Question: Consider a $2025 \\times 2025$ grid of unit squares. Matilda wishes to place on the grid some rectangular tiles, possibly of different sizes, such that each side of every tile lies on a grid line and every unit square is covered by at most one tile. Determine the minimum number of tiles Matilda needs to place so that each row and each column of the grid has exactly one unit square that is not covered by any tile.

Instruction:
For the specific case where the grid size is $529 \times 529$ (corresponding to $G=23$), provide an explicit construction of the tiles.

Construction requirements:
- Place the minimum possible number of rectangles such that each row and each column of the grid has exactly one unit square not covered by any tile.
- Output a single Python expression that evaluates to a list of rectangles.
- Each rectangle must be represented as a list of four integers `[r1, r2, c1, c2]`, covering all cells `(r, c)` such that `r1 <= r <= r2` and `c1 <= c <= c2`.
- Use 0-based indexing for coordinates, so rows and columns range from 0 to 528.
- The rectangles must not overlap, and the union of all rectangles must leave exactly one cell uncovered in each of the 529 rows and 529 columns.
- The output must be a single, valid Python expression (e.g., using list comprehensions and list concatenation) that can be directly evaluated by `eval()`.
- Do NOT include any explanations, comments, markdown formatting (like ```python), assignments, imports, or any non-expression content.
- Put ONLY the raw Python expression inside the construction block.

import sys

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

    try:
        # The instruction explicitly requires a Python expression that can be evaluated by eval(),
        # such as list comprehensions and list concatenations.
        rects = eval(payload)
    except Exception as e:
        print(f"Failed to parse payload: {e}")
        return

    if not isinstance(rects, list):
        print("Parsed object is not a list")
        return

    if len(rects) != 572:
        print(f"Expected exactly 572 rectangles, got {len(rects)}")
        return

    grid = [[False] * 529 for _ in range(529)]

    for idx, rect in enumerate(rects):
        if not (isinstance(rect, (list, tuple)) and len(rect) == 4):
            print(f"Rectangle at index {idx} is not a list or tuple of 4 elements")
            return

        r1, r2, c1, c2 = rect
        if not (type(r1) is int and type(r2) is int and type(c1) is int and type(c2) is int):
            print(f"Rectangle at index {idx} contains non-integer coordinates")
            return

        if not (0 <= r1 <= r2 < 529 and 0 <= c1 <= c2 < 529):
            print(f"Rectangle at index {idx} has out-of-bounds or invalid coordinates: {rect}")
            return

        for r in range(r1, r2 + 1):
            for c in range(c1, c2 + 1):
                if grid[r][c]:
                    print(f"Rectangles overlap at cell ({r}, {c})")
                    return
                grid[r][c] = True

    for r in range(529):
        uncovered_count = grid[r].count(False)
        if uncovered_count != 1:
            print(f"Row {r} has {uncovered_count} uncovered cells, expected exactly 1")
            return

    for c in range(529):
        uncovered_count = sum(not grid[r][c] for r in range(529))
        if uncovered_count != 1:
            print(f"Column {c} has {uncovered_count} uncovered cells, expected exactly 1")
            return

    print("True")

if __name__ == "__main__":
    verify()

(Partial) 1. Gave a valid construction using 32 tiles (or the general pattern for n=k^2) and correctly verified that it leaves exactly one uncovered square in each row and each column. 2. Observed that the uncovered squares form a permutation of {1,…,25} (one per row and per column), and introduced an LIS of length a and an LDS of length b with the correct inequality ab≥25 (or max(a,b)≥5). 3. Set up a meaningful lower-bound framework by marking letters N/E/S/W adjacent to uncovered squares using the LIS/LDS partition, and stated (with some justification) that no tile can contain two marked letters. (Almost) 1. Proved the lower bound ≥32 via the LIS/LDS letter-counting method, but made a localized counting slip (e.g. miscounted by 1 the total number of letters C, or mishandled the case of LIS and LDS intersecting vs disjoint). 2. Completed the construction and essentially all of the lower-bound proof, but did not fully justify Claim 2 (that no tile can contain more than one letter) in one of the four regions or along the LIS/LDS boundary. 3. Gave the full LIS/LDS framework and derived ≥ n+2√n−3 for n=25, but omitted the off-board-letter correction (subtracting at most 4) or the extra-tile correction in the disjoint case, leaving the final bound at 31 or 33 without resolving the discrepancy.

The answer is $32 = 25 + 2(5) - 3$.

In general, for an $n \times n$ grid, the answer turns out to be $\left\lceil n + 2 \sqrt n - 3 \right\rceil$, but when $n$ is not a perfect square the solution is more complicated. Here $n=25$.

**Construction.**
We describe a general construction when $n = k^2$ (for this problem, $k=5$); it generalizes readily. There are a total of $(k-1)^2$ tiles which are $k \times k$ squares and another $4(k-1)$ tiles on the boundary, giving a total of
\[ (k-1)^2 + 4(k-1) = k^2 + 2k - 3 \]
tiles as promised. For $k=5$, this yields $16 + 16 = 32$ tiles.

**Bound.**
There are several approaches, but the shortest proof seems to be the following one that exploits the Erdős–Szekeres theorem. The theorem being quoted is:

**Theorem (Erdős–Szekeres).**
*Let $n \ge 1$ be an integer. Given a permutation of $(1, \dots, n)$, if a longest increasing subsequence (LIS) has length $a$ and a longest decreasing subsequence (LDS) has length $b$, then $ab \ge n$.*

This is a stronger version of the theorem compared to another version which instead just asserts that $\max(a,b) \ge \sqrt n$.

