Source: CGMO-2007-P8
Category: Extremal Problems

Problem: $n$ chess players participate in a chess tournament, where every two players play exactly one game. It is stipulated that a win earns $1$ point, a loss earns $0$ points, and a draw earns $0.5$ points for each player. If after the tournament, for any group of $m$ players, there is always one player who has defeated the other $m-1$ players, and also one player who has lost to the other $m-1$ players, then the tournament result is said to have property $P(m)$. For a given integer $m \\geq 4$, find the minimum value $f(m)$ of $n$ such that for any tournament result with property $P(m)$, the scores of all $n$ players are distinct.

Answer: 2m-3

Reference solution:
First, we prove a lemma.

**Lemma:** For $n \ge m$, if the results of a tournament with $n$ players have property $P(m)$, then there must be a player who wins all his games.

**Proof of the lemma:** We proceed by induction on $n$. 
When $n = m$, the statement follows immediately from the definition of property $P(m)$. 
Assume the statement holds for $n$, and consider a tournament of $n+1$ players. Choose any subset of $n$ players; by the induction hypothesis, there is a player among them who wins all his games within this subset. Suppose that among $A_1, A_2, \dots, A_n$, the player $A_1$ is undefeated.

Now consider the last player $A_{n+1}$ and his match against $A_1$:
- If $A_1$ defeats $A_{n+1}$, then $A_1$ is undefeated among all $n+1$ players.
- If $A_1$ draws with $A_{n+1}$, consider the subset of $n$ players $\{A_1, A_2, \dots, A_{n-1}, A_{n+1}\}$. In this subset, $A_1$ is not undefeated because he drew with $A_{n+1}$; $A_{n+1}$ is not undefeated because he did not defeat $A_1$; and for any $k \in \{2, \dots, n-1\}$, $A_k$ is not undefeated because he lost to $A_1$. Thus, nobody in this subset is undefeated, which contradicts the induction hypothesis.
- If $A_1$ loses to $A_{n+1}$, consider the subset of $n$ players $\{A_1, A_3, A_4, \dots, A_n, A_{n+1}\}$. By the induction hypothesis, there must be an undefeated player in this subset. Since $A_1$ lost to $A_{n+1}$, and $A_3, \dots, A_n$ all lost to $A_1$, none of $A_1, A_3, \dots, A_n$ can be undefeated. The only possible undefeated player is $A_{n+1}$, which implies $A_{n+1}$ defeats $A_3, A_4, \dots, A_n$. Similarly, by considering the subset $\{A_1, A_2, A_4, \dots, A_n, A_{n+1}\}$, we deduce that $A_{n+1}$ defeats $A_2$. Hence, $A_{n+1}$ defeats all of $A_1, A_2, \dots, A_n$, meaning $A_{n+1}$ is undefeated among the $n+1$ players.

Thus, the statement holds for $n+1$. This completes the induction and the proof of the lemma.

By a symmetric argument, one can prove: for $n \ge m$, if the results of $n$ players have property $P(m)$, then there must be a player who loses all his games.

Returning to the original problem, we first prove that when $n \ge 2m-3$, all players' scores must be distinct. 
By repeatedly applying the first lemma, we can order the top players:
- There is a player $A_1$ who defeats the other $n-1$ players.
- Removing $A_1$, the remaining $n-1$ players still satisfy property $P(m)$ (since $n-1 \ge m$). Thus, there is a player $A_2$ who defeats all remaining $n-2$ players.
- Continuing this process, we can extract players $A_1, A_2, \dots, A_{n-m+1}$ such that each defeats all players below them. The extraction of $A_{n-m+1}$ is valid because the number of remaining players before his extraction is $n - (n-m) = m$, which is sufficient to apply the lemma.

Similarly, by repeatedly applying the second lemma from the bottom:
- There is a player $A_n$ who loses to the other $n-1$ players.
- Removing $A_n$, there is a player $A_{n-1}$ who loses to all remaining $n-2$ players.
- Continuing this process, we can extract players $A_n, A_{n-1}, \dots, A_{n-m+3}$ such that each loses to all players above them. The extraction of $A_{n-m+3}$ is valid because the number of remaining players before his extraction is $n - (m-3) = n - m + 3 \ge m$ (since $n \ge 2m-3$).

