Question: A draws a regular 2021-gon in the plane. B labels each vertex with a real number such that the labels on any two adjacent vertices differ by at most 1. For any two non-adjacent vertices whose labels differ by at most 1, A draws a diagonal connecting them. Let $d$ denote the number of diagonals drawn by A. Find the minimum possible value of $d$.

The minimum possible value of $d$ is $2018$.

There exists a valid numbering of the vertices such that exactly $d = 2018$ diagonals are drawn.

We prove by induction that for a general $n$-gon ($n \ge 3$), A must draw at least $n-3$ diagonals.
When $n=3$, the number of diagonals drawn is at least $3-3=0$, which is obvious.
Assume the statement holds for $n-1$, and consider the case $n$. Let the label on vertex $v$ be $a$, which is the smallest among all labels, and let the labels on the two vertices adjacent to $v$, namely $v_1$ and $v_2$, be $b$ and $c$, respectively. Since the labels on adjacent vertices differ by at most $1$, we have $0 \le b-a \le 1$ and $0 \le c-a \le 1$. This implies that $|b-c| \le 1$, so $v_1$ and $v_2$ are connected by a diagonal.

Remove vertex $v$ to obtain an $(n-1)$-gon; the segment $v_1v_2$ is an edge of this smaller polygon. By the induction hypothesis, this $(n-1)$-gon has at least $n-4$ diagonals. Adding the diagonal $v_1v_2$ shows that the original $n$-gon has at least $(n-4) + 1 = n-3$ diagonals. The induction is complete.

For $n=2021$, A must draw at least $2021-3 = 2018$ diagonals. Thus, the minimum possible value of $d$ is indeed $2018$.

(Partial) 1. Gave a labeling achieving $d=2018$ (or in general $d=n-3$), and correctly checked that adjacent vertices differ by at most 1 and that exactly those $2018$ non-adjacent pairs with label difference at most 1 occur. 2. Proved the key structural step: if $v$ has minimal label $a$ and its neighbors have labels $b,c$, then $0\le b-a\le1$ and $0\le c-a\le1$, hence $|b-c|\le1$ and therefore $v_1v_2$ must be drawn. 3. Correctly set up the induction/reduction $n\to n-1$ by removing a vertex of minimal label so that $v_1v_2$ becomes an edge of the smaller polygon, even if the diagonal-count conclusion is not fully justified. (Almost) 1. Proved the lower bound $d\ge n-3$ (hence $d\ge2018$) by induction using the minimal-label vertex and deletion, but did not provide (or incorrectly verified) a construction achieving $2018$. 2. Gave both the construction and the inductive lower bound idea, but left a small gap in the induction step (e.g. not justifying that diagonals in the $(n-1)$-gon are also diagonals of the original $n$-gon, or double-counting/omitting the added diagonal $v_1v_2$). 3. Solution is almost complete, but made minor mistakes which are not negligible.

Instruction: Provide an explicit labeling for the original regular 2021-gon: output exactly one Python expression which, when eval-ed, returns a list of 2021 real numbers (int/float), where the i-th entry is the label of vertex i (in cyclic order). The labels must satisfy that for every pair of adjacent vertices (including vertices 2021 and 1), the absolute difference of their labels is at most 1. Among all such labelings, your labeling must make the number of diagonals drawn by A (i.e., the number of unordered pairs of non-adjacent vertices whose labels differ by at most 1) as small as possible. Put the expression as the only content inside <construct>...</construct>.

import sys, math, ast

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

s = sys.stdin.read()
if s is None:
    fail("Parse error: empty input")
s = s.strip()
if not s:
    fail("Parse error: empty input")

# Parse as a Python expression (strict), then eval in restricted environment
try:
    tree = ast.parse(s, mode="eval")
except Exception as e:
    fail(f"Parse error: {e}")

try:
    codeobj = compile(tree, "<stdin>", "eval")
    val = eval(codeobj, {"__builtins__": {}, "range": range}, {})
except Exception as e:
    fail(f"Eval error: {e}")

# Check list length
if not isinstance(val, list):
    fail(f"Type error: expected a list, got {type(val).__name__}")
if len(val) != 2021:
    fail(f"Length error: expected 2021 labels, got {len(val)}")

# Check each element is int/float (reject bool and non-finite)
a = val
for idx, x in enumerate(a):
    if isinstance(x, bool) or not isinstance(x, (int, float)):
        fail(f"Element type error at index {idx}: expected int/float, got {type(x).__name__}")
    if not math.isfinite(float(x)):
        fail(f"Element value error at index {idx}: non-finite value {x!r}")

# Check adjacency constraints
n = 2021
for i in range(n):
    j = (i + 1) % n
    diff = abs(float(a[i]) - float(a[j]))
    if diff > 1 + 1e-12:
        fail(f"Adjacency constraint violated at indices ({i},{j}): values {a[i]!r}, {a[j]!r}, diff={diff}")

# Count diagonals
d = 0
for i in range(n):
    ai = float(a[i])
    for j in range(i + 1, n):
        # adjacent if consecutive or (0, n-1)
        if j == i + 1 or (i == 0 and j == n - 1):
            continue
        if abs(ai - float(a[j])) <= 1 + 1e-12:
            d += 1

if d != 2018:
    fail(f"Diagonal count incorrect: computed d={d}, expected 2018")

print("True")


Question:
"Given an integer $n>1$. An $n\\times n$ chessboard is called a direction diagram if each of its $n^2$ squares contains an arrow pointing to one of the four directions: up, down, left, or right. For an initial direction diagram, a turbo snail starts moving from a certain square on the chessboard. In each move, the snail advances one square in the direction of the arrow in its current square (it may move off the board). After each move, all arrows on the chessboard rotate $90^\\circ$ counterclockwise simultaneously. A square is called good if, starting from this square and moving according to the above rules, the snail passes through each square of the chessboard exactly once, never moves off the board, and finally returns to the initial square. For a given $n$, determine the maximum number of good squares over all possible initial direction diagrams.
"

It is $0$ when $n$ is odd, and $\\frac{n^2}{4}$ when $n$ is even.

The maximum number of good squares is $0$ when $n$ is odd, and $\frac{n^2}{4}$ when $n$ is even.

For odd $n$, coloring the board black and white demonstrates that the snail moves between cells of alternating colors on each turn. Since the total number of cells is odd, a Hamiltonian cycle cannot exist, yielding exactly $0$ good cells.

For even $n$, choosing a Hamiltonian cycle and setting the initial arrow directions accordingly allows the starting cell to be good. Along this cycle, exactly one in every four consecutive cells is a good cell, because the arrow direction aligns with the required path only every four moves due to the simultaneous $90^\circ$ rotations. This establishes a lower bound of $\frac{n^2}{4}$ good cells.

At the top-left corner of the board, there are exactly four valid local configurations for the sequence of moves entering and exiting the corner cell. Because one cannot obtain another configuration by simultaneously rotating the three arrows involved in a given configuration, any fixed initial arrow diagram uniquely determines how a Hamiltonian cycle passes through the top-left corner. This uniquely dictates the subsequent route along the arrow directions, meaning there is only one valid Hamiltonian cycle. Since the number of moves from any good cell to the top-left corner must be congruent modulo $4$, the number of good cells is exactly $\frac{n^2}{4}$.

(Partial) 1. For odd n, used a correct parity/colouring argument (e.g. checkerboard colouring) to show the snail alternates colours each move and hence a Hamiltonian cycle visiting all n^2 cells is impossible, concluding there are no good squares. 2. For even n, gave a valid construction from a Hamiltonian cycle: defined initial arrows so that starting from some chosen cell the snail follows the cycle while arrows rotate, producing at least one good square (or a positive proportion of them). 3. Established the key “mod 4” alignment fact: along any fixed cycle, a starting square is good only when the local required direction matches the current arrow orientation, which occurs with period 4, implying at most one in every four consecutive vertices on the cycle can be good. (Almost) 1. Correctly obtained the lower bound n^2/4 for even n via a Hamiltonian-cycle construction, but did not fully justify the matching upper bound (e.g. stated “at most one quarter are good” without a complete argument). 2. Proved the upper bound n^2/4 for even n using the mod-4 restriction, but left a small gap in showing that all good starting squares must lie in the same congruence class (e.g. did not rigorously justify why the path/cycle is uniquely forced from a corner or another anchoring configuration). 3. Gave a complete odd-n impossibility proof and the even-n framework, but made a localized mistake in the corner/local-configuration forcing step (or omitted checking a boundary/local case), while the rest of the solution is correct and would yield the stated maximum.

Instruction:
For the specific case $n=64$, determine a valid initial direction diagram that maximizes the number of "good squares". Provide an explicit construction for this diagram.

Construction requirements:
- Output a single Python expression that evaluates to a $64 \times 64$ list of lists of integers.
- The integers must be in $\{0, 1, 2, 3\}$, representing the initial arrow directions: 0 for right, 1 for up, 2 for left, and 3 for down.
- The output must be a pure Python expression (e.g., a list comprehension). Do not include any explanations, comments, markdown formatting, print statements, assignments, function definitions, imports, classes, asserts, or loops outside of comprehensions.
- Prioritize using list comprehensions, conditional expressions, and integer constants to generate a compact, regular, and reproducible expression.
- Put ONLY the Python expression inside the construction block.

