Question: A meeting has $30$ participants. It is known that each person has at most $5$ acquaintances among the others, and in any group of $5$ people, there exist two who are not acquaintances. Find the largest positive integer $k$ such that, under the given conditions, there always exist $k$ people who are pairwise non-acquaintances.

The largest positive integer $k$ is $6$.

We model the problem using a simple graph $G$. Let the $30$ participants be represented by $30$ vertices, and draw an edge between two vertices if the corresponding people are acquaintances. Let $V$ denote the vertex set of $G$. By the given conditions, every vertex in $G$ has degree at most $5$, and any set of $5$ vertices contains at least two vertices that are not adjacent. We are looking for the largest integer $k$ such that $G$ always contains an independent set of size at least $k$.

First, we prove that the size of a maximum independent set in $G$ is at least $6$.
Let $X$ be a maximum independent set of $G$. By the maximality of $X$, every vertex in $V \setminus X$ must be adjacent to at least one vertex in $X$. (Otherwise, if some $a \in V \setminus X$ were not adjacent to any vertex in $X$, we could add $a$ to $X$ to obtain a strictly larger independent set, a contradiction.) Since each vertex in $X$ has degree at most $5$, the number of edges between $X$ and $V \setminus X$ is at most $5|X|$. Thus, we have
$$ 5|X| \ge |V \setminus X| = 30 - |X|, $$
which yields $6|X| \ge 30$, so $|X| \ge 5$.

If $|X| = 5$, then the inequality $5|X| \ge 30 - |X|$ becomes an equality. This implies that each vertex in $X$ is adjacent to exactly $5$ vertices in $V \setminus X$, and the neighbor sets of any two distinct vertices in $X$ are disjoint. Let $X = \{a, b, c, d, e\}$, and let the neighbor set of $a$ be
$$ Y_a = \{y_1, y_2, y_3, y_4, y_5\}. $$
By the problem condition, among the $5$ vertices in $Y_a$, there must be two that are not adjacent; without loss of generality, let them be $y_1$ and $y_2$. Since the neighbor sets of the vertices in $X$ are disjoint, neither $y_1$ nor $y_2$ is adjacent to any of $b, c, d, e$. Therefore,
$$ \{y_1, y_2, b, c, d, e\} $$
is an independent set in $G$ of size $6$, contradicting the assumption that the maximum independent set $X$ has size $5$. Hence, we must have $|X| \ge 6$.

Next, we prove that there exists a graph $G$ satisfying the conditions for which the maximum independent set has exactly $6$ vertices.
Partition the vertex set $V$ into three subsets $V_1, V_2, V_3$ such that
$$ |V_1| = |V_2| = |V_3| = 10. $$
Let $V_1 = \{a_1, a_2, \dots, a_{10}\}$. We define the edges within $V_1$ by declaring that $a_i$ is adjacent to
$$ a_{i+1}, \; a_{i+4}, \; a_{i+5}, \; a_{i+6}, \; a_{i+9}, $$
where the indices are taken modulo $10$. We define the edges within $V_2$ and $V_3$ in exactly the same way. For $1 \le i < j \le 3$, there are no edges between $V_i$ and $V_j$.

We now verify that this graph $G$ meets all the required conditions:

1. Every vertex of $G$ has degree exactly $5$.

2. Any $5$ vertices in $G$ contain two vertices that are not adjacent. It suffices to check that there is no clique of size $5$. Any such clique would have to be contained entirely within one component, say $V_1$. If it contains $a_i$, the other $4$ vertices must be chosen from its neighbors $a_{i+1}, a_{i+4}, a_{i+5}, a_{i+6}, a_{i+9}$. However, $a_{i+1}$ is not adjacent to $a_{i+4}$, and $a_{i+6}$ is not adjacent to $a_{i+9}$. Thus, it is impossible to choose $4$ pairwise adjacent neighbors of $a_i$.

3. The maximum independent set in $G$ has size $6$. Suppose $a_{i_1}, a_{i_2}, a_{i_3}$ are three distinct vertices in $V_1$ such that $a_{i_1}$ is not adjacent to $a_{i_2}$, and $a_{i_2}$ is not adjacent to $a_{i_3}$. By the construction of $V_1$, non-adjacent vertices must have index differences of $2$ or $3$ modulo $10$. Thus,
$$ |i_1 - i_2| \equiv 2 \text{ or } 3 \pmod{10}, \qquad |i_2 - i_3| \equiv 2 \text{ or } 3 \pmod{10}. $$
It follows that the non-zero difference $|i_1 - i_3| \pmod{10}$ must be one of $\pm 1, \pm 4, \text{ or } 5$. Since all these differences correspond to adjacent vertices, $a_{i_1}$ and $a_{i_3}$ must be adjacent. This shows that the largest independent set in $V_1$ has size at most $2$. Consequently, the largest independent set in $G$ has size $2 \times 3 = 6$.

We conclude that the largest positive integer $k$ is $6$.

(Partial) 1. Modeled the problem as a graph and identified the goal as finding a lower bound on the independence number (and/or an upper bound via a construction). 2. Proved the basic bound using a maximal independent set X and degree ≤5: every vertex outside X is adjacent to X, hence 5|X| ≥ 30−|X| and so |X| ≥ 5. 3. Used the “any 5 vertices contain two non-adjacent” condition to exclude some small-clique configuration (e.g. observed that the graph has no K5). 4. Gave a plausible extremal construction idea (e.g. disjoint union of 3 graphs on 10 vertices each with maximum independent set 2), but did not fully verify the required properties. (Almost) 1. Proved |X|≥6 via the maximal independent set counting argument and the equality case |X|=5 leading to disjoint neighbor sets, but left a small gap in deriving the disjointness/forcing the equality structure. 2. Completed the lower bound and presented a construction intended to show k≤6, but did not fully check one of: degree ≤5, absence of a 5-clique (the “any 5 contain two non-adjacent” condition), or that the maximum independent set in the construction is exactly 6. 3. Solution is almost complete, but made minor mistakes which are not negligible.

Instruction:
Construct an explicit 30x30 adjacency matrix representing the acquaintance graph that proves the boundary value $k$ is tight.

Output requirements:
- The construction must be a single Python expression. Evaluating this expression must directly return the required 30x30 adjacency matrix.
- The value of the expression must be a single Python list of lists representing the 30x30 adjacency matrix.
- Use 0 and 1 for the matrix elements.
- The matrix must represent an undirected simple graph (it must be symmetric, with 0s on the main diagonal).
- The graph must satisfy the constraints: max degree $\le 5$, and no clique of size 5 ($K_5$).
- The maximum independent set size of the graph must be exactly the boundary value $k$ implied by the problem.
- Do not use explanations, comments, assignments, imports, function definitions, or any non-expression content.
- Put ONLY the Python expression inside the `<construct>` block.

import sys
import itertools

def solve():
    payload = sys.stdin.read().strip()
    if not payload:
        print("Empty input.")
        return
    try:
        matrix = eval(payload, {"__builtins__": {}, "range": range, "int": int}, {})
    except Exception as e:
        print(f"Failed to evaluate construction: {e}")
        return

    if not isinstance(matrix, list) or len(matrix) != 30:
        print("Matrix must be a list of 30 rows.")
        return
    for i, row in enumerate(matrix):
        if not isinstance(row, list) or len(row) != 30:
            print(f"Row {i} must be a list of 30 elements.")
            return
        for j, val in enumerate(row):
            if val not in (0, 1):
                print(f"Element at ({i}, {j}) is not 0 or 1.")
                return

    for i in range(30):
        if matrix[i][i] != 0:
            print(f"Diagonal element at ({i}, {i}) is not 0.")
            return
        for j in range(30):
            if matrix[i][j] != matrix[j][i]:
                print(f"Matrix is not symmetric at ({i}, {j}).")
                return

    for i in range(30):
        deg = sum(matrix[i])
        if deg > 5:
            print(f"Vertex {i} has degree {deg} > 5.")
            return

    for i in range(30):
        neighbors = [j for j in range(30) if matrix[i][j] == 1]
        if len(neighbors) >= 4:
            for combo in itertools.combinations(neighbors, 4):
                is_clique = True
                for u, v in itertools.combinations(combo, 2):
                    if matrix[u][v] == 0:
                        is_clique = False
                        break
                if is_clique:
                    print(f"Found a clique of size 5 including vertex {i}: {tuple([i] + list(combo))}")
                    return

    def get_max_independent_set_size(adj):
        n = len(adj)
        max_size = 0
        comp_neighbors = [0] * n
        for i in range(n):
            for j in range(n):
                if i != j and adj[i][j] == 0:
                    comp_neighbors[i] |= (1 << j)

        def bk(r, p, x):
            nonlocal max_size
            r_count = bin(r).count('1')
            p_count = bin(p).count('1')

            if p == 0 and x == 0:
                if r_count > max_size:
                    max_size = r_count
                return

            if r_count + p_count <= max_size:
                return

            pivot = -1
            max_p_intersect = -1
            union_px = p | x
            temp = union_px
            while temp > 0:
                lsb = temp & -temp
                i = lsb.bit_length() - 1
                intersect = p & comp_neighbors[i]
                c = bin(intersect).count('1')
                if c > max_p_intersect:
                    max_p_intersect = c
                    pivot = i
                temp &= temp - 1

            explore = p & ~comp_neighbors[pivot] if pivot != -1 else p
            while explore > 0:
                v = explore & -explore
                v_idx = v.bit_length() - 1
                bk(r | v, p & comp_neighbors[v_idx], x & comp_neighbors[v_idx])
                p &= ~v
                x |= v
                explore &= ~v

        bk(0, (1 << n) - 1, 0)
        return max_size

    mis_size = get_max_independent_set_size(matrix)
    if mis_size != 6:
        print(f"Maximum independent set size is {mis_size}, expected exactly 6.")
        return

    print("True")

if __name__ == "__main__":
    solve()


Question:
"Given a prime number $p$. Let $A = (a_{ij})$ be a $p \\times p$ matrix such that $\\{a_{ij} \\mid 1 \\leq i, j \\leq p\\} = \\{1, 2, \\cdots, p^2\\}$. The following operation is allowed: choose a row or a column, and simultaneously add $1$ or subtract $1$ to each number in that row or column. If it is possible to turn all elements of $A$ into $0$ after finitely many such operations, then $A$ is called a \""good matrix\"". Find the number of \""good matrices\"".
"

Let the operations applied to the rows and columns be represented by subtracting $x_i$ from row $i$ and $y_j$ from column $j$. The matrix can be turned into the zero matrix if and only if the original entries satisfy $a_{ij} = x_i + y_j$ for some integers $x_i$ and $y_j$ for all $1 \le i, j \le p$.

