Question: Given an integer $n \\geq 2$. Let the grid tables \n$A = \\begin{pmatrix} a_{11} & a_{12} & \\cdots & a_{1n} \\\\ a_{21} & a_{22} & \\cdots & a_{2n} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ a_{n1} & a_{n2} & \\cdots & a_{nn} \\end{pmatrix}, \\quad B = \\begin{pmatrix} b_{11} & b_{12} & \\cdots & b_{1n} \\\\ b_{21} & b_{22} & \\cdots & b_{2n} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ b_{n1} & b_{n2} & \\cdots & b_{nn} \\end{pmatrix}$ \nsatisfy $\\{a_{ij} \\mid 1 \\leq i,j \\leq n\\} = \\{b_{ij} \\mid 1 \\leq i,j \\leq n\\} = \\{1, 2, \\cdots, n^2\\}.$ \nThe following operation can be performed on $A$: select two numbers in the same row or the same column, swap their positions, while the remaining $n^2 - 2$ numbers remain unchanged. This operation is called a transposition. \nFind the smallest positive integer $m$ such that for any $A, B$, it is always possible to transform $A$ into $B$ through at most $m$ transpositions.

First, consider the one-dimensional problem: let arrays $X=(x_1,x_2,\dots,x_m)$ and $Y=(y_1,y_2,\dots,y_m)$ be permutations of the same $m$ elements. Then $X$ can be transformed into $Y$ by at most $m-1$ swaps. In fact, for $i=1,2,\dots,m-1$, we can sequentially swap the element $y_i$ into the $i$-th position of $X$; after $m-1$ such swaps, the $m$-th entry must naturally be $y_m$.

For the two-dimensional problem, since the allowed operations (row and column transpositions) are invariant under relabeling the elements, we can assume without loss of generality that the target matrix is the standard row-major table:
$$B=\begin{pmatrix}
1&2&\cdots&n\\
n+1&n+2&\cdots&2n\\
\vdots&\vdots&\ddots&\vdots\\
n^2-n+1&n^2-n+2&\cdots&n^2
\end{pmatrix}.$$

We first prove that any matrix $A$ can be transformed into $B$ by at most $2n^2-2n$ transpositions. We begin by showing a more general claim: for any $m \le n$, any $m$ specified target elements located in $m$ columns can be moved to the first row of these $m$ columns in a specified target permutation using at most $2m-1$ transpositions, without changing the positions of any elements outside these $m$ columns. We proceed by induction on $m$.
When $m=1$, the statement is obvious: $1$ element in $1$ column can be moved to the first row of that column by at most $1$ column transposition, and $1 \le 2(1)-1$.
Assume the claim holds for $m-1$, and consider the case $m$.
**Case 1:** If each of the $m$ columns contains exactly one of the $m$ target elements, then we can first move the $m$ elements to the first row by at most $m$ column transpositions. Then, by the one-dimensional problem, we can sort them into the target permutation using at most $m-1$ row transpositions within the first row. Hence at most $m+(m-1)=2m-1$ transpositions are needed.
**Case 2:** If there exists a column $k$ among the $m$ columns that does not contain any of the $m$ target elements, let $x$ be the element whose target position is the first row of column $k$. Since column $k$ does not contain $x$, $x$ is in some other column among the $m$ columns. Swap $x$ into column $k$ (via a row transposition) and then swap it into the first row (via a column transposition). Since column $k$ initially contained none of the $m$ elements, the elements swapped with $x$ are not among the $m$ target elements, so these two transpositions do not change the positions of the remaining $m-1$ target elements. After removing column $k$ from consideration, the induction hypothesis gives that we can finish placing the remaining $m-1$ elements with at most $2(m-1)-1 = 2m-3$ more transpositions. Thus we need at most $2+(2m-3)=2m-1$ transpositions. The induction is complete.

Applying this claim for $m=n$, the first row of $A$ can be turned into the first row of $B$ (i.e., $(1,2,\dots,n)$) by at most $2n-1$ transpositions.
Similarly, for each $i$ with $2\le i\le n-1$, we can treat the subgrid consisting of rows $i$ through $n$ as a new grid. The target elements for row $i$ are all located in this subgrid. By the same logic, the $i$-th row of $A$ can be turned into the $i$-th row of $B$ by at most $2n-1$ transpositions. Because these transpositions are performed entirely within the subgrid (i.e., row transpositions within rows $\ge i$, and column transpositions between rows $\ge i$), they do not disturb the already fixed rows $1$ through $i-1$.
At this point, the $n$-th row of $A$ is a permutation of $n^2-n+1,n^2-n+2,\dots,n^2$, so it can be turned into the $n$-th row of $B$ by at most $n-1$ row transpositions.
Consequently, $A$ can be transformed into $B$ by at most $(n-1)(2n-1)+(n-1)=2n^2-2n$ transpositions.

To show that $2n^2-2n$ transpositions are necessary, consider the matrix
$$A=\begin{pmatrix}
n+2&n+3&\cdots&2n&n+1\\
2n+2&2n+3&\cdots&3n&2n+1\\
\vdots&\vdots&\ddots&\vdots&\vdots\\
n^2-n+2&n^2-n+3&\cdots&n^2&n^2-n+1\\
2&3&\cdots&n&1
\end{pmatrix}.$$
Let $D$ be the number of elements in the grid that are located in a row strictly above their target row in $B$ (i.e., current row index $<$ target row index). In $A$, for $1 \le i \le n-1$, the $n$ elements in row $i$ all belong to row $i+1$ in $B$, so their current row index is less than their target row index. The elements in row $n$ belong to row $1$, so they are not above their target rows. Thus, initially exactly $n(n-1)$ elements are above their target rows, so $D = n(n-1)$. In $B$, $D = 0$.
A row transposition does not change the row index of any element, so it changes $D$ by $0$. A column transposition swaps an element in row $r_1$ with an element in row $r_2$. Assuming $r_1 < r_2$, the element at $r_1$ moves downward to $r_2$, and the element at $r_2$ moves upward to $r_1$. The element moving upward has its row index decreased; if it was already above its target row, it remains above its target row, so it cannot decrease $D$. The element moving downward can change from being above its target row to not being above its target row, which decreases $D$ by at most $1$. Therefore, $D$ decreases by at most $1$ per column transposition. Thus, at least $n(n-1)$ column transpositions are required to reduce $D$ to $0$.

Similarly, let $R$ be the number of elements in the grid that are located in a column strictly to the left of their target column in $B$. In $A$, for $1 \le j \le n-1$, the $n$ elements in column $j$ belong to column $j+1$ in $B$, so they are to the left of their target columns. The elements in column $n$ belong to column $1$. Thus, initially $R = n(n-1)$. In $B$, $R = 0$.
A column transposition does not change the column of any element, so it changes $R$ by $0$. A row transposition swaps two elements in the same row, moving at most one element to the right, so it can decrease $R$ by at most $1$. Therefore, at least $n(n-1)$ row transpositions are required to reduce $R$ to $0$.