To apply this to the present problem, take the $n$ uncovered squares, which we henceforth call "black", as a permutation. Then consider both an LIS of length $a$ and an LDS of length $b$. We do the following illustration:
- Draw the LIS as a broken line, then connect it to the southwest and northeast corners of the board.
- Similarly, draw the LDS as a broken line, then connect it to the northwest and southeast corners of the board.
- These two steps partition the board into four quadrants, which we call north, east, south, and west.
- For each black cell in the north quadrant, write an $\mathbf{N}$ in the cell above it (for a cell in the first row, this will be off the board). Do the same for $\mathbf{E}$ (east, in the cell to the right), $\mathbf{S}$ (south, in the cell below), and $\mathbf{W}$ (west, in the cell to the left).
- Some black cells are in multiple quadrants (i.e., part of the LIS/LDS). Write all corresponding letters in that case.

We observe that:

**Claim 1.**
In this algorithm, the total number of letters written is exactly
\[
  C \coloneq \begin{cases}
    n + a + b + 1 & \text{if the LIS and LDS intersect} \\
    n + a + b & \text{otherwise}.
  \end{cases}
\]

*Proof.*
This is obvious. Each black square contributes at least one letter. Each black square on exactly one of the LIS and LDS contributes one extra letter. And a black square on both contributes $4$ letters instead of $1+1+1$.

Note by AM–GM we have $a + b \ge 2 \sqrt{ab} \ge 2 \sqrt n$, so we have a bound of
\[ C \ge n + 2 \sqrt n + \varepsilon \quad\text{where}\quad
  \varepsilon \coloneq \begin{cases}
    1 & \text{if the LIS and LDS intersect} \\
    0 & \text{otherwise}.
  \end{cases} \]

To relate $C$ to the number of tiles, the critical claim is the following, which is by construction:

**Claim 2.**
None of Matilda's tiles can have more than one letter written on it.

*Proof.*
This follows from the construction.

We split into two cases based on $\varepsilon$:
- When $\varepsilon = 1$, at most four letters go off the grid (one for each direction), so the number of tiles is at least $C - 4 \ge n + 2 \sqrt n - 3$.
- Suppose $\varepsilon = 0$. Then $C - 4 \ge n + 2 \sqrt n - 4$. However, we make the additional observation here that the tile where the LIS and LDS meet has no letters on it either; hence there are at least $1 + (C-4) \ge n + 2 \sqrt n - 3$ tiles.

For $n=25$, this yields a minimum of $25 + 2(5) - 3 = 32$ tiles.


Question:
Let $n$ be a positive integer. A Japanese triangle consists of $1 + 2 + \cdots + n$ circles arranged in an equilateral triangular shape such that for each $1 \leq i \leq n$, the ithrow contains exactly $i$ circles, exactly one of which is colored red. A ninja path in a Japanese triangle is a sequence of $n$ circles obtained by starting in the top row, then repeatedly going from a circle to one of the two circles immediately below it and finishing in the bottom row. In terms of $n$, find the greatest $k$ such that in each Japanese triangle there is a ninja
path containing at least $k$ red circles.

The answer is
\[
k = \lfloor \log_2(n) \rfloor + 1.
\]

**Construction (Upper Bound):**
It suffices to construct a Japanese triangle for any $n$ such that no ninja path contains more than $\lfloor \log_2(n) \rfloor + 1$ red circles.
Let the circles be indexed by $(i, j)$, where $i$ is the row number ($1 \le i \le n$) and $j$ is the column number ($1 \le j \le i$). A ninja path moves from $(i, j)$ to either $(i+1, j)$ or $(i+1, j+1)$ at each step.
Partition the rows $1, 2, \dots, n$ into $e = \lfloor \log_2(n) \rfloor + 1$ blocks. For $1 \le m \le e$, the $m$-th block consists of rows $i$ such that $2^{m-1} \le i \le \min(n, 2^m - 1)$.
For each row $i$ in the $m$-th block, we can write $i = 2^{m-1} + d$ for some integer $0 \le d < 2^{m-1}$. We place the red circle in row $i$ at column $2d + 1$. (This is a valid position because $2d + 1 \ge 1$ and $i - (2d + 1) = 2^{m-1} - d - 1 \ge 0$, so $2d + 1 \le i$.)

Suppose a ninja path passes through two red circles in the same block $m$, say at rows $i = 2^{m-1} + d$ and $i' = 2^{m-1} + d'$ with $d < d'$. The red circle at row $i$ is at column $2d+1$. At each step, a ninja path can increase its column index by at most $1$. Thus, moving from row $i$ to row $i'$, the column index can increase by at most $i' - i = d' - d$. Therefore, at row $i'$, the path can reach a column index of at most $(2d+1) + (d' - d) = d + d' + 1$. However, the red circle at row $i'$ is located at column $2d' + 1$. Since $d < d'$, we have $d + d' + 1 < 2d' + 1$, meaning the ninja path cannot reach the red circle at row $i'$.
This proves that any ninja path contains at most one red circle per block. Since there are $e$ blocks, any ninja path contains at most $e = \lfloor \log_2(n) \rfloor + 1$ red circles.

**Lower Bound (First Proof):**
Conversely, we show that in any Japanese triangle, one can find a ninja path containing at least $\lfloor \log_2(n) \rfloor + 1$ red circles. Let $(i, r_i)$ denote the unique red circle in row $i$.

We construct a directed graph $T_1$ on the set of all circles by defining the outgoing edges from row $i$ to row $i+1$ (for $1 \le i < n$) as follows:
- The red circle $(i, r_i)$ has two outgoing edges, connecting to both circles immediately below it: $(i+1, r_i)$ and $(i+1, r_i+1)$.
- Any white circle $(i, j)$ to the left of the red circle ($j < r_i$) has one outgoing edge, connecting to $(i+1, j)$.
- Any white circle $(i, j)$ to the right of the red circle ($j > r_i$) has one outgoing edge, connecting to $(i+1, j+1)$.