Since the entries of the matrix are a permutation of $\{1, 2, \dots, p^2\}$, we have
\[
\{x_i + y_j \mid 1 \le i, j \le p\} = \{1, 2, \dots, p^2\}.
\]
Let us define the polynomials
\[
f(z) = \sum_{i=1}^{p} z^{x_i} \quad \text{and} \quad g(z) = \sum_{j=1}^{p} z^{y_j}.
\]
Then, the condition on the sums $x_i + y_j$ translates to the polynomial identity
\[
f(z)g(z) = z + z^2 + \cdots + z^{p^2}.
\]
We can factor the right-hand side as
\[
z + z^2 + \cdots + z^{p^2} = z(1 + z + \cdots + z^{p-1})(1 + z^p + \cdots + z^{p(p-1)}).
\]
Because the cyclotomic polynomials $\Phi_p(z) = 1 + z + \cdots + z^{p-1}$ and $\Phi_{p^2}(z) = 1 + z^p + \cdots + z^{p(p-1)}$ are both irreducible, and $f(1) = g(1) = p$, each of $f(z)$ and $g(z)$ must be the product of a monomial $z^a$ and exactly one of $\Phi_p(z)$ or $\Phi_{p^2}(z)$. Consequently, the set $\{x_1, x_2, \dots, x_p\}$ must be one of the following:
\[
\{0, 1, \dots, p-1\} \quad \text{or} \quad \{1, p+1, \dots, p(p-1)+1\} \quad \text{or} \quad \{1, 2, \dots, p\} \quad \text{or} \quad \{0, p, \dots, p(p-1)\}.
\]
From this, it is easy to obtain the solution to the original problem, yielding a total of $2(p!)^2$ good matrices.

(Partial) 1. Reformulated the operations: showed that $A$ is good iff there exist integers $(x_i)_{i=1}^p,(y_j)_{j=1}^p$ with $a_{ij}=x_i+y_j$ for all $i,j$ (equivalently, all $2\times2$ minors satisfy $a_{ij}+a_{kl}=a_{il}+a_{kj}$). 2. Set up the generating-function identity $f(z)=\sum_{i=1}^p z^{x_i},\ g(z)=\sum_{j=1}^p z^{y_j}$ and derived $f(z)g(z)=z+z^2+\cdots+z^{p^2}$ from the multiset condition $\{x_i+y_j\}=\{1,2,\dots,p^2\}$. 3. From the identity (or from modular/spacing considerations), correctly obtained at least one of the two structural possibilities: either $\{x_i\}$ is a translate of $\{0,1,\dots,p-1\}$ and $\{y_j\}$ is a translate of $\{0,p,2p,\dots,(p-1)p\}$, or the swapped roles of rows/columns. (Almost) 1. Completed the classification of possible exponent sets (up to translation): proved that $\{x_i\}$ and $\{y_j\}$ must correspond to the two cyclotomic factors (hence to one of the two “arithmetic progression mod $p$ / progression step $p$” patterns), but did not finish converting this into the final count $2(p!)^2$. 2. Counted the matrices once the two patterns for $(x_i),(y_j)$ are identified, but missed a global shift/normalization issue (e.g. double-counted or forgot the symmetry swapping rows and columns), leading to an answer off by a factor of $2$ or by a constant factor. 3. Solution is almost complete, but made minor mistakes which are not negligible.

Instruction:
For the specific case $p=3$, provide an explicit construction of all valid "good matrices".

Construction requirements:
- Output a single Python list containing all such matrices.
- Each matrix must be represented as a list of 3 lists, where each inner list represents a row of 3 integers.
- The matrices can be listed in any order, but there must be no duplicates.
- Your output must be exhaustive, including every possible good matrix exactly once.
- Each matrix must contain all integers from 1 to 9 exactly once.
- Each matrix must be a "good matrix" according to the problem definition (i.e., it can be reduced to the zero matrix by the allowed operations).
- Put ONLY the Python literal list of lists of lists inside the construction block.

import sys
import ast

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

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

    if not isinstance(matrices, list):
        print("The payload must be a list.")
        return

    if len(matrices) != 72:
        print(f"Expected exactly 72 matrices, but got {len(matrices)}.")
        return

    unique_matrices = set()

    for idx, matrix in enumerate(matrices):
        if not isinstance(matrix, list) or len(matrix) != 3:
            print(f"Matrix at index {idx} is not a list of exactly 3 elements.")
            return

        elements = []
        tuple_matrix = []
        for r_idx, row in enumerate(matrix):
            if not isinstance(row, list) or len(row) != 3:
                print(f"Row {r_idx} of matrix at index {idx} is not a list of exactly 3 elements.")
                return
            for val in row:
                if type(val) is not int:
                    print(f"Element {val} in matrix {idx} is not an integer.")
                    return
                elements.append(val)
            tuple_matrix.append(tuple(row))

        unique_matrices.add(tuple(tuple_matrix))

        if sorted(elements) != [1, 2, 3, 4, 5, 6, 7, 8, 9]:
            print(f"Matrix at index {idx} does not contain exactly the integers 1 to 9.")
            return

        for i in range(3):
            for j in range(3):
                if matrix[i][j] - matrix[i][0] != matrix[0][j] - matrix[0][0]:
                    print(f"Matrix at index {idx} is not a 'good matrix'. Failed property at i={i}, j={j}.")
                    return

    if len(unique_matrices) != 72:
        print(f"Expected 72 unique matrices, but found {len(unique_matrices)}.")
        return

    print("True")

if __name__ == "__main__":
    verify()


Question: A garden consists of a $10\times10$ grid of squares. On New Year's Eve, it begins to snow, and from that point on, every night $10$ cm of snow falls on each square (snow only falls at night). Every morning, the gardener chooses one row or one column and shovels all the snow from it onto an adjacent row or column, with the rule that snow cannot be shoveled outside the garden. On the evening of the $100$th day of the new year, an inspector will arrive and find the highest snow pile. The gardener's goal is to minimize this height. Find the height of the snow pile found by the inspector. Note: The snow does not melt during these $100$ days.

On the morning of day $100$, the caretaker clears the snow from an entire row or column. Since snow only falls at night, by the evening of day $100$ (when the inspector arrives), there are $10$ cells in that cleared row or column with $0$ cm of snow.

Over the $100$ days, a total of $100 \times 100$ units of $10$-centimeter snow has fallen on the garden. Since the total amount of snow is conserved (none is shoveled outside or melts), these $10,000$ units of snow must be distributed among the remaining at most $90$ cells. By the averaging principle, there must be at least one cell containing at least
$$ 10 \left\lceil \frac{100 \times 100}{90} \right\rceil = 10 \lceil 111.11\dots \rceil = 1120 \text{ cm} $$
of snow. Thus, the maximum height of a snow pile is at least $1120$ cm.

We can achieve exactly $1120$ cm with the following strategy:
- In the first $11$ days (days $1$ to $11$), the caretaker sweeps all the snow from the second column into the first column.
- In the next $11$ days (days $12$ to $22$), the caretaker sweeps all the snow from the third column into the second column.
- In general, for each $k = 1, 2, \dots, 9$, during the $11$ days from day $11k - 10$ to day $11k$, the caretaker sweeps all the snow from column $k+1$ into column $k$.

Let us track the amount of snow in each column. We claim that after $11k$ days, each cell in columns $1$ through $k$ has $110(k+1)$ cm of snow, each cell in column $k+1$ has $0$ cm, and each cell in columns $k+2$ through $10$ has $110k$ cm. We proceed by induction on $k$:
- **Base case ($k=1$)**: During the first $11$ days, column $2$ is swept to column $1$ every morning. Each day, column $1$ receives its own $10$ cm of snowfall plus the $10$ cm of snow swept from column $2$, accumulating $20 \times 11 = 220$ cm. Column $2$ is cleared every morning and thus ends with $0$ cm. Columns $3$ through $10$ are untouched and each accumulate $10 \times 11 = 110$ cm. The claim holds.
- **Inductive step**: Suppose the claim holds for $k$. During the next $11$ days (days $11k+1$ to $11k+11$), column $k+2$ is swept into column $k+1$.
  On the morning of day $11k+1$, column $k+2$ has its previous $110k$ cm plus $10$ cm from the overnight snowfall, totaling $110k + 10$ cm. This is swept into column $k+1$, which also received $10$ cm overnight. Thus, column $k+1$ now holds $10 + 110k + 10 = 110k + 20$ cm.
  For the remaining $10$ days of this $11$-day period, column $k+1$ receives its own $10$ cm plus the $10$ cm from column $k+2$ each day, gaining an additional $20 \times 10 = 200$ cm. Thus, column $k+1$ reaches $110k + 20 + 200 = 110(k+2)$ cm.
  Column $k+2$ is cleared every morning and ends with $0$ cm.
  Columns $1$ through $k$ simply accumulate $10$ cm each night for $11$ nights, adding $110$ cm to reach $110(k+1) + 110 = 110(k+2)$ cm.
  Columns $k+3$ through $10$ also accumulate $110$ cm, reaching $110k + 110 = 110(k+1)$ cm.
  This completes the induction.

Applying this for $k=9$, after $99$ days, each cell in columns $1$ through $9$ has $110(10) = 1100$ cm of snow, and each cell in column $10$ has $0$ cm.

On day $100$, $10$ cm of snow falls on all columns overnight. Columns $1$ through $9$ now have $1110$ cm, and column $10$ has $10$ cm. In the morning, the caretaker sweeps the snow from the tenth column into the ninth column. Each cell in column $9$ receives the $10$ cm from column $10$, reaching $1110 + 10 = 1120$ cm. Column $10$ is left with $0$ cm, and columns $1$ through $8$ remain at $1110$ cm.

Therefore, on the evening of the $100$th day, the maximum snow pile height found by the inspector is exactly $1120$ cm.

(Partial) 1. Observed that on day 100 the gardener can create (at least) 10 cells with 0 cm by choosing a whole row/column to be cleared that morning, and used this to reduce the final distribution to at most 90 nonzero cells. 2. Used conservation of total snow and an averaging/pigeonhole argument to obtain a nontrivial lower bound for the maximum pile (e.g. deriving the bound 1120 cm, or correctly setting up 10000 units over 90 cells). 3. Gave a concrete, checkable shoveling strategy that keeps the maximum height bounded by a specific value close to the optimum (even if not proved optimal), e.g. a systematic column-by-column shifting plan. (Almost) 1. Proved the correct lower bound 1120 cm, but the construction is incomplete (e.g. strategy described but not tracked rigorously to ensure no cell exceeds 1120). 2. Presented the intended 11-day-per-column shifting strategy (or equivalent) and tracked column heights to reach a maximum of 1120 cm, but with a small arithmetic/indexing error (such as miscounting 99 vs 100 days) that is localized and fixable. 3. Gave both a correct lower bound via averaging and a near-complete matching strategy, but omitted a boundary check (e.g. the final day’s move or the fact that shoveling is to an adjacent row/column only) while the rest of the argument is correct.

Instruction:
Provide an explicit 100-day morning operation schedule that achieves the optimal minimization objective from the original problem.

Output exactly one construction block with ONLY a Python literal list of length 100 between the tags <construct> and </construct>. Each element must be a 3-tuple (kind, idx, dir) where:
- kind is the string "R" (row) or "C" (column),
- idx is an integer in {0,1,...,9} indicating which row/column is selected,
- dir is either -1 or +1 indicating shoveling to the adjacent row/column with index idx+dir.