Since column transpositions do not decrease $R$ and row transpositions do not decrease $D$, the total number of transpositions must be at least the sum of the minimum required column and row transpositions. This requires at least $n(n-1) + n(n-1) = 2n^2 - 2n$ transpositions.

Thus, the smallest positive integer $m$ is $2n^2-2n$.

(Partial) 1. Observed that one can assume WLOG that the target table is the standard row-major table $B$ (relabeling/invariance argument). 2. Proved the 1D lemma: any permutation of $m$ elements can be transformed into any other using at most $m-1$ swaps (and applied it correctly to a row). 3. Established a correct lower-bound invariant for some explicit “shifted” matrix $A$, e.g. defining $D$ (elements above their target row) or $R$ (elements left of their target column) and showing that one type of transposition does not change it while the other decreases it by at most 1, hence needing at least $n(n-1)$ transpositions of that type. (Almost) 1. Gave a complete upper-bound strategy leading to $2n^2-2n$ (e.g. the induction claim giving $2m-1$ to place $m$ targets in $m$ columns, then applying it row-by-row), but with a minor gap such as not fully justifying that previously fixed rows are not disturbed or not fully proving the induction step in Case 2. 2. Gave the correct extremal example matrix and essentially the full lower-bound argument $D$ and $R$ leading to $2n^2-2n$, but with a localized mistake (e.g. unclear why a column transposition decreases $D$ by at most 1, or why the required row/column transpositions add without interference). 3. Solution reaches the correct value $m=2n^2-2n$ with the main ideas present, but has a small number of boundary cases unchecked (e.g. handling the last row requiring only $n-1$ row swaps, or treating $n=2$ separately).

Instruction: Provide exactly one explicit 10x10 integer matrix A as a single Python expression (Python-literal-friendly), such that evaluating it yields a nested list of 10 lists each of length 10. The entries of A must be exactly the integers 1 through 100, each appearing once. A legal transposition is swapping two entries in the same row or in the same column. Let the target matrix B be the 10x10 row-major matrix with entries 1,2,...,100. Your constructed A must be such that the minimum number of legal transpositions required to transform A into this B attains the maximum possible value among all such A for n=10. Put only the Python expression inside <construct>...</construct> and nothing else.

import sys

def fail(msg):
    print(msg)
    sys.exit(0)

def safe_eval(expr: str):
    allowed_builtins = {
        "range": range,
        "len": len,
        "list": list,
        "int": int,
    }
    env = {"__builtins__": allowed_builtins}
    return eval(expr, env, {})

def main():
    s = sys.stdin.read()
    if not s.strip():
        fail("Parsing error: empty input.")
    try:
        obj = safe_eval(s)
    except Exception as e:
        fail(f"Parsing error: {e}")

    # Dimension/type checks
    if not isinstance(obj, list):
        fail("Dimensions error: top-level object is not a list.")
    if len(obj) != 10:
        fail(f"Dimensions error: expected 10 rows, got {len(obj)}.")
    for i, row in enumerate(obj):
        if not isinstance(row, list):
            fail(f"Dimensions error: row {i} is not a list.")
        if len(row) != 10:
            fail(f"Dimensions error: expected row {i} length 10, got {len(row)}.")
        for j, v in enumerate(row):
            if type(v) is not int:
                fail(f"Type error: entry at ({i},{j}) is not a strict int (got {type(v).__name__}).")

    A = obj

    # Permutation check
    flat = [A[i][j] for i in range(10) for j in range(10)]
    expected_set = set(range(1, 101))
    got_set = set(flat)
    if got_set != expected_set:
        missing = sorted(expected_set - got_set)
        extra = sorted(got_set - expected_set)
        if missing:
            fail(f"Permutation error: missing values (first few): {missing[:10]}")
        if extra:
            fail(f"Permutation error: extra values (first few): {extra[:10]}")
        fail("Permutation error: entries are not exactly a permutation of 1..100.")
    if len(flat) != len(got_set):
        # duplicates exist
        # find one duplicate
        seen = set()
        dup = None
        for v in flat:
            if v in seen:
                dup = v
                break
            seen.add(v)
        fail(f"Permutation error: duplicate value detected ({dup}).")

    # Cyclic-shift pattern check
    for i in range(10):
        for j in range(10):
            exp = 10 * ((i + 1) % 10) + ((j + 1) % 10) + 1
            if A[i][j] != exp:
                fail(
                    f"Cyclic pattern error at ({i},{j}): got {A[i][j]}, expected {exp} "
                    f"for A[i][j]=10*((i+1)%10)+((j+1)%10)+1."
                )

    # Invariant checks: D and R
    D = 0  # number of entries in a row strictly above their target row
    R = 0  # number of entries in a column strictly left of their target column
    for i in range(10):
        for j in range(10):
            v = A[i][j]
            tr = (v - 1) // 10
            tc = (v - 1) % 10
            if i < tr:
                D += 1
            if j < tc:
                R += 1
    if D != 90:
        fail(f"Invariant error: D should be 90, got {D}.")
    if R != 90:
        fail(f"Invariant error: R should be 90, got {R}.")

    print("True")

if __name__ == "__main__":
    main()


Question: Consider a $12\\times12$ grid where each cell is colored either black or white, such that every $3\\times4$ or $4\\times3$ subgrid contains at least one black cell. Find the minimum number of black cells required.

Because a $12\times 12$ grid can be partitioned into $\frac{12\times 12}{3\times 4}=12$ disjoint $3\times 4$ subgrids, there must be at least $12$ black cells.

Multiple valid configurations with exactly $12$ black cells that satisfy the requirements can be constructed. Therefore, the minimum number of black cells required is $12$.

(Partial) 1. Observed that the board can be tiled by disjoint 3×4 rectangles (e.g. 4 by 3 array of blocks), and concluded at least one black cell is needed in each block. 2. Gave an explicit placement pattern with 12 black cells and argued it meets the condition for all 3×4 (and/or 4×3) subgrids, but did not justify minimality. 3. Proved a weaker correct lower bound (e.g. ≥10 or ≥11) using any valid disjoint-family/covering argument, even if not sharp. (Almost) 1. Correctly proved the lower bound 12 via a partition into 12 disjoint 3×4 subgrids, but did not supply (or did not fully verify) a configuration achieving 12. 2. Gave a correct construction with 12 black cells, and a correct lower bound argument, but left a small gap such as not checking one orientation (3×4 vs 4×3) or not clearly justifying that the chosen 12 rectangles are disjoint and cover the whole board. 3. Solution is almost complete, but made minor mistakes which are not negligible.

Instruction:
Provide an explicit construction for the grid that achieves this minimum number of black cells.