[[(((0 if c < 63 else 3) if r == 0 else (1 if c == 0 else ((2 if c > 1 else (3 if r < 63 else 2)) if r % 2 == 1 else (0 if c < 63 else 3)))) - (c if r == 0 else (4096 - r if c == 0 else (r * 63 + 64 - c if r % 2 == 1 else r * 63 + c)))) % 4 for c in range(64)] for r in range(64)]

import sys

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

    try:
        # Evaluate the payload to obtain the 64x64 grid
        # eval is used here to properly parse list comprehensions as required by the instruction
        grid = eval(payload)
    except Exception as e:
        print(f"Failed to parse input: {e}")
        return

    # Validate structural correctness
    if type(grid) is not list:
        print("Grid must be a list.")
        return
    if len(grid) != 64:
        print(f"Grid must have exactly 64 rows, got {len(grid)}.")
        return

    flat_grid = [0] * 4096
    for i, row in enumerate(grid):
        if type(row) is not list:
            print(f"Row {i} is not a list.")
            return
        if len(row) != 64:
            print(f"Row {i} must have exactly 64 elements, got {len(row)}.")
            return
        for j, val in enumerate(row):
            if type(val) is not int or val not in {0, 1, 2, 3}:
                print(f"Element at ({i}, {j}) is not an integer in {{0, 1, 2, 3}}.")
                return
            flat_grid[(i << 6) | j] = val

    # Simulation setup
    visited = [0] * 4096
    attempt = 0
    good_squares_count = 0

    # Simulate snail movement from every possible starting square
    for r in range(64):
        for c in range(64):
            attempt += 1
            curr_r, curr_c = r, c
            success = True

            for t in range(4096):
                # Check out-of-bounds
                if curr_r < 0 or curr_r > 63 or curr_c < 0 or curr_c > 63:
                    success = False
                    break

                idx = (curr_r << 6) | curr_c

                # Check if already visited in the current attempt
                if visited[idx] == attempt:
                    success = False
                    break

                # Mark as visited
                visited[idx] = attempt

                # Calculate current direction with 90-degree counterclockwise rotation (adding 1 mod 4 per step)
                d = (flat_grid[idx] + t) & 3

                # Update coordinates
                if d == 0:     # right
                    curr_c += 1
                elif d == 1:   # up
                    curr_r -= 1
                elif d == 2:   # left
                    curr_c -= 1
                else:          # down
                    curr_r += 1

            # After 4096 steps, check if it successfully returned to the start
            if success and curr_r == r and curr_c == c:
                good_squares_count += 1

    # Verify that the maximum number of good squares is achieved
    if good_squares_count != 1024:
        print(f"Expected 1024 good squares, but got {good_squares_count}.")
        return

    print("True")

if __name__ == "__main__":
    verify()


Question:
"Given a positive integer $n$. Place several dominoes on a $2n\\times 2n$ grid such that every cell of the grid is adjacent to exactly one cell covered by a domino. Find the maximum number of dominoes that can be placed. Note: A domino is a $1\\times2$ or $2\\times1$ grid. Each domino must be placed on two cells of the grid, and different dominoes cannot overlap. Two distinct cells are called adjacent if they share a common edge.
"

We will prove that the maximum number of dominoes that can be placed is $\frac{n^2+n}{2}$.

Let the cells of the $2n \times 2n$ grid be denoted by coordinates $(x,y)$ for $1 \le x, y \le 2n$. We define the "layer" or "depth" of a cell as its distance from the boundary of the grid:
$$ d(x,y) = \min(x, 2n+1-x, y, 2n+1-y) $$
Note that $1 \le d(x,y) \le n$. We color the grid with two colors, black and white, by defining the set of black cells $B$ as those belonging to odd layers:
$$ B = \{ (x,y) \mid d(x,y) \text{ is odd} \} $$

**Claim 1:** Every cell in the grid is adjacent to exactly two black cells.
*Proof.* Let $(x,y)$ be a cell in layer $m$, so $d(x,y) = m$. Its neighbors can only belong to layers $m-1$, $m$, or $m+1$.
- If $(x,y)$ is not a corner of its layer, it has exactly two neighbors in layer $m$, one neighbor in layer $m-1$, and one neighbor in layer $m+1$. Since $m-1$ and $m+1$ have the same parity, which is opposite to the parity of $m$, the cell $(x,y)$ has either 2 neighbors in an odd layer (if $m$ is odd) and 2 in an even layer, or 2 neighbors in an even layer (if $m$ is even) and 2 in an odd layer. In either case, it has exactly 2 neighbors in $B$.
- If $(x,y)$ is a corner of layer $m$, it has exactly two neighbors in layer $m$, two neighbors in layer $m-1$, and zero neighbors in layer $m+1$. By the same parity argument, exactly two of its neighbors belong to $B$. (For $m=1$, the layer $m-1$ is empty, meaning the corner cell has only 2 neighbors in total, both of which are in layer $1 \subseteq B$).
Thus, every cell in the grid has exactly two black neighbors.

**Claim 2:** The total number of black cells is $2n^2 + 2n$.
*Proof.* The layer $L_m = \{(x,y) \mid d(x,y) = m\}$ forms a square border of size $(2n-2m+2) \times (2n-2m+2)$. The number of cells in $L_m$ is $4(2n-2m+1) = 8n - 8m + 4$. Summing this over all odd $m \in \{1, 3, \dots\}$ up to $n$ yields exactly $2n^2 + 2n$ black cells.

**Upper Bound:**
Let $D$ be the set of cells covered by the dominoes. Suppose there are $k$ dominoes; since each domino covers exactly 2 cells, $|D| = 2k$.
The problem requires that every cell in the grid is adjacent to exactly one cell in $D$. Therefore, the total number of adjacencies (edges) between the set of all grid cells and $D$ must equal the total number of grid cells.
By restricting our focus to the black cells, the number of edges between $B$ and $D$ is exactly $|B| = 2n^2 + 2n$, because each black cell has exactly one neighbor in $D$.
On the other hand, by Claim 1, every cell in $D$ is adjacent to exactly two black cells. Thus, the number of edges between $B$ and $D$ can also be counted as $2|D| = 4k$.
Equating the two counts gives $4k = 2n^2 + 2n$, which implies $k = \frac{n^2+n}{2}$. This proves that any valid placement must consist of exactly $\frac{n^2+n}{2}$ dominoes.

**Construction of exactly $\frac{n^2+n}{2}$ dominoes:**
To show that such a placement is always possible, we construct it layer by layer. For each odd layer $m \in \{1, 3, \dots\}$ with $m \le n$, the subgrid bounded by $L_m$ has side length $S = 2n - 2m + 2$. We place exactly $S - 1$ dominoes on $L_m$ using local coordinates $(i,j)$ for $1 \le i, j \le S$ (which correspond to global coordinates $(i+m-1, j+m-1)$).

- **If $S \equiv 2 \pmod 4$:**
  - **Top edge:** Horizontal dominoes covering $(1, j)$ and $(1, j+1)$ for $j \in \{1, 5, 9, \dots, S-1\}$.
  - **Bottom edge:** Horizontal dominoes covering $(S, j)$ and $(S, j+1)$ for $j \in \{3, 7, 11, \dots, S-3\}$.
  - **Left edge:** Vertical dominoes covering $(i, 1)$ and $(i+1, 1)$ for $i \in \{4, 8, 12, \dots, S-2\}$.
  - **Right edge:** Vertical dominoes covering $(i, S)$ and $(i+1, S)$ for $i \in \{4, 8, 12, \dots, S-2\}$.

- **If $S \equiv 0 \pmod 4$:**
  - **Top edge:** Horizontal dominoes covering $(1, j)$ and $(1, j+1)$ for $j \in \{2, 6, 10, \dots, S-2\}$.
  - **Bottom edge:** Horizontal dominoes covering $(S, j)$ and $(S, j+1)$ for $j \in \{4, 8, 12, \dots, S-4\}$.
  - **Left edge:** Vertical dominoes covering $(i, 1)$ and $(i+1, 1)$ for $i \in \{3, 7, 11, \dots, S-1\}$.
  - **Right edge:** Vertical dominoes covering $(i, S)$ and $(i+1, S)$ for $i \in \{3, 7, 11, \dots, S-1\}$.

This staggered boundary placement ensures that every cell in the grid is adjacent to exactly one covered cell. Summing the dominoes placed across all odd layers gives $\sum_{m \text{ odd}} (S - 1) = \frac{n^2+n}{2}$.

Thus, the maximum number of dominoes that can be placed is $\frac{n^2+n}{2}$.

(Partial) 1. Introduced the layer depth d(x,y)=min(x,2n+1−x,y,2n+1−y) and the colouring by parity of d, and correctly proved the key local fact that every cell has exactly two neighbours in odd layers (equivalently: exactly two black neighbours). 2. Counted the number of black cells by summing the sizes of the odd layers (or an equivalent correct counting), obtaining |B|=2n^2+2n (or an equivalent explicit formula). 3. Set up a correct double-counting of edges between black cells and covered cells D to derive a linear relation 4k=|B| (or equivalent), thus giving the upper bound k≤(n^2+n)/2, even if the details are incomplete. (Almost) 1. Gave the correct double-counting argument and obtained k=(n^2+n)/2 as a necessary value, but did not provide a complete existence construction (or only sketched it without verifying the adjacency condition everywhere). 2. Provided a near-complete layer-by-layer construction achieving (n^2+n)/2 dominoes, but left a small gap (e.g. one boundary/parity case S≡0 or 2 (mod 4) not checked, or a corner/last-layer adjacency not fully verified). 3. Solution is almost complete, but made minor mistakes which are not negligible.

Instruction:
For the specific case $n=80$, provide an explicit construction for the maximum number of dominoes.

Construction requirements:
- Output a single Python expression (using list comprehensions) that evaluates to a flat list of tuples.
- Each tuple must be a 1-indexed coordinate `(r, c)` representing a cell on the $160 \times 160$ grid.
- Adjacent tuples at indices $2i$ and $2i+1$ represent a single domino covering those two cells.
- The configuration must maximize the number of dominoes placed, while satisfying the condition that every cell in the $160 \times 160$ grid is adjacent to exactly one cell covered by a domino.

Output formatting requirements:
- The output must be a single, valid Python expression.
- Do NOT wrap the output in markdown code blocks (e.g., no ```python ... ```).
- Do NOT include any explanations, comments, print statements, or assignments.
- 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 to support list comprehensions
        dominoes_cells = eval(payload)
    except Exception as e:
        print(f"Failed to parse input: {e}")
        return

    if not isinstance(dominoes_cells, list):
        print("Output must be a list.")
        return

    if len(dominoes_cells) != 6480:
        print(f"Expected exactly 6480 cells (3240 dominoes), got {len(dominoes_cells)}.")
        return

    for cell in dominoes_cells:
        if not (isinstance(cell, tuple) and len(cell) == 2):
            print(f"Invalid cell format: {cell}. Expected a 2-tuple.")
            return
        r, c = cell
        if type(r) is not int or type(c) is not int:
            print(f"Coordinates must be integers: {cell}.")
            return
        if not (1 <= r <= 160 and 1 <= c <= 160):
            print(f"Coordinates out of bounds: {cell}. Expected range [1, 160].")
            return

    for i in range(0, 6480, 2):
        c1 = dominoes_cells[i]
        c2 = dominoes_cells[i+1]
        dist = abs(c1[0] - c2[0]) + abs(c1[1] - c2[1])
        if dist != 1:
            print(f"Cells at indices {i} and {i+1} do not form a valid domino: {c1}, {c2}. Manhattan distance is {dist}, expected 1.")
            return

    covered_cells = set(dominoes_cells)
    if len(covered_cells) != 6480:
        print("Overlapping dominoes detected. Covered cells are not completely distinct.")
        return

    # Check neighbor count for every cell in the 160x160 grid
    for r in range(1, 161):
        for c in range(1, 161):
            neighbors = [
                (r-1, c),
                (r+1, c),
                (r, c-1),
                (r, c+1)
            ]
            count = sum(1 for nr, nc in neighbors if (nr, nc) in covered_cells)
            if count != 1:
                print(f"Cell ({r}, {c}) is adjacent to {count} covered cells, expected exactly 1.")
                return

    print("True")

if __name__ == "__main__":
    verify()


Question:
"Let integers $k,n$ satisfy $k\\geq2$, $k\\leq n\\leq 2k-1$. Place some $1\\times k$ or $k\\times 1$ rectangular cards on an $n\\times n$ grid such that each card covers exactly $k$ cells, no two cards overlap, and no additional card can be placed. Determine the minimum possible number of cards placed among all such placements.
"

When $n=k$ it is $n$, and when $k+1\le n\le 2k-1$ it is $\min\{n,2n-2k+2\}$

The minimum possible number of cards is $n$ when $n=k$, and $\min\{n, 2n-2k+2\}$ when $k+1 \le n \le 2k-1$.