Operational meaning to be verified (do not restate as extra text inside the construction):
- Nights: for each day d = 1..100, before the morning operation, every one of the 100 squares gains +10 cm of snow.
- Mornings: then execute the scheduled operation for that day: choose the specified full row/column, remove all its snow and add it cellwise onto the adjacent target row/column (same columns for a row move; same rows for a column move). The source row/column becomes 0 afterwards. Snow may not be shoveled outside the 10x10 grid, so idx+dir must be in {0..9}.

The verifier will simulate all 100 days in this time order and then compute, on the evening of day 100 (i.e., after the 100th morning operation), the maximum snow height over all squares; your schedule must attain the optimal bound for this instance.

import sys, ast

TARGET_MAX = 1120
N = 10
DAYS = 100
FALL = 10

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

def main():
    s = sys.stdin.read()
    if s.strip() == "":
        fail("Parse error: empty input (expected a Python literal list).")
    try:
        obj = ast.literal_eval(s)
    except Exception as e:
        fail(f"Parse error: {e}")

    if not isinstance(obj, list):
        fail(f"Validation error: expected a list, got {type(obj).__name__}.")
    if len(obj) != DAYS:
        fail(f"Validation error: expected list of length {DAYS}, got {len(obj)}.")

    # grid[r][c] in centimeters
    grid = [[0 for _ in range(N)] for _ in range(N)]

    for day in range(DAYS):
        step = obj[day]
        if not isinstance(step, (tuple, list)) or len(step) != 3:
            fail(f"Validation error (day {day+1}): expected a 3-tuple/list (kind, idx, dir).")
        kind, idx, direction = step

        if kind not in ("R", "C"):
            fail(f"Validation error (day {day+1}): kind must be 'R' or 'C'.")
        if not isinstance(idx, int):
            fail(f"Validation error (day {day+1}): idx must be an int.")
        if idx < 0 or idx >= N:
            fail(f"Validation error (day {day+1}): idx out of range 0..9.")
        if direction not in (-1, 1):
            fail(f"Validation error (day {day+1}): dir must be -1 or +1.")
        dst = idx + direction
        if not isinstance(dst, int) or dst < 0 or dst >= N:
            fail(f"Validation error (day {day+1}): illegal destination idx+dir={dst} (must be 0..9).")

        # Night snowfall
        for r in range(N):
            row = grid[r]
            for c in range(N):
                row[c] += FALL

        # Morning operation
        if kind == "R":
            src = idx
            tgt = dst
            src_row = grid[src]
            tgt_row = grid[tgt]
            for c in range(N):
                tgt_row[c] += src_row[c]
                src_row[c] = 0
        else:  # kind == "C"
            src = idx
            tgt = dst
            for r in range(N):
                grid[r][tgt] += grid[r][src]
                grid[r][src] = 0

        # Nonnegativity sanity
        for r in range(N):
            for c in range(N):
                if grid[r][c] < 0:
                    fail(f"Validation error (day {day+1}): negative snow height encountered.")

    mx = max(max(row) for row in grid)
    if mx != TARGET_MAX:
        fail(f"Validation failure: final maximum snow height is {mx} cm, expected {TARGET_MAX} cm.")
    print("True")

if __name__ == "__main__":
    main()


Question: A math group has 24 students. For every team of 6 students, the supervisor gives one of two evaluations: "can cooperate" or "cannot cooperate." For training in a math competition, the supervisor plans to divide the students into 4 teams, each with 6 students. Is it possible that for any division into 4 teams, either exactly three teams can cooperate or exactly one team can cooperate, and both cases occur?

Yes, it is possible.

Select three specific students from the 24. Evaluate any team of 6 students that contains exactly one or all three of these selected students as "can cooperate", and evaluate all other teams (those containing exactly zero or exactly two of the selected students) as "cannot cooperate".

Below we verify that this evaluation method satisfies the requirement. When the 24 students are divided into 4 teams of 6, the three selected students must be distributed among the four teams. The possible distributions of the three selected students are as follows:

1. If the three selected students are placed in three different teams (a 1-1-1-0 distribution), then exactly three teams contain one selected student and one team contains zero. The three teams with one selected student are evaluated as "can cooperate", and the remaining team is evaluated as "cannot cooperate". Thus, exactly three teams can cooperate.
2. If all three selected students are placed in the same team (a 3-0-0-0 distribution), then exactly one team contains three selected students and the other three contain zero. The team with three selected students is evaluated as "can cooperate", and the other three are evaluated as "cannot cooperate". Thus, exactly one team can cooperate.
3. If two of the selected students are placed in the same team and the remaining one is placed in another team (a 2-1-0-0 distribution), then exactly one team contains one selected student, one team contains two, and two teams contain zero. The team containing exactly one selected student is evaluated as "can cooperate", while the team with two and the teams with zero are evaluated as "cannot cooperate". Thus, exactly one team can cooperate.

Since the total number of selected students across the four teams is exactly 3, these are all the possible cases. In every division into 4 teams, either exactly three teams can cooperate or exactly one team can cooperate.

Furthermore, since each team has a capacity of 6 students (which is sufficient to hold up to 3 selected students) and there are enough remaining students to fill the teams, we can easily construct divisions that achieve the 1-1-1-0 distribution and the 3-0-0-0 distribution. Thus, both outcomes (exactly 3 teams can cooperate, and exactly 1 team can cooperate) actually occur.

Hence, the proposition is proved.

(Partial) 1. Proposed a concrete evaluation rule depending only on how a 6-person team intersects a fixed small distinguished set of students (e.g. 3 chosen students), in a way intended to force the number of “can cooperate” teams in any 4-way partition to be 1 or 3. 2. Correctly analysed at least one nontrivial distribution pattern of the distinguished students among the four teams (e.g. 1-1-1-0 or 3-0-0-0 or 2-1-0-0) and deduced the corresponding number of “can cooperate” teams under the proposed rule. 3. Exhibited at least one explicit partition of the 24 students into four 6-person teams that achieves one of the two required outcomes (either exactly 1 cooperative team or exactly 3 cooperative teams). (Almost) 1. Gave essentially the intended construction (e.g. choose 3 special students; declare teams with exactly 1 or 3 of them as “can cooperate”, others as “cannot”) and checked the cooperative-team count for all distributions of the 3 special students among the four teams, but omitted a short justification that these are the only possible distributions. 2. Proved that in every 4-team division the number of cooperative teams is always either 1 or 3, but did not explicitly show that both outcomes actually occur (missing an explicit partition for one of the two cases). 3. Solution is almost complete, but made minor mistakes which are not negligible.

Instruction:
For the 24 students (numbered 1 through 24), provide an explicit rule to determine whether any given 6-student team "can cooperate" or "cannot cooperate", and construct two specific partitions to demonstrate the required outcomes.

Requirements:
1. Output a single Python dictionary with exactly three keys: `"rule"`, `"partition_1"`, and `"partition_3"`.
2. `"rule"` must be a string containing a valid Python `lambda` function. This function must take a single argument `team` (a list of 6 distinct integers from 1 to 24) and return a boolean (`True` if the team can cooperate, `False` otherwise).
3. `"partition_1"` must be a list of 4 lists, representing a valid partition of the 24 students into four 6-person teams, such that exactly ONE team evaluates to `True` under your rule.
4. `"partition_3"` must be a list of 4 lists, representing a valid partition of the 24 students into four 6-person teams, such that exactly THREE teams evaluate to `True` under your rule.
5. Your rule must mathematically guarantee that for *any* valid partition of the 24 students into four 6-person teams, the number of cooperating teams is always exactly 1 or exactly 3.

Format:
- Put ONLY the Python dictionary literal inside the `<construct>...</construct>` block.
- Do not use markdown code fences around the dictionary.

import sys
import ast
from itertools import combinations
import builtins

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

    # Parse payload using ast.literal_eval
    try:
        data = ast.literal_eval(input_str)
    except Exception as e:
        print(f"Parse error: {e}")
        return

    if not isinstance(data, dict):
        print("Payload must be a dictionary")
        return

    expected_keys = {"rule", "partition_1", "partition_3"}
    if set(data.keys()) != expected_keys:
        print(f"Dictionary must contain exactly keys {expected_keys}")
        return

    rule_str = data["rule"]
    if not isinstance(rule_str, str):
        print("'rule' must be a string")
        return

    # Evaluate the lambda function in a safe environment
    allowed_names = {'sum', 'len', 'set', 'list', 'tuple', 'any', 'all', 'min', 'max', 'abs', 'bool', 'int', 'map', 'filter'}
    safe_builtins = {k: getattr(builtins, k) for k in allowed_names if hasattr(builtins, k)}
    safe_env = {"__builtins__": safe_builtins}

    try:
        rule = eval(rule_str, safe_env, {})
    except Exception as e:
        print(f"Error evaluating rule: {e}")
        return

    if not callable(rule):
        print("'rule' must evaluate to a callable function")
        return

    # Helper to validate partition structures
    def validate_partition(part, name):
        if not isinstance(part, list) or len(part) != 4:
            return False, f"'{name}' must be a list of 4 lists"
        seen = set()
        for team in part:
            if not isinstance(team, list) or len(team) != 6:
                return False, f"Each team in '{name}' must be a list of 6 integers"
            for student in team:
                if not isinstance(student, int) or student < 1 or student > 24:
                    return False, f"Students in '{name}' must be integers from 1 to 24"
                if student in seen:
                    return False, f"Student {student} appears multiple times in '{name}'"
                seen.add(student)
        if len(seen) != 24:
            return False, f"'{name}' does not contain all 24 students"
        return True, ""

    # Validate partition_1
    ok, msg = validate_partition(data["partition_1"], "partition_1")
    if not ok:
        print(msg)
        return

    # Validate partition_3
    ok, msg = validate_partition(data["partition_3"], "partition_3")
    if not ok:
        print(msg)
        return

    # Apply rule to partition_1 and verify exactly 1 True
    try:
        p1_evals = [bool(rule(list(team))) for team in data["partition_1"]]
    except Exception as e:
        print(f"Error applying rule to partition_1: {e}")
        return

    if sum(p1_evals) != 1:
        print(f"partition_1 must have exactly 1 cooperating team, got {sum(p1_evals)}")
        return

    # Apply rule to partition_3 and verify exactly 3 True
    try:
        p3_evals = [bool(rule(list(team))) for team in data["partition_3"]]
    except Exception as e:
        print(f"Error applying rule to partition_3: {e}")
        return

    if sum(p3_evals) != 3:
        print(f"partition_3 must have exactly 3 cooperating teams, got {sum(p3_evals)}")
        return

    # Mathematical verification of the universal property
    # Any rule that guarantees a constant sum modulo 2 over all partitions MUST be a degree-1 function.
    # We evaluate the rule on all 134,596 possible teams and verify it is of the form:
    # f(S) = sum_{x in S} w_x + C (mod 2)
    elements = list(range(1, 25))
    subsets = list(combinations(elements, 6))

    f_vals = {}
    try:
        for S in subsets:
            f_vals[S] = bool(rule(list(S)))
    except Exception as e:
        print(f"Error applying rule to subset {S}: {e}")
        return