Construction requirements:
- Output a single Python list of lists representing the $12 \times 12$ grid.
- Use the integer `0` to represent a white cell and `1` to represent a black cell.
- Your grid must contain the minimum possible number of black cells such that every $3 \times 4$ subgrid and every $4 \times 3$ subgrid contains at least one black cell.
- Put ONLY the Python literal inside the construction block.

import sys
import ast

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

    try:
        grid = ast.literal_eval(payload)
    except Exception as e:
        print(f"Failed to parse python literal: {e}")
        return

    if not isinstance(grid, list):
        print("Grid must be a list.")
        return

    if len(grid) != 12:
        print(f"Grid must have exactly 12 rows, got {len(grid)}.")
        return

    for i, row in enumerate(grid):
        if not isinstance(row, list):
            print(f"Row {i} is not a list.")
            return
        if len(row) != 12:
            print(f"Row {i} must have exactly 12 elements, got {len(row)}.")
            return
        for j, val in enumerate(row):
            if type(val) is not int or val not in (0, 1):
                print(f"Element at ({i}, {j}) is not the integer 0 or 1, got {val!r}.")
                return

    total_ones = sum(sum(row) for row in grid)
    if total_ones != 12:
        print(f"Total number of 1s must be exactly 12, got {total_ones}.")
        return

    # Check 3x4 subgrids
    for r in range(10):
        for c in range(9):
            subgrid_sum = sum(grid[r+i][c+j] for i in range(3) for j in range(4))
            if subgrid_sum < 1:
                print(f"3x4 subgrid starting at ({r}, {c}) has no black cells.")
                return

    # Check 4x3 subgrids
    for r in range(9):
        for c in range(10):
            subgrid_sum = sum(grid[r+i][c+j] for i in range(4) for j in range(3))
            if subgrid_sum < 1:
                print(f"4x3 subgrid starting at ({r}, {c}) has no black cells.")
                return

    print("True")

if __name__ == "__main__":
    verify()


Question: $n$ 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.

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 $f(m) = 2m-3$.

(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.

Instruction:
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.

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()


Question:
"Let $m$ and $n$ be integers greater than $1$. On an $m\\times n$ grid, each cell contains a coin placed face down. In each operation, one may choose a $2\\times2$ square, flip the coins at the top-left and bottom-right corners, and flip one of the coins at the bottom-left or top-right corner. Determine all pairs $(m,n)$ for which it is possible to turn all coins face up after finitely many operations.
"

The possible pairs are all $(m,n)$ such that $3\mid mn$.

When $3\mid mn$, the $m\times n$ grid can be partitioned into several $1\times3$ or $3\times1$ small rectangles. By applying a sequence of valid operations to flip the coins in a single small rectangle, all coins on the board can be turned over in a finite number of steps.

In the cell $(i,j)$, write the remainder of $i+j$ modulo $3$. For $k=0,1,2$, let $A_k$ denote the number of coins showing heads in the cells containing $k$. Initially, $A_0=A_1=A_2=0$. Because each operation simultaneously flips exactly one coin from each of the three types of cells, $A_0$, $A_1$, and $A_2$ change parity together, so the parities of $A_0$, $A_1$, and $A_2$ are always the same. If finally all coins are heads up, then the total numbers of the three types of cells in the board must have the same parity. But when $3\nmid mn$, one type of cell has one more or one fewer than the other two types, a contradiction!

If we assign to the cell $(i,j)$ the value $x^i y^j$, then there exist polynomials $P(x,y)$ and $Q(x,y)$ such that
\begin{aligned}
&P(x,y)(x+y+1)+Q(x,y)(x+y+xy)\\
&\equiv (1+x+\cdots+x^{m-1})(1+x+\cdots+x^{n-1})\pmod{2}.
\end{aligned}

(Partial) 1. Introduced the 3-colouring by i+j (mod 3) and correctly observed that each move flips exactly one coin of each colour. 2. Derived a correct invariant/congruence consequence of the move, e.g. that the parities of the numbers A0,A1,A2 of heads in the three colour classes always stay equal (or an equivalent mod-2 linear invariant). 3. From the invariant, proved a necessary condition excluding some boards (e.g. showed impossibility when 3∤mn by comparing the parities of the sizes of the three colour classes in an m×n grid). (Almost) 1. Proved the necessity (3|mn) via the 3-colouring parity argument, and for sufficiency gave a valid decomposition into 1×3/3×1 blocks but did not fully justify that a single 1×3 (or 3×1) block can always be solved by legal 2×2 moves (missing an explicit sequence or a local argument). 2. Proved sufficiency by an explicit local procedure (e.g. solving a 1×3/3×1 block and tiling the board when 3|mn), but the impossibility part is left with only a sketchy/incorrect parity count for the case 3∤mn (e.g. did not fully justify the distribution of i+j mod 3 in the m×n grid). 3. Established both directions and the final condition 3|mn, but with a small gap such as not handling one boundary case in the tiling (e.g. when only one of m,n is divisible by 3) or an unverified final step in the invariant-to-contradiction argument.

Instruction: Give, for the 12 \times 10 grid (rows 0..11, cols 0..9), an explicit finite sequence of allowed operations that turns all coins from face-down to face-up. Output exactly one Python expression (Python-literal-friendly) that evals to a list of operations. Each operation must be a length-3 list of distinct [r,c] integer coordinates, 0-based, within the stated ranges, and must correspond to choosing some 2 \times 2 square with top-left corner (r,c) and flipping exactly: [r,c] (top-left), [r+1,c+1] (bottom-right), and exactly one of [r+1,c] (bottom-left) or [r,c+1] (top-right). The returned operation list must achieve that after applying all operations (toggling coin state), every cell is face-up.

import sys, ast

def fail(msg):
    print(msg)
    raise SystemExit(0)

data = sys.stdin.read()
try:
    tree = ast.parse(data, mode="eval")
    obj = eval(compile(tree, "<construction>", "eval"), {"__builtins__": {}, "range": range}, {})
except Exception as e:
    fail(f"Parse/eval error: {e}")

if not isinstance(obj, list):
    fail("Top-level must be a list of operations.")

ROWS, COLS = 12, 10
grid = [[0 for _ in range(COLS)] for _ in range(ROWS)]

def coord_to_tuple(x, op_idx, coord_idx):
    if not isinstance(x, (list, tuple)) or len(x) != 2:
        fail(f"Operation {op_idx}: coordinate {coord_idx} must be a list/tuple of length 2.")
    r, c = x[0], x[1]
    if not isinstance(r, int) or not isinstance(c, int):
        fail(f"Operation {op_idx}: coordinate {coord_idx} entries must be integers.")
    if not (0 <= r <= ROWS - 1 and 0 <= c <= COLS - 1):
        fail(f"Operation {op_idx}: coordinate {coord_idx} out of bounds: [{r},{c}].")
    return (r, c)

for i, op in enumerate(obj):
    if not isinstance(op, list) or len(op) != 3:
        fail(f"Operation {i}: must be a list of length 3.")

    coords = [coord_to_tuple(op[j], i, j) for j in range(3)]

    if len(set(coords)) != 3:
        fail(f"Operation {i}: coordinates must be pairwise distinct.")

    pts = set(coords)

    legal = False
    legal_rtct = None
    for (rt, ct) in pts:
        if 0 <= rt <= ROWS - 2 and 0 <= ct <= COLS - 2:
            pattern1 = {(rt, ct), (rt + 1, ct + 1), (rt, ct + 1)}
            pattern2 = {(rt, ct), (rt + 1, ct + 1), (rt + 1, ct)}
            if pts == pattern1 or pts == pattern2:
                legal = True
                legal_rtct = (rt, ct)
                break
    if not legal:
        fail(f"Operation {i}: illegal operation; does not match required 2x2 pattern with a valid top-left.")