In row $i+1$, circles $(i+1, 1)$ through $(i+1, r_i-1)$ each receive exactly one incoming edge (from the white circles to the left of $r_i$). Circles $(i+1, r_i)$ and $(i+1, r_i+1)$ each receive exactly one incoming edge (from the red circle). Circles $(i+1, r_i+2)$ through $(i+1, i+1)$ each receive exactly one incoming edge (from the white circles to the right of $r_i$). Thus, every circle in row $i+1$ has exactly one incoming edge from row $i$. Since row 1 contains only the root circle $(1,1)$ (which must be red), $T_1$ is a rooted binary tree where every path from the root to a leaf in the bottom row is a valid ninja path.

Next, we build a new rooted tree $T_2$ consisting only of the $n$ red circles. We define the edges of $T_2$ by connecting each red circle to its closest red descendant(s) in $T_1$. That is, a red circle $v$ is a child of a red circle $u$ in $T_2$ if there is a directed path in $T_1$ from $u$ to $v$ containing only white circles as intermediate nodes.
In $T_1$, each red circle has exactly two children, and each white circle has exactly one child. Therefore, any path in $T_1$ starting from a child of a red circle $u$ does not branch until it reaches another red circle (or a leaf). This implies that $u$ can have at most two children in $T_2$.
Thus, $T_2$ is a binary tree with $n$ nodes. Let $k$ be the maximum number of nodes on a path from the root to a leaf in $T_2$. Since a binary tree of depth $k$ has at most $2^k - 1$ nodes, we must have $2^k - 1 \ge n$, which implies $k \ge \lfloor \log_2(n) \rfloor + 1$.
This path of length $k$ in $T_2$ corresponds to a sequence of $k$ red circles that lie on a single path in $T_1$. Since any path in $T_1$ is a valid ninja path, there exists a ninja path containing at least $\lfloor \log_2(n) \rfloor + 1$ red circles.

**Lower Bound (Alternative Recursive Proof):**
Define $f(i, j)$ as the maximum number of red circles contained in the portion of a ninja path from $(1, 1)$ down to $(i, j)$, including the endpoints. If $j < 1$ or $j > i$, we define $f(i, j) = 0$. Since every ninja path reaching $(i, j)$ must pass through either $(i-1, j-1)$ or $(i-1, j)$, we have the recurrence:
\[
f(i, j) = \max \{f(i-1, j-1), f(i-1, j)\} + \begin{cases} 1 & \text{if } (i, j) \text{ is red} \\ 0 & \text{otherwise} \end{cases}
\]
Define $S_i = \sum_{j=1}^i f(i, j)$.

**Claim 1:** $S_{i+1} \ge S_i + \left\lceil \frac{S_i}{i} \right\rceil + 1$.
*Proof.* We evaluate $S_{i+1}$:
\[
S_{i+1} = \sum_{j=1}^{i+1} \max \{f(i, j-1), f(i, j)\} + \sum_{j=1}^{i+1} \begin{cases} 1 & \text{if } (i+1, j) \text{ is red} \\ 0 & \text{otherwise} \end{cases}
\]
Since exactly one circle in row $i+1$ is red, the second sum is exactly $1$. Let $f(i, m)$ be the maximum element among $\{f(i, 1), \dots, f(i, i)\}$. We split the first sum at $m$:
\begin{align*}
S_{i+1} &= \sum_{j=1}^m \max \{f(i, j-1), f(i, j)\} + \sum_{j=m+1}^{i+1} \max \{f(i, j-1), f(i, j)\} + 1 \\
&\ge \sum_{j=1}^m f(i, j) + \sum_{j=m+1}^{i+1} f(i, j-1) + 1 \\
&= \sum_{j=1}^m f(i, j) + \sum_{j=m}^i f(i, j) + 1 \\
&= S_i + f(i, m) + 1
\end{align*}
Since $f(i, m)$ is the maximum of $i$ integers whose sum is $S_i$, by the Pigeonhole Principle, $f(i, m) \ge \lceil \frac{S_i}{i} \rceil$. Thus, $S_{i+1} \ge S_i + \lceil \frac{S_i}{i} \rceil + 1$. $\blacksquare$

**Claim 2:** Write $n = 2^c + r$, where $0 \le r < 2^c$. Then $S_n \ge cn + 2r + 1$.
*Proof.* We proceed by induction on $n$.
Base case $n=1$: We have $1 = 2^0 + 0$, so $c=0$ and $r=0$. The only circle in row $1$ is red, so $S_1 = f(1, 1) = 1$. The formula gives $0 \cdot 1 + 2(0) + 1 = 1$, so the claim holds for $n=1$.
Assuming the claim holds for some $n=j$, let $j = 2^c + r$. We have:
\begin{align*}
S_{j+1} &\ge S_j + \left\lceil \frac{S_j}{j} \right\rceil + 1 \\
&\ge cj + 2r + 1 + \left\lceil \frac{cj + 2r + 1}{j} \right\rceil + 1 \\
&= cj + 2r + 1 + c + \left\lceil \frac{2r + 1}{j} \right\rceil + 1
\end{align*}
Since $j = 2^c + r$ and $r \le 2^c - 1$, we have $j - (2r+1) = 2^c - r - 1 \ge 0$, which means $2r+1 \le j$. Thus, $0 < \frac{2r+1}{j} \le 1$, which implies $\lceil \frac{2r+1}{j} \rceil = 1$. Therefore:
\[
S_{j+1} \ge cj + 2r + 1 + c + 1 + 1 = c(j+1) + 2(r+1) + 1
\]
If $j+1$ is not a power of $2$, then $j+1 = 2^c + (r+1)$ with $r+1 < 2^c$. The formula matches exactly.
If $j+1$ is a power of $2$, then $r+1 = 2^c$, so $j+1 = 2^{c+1}$. We can rewrite the bound as:
\[
c(j+1) + 2(2^c) + 1 = c(j+1) + (j+1) + 1 = (c+1)(j+1) + 2(0) + 1
\]
This matches the formula for $j+1$ with parameters $c+1$ and $0$. The induction is complete. $\blacksquare$