The top players extracted are $A_1, \dots, A_{n-m+1}$ (which is $n-m+1$ players) and the bottom players extracted are $A_{n-m+3}, \dots, A_n$ (which is $m-2$ players). The total number of extracted players is $(n-m+1) + (m-2) = n-1$. The only remaining unassigned player is $A_{n-m+2}$.
Thus, among the $n$ players $A_1, A_2, \dots, A_n$, those with smaller indices defeat those with larger indices. Their scores are $n-1, n-2, \dots, 1, 0$, which are all distinct.

Next, we prove that when $n = 2m-4$, there exists a tournament with property $P(m)$ in which two players have the same score. 
Let the $n$ players be assigned strengths $1, 2, \dots, m-3, m-2, m-2, m-1, \dots, 2m-5$. Assume that a player with a numerically smaller strength defeats a player with a larger strength, and players with equal strengths draw. Thus, the two players with strength $m-2$ draw against each other, and all other matches are decisive wins or losses. Because the two players with strength $m-2$ have identical match results against all other players and draw with each other, their scores are equal.

We now verify that this tournament satisfies property $P(m)$. For any chosen group of $m$ players:
- The number of players in the entire tournament with strength $\ge m-2$ is $2 + ((2m-5) - (m-1) + 1) = m-1$. Thus, the chosen $m$ players cannot all have strengths $\ge m-2$. The smallest strength $i$ in the group must satisfy $i \le m-3$. Since there is only one player with strength $i$ in the tournament, this player is unique in the group and wins all games against the other $m-1$ players.
- The number of players in the entire tournament with strength $\le m-2$ is $(m-3) + 2 = m-1$. Thus, the chosen $m$ players cannot all have strengths $\le m-2$. The largest strength $j$ in the group must satisfy $j \ge m-1$. Since there is only one player with strength $j$ in the tournament, this player is unique in the group and loses all games against the other $m-1$ players.

Hence, the tournament results have property $P(m)$, but there are two players whose scores are equal. 

Therefore, the minimum value of $n$ is $f(m) = 2m-3$.

Grading guidelines:
(Partial) 1. Proved (or correctly stated with proof) the key lemma: for any n\ge m, property P(m) forces the existence of a player who defeats all other players (and/or symmetrically, a player who loses to all other players). 2. Gave the iterative removal idea: repeatedly delete an all-winning player (or all-losing player) and argued that P(m) is inherited by the remaining players as long as the remaining number is still \ge m. 3. Constructed a nontrivial example for n=2m-4 with property P(m) where two players have equal score (e.g., a ‘strength ordering’ with exactly two equal strengths producing one draw), and verified at least one of the two required extremal conditions in every m-subset (existence of a winner-all / loser-all). (Almost) 1. Correctly proved the upper bound f(m)\le 2m-3 by extracting a long top chain of all-winners and a long bottom chain of all-losers, but made a small counting/indexing slip (e.g., off by 1 in how many players can be extracted, or did not fully justify why the chains cover n-1 players), while the main structure is correct. 2. Correctly constructed the n=2m-4 counterexample with two equal scores and essentially checked property P(m), but left a minor gap in the verification for an arbitrary chosen m-subset (e.g., did not fully justify uniqueness of the minimal/maximal strength or missed one boundary case). 3. Solution is almost complete, but made minor mistakes which are not negligible.

Construction task:
For the specific case $m=7$, determine the largest integer $n$ for which it is possible that a tournament satisfies property $P(7)$ but the scores of the $n$ players are NOT all distinct. Provide an explicit construction of such a tournament.

Construction requirements:
- Output a single Python list of lists representing the $n \times n$ tournament matrix $A$.
- $A[i][j]$ represents the points player $i$ gets against player $j$.
- Entries must be $1$ (win), $0$ (loss), or $0.5$ (draw).
- The diagonal $A[i][i]$ must be $0$.
- The matrix must satisfy $A[i][j] + A[j][i] = 1$ for all $i \neq j$.
- The tournament must satisfy Property $P(7)$: for every subset of 7 players, there is one player who defeated the other 6 (score 1 against each), and one player who lost to the other 6 (score 0 against each).
- The row sums (total scores) must NOT be all distinct (i.e., at least two players must have the same total score).
- Put ONLY the Python literal matrix inside the construction block.