    # Determine the weights w_x for each element
    w = {1: 0}
    for v in range(2, 25):
        # Pick a base set A containing 1 but not v
        if v > 6:
            A = (1, 2, 3, 4, 5, 6)
        else:
            A = (1, 7, 8, 9, 10, 11)
        A_minus_1_plus_v = tuple(sorted((set(A) - {1}) | {v}))
        w[v] = (f_vals[A] + f_vals[A_minus_1_plus_v]) % 2

    # Determine the constant C
    S0 = (1, 2, 3, 4, 5, 6)
    sum_w_S0 = sum(w[x] for x in S0) % 2
    C = (f_vals[S0] + sum_w_S0) % 2

    # Verify that the function exactly matches the degree-1 polynomial for ALL subsets
    for S in subsets:
        expected = (w[S[0]] + w[S[1]] + w[S[2]] + w[S[3]] + w[S[4]] + w[S[5]] + C) % 2
        if f_vals[S] != expected:
            print("Rule does not mathematically guarantee the universal property (not a degree-1 function).")
            return

    # Verify that the sum over any partition is exactly 1 or 3
    # The sum over a partition modulo 2 is equivalent to the sum of all w_x modulo 2.
    sum_all_w = sum(w.values()) % 2
    if sum_all_w != 1:
        print("Rule guarantees a constant parity of cooperating teams, but the sum is not odd (so it cannot be exactly 1 or 3).")
        return

    print("True")

if __name__ == "__main__":
    verify()


Question:
"In an $n\\times n$ grid, some squares are painted black such that the bottom-left square $L$ and the top-right square $R$ are not painted black, and every possible path of a knight from $L$ to $R$ passes through at least one black square. Determine all integers $n\\geq4$ for which there must exist three diagonally adjacent squares, at least two of which are painted black.
"

All integers that are congruent to $1$ modulo $3$ and greater than $1$

The answer is all integers $n \equiv 1 \pmod 3$ with $n > 1$.

Number the rows from bottom to top and the columns from left to right consecutively as $1,2,\dots,n$. Let the coordinates of the bottom-left square $L$ be $(1,1)$ and the top-right square $R$ be $(n,n)$.

**Case 1: $n = 3k$**
Color black all squares in rows whose numbers are congruent to $2 \pmod 3$.
In this coloring, there is no set of three diagonally adjacent squares with at least two black squares, because out of any three consecutive rows, exactly one is congruent to $2 \pmod 3$.
If a knight starts from $L(1,1)$ and does not pass through any black square, it can only move between the uncolored rows $y \equiv 1 \pmod 3$ and $y \equiv 0 \pmod 3$. A knight move from $y \equiv 1 \pmod 3$ must land in $y' \equiv 0 \pmod 3$, and vice versa. By checking the possible changes in coordinates, one can verify the invariant that for any reachable square $(x,y)$, $x \equiv y \pmod 2$ if $y \equiv 1 \pmod 3$, and $x \not\equiv y \pmod 2$ if $y \equiv 0 \pmod 3$. Since $R=(3k, 3k)$ has $y \equiv 0 \pmod 3$ but $x \equiv y \pmod 2$, it violates the invariant and cannot be reached. Thus, this coloring satisfies all requirements.

**Case 2: $n = 3k+2$**
Color black all squares in rows whose numbers are congruent to $0 \pmod 3$.
Analogous to the previous case, no three diagonally adjacent squares contain more than one black square. The uncolored rows are $y \equiv 1 \pmod 3$ and $y \equiv 2 \pmod 3$. The invariant for reachable squares becomes $x \equiv y \pmod 2$ for $y \equiv 1 \pmod 3$, and $x \not\equiv y \pmod 2$ for $y \equiv 2 \pmod 3$. The square $R=(3k+2, 3k+2)$ has $y \equiv 2 \pmod 3$ but $x \equiv y \pmod 2$, so it is unreachable.

**Case 3: $n = 3k+1$**
We prove by induction that the statement holds (i.e., no such coloring exists) when $n=3k+1$. It suffices to show that if the statement were false, there would exist a path for the knight from $L$ to $R$ containing no black squares, contradicting the hypothesis.

*Base case $k=1$ ($n=4$):* By the contradiction assumption, the diagonally adjacent squares $(1,4), (2,3), (3,2)$ cannot contain two black squares. Thus $(2,3)$ and $(3,2)$ are not both black. Assume without loss of generality that $(2,3)$ is unpainted; then the path $L(1,1) \rightarrow (2,3) \rightarrow R(4,4)$ contains no black squares, a contradiction.

*Inductive step:* Assume $k>1$ and that the statement holds for all smaller values of $k$; we consider the case $k$. For a square $(a,b)$, denote $(a',b')=(a+3k-3,\;b+3k-3)$.
By the contradiction assumption, the squares $(2,3)$ and $(3,2)$ are not both black; assume without loss of generality that $(2,3)$ is unpainted.
If the square $(2',3')$ is also unpainted, then by the induction hypothesis (applied to the $(3k-2) \times (3k-2)$ subgrid from $(2,3)$ to $(2',3')$) there exists a path from $(2,3)$ to $(2',3')$ containing no black squares. Consequently, the path $L \rightarrow (2,3) \rightarrow \dots \rightarrow (2',3') \rightarrow R$ contains no black squares.

If the square $(2',3')$ is painted black, then by the contradiction assumption the square $(3',2')$ is unpainted. To block the symmetric unpainted path $L \rightarrow (3,2) \rightarrow \dots \rightarrow (3',2') \rightarrow R$, we may assume that $(3,2)$ is painted black.
Furthermore, if $(4,4)$ were unpainted, the induction hypothesis applied to the subgrid from $(4,4)$ to $(4',4')$ would yield an unpainted path from $(4,4)$ to $R=(4',4')$, creating the unpainted path $L \rightarrow (2,3) \rightarrow (4,4) \rightarrow \dots \rightarrow R$. Thus, we may assume the square $(4,4)$ is painted black. Similarly, to block the path $L \rightarrow \dots \rightarrow (1',1') \rightarrow (3',2') \rightarrow R$, we may assume the square $(1',1')$ is painted black.

By the contradiction assumption, any three consecutive squares on a diagonal can contain at most one black square. Thus, the black squares $(3,2)$ and $(4,4)$ force their diagonally adjacent squares to be unpainted. This ensures that the knight can go from $L$ to the square $(1,4)$ entirely via forced unpainted squares (e.g., $L \rightarrow (2,3) \rightarrow (3,5) \rightarrow (1,4)$). Similarly, the black squares $(1',1')$ and $(2',3')$ force a sequence of their diagonal neighbors to be unpainted, providing an unpainted path from $(1',4')$ to $R$.

Finally, by the induction hypothesis applied to the $(3k-2) \times (3k-2)$ subgrid from $(1,4)$ to $(1',4')$, the knight can travel from $(1,4)$ to $(1',4')$ without landing on any black squares. Concatenating these segments yields a valid knight path from $L$ to $R$ with no black squares, which completes the induction and establishes the contradiction.

(Partial) 1. For n≡0 or 2 (mod 3), gave an explicit coloring (e.g. all squares in one residue class of rows mod 3) and correctly argued that it blocks all knight paths from L to R. 2. Proved the “no forced triple” property for the construction: in any three diagonally adjacent squares, at most one can be black (e.g. because among three consecutive rows exactly one is a black row). 3. Established a correct invariant for knight moves restricted to the unpainted rows (typically combining y mod 3 with x≡y (mod 2) / x≢y (mod 2)) and used it to show R is unreachable under the proposed construction. (Almost) 1. Correctly proved the positive direction for n≡1 (mod 3) by induction/reduction to a smaller subgrid, but left a localized gap in one of the forced-unpainted path concatenations near the boundary (e.g. reaching (1,4) or (1',4') without hitting black squares). 2. Solved the base case n=4 and set up the inductive step for n=3k+1, including the necessary WLOG choices (e.g. assuming (2,3) unpainted, then considering (2',3') black/unpainted), but did not fully justify one of the “to block the symmetric path, we may assume … is black” implications. 3. Completed all three cases and obtained the correct answer set, but made a minor non-negligible mistake in one congruence/invariant check (e.g. the parity condition at R in Case 1/2) while the overall method is clearly correct.

Instruction:
For the specific case $n=81$, provide an explicit construction of the grid coloring that satisfies all the problem's conditions (i.e., $L$ and $R$ are white, every knight path from $L$ to $R$ contains at least one black square, and no three diagonally adjacent squares have two or more black squares).

Construction requirements:
- Output a single Python expression that evaluates to a list of lists of integers, representing the $81 \times 81$ grid.
- Use `0` for white squares and `1` for black squares.
- The outer list should contain 81 inner lists (rows), ordered from top to bottom. Each inner list should contain 81 integers representing the columns from left to right.
- The expression must be self-contained and evaluate directly to the target data structure when passed to `eval()`.
- You may use list comprehensions, conditional expressions, and integer constants. Do not use assignment statements, imports, or any non-expression syntax.
- Put ONLY the raw Python expression inside the construction block. Do not wrap it in markdown code blocks or add any other text.

[[1 if i % 3 == 1 else 0 for j in range(81)] for i in range(81)]

import sys
from collections import deque

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

    try:
        # The instruction allows list comprehensions, so we use eval instead of ast.literal_eval
        grid = eval(raw_input)
    except Exception as e:
        print(f"Failed to parse input: {e}")
        return

    # 2. Verify grid dimensions and elements
    if not isinstance(grid, list):
        print("Grid must be a list.")
        return
    if len(grid) != 81:
        print(f"Grid must have exactly 81 rows, got {len(grid)}.")
        return

    for r_idx, row in enumerate(grid):
        if not isinstance(row, list):
            print(f"Row {r_idx} is not a list.")
            return
        if len(row) != 81:
            print(f"Row {r_idx} must have exactly 81 elements, got {len(row)}.")
            return
        for c_idx, val in enumerate(row):
            if type(val) is not int or val not in (0, 1):
                print(f"Element at ({r_idx}, {c_idx}) is not strictly the integer 0 or 1.")
                return

    # 3. Verify L and R are 0
    if grid[80][0] != 0:
        print("Bottom-left square (80, 0) must be 0 (white).")
        return
    if grid[0][80] != 0:
        print("Top-right square (0, 80) must be 0 (white).")
        return

    # 4. Iterate through all valid starting positions for a diagonal of length 3
    # Main diagonals
    for r in range(79):
        for c in range(79):
            count = grid[r][c] + grid[r+1][c+1] + grid[r+2][c+2]
            if count >= 2:
                print(f"Main diagonal starting at ({r}, {c}) has {count} black squares (must be strictly less than 2).")
                return

    # Anti-diagonals
    for r in range(79):
        for c in range(2, 81):
            count = grid[r][c] + grid[r+1][c-1] + grid[r+2][c-2]
            if count >= 2:
                print(f"Anti-diagonal starting at ({r}, {c}) has {count} black squares (must be strictly less than 2).")
                return

    # 5. Initialize a BFS queue to find all reachable squares from (80, 0) using valid knight moves
    queue = deque([(80, 0)])
    visited = {(80, 0)}

    knight_moves = [
        (-2, -1), (-2, 1), (-1, -2), (-1, 2),
        (1, -2), (1, 2), (2, -1), (2, 1)
    ]

    while queue:
        r, c = queue.popleft()

        # 6. Assert that the top-right square (0, 80) is not reachable
        if (r, c) == (0, 80):
            print("Top-right square (0, 80) is reachable from bottom-left square without hitting a black square.")
            return

        for dr, dc in knight_moves:
            nr, nc = r + dr, c + dc
            if 0 <= nr < 81 and 0 <= nc < 81:
                if (nr, nc) not in visited:
                    # Restricted only to squares where grid[r][c] == 0
                    if grid[nr][nc] == 0:
                        visited.add((nr, nc))
                        queue.append((nr, nc))

    print("True")

if __name__ == "__main__":
    verify()


Question: On a $200\\times200$ chessboard, some squares contain a red or blue piece, while the remaining squares are empty. Two pieces are said to \"see\" each other if they are in the same row or the same column. Suppose each piece sees exactly five pieces of the opposite color (and may also see some pieces of the same color). Find the maximum possible total number of pieces on the chessboard.