Finally, we show that this implies the existence of a path with at least $c+1$ red circles. The average value of $f(n, j)$ for $1 \le j \le n$ is $\frac{S_n}{n}$. Thus, the maximum value is at least $\lceil \frac{S_n}{n} \rceil$. Using Claim 2:
\[
\left\lceil \frac{S_n}{n} \right\rceil \ge \left\lceil \frac{cn + 2r + 1}{n} \right\rceil = c + \left\lceil \frac{2r + 1}{n} \right\rceil = c + 1
\]
Therefore, there exists some $j$ such that $f(n, j) \ge c+1$. Since $c = \lfloor \log_2(n) \rfloor$, there is a ninja path containing at least $\lfloor \log_2(n) \rfloor + 1$ red circles.

(Partial) 1. Gave the correct candidate value k = ⌊log2(n)⌋ + 1, with some justification from small cases or a heuristic (e.g. “at most one per power-of-2 block” or “binary tree depth”). 2. Proved a nontrivial upper bound by an explicit coloring construction showing that some Japanese triangle forces every ninja path to have O(log n) red circles (e.g. partitioning rows into blocks [2^{m-1},2^m−1] and arguing a path cannot hit two reds within one block). 3. Proved a nontrivial lower bound of the form k ≥ ⌊log2(n)⌋ (or k ≥ c+1 where 2^c ≤ n), e.g. by introducing the rooted/binary-tree encoding of paths or a dynamic-programming recurrence f(i,j), but without completing the final counting step. (Almost) 1. Proved the upper bound construction correctly (at most one red per block, hence ≤ ⌊log2(n)⌋+1), and proved the lower bound via the binary-tree argument, but left a minor gap in defining/justifying the tree T2 (e.g. why each red has ≤2 red-children, or why a path in T2 corresponds to a valid ninja path). 2. Proved the lower bound via the f(i,j), S_i method up to a correct inequality such as S_{i+1} ≥ S_i + ⌈S_i/i⌉ + 1 (or an equivalent growth estimate), but made a small induction/boundary slip when converting it into max_j f(n,j) ≥ ⌊log2(n)⌋+1. 3. Solution is almost complete, but made minor mistakes which are not negligible.

Instruction:
For $n=8$, provide an explicit construction of a Japanese triangle that minimizes the maximum number of red circles any ninja path can contain.

Requirements:
- Represent the triangle as an 8x8 matrix $M$, where $M[i][j]$ is the circle in row $i$, column $j$ (0-indexed).
- Use `0` for cells outside the triangle (where $j > i$).
- Use `1` for white circles and `2` for red circles.
- Each row $i \in \{0, \dots, 7\}$ must contain exactly one red circle (`2`) among its $i+1$ circles.
- A ninja path is a sequence of cells $(r, c_r)$ for $r=0, \dots, 7$ where $c_0=0$ and $c_{r+1} \in \{c_r, c_r+1\}$.
- Provide exactly one construction block and put ONLY the Python literal list of lists between the tags:
<construct>
[...matrix...]
</construct>
- Do not use code fences. Do not put any extra text inside <construct>...</construct>.

import sys
import ast

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

        # Parse the input using a strict Python-literal strategy
        try:
            matrix = ast.literal_eval(data)
        except Exception as e:
            print(f"Error: Failed to parse input as a Python literal. {e}")
            return

        # 1. Structural Verification: Matrix must be 8x8
        if not isinstance(matrix, list) or len(matrix) != 8:
            print(f"Error: Matrix must be a list of 8 rows, but found {len(matrix) if isinstance(matrix, list) else 'non-list'}.")
            return

        for i in range(8):
            if not isinstance(matrix[i], list) or len(matrix[i]) != 8:
                print(f"Error: Row {i} must be a list of 8 elements.")
                return

        # 2. Constraint Verification: Lower triangular property and red circle constraint
        for i in range(8):
            red_count = 0
            for j in range(8):
                val = matrix[i][j]
                if j > i:
                    # Outside the triangle (j > i)
                    if val != 0:
                        print(f"Error: Cell ({i}, {j}) is outside the triangle (j > i) and must be 0.")
                        return
                else:
                    # Inside the triangle (j <= i)
                    if val == 2:
                        red_count += 1
                    elif val == 1:
                        pass
                    else:
                        print(f"Error: Cell ({i}, {j}) must be 1 (white) or 2 (red).")
                        return

            # Exactly one red circle per row
            if red_count != 1:
                print(f"Error: Row {i} must contain exactly one red circle (value 2), but found {red_count}.")
                return

        # 3. Mathematical Verification: Dynamic Programming to find the max red circles on any ninja path
        # dp[r][c] stores the maximum number of red circles on a ninja path starting at (0,0) and ending at (r,c)
        dp = [[-1] * 8 for _ in range(8)]

        # Initial condition (Row 0)
        # We already verified there is exactly one '2' in row 0, which must be at (0,0)
        dp[0][0] = (1 if matrix[0][0] == 2 else 0)

        for r in range(1, 8):
            for c in range(r + 1):
                # A ninja path can reach (r, c) from (r-1, c) or (r-1, c-1)
                prev_max = -1