Reference construction:
[[0, 1, 1, 1, 1, 1, 1, 1, 1, 1],
 [0, 0, 1, 1, 1, 1, 1, 1, 1, 1],
 [0, 0, 0, 1, 1, 1, 1, 1, 1, 1],
 [0, 0, 0, 0, 1, 1, 1, 1, 1, 1],
 [0, 0, 0, 0, 0, 0.5, 1, 1, 1, 1],
 [0, 0, 0, 0, 0.5, 0, 1, 1, 1, 1],
 [0, 0, 0, 0, 0, 0, 0, 1, 1, 1],
 [0, 0, 0, 0, 0, 0, 0, 0, 1, 1],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]

Verifier code:
import sys
import ast
import itertools

def verify():
    """
    Verifies the construction of a 10x10 tournament matrix satisfying Property P(7)
    where the scores of the players are NOT all distinct.
    """
    # 1. Parse the input
    try:
        input_data = sys.stdin.read().strip()
        if not input_data:
            print("Error: Input is empty.")
            return
        matrix = ast.literal_eval(input_data)
    except Exception as e:
        print(f"Error: Failed to parse input as a Python literal. {e}")
        return

    # 2. Verify matrix structure and size
    if not isinstance(matrix, list):
        print("Error: Input must be a list of lists.")
        return

    n = len(matrix)
    # Instruction implies n=10 for the "largest integer n" case where scores might not be distinct (m=7 -> 2m-4=10)
    if n != 10:
        print(f"Error: Expected a 10x10 matrix, but found {n} rows.")
        return

    valid_entries = {0, 0.5, 1}

    # 3. Verify entries, diagonal, and symmetry
    for r, row in enumerate(matrix):
        if not isinstance(row, list):
            print(f"Error: Row {r} is not a list.")
            return
        if len(row) != n:
            print(f"Error: Row {r} has length {len(row)}, expected {n}.")
            return
        
        for c, val in enumerate(row):
            if val not in valid_entries:
                print(f"Error: Invalid entry at A[{r}][{c}]: {val}. Must be 0, 0.5, or 1.")
                return
            
            # Diagonal check
            if r == c and val != 0:
                print(f"Error: Diagonal entry A[{r}][{c}] must be 0.")
                return

    # Symmetry check: A[i][j] + A[j][i] = 1 for i != j
    for i in range(n):
        for j in range(i + 1, n):
            val_ij = matrix[i][j]
            val_ji = matrix[j][i]
            # 0.5 + 0.5 == 1.0 (float), which equals 1 (int). Strict equality is safe here.
            if val_ij + val_ji != 1:
                print(f"Error: Consistency violation for players {i} and {j}. A[{i}][{j}] + A[{j}][{i}] = {val_ij} + {val_ji} != 1.")
                return

    # 4. Verify Property P(7)
    # For every subset of 7 players, there must be:
    # - One player who defeated all others in the subset (score 1 against them).
    # - One player who lost to all others in the subset (score 0 against them).
    m = 7
    players = range(n)
    
    for subset in itertools.combinations(players, m):
        # Check for a 'winner' in the subset
        has_winner = False
        for p in subset:
            # p is a winner if A[p][other] == 1 for all other in subset (other != p)
            if all(matrix[p][other] == 1 for other in subset if other != p):
                has_winner = True
                break
        
        if not has_winner:
            print(f"Error: Property P(7) failed. In subset {subset}, no player defeated all other 6 players.")
            return

        # Check for a 'loser' in the subset
        has_loser = False
        for p in subset:
            # p is a loser if A[p][other] == 0 for all other in subset (other != p)
            if all(matrix[p][other] == 0 for other in subset if other != p):
                has_loser = True
                break
        
        if not has_loser:
            print(f"Error: Property P(7) failed. In subset {subset}, no player lost to all other 6 players.")
            return

    # 5. Calculate scores
    scores = [sum(row) for row in matrix]

    # 6. Verify scores are NOT all distinct
    if len(set(scores)) == n:
        print(f"Error: All {n} players have distinct scores: {sorted(scores)}. The construction requires that at least two players have the same score.")
        return

    print("True")

if __name__ == "__main__":
    verify()