Source: USAMO-2001-P1
Category: Extremal Problems

Problem: Each of eight boxes contains six balls. Each ball has been colored with one of $n$ colors, such that no two balls in the same box are the same color, and no two colors occur together in more than one box. Find with proof the smallest possible $n$.

Answer: The answer is $n = 23$.

Reference solution:
The answer is $n = 23$. 

Shown below is a construction using 23 colors, which we denote by $\{1,2,\dots,15,a,\dots,f,X,Y\}$. The 8 columns represent the 8 boxes, each containing 6 balls:
\[
\begin{bmatrix}
X & X & X & 1 & 2 & 3 & 4 & 5 \\
1 & 6 & 11 & 6 & 7 & 8 & 9 & 10 \\
2 & 7 & 12 & 11 & 12 & 13 & 14 & 15 \\
3 & 8 & 13 & Y & Y & Y & a & b \\
4 & 9 & 14 & a & c & e & c & d \\
5 & 10 & 15 & b & d & f & e & f
\end{bmatrix}
\]
It is straightforward to verify that no two balls in the same box have the same color, and any two boxes share at most one color.

We present two proofs that $n = 23$ is the best possible. The first proof also motivates the construction above. We say a color $x$ is *overrated* if it is used at least three times overall. 

First, we make the following smoothing argument. Suppose some box $A$ contains a ball of an overrated color $x$ and a ball of color $y$ used only once overall. Then we can change a ball of color $x$ in a different box $B$ to color $y$ while preserving all conditions. (Since $y$ was only in $A$, the only new pairs of balls with the same color are between $A$ and $B$, which previously intersected at $x$.) By applying this operation as many times as possible, we arrive at a configuration where whenever a box contains an overrated color, all other colors in that box are used twice or more overall.

Assume that $n \le 23$ and that we have applied this smoothing. We will show that $n$ must be exactly 23 and the configuration must be of the form given above. First, observe that no color can be used 4 or more times. If a color $X$ appears in $k$ boxes, then any such box contains 5 other colors. Since $X$ is overrated, our smoothing ensures these 5 colors are used at least twice each, so they must appear at least 5 times in total among the remaining $8 - k$ boxes. However, to avoid two boxes sharing more than one color, these 5 colors can appear at most once in each of the remaining $8 - k$ boxes. Thus, $5 \le 8 - k$, which implies $k \le 3$. Therefore, no color is used more than 3 times.

Since there are a total of 48 balls, and each of the $n \le 23$ colors is used at most 3 times, at least two colors must be overrated (used exactly 3 times). Let $X$ be an overrated color and consider the three boxes where it appears. There are $3 \times 5 = 15$ other balls in these boxes. Since these three boxes already pairwise intersect at $X$, all 15 of these balls must have distinct colors, say $\{1, \dots, 15\}$. Because $X$ is overrated, the smoothing condition requires that each of these 15 colors must appear at least once more in the remaining 5 boxes. To avoid sharing more than one color with the first three boxes, each of the remaining 5 boxes must contain exactly one color from the first box, one from the second, and one from the third. Thus, we arrive at the following configuration for the first three boxes and the partial configuration for the remaining five (up to a harmless permutation of the colors and boxes):
\[
\begin{bmatrix}
X & X & X & 1 & 2 & 3 & 4 & 5 \\
1 & 6 & 11 & 6 & 7 & 8 & 9 & 10 \\
2 & 7 & 12 & 11 & 12 & 13 & 14 & 15 \\
3 & 8 & 13 & & & & & \\
4 & 9 & 14 & & & & & \\
5 & 10 & 15 & & & & & 
\end{bmatrix}
\]
Now, note that none of these 15 colors can reappear in the remaining slots of the last five boxes, as that would force a box to contain two colors from the same group of 5, creating an intersection of size two with one of the first three boxes. 