                # Path from (r-1, c)
                if c < r:
                    if dp[r-1][c] != -1:
                        prev_max = max(prev_max, dp[r-1][c])

                # Path from (r-1, c-1)
                if c > 0:
                    if dp[r-1][c-1] != -1:
                        prev_max = max(prev_max, dp[r-1][c-1])

                # If reachable, calculate max red circles for current cell
                if prev_max != -1:
                    dp[r][c] = prev_max + (1 if matrix[r][c] == 2 else 0)
                else:
                    # Every cell in the triangle should be reachable under ninja path rules
                    print(f"Error: Cell ({r}, {c}) is unreachable, which violates triangle structure.")
                    return

        # The objective is the maximum value in the final row
        max_path_reds = max(dp[7])

        # The theoretical minimum of the maximum number of red circles for n=8 is floor(log2(8)) + 1 = 4.
        # The instruction asks for a construction that minimizes this maximum.
        if max_path_reds == 4:
            print("True")
        else:
            print(f"Error: The maximum number of red circles on any ninja path is {max_path_reds}, but it should be 4.")

    except Exception as e:
        print(f"Error: An unexpected error occurred during verification. {e}")

if __name__ == "__main__":
    verify()


Question: Let $n$ be a positive integer. A Nordic square is an $n \\times n$ board containing all the integers from $1$ to $n^2$ so that each cell contains exactly one number. An uphill pathis a sequence of one or more cells such that:\n1. the first cell in the sequence is a valley, meaning the number written is less than all its orthogonal neighbors,\n2. each subsequent cell in the sequence is orthogonally adjacent to the previous cell, and\n3. the numbers written in the cells in the sequence are in increasing order. \nFind, as a function of n, the smallest possible total number of uphill paths in a Nordic square.

Instruction:
For the specific case $n=64$, provide an explicit construction of the $64 \times 64$ Nordic square that minimizes the total number of uphill paths.

Construction requirements:
- Output a single Python expression that evaluates to a list of 64 lists, representing the $64 \times 64$ grid.
- Each inner list should correspond to a row in the grid, containing 64 integers.
- The grid must contain all integers from 1 to $64^2$ exactly once.
- The configuration must minimize the total number of uphill paths.
- The output must be a single, valid Python expression (e.g., using list comprehensions). Do not include any assignments, imports, function definitions, or extra text.
- Put ONLY the Python expression inside the construction block.

[
    [val_dict[(R, C)] for C in range(64)]
    for dist_dict in [{
        (r, c): (
            8*(r//6) + (0 if c==1 else c-1) if r%6==0 and c>=1 else
            8*(r//6) + 1 if r%6==1 and c==1 else
            8*(r//6) + 2 if r%6==1 and c==0 else
            8*(r//6) + c if r%6==1 and c>=3 and c%2==1 else
            8*(r//6) + 3 if r%6==2 and c==0 else
            8*(r//6) + c + 5 if r%6==2 and c>=2 and c%2==0 else
            8*(r//6) + 4 if r%6==3 and c==0 else
            8*(r//6) + c + 4 if r%6==3 and c>=1 else
            8*(r//6) + 5 if r%6==4 and c==0 else
            8*(r//6) + c + 5 if r%6==4 and c>=2 and c%2==0 else
            8*(r//6) + 6 if r%6==5 and c==0 else
            8*(r//6) + 7 if r%6==5 and c==1 else
            8*(r//6+1) + c if r%6==5 and c>=3 and c%2==1 else
            -1
        ) for r in range(64) for c in range(64)
    }]
    for val_dict in [{
        p: i+1 for i, p in enumerate(
            [p for d, p in sorted([(d, p) for p, d in dist_dict.items() if d != -1])] +
            [p for d, p in sorted([(d, p) for p, d in dist_dict.items() if d == -1])]
        )
    }]
    for R in range(64)
]

import sys

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

    # 1. Evaluate the raw string payload using eval() to obtain the Python object.
    try:
        grid = eval(input_text)
    except Exception as e:
        print(f"Error parsing input: {e}")
        return

    # 2. Verify that the parsed object is a list of exactly 64 elements,
    # and each element is a list of exactly 64 integers.
    if type(grid) is not list:
        print("Error: The parsed object is not a list.")
        return
    if len(grid) != 64:
        print(f"Error: Expected 64 rows, got {len(grid)}.")
        return
    for i, row in enumerate(grid):
        if type(row) is not list:
            print(f"Error: Row {i} is not a list.")
            return
        if len(row) != 64:
            print(f"Error: Row {i} does not have exactly 64 elements.")
            return
        for j, val in enumerate(row):
            if type(val) is not int:
                print(f"Error: Element at ({i}, {j}) is not an integer.")
                return

    # 3. Flatten the grid and verify that it contains all integers from 1 to 4096 exactly once.
    flattened = [val for row in grid for val in row]
    if len(flattened) != 4096:
        print("Error: Total number of elements is not 4096.")
        return

    counts = set(flattened)
    if len(counts) != 4096:
        print("Error: Elements in the grid are not distinct.")
        return

    min_val, max_val = min(counts), max(counts)
    if min_val != 1 or max_val != 4096:
        print(f"Error: Grid values must be from 1 to 4096. Found min {min_val}, max {max_val}.")
        return

    def get_neighbors(r, c):
        for nr, nc in ((r-1, c), (r+1, c), (r, c-1), (r, c+1)):
            if 0 <= nr < 64 and 0 <= nc < 64:
                yield nr, nc

    # 4. Iterate over all cells in the grid to identify all "valleys".
    valleys = []
    for r in range(64):
        for c in range(64):
            is_valley = True
            for nr, nc in get_neighbors(r, c):
                if grid[nr][nc] <= grid[r][c]:
                    is_valley = False
                    break
            if is_valley:
                valleys.append((r, c))