    # Apply toggles
    for (r, c) in coords:
        grid[r][c] ^= 1

bad = [(r, c) for r in range(ROWS) for c in range(COLS) if grid[r][c] == 0]
if bad:
    preview = ", ".join([f"[{r},{c}]" for (r, c) in bad[:20]])
    more = "" if len(bad) <= 20 else f" ... and {len(bad)-20} more"
    fail(f"Not all coins are face-up; face-down cells include: {preview}{more}")

print("True")


Question: Given an integer $n \\geq 3$. There are $m \\g We want to color the beads using $n$ colors such that in any consecutive $n + 1$ beads, all colors appear at least once. Find the largest positive integer $m$ for which this is impossible.

We claim that the largest positive integer $m$ for which such a coloring is impossible is $m = n^2-n-1$.

First, we prove that when $m=n^{2}-n-1$, the coloring cannot be achieved. Assume for the sake of contradiction that a valid coloring exists. By the Pigeonhole Principle, there exists a color that appears at most $\left\lfloor\frac{n^{2}-n-1}{n}\right\rfloor=n-2$ times. Let $c \le n-2$ be the number of times this color appears. These $c$ beads divide the remaining $m-c$ beads into $c$ consecutive blocks (intervals) around the necklace. By the Pigeonhole Principle, the maximum length of such a block is at least
$$ \left\lceil\frac{m-c}{c}\right\rceil \ge \left\lceil\frac{m-(n-2)}{n-2}\right\rceil = \left\lceil\frac{n^{2}-2n+1}{n-2}\right\rceil = \left\lceil n + \frac{1}{n-2}\right\rceil = n+1, $$
where we used the fact that $n \ge 3$. This means there must be a consecutive block of at least $n+1$ beads without that color, which contradicts the requirement that any consecutive $n+1$ beads contain all $n$ colors. Thus, $m = n^2-n-1$ is impossible.

Next, we prove that when $m \ge n^{2}-n$, a valid coloring can always be achieved. Let $m=kn+j$, where $0\le j\le n-1$. Since $m \ge n^2-n = (n-1)n$, we have $k \ge n-1$, which implies $k \ge j$. We divide the $m$ beads into $k$ groups as follows:
- $k-j$ groups of length $n$, coloring each group with the sequence of colors $1, 2, 3, \dots, n$;
- $j$ groups of length $n+1$, coloring each group with the sequence of colors $1, 1, 2, 3, \dots, n$.

It is easy to verify that this coloring satisfies the requirement. Since the minimum group length is $n$, any $n+1$ consecutive beads can span at most two adjacent groups. If the $n+1$ beads are entirely contained within a single group, it must be a group of length $n+1$, which is colored $1, 1, 2, \dots, n$ and thus contains all $n$ colors. If they span two adjacent groups, they consist of the last $x$ beads of the first group and the first $y$ beads of the second group, where $x, y \ge 1$ and $x+y = n+1$. The first group ends with the colors $n-x+1, n-x+2, \dots, n$. The second group begins with the colors $1, 2, \dots, y$ (if its length is $n$) or $1, 1, 2, \dots, y-1$ (if its length is $n+1$). In either case, the second group provides the colors $1, 2, \dots, y-1$ (and also $1$ if $y=1$). Since $x+y = n+1$, we have $y-1 = n-x$. Therefore, the union of the colors from both parts contains $\{n-x+1, \dots, n\} \cup \{1, \dots, n-x\}$, which is exactly the set of all $n$ colors.

In general, let $m = kn + j$, where $0 \le j \le n-1$; then a valid coloring is achievable if and only if $k \ge j$.

(Partial) 1. Showed an upper bound by a pigeonhole/block argument: picked a least frequent color, bounded its number of occurrences by at most n−2 when m=n^2−n−1, and concluded there is a gap of at least n+1 consecutive beads missing that color, contradicting the condition. 2. Gave a workable construction idea for large m (e.g. repeating 1,2,…,n in blocks, possibly inserting one extra repeated color to make length n+1 blocks) and correctly argued that any n+1 consecutive beads lie in at most two adjacent blocks. 3. Correctly reduced the problem to writing m=kn+j (0≤j≤n−1) and stated/derived the key feasibility condition k≥j (equivalently m≥n^2−n), even if the proof or construction is incomplete. (Almost) 1. Correctly proved impossibility for m=n^2−n−1 using the least-frequent-color and longest-gap estimate, and correctly described a valid coloring for all m≥n^2−n, but left a small gap in checking the boundary/wrap-around case for the necklace (windows crossing the end/start). 2. Gave the block construction for m=kn+j with k≥j and verified the two-block window check, but made a localized counting error (floor/ceiling or off-by-one) in the impossibility part that does not affect the main strategy. 3. Solution is almost complete, but made minor mistakes which are not negligible.

Instruction:
For the specific case $n=100$, output a valid coloring of a circular necklace with $m=12345$ beads using the $n$ colors, such that any consecutive $n+1$ beads contain all $n$ colors at least once.

Construction requirements:
- Output a single Python expression that evaluates to a list of 12345 integers.
- The integers must be from 1 to 100, representing the colors of the beads in order around the necklace.
- The configuration must satisfy the condition that any window of 101 consecutive beads (including wrap-around) contains all 100 colors.
- The construction must be a single, raw Python expression (e.g., using list concatenation, multiplication, and `range()`).
- Do not include any explanations, comments, markdown formatting, assignments, or any non-expression content.
- Put ONLY the Python expression inside the construction block.

import sys

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

    try:
        # Evaluate the raw construction payload using Python's eval()
        # as requested by user intent to support list concatenation, multiplication, and range()
        beads = eval(payload)
    except Exception as e:
        print(f"Failed to evaluate expression: {e}")
        return

    if not isinstance(beads, list):
        print("Evaluated result is not a list.")
        return

    if len(beads) != 12345:
        print(f"List length is {len(beads)}, expected 12345.")
        return

    for idx, bead in enumerate(beads):
        if not isinstance(bead, int) or isinstance(bead, bool):
            print(f"Element at index {idx} is not an integer.")
            return
        if not (1 <= bead <= 100):
            print(f"Element at index {idx} is {bead}, expected between 1 and 100.")
            return

    # Simulate the circular nature of the necklace by appending the first 100 elements
    extended_beads = beads + beads[:100]

    # Iterate over all sliding windows of length 101
    for i in range(len(beads)):
        window = extended_beads[i:i+101]
        distinct_colors = set(window)
        if len(distinct_colors) != 100:
            print(f"Window starting at index {i} does not contain all 100 colors.")
            return

    print("True")

if __name__ == "__main__":
    verify()


Question: Given an integer $n\\geq2$. Paul has a $1\\times n^2$ strip of paper, with squares labeled $1, 2, \\cdots, n^2$ in order. Paul wants to cut the strip into $m$ consecutive segments such that, after translating these segments (rotation and flipping are not allowed), they can form an $n\\times n$ grid. In this grid, for each $1\\leq i,j\\leq n$, the number $a_{i,j}$ in the $i$-th row and $j$-th column satisfies $a_{i,j}\\equiv i+j-1\\pmod{n}$. If Paul can achieve this, find the minimum value of $m$.

