Source: IMO-2024-P5
Category: Operations and Strategies

Problem:
Turbo the snail is in the top row of a grid with $2024$ rows and $2023$ columns and wants to get to the bottom row. However, there are $2022$ hidden monsters, one in every row except the first and last, with no two monsters in the same column.

Turbo makes a series of attempts to go from the first row to the last row. On each attempt, he chooses to start on any cell in the first row, then repeatedly moves to an orthogonal neighbor. If Turbo reaches a cell with a monster, his attempt ends and he is transported back to the first row to start a new attempt. The monsters do not move between attempts, and Turbo remembers whether or not each cell he has visited contains a monster. If he reaches any cell in the last row, his attempt ends and Turbo wins.

Find the smallest integer $n$ such that Turbo has a strategy which guarantees being able to reach the bottom row in at most $n$ attempts, regardless of how the monsters are placed.

Solution:
The minimum number of attempts is $n=3$. In fact, the same argument works for any $s \times (s-1)$ grid with $s \ge 4$; here the grid is $2024 \times 2023$, so $s = 2024$.

First, we show that at least three attempts are necessary. On the first attempt, when Turbo first enters the second row, he may immediately hit a monster $M_1$. Because his attempt ends immediately, he learns nothing else about the grid. On his second attempt, he must avoid the cell containing $M_1$, so he must enter the third row in a different column. However, upon entering the third row, he may again immediately hit a monster $M_2$. In this scenario, his second attempt ends, and he still has not reached the bottom row. Thus, no strategy can guarantee success in only two attempts.

Now we prove that three attempts always suffice by providing a strategy that guarantees reaching the bottom row in at most three attempts.

On the first attempt, Turbo moves into the second row and explores it horizontally until he discovers the unique monster $M_1$ located in that row. Upon hitting $M_1$, his first attempt ends. There are two cases to consider based on the position of $M_1$.

Case 1: $M_1$ is not on an edge cell of the second row. 
Since $M_1$ is not in the leftmost or rightmost column, Turbo knows that the two cells adjacent to $M_1$ in the second row (one on the left and one on the right) are safe. Starting from either of these safe cells, he can attempt to follow a path going downward on the corresponding side of $M_1$, and then continue to the bottom row. These two candidate paths lie on opposite sides of the column containing $M_1$. Because there is only one monster in each lower row and no two monsters share a column, at least one of the two downward routes avoids all monsters. On his second attempt, Turbo tries one of these routes. If he succeeds, he reaches the bottom row. If he encounters a monster, his second attempt ends, but he now knows that the other route must be completely free of monsters. Hence, on his third attempt, Turbo can take the other route and is guaranteed to succeed. Thus, he succeeds within the next two attempts.

Case 2: $M_1$ lies on an edge cell of the second row. 
By symmetry, suppose $M_1$ is in the leftmost cell of the second row (column 1). On the second attempt, Turbo follows a staircase path heading down and gradually to the right. If this staircase path reaches the last row without hitting any monsters, he wins immediately on the second attempt. Otherwise, he encounters a second monster $M_2$ somewhere on the staircase path. At this point, his second attempt ends. However, because he reached $M_2$, everything strictly to the left of that encountered point on the staircase path has been verified as safe. Furthermore, Turbo also knows that the entire first column below $M_1$ is completely free of monsters, because $M_1$ already occupies the unique monster position for the first column (recall that no two monsters share a column). Using this combined information, on the third attempt Turbo can safely travel down along the verified safe region to the left of $M_2$, move horizontally into the first column (the column of $M_1$), and then continue straight down to the bottom row. This route avoids every monster.

Thus, Turbo always has a strategy that guarantees success in at most three attempts, and two attempts are impossible. Therefore, the answer is
\[
\boxed{3}
\]

Answer: $3$

Grading guidelines:
(Partial) 1. Proved the lower bound n\ge 3 by exhibiting an adversarial placement causing failure in 2 attempts (e.g. first hit a monster upon first entering row 2, then forced to enter row 3 in a different column and can be hit immediately again). 2. Gave the key idea for the upper bound strategy: first locate the monster in row 2 by sweeping along that row until hitting it, and then use information from this to plan later attempts. 3. Solved Case 1 (monster in row 2 not on an edge): proposed the two “routes on opposite sides of the row-2 monster” and correctly argued that, since monsters occupy distinct columns, at least one route is monster-free so success is guaranteed within 3 attempts. (Almost) 1. Correctly proved n\ge 3 and presented a complete 3-attempt strategy for the non-edge Case 1, but handled the edge Case 2 only heuristically (missing a rigorous construction of the final safe path after the staircase hits a monster). 2. Gave the full two-case strategy (non-edge and edge) and the distinct-columns argument, but left one localized gap, such as not justifying why the “other side” route must be completely free after one failure, or not proving that the claimed safe region in the edge case can always be reached without stepping on unknown cells. 3. Solution is almost complete, but made minor mistakes which are not negligible.