**Constructions:**
When $n=k$, a total of $n$ pieces can be placed by filling the grid with $n$ horizontal pieces (one in each row) or $n$ vertical pieces (one in each column). Because a piece of length $k$ in a $k\times k$ grid spans an entire row or column, placing even one horizontal piece prevents any vertical piece from being placed, and vice versa. Thus, all pieces must be of the same orientation, and a maximal placement requires exactly $n$ pieces.

When $k+1 \le n \le 2k-2$, $2n-2k+2$ pieces can be placed as follows:
In the upper-left $(k+1)\times(k+1)$ square, place $4$ pieces arranged in a pinwheel fashion: a $1\times k$ piece in row $1$, columns $1$ to $k$; a $k\times 1$ piece in column $k+1$, rows $1$ to $k$; a $1\times k$ piece in row $k+1$, columns $2$ to $k+1$; and a $k\times 1$ piece in column $1$, rows $2$ to $k+1$. This leaves a central $(k-1)\times(k-1)$ empty square. For the remaining grid, the next $n-k-1$ rows (rows $k+2$ to $n$) each contain one $1\times k$ horizontal piece in columns $1$ to $k$, and the next $n-k-1$ columns (columns $k+2$ to $n$) each contain one $k\times 1$ vertical piece in rows $1$ to $k$. It is easy to verify that no additional piece can be inserted, as the remaining empty spaces are too small to accommodate a $k$-length piece. This uses exactly $4 + 2(n-k-1) = 2n-2k+2$ pieces.

When $n=2k-1$, $n$ pieces can be placed as follows: all pieces are placed vertically, alternating between the topmost and bottommost positions of each column. Specifically, for each column $j$ from $1$ to $n$, place a $k\times 1$ piece in rows $1$ to $k$ if $j$ is odd, and in rows $k$ to $2k-1$ if $j$ is even. It is easy to verify that no additional piece can be inserted at this stage, as each column has only $k-1$ empty cells, and any row has at most $1$ consecutive empty cell (row $k$ has none), meaning no $1\times k$ or $k\times 1$ piece can fit.

**Proof of the Lower Bound:**
When $k+1 \le n \le 2k-1$, we prove that any placement satisfying the requirements contains at least $\min\{n, 2n-2k+2\}$ pieces. Let $r$ be the number of rows that do not contain any horizontal piece, and $c$ be the number of columns that do not contain any vertical piece.

If $r=0$ or $c=0$, then at least $n$ pieces are placed. (For instance, if $r=0$, every row contains at least one horizontal piece. Since a horizontal piece is contained entirely in one row, there must be at least $n$ horizontal pieces.)

If both $r$ and $c$ are non-zero, we prove that at least $2n-2k+2$ pieces are placed. Consider a horizontal piece $T$. Since $n < 2k$, the top or bottom boundary of the board is at most $k-1$ rows away from the row containing $T$. Assume without loss of generality that the top boundary is closer. Let $y$ be the row containing $T$; then $y-1 \le k-1$. The rows strictly between $T$ and the top boundary (rows $1$ to $y-1$) intersect the $k$ columns occupied by $T$, forming a rectangular region of size $(y-1) \times k$.

This region cannot accommodate a vertical piece, because any vertical piece intersecting this region must not overlap with $T$, and thus would be restricted to rows $1$ to $y-1$. However, $y-1 < k$, so a $k \times 1$ vertical piece cannot fit. Therefore, no vertical piece intersects this region. Since the placement is maximal, no additional horizontal piece can be placed. Thus, every row in this region must already share at least one cell with a horizontal piece (otherwise, all $k$ cells in that row within the region would be empty, allowing a new horizontal piece to be placed).

Since a horizontal piece is entirely contained in a single row, this means every row from $1$ to $y-1$ must contain a horizontal piece. This argument shows that for any row containing a horizontal piece, all rows between it and the closest boundary (top or bottom) must also contain a horizontal piece. Consequently, the set of rows containing a horizontal piece forms a union of a prefix of rows and/or a suffix of rows. Therefore, the $r$ rows lacking a horizontal piece must be strictly consecutive.

Similarly, all $c$ columns lacking a vertical piece are consecutive. The $r$ rows and $c$ columns thus intersect to form an $r \times c$ rectangular region. Because no horizontal piece is in these $r$ rows and no vertical piece is in these $c$ columns, this region contains no pieces at all. Since no new piece can be placed in this empty region, we must have $r < k$ and $c < k$.

Consequently, there are $n-r \ge n-k+1$ rows that each contain at least one horizontal piece, implying there are at least $n-k+1$ horizontal pieces. Similarly, there are $n-c \ge n-k+1$ columns that each contain at least one vertical piece, implying there are at least $n-k+1$ vertical pieces. Hence, at least $(n-k+1) + (n-k+1) = 2n-2k+2$ pieces are placed in total. This completes the proof.

(Partial) 1. Gave a correct construction achieving the upper bound in at least one nontrivial case (e.g. the pinwheel-based construction with exactly $2n-2k+2$ cards for $k+1\le n\le 2k-2$, or the alternating-columns construction with $n$ cards for $n=2k-1$), and justified maximality of that construction. 2. Proved the structural claim that if both orientations occur in a maximal placement then the set of rows containing a horizontal card is a union of a top prefix and a bottom suffix (equivalently, the rows with no horizontal card form one consecutive block), and analogously for columns/vertical cards. 3. Established a valid lower bound of the right order, e.g. showed that if $r$ rows have no horizontal card and $c$ columns have no vertical card, then the $r\times c$ intersection must be empty and hence $r<k$ and $c<k$, leading to at least $(n-k+1)+(n-k+1)=2n-2k+2$ cards (even if some details/cases are incomplete). (Almost) 1. Correctly derived the lower bound $\ge \min\{n,2n-2k+2\}$ for all $k\le n\le 2k-1$ via the $r,c$ argument (including the consecutive-block property and the conclusion $r,c<k$), but did not fully present/verify a matching construction in one of the subranges. 2. Provided matching constructions for all required ranges and proved the lower bound, but left a small gap in the maximality check (e.g. did not fully justify why no additional horizontal/vertical card fits in the remaining empty region, or omitted a boundary subcase such as $n=k$ or $n=2k-1$). 3. Solution is almost complete, but made minor mistakes which are not negligible.

Instruction:
For the specific case $n=19$ and $k=10$, provide an explicit construction of a valid card placement that achieves the minimum possible number of cards.

Construction requirements:
- Output a single Python list of lists, where each inner list represents a single card.
- Each card should be represented by its two endpoints as a list of two tuples: `[(r1, c1), (r2, c2)]`, where `r` is the row index and `c` is the column index, with `r1 <= r2` and `c1 <= c2`.
- Use 0-based indexing for coordinates on the $19 \times 19$ grid, so `r` and `c` are integers from 0 to 18.
- For example, a card covering from row 0 to 9 in column 0 is `[(0, 0), (9, 0)]`.
- The placement must be unextendable (no additional $1 \times 10$ or $10 \times 1$ card can be placed anywhere on the grid without overlapping existing cards).
- The number of cards must be exactly the minimum possible for this unextendable condition.
- Put ONLY the Python expression (e.g., a literal list or a list comprehension) inside the construction block.

import sys

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

    # Parse the construction payload safely
    try:
        # We supply 'range' in locals to support list comprehensions as seen in the ref_construction
        cards = eval(payload, {"__builtins__": {}}, {"range": range})
    except Exception as e:
        print(f"Failed to parse construction: {e}")
        return

    # 1. Verify the parsed object is a list
    if type(cards) is not list:
        print("Construction must be a list of lists.")
        return

    # 2. Check total number of cards
    if len(cards) != 19:
        print(f"Expected exactly 19 cards, but got {len(cards)}.")
        return

    # Initialize a 19x19 boolean grid to track occupied cells
    grid = [[False] * 19 for _ in range(19)]

    for i, card in enumerate(cards):
        # 3. Verify each card is a list containing exactly two endpoints
        if type(card) is not list:
            print(f"Card at index {i} is not a list.")
            return
        if len(card) != 2:
            print(f"Card at index {i} does not contain exactly two endpoints.")
            return

        ep1, ep2 = card

        # 4. Verify endpoints are tuples of two integers
        if type(ep1) is not tuple or len(ep1) != 2 or type(ep2) is not tuple or len(ep2) != 2:
            print(f"Endpoints of card {i} must be tuples of length 2.")
            return

        r1, c1 = ep1
        r2, c2 = ep2

        if not (type(r1) is int and type(c1) is int and type(r2) is int and type(c2) is int):
            print(f"Coordinates of card {i} must be integers.")
            return

        # 5. Verify coordinates are within bounds and properly ordered
        if not (0 <= r1 <= r2 <= 18 and 0 <= c1 <= c2 <= 18):
            print(f"Coordinates of card {i} are out of bounds or improperly ordered.")
            return

        # 6. Verify dimensions to ensure it covers exactly 10 cells
        if not ((r1 == r2 and c2 - c1 == 9) or (c1 == c2 and r2 - r1 == 9)):
            print(f"Card {i} does not cover exactly 10 cells.")
            return

        # 7. Mark cells on the grid and check for overlaps
        for r in range(r1, r2 + 1):
            for c in range(c1, c2 + 1):
                if grid[r][c]:
                    print(f"Card {i} overlaps with an existing card at ({r}, {c}).")
                    return
                grid[r][c] = True

    # 8. Verify the placement is strictly unextendable
    # Scan all 19 rows
    for r in range(19):
        consecutive_false = 0
        for c in range(19):
            if not grid[r][c]:
                consecutive_false += 1
                if consecutive_false >= 10:
                    print(f"Placement is extendable: a 1x10 card can be placed in row {r}.")
                    return
            else:
                consecutive_false = 0

    # Scan all 19 columns
    for c in range(19):
        consecutive_false = 0
        for r in range(19):
            if not grid[r][c]:
                consecutive_false += 1
                if consecutive_false >= 10:
                    print(f"Placement is extendable: a 10x1 card can be placed in column {c}.")
                    return
            else:
                consecutive_false = 0

    # All checks passed
    print("True")

if __name__ == "__main__":
    verify()


Question: In a $100\\times100$ grid, let $(a,b)$ denote the cell in the $a$-th row and $b$-th column. For an integer $51\\leq k\\leq 99$, a $k$-knight moves one step horizontally or vertically and $k$ steps in the other direction. The $k$-knight starts at cell $(1,1)$. A moving sequence is a sequence of cells $(x_0,y_0)=(1,1)$, $(x_1,y_1)$, $\\cdots$, $(x_n,y_n)$ such that for $i=1$, $2$, $\\cdots$, $n$, $1\\leq x_i,y_i\\leq 100$, and the $k$-knight can move from $(x_{i-1},y_{i-1})$ to $(x_i,y_i)$. The cells $(x_i,y_i)$ are called reachable. For each $k$, find the number of reachable cells $L(k)$.

