Source: IMO-2025-P6
Category: Extremal Problems

Problem: Consider a $2025 \\times 2025$ grid of unit squares. Matilda wishes to place on the grid some rectangular tiles, possibly of different sizes, such that each side of every tile lies on a grid line and every unit square is covered by at most one tile. Determine the minimum number of tiles Matilda needs to place so that each row and each column of the grid has exactly one unit square that is not covered by any tile.

Answer: 2112

Reference solution:
The answer is $32 = 25 + 2(5) - 3$.

In general, for an $n \times n$ grid, the answer turns out to be $\left\lceil n + 2 \sqrt n - 3 \right\rceil$, but when $n$ is not a perfect square the solution is more complicated. Here $n=25$.

**Construction.**
We describe a general construction when $n = k^2$ (for this problem, $k=5$); it generalizes readily. There are a total of $(k-1)^2$ tiles which are $k \times k$ squares and another $4(k-1)$ tiles on the boundary, giving a total of
\[ (k-1)^2 + 4(k-1) = k^2 + 2k - 3 \]
tiles as promised. For $k=5$, this yields $16 + 16 = 32$ tiles.

**Bound.**
There are several approaches, but the shortest proof seems to be the following one that exploits the Erdős–Szekeres theorem. The theorem being quoted is:

**Theorem (Erdős–Szekeres).**
*Let $n \ge 1$ be an integer. Given a permutation of $(1, \dots, n)$, if a longest increasing subsequence (LIS) has length $a$ and a longest decreasing subsequence (LDS) has length $b$, then $ab \ge n$.*

This is a stronger version of the theorem compared to another version which instead just asserts that $\max(a,b) \ge \sqrt n$.

To apply this to the present problem, take the $n$ uncovered squares, which we henceforth call "black", as a permutation. Then consider both an LIS of length $a$ and an LDS of length $b$. We do the following illustration:
- Draw the LIS as a broken line, then connect it to the southwest and northeast corners of the board.
- Similarly, draw the LDS as a broken line, then connect it to the northwest and southeast corners of the board.
- These two steps partition the board into four quadrants, which we call north, east, south, and west.
- For each black cell in the north quadrant, write an $\mathbf{N}$ in the cell above it (for a cell in the first row, this will be off the board). Do the same for $\mathbf{E}$ (east, in the cell to the right), $\mathbf{S}$ (south, in the cell below), and $\mathbf{W}$ (west, in the cell to the left).
- Some black cells are in multiple quadrants (i.e., part of the LIS/LDS). Write all corresponding letters in that case.

We observe that:

**Claim 1.**
In this algorithm, the total number of letters written is exactly
\[
  C \coloneq \begin{cases}
    n + a + b + 1 & \text{if the LIS and LDS intersect} \\
    n + a + b & \text{otherwise}.
  \end{cases}
\]

*Proof.*
This is obvious. Each black square contributes at least one letter. Each black square on exactly one of the LIS and LDS contributes one extra letter. And a black square on both contributes $4$ letters instead of $1+1+1$.

Note by AM–GM we have $a + b \ge 2 \sqrt{ab} \ge 2 \sqrt n$, so we have a bound of
\[ C \ge n + 2 \sqrt n + \varepsilon \quad\text{where}\quad
  \varepsilon \coloneq \begin{cases}
    1 & \text{if the LIS and LDS intersect} \\
    0 & \text{otherwise}.
  \end{cases} \]

To relate $C$ to the number of tiles, the critical claim is the following, which is by construction:

**Claim 2.**
None of Matilda's tiles can have more than one letter written on it.

*Proof.*
This follows from the construction.

We split into two cases based on $\varepsilon$:
- When $\varepsilon = 1$, at most four letters go off the grid (one for each direction), so the number of tiles is at least $C - 4 \ge n + 2 \sqrt n - 3$.
- Suppose $\varepsilon = 0$. Then $C - 4 \ge n + 2 \sqrt n - 4$. However, we make the additional observation here that the tile where the LIS and LDS meet has no letters on it either; hence there are at least $1 + (C-4) \ge n + 2 \sqrt n - 3$ tiles.

For $n=25$, this yields a minimum of $25 + 2(5) - 3 = 32$ tiles.