The maximum possible total number of pieces on the chessboard is 3800.

There exists a valid placement of pieces that satisfies the requirements, yielding a total of $10\times190\times2=3800$ pieces.

Note that when a row contains pieces of only one colour, the row can have at most $200$ pieces. When a row contains pieces of two colours, each colour appears at most $5$ times, so the row can have at most $10$ pieces. If there are $191$ rows that contain two colours, then the total number of pieces does not exceed $9\times200+191\times10=3710<3800$. Suppose at most $190$ rows contain two colours; similarly, we may assume at most $190$ columns contain two colours.

Consider the number $S$ of pairs of pieces of different colours that lie in the same row or the same column. On the one hand, if a row contains only one colour, it contributes no heterochromatic pairs; if a row contains two colours, the number of heterochromatic pairs in that row is at most $5\times5=25$. The same holds for columns. Hence, $S\le 25\times190\times2$. On the other hand, each piece forms heterochromatic pairs with exactly $5$ other pieces, so the total number of pieces is at most $\frac{2S}{5}\le \frac{2 \times (25 \times 190 \times 2)}{5} = 3800$.

(Partial) 1. Observed that any row (or column) containing both colours has at most 10 pieces (at most 5 of each colour), since each piece must see exactly five opposite-colour pieces in its row/column. 2. Introduced the double-counting parameter S = number of opposite-colour pairs in a common row or column, and correctly deduced that the total number of pieces equals 2S/5. 3. Obtained a nontrivial global upper bound by combining a bound on heterochromatic pairs per mixed row/column (e.g. ≤25) with a bound on the number of mixed rows/columns (e.g. ≤200), even if the final numeric maximum is not reached. (Almost) 1. Proved the upper bound 3800 via the S double count, but made a small slip in justifying that one may assume at most 190 mixed rows and at most 190 mixed columns (or handled only rows but not columns), while the rest of the argument is correct. 2. Gave a construction intended to achieve 3800 pieces (with 190 mixed rows and 190 mixed columns, 5 of each colour in each mixed line), but did not fully verify the condition “each piece sees exactly five opposite-colour pieces” for all pieces (e.g. missed checking edge rows/columns or symmetry cases). 3. Solution is almost complete, but made minor mistakes which are not negligible.

Instruction:
Provide an explicit construction of the chessboard that achieves this maximum possible total number of pieces.

Construction requirements:
- Output a single Python expression that evaluates to a 200x200 list of lists of integers representing the chessboard.
- Use 1 for a red piece, 2 for a blue piece, and 0 for an empty square.
- The configuration must satisfy all problem constraints: every piece (1 or 2) must share its row and column with exactly 5 pieces in total of the opposite color.
- The total number of pieces (1s and 2s) must be maximized.
- The output must be a pure Python expression (e.g., using list comprehensions and conditional expressions). Do not include any explanations, comments, variable assignments, or markdown formatting (such as ```python). Put ONLY the raw Python expression inside the construction block.

[[1 if (r < 5 and c >= 10) or (c < 5 and r >= 10) else 2 if (5 <= r < 10 and c >= 10) or (5 <= c < 10 and r >= 10) else 0 for c in range(200)] for r in range(200)]

import sys

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

    try:
        # The instruction allows pure Python expressions including list comprehensions,
        # so we use eval() as specified in the user intent.
        board = eval(payload)
    except Exception as e:
        print(f"Failed to evaluate python expression: {e}")
        return

    if type(board) is not list or len(board) != 200:
        print("Board must be a list of exactly 200 elements.")
        return

    for r, row in enumerate(board):
        if type(row) is not list or len(row) != 200:
            print(f"Row {r} is not a list of exactly 200 elements.")
            return
        for c, val in enumerate(row):
            if type(val) is not int or val not in {0, 1, 2}:
                print(f"Invalid value {repr(val)} at ({r}, {c}). Expected 0, 1, or 2.")
                return

    total_pieces = 0
    row_counts = {1: [0]*200, 2: [0]*200}
    col_counts = {1: [0]*200, 2: [0]*200}

    # Count pieces and precompute the number of 1s and 2s in each row and column
    for r in range(200):
        for c in range(200):
            val = board[r][c]
            if val in {1, 2}:
                total_pieces += 1
                row_counts[val][r] += 1
                col_counts[val][c] += 1

    if total_pieces != 3800:
        print(f"Total pieces is {total_pieces}, expected exactly 3800.")
        return

    # Verify that each piece sees exactly 5 pieces of the opposite color
    for r in range(200):
        for c in range(200):
            val = board[r][c]
            if val in {1, 2}:
                opp = 3 - val
                sees = row_counts[opp][r] + col_counts[opp][c]
                if sees != 5:
                    print(f"Piece at ({r}, {c}) of color {val} sees {sees} opposite color pieces in its row and column combined, expected exactly 5.")
                    return

    print("True")

if __name__ == "__main__":
    verify()


Question:
"On a chessboard, the eight squares on a diagonal are called a fence. A rook starts moving from a square outside the fence. In each move, it goes from one square to another, satisfying: (1) It cannot move onto the fence; (2) Each square is moved to at most once. Find the maximum number of moves that cross the fence.
"

The answer is $47$.

Let the square in row $i$ and column $j$ be denoted by $(i,j)$. The fence consists of the eight squares $(i,i)$ for $1\le i\le 8$. We divide the remaining squares into four classes:
\[
\begin{aligned}
A&=\{(i,j):2\le j+1\le i\le 4\}\cup\{(i,j):5\le j\le i-1\le 7\},\\
B&=\{(i,j):2\le i+1\le j\le 4\}\cup\{(i,j):5\le i\le j-1\le 7\},\\
C&=\{(i,j):5\le i\le 8,\ 1\le j\le 4\},\\
D&=\{(i,j):1\le i\le 4,\ 5\le j\le 8\}.
\end{aligned}
\]
Thus $A$ and $C$ are below the fence, while $B$ and $D$ are above the fence. Also $|A|=|B|=12$, so $|A\cup B|=24$.

A rook move crosses the fence exactly when its two endpoints lie on opposite sides of the main diagonal. With the above coloring, every crossing is of one of the following two types:
\[
A\leftrightarrow D \qquad\text{or}\qquad B\leftrightarrow C.
\]
Indeed, a horizontal or vertical segment crossing the diagonal in the top-left or bottom-right part of the board has its near-diagonal endpoint in $A$ or $B$, and the other endpoint in $D$ or $C$ respectively.

Therefore every crossing has one endpoint in $A\cup B$. Since the rook never visits a square more than once, any fixed square in $A\cup B$ can be an endpoint of at most two crossing moves: one when the rook enters the square and one when it leaves. Hence the total number of crossings is at most
\[
2|A\cup B|=48.
\]

We now rule out equality. Suppose that $48$ crossings were possible. Then every square in $A\cup B$ would have to be used as an endpoint of exactly two crossing moves. In particular, no square in $A\cup B$ could be incident to a non-crossing move in the rook's path.

Consequently, once the path uses crossings of type $A\leftrightarrow D$, it cannot switch to crossings of type $B\leftrightarrow C$ through a non-crossing move involving $A$ or $B$. The only possible switch would have to pass between a square in $C$ and a square in $D$ without using $A$ or $B$. But this is impossible: squares of $C$ have rows $5,6,7,8$ and columns $1,2,3,4$, while squares of $D$ have rows $1,2,3,4$ and columns $5,6,7,8$, so no rook move can go directly from $C$ to $D$.

Thus, under the equality assumption, all crossings would have to be of a single type. If all crossings are of type $A\leftrightarrow D$, then only the $12$ squares of $A$ can serve as the required near-diagonal endpoints, giving at most $2|A|=24$ crossings. Similarly, if all crossings are of type $B\leftrightarrow C$, there are at most $2|B|=24$ crossings. This contradicts the assumption of $48$ crossings. Hence at most $47$ crossings are possible.

It remains to show that $47$ crossings can be achieved. The following sequence of visited squares does so:
\[
\begin{aligned}
&(8,2)\to(1,2)\to(7,2)\to(7,8)\to(7,3)\to(1,3)\to(8,3)\to(2,3)\\
&\to(6,3)\to(6,8)\to(6,2)\to(6,7)\to(6,4)\to(1,4)\to(8,4)\to(2,4)\\
&\to(7,4)\to(3,4)\to(5,4)\to(5,8)\to(5,3)\to(5,7)\to(5,2)\to(5,6)\\
&\to(8,6)\to(1,6)\to(7,6)\to(2,6)\to(2,1)\to(2,7)\to(8,7)\to(3,7)\\
&\to(3,1)\to(3,6)\to(3,2)\to(3,5)\to(8,5)\to(2,5)\to(7,5)\to(1,5)\\
&\to(6,5)\to(4,5)\to(4,1)\to(4,6)\to(4,2)\to(4,7)\to(4,3)\to(4,8).
\end{aligned}
\]
Each listed square is off the fence, no square is repeated, and each consecutive pair lies in a common row or column. Moreover, for every consecutive pair, the two endpoints lie on opposite sides of the main diagonal, so every one of the $47$ moves crosses the fence. Therefore the maximum number of crossings is exactly $47$.

(Partial) 1. Observed/proved that a rook move crosses the fence iff its endpoints lie on opposite sides of the main diagonal (equivalently, one has i<j and the other i>j), since the rook moves horizontally/vertically and cannot land on the diagonal. 2. Established a global upper bound of the form “#crossings ≤ 48” by identifying a set of 24 off-diagonal squares near the fence (e.g. A∪B in the official partition) such that every crossing has an endpoint in this set, and arguing each such square can be an endpoint of at most two crossing moves (enter/leave) because squares are not revisited. 3. Gave an explicit long non-repeating rook path avoiding the fence that realizes at least 40 crossing moves (or, more generally, a concrete construction with a stated number of crossings and a verifiable check that every move crosses). (Almost) 1. Proved the upper bound 48 using the “each crossing uses A∪B, hence ≤2|A∪B|” endpoint-counting argument, and attempted to rule out equality by a switching/connectedness argument, but left a small gap in justifying why a 48-crossing path cannot mix the two crossing types (A↔D and B↔C). 2. Correctly proved that 48 is impossible (hence the maximum is ≤47), but did not supply a complete explicit path achieving 47, or gave a construction that is off by one move / has one repeated square / has one move that does not cross. 3. Presented a construction intended to achieve 47 crossings and verified legality (no fence, no repeats, rook moves), but did not fully check that every consecutive pair lies on opposite sides of the diagonal (or missed a boundary case where a move would cross through a fence square without landing on it).