We prove that for every integer $51 \le k \le 99$,
\[
L(k)=
\begin{cases}
4k(100-k), & k \text{ even},\\
2k(100-k), & k \text{ odd}.
\end{cases}
\]

First determine which cells can possibly be reached by one move from another cell. A cell $(x,y)$ is directly reachable from some other cell if and only if at least one of
\[
x-k\ge 1,\qquad x+k\le 100,\qquad y-k\ge 1,\qquad y+k\le 100
\]
holds. Equivalently,
\[
x\ge k+1 \quad\text{or}\quad x\le 100-k \quad\text{or}\quad y\ge k+1 \quad\text{or}\quad y\le 100-k. \tag{1}
\]
Indeed, a $k$-knight move changes one coordinate by $k$ and the other coordinate by $1$; hence, in order to land at $(x,y)$, one of the four cells obtained by changing either coordinate by $k$ must still lie on the board.

Therefore every cell satisfying
\[
101-k\le x\le k,
\qquad
101-k\le y\le k
\]
is unreachable. Let
\[
S=\{(x,y):101-k\le x\le k,\ 101-k\le y\le k\}.
\]
This is the central square of unreachable cells. Its side length is
\[
k-(101-k)+1=2k-100,
\]
so the number of cells outside $S$ is
\[
100^2-(2k-100)^2=4k(100-k). \tag{2}
\]
Thus $L(k)\le 4k(100-k)$.

We now show that all cells outside $S$ are reachable when $k$ is even, and exactly the cells outside $S$ with the same chessboard color as $(1,1)$ are reachable when $k$ is odd.

Color the board like a chessboard, according to the parity of $x+y$. A $k$-knight move changes $x+y$ by one of $\pm k\pm1$. Hence if $k$ is odd, every move preserves the color, while if $k$ is even, every move changes the color.

The key observation is that, outside $S$, one can move between same-colored neighboring cells by two $k$-knight moves. More precisely, suppose $(x,y)$ and one of $(x,y\pm2)$, $(x\pm2,y)$ both lie on the board and both satisfy condition (1). If the usable coordinate is on the lower side, for example $x\le 100-k$, then
\[
(x,y)\to (x+k,y\pm1)
       \to (x,y\pm2)
\]
is a legal two-move path whenever the displayed cells are on the board. If $x\ge k+1$, we instead use
\[
(x,y)\to (x-k,y\pm1)
       \to (x,y\pm2).
\]
Similarly, if $y\le 100-k$, then
\[
(x,y)\to (x\pm1,y+k)
       \to (x\pm2,y),
\]
and if $y\ge k+1$, then
\[
(x,y)\to (x\pm1,y-k)
       \to (x\pm2,y).
\]
Thus any allowed step by $(\pm2,0)$ or $(0,\pm2)$ between cells outside $S$ can be realized by two legal $k$-knight moves.

Since $S$ is a central square and $k\ge 51$, the cells outside $S$ form a frame around it. Within each chessboard color, this frame is connected by steps of the form $(\pm2,0)$ and $(0,\pm2)$. Starting from $(1,1)$, which lies outside $S$, the previous paragraph shows that every cell outside $S$ with the same color as $(1,1)$ is reachable.

If $k$ is odd, the color is preserved by every move, so no cell of the opposite color can be reached. The board has even side length, and the square $S$ also has even side length $2k-100$, so the cells outside $S$ are split equally between the two colors. Therefore
\[
L(k)=\frac12\bigl(100^2-(2k-100)^2\bigr)=2k(100-k).
\]

If $k$ is even, every move changes color. Let $(x,y)$ be any cell outside $S$ whose color is opposite to that of $(1,1)$. Since $(x,y)$ satisfies condition (1), it has a legal predecessor or successor by a $k$-knight move inside the board; that adjacent cell has the same color as $(1,1)$ and is also outside $S$. As shown above, that adjacent cell is reachable, so one more move reaches $(x,y)$. Hence every cell outside $S$ is reachable, and by (2)
\[
L(k)=100^2-(2k-100)^2=4k(100-k).
\]

This proves the desired formula.

(Partial) 1. Identified the central square S of cells with 101−k≤x≤k and 101−k≤y≤k as unreachable (e.g. by proving a necessary condition for landing at (x,y): x≥k+1 or x≤100−k or y≥k+1 or y≤100−k). 2. Correctly counted the number of cells outside S as 100^2−(2k−100)^2=4k(100−k), obtaining the upper bound L(k)≤4k(100−k). 3. Made a correct parity/color observation: a k-knight move preserves chessboard color iff k is odd, and flips color iff k is even. (Almost) 1. Established reachability of all cells outside S of the same color as (1,1) by exhibiting the key two-move simulation of (±2,0) and (0,±2) steps between admissible cells, but did not fully justify that the outside-of-S frame is connected (within a color) by these steps. 2. Correctly derived the odd-k case L(k)=2k(100−k) from color preservation and the count of outside-of-S cells, but the argument that the outside-of-S cells split equally between the two colors is missing or flawed. 3. Correctly derived the even-k case L(k)=4k(100−k) from reachability of one color plus one extra move to the opposite color, but left a localized gap in proving every opposite-colored cell outside S has a valid adjacent (by a k-knight move) same-colored cell still outside S. 4. Solution is almost complete, but made minor mistakes which are not negligible.

Instruction:
For the specific case $k=68$, provide an explicit construction of a path from the starting cell $(1,1)$ to every reachable cell.

Construction requirements:
- The output must be a SINGLE valid Python expression that evaluates to a dictionary.
- The dictionary should map each reachable cell `(x, y)` to its corresponding path.
- A path must be represented as a list of tuples `[(1,1), (x_1, y_1), ..., (x, y)]`, where each consecutive pair of cells represents a valid move for a 68-knight on the $100 \times 100$ grid.
- Do not use any assignment statements, `import` statements, `for` loops (outside of comprehensions), or any non-expression syntax.
- You may use lambda functions, comprehensions, and dictionary unpacking to generate the paths algorithmically.
- Put ONLY the Python expression inside the `<construct>` block.

(lambda adj: (lambda bfs: bfs(bfs, {}, {(1, 1): [(1, 1)]}))(
    lambda f, v, frontier: {**v, **frontier} if not frontier else f(
        f,
        {**v, **frontier},
        {n: frontier[u] + [n] for u in frontier for n in adj(u) if n not in v and n not in frontier}
    )
))(lambda u: [(u[0]+dx, u[1]+dy) for dx, dy in [(1,68), (1,-68), (-1,68), (-1,-68), (68,1), (68,-1), (-68,1), (-68,-1)] if 1<=u[0]+dx<=100 and 1<=u[1]+dy<=100])

import sys

def verify():
    # Increase recursion limit to support deep recursive lambda functions (e.g., BFS implementations)
    sys.setrecursionlimit(20000)

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

    try:
        # The instruction specifies the output must be a SINGLE valid Python expression.
        # eval() enforces this and supports algorithmic generation (lambdas, comprehensions).
        paths_dict = eval(payload)
    except Exception as e:
        print(f"Failed to evaluate payload: {e}")
        return

    if not isinstance(paths_dict, dict):
        print(f"Payload evaluated to {type(paths_dict).__name__}, expected dict.")
        return

    # Check for the exact number of reachable cells for k=68 on a 100x100 grid
    # Formula: L(k) = 4k(100-k) -> 4 * 68 * 32 = 8704
    if len(paths_dict) != 8704:
        print(f"Dictionary contains {len(paths_dict)} keys, expected exactly 8704.")
        return

    for target, path in paths_dict.items():
        # Validate key format
        if not isinstance(target, tuple) or len(target) != 2:
            print(f"Key {target} is not a 2-tuple.")
            return

        tx, ty = target
        if type(tx) is not int or type(ty) is not int:
            print(f"Key {target} contains non-integer coordinates.")
            return

        # Validate path format
        if not isinstance(path, list):
            print(f"Path for {target} is not a list.")
            return

        if not path:
            print(f"Path for {target} is empty.")
            return

        if path[0] != (1, 1):
            print(f"Path for {target} does not start with (1, 1).")
            return

        if path[-1] != target:
            print(f"Path for {target} does not end with its key {target}.")
            return

        # Validate all cells in the path
        for cell in path:
            if not isinstance(cell, tuple) or len(cell) != 2:
                print(f"Cell {cell} in path for {target} is not a 2-tuple.")
                return
            x, y = cell
            if type(x) is not int or type(y) is not int:
                print(f"Coordinates of cell {cell} in path for {target} are not integers.")
                return
            if not (1 <= x <= 100 and 1 <= y <= 100):
                print(f"Cell {cell} in path for {target} is out of 100x100 grid bounds.")
                return

        # Validate consecutive moves in the path
        for i in range(len(path) - 1):
            x1, y1 = path[i]
            x2, y2 = path[i+1]
            dx = abs(x2 - x1)
            dy = abs(y2 - y1)

            # A valid 68-knight move must have absolute coordinate differences of exactly {1, 68}
            if {dx, dy} != {1, 68}:
                print(f"Invalid 68-knight move from {path[i]} to {path[i+1]} in path for {target}.")
                return

    print("True")

if __name__ == "__main__":
    verify()


Question: In a $2012\\times2012$ grid, each cell is filled with a real number from the interval $[0,1]$. The grid is divided into two smaller rectangular grids by drawing a straight line parallel to the edges. If for every possible division, at least one of the smaller grids has the sum of all its numbers not greater than $1$, find the maximum possible sum of all numbers in the entire grid.

