Partial) 1. Makes the standard reduction to an equivalent "+1/-1/0" scoring (or otherwise argues that only score differences matter) and assumes WLOG that only other contestants’ scores need to be controlled while your score stays fixed. 2. States and correctly justifies the key two-round cancellation idea: if a set S gained in some earlier round and later all contestants in S give the same answer, then by copying that answer in the later round one can ensure the earlier gain of S is neutralized while no other contestant’s total increases across the two rounds. 3. Introduces the bookkeeping strategy of maintaining a list/multiset of previously correct sets S and: (i) when a current equal-answer group matches an S on the list, copy them and delete S; (ii) otherwise copy some large equal-answer group and add the current correct set S to the list, with an explanation why the added sets have size at most n/2. (Almost) 1. Presents the full list-of-sets strategy and proves an upper bound on every opponent’s score via counting appearances in the list, but has a small gap in the combinatorial estimate (e.g., bounding the number of subsets containing a fixed contestant of size ≤ n/2) or mishandles the parity/boundary case when n is odd. 2. Completes the argument that no opponent can ever exceed your initial lead (hence you win) but does not fully justify one reduction step (e.g., “ignore rounds where all answers are the same” or “ignore rounds where you answer correctly”) or does not clearly argue that duplicates cannot appear in the maintained list. 3. Solution is almost complete, but made minor mistakes which are not negligible.


Question: A physicist encounters $2015$ atoms called usamons. Each usamon either has one electron or zero electrons, and the physicist can't tell the difference. The physicist's only tool is a diode. The physicist may connect the diode from any usamon $A$ to any other usamon $B$. (This connection is directed.) When she does so, if usamon $A$ has an electron and usamon $B$ does not, then the electron jumps from $A$ to $B$. In any other case, nothing happens. In addition, the physicist cannot tell whether an electron jumps during any given step. The physicist's goal is to isolate two usamons that she is 100 % sure are currently in the same state. Is there any series of diode usage that makes this possible?

The answer is no.

Let $m = 2015$ and label the usamons $U_1, \dots, U_m$. Consider a family of $m+1$ initial configurations, which we will call models $M_0, M_1, \dots, M_m$. In model $M_k$ (for $0 \le k \le m$), the usamons $U_1, \dots, U_k$ are charged (each has one electron) and the remaining usamons $U_{k+1}, \dots, U_m$ are uncharged. Note that in this set of models, for any pair of distinct usamons, there is a model where they are in different states by construction. Specifically, if $a < b$, then in model $M_a$, $U_a$ is charged and $U_b$ is uncharged.

We can consider the physicist as acting on these $m+1$ models simultaneously. To be 100% sure that two usamons are in the same state, the physicist must be able to reach a state where some pair of usamons has the same charge across all possible models. Thus, reaching a state where a pair is identical across our $m+1$ models is a necessary condition for a winning strategy to exist.

We claim that any diode operation $U_i \to U_j$ results in the $m+1$ models being an isomorphic copy of the previous set. Suppose the current set of models is defined by some ordering $V_1, \dots, V_m$ of the usamons, where in model $M_k$, the first $k$ usamons $V_1, \dots, V_k$ are charged and the rest are uncharged. Consider a diode operation directed from $V_i$ to $V_j$:
- If $i < j$:
  - In models $M_0, \dots, M_{i-1}$, neither $V_i$ nor $V_j$ is charged, so nothing happens.
  - In models $M_i, \dots, M_{j-1}$, $V_i$ is charged and $V_j$ is uncharged. The electron jumps from $V_i$ to $V_j$, meaning $V_i$ becomes uncharged and $V_j$ becomes charged.
  - In models $M_j, \dots, M_m$, both $V_i$ and $V_j$ are charged, so nothing happens.
  The net effect across all models is exactly the same as swapping the positions of $V_i$ and $V_j$ in the ordering. The new set of models is simply defined by the permuted ordering, which preserves the overall structure of the set of models.
- If $i > j$:
  - In models $M_0, \dots, M_{j-1}$, neither is charged, so nothing happens.
  - In models $M_j, \dots, M_{i-1}$, $V_j$ is charged and $V_i$ is uncharged. The diode allows an electron to jump from $V_i$ to $V_j$, but since $V_i$ is empty, nothing happens.
  - In models $M_i, \dots, M_m$, both are charged, so nothing happens.
  Thus, if $i > j$, the operation never does anything in any of the $m+1$ models.

By induction, after any sequence of diode operations, the set of $m+1$ possible states of the system is isomorphic to the initial set of prefix models, simply corresponding to a permuted ordering of the usamons. Because this structure is preserved, it remains true that for any pair of usamons, there is always some model where one is charged and the other is uncharged. Therefore, no pair of usamons can ever be guaranteed to be in the same state, and the physicist cannot achieve her goal.

(Partial) 1. Introduced (explicitly) a finite family of candidate initial configurations/models, e.g. the “prefix” models with 0,1,2,…,2015 electrons in a fixed order, and observed that for any fixed pair of usamons there is a model in which they are in different states. 2. Showed that to be 100% sure two given usamons are equal, they must be equal in every configuration still consistent with the physicist’s observations (equivalently, across all tracked models, since no feedback is received from diode uses). 3. Proved a correct invariant/transition rule for these models under one diode operation, e.g. that applying a diode between positions i<j swaps the two positions in the prefix-order family (or otherwise preserves the “all prefixes of an ordering” structure), and/or that for i>j the operation does nothing on this family. (Almost) 1. Gave the prefix-model counterexample and correctly argued that every diode operation preserves the family up to relabeling (permuting the underlying ordering), but did not fully justify one of the two direction cases (i<j vs i>j) or did not clearly state why the family remains of the same form after the operation. 2. Proved the preservation up to permutation and concluded that no pair can ever be forced equal, but left a small logical gap connecting “no pair is equal in all models” to “physicist can never be 100% sure”, or did not explicitly argue that the absence of feedback means all tracked models remain possible. 3. Solution is almost complete, but made minor mistakes which are not negligible.