So it remains to fill the remaining 15 slots in the last five boxes. We established that there is at least one more overrated color, distinct from any we have seen (since $X$ is used 3 times and $1, \dots, 15$ are used 2 times each); call it $Y$. Since the first three boxes are full, $Y$ must appear in three of the last five boxes. In the three boxes where $Y$ appears, there are $3 \times 6 = 18$ total slots. Nine of these slots are filled by colors from $\{1, \dots, 15\}$, and three are filled by $Y$. This leaves $18 - 9 - 3 = 6$ slots. These 6 slots must be filled with completely new colors, and they must all be distinct to avoid creating multiple intersections among the three boxes containing $Y$. This gives a strict lower bound on the total number of colors: $n \ge 1 \text{ (for } X) + 15 \text{ (for } 1 \dots 15) + 1 \text{ (for } Y) + 6 \text{ (new colors)} = 23$. This is exactly the bound we sought, and equality occurs exactly as in our construction.

Here is a short second proof of the bound. Consider the $8 \times 6$ grid of colors as before. For each ball $b$, let $n_b$ be the total number of times its color is used overall. We assign each ball a weight of $\frac{1}{n_b}$. Summing these weights over all 48 balls gives exactly the total number of colors $n$:
\[
n = \sum_{\text{all } 48 \text{ balls } b} \frac{1}{n_b}.
\]
On the other hand, consider any given box $B$. For each ball $b \in B$, its color appears in $n_b - 1$ other boxes. Because any two boxes share at most one color, the sets of other boxes containing the colors of the balls in $B$ must be pairwise disjoint. Since there are only 7 other boxes in total, we must have:
\[
\sum_{b \in B} (n_b - 1) \le 7 \implies \sum_{b \in B} n_b \le 13.
\]
We want to minimize the sum of the weights $\sum_{b \in B} \frac{1}{n_b}$ in box $B$ subject to the constraints that the $n_b$ are positive integers and $\sum_{b \in B} n_b \le 13$. Since the function $f(x) = \frac{1}{x}$ is strictly convex, the sum is minimized when the $n_b$ are as large as possible and as nearly equal to each other as possible (a smoothing argument). The optimal integer distribution for a maximum sum of 13 among 6 variables is one 3 and five 2s. Therefore,
\[
\sum_{b \in B} \frac{1}{n_b} \ge \frac{1}{3} \cdot 1 + \frac{1}{2} \cdot 5 = \frac{17}{6}.
\]
Summing this inequality over all 8 boxes yields:
\[
n = \sum_{\text{all } 48 \text{ balls } b} \frac{1}{n_b} = \sum_{B} \sum_{b \in B} \frac{1}{n_b} \ge 8 \cdot \frac{17}{6} = \frac{68}{3} = 22.66\dots
\]
Since $n$ must be an integer, we conclude that $n \ge 23$.

Grading guidelines:
(Partial) 1. Gave a valid construction achieving n=23 (or any explicit construction with n≤23) and checked that no box repeats a color and no pair of boxes shares more than one color. 2. Proved a genuine lower bound n≥23 by a counting/weighting argument, e.g. defining n_b as the multiplicity of a color and showing for each box B that ∑_{b∈B}(n_b−1)≤7 (hence ∑_{b∈B} n_b≤13), then deriving a global bound on n. 3. Established a key structural step from the “overrated color” approach, e.g. proved that no color can occur in 4 or more boxes (hence every color occurs at most 3 times), or carried out the smoothing operation that eliminates a color used once from any box containing a color used ≥3 times. (Almost) 1. Proved n≥23 via the weighting method but left a minor gap in the optimization step (why the minimum of ∑_{b∈B}1/n_b under integer constraints and ∑ n_b≤13 is attained at {3,2,2,2,2,2}), or made a small arithmetic slip after obtaining a bound >22. 2. Completed the overrated-color/smoothing lower-bound strategy up to the final count (identifying one 3-times color X, forcing 15 distinct other colors and their forced second appearances, introducing a second 3-times color Y and concluding 6 new colors), but omitted/blurred one localized justification (e.g. why the 15 colors cannot reappear in the remaining slots of the last five boxes). 3. Produced both the lower bound n≥23 and a construction with 23 colors, but the verification of the construction has a small unchecked intersection case or a minor bookkeeping error; the intended configuration clearly satisfies the conditions.