We first construct a grid where the sum of all numbers is $5$. Let the top-left $3\times3$ cells of the grid be filled such that the cells $(1,2), (2,1), (2,2), (2,3),$ and $(3,2)$ contain $1$, and the remaining four cells in this $3\times 3$ block contain $0$. All other cells in the $2012\times 2012$ grid are filled with $0$. It is easy to verify that the requirement is satisfied: for any horizontal or vertical division, the sum of the numbers in one of the resulting parts is at most $1$ (in fact, exactly $1$ or $0$). The sum of all numbers in this grid is $5$.

Next, we prove that for every grid satisfying the conditions, the sum of all numbers does not exceed $5$. Assume the total sum of the grid is greater than $1$ (otherwise the bound of $5$ trivially holds). Consider the division after the first row. By the problem condition, either the first row has sum $\le 1$, or the remaining $2011$ rows have sum $\le 1$. In the latter case, since all entries are non-negative, the last row must also have sum $\le 1$. Thus, the sum of the numbers in the first row or the last row does not exceed $1$; by symmetry, we may assume it is the former.

Since the sum of the first row is at most $1$ and the total sum is greater than $1$, there exists a largest positive integer $i$ such that the sum of the numbers in the first $i-1$ rows does not exceed $1$. By the maximality of $i$, the sum of the numbers in the first $i$ rows is greater than $1$. Therefore, by the problem's condition applied to the cut after the $i$-th row, the sum of the numbers in the last $2012-i$ rows does not exceed $1$.

Similarly, by considering vertical divisions, there exists a positive integer $j$ such that the sum of the numbers in the first $j-1$ columns does not exceed $1$, and the sum of the numbers in the last $2012-j$ columns does not exceed $1$.

Notice that every cell in the grid is covered by at least one of the following five regions:
- The first $i-1$ rows
- The last $2012-i$ rows
- The first $j-1$ columns
- The last $2012-j$ columns
- The single intersecting cell at row $i$, column $j$

To see this, observe that any cell not in the first $i-1$ rows and not in the last $2012-i$ rows must lie in row $i$. If it is in row $i$, and it is not in the first $j-1$ columns and not in the last $2012-j$ columns, it must be exactly the cell at row $i$, column $j$.

Since all numbers in the grid are non-negative, the total sum of all numbers is bounded by the sum of the totals of these five regions. The sum of each of the first four regions does not exceed $1$, and the value in the cell at row $i$, column $j$ is in $[0,1]$, so it also does not exceed $1$. Therefore, the total sum of all numbers in the grid does not exceed $1+1+1+1+1=5$.

(Partial) 1. Gave a concrete configuration with total sum 5 (e.g. the cross of five 1's in a 3×3 block) and checked that every horizontal/vertical cut leaves a part with sum ≤1. 2. From considering the cut after the first row (or first column), correctly deduced that at least one of the two boundary rows (or boundary columns) has sum ≤1 (using nonnegativity). 3. Defined the maximal i with sum of first i−1 rows ≤1 (or the analogous j for columns), and concluded from the condition that the complementary block of last 2012−i rows (resp. last 2012−j columns) has sum ≤1. (Almost) 1. Obtained the correct 5-region cover idea (first i−1 rows, last 2012−i rows, first j−1 columns, last 2012−j columns, and the single cell (i,j)) and bounded each of the four strips by 1, but did not fully justify that every cell is covered (or mishandled the boundary indices i=1 or i=2012, etc.). 2. Proved the upper bound ≤5 via the above row/column maximality argument but left one localized gap (e.g. did not justify existence of i,j from the assumption total sum >1, or forgot to bound the central cell by ≤1). 3. Solution is almost complete, but made minor mistakes which are not negligible.

Instruction:
Provide an explicit construction for the $2012 \times 2012$ grid that achieves the maximum possible sum of all numbers.

Construction requirements:
- Output a single Python expression that evaluates to a list of lists of numbers representing the $2012 \times 2012$ grid.
- Each number must be in the interval $[0, 1]$.
- For every possible division of the grid into two smaller rectangular grids by a straight line parallel to the edges, at least one of the smaller grids must have the sum of all its numbers not greater than $1$.
- The sum of all numbers in the grid must be exactly the maximum possible sum.
- The output must be a single, valid Python expression (e.g., using list comprehensions) that evaluates directly to the target data structure.
- Do not include any explanations, comments, print statements, assignments, or imports.

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

[[1 if (i == 1 and j <= 2) or (j == 1 and i <= 2) else 0 for j in range(2012)] for i in range(2012)]

import sys
import math

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

    # Allow safe builtins to evaluate list comprehensions
    allowed_builtins = {
        'range': range,
        'len': len,
        'min': min,
        'max': max,
        'sum': sum,
        'int': int,
        'float': float,
        'bool': bool,
        'list': list,
        'tuple': tuple,
        'zip': zip,
        'abs': abs,
        'round': round,
        'enumerate': enumerate,
        'map': map,
        'filter': filter,
        'any': any,
        'all': all,
    }

    try:
        grid = eval(payload, {"__builtins__": allowed_builtins}, {})
    except Exception as e:
        print(f"Failed to evaluate payload: {e}")
        return

    if not isinstance(grid, list):
        print("Grid must be a list.")
        return
    if len(grid) != 2012:
        print(f"Grid must have exactly 2012 rows, got {len(grid)}.")
        return

    for i in range(2012):
        row = grid[i]
        if not isinstance(row, list):
            print(f"Row {i} is not a list.")
            return
        if len(row) != 2012:
            print(f"Row {i} must have exactly 2012 elements, got {len(row)}.")
            return
        for j in range(2012):
            val = row[j]
            if not isinstance(val, (int, float)):
                print(f"Element at ({i}, {j}) is not a number: {type(val).__name__}")
                return
            if not (0 <= val <= 1):
                print(f"Element at ({i}, {j}) is out of bounds [0, 1]: {val}")
                return

    # Precalculate row sums and column sums using math.fsum for numerical stability
    row_sums = [math.fsum(row) for row in grid]
    col_sums = [math.fsum(grid[i][j] for i in range(2012)) for j in range(2012)]
    total_sum = math.fsum(row_sums)

    if not math.isclose(total_sum, 5.0, rel_tol=0.0, abs_tol=1e-9):
        print(f"Total sum must be exactly 5, got {total_sum}.")
        return

    # Verify horizontal division constraint
    for k in range(1, 2012):
        top_sum = math.fsum(row_sums[:k])
        bottom_sum = math.fsum(row_sums[k:])
        if top_sum > 1 + 1e-9 and bottom_sum > 1 + 1e-9:
            print(f"Horizontal division at row {k} fails constraint: top_sum={top_sum}, bottom_sum={bottom_sum}")
            return

    # Verify vertical division constraint
    for k in range(1, 2012):
        left_sum = math.fsum(col_sums[:k])
        right_sum = math.fsum(col_sums[k:])
        if left_sum > 1 + 1e-9 and right_sum > 1 + 1e-9:
            print(f"Vertical division at column {k} fails constraint: left_sum={left_sum}, right_sum={right_sum}")
            return

    print("True")

if __name__ == "__main__":
    verify()


Question: In a faulty $5\\times5$ grid of $25$ lights, pressing the switch of one light changes the state (from off to on, or from on to off) of that light and all adjacent lights in its row and column. Initially, all lights are off. After pressing the switches several times, exactly one light is on. Determine all possible positions of this light that is on.

Let the lights be represented by a $5 \times 5$ grid of cells $(i,j)$ for $1 \le i, j \le 5$. Pressing a switch at $(i,j)$ toggles the state of the light at $(i,j)$ and all its orthogonal neighbors. We want to find all cells $(r,c)$ such that there exists a sequence of presses leaving exactly the light at $(r,c)$ on, with all other lights off.

First, we define two specific sets of cells:
$$S_1 = \{ (i,j) \mid i \in \{1,2,4,5\} \text{ and } j \in \{1,3,5\} \}$$
$$S_2 = \{ (i,j) \mid i \in \{1,3,5\} \text{ and } j \in \{1,2,4,5\} \}$$
Notice that $S_2$ is the reflection of $S_1$ across the main diagonal. We claim that for any switch pressed, the number of toggled lights belonging to $S_1$ is always even. Let $N(r,c)$ be the set of cells toggled by pressing $(r,c)$ (which includes $(r,c)$ and its orthogonal neighbors). We can exhaustively check the size of $N(r,c) \cap S_1$:
- If $r \in \{1,5\}$ and $c \in \{1,5\}$, $|N(r,c) \cap S_1| = 2$.
- If $r \in \{1,5\}$ and $c = 3$, $|N(r,c) \cap S_1| = 2$.
- If $r \in \{2,4\}$ and $c \in \{1,5\}$, $|N(r,c) \cap S_1| = 2$.
- If $r \in \{2,4\}$ and $c = 3$, $|N(r,c) \cap S_1| = 2$.
- If $r = 3$ and $c \in \{1,5\}$, $|N(r,c) \cap S_1| = 2$.
- If $r = 3$ and $c = 3$, $|N(r,c) \cap S_1| = 2$.
- If $c \in \{2,4\}$ and $r \in \{1,5\}$, $|N(r,c) \cap S_1| = 2$.
- If $c \in \{2,4\}$ and $r \in \{2,4\}$, $|N(r,c) \cap S_1| = 2$.
- If $c \in \{2,4\}$ and $r = 3$, $|N(r,c) \cap S_1| = 0$.

In all cases, pressing any switch changes the state of an even number of lights in $S_1$. Since initially all lights are off (so $0$ lights in $S_1$ are on), the parity of the number of lights turned on in $S_1$ is invariant and must remain even. Therefore, if exactly one light is finally on, its position cannot be in $S_1$.

By an identical symmetric argument, any switch press toggles an even number of lights in $S_2$. Thus, the single light that is finally on cannot be in $S_2$ either.

The position of the single light must therefore lie in the complement of $S_1$ and the complement of $S_2$.
The complement of $S_1$ is $Z_1 = \{ (i,j) \mid i = 3 \text{ or } j \in \{2,4\} \}$.
The complement of $S_2$ is $Z_2 = \{ (i,j) \mid j = 3 \text{ or } i \in \{2,4\} \}$.
Taking the intersection $Z_1 \cap Z_2$:
- If $i = 3$, then for the cell to be in $Z_2$, we must have $j = 3$ (since $i \notin \{2,4\}$). This gives the position $(3,3)$.
- If $j \in \{2,4\}$, then for the cell to be in $Z_2$, we must have $i \in \{2,4\}$ (since $j \neq 3$). This gives the positions $(2,2), (2,4), (4,2), (4,4)$.