Instruction:
Provide an explicit construction for the sequence of squares visited by the rook to achieve the maximum possible number of moves that cross the fence.

Construction format requirements:
- Output a single Python list of lists, where each inner list represents a square as `[r, c]`.
- Use 0-based indexing for the 8x8 chessboard, so `0 <= r <= 7` and `0 <= c <= 7`.
- The fence is defined as the main diagonal where `r == c`.
- The sequence must start with the initial square and contain all subsequently visited squares in order.
- The rook cannot move onto the fence (i.e., no square in the sequence can have `r == c`).
- Each square can be visited at most once.
- Every move must be a valid rook move (either `r` or `c` remains constant, and the other strictly changes).
- To maximize efficiency in your output, every move in your sequence must strictly cross the fence. A move from `[r1, c1]` to `[r2, c2]` crosses the fence if `(r1 - c1) * (r2 - c2) < 0`.
- The sequence must achieve the maximum possible number of fence-crossing moves for this board.
- Put ONLY the Python literal list of lists inside the `<construct>` block.

import sys
import ast

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

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

    if not isinstance(points, list):
        print("The construction must be a list.")
        return

    if len(points) != 48:
        print(f"Expected exactly 48 points, got {len(points)}.")
        return

    seen = set()
    for i, pt in enumerate(points):
        if not isinstance(pt, (list, tuple)):
            print(f"Point at index {i} is not a list or tuple.")
            return
        if len(pt) != 2:
            print(f"Point at index {i} does not have exactly 2 elements.")
            return

        r, c = pt
        if type(r) is not int or type(c) is not int:
            print(f"Coordinates of point at index {i} are not strictly integers.")
            return

        if not (0 <= r <= 7 and 0 <= c <= 7):
            print(f"Coordinates of point {pt} at index {i} are out of bounds (0-7).")
            return

        if r == c:
            print(f"Point {pt} at index {i} lies on the fence (r == c).")
            return

        tup = (r, c)
        if tup in seen:
            print(f"Point {pt} at index {i} is visited more than once.")
            return
        seen.add(tup)

    for i in range(len(points) - 1):
        r1, c1 = points[i]
        r2, c2 = points[i+1]

        # Check valid rook move
        if not ((r1 == r2 and c1 != c2) or (c1 == c2 and r1 != r2)):
            print(f"Move from {points[i]} to {points[i+1]} is not a valid rook move.")
            return

        # Check fence crossing
        if (r1 - c1) * (r2 - c2) >= 0:
            print(f"Move from {points[i]} to {points[i+1]} does not strictly cross the fence.")
            return

    print("True")

if __name__ == "__main__":
    verify()


Question: On a $10\\times10$ chessboard there are $k$ rooks. A square that can be attacked by at least one rook is called dangerous (including the square where a rook is located). If after removing any one rook, at least one dangerous square becomes not dangerous, determine the maximum possible value of $k$.

First, we prove that $k \le 16$.
Suppose that every row contains at least one rook. Then all squares on the board are dangerous. If some row contains at least two rooks, removing one of them leaves every row with at least one rook, meaning all squares are still dangerous, which contradicts the problem condition. Hence, each row must contain exactly one rook, yielding $k=10$. By symmetry, if every column contains at least one rook, we also have $k=10$.

Now assume there is at least one row and at least one column without a rook.
For each rook, it must be the unique rook in its row or the unique rook in its column; otherwise, removing it would leave at least one rook in its row and at least one rook in its column, meaning no dangerous square would become safe.
We associate each rook with a row or column in which it is the unique rook.
Let $R$ be the set of rows containing exactly one rook, and $C$ be the set of columns containing exactly one rook. Since every rook is associated with a row in $R$ or a column in $C$, we have $k \le |R| + |C|$.

Since there is at least one empty row, $|R| \le 9$. If $|R| = 9$, then exactly 9 rows contain exactly one rook, and the remaining row is empty. This implies the total number of rooks on the board is exactly 9, so $k=9$. By symmetry, if $|C| = 9$, then $k=9$.
Otherwise, we must have $|R| \le 8$ and $|C| \le 8$. It follows that the total number of rooks is bounded by $k \le |R| + |C| \le 8 + 8 = 16$.

Finally, we construct an example with $k=16$. Let the squares occupied by the 16 rooks be $(1,i)$ and $(i,1)$ for $i=2,3,\dots,9$.
In this configuration, after removing the rook from square $(1,i)$, the square $(10,i)$ changes from dangerous to safe (since row 10 is empty and the only rook in column $i$ was at $(1,i)$).
Similarly, after removing the rook from square $(i,1)$, the square $(i,10)$ changes from dangerous to safe.
This configuration satisfies the problem's requirement. Therefore, the maximum possible value of $k$ is $16$.

(Partial) 1. Observed the key necessity: for every rook, it must be the unique rook in its row or the unique rook in its column (otherwise removing it leaves its whole row and column still dangerous, so nothing changes). 2. Derived an upper bound of the form k ≤ |R|+|C| where R is the set of rows with exactly one rook and C is the set of columns with exactly one rook, by assigning each rook to one such unique row/column. 3. Gave a concrete configuration with 16 rooks (e.g. on (1,i) and (i,1) for i=2,…,9) and correctly argued that removing any rook makes at least one previously dangerous square safe. (Almost) 1. Proved the upper bound k ≤ 16 via the |R|+|C| argument, but did not handle the special subcase “all rows occupied” or “all columns occupied” correctly (or omitted it), leaving a small logical gap. 2. Found the correct extremal configuration with k=16 and checked the condition for most rooks, but missed/incorrectly justified one boundary square becoming safe after deletion (e.g. the need for an empty row/column in the witness square). 3. Solution is almost complete, but made minor mistakes which are not negligible.

Instruction:
Provide an explicit construction for the maximum possible value of $k$.

Construction requirements:
- Output a single Python expression (e.g., a list comprehension) that evaluates to a 10x10 matrix (a list of 10 lists, each containing 10 integers).
- The matrix should contain only 0s and 1s, where 1 represents a rook and 0 represents an empty square.
- The configuration must satisfy the problem constraint: for every rook, there must be at least one square (which could be its own square) that is attacked ONLY by this rook.
- The number of rooks $k$ (i.e., the number of 1s in the matrix) must be exactly the maximum possible value.
- Put ONLY the Python expression inside the construction block, without any additional text, markdown, or code fences.

[[1 if (r == 1 and c >= 2) or (c == 1 and r >= 2) else 0 for c in range(10)] for r in range(10)]

import sys

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

    try:
        # Evaluate the payload as a Python expression (e.g., list comprehension)
        matrix = eval(payload)
    except Exception as e:
        print(f"Failed to evaluate python expression: {e}")
        return

    if type(matrix) is not list:
        print("Matrix must be a list.")
        return

    if len(matrix) != 10:
        print(f"Matrix must have exactly 10 rows, got {len(matrix)}.")
        return

    for i, row in enumerate(matrix):
        if type(row) is not list:
            print(f"Row {i} is not a list.")
            return
        if len(row) != 10:
            print(f"Row {i} must have exactly 10 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 0 or 1.")
                return

    rooks = []
    for r in range(10):
        for c in range(10):
            if matrix[r][c] == 1:
                rooks.append((r, c))

    if len(rooks) != 16:
        print(f"Expected exactly 16 rooks, got {len(rooks)}.")
        return

    # Count the number of rooks in each row and column
    row_counts = [0] * 10
    col_counts = [0] * 10
    for r, c in rooks:
        row_counts[r] += 1
        col_counts[c] += 1

    # Calculate how many rooks attack each square (r, c)
    attack_counts = [[0]*10 for _ in range(10)]
    for r in range(10):
        for c in range(10):
            # A rook at (r, c) would be counted twice (once in row_counts, once in col_counts)
            attack_counts[r][c] = row_counts[r] + col_counts[c] - matrix[r][c]

    # Verify that every rook has at least one uniquely attacked square
    for rr, cc in rooks:
        has_unique = False

        # Check all squares in the same column
        for r in range(10):
            if attack_counts[r][cc] == 1:
                has_unique = True
                break

        if has_unique:
            continue

        # Check all squares in the same row
        for c in range(10):
            if attack_counts[rr][c] == 1:
                has_unique = True
                break

        if not has_unique:
            print(f"Rook at ({rr}, {cc}) does not have any uniquely attacked square.")
            return

    print("True")

if __name__ == "__main__":
    verify()


Question:
"A $100\\times100$ grid is colored using $101$ colors. A cell is called \""colorful\"" if in its row and column (a total of $199$ cells), each of the $101$ colors appears at least once. Find the maximum possible number of colorful cells.
"

We claim that the maximum possible number of colorful cells is $9996$.

We first provide a construction achieving $9996$ colorful cells. For $1\le i\le 101$, filling a cell with $i$ means coloring that cell with the $i$-th color. Consider the following coloring of the $100 \times 100$ grid:
$$
\begin{array}{|c|c|c|c|c|c|}
\hline
4 & 5 & \cdots & 101 & 3 & 2 \\
\hline
5 & 6 & \cdots & 4   & 3 & 2 \\
\hline
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\
\hline
101 & 4 & \cdots & 100 & 3 & 2 \\
\hline
3 & 3 & \cdots & 3   & 1 & 2 \\
\hline
1 & 1 & \cdots & 1   & 2 & 1 \\
\hline
\end{array}
$$
In this table, the upper-left $98\times 98$ square is colored cyclically with colors $4, 5, \dots, 101$. It is easy to verify that, except for the $4$ cells in the lower-right $2\times 2$ square (which only see colors $1, 2,$ and $3$ in their rows and columns), the remaining $9996$ cells are colorful. Thus, $9996$ colorful cells are attainable.

Now we prove that $9996$ is the maximum. We consider two cases based on the frequencies of the colors used.

**Case 1:** A certain color appears at most $98$ times.
If a color appears at most $98$ times, it can occupy at most $98$ rows and $98$ columns. Thus, there must exist at least two rows and two columns that contain no cell of that color. The four cells at the intersection of these rows and columns are then not colorful, because neither their row nor their column contains this color. Therefore, the number of colorful cells is at most $100^2 - 4 = 9996$.

**Case 2:** Each color appears at least $99$ times.
Because there are $101$ colors and $100^2 = 101 \times 99 + 1$ cells, there must be exactly $100$ colors that appear exactly $99$ times, and one color that appears exactly $100$ times. Denote the color appearing $100$ times by color $101$. We will prove that in this case, the number of colorful cells does not exceed $9900$.