    # 5. Traverse and count all valid "uphill paths" originating from each valley.
    # We use a Dynamic Programming approach processing cells in descending order
    # to efficiently count paths and avoid recursion limits.
    cells = []
    for r in range(64):
        for c in range(64):
            cells.append((grid[r][c], r, c))
    cells.sort(reverse=True)

    paths_from = [[1] * 64 for _ in range(64)]
    for val, r, c in cells:
        for nr, nc in get_neighbors(r, c):
            if grid[nr][nc] > grid[r][c]:
                paths_from[r][c] += paths_from[nr][nc]

    # 6. Sum the total number of uphill paths across all valleys.
    total_uphill_paths = 0
    for vr, vc in valleys:
        total_uphill_paths += paths_from[vr][vc]

    # 7. Verify that the total number of uphill paths is exactly 8065.
    if total_uphill_paths != 8065:
        print(f"Error: Expected exactly 8065 uphill paths, but found {total_uphill_paths}.")
        return

    print("True")

if __name__ == "__main__":
    verify()

(Partial) 1. Established the lower bound idea: for each adjacent pair of cells with values a>b, following a strictly decreasing walk to a valley and reversing it yields an uphill path of length ≥2 ending with the step b→a. 2. Correctly counted the number of orthogonally adjacent pairs in an n×n grid as 2n(n−1), and concluded there are at least 2n^2−2n uphill paths of length ≥2. 3. Noted that at least one uphill path of length 1 always exists (e.g. the single cell containing 1, or any valley), and combined with the previous bound to get ≥2n^2−2n+1. (Almost) 1. Gave the full lower bound 2n^2−2n+1 and a construction intended to match it using a set/tree T so that no two cells outside T are adjacent, but did not fully justify the existence of such a T for all n (or left a gap in the “cropping/adaptation” argument). 2. Provided the construction and argued (correctly in outline) that 1 is the unique valley and every uphill path starts at 1, but did not rigorously justify that each adjacent step corresponds to a unique uphill path (e.g. the uniqueness of the increasing path inside T, or why no extra uphill paths arise). 3. Solution is almost complete, but made minor mistakes which are not negligible.

Answer: $2n^2-2n+1$

\paragraph{Bound.}
The lower bound is established as follows:
\begin{itemize}
    \item For any pair of orthogonally adjacent cells, say $a > b$, one can extend the step $a \to b$ to a downhill path (the reverse of an uphill path) by repeatedly moving to a smaller adjacent cell until a valley is reached. Reversing this sequence gives a distinct uphill path of length $\ge 2$ ending with the step $b \to a$. Since there are $2n(n-1)$ pairs of adjacent cells in the grid, this gives at least $2n(n-1)=2n^2-2n$ uphill paths of length $\ge 2$.
    \item There is always at least one uphill path of length $1$, namely the single cell containing $1$ (or indeed any valley).
\end{itemize}
Summing these yields at least $2n^2-2n+1$ uphill paths.

\paragraph{Construction.}
For the construction, the idea is to build a tree $T$ on the grid such that no two cells not in $T$ are adjacent. (Such a tree can be explicitly constructed for any $n$; for example, one can design a periodic pattern that forms a tree for grids where $3 \mid n$, and then adapt it to any $n$ by cropping the last $n \bmod 3$ rows and the necessary columns, ensuring that the cells not in $T$ remain strictly isolated from one another.)

Place $1$ anywhere in $T$, and then place all the small numbers up to $|T|$ inside $T$ such that the numbers strictly increase along the unique paths in $T$ branching out from $1$. Then, place the remaining numbers outside $T$ arbitrarily.

Because no two cells outside $T$ are adjacent, every neighbor of a cell outside $T$ belongs to $T$. Since all numbers in $T$ are at most $|T|$ and all numbers outside $T$ are strictly greater than $|T|$, any cell outside $T$ contains a strictly greater number than all its neighbors. Consequently, no cell outside $T$ can be a valley, making $1$ the unique valley in the entire grid.

By construction, any uphill path must start from $1$. Furthermore, because $T$ is a tree and the numbers increase along its branches, the sequence of cells from $1$ to any cell $b \in T$ is unique. Thus, an uphill path is uniquely determined by its final step $b \to a$ (where $b \in T$, and $a$ is either in $T$ or outside $T$). This implies that the only uphill paths are exactly the length-$1$ path at $\{1\}$ and the reverses of the uniquely determined downhill paths we counted in the lower bound. This achieves exactly $2n^2-2n+1$ uphill paths.


Question: There is an integer $n > 1$. There are $n^2$ stations on a slope of a mountain, all at different altitudes. Each of two cable car companies, A and B, operates $k$ cable cars; each cable car provides a transfer from one of the stations to a higher one (with no intermediate stops). The $k$ cable cars of A have $k$ different starting points and kdifferent finishing points, and a cable car which starts higher also finishes higher. The same conditions hold for B. We say that two stations are linked by a company if one can start from the lower station and reach the higher one by using one or more cars of that company (no other movements between stations are allowed). Determine the smallest positive integer $k$ for which one can guarantee that there are\ntwo stations that are linked by both companies.

Instruction:
For the specific case $n=33$, determine the maximum number of cable cars $k$ that each company can operate without creating any pair of stations linked by both companies. Provide an explicit construction for this maximum case.