Hence, the only possible positions for the final single light are the five cells $(2,2), (2,4), (3,3), (4,2),$ and $(4,4)$.

Below we provide explicit constructions for the middle position $(3,3)$ and the upper-right position $(2,4)$. We represent the presses as a $5 \times 5$ matrix where a $1$ indicates the switch is pressed once, and a $0$ indicates it is not pressed.

For the middle position $(3,3)$, the construction is:
$$ \begin{pmatrix}
0 & 0 & 0 & 1 & 1 \\
0 & 0 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 & 1 \\
1 & 0 & 0 & 0 & 1 \\
1 & 0 & 1 & 1 & 0
\end{pmatrix} $$
One can verify that this sequence of presses toggles the light at $(3,3)$ an odd number of times (three times) and every other light an even number of times.

For the upper-right position $(2,4)$, the construction is:
$$ \begin{pmatrix}
0 & 0 & 1 & 0 & 0 \\
0 & 1 & 1 & 1 & 0 \\
1 & 0 & 0 & 1 & 1 \\
1 & 0 & 0 & 1 & 0 \\
0 & 1 & 1 & 0 & 0
\end{pmatrix} $$
Similarly, this toggles the light at $(2,4)$ an odd number of times and all other lights an even number of times.

For the other three positions $(2,2), (4,4),$ and $(4,2)$, one can obtain the required switch presses by rotating the construction matrix for $(2,4)$ by $90^\circ, 180^\circ,$ and $270^\circ$, respectively.

Thus, the possible positions of the single light that is on are exactly $(2,2), (2,4), (3,3), (4,2),$ and $(4,4)$.

(Partial) 1. Used a parity/invariant argument to obtain a nontrivial restriction on the final position (e.g. defined a subset of cells and proved that every press toggles an even number of lights in it, hence the final single on-light cannot lie in that subset). 2. Correctly identified at least one of the two key forbidden sets (equivalently, proved that the final on-light cannot be in all cells with i∈{1,2,4,5}, j∈{1,3,5}, or the symmetric set with i∈{1,3,5}, j∈{1,2,4,5}), even if the full classification is not finished. 3. Gave a correct explicit press pattern (or a verifiable constructive procedure) that results in exactly one light on for at least one of the five correct target cells. (Almost) 1. Correctly proved (typically via two parity invariants) that the final on-light must belong to the intersection of the two complements, yielding the candidate set {(2,2),(2,4),(3,3),(4,2),(4,4)}, but did not provide constructions showing these candidates are all attainable. 2. Gave constructions for (3,3) and one corner-adjacent candidate (e.g. (2,4)) and indicated obtaining the other three by rotations/symmetry, but left the symmetry reduction or the verification of the constructions insufficiently justified. 3. Solution is almost complete, but made minor mistakes which are not negligible.

Instruction:
Find all possible positions of the single light that is on. For each possible position, determine a sequence of switch presses that achieves this state starting from all lights off.

Construction format requirements:
- Output a single Python dictionary.
- The keys must be all valid 0-based coordinates `(r, c)` (tuples of integers) of a light that can be the only one turned on.
- The value for each key must be a list of 0-based coordinates `(r, c)` representing a sequence of switch presses that results in exactly that light being on.
- Put ONLY the Python dictionary literal inside the `<construct>` block. Do not include any other text or markdown formatting.

import ast
import sys

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

    # 1. Parse using ast.literal_eval
    try:
        data = ast.literal_eval(payload)
    except Exception as e:
        print(f"Failed to parse input as Python literal: {e}")
        return

    # Verify it's a dictionary
    if not isinstance(data, dict):
        print("Input must be a Python dictionary.")
        return

    # 2. Verify keys
    expected_keys = {(1, 1), (1, 3), (2, 2), (3, 1), (3, 3)}
    if set(data.keys()) != expected_keys:
        print(f"Dictionary keys must be exactly {expected_keys}.")
        return

    # 3-7. Verify moves for each key
    for target, moves in data.items():
        if not isinstance(moves, list):
            print(f"Value for key {target} must be a list of moves.")
            return

        # 3. Initialize grid
        grid = [[False] * 5 for _ in range(5)]

        for move in moves:
            if not (isinstance(move, tuple) and len(move) == 2):
                print(f"Move {move} for key {target} is not a 2-tuple.")
                return
            r, c = move
            if type(r) is not int or type(c) is not int:
                print(f"Move {move} for key {target} contains non-integers.")
                return

            # 4. Verify bounds
            if not (0 <= r < 5 and 0 <= c < 5):
                print(f"Move {move} for key {target} is out of bounds.")
                return

            # 5. Toggle state of the cell and its valid orthogonal neighbors
            for dr, dc in [(0, 0), (-1, 0), (1, 0), (0, -1), (0, 1)]:
                nr, nc = r + dr, c + dc
                if 0 <= nr < 5 and 0 <= nc < 5:
                    grid[nr][nc] = not grid[nr][nc]

        # 6. Count True cells
        true_cells = []
        for r in range(5):
            for c in range(5):
                if grid[r][c]:
                    true_cells.append((r, c))

        # 7. Verify exactly one cell is True and matches target
        if len(true_cells) != 1:
            print(f"For key {target}, expected exactly 1 light on, but found {len(true_cells)} lights on.")
            return
        if true_cells[0] != target:
            print(f"For key {target}, the light on is at {true_cells[0]}, which does not match the target.")
            return

    print("True")

if __name__ == "__main__":
    verify()


Question: In a $5\\times5$ matrix $X$, each element is either $0$ or $1$. Consider all $5$-tuples (ordered arrays) from the rows, columns, and diagonals of $X$ (a total of $24$): $(x_{i,1},x_{i,2},\\cdots,x_{i,5})$, $(x_{i,5},x_{i,4},\\cdots,x_{i,1})$ for $i=1,2,\\cdots,5$; $(x_{1,j},x_{2,j},\\cdots,x_{5,j})$, $(x_{5,j},x_{4,j},\\cdots,x_{1,j})$ for $j=1,2,\\cdots,5$; $(x_{1,1},x_{2,2},\\cdots,x_{5,5})$, $(x_{5,5},x_{4,4},\\cdots,x_{1,1})$, $(x_{1,5},x_{2,4},\\cdots,x_{5,1})$, $(x_{5,1},x_{4,2},\\cdots,x_{1,5})$. If all these arrays are pairwise distinct, find the possible values of the sum of all elements in matrix $X$.

Let the set of all $24$ five-tuples in the problem be denoted by $T$. Call a five-tuple $(a,b,c,d,e)$ "palindromic" if $a=e$ and $b=d$. Note that for each tuple $\alpha=(a_{1},a_{2},\dots,a_{5})\in T$, its reverse $\beta=(a_{5},a_{4},\dots,a_{1})$ is also in $T$. By the problem condition, all tuples in $T$ are distinct, so $\alpha\neq\beta$, which means $\alpha$ is not palindromic. Because the number of five-tuples with entries $0$ or $1$ that are non-palindromic is exactly $2^{5}-2^{3}=24$, $T$ is precisely the set of all non-palindromic five-tuples with entries $0$ or $1$.

For $\alpha\in T$, denote by $S_{\alpha}$ the sum of the five components of $\alpha$, and consider the total sum $\displaystyle\sum_{\alpha\in T} S_{\alpha}$.

On the one hand, pairing each $\alpha=(a_{1},a_{2},\dots,a_{5})\in T$ with its bitwise complement $\alpha'=(1-a_{1},1-a_{2},\dots,1-a_{5})$ shows that $\alpha'\in T$ and $S_{\alpha}+S_{\alpha'}=5$. Since the third components of $\alpha$ and $\alpha'$ differ, $\alpha\neq\alpha'$, meaning no $\alpha$ is paired with itself. Thus, the $24$ tuples in $T$ can be partitioned into $12$ pairs, giving $\displaystyle\sum_{\alpha\in T} S_{\alpha} = 12 \times 5 = 60$.

On the other hand, in the sum $\displaystyle\sum_{\alpha\in T} S_{\alpha}$, the central element $x_{3,3}$ of $X$ is counted $8$ times, each off-center element on the two main diagonals is counted $6$ times, and each element not on the diagonals is counted $4$ times. Let $S$ be the sum of all elements of $X$. Then we can express the total sum as:
\[
4x_{3,3}+2\bigl(x_{1,1}+x_{2,2}+x_{4,4}+x_{5,5}+x_{1,5}+x_{2,4}+x_{4,2}+x_{5,1}\bigr)+4S=60.
\]
From this equation and the constraints on the non-palindromic tuples, it follows that the possible values for the sum $S$ of all elements in matrix $X$ are $12$ and $13$.

(Partial) 1. Observed that for each listed 5-tuple its reverse also appears in the list, hence every occurring 5-tuple must be non-palindromic (i.e. not equal to its reverse). 2. Counted the number of non-palindromic binary 5-tuples as 24, and concluded that the 24 arrays must be exactly the set of all non-palindromic binary 5-tuples. 3. Correctly set up a double-counting equation for \(\sum_{\alpha\in T} S_\alpha\) by counting how many times each entry \(x_{i,j}\) appears among the 24 arrays (center 8 times, other diagonal entries 6 times, off-diagonal entries 4 times), leading to a linear relation between this total and \(S=\sum x_{i,j}\). 4. Computed \(\sum_{\alpha\in T} S_\alpha=60\) (e.g. by pairing each tuple with its bitwise complement, or by another correct global symmetry argument). (Almost) 1. Derived the correct global equation (total sum over tuples equals 60, and also equals the weighted count of matrix entries) and reduced the problem to a small finite check on the diagonal/center variables, but did not complete the final casework needed to conclude \(S\in\{12,13\}\). 2. Correctly concluded that the 24 tuples are exactly all non-palindromic binary 5-tuples and that \(\sum_{\alpha\in T} S_\alpha=60\), and obtained a near-final formula giving \(S\) up to an unresolved parity/boundary issue (e.g. mishandled the contribution of \(x_{3,3}\) or one diagonal cell), so the final set of possible values for \(S\) is not fully justified. 3. Gave a complete derivation that forces \(S=12\) or \(S=13\), but provided only one explicit construction (or omitted verifying that both values are attainable under the distinctness condition). 4. Solution is almost complete, but made minor mistakes which are not negligible.

Instruction:
For every possible valid sum of all elements in the matrix, you must provide exactly one example of a $5 \times 5$ matrix that satisfies all the conditions.