Suppose for the sake of contradiction that at most $99$ cells are not colorful. Since there are $100$ rows and $100$ columns, there must exist at least one row and at least one column that consist entirely of colorful cells. By permuting the rows and columns, we may assume that row $100$ and column $100$ consist entirely of colorful cells, so that all non-colorful cells lie inside the $99 \times 99$ square in the upper-left corner.

For each $1 \le i \le 100$, color $i$ appears exactly $99$ times in the grid. Thus, there must be at least one column that contains no cell of color $i$. Since every cell in row $100$ is colorful, the cell in row $100$ of this column must still see color $i$ in either its row or its column. Because its column lacks color $i$, row $100$ must contain color $i$. This holds for all $i \in \{1, 2, \dots, 100\}$. Consequently, row $100$ contains all $100$ colors from $1$ to $100$. Since row $100$ has exactly $100$ cells, it cannot contain any cell of color $101$.

By symmetric reasoning, for each $1 \le i \le 100$, there must be a row that contains no cell of color $i$. Since every cell in column $100$ is colorful, column $100$ must contain color $i$. Thus, column $100$ also contains all colors from $1$ to $100$, and consequently contains no cell of color $101$.

Finally, consider the cell at the intersection of row $100$ and column $100$. Neither its row nor its column contains color $101$. Thus, this cell is not colorful, which contradicts our assumption that row $100$ and column $100$ consist entirely of colorful cells.

Therefore, our assumption was false, and there must be at least $100$ non-colorful cells. Hence, the number of colorful cells is at most $10000 - 100 = 9900 \le 9996$.

In all cases, the maximum possible number of colorful cells is $9996$.

(Partial) 1. Gave a construction with at least 9996 colorful cells (e.g. a cyclic 98×98 block using 98 colors plus a completion), with a correct explanation that only 4 specified cells fail to be colorful. 2. Proved the key upper-bound lemma: if some color appears at most 98 times, then there exist two rows and two columns missing this color, hence at least 4 cells are not colorful and the number of colorful cells is at most 9996. 3. In the complementary frequency case, correctly derived the count constraint: if every color appears at least 99 times, then exactly one color appears 100 times and the other 100 colors appear 99 times each. (Almost) 1. Proved the upper bound 9996 in Case 1 (a color used ≤98 times) and correctly set up Case 2 (all colors used ≥99 times), but did not complete the Case 2 contradiction (e.g. failed to force a row/column with all colors 1–100 and hence no 101). 2. Completed both cases to obtain the bound 9996, but the proof that “if fewer than 100 non-colorful cells then there is a fully colorful row and a fully colorful column” is not fully justified or mishandles a boundary case. 3. Gave a correct construction achieving 9996 and the correct overall strategy for the upper bound, but left a localized gap (such as not rigorously arguing why row 100/column 100 cannot contain color 101, or not justifying the needed row/column permutations).

Instruction:
Construct a $100 \times 100$ grid of integers representing the colors.

Construction requirements:
- The grid must be a list of lists of integers.
- Use colors $1, 2, \dots, 101$.
- The grid must find the maximum possible number of colorful cells.  A cell $(r, c)$ is colorful if the union of colors in row $r$ and column $c$ contains all 101 colors.
- Provide the answer as a single valid Python expression (using list comprehensions) that evaluates to the matrix. Do not list all 10,000 numbers explicitly; use logic to generate them.
- Put ONLY the Python expression inside the construction block.Output formatting requirements:
- Include exactly one construction block and put ONLY the JSON object between the tags:
<construct>
<json_object>
</construct>
- Do not use code fences. Do not put any extra text inside <construct>...</construct>.

[
    [
        ((r + c) % 98 + 4) if c < 98 else (3 if c == 98 else 2)
        for c in range(100)
    ]
    if r < 98 else
    (
        [3 if c < 98 else (1 if c == 98 else 2) for c in range(100)]
        if r == 98 else
        [1 if c < 98 else (2 if c == 98 else 1) for c in range(100)]
    )
    for r in range(100)
]

import sys

def verify():
    # Read the raw content from stdin.
    # The instruction guarantees the input is a Python expression.
    content = sys.stdin.read().strip()

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

    # Evaluate the string as a Python expression.
    # We use eval() because the instruction allows list comprehensions,
    # which ast.literal_eval() does not support.
    try:
        # We allow standard builtins (e.g., range, list) as they are needed for comprehensions.
        grid = eval(content, {"__builtins__": __builtins__})
    except SyntaxError:
        print("Error: Input is not a valid Python expression.")
        return
    except Exception as e:
        print(f"Error: Failed to evaluate expression. {e}")
        return

    # Validate outer structure (List of Lists)
    if not isinstance(grid, list):
        print("Error: The evaluated object is not a list.")
        return

    if len(grid) != 100:
        print(f"Error: Grid must have exactly 100 rows, found {len(grid)}.")
        return

    # Validate inner structure and values, and precompute row sets
    row_sets = []

    for r_idx, row in enumerate(grid):
        if not isinstance(row, list):
            print(f"Error: Row {r_idx} is not a list.")
            return
        if len(row) != 100:
            print(f"Error: Row {r_idx} must have exactly 100 columns, found {len(row)}.")
            return

        current_row_set = set()
        for c_idx, val in enumerate(row):
            if not isinstance(val, int):
                print(f"Error: Cell ({r_idx}, {c_idx}) is not an integer.")
                return
            if not (1 <= val <= 101):
                print(f"Error: Cell ({r_idx}, {c_idx}) value {val} is out of valid range [1, 101].")
                return
            current_row_set.add(val)
        row_sets.append(current_row_set)

    # Precompute column sets
    col_sets = []
    for c_idx in range(100):
        current_col_set = set()
        for r_idx in range(100):
            current_col_set.add(grid[r_idx][c_idx])
        col_sets.append(current_col_set)

    # Count "colorful" cells
    # A cell (r, c) is colorful if the union of colors in row r and column c contains all 101 colors.
    colorful_count = 0
    target_colors = 101

    for r in range(100):
        r_set = row_sets[r]
        for c in range(100):
            c_set = col_sets[c]
            # Check if union covers all 101 colors
            if len(r_set | c_set) == target_colors:
                colorful_count += 1

    # Verify the count against the requirement
    if colorful_count >= 9996:
        print("True")
    else:
        print(f"Error: Expected at least 9996 colorful cells, but found {colorful_count}.")

if __name__ == "__main__":
    verify()


Question: A $2^{2014} + 1$ by $2^{2014} + 1$ grid has some black squares filled. The filled black squares form one or more snakes on the plane, each of whose heads splits at some points but never comes back together. In other words, for every positive integer $n$ greater than $2$, there do not exist pairwise distinct black squares $s_1, s_2, \dots, s_n$ such that $s_i$ and $s_{i+1}$ share an edge for $i = 1, 2, \dots, n$ (here $s_{n+1} = s_1$).What is the maximum possible number of filled black squares?

Let $n = 2^m + 1$ be the dimension of the grid (for the given problem, $m = 2014$). Let $v$ be the number of filled black squares, and $e$ be the number of shared edges between black squares.

Since the graph formed by the black squares contains no cycles, it is a forest, which implies $e \le v - 1$.
Double-counting the edges of the black squares yields $3v + 1 \le 4v - e$.
The total number of edges in the $n \times n$ grid is $2n(n+1)$. Since $4v - e$ is the number of grid edges covered by at least one black square, we have $4v - e \le 2n(n+1)$.
Thus, $3v + 1 \le 2n(n+1)$.

Because $n = 2^m + 1$, for any $m \ge 1$ we have $n \not\equiv 1 \pmod 3$, which means $n(n+1)$ is a multiple of $3$.
Therefore, $2n(n+1)$ is a multiple of $3$, and $3v + 1 \le 2n(n+1)$ implies $3v \le 2n(n+1) - 3$, yielding
$$v \le \frac{2n(n+1)}{3} - 1.$$
Observe that for $n \not\equiv 1 \pmod 3$, this equality is achieved if and only if:
(a) the graph formed by the black squares is a connected forest (i.e., a tree), and
(b) all but two square edges belong to at least one black square.

We prove by induction on $m \ge 1$ that equality can in fact be achieved.
For $m = 1$, take an “H-shape” on a $3 \times 3$ grid (so if we set the center at $(0,0)$ in the coordinate plane, everything but $(0,\pm 1)$ is black); call this configuration $G_1$. It is easy to verify that $G_1$ satisfies both conditions.

To go from $G_m$ to $G_{m+1}$, assume $G_m$ is defined on $[-2^{m-1}, 2^{m-1}]^2$. We define $G_{m+1}$ on $[-2^m, 2^m]^2$ as follows:
fill in $(2x,2y)$ in $G_{m+1}$ if and only if $(x,y)$ is filled in $G_m$, and fill in $(x,y)$ with $x,y$ not both even if and only if $x + y$ is odd (so exactly one of $x,y$ is odd and the other is even).

Each “newly-created” white square has both coordinates odd. Its four neighbors each have one even and one odd coordinate, so their coordinate sum is odd, meaning they are newly-created black squares. Thus, all four edges of any newly-created white square are covered by black squares.
The boundary of $G_{m+1}$ consists of squares with at least one coordinate equal to $\pm 2^m$ (which is even), so no newly-created white square lies on the boundary.
Furthermore, no two white squares can share an edge in $G_{m+1}$, because adjacent squares have different coordinate sum parities, and no square with an odd coordinate sum is white.
Therefore, the only uncovered edges in $G_{m+1}$ are the boundary edges of the “old” white squares that lie on the boundary of the grid. By the inductive hypothesis, there are exactly two such uncovered edges (originating from the two white squares $(0, \pm 1)$ in $G_1$). Thus $G_{m+1}$ satisfies (b).

To check that (a) holds, first we show that $(2x_1,2y_1)$ and $(2x_2,2y_2)$ are connected in $G_{m+1}$ if and only if $(x_1,y_1)$ and $(x_2,y_2)$ are black squares (and thus connected) in $G_m$ (the new black squares are essentially just “bridges”).
Indeed, every path of black squares in $G_{m+1}$ must alternate between squares with odd and even coordinate sums, which corresponds to alternating between new and old black squares. Two old black squares $(2x_1,2y_1)$ and $(2x_2,2y_2)$ are adjacent to the same new black square $(x_1 + x_2, y_1 + y_2)$ in $G_{m+1}$ if and only if $(x_1,y_1)$ and $(x_2,y_2)$ are adjacent in $G_m$, whence the claim readily follows.

The rest is clear: the set of old black squares must remain connected in $G_{m+1}$, and all new black squares (including those on the boundary) border at least one old black square (otherwise, its two even-sum neighbors would both be white, meaning two white squares would be adjacent in $G_m$, which violates condition (b) for $G_m$). Thus, $G_{m+1}$ is fully connected.

On the other hand, $G_{m+1}$ cannot have any cycles. Since each new black square has at most two old black neighbors (its other two neighbors have both coordinates odd and are thus white), any cycle in $G_{m+1}$ would alternate between new and old black squares. Removing the new black squares from such a cycle would yield a corresponding cycle of old black squares in $G_m$, which contradicts the inductive hypothesis.