Construction task:
Find the smallest possible $n$ and provide a construction of the boxes and balls for this $n$. Represent the arrangement as a $6 \times 8$ matrix where each column corresponds to one of the eight boxes and each row corresponds to one of the six balls in that box. Use integers $1, \dots, n$ to represent different colors.

Requirements for the construction:
- The matrix must have exactly 6 rows and 8 columns.
- Each column must contain 6 distinct integers.
- For any two distinct columns, the intersection of their sets of integers must contain at most one element.
- The total number of unique integers $n$ must be the minimum possible.
- Include exactly one construction block and put ONLY the Python literal matrix between the tags:
<construct>
</construct>
- Do not use code fences. Do not put any extra text inside <construct>...</construct>.

Reference construction:
[[16, 16, 16, 1, 2, 3, 4, 5], [1, 6, 11, 6, 7, 8, 9, 10], [2, 7, 12, 11, 12, 13, 14, 15], [3, 8, 13, 17, 17, 17, 18, 19], [4, 9, 14, 18, 20, 22, 20, 21], [5, 10, 15, 19, 21, 23, 22, 23]]

Verifier code:
import sys
import ast

def verify():
    try:
        # Read the raw input from stdin
        input_data = sys.stdin.read().strip()
        if not input_data:
            print("Error: No input provided.")
            return

        # Parse the input as a Python literal
        try:
            # ast.literal_eval is used to safely evaluate the Python literal matrix
            matrix = ast.literal_eval(input_data)
        except Exception as e:
            print(f"Error: Failed to parse construction: {e}")
            return

        # Check if the parsed object is a 2D list
        if not isinstance(matrix, list) or not all(isinstance(row, list) for row in matrix):
            print("Error: Construction must be a 2D list.")
            return

        # Check dimensions: 6 rows and 8 columns as per instructions
        if len(matrix) != 6:
            print(f"Error: Matrix must have exactly 6 rows, but has {len(matrix)}.")
            return
        for i, row in enumerate(matrix):
            if len(row) != 8:
                print(f"Error: Row {i} must have exactly 8 columns, but has {len(row)}.")
                return

        # Each column represents a box. Transpose the matrix to get the boxes.
        boxes = []
        for c in range(8):
            box = [matrix[r][c] for r in range(6)]
            boxes.append(box)

        # 1. Verify that each box (column) contains 6 distinct integers.
        for c, box in enumerate(boxes):
            if len(set(box)) != 6:
                print(f"Error: Column {c} (box) contains duplicate colors or fewer than 6 elements.")
                return
            for color in box:
                if not isinstance(color, int):
                    print(f"Error: Color {color} in column {c} is not an integer.")
                    return

        # 2. Verify that no two colors occur together in more than one box.
        # This is equivalent to checking that any two distinct columns share at most one color.
        for i in range(8):
            for j in range(i + 1, 8):
                intersection = set(boxes[i]) & set(boxes[j])
                if len(intersection) > 1:
                    print(f"Error: Columns {i} and {j} share {len(intersection)} colors: {intersection}.")
                    return

        # 3. Identify all unique colors used in the matrix.
        all_colors = set()
        for box in boxes:
            all_colors.update(box)
        
        # Verify that the total number of unique integers n is 23 (the minimum possible).
        if len(all_colors) != 23:
            print(f"Error: Total number of unique colors is {len(all_colors)}, but 23 is required.")
            return

        # 4. Verify that the colors used are exactly the integers 1, ..., 23.
        if all_colors != set(range(1, 24)):
            print(f"Error: Colors must be exactly the integers from 1 to 23.")
            return

        # If all structural and mathematical constraints are satisfied
        print("True")

    except Exception as e:
        # Catch any unexpected errors during the verification process
        print(f"Error: An unexpected error occurred: {e}")

if __name__ == "__main__":
    verify()