Grading guidelines:
(Partial) 1. Gave a valid construction using 32 tiles (or the general pattern for n=k^2) and correctly verified that it leaves exactly one uncovered square in each row and each column. 2. Observed that the uncovered squares form a permutation of {1,…,25} (one per row and per column), and introduced an LIS of length a and an LDS of length b with the correct inequality ab≥25 (or max(a,b)≥5). 3. Set up a meaningful lower-bound framework by marking letters N/E/S/W adjacent to uncovered squares using the LIS/LDS partition, and stated (with some justification) that no tile can contain two marked letters. (Almost) 1. Proved the lower bound ≥32 via the LIS/LDS letter-counting method, but made a localized counting slip (e.g. miscounted by 1 the total number of letters C, or mishandled the case of LIS and LDS intersecting vs disjoint). 2. Completed the construction and essentially all of the lower-bound proof, but did not fully justify Claim 2 (that no tile can contain more than one letter) in one of the four regions or along the LIS/LDS boundary. 3. Gave the full LIS/LDS framework and derived ≥ n+2√n−3 for n=25, but omitted the off-board-letter correction (subtracting at most 4) or the extra-tile correction in the disjoint case, leaving the final bound at 31 or 33 without resolving the discrepancy.

Construction task:
For the specific case where the grid size is $529 \times 529$ (corresponding to $G=23$), provide an explicit construction of the tiles.

Construction requirements:
- Place the minimum possible number of rectangles such that each row and each column of the grid has exactly one unit square not covered by any tile.
- Output a single Python expression that evaluates to a list of rectangles.
- Each rectangle must be represented as a list of four integers `[r1, r2, c1, c2]`, covering all cells `(r, c)` such that `r1 <= r <= r2` and `c1 <= c <= c2`.
- Use 0-based indexing for coordinates, so rows and columns range from 0 to 528.
- The rectangles must not overlap, and the union of all rectangles must leave exactly one cell uncovered in each of the 529 rows and 529 columns.
- The output must be a single, valid Python expression (e.g., using list comprehensions and list concatenation) that can be directly evaluated by `eval()`.
- Do NOT include any explanations, comments, markdown formatting (like ```python), assignments, imports, or any non-expression content.
- Put ONLY the raw Python expression inside the construction block.

Reference construction:
[[23 * i + j + 1, 23 * i + j + 23, 23 * j - i + 22, 23 * j - i + 44] for i in range(22) for j in range(22)] + [[0, j, 23 * j + 23, 23 * j + 45] for j in range(22)] + [[23 * i, 23 * i + 22, 0, 21 - i] for i in range(22)] + [[507 + j, 528, 23 * j, 23 * j + 22] for j in range(22)] + [[23 * i + 23, 23 * i + 45, 528 - i, 528] for i in range(22)]

Verifier code:
import sys

def verify():
    payload = sys.stdin.read().strip()
    if not payload:
        print("Empty payload")
        return

    try:
        # The instruction explicitly requires a Python expression that can be evaluated by eval(),
        # such as list comprehensions and list concatenations.
        rects = eval(payload)
    except Exception as e:
        print(f"Failed to parse payload: {e}")
        return

    if not isinstance(rects, list):
        print("Parsed object is not a list")
        return

    if len(rects) != 572:
        print(f"Expected exactly 572 rectangles, got {len(rects)}")
        return

    grid = [[False] * 529 for _ in range(529)]

    for idx, rect in enumerate(rects):
        if not (isinstance(rect, (list, tuple)) and len(rect) == 4):
            print(f"Rectangle at index {idx} is not a list or tuple of 4 elements")
            return
        
        r1, r2, c1, c2 = rect
        if not (type(r1) is int and type(r2) is int and type(c1) is int and type(c2) is int):
            print(f"Rectangle at index {idx} contains non-integer coordinates")
            return
            
        if not (0 <= r1 <= r2 < 529 and 0 <= c1 <= c2 < 529):
            print(f"Rectangle at index {idx} has out-of-bounds or invalid coordinates: {rect}")
            return
            
        for r in range(r1, r2 + 1):
            for c in range(c1, c2 + 1):
                if grid[r][c]:
                    print(f"Rectangles overlap at cell ({r}, {c})")
                    return
                grid[r][c] = True

    for r in range(529):
        uncovered_count = grid[r].count(False)
        if uncovered_count != 1:
            print(f"Row {r} has {uncovered_count} uncovered cells, expected exactly 1")
            return

    for c in range(529):
        uncovered_count = sum(not grid[r][c] for r in range(529))
        if uncovered_count != 1:
            print(f"Column {c} has {uncovered_count} uncovered cells, expected exactly 1")
            return

    print("True")

if __name__ == "__main__":
    verify()