Source: USAMO-2000-P4
Category: Extremal Problems

Problem: Find the smallest positive integer $n$ such that if $n$ squares of a $m \\times m$ chessboard are colored, then there will exist three colored squares whose centers form a right triangle with sides parallel to the edges of the board.

Answer: The answer is $n = 2m - 1$.

Reference solution:
The answer is $n = 2m - 1$. We present the proof for a $1000 \times 1000$ board, for which $n = 1999$. 

For a construction with $n = 1998$, take a punctured L as illustrated below (with $1000$ replaced by $4$):
\[
\begin{bmatrix}
1 \\ 
1 \\ 
1 \\ 
& 1 & 1 & 1
\end{bmatrix}.
\]

We now show that if there is no right triangle, there are at most $1998$ tokens (colored squares). In every column with more than one token, we have each token emit a bidirectional horizontal death ray (laser) covering its entire row: the hypothesis is that the death ray won't hit any other tokens.

Assume there are $n$ tokens and that $n > 1000$. Then obviously some column has more than one token, so at most $999$ tokens don't emit a death ray (namely, any token that is the only token in its column). Thus there are at least $n-999$ death rays. On the other hand, we can have at most $999$ death rays total (since it would not be okay for the whole board to have death rays, as some row should have more than one token, which contradicts the death ray property). Therefore, $n \le 999 + 999 = 1998$ as desired.

By replacing $1000$ with $m$, the exact same argument shows that $n \le 2m - 2$ when no right triangle is formed, so $n = 2m - 1$ is the minimum number of tokens required.

Grading guidelines:
(Partial) 1. Gave the extremal construction with 2m-2 colored squares (e.g. a “punctured L”: all squares in one column and one row except the corner), and correctly argued it contains no such right triangle. 2. Made the key observation that a forbidden right triangle is equivalent to: there exist two colored squares in the same column and a third colored square in the same row as one of them but in a different column (or the symmetric statement with rows/columns swapped). 3. Set up a valid counting reduction along the intended lines: classify columns with exactly one colored square versus columns with at least two, and derive a nontrivial bound relating the total number of colored squares to the number of rows that can contain “blocked” positions (even if the final inequality is not completed). (Almost) 1. Proved the upper bound n<=2m-2 for configurations with no right triangle using the row/column blocking ("death ray") idea, but left a small gap in justifying why the number of rays/blocked rows is at most m-1 (or made an off-by-one slip that affects only the final constant). 2. Gave both the construction for 2m-2 and a near-complete proof that 2m-1 forces a right triangle, but omitted a boundary case (e.g. handling n<=m separately, or the step that ensures existence of a column with >=2 colored squares when n>m). 3. Solution is almost complete, but made minor mistakes which are not negligible.

Construction task:
For the specific case $m=1000$, determine the maximum number of squares that can be colored without creating any three colored squares whose centers form a right triangle with sides parallel to the edges of the board. Provide an explicit construction for this maximum case.

Construction format requirements:
- Output a single Python expression (e.g., a list comprehension) that evaluates to a $1000 \times 1000$ 2D list of integers.
- The list represents the chessboard, where 1 indicates a colored square and 0 indicates an uncolored square.
- The construction must strictly avoid any three 1s forming a right triangle with sides parallel to the edges.
- The total number of colored squares (1s) must be exactly the maximum possible for a $1000 \times 1000$ board without violating the right triangle constraint.
- The output must be a pure Python literal/expression that can be directly evaluated by `eval()`. Do not include explanations, comments, print statements, assignments, or any non-expression content.

Output formatting requirements:
- Include exactly one construction block and put ONLY the Python expression between the tags:
<construct>
</construct>
- Do not use markdown code fences (like ```python). Do not put any extra text inside <construct>...</construct>.

Reference construction:
[[1 if (i == 0 or j == 0) and (i != j) else 0 for j in range(1000)] for i in range(1000)]

Verifier code:
import sys

def verify():
    payload = sys.stdin.read().strip()
    if not payload:
        print("Empty input.")
        return
    
    try:
        # Use eval as per instruction to allow list comprehensions
        grid = eval(payload)
    except Exception as e:
        print(f"Failed to evaluate input: {e}")
        return
        
    # Verify the parsed object is a list
    if not isinstance(grid, list):
        print(f"Expected a list, got {type(grid).__name__}.")
        return
        
    # Verify exactly 1000 rows
    if len(grid) != 1000:
        print(f"Expected exactly 1000 rows, got {len(grid)}.")
        return
        
    # Verify each row is a list of exactly 1000 integers (0 or 1)
    for i, row in enumerate(grid):
        if not isinstance(row, list):
            print(f"Row {i} is not a list.")
            return
        if len(row) != 1000:
            print(f"Row {i} does not have exactly 1000 elements (has {len(row)}).")
            return
        for j, val in enumerate(row):
            if val not in (0, 1):
                print(f"Element at ({i}, {j}) is not 0 or 1 (got {repr(val)}).")
                return
                
    # Precompute the sum of 1s in each row and each column
    row_sums = [sum(row) for row in grid]
    col_sums = [sum(grid[r][c] for r in range(1000)) for c in range(1000)]
    
    # Verify the total number of colored squares is exactly 1998
    total_colored = sum(row_sums)
    if total_colored != 1998:
        print(f"Total number of colored squares must be exactly 1998, but got {total_colored}.")
        return
        
    # Verify no right triangles with sides parallel to the edges exist
    for r in range(1000):
        for c in range(1000):
            if grid[r][c] == 1:
                if row_sums[r] > 1 and col_sums[c] > 1:
                    print(f"Forbidden right triangle right-angle vertex found at ({r}, {c}). "
                          f"Row {r} has {row_sums[r]} ones, Column {c} has {col_sums[c]} ones.")
                    return
                    
    print("True")

if __name__ == "__main__":
    verify()