It suffices to work modulo $n$. In this case, the numbers on the tape are $1, 2, \dots, n$ repeated $n$ times, and the target grid table is given by $a_{i,j} \equiv i+j-1 \pmod n$:
$$
\def\arraystretch{1.5}
\begin{array}{|c|c|c|c|c|}
\hline
1 & 2 & \cdots & n-1 & n \\ \hline
2 & 3 & \cdots & n & 1 \\ \hline
\vdots & \vdots & \ddots & \vdots & \vdots \\ \hline
n & 1 & \cdots & n-2 & n-1 \\ \hline
\end{array}
$$

**Construction for $m = 2n-1$:**
Cut the tape into $2n-1$ pieces whose lengths, in order, are $n, 1, n-1, 2, n-2, \dots, n-1, 1$.
Specifically, the first piece has length $n$. For $2 \le i \le n$, the $(2i-2)$-th piece has length $i-1$ and the $(2i-1)$-th piece has length $n-i+1$.

Because the tape consists of repeating blocks of $1, 2, \dots, n$, the first piece is exactly $1, 2, \dots, n$. For $2 \le i \le n$, the $(2i-2)$-th piece starts at the beginning of a block and thus contains $1, 2, \dots, i-1$, while the $(2i-1)$-th piece contains the remainder of that block, $i, i+1, \dots, n$.

Use the first piece as the first row of the grid table. For $2 \le i \le n$, swap the $(2i-2)$-th piece with the $(2i-1)$-th piece to form the $i$-th row of the grid table. This places $i, i+1, \dots, n$ followed by $1, 2, \dots, i-1$, which exactly matches the $i$-th row of the grid. Hence, this cutting satisfies the requirement.

**Lower bound $m \ge 2n-1$:**
Construct a directed graph $G$ whose vertices are $1, 2, \dots, n$. For each of the $m$ segments, if the numbers on the segment are $a, a+1, \dots, b-1 \pmod n$, add a directed edge from $a$ to $b$. The graph $G$ has a total of $m$ edges, allowing multiple edges and loops.

Since the segments are contiguous on the tape, the end of one segment is immediately followed by the start of the next segment. Thus, the sequence of $m$ segments forms a continuous walk. Since the tape starts at $1$ and ends at $n \equiv 0 \pmod n$, the walk starts at $1$ and ends at $1$, meaning the whole tape corresponds to an Eulerian circuit of $G$. Because the tape contains all $n$ numbers, $G$ is connected and spans all $n$ vertices.

The $i$-th row of the grid is formed by a sequence of segments that starts at $i$ and ends at $i-1 \pmod n$. Thus, the edges corresponding to the segments of the $i$-th row form a directed cycle (or loop) in $G$ that starts and ends at vertex $i$. Since there are $n$ rows, the $m$ edges of $G$ can be partitioned into $n$ such cycles.

Observe that deleting one edge from a cycle leaves a path that still connects all the vertices of that cycle. Therefore, after deleting one edge from each of the $n$ cycles corresponding to the rows, the vertex set of each cycle remains connected. Since the original graph $G$ is the connected union of these cycles, the remaining graph remains connected and spans all $n$ vertices.

At this point, the connected spanning subgraph must have at least $n-1$ edges. Since we deleted exactly $n$ edges (one from each row's cycle), the total number of edges in $G$ must be at least $n + (n-1) = 2n-1$.

Hence, the minimum value of $m$ is $2n-1$.

(Partial) 1. Gave the construction achieving m=2n-1 (e.g. cut lengths n,1,n-1,2,n-2,...,n-1,1) and correctly explained how to place the pieces row-by-row to obtain rows i,i+1,...,n,1,...,i-1 modulo n. 2. Reduced the problem to working modulo n (tape is 1,2,...,n repeated n times; target grid is the cyclic Latin square a_{i,j}≡i+j-1 (mod n)), and used this to justify that each row must be formed from a cyclic shift of 1,2,...,n. 3. Set up a valid lower-bound framework by encoding each segment as an edge a→b in a directed multigraph on {1,...,n} (where the segment reads a,a+1,...,b-1 mod n), and observed that the segments in each row correspond to a directed cycle. (Almost) 1. Proved m≥2n-1 via the directed-graph/cycle decomposition idea, and gave the construction for m=2n-1, but left a small gap in justifying that after deleting one edge from each row-cycle the remaining spanning subgraph is still connected (or equivalently that the union of the resulting paths must connect all n vertices). 2. Completed the lower bound m≥2n-1 except for a localized issue in translating the geometry of the grid into the statement that the m edges can be partitioned into n cycles (one per row), or in the argument that the graph spans all n vertices. 3. Solution is almost complete, but made minor mistakes which are not negligible.

Instruction:
For the specific case $n=150$, provide an explicit construction of the segments and their arrangement that achieves the minimum number of segments $m$.

Construction format requirements:
- Output a single Python dictionary with two keys: `'endpoints'` and `'rows'`.
- `'endpoints'` must be a list of strictly increasing positive integers $a_1 < a_2 < \dots < a_m = 150^2$, representing the endpoints of the segments. The $k$-th segment contains the numbers from $a_{k-1} + 1$ to $a_k$ (with $a_0 = 0$).
- `'rows'` must be a list of 150 lists. The $i$-th list specifies the order of segments used to form the $i$-th row of the grid. Segments are 1-indexed (i.e., $1, 2, \dots, m$).
- The total number of segments $m$ must be the minimum possible for $n=150$.
- The resulting $150 \times 150$ grid must satisfy the condition $a_{i,j} \equiv i+j-1 \pmod{150}$ for all $1 \le i, j \le 150$.
- Put ONLY the Python expression inside the construction block. The expression will be evaluated using `eval()` to obtain the dictionary, so you may use list comprehensions to keep the output compact. Do not include any imports, assignments, or markdown formatting.