Thus, the maximum possible number of filled black squares is achieved. For the given problem where $n = 2^{2014} + 1$, the maximum number of filled black squares is $\frac{2n(n+1)}{3} - 1$.

(Partial) 1. Interpreted the filled squares as a graph (vertices = black squares, edges = shared sides) and observed that the “no cycle” condition implies the graph is a forest, hence e ≤ v−1. 2. Double-counted square-sides to obtain an inequality of the form 3v+1 ≤ 4v−e (or an equivalent bound relating v and e via boundary/perimeter counting). 3. Combined the side-count with the grid-edge bound 4v−e ≤ 2n(n+1) to deduce 3v+1 ≤ 2n(n+1), hence v ≤ 2n(n+1)/3 − 1 after the divisibility observation for n=2^m+1. (Almost) 1. Derived the sharp upper bound v ≤ 2n(n+1)/3 − 1 and correctly identified the equality conditions (connected forest and exactly two uncovered grid-edges), but did not provide a full construction achieving equality. 2. Gave the recursive/inductive construction (from a 3×3 base pattern to size 2^m+1) intended to achieve equality, but left a localized gap in verifying either (i) no cycles are created, or (ii) only two edges remain uncovered (e.g. boundary case). 3. Provided a complete construction and matching-count argument for general n=2^m+1 but made a small arithmetic/parity slip that affects only the final −1 or a boundary case, while the main method is correct.

Instruction:
Provide a concrete construction for the 17x17 grid (the $N = 2^4+1$ case) that achieves the maximum possible number of filled black squares without forming any cycles.

Your output must be a single, valid Python expression. When evaluated, this expression must return a 17x17 matrix (a list of lists) of strings, where "1" represents a filled black square and "0" represents an empty square. The orthogonally adjacent "1"s must form a forest (zero cycles), and the total count of "1"s must exactly equal the theoretical maximum for this 17x17 instance.

Do not state the specific numerical maximum. Do not use hardcoded full-grid enumeration; use compact Python constructs like list comprehensions and conditional expressions. Do not include any imports, assignments, function definitions, markdown formatting, or explanatory text. Put ONLY the Python expression inside the `<construct>` block.

import sys

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

    try:
        grid = eval(payload, {"__builtins__": {}, "range": range}, {})
    except Exception as e:
        print(f"Parse error: {e}")
        return

    if not isinstance(grid, list) or len(grid) != 17:
        print("Grid must be a list of exactly 17 elements.")
        return

    for row in grid:
        if not isinstance(row, list) or len(row) != 17:
            print("Each row must be a list of exactly 17 elements.")
            return
        for cell in row:
            if not isinstance(cell, str) or cell not in ["0", "1"]:
                print("Each cell must be the string '0' or '1'.")
                return

    count_ones = sum(cell == "1" for row in grid for cell in row)
    if count_ones != 203:
        print(f"Total number of '1's must be exactly 203, but got {count_ones}.")
        return

    visited = set()
    for i in range(17):
        for j in range(17):
            if grid[i][j] == "1" and (i, j) not in visited:
                stack = [(i, j, -1, -1)]
                while stack:
                    curr_r, curr_c, parent_r, parent_c = stack.pop()
                    if (curr_r, curr_c) in visited:
                        print("Cycle detected among the '1's.")
                        return
                    visited.add((curr_r, curr_c))

                    for dr, dc in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
                        nr, nc = curr_r + dr, curr_c + dc
                        if 0 <= nr < 17 and 0 <= nc < 17 and grid[nr][nc] == "1":
                            if nr == parent_r and nc == parent_c:
                                continue
                            stack.append((nr, nc, curr_r, curr_c))

    print("True")

if __name__ == "__main__":
    verify()


Question:
A football match lasts $90$ minutes. A team always has $11$ players on the field at the same time. The coach wants each player's playing time to be the same integer number of minutes.

(1) If each player's playing time is less than $60$ minutes, what is the minimum number of players on the team?

(2) Under the condition of the minimum number of players in (1), what is the minimum number of substitutions required?

Note: Substitutions can only occur at integer minutes, the time for substitution is negligible, and substituted players can return to the field.

**Final Answers:**
(1) $18$ players
(2) $17$ substitutions

**Part (1)**
Suppose there are $n$ players on the team, and each player's playing time is exactly $k$ minutes. Both $n$ and $k$ must be positive integers. Since there are always $11$ players on the field during the $90$-minute match, the total playing time for all players combined is $11 \times 90 = 990$ minutes. Therefore, we have the equation:
$nk = 990$

We are given that each player's playing time is strictly less than $60$ minutes, so $k < 60$. Since $k$ must be an integer that divides $990$, the maximum possible value for $k$ is $55$.
Using $k \le 55$, we find:
$n = \frac{990}{k} \ge \frac{990}{55} = 18$
Thus, the team must have a minimum of $18$ players.

**Part (2)**
Assume the team has the minimum of $18$ players, meaning each player plays exactly $55$ minutes.

We model the substitution process as a graph. Let each of the $18$ players be represented by a vertex. An edge is drawn between two vertices if one of the corresponding players directly substitutes the other.

Suppose, for the sake of contradiction, that this graph is disconnected. Let the smallest connected component of the graph contain $l$ vertices. Since the graph is disconnected and has $18$ vertices in total, the smallest component can have at most half the vertices, meaning $l \le \frac{18}{2} = 9$.

Let these $l$ vertices correspond to a specific group of players. Suppose that at any given time, there are $m$ players from this group on the field. Because substitutions within this component only happen among these $l$ players, the number of players $m$ from this group on the field must remain constant throughout the entire $90$-minute match.

The total playing time contributed by these $l$ players is $55l$ minutes. Since exactly $m$ players from this group are on the field at all times during the $90$-minute match, their total playing time can also be expressed as $90m$ minutes. Equating the two gives:
$55l = 90m$
$11l = 18m$

Since $11$ and $18$ are coprime, $18$ must divide $l$. However, this implies $l \ge 18$, which directly contradicts our earlier deduction that $l \le 9$.

Therefore, our assumption that the graph is disconnected must be false. The graph must be connected. A connected graph with $18$ vertices must have at least $18 - 1 = 17$ edges. Since each edge represents a substitution, there must be at least $17$ substitutions.

This theoretical minimum of $17$ substitutions can indeed be achieved in practice. It is possible to arrange the $11$ players on the field at each time such that exactly $17$ substitutions are made, with one substitution occurring every $5$ minutes.

(Partial) 1. Set up the counting equation total player-minutes = 11·90 = 990, hence if there are n players each playing k minutes then nk = 990 with n,k integers. 2. Used the constraint k < 60 to deduce a lower bound on n via a divisor bound for k (e.g. noting k ≤ 55 since 990 has no divisor in {56,57,58,59}), giving n ≥ 18. 3. For part (2), introduced a substitution graph (vertices = players, edge when one directly substitutes the other) and correctly argued that if the graph had a component of size l with m of its players always on the field, then 55l = 90m (or equivalent invariant), leading to a divisibility restriction on l. (Almost) 1. Correctly proved n = 18 is minimal in (1), and for (2) proved the substitution graph must be connected and hence at least 17 substitutions are necessary, but did not provide (or gave an incomplete) explicit schedule achieving 17 substitutions. 2. Gave a construction intending to realize 17 substitutions (e.g. substitutions every 5 minutes) and checked most player-times equal 55, but left a small number of players’ minutes (or a boundary time at start/end) unchecked/miscounted. 3. Solution is almost complete, but made minor mistakes which are not negligible.

Instruction:
Provide an explicit 90-minute playing schedule for the minimum number of players $N$ found in part (1), such that each player's total playing time is the same integer and the total number of substitutions is minimized as in part (2).

Output requirements:
- The construction must be a single Python expression. Evaluating this expression must directly return the required schedule list.
- The value of the expression must be exactly one Python list of entries `[start, end, lineup]`.
- `start` and `end` are the start and end times (in minutes) of a period where the lineup remains constant.
- `lineup` is a sorted list of the 11 players on the field during that period.
- Players must be numbered $1, 2, \dots, N$.
- The periods must be contiguous and cover the full 90 minutes.
- All players must have the same integer total playing time, and this time must be less than 60 minutes.
- A substitution is defined as a player entering the field who was not in the previous period's lineup. The total number of substitutions is the sum of these entrance counts across all transitions. This total must be minimized.
- Do not use explanations, comments, assignments, imports, function definitions, or any non-expression content.
- Put ONLY the Python expression inside the `<construct>` block.

import sys

def verify():
    try:
        raw_input = sys.stdin.read().strip()
        if not raw_input:
            print("No input provided.")
            return
        safe_globals = {"__builtins__": {}, "range": range, "sorted": sorted}
        data = eval(raw_input, safe_globals, {})
    except Exception as e:
        print(f"Failed to evaluate construction: {e}")
        return

    if not isinstance(data, list):
        print("Construction must evaluate to a list.")
        return

    current_time = 0
    player_times = {}
    total_subs = 0
    previous_lineup = None

    for i, entry in enumerate(data):
        if not (isinstance(entry, list) and len(entry) == 3):
            print(f"Entry {i} must be a list of [start, end, lineup].")
            return

        start, end, lineup = entry

        if not (isinstance(start, int) and isinstance(end, int)):
            print(f"Start and end times must be integers in entry {i}.")
            return

        if start != current_time:
            print(f"Gap or overlap in time at entry {i}: expected start {current_time}, got {start}.")
            return

        if end <= start:
            print(f"End time must be greater than start time in entry {i}.")
            return

        if not isinstance(lineup, list) or len(lineup) != 11:
            print(f"Lineup must be a list of 11 players in entry {i}.")
            return

        if lineup != sorted(list(set(lineup))):
            print(f"Lineup must be sorted and contain 11 distinct player IDs in entry {i}.")
            return

        for p in lineup:
            if not isinstance(p, int) or p <= 0:
                print(f"Player IDs must be positive integers in entry {i}.")
                return
            player_times[p] = player_times.get(p, 0) + (end - start)

        if previous_lineup is not None:
            subs = len(set(lineup) - set(previous_lineup))
            total_subs += subs

        previous_lineup = lineup
        current_time = end

    if current_time != 90:
        print(f"Schedule must cover exactly 90 minutes, but ends at {current_time}.")
        return

    if not player_times:
        print("No players found in the schedule.")
        return

    times = set(player_times.values())
    if len(times) != 1:
        print(f"Not all players have the same playing time. Playing times: {player_times}")
        return

    T = list(times)[0]
    if T >= 60:
        print(f"Total playing time {T} must be less than 60 minutes.")
        return

    N = len(player_times)
    if N != 18:
        print(f"Number of players {N} is not the minimum required (18).")
        return

    if set(player_times.keys()) != set(range(1, 19)):
        print(f"Player IDs must be 1, 2, ..., 18. Found IDs: {sorted(player_times.keys())}")
        return

    if total_subs != 17:
        print(f"Total substitutions {total_subs} is not the minimum required (17).")
        return

    print("True")

if __name__ == "__main__":
    verify()