Construction requirements:
- Output a single Python list containing lists of strings.
- Each inner list represents one $5 \times 5$ matrix and must contain exactly 5 strings of length 5, consisting only of the characters '0' and '1'.
- You must include exactly one matrix for each possible sum.
- For each matrix, the 24 tuples extracted (rows, columns, and diagonals, in both forward and reverse directions) must be pairwise distinct.
- Put ONLY the Python literal inside the construction block. Do not include any other text.

import ast
import sys

def verify():
    try:
        payload = sys.stdin.read().strip()
        if not payload:
            print("Empty input.")
            return
        data = ast.literal_eval(payload)
    except Exception as e:
        print(f"Parse error: {e}")
        return

    if not isinstance(data, list):
        print("Output must be a Python list.")
        return

    if len(data) != 2:
        print(f"Expected exactly 2 matrices, got {len(data)}.")
        return

    sums = set()

    for idx, M in enumerate(data):
        if not isinstance(M, list):
            print(f"Matrix at index {idx} must be a list.")
            return
        if len(M) != 5:
            print(f"Matrix at index {idx} must contain exactly 5 strings.")
            return

        total_sum = 0
        for row_idx, row in enumerate(M):
            if not isinstance(row, str):
                print(f"Row {row_idx} in matrix {idx} must be a string.")
                return
            if len(row) != 5:
                print(f"Row {row_idx} in matrix {idx} must be of length 5.")
                return
            if not all(c in '01' for c in row):
                print(f"Row {row_idx} in matrix {idx} must contain only '0' and '1'.")
                return
            total_sum += row.count('1')

        sums.add(total_sum)

        # Extract the 12 base tuples
        rows = [tuple(M[i]) for i in range(5)]
        cols = [tuple(M[i][j] for i in range(5)) for j in range(5)]
        diag1 = [tuple(M[i][i] for i in range(5))]
        diag2 = [tuple(M[i][4-i] for i in range(5))]

        base_tuples = rows + cols + diag1 + diag2

        # Generate the reversed counterparts, resulting in 24 tuples
        all_tuples = []
        for t in base_tuples:
            all_tuples.append(t)
            all_tuples.append(t[::-1])

        if len(set(all_tuples)) != 24:
            print(f"Matrix at index {idx} does not have 24 pairwise distinct tuples.")
            return

    if sums != {12, 13}:
        print(f"The set of sums must be exactly {{12, 13}}, got {sums}.")
        return

    print("True")

if __name__ == "__main__":
    verify()


Question:
"Given an integer $m \\geq 2$, a conference has $3m$ delegates. Every two delegates either shake hands once or do not shake hands. For a positive integer $n \\leq 3m - 1$, if there exist $n$ delegates whose numbers of handshakes are $1, 2, \\cdots, n$ respectively, then the conference is called \""$n$-interesting\"". If for every possible $n$-interesting conference, there always exist three delegates who have shaken hands pairwise, find the minimum value of $n$.
"

Treat the delegates as vertices of a graph; if two delegates have shaken hands, draw an edge between their corresponding vertices.

First, we show that $n \le 2m$ is not sufficient to guarantee a triangle. Construct a bipartite graph whose vertices are partitioned into two sets $A = \{A_1, A_2, \dots, A_m\}$ and $B = \{B_1, B_2, \dots, B_{2m}\}$. Add edges $A_iB_j$ for all $1\le i \le m$ and $1 \le j \le 2m$ such that $i \le j$.
In this graph, the degree of $B_j$ is the number of $A_i$ such that $i \le j$, which is $\min(m, j)$. Thus, the degrees of $B_1, B_2, \dots, B_m$ are $1, 2, \dots, m$, respectively.
The degree of $A_i$ is the number of $B_j$ such that $j \ge i$, which is $2m - i + 1$. Thus, the degrees of $A_m, A_{m-1}, \dots, A_1$ are $m+1, m+2, \dots, 2m$, respectively.
Therefore, the graph contains vertices with degrees $1, 2, \dots, 2m$, making it $n$-interesting for any $n \le 2m$. Since the graph is bipartite, it contains no triangles. This shows that $n$ must be strictly greater than $2m$.

Now we prove that when $n=2m+1$, the graph must contain a triangle. Suppose for the sake of contradiction that there exists a $(2m+1)$-interesting graph on $3m$ vertices that contains no triangles.
By the definition of a $(2m+1)$-interesting graph, there exist vertices with degrees $1, 2, \dots, 2m+1$. In particular, there exists a vertex of degree $2m+1$; without loss of generality, let this vertex be $A_1$, and let its neighbors be $B_1, B_2, \dots, B_{2m+1}$.
Suppose the graph contains no triangles; then the neighbors $B_1, \dots, B_{2m+1}$ must be pairwise nonadjacent. This implies that any vertex $B_k$ can only be adjacent to vertices outside the set $\{B_1, \dots, B_{2m+1}\}$. Since the total number of vertices is $3m$, the number of such available vertices is $3m - (2m+1) = m-1$. Consequently, the degree of each $B_k$ is at most $m-1$.

The $(2m+1)$-interesting condition requires the graph to have vertices with degrees $m, m+1, \dots, 2m$, which constitute $m+1$ distinct values strictly greater than $m-1$. Since the degrees of $B_1, \dots, B_{2m+1}$ are at most $m-1$, the vertices achieving these $m+1$ degrees must be among the remaining vertices.
Excluding $A_1$ (which has degree $2m+1$) and its neighbors $B_1, \dots, B_{2m+1}$, the number of remaining vertices is exactly $3m - 1 - (2m+1) = m-2$.
However, it is impossible for $m-2$ vertices to cover the $m+1$ required degrees $m, m+1, \dots, 2m$. This contradicts the graph being $(2m+1)$-interesting.

Hence, the graph must contain a triangle, satisfying the requirement. The minimum value of $n$ is $2m+1$.

(Partial) 1. Reformulated the problem in graph language (delegates = vertices, handshakes = edges) and correctly identified the goal as forcing a triangle in any triangle-free graph having degrees 1,2,...,n among its vertices. 2. Gave a correct triangle-free construction on 3m vertices which is n-interesting for all n≤2m (e.g. an explicit bipartite graph with vertices of degrees 1,2,...,2m), thereby proving n must be >2m. 3. In the n=2m+1 case, proved the key local fact: if a vertex has degree 2m+1 in a triangle-free graph, then its 2m+1 neighbors form an independent set, hence each of those neighbors has degree at most m−1. (Almost) 1. Proved the lower bound via a correct triangle-free construction for all n≤2m, and for n=2m+1 set up the contradiction using a degree-(2m+1) vertex and the bound deg(neighbor)≤m−1, but made a small counting error in the final pigeonhole step (e.g. miscounted remaining vertices or the number of required degree values). 2. Completed the n=2m+1 contradiction argument in essence (neighbors independent, too few remaining vertices to realize degrees m,m+1,...,2m), but did not fully justify one boundary case in the counting (such as treating m=2 separately or checking that the degree-2m+1 vertex is distinct from the vertices intended to realize degrees m..2m). 3. Derived that n≥2m+1 and attempted to show n=2m+1 works with the correct strategy, but left a minor gap in arguing why vertices of degrees m,m+1,...,2m cannot lie among the 2m+1 neighbors (i.e. did not explicitly use the deg≤m−1 bound to exclude them).

Instruction:
For the specific case $m = 789$, determine the maximum integer $K$ such that there exists a $K$-interesting conference where no three delegates have shaken hands pairwise. Provide an explicit construction of the handshake graph for this maximum $K$.

Construction requirements:
- Output a single Python expression (e.g., a list comprehension) that evaluates to a $2367 \times 2367$ adjacency matrix (a list of lists of integers) representing the handshakes between the $3m = 2367$ delegates.
- The matrix must be symmetric, with 0s on the main diagonal and 0s or 1s elsewhere.
- The graph must be triangle-free (no three delegates shake hands pairwise).
- There must exist $K$ delegates whose degrees (number of handshakes) are exactly $1, 2, \dots, K$.
- $K$ must be the maximum possible value for $m = 789$.
- Put ONLY the valid Python expression inside the construction block, without any markdown formatting, explanations, or assignments.

[[1 if (i < 789 and j >= i + 789) or (j < 789 and i >= j + 789) else 0 for j in range(2367)] for i in range(2367)]

import sys

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

    try:
        # Evaluate the payload directly to support list comprehensions
        M = eval(payload)
    except Exception as e:
        print(f"Failed to evaluate payload: {e}")
        return

    m = 789
    N = 3 * m  # 2367
    K = 2 * m  # 1578

    # 1 & 2. Check that M is a list of 2367 lists of length 2367, containing only 0 and 1
    if not isinstance(M, list):
        print("M is not a list")
        return
    if len(M) != N:
        print(f"M should have length {N}, got {len(M)}")
        return

    for i, row in enumerate(M):
        if not isinstance(row, list):
            print(f"Row {i} is not a list")
            return
        if len(row) != N:
            print(f"Row {i} should have length {N}, got {len(row)}")
            return
        for j, val in enumerate(row):
            # Strictly check for int to reject booleans or other types
            if type(val) is not int:
                print(f"M[{i}][{j}] is not an integer")
                return
            if val not in (0, 1):
                print(f"M[{i}][{j}] is not 0 or 1")
                return

    # 3. Verify symmetry and 0s on the main diagonal
    for i in range(N):
        if M[i][i] != 0:
            print(f"M[{i}][{i}] is not 0")
            return
        for j in range(i + 1, N):
            if M[i][j] != M[j][i]:
                print(f"M is not symmetric at ({i}, {j})")
                return

    # 4 & 5. Verify the degrees
    degrees = [sum(row) for row in M]
    degree_set = set(degrees)

    for d in range(1, K + 1):
        if d not in degree_set:
            print(f"Degree {d} is missing from the graph")
            return

    # 6. Verify that the graph contains no triangles
    neighbors = [set() for _ in range(N)]
    for i in range(N):
        for j in range(N):
            if M[i][j] == 1:
                neighbors[i].add(j)

    for i in range(N):
        ni = neighbors[i]
        for j in ni:
            if i < j:
                # If the intersection of neighbors[i] and neighbors[j] is not empty, a triangle exists
                if not ni.isdisjoint(neighbors[j]):
                    print(f"Triangle found containing edge ({i}, {j})")
                    return

    print("True")

if __name__ == "__main__":
    verify()


Question:
"Given an integer $n > 2004$. Fill the numbers $1, 2, \\cdots, n^2$ into an $n \\times n$ grid, with one number in each cell. If a cell contains a number that is greater than at least $2004$ numbers in its row and greater than at least $2004$ numbers in its column, then the cell is called an \""excellent cell\"". Find the maximum possible number of excellent cells.
"

The maximum possible number of excellent cells is $n(n-2004)$.