import sys

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

    try:
        data = eval(raw_str)
    except Exception as e:
        print(f"Failed to parse input with eval: {e}")
        return

    if type(data) is not dict:
        print("Parsed object is not a dictionary")
        return

    if set(data.keys()) != {'endpoints', 'rows'}:
        print("Dictionary must contain exactly the keys 'endpoints' and 'rows'")
        return

    endpoints = data['endpoints']
    rows = data['rows']

    if type(endpoints) is not list:
        print("'endpoints' is not a list")
        return

    if len(endpoints) != 299:
        print(f"'endpoints' must have exactly 299 elements, got {len(endpoints)}")
        return

    if endpoints[-1] != 22500:
        print(f"The last element of 'endpoints' must be 22500, got {endpoints[-1]}")
        return

    prev = 0
    for i, val in enumerate(endpoints):
        if type(val) is not int:
            print(f"Element at index {i} in 'endpoints' is not an integer")
            return
        if val <= prev:
            print(f"'endpoints' must be strictly increasing positive integers, but found {prev} followed by {val}")
            return
        prev = val

    if type(rows) is not list:
        print("'rows' is not a list")
        return

    if len(rows) != 150:
        print(f"'rows' must have exactly 150 lists, got {len(rows)}")
        return

    seen_segments = set()
    for i, r in enumerate(rows, 1):
        if type(r) is not list:
            print(f"Row {i} in 'rows' is not a list")
            return
        for idx in r:
            if type(idx) is not int:
                print(f"Segment index {idx} in row {i} is not an integer")
                return
            if idx < 1 or idx > 299:
                print(f"Segment index {idx} in row {i} is out of bounds (1 to 299)")
                return
            if idx in seen_segments:
                print(f"Segment index {idx} appears more than once")
                return
            seen_segments.add(idx)

    if len(seen_segments) != 299:
        missing = set(range(1, 300)) - seen_segments
        print(f"Not all segment indices from 1 to 299 are used. Missing: {missing}")
        return

    segments = {}
    prev = 0
    for k, val in enumerate(endpoints, 1):
        segments[k] = list(range(prev + 1, val + 1))
        prev = val

    grid = []
    for i, r in enumerate(rows, 1):
        row_elements = []
        for idx in r:
            row_elements.extend(segments[idx])
        if len(row_elements) != 150:
            print(f"Reconstructed row {i} does not have exactly 150 elements (has {len(row_elements)})")
            return
        grid.append(row_elements)

    for i in range(150):
        for j in range(150):
            val = grid[i][j]
            expected = (i + 1 + j + 1 - 1) % 150
            if val % 150 != expected:
                print(f"Grid element at row {i+1}, col {j+1} is {val}, which is {val % 150} mod 150, expected {expected}")
                return

    print("True")

if __name__ == "__main__":
    verify()


Question: A \"stick\" is a $1\\times k$ or $k\\times 1$ rectangle, where $k$ is any positive integer. Find the smallest positive integer $m$ such that a $2022\\times2022$ board can be partitioned into $m$ disjoint sticks, and any two sticks share at most one unit-length common edge.

Let the $2022 \times 2022$ board be partitioned into $m$ disjoint sticks. Each stick is a rectangle of dimensions $1 \times k$ or $k \times 1$ for some integer $k \ge 1$, and thus has exactly $4$ vertices. Therefore, the total number of vertices of the $m$ sticks is $4m$.

We count the number of stick vertices located at the grid points of the board.
- **Interior points:** There are $2021^2$ grid points strictly inside the board. Consider any such interior point $P$. It is shared by $4$ unit squares. If $P$ were a vertex of $0$ sticks, the $4$ squares meeting at $P$ would be covered by sticks that pass straight through $P$. Because sticks have width $1$, this requires exactly two sticks to cover the $4$ squares (either two horizontal sticks or two vertical sticks). These two sticks would then share a boundary segment of length $2$ centered at $P$, meaning they share two unit-length common edges. This contradicts the condition that any two sticks share at most one unit-length common edge. If $P$ were a vertex of exactly $1$ stick, that stick would cover exactly $1$ of the $4$ squares meeting at $P$. The remaining $3$ squares cannot be covered by a single stick passing straight through $P$ (which would cover exactly $2$ squares), meaning at least one more stick must have a corner at $P$, making it a vertex of at least $2$ sticks. Thus, every interior point must be a vertex of at least $2$ sticks.
- **Corner points of the board:** There are $4$ corner points of the board. Each is incident to exactly $1$ square of the board, so each must be a vertex of exactly $1$ stick.
- **Boundary points (non-corners):** Consider the $4$ corner squares of the board. Each is covered by some stick. Since a stick has width $1$ and cannot extend outside the board, the stick covering a corner square must end at one or both of the boundary points adjacent to the corner. Thus, among the two boundary points adjacent to any corner of the board, at least one is a vertex of this stick. Because a non-corner boundary point is incident to exactly $2$ squares of the board, any stick vertex at this point must be paired with another stick vertex (a stick cannot pass straight through the boundary and exit the board). Thus, this point is a vertex of exactly $2$ sticks. Since the board is $2022 \times 2022$, these $4$ boundary points (at least one for each corner) are distinct. Hence, there are at least $4$ non-corner boundary points that are vertices of exactly $2$ sticks.

Summing the number of stick vertices over all grid points, we obtain:
$$ 4m \ge 2 \times 2021^2 + 1 \times 4 + 2 \times 4 $$
$$ 4m \ge 2(4084441) + 4 + 8 $$
$$ 4m \ge 8168882 + 12 = 8168894 $$

Dividing by $4$ yields:
$$ m \ge \frac{8168894}{4} = 2042223.5 $$

Since $m$ must be an integer, we conclude that the smallest positive integer $m$ is $2042224$.

(Partial) 1. Introduced counting of stick corners/vertices and the identity that each stick contributes exactly 4 vertices, so total stick-vertex count is 4m. 2. Proved the key local restriction at an interior grid point P: P cannot be a vertex of 0 sticks (else two sticks must pass straight and share a length-2 boundary, violating the “at most one unit edge” condition). 3. Proved that every interior grid point must be a vertex of at least 2 sticks (ruled out “exactly 1 stick has a corner at P” by showing the remaining 3 unit squares around P cannot be covered without another corner at P). (Almost) 1. Derived the lower bound 4m ≥ 2·2021^2 + 4 + (boundary contribution), and correctly concluded m ≥ 2042224, but the argument for the needed boundary contribution (at least four non-corner boundary points that are vertices of exactly 2 sticks) is incomplete or not fully justified. 2. Carried out the full vertex-counting approach and reached m ≥ 2042223.5, but made a localized counting slip (e.g., miscounted interior points as 2022^2 or mishandled corner/boundary points), while the overall method and use of the “share at most one edge” condition are correct. 3. Solution is almost complete, but made minor mistakes which are not negligible.