Construction requirements:
- The $33^2 = 1089$ stations are numbered 1 to 1089 from lowest to highest altitude.
- Output a single Python expression that evaluates to a tuple of two sets: `(A_lines, B_lines)`.
- `A_lines` and `B_lines` are sets of tuples `(i, j)`, representing a cable car line from station `i` to station `j` for company A and B respectively.
- The configuration must satisfy all problem constraints for valid cable car operations (distinct start/end points, and a cable car starting higher finishes higher).
- The number of cars for each company must be exactly $k$ (the maximum possible counter-example size).
- Put ONLY the Python expression inside the construction block. Do not include any markdown formatting, explanations, or surrounding tags other than `<construct>...</construct>`.

import sys

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

    try:
        # Parse the payload directly as a Python expression.
        # eval is used because the reference construction uses comprehensions and range().
        data = eval(payload)
    except Exception as e:
        print(f"Failed to parse Python expression: {e}")
        return

    if type(data) is not tuple or len(data) != 2:
        print("Parsed object is not a tuple of exactly two elements")
        return

    A_lines, B_lines = data

    for name, lines in [("A_lines", A_lines), ("B_lines", B_lines)]:
        if type(lines) is not set:
            print(f"{name} is not a set")
            return
        for item in lines:
            if type(item) is not tuple or len(item) != 2:
                print(f"An item in {name} is not a tuple of exactly two elements")
                return
            u, v = item
            if type(u) is not int or type(v) is not int:
                print(f"An item in {name} contains non-integers")
                return

    if len(A_lines) != 1056:
        print(f"A_lines has {len(A_lines)} elements, expected 1056")
        return
    if len(B_lines) != 1056:
        print(f"B_lines has {len(B_lines)} elements, expected 1056")
        return

    for name, lines in [("A_lines", A_lines), ("B_lines", B_lines)]:
        u_set = set()
        v_set = set()
        sorted_lines = sorted(list(lines), key=lambda x: x[0])
        prev_v = -1

        for u, v in sorted_lines:
            if not (1 <= u <= 1089 and 1 <= v <= 1089):
                print(f"Station out of range in {name}: ({u}, {v})")
                return
            if u >= v:
                print(f"Invalid transfer in {name}: ({u}, {v}), must be u < v")
                return
            if u in u_set:
                print(f"Duplicate starting point {u} in {name}")
                return
            u_set.add(u)
            if v in v_set:
                print(f"Duplicate finishing point {v} in {name}")
                return
            v_set.add(v)
            if v <= prev_v:
                print(f"Monotonicity violated in {name}: sequence of v is not strictly increasing.")
                return
            prev_v = v

    def get_transitive_closure(lines):
        adj = {u: v for u, v in lines}
        closure = set()
        for start_node in adj:
            curr = start_node
            while curr in adj:
                nxt = adj[curr]
                closure.add((start_node, nxt))
                curr = nxt
        return closure

    A_closure = get_transitive_closure(A_lines)
    B_closure = get_transitive_closure(B_lines)

    intersection = A_closure.intersection(B_closure)
    if intersection:
        example = list(intersection)[0]
        print(f"Intersection is not empty. Found linked pairs in both companies, e.g., {example}.")
        return

    print("True")

if __name__ == "__main__":
    verify()

(Partial) 1. Observed/modelled each company’s system as a directed graph on the stations, and correctly deduced from the “order-preserving” and “distinct start/endpoints” conditions that every vertex has indegree ≤1 and outdegree ≤1, hence each connected component is a path (or isolated vertex). 2. Derived the component count formula for one company: with k edges on n^2 vertices and components being paths, the number of connected components equals n^2−k. 3. Gave a correct extremal/pigeonhole step in one graph: from n^2=(n−1)(n+1)+1 and having only n−1 components, concluded that some component has at least n+1 vertices. (Almost) 1. Proved that k=n^2−n+1 is sufficient using the path-component structure and pigeonhole principle, but made a small counting slip (e.g. wrote “at most n−1 components” or miscomputed n^2−k) while the intended argument clearly yields the correct bound. 2. Presented the full sufficiency proof and the idea of a counterexample for k=n^2−n, but did not fully describe/verify the construction (or verified it only for one small n) while the generalization is immediate. 3. Solution is almost complete, but made minor mistakes which are not negligible.

Answer: $k = n^2 - n + 1$.

When $k = n^2-n$, a valid construction exists for $n=4$ which generalizes readily, demonstrating that this value is not sufficient.

To see that $k = n^2-n+1$ is sufficient, view $A$ and $B$ as graphs whose vertices are the $n^2$ stations and whose edges are the cable cars. For each company, the condition that a cable car starting higher also finishes higher, with all starting points and finishing points being distinct, implies that the connected components of these graphs are paths (possibly with $0$ edges; the direction of these edges is irrelevant).

Now, if $k = n^2-n+1$, it follows that $A$ and $B$ each have exactly $n^2 - k = n-1$ connected components. Since there are $n^2$ vertices in total and $n^2 = (n-1)(n+1) + 1$, by the Pigeonhole Principle, some component of $A$ has at least $n+1$ vertices. Because graph $B$ has only $n-1$ connected components, this component of $A$ must contain two vertices that belong to the same component of $B$. These two stations are therefore linked by both companies, as desired.


Question:
Find all integers n for which each cell of $n \times n$ table can be filled with one of the letters I, M and O in such a way that:
• In each row and column, one third of the entries are I, one third are M and one third are O; and
• in any diagonal, if the number of entries on the diagonal is a multiple of three, then one third of the entries are I, one third are M and one third are O.
Note that an $n \times n$ table has $4n − 2$ diagonals.

The answer is that $n$ must be divisible by $9$.