By evaluating row constraints, the $2004$ smallest numbers in any given row cannot form excellent cells, restricting each row to at most $n-2004$ excellent cells and establishing the total upper bound of $n(n-2004)$.

Applying the optimal assignment directly: for $1 \le i \le n$, mark cells in columns $i, i+1, \dots, i+2003 \pmod n$ of row $i$ with $\ast$. Allocating the numbers $1, 2, \dots, 2004n$ to the $\ast$ cells and the remaining larger numbers to the unmarked cells guarantees that all unmarked cells strictly exceed the required threshold of $2004$ elements in both their respective rows and columns. This configuration achieves exactly $n(n-2004)$ excellent cells.

(Partial) 1. Proved the per-row upper bound: in any row, the 2004 smallest entries cannot be excellent, hence each row has at most n-2004 excellent cells (or an equivalent counting argument giving the same bound). 2. Proved the per-column upper bound analogously (each column has at most n-2004 excellent cells), and used it to justify an O(n^2) global upper bound (even if not simplified to n(n-2004)). 3. Gave a plausible construction scheme aiming for n(n-2004) excellent cells (e.g. selecting 2004 “small” positions in each row/column and placing small numbers there), but did not fully verify the column condition or the strict “>2004” requirement. (Almost) 1. Correctly proved the upper bound n(n-2004), but the construction is only sketched or has a small gap (e.g. not fully justifying that every non-star cell has at least 2004 smaller numbers in its column, or mishandling the cyclic shift argument). 2. Gave the cyclic-star construction (2004 marked cells per row, shifted so each column has exactly 2004 marked cells) and the intended number placement, but made a localized mistake such as using 2004 vs 2003 shifts, or failing to check the boundary case n=2005 / the strictness of inequalities; overall method would work after minor fixes. 3. Achieved the value n(n-2004) with a nearly complete proof, but left one nontrivial verification step insufficiently justified (typically: why all unmarked entries are simultaneously among the largest n-2004 in their row and among the largest n-2004 in their column).

Instruction:
For the specific case $n = 35$ and using a threshold of $24$ instead of $2004$ (i.e., a cell is "excellent" if its value is strictly greater than at least $24$ numbers in its row and strictly greater than at least $24$ numbers in its column), provide an explicit construction of the $35 \times 35$ grid that maximizes the number of excellent cells.

Construction requirements:
- Output a single Python list of lists representing the $35 \times 35$ grid.
- The grid must contain every integer from $1$ to $1225$ exactly once.
- The number of excellent cells in your grid must be the maximum possible for $n=35$ and threshold $24$.
- Put ONLY the Python literal (the list of lists) inside the construction block.

import sys
import ast

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

    try:
        grid = ast.literal_eval(input_text)
    except Exception as e:
        print(f"Parse error: {e}")
        return

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

    if len(grid) != 35:
        print(f"Expected exactly 35 rows, got {len(grid)}.")
        return

    elements = []
    for i, row in enumerate(grid):
        if not isinstance(row, list):
            print(f"Row {i} is not a list.")
            return
        if len(row) != 35:
            print(f"Row {i} must have exactly 35 elements, got {len(row)}.")
            return
        for j, val in enumerate(row):
            if type(val) is not int:
                print(f"Element at row {i}, col {j} is not an integer.")
                return
            elements.append(val)

    if sorted(elements) != list(range(1, 1226)):
        print("The grid must contain every integer from 1 to 1225 exactly once.")
        return

    excellent_count = 0
    for i in range(35):
        for j in range(35):
            val = grid[i][j]
            # Count strictly smaller numbers in the same row
            row_smaller = sum(1 for c in range(35) if grid[i][c] < val)
            # Count strictly smaller numbers in the same column
            col_smaller = sum(1 for r in range(35) if grid[r][j] < val)

            if row_smaller >= 24 and col_smaller >= 24:
                excellent_count += 1

    if excellent_count != 385:
        print(f"Expected exactly 385 excellent cells, but found {excellent_count}.")
        return

    print("True")

if __name__ == "__main__":
    verify()


Question:
"$n$ teams play a double round-robin tournament (i.e., every two teams play two matches, with each team having one home match). It is known that each team can play multiple away matches in one week (from Sunday to Saturday, seven days), but if a team has a home match in a certain week, then no away matches can be scheduled for that team in that week. If all matches can be completed within $4$ weeks, find the maximum value of $n$.
"

For $1\le i\le n$, let $A_i$ be the set of weeks in which the $i$-th team plays home games. According to the problem's conditions, the different sets $A_i$ must be mutually non-contained. This is because if $A_i\subseteq A_j$, then a home game of team $i$ against team $j$ could not occur: the week of this match would have to be in $A_i$, and hence in $A_j$, meaning team $j$ would have a home game that week. This violates the condition that a team cannot play an away game in a week it has a home game.

Since $A_1, A_2, \dots, A_n$ are all subsets of the $4$ available weeks, meaning $A_i \subseteq \{1,2,3,4\}$, Sperner's theorem gives the upper bound:
$$ n \le \binom{4}{2} = 6 $$

When $n=6$, the schedule can be arranged as shown in the following table, where rows represent teams, columns represent weeks, and "$*$" indicates that the team has a home game in that week.

$$
\def\arraystretch{1.5}
\begin{array}{|c|c|c|c|c|}
\hline
\text{Team \textbackslash\ Week} & 1 & 2 & 3 & 4 \\
\hline
1 & * & * & & \\
2 & * & & * & \\
3 & * & & & * \\
4 & & * & * & \\
5 & & * & & * \\
6 & & & * & * \\
\hline
\end{array}
$$

For any two distinct teams $i$ and $j$, team $i$ can schedule its home game against team $j$ in the unique week that is in $A_i$ but not in $A_j$. Since each team has exactly two home weeks and their sets are distinct, there is exactly one such week for each ordered pair, allowing all matches to be successfully scheduled without violating the weekly restrictions.

Thus, the maximum value of $n$ is $6$.

(Partial) 1. Introduced the sets $A_i\subseteq\{1,2,3,4\}$ of weeks in which team $i$ plays at home, and correctly derived that for any $i\ne j$ one must have $A_i\nsubseteq A_j$ (otherwise the match $i$ at home vs $j$ cannot be scheduled). 2. Obtained a nontrivial upper bound for $n$ from the non-containment property of the $A_i$ (e.g. by noting the family is an antichain and invoking/quoting Sperner for $4$ weeks, or by an equivalent counting argument). 3. Gave a concrete schedule pattern for some $n\ge 5$ consistent with the rule “home week implies no away matches that week”, even if not fully justified for all ordered pairs. (Almost) 1. Correctly proved the upper bound $n\le 6$ via Sperner’s theorem (or an equivalent antichain argument), but did not provide a working construction for $n=6$. 2. Gave the $n=6$ construction (e.g. six distinct 2-element home-week sets) and explained how to place each home match $i$ vs $j$ in a week in $A_i\setminus A_j$, but left a minor gap in justifying existence/uniqueness of such a week for every ordered pair. 3. Presented a full-looking schedule/table for $n=6$ but failed to check a small number of pairwise constraints (e.g. one ordered pair not assigned, or one week-choice accidentally conflicts with the home/away restriction), while the main Sperner/antichain idea is correct.

Instruction:
For the maximum possible number of teams $n$, provide an explicit construction of the schedule matrix.

Construction requirements:
- Output a single Python list of lists representing a $n \times 4$ matrix.
- The rows represent the n teams, and the columns represent the 4 weeks.
- Matrix entry `M[i][j]` should be:
  - `1` if team `i` plays home matches in week `j`.
  - `0` if team `i` does not play home matches in week `j` (meaning it is available to play away).
- The matrix must satisfy the condition that for every pair of distinct teams $(A, B)$, there is at least one week where $A$ is home (`1`) and $B$ is not (`0`), and at least one week where $B$ is home (`1`) and $A$ is not (`0`).
- Put ONLY the Python literal matrix inside the construction block.

import sys
import ast

def verify():
    # Read the full input from stdin
    input_str = sys.stdin.read().strip()

    if not input_str:
        print("Error: Input is empty.")
        return

    # 1. Parse the input string as a Python list of lists.
    # We use ast.literal_eval for strict parsing of Python literals.
    try:
        matrix = ast.literal_eval(input_str)
    except Exception as e:
        print(f"Error: Failed to parse input as a Python literal. {e}")
        return

    # 2. Verify that the outer list has length n (teams) and inner lists length 4 (weeks).
    if not isinstance(matrix, list):
        print("Error: Input must be a list.")
        return

    n_teams = 6
    n_weeks = 4

    if len(matrix) != n_teams:
        print(f"Error: Matrix must have {n_teams} rows (teams), found {len(matrix)}.")
        return

    for r_idx, row in enumerate(matrix):
        if not isinstance(row, list):
            print(f"Error: Row {r_idx} must be a list.")
            return
        if len(row) != n_weeks:
            print(f"Error: Row {r_idx} must have {n_weeks} columns (weeks), found {len(row)}.")
            return

        # 3. Verify that all entries in the matrix are integers 0 or 1.
        for c_idx, val in enumerate(row):
            if val not in (0, 1):
                print(f"Error: Matrix entry at row {r_idx}, column {c_idx} must be 0 or 1. Found {val}.")
                return

    # 4. Iterate through all unique pairs of row indices (i, j) where i != j.
    # We iterate i from 0 to n-1 and j from i+1 to n-1.
    for i in range(n_teams):
        for j in range(i + 1, n_teams):

            # 5. Verify that there exists at least one column c1 such that row[i][c1] == 1 and row[j][c1] == 0
            # This ensures Team i can host Team j.
            i_hosts_j = False
            for w in range(n_weeks):
                if matrix[i][w] == 1 and matrix[j][w] == 0:
                    i_hosts_j = True
                    break

            if not i_hosts_j:
                print(f"Error: Condition failed for pair ({i}, {j}). Team {i} never hosts Team {j} "
                      f"(requires a week where Team {i} is home/1 and Team {j} is away/0).")
                return

            # 6. Verify that there exists at least one column c2 such that row[j][c2] == 1 and row[i][c2] == 0
            # This ensures Team j can host Team i.
            j_hosts_i = False
            for w in range(n_weeks):
                if matrix[j][w] == 1 and matrix[i][w] == 0:
                    j_hosts_i = True
                    break

            if not j_hosts_i:
                print(f"Error: Condition failed for pair ({i}, {j}). Team {j} never hosts Team {i} "
                      f"(requires a week where Team {j} is home/1 and Team {i} is away/0).")
                return

    # 7. If all checks pass, return True.
    print("True")

if __name__ == "__main__":
    verify()