Instruction: Solve the original problem, but for a concrete 10 \times 10 board: output exactly one explicit partition achieving the minimum possible number of sticks under the same rules. Your final answer must consist of exactly one construction enclosed in <construct>...</construct>, where the payload is a single Python expression that evals to a 10\times 10 matrix (list of 10 lists of length 10) of integers. Equal integers denote cells belonging to the same stick; different integers denote different sticks. Each label's cells must form a single 1\times k or k\times 1 rectangle (k\geq 1) and the sticks must partition the grid. Any two distinct sticks may share at most one unit-length common edge in total. The set of labels used must be exactly {1,2,\cdots ,M} for some M, each used at least once, and M must be as small as possible for the 10×10 case.

import sys
import ast
from collections import defaultdict

def fail(msg):
    print(msg)
    sys.exit(0)

src = sys.stdin.read()

# 1) Evaluate payload in restricted environment
safe_globals = {
    "__builtins__": {},
    "range": range,
    "list": list,
    "dict": dict,
    "set": set,
    "tuple": tuple,
    "len": len,
    "min": min,
    "max": max,
    "sum": sum,
    "abs": abs,
    "sorted": sorted,
    "enumerate": enumerate,
    "zip": zip,
}
try:
    tree = ast.parse(src, mode="eval")
except Exception as e:
    fail(f"Parse error: {e}")

try:
    code = compile(tree, "<stdin>", "eval")
    obj = eval(code, safe_globals, {})
except Exception as e:
    fail(f"Eval error: {e}")

# 2) Confirm 10x10 matrix of strict ints
if not isinstance(obj, list):
    fail("Dimensions/type error: top-level must be a list (10 rows).")
if len(obj) != 10:
    fail(f"Dimensions error: expected 10 rows, got {len(obj)}.")
grid = obj
for r, row in enumerate(grid):
    if not isinstance(row, list):
        fail(f"Dimensions/type error: row {r} must be a list.")
    if len(row) != 10:
        fail(f"Dimensions error: row {r} expected length 10, got {len(row)}.")
    for c, v in enumerate(row):
        if type(v) is not int:
            fail(f"Cell type error: grid[{r}][{c}] must be a strict int, got {type(v).__name__}.")

# 3) Labels must be exactly {1..44}
labels = set()
for row in grid:
    labels.update(row)

if not labels:
    fail("Label error: no labels found.")
mn, mx = min(labels), max(labels)
if mn != 1:
    fail(f"Label set error: minimum label must be 1, got {mn}.")
if mx != 44:
    fail(f"Label set error: maximum label must be 44, got {mx}.")
expected = set(range(1, 45))
missing = sorted(expected - labels)
extra = sorted(labels - expected)
if missing or extra:
    if missing and extra:
        fail(f"Label set error: missing {missing}, extra {extra}.")
    if missing:
        fail(f"Label set error: missing labels {missing}.")
    fail(f"Label set error: extra labels {extra}.")

# 4) Each label forms one stick: fills bounding box, and box is 1xk or kx1
coords = {t: [] for t in range(1, 45)}
for r in range(10):
    for c in range(10):
        coords[grid[r][c]].append((r, c))

for t in range(1, 45):
    cells = coords[t]
    if not cells:
        fail(f"Stick error: label {t} is unused.")
    rs = [p[0] for p in cells]
    cs = [p[1] for p in cells]
    rmin, rmax = min(rs), max(rs)
    cmin, cmax = min(cs), max(cs)
    h = rmax - rmin + 1
    w = cmax - cmin + 1

    if not (h == 1 or w == 1):
        fail(f"Stick shape error: label {t} bounding box is {h}x{w}, not a 1xk or kx1 stick.")

    cellset = set(cells)
    expected_box = {(rr, cc) for rr in range(rmin, rmax + 1) for cc in range(cmin, cmax + 1)}
    if cellset != expected_box:
        missing_in_box = expected_box - cellset
        extra_outside = cellset - expected_box
        if missing_in_box:
            ex = next(iter(missing_in_box))
            fail(f"Stick fill error: label {t} does not fill its bounding box; missing cell {ex}.")
        ex = next(iter(extra_outside))
        fail(f"Stick fill error: label {t} has cell outside its bounding box? example {ex}.")

# 5) Any unordered label pair shares at most one unit-length common edge
shared = defaultdict(int)

for r in range(10):
    for c in range(10):
        a = grid[r][c]
        if c + 1 < 10:
            b = grid[r][c + 1]
            if a != b:
                x, y = (a, b) if a < b else (b, a)
                shared[(x, y)] += 1
        if r + 1 < 10:
            b = grid[r + 1][c]
            if a != b:
                x, y = (a, b) if a < b else (b, a)
                shared[(x, y)] += 1

for (x, y), cnt in shared.items():
    if cnt > 1:
        fail(f"Shared-boundary error: labels {x} and {y} share {cnt} unit edges (>1).")

print("True")


Question: A \"doughnut\" is a ring formed by removing an $a \\times b$ rectangle from the center of an $(a+2) \\times (b+2)$ rectangle, consisting of $2a + 2b + 4$ unit squares, where $a$ and $b$ are arbitrary positive integers. Consider an infinite grid. For each even $n \\geq 8$, an \"$n$-order bakery\" is a finite set $S$ of squares such that for any doughnut $B$ composed of $n$ squares, there exists a doughnut congruent to $B$ (by translation and rotation) whose squares all belong to $S$. Let $f(n)$ denote the minimum possible number of squares in an $n$-order bakery. Find the real number $\\alpha$ such that for all sufficiently large even $n$, we have $\\frac{1}{100} < \\frac{f(n)}{n^\\alpha} < 100$.

We say that $g(n)$ is "at least arbitrarily close" to $h(n)$ if
\[
\lim_{n\to\infty}\frac{g(n)}{h(n)}\ge 1.
\]
A similar definition gives "at most arbitrarily close." Since this problem concerns only sufficiently large integers, we may work in the above sense.

We first prove that $f(n)$ is at least arbitrarily close to
\[
\frac{1}{8}n^{\frac{3}{2}}.
\]

Call a doughnut "horizontal" if its longer side lies in a row of the grid. Assume that at least half of the doughnuts in $S$ are horizontal. Because there are $\lfloor \frac{n}{4}-1 \rfloor$ distinct doughnuts, there are at least $\frac{n}{8}-1$ horizontal doughnuts.

Let the longer sides of these horizontal doughnuts lie in $k$ rows of $S$. Since their shorter sides are all different, we must choose pairs of rows to form the horizontal boundaries. Thus, we have
\[
\binom{k}{2}\ge \frac{n}{8}-1,
\]
which implies that $k$ is at least arbitrarily close to
\[
\frac{\sqrt{n}}{2}.
\]

Each longer side contains at least $\frac{n}{4}+1$ squares of $S$, so
\[
f(n)\text{ is at least arbitrarily close to }\frac{\sqrt{n}}{2}\cdot\frac{n}{4}
= \frac{1}{8}n^{\frac{3}{2}}.
\]

We now prove that $f(n)$ is at most arbitrarily close to $\sqrt{2}\,n^{\frac{3}{2}}$.
Let $l=\left\lceil\sqrt{\frac{n}{2}}\right\rceil$, and define $S$ to consist of grid squares $(x,y)$ with $1\le x,y\le l^{2}$ such that at least one of $x,y$ belongs to the set
\[
A=\{1,2,\dots,l\}\cup\{2l,3l,\dots,l^{2}\}.
\]
Note that $\{1,2,\dots,l^{2}-1\}\subseteq\{a_{1}-a_{2}\mid a_{1},a_{2}\in A\}$, i.e., every integer from $1$ to $\frac{n}{2}-1$ can be expressed as the difference of two elements of $A$.
Moreover, each doughnut consisting of $n$ squares has its longer side containing at most $\frac{n}{2}-1$ squares, so any such doughnut can be placed entirely inside $S$.