First, we construct a valid table for $n=9$. We can fill a $9 \times 9$ table as follows:
$$
\begin{array}{|ccc|ccc|ccc|}
\hline
I & I & I & M & M & M & O & O & O \\
M & M & M & O & O & O & I & I & I \\
O & O & O & I & I & I & M & M & M \\
\hline
I & I & I & M & M & M & O & O & O \\
M & M & M & O & O & O & I & I & I \\
O & O & O & I & I & I & M & M & M \\
\hline
I & I & I & M & M & M & O & O & O \\
M & M & M & O & O & O & I & I & I \\
O & O & O & I & I & I & M & M & M \\
\hline
\end{array}
$$
This pattern can be extended to any $n$ that is a multiple of $9$ by tiling the grid with this $9 \times 9$ block.

Now we prove that $9 \mid n$ is a necessary condition. From the condition on the rows and columns, $n$ must be a multiple of $3$. Let $n = 3k$. We can partition the $n \times n$ grid into $k^2$ sub-boxes, each of size $3 \times 3$.

We say a multiset of squares is *clean* if the letters I, M, and O appear an equal number of times among them (i.e., each letter constitutes exactly one third of the entries). Note that the multiset union of clean multisets is also clean.

Consider the following sets of squares, all of which are clean by the problem statement:
1. All columns with indices congruent to $2 \pmod 3$.
2. All rows with indices congruent to $2 \pmod 3$.
3. All $4k-2$ diagonals whose lengths are multiples of $3$.

Take the multiset union of all these clean sets. We determine how many times each cell in the $n \times n$ grid is covered by this union.
- The center cell of each $3 \times 3$ sub-box (which lies at the intersection of a row indexed $2 \pmod 3$ and a column indexed $2 \pmod 3$) is covered exactly four times: once by its row, once by its column, and twice by the two diagonals passing through it. (The diagonals passing through $(x, y)$ where $x \equiv 2 \pmod 3$ and $y \equiv 2 \pmod 3$ have lengths that are multiples of $3$).
- Every other cell in the grid is covered exactly once by this union.

Since the entire grid is clean (for instance, as the union of all its columns), the multiset consisting of the entire grid covered once is clean. The multiset union we constructed is also clean. Removing the entire grid from our multiset union leaves exactly the center squares of the $k^2$ sub-boxes, each covered exactly three times.

Because removing a clean multiset from another clean multiset yields a clean multiset, the set of the $k^2$ center squares (taken three times) is clean. This implies that the set of the $k^2$ center squares taken once is also clean. For this set of $k^2$ squares to be clean, its total number of elements must be divisible by $3$. Thus, $3 \mid k^2$, which implies $3 \mid k$. Since $n = 3k$, we conclude that $9 \mid n$, as desired.

(Partial) 1. Observed from the row/column condition that 3 | n (each row/column has n/3 of each letter). 2. Gave a correct construction for n = 9 (or another explicit n divisible by 9), and indicated how to tile/extend it to all n multiple of 9. 3. Introduced the key “clean multiset” (equal numbers of I,M,O) viewpoint and correctly noted that multiset unions/differences of clean multisets are clean. (Almost) 1. Proved the necessity 9 | n using the 3×3 block/clean-multiset counting idea, but made a localized counting error in the coverage multiplicities (e.g. missed that the special cells are covered 4 times / others once) while the intended conclusion 9 | n is still clearly supported. 2. Completed the main reduction to showing the k^2 center cells form a clean set and hence 3 | k, but did not fully justify one of: (i) why the relevant diagonals through those cells have length multiple of 3, or (ii) why subtracting the whole grid from the union preserves cleanliness. 3. Gave the construction for all n multiple of 9 and the argument that n must be a multiple of 3, and outlined the final step to force 9 | n via the center-cell multiset, but left the last divisibility inference (3 | k^2 ⇒ 3 | k) or an equivalent concluding step insufficiently written.

Instruction:
For the smallest feasible integer $n > 0$, construct a valid $n \times n$ table satisfying the given conditions.

Requirements for the construction:
- Represent the table as a Python-literal list of lists of strings, where each string is "I", "M", or "O".
- The table must satisfy all row, column, and diagonal conditions for the chosen $n$.
- Include exactly one construction block and put ONLY the Python literal matrix between the tags:
<construct>
</construct>
- Do not use code fences. Do not put any extra text inside <construct>...</construct>.

import sys
import ast

def main():
    # Read the raw input from stdin
    try:
        input_data = sys.stdin.read().strip()
        if not input_data:
            print("Error: Empty input")
            return
        # Parse the input as a Python literal (list of lists of strings)
        matrix = ast.literal_eval(input_data)
    except Exception as e:
        print(f"Error: Could not parse input as Python literal: {e}")
        return

    # Basic structure check
    if not isinstance(matrix, list) or not all(isinstance(row, list) for row in matrix):
        print("Error: Input is not a list of lists")
        return

    # Smallest feasible n is 9
    n = len(matrix)
    if n != 9:
        print(f"Error: The smallest feasible integer n is 9, but the provided matrix is {n}x{len(matrix[0]) if n > 0 else 0}")
        return

    # Check that all rows have length n and all entries are one of "I", "M", "O"
    for r, row in enumerate(matrix):
        if len(row) != n:
            print(f"Error: Row {r} has length {len(row)} instead of {n}")
            return
        for c, val in enumerate(row):
            if val not in ("I", "M", "O"):
                print(f"Error: Invalid character '{val}' at row {r}, column {c}")
                return

    # Row and column requirements: each letter appears n/3 times (3 times for n=9)
    target_count = n // 3
    for i in range(n):
        # Row check
        row = matrix[i]
        for letter in ("I", "M", "O"):
            if row.count(letter) != target_count:
                print(f"Error: Row {i} does not have exactly {target_count} of letter '{letter}'")
                return

        # Column check
        col = [matrix[j][i] for j in range(n)]
        for letter in ("I", "M", "O"):
            if col.count(letter) != target_count:
                print(f"Error: Column {i} does not have exactly {