At this point, the number of coordinates in $\{1, 2, \dots, l^2\}$ not in $A$ is $l^2 - (2l-1) = (l-1)^2$. The number of squares $(x,y)$ where neither coordinate is in $A$ is therefore $((l-1)^2)^2 = (l-1)^4$. Thus, $S$ contains exactly $l^{4}-(l-1)^{4}$ squares. Because $l^{4}-(l-1)^{4} = 4l^3 - 6l^2 + 4l - 1$, $f(n)$ is at most arbitrarily close to $4l^{3}$, i.e., $\sqrt{2}\,n^{\frac{3}{2}}$.

Since $f(n)$ is bounded between $\sim \frac{1}{8}n^{3/2}$ and $\sim \sqrt{2}n^{3/2}$ for large $n$, we conclude that the real number $\alpha$ satisfying $\frac{1}{100} < \frac{f(n)}{n^\alpha} < 100$ for all sufficiently large $n$ is $\alpha = \frac{3}{2}$.

(Partial) 1. Established a nontrivial lower bound of the form f(n) \ge c\,n^{3/2} (e.g. by counting many distinct doughnut shapes and relating them to pairs of rows/columns, obtaining a \binom{k}{2} bound and hence k \gg \sqrt n). 2. Established a nontrivial upper bound of the form f(n) \le C\,n^{3/2} by explicitly constructing a bakery S with |S| = O(n^{3/2}) and explaining how any n-square doughnut can be translated/rotated into S. 3. Correctly reduced the problem to determining the growth exponent by proving both f(n)=\Omega(n^{3/2}) and f(n)=O(n^{3/2}) (even with weak/unspecified constants), hence identifying \alpha=3/2 but with major gaps in at least one direction. (Almost) 1. Proved the lower bound via the “horizontal doughnuts” counting argument (many distinct horizontal shapes; their long sides lie in k rows; need \binom{k}{2} at least the number of shapes; conclude k\ge(1/2+o(1))\sqrt n; then |S|\ge k\cdot(n/4+O(1)) ), but made a small slip in the exact count of distinct doughnuts or the ‘at least half are horizontal’ step. 2. Gave essentially the full explicit construction with l=\lceil\sqrt{n/2}\rceil and a set A so that every length \le n/2-1 is a difference of two elements of A, and computed |S|=l^4-(l-1)^4=4l^3+O(l^2), but left a minor gap in justifying that every doughnut’s longer side length is \le n/2-1 or in the “difference set covers all needed side-lengths” claim. 3. Completed both bounds and concluded \alpha=3/2, but omitted a boundary case (e.g. rotations/which side is longer, or small exceptional n) or had a localized non-negligible mistake in the constant-factor estimates while the exponent argument is correct.

Instruction:
Provide an explicit construction of a bakery for $n=448$.

Construction format requirements:
- Output a single Python list of lists of integers representing a 2D matrix of 0s and 1s.
- The matrix can be of any size, but it must contain at most 12300 ones.
- For every possible valid doughnut composed of exactly $n=448$ unit squares, the matrix must contain at least one congruent copy (up to translation and rotation) made entirely of 1s.
- A doughnut is formed by removing an $a \times b$ rectangle from the center of an $(a+2) \times (b+2)$ rectangle, consisting of $2a + 2b + 4$ unit squares ($a, b \ge 1$).

Output formatting requirements:
- Include exactly one construction block and put ONLY the Python list comprehension or literal between the tags:
<construct>
...
</construct>
- Do not use markdown formatting (no ```python ... ```).
- The payload must be a single evaluable Python expression.

import sys

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

    try:
        # Use eval to support list comprehensions as seen in the reference construction.
        # This is necessary because ast.literal_eval does not support comprehensions.
        matrix = eval(payload)
    except Exception as e:
        print(f"Failed to parse input: {e}")
        return

    if not isinstance(matrix, list):
        print("Input must be a list of lists.")
        return

    if len(matrix) == 0:
        print("Matrix cannot be empty.")
        return

    num_rows = len(matrix)
    if not all(isinstance(row, list) for row in matrix):
        print("Input must be a list of lists.")
        return

    num_cols = len(matrix[0])
    if num_cols == 0:
        print("Matrix rows cannot be empty.")
        return

    for row in matrix:
        if len(row) != num_cols:
            print("Matrix must be rectangular.")
            return
        for cell in row:
            if cell not in (0, 1):
                print("Matrix must contain only 0s and 1s.")
                return

    total_ones = sum(sum(row) for row in matrix)
    if total_ones > 12300:
        print(f"Matrix contains {total_ones} ones, which exceeds the maximum allowed of 12300.")
        return

    # Precompute contiguous horizontal 1s to the right
    horiz = [[0] * num_cols for _ in range(num_rows)]
    for r in range(num_rows):
        cnt = 0
        for c in range(num_cols - 1, -1, -1):
            if matrix[r][c] == 1:
                cnt += 1
            else:
                cnt = 0
            horiz[r][c] = cnt

    # Precompute contiguous vertical 1s downwards
    vert = [[0] * num_cols for _ in range(num_rows)]
    for r in range(num_rows - 1, -1, -1):
        for c in range(num_cols):
            if matrix[r][c] == 1:
                vert[r][c] = 1 + (vert[r+1][c] if r + 1 < num_rows else 0)
            else:
                vert[r][c] = 0

    # Collect all coordinates of 1s to optimize the search
    ones = [(r, c) for r in range(num_rows) for c in range(num_cols) if matrix[r][c] == 1]

    def check_doughnut_fast(a, b):
        """
        Checks if a doughnut of size (a+2) x (b+2) exists in the matrix.
        The top-left corner of the doughnut must be at some (r, c).
        """
        for r, c in ones:
            if r + b + 1 < num_rows and c + a + 1 < num_cols:
                if horiz[r][c] >= a + 2 and \
                   horiz[r+b+1][c] >= a + 2 and \
                   vert[r][c] >= b + 2 and \
                   vert[r][c+a+1] >= b + 2:
                    return True
        return False

    # Iterate through all possible integer pairs (a, b) such that a >= 1, b >= 1, and a + b = 222
    for a in range(1, 222):
        b = 222 - a

        # Verify that the matrix contains the corresponding doughnut.
        # We check both (a+2)x(b+2) and (b+2)x(a+2) to account for valid rotations.
        if not (check_doughnut_fast(a, b) or check_doughnut_fast(b, a)):
            print(f"Missing doughnut for a={a}, b={b}.")
            return

    print("True")

if __name__ == "__main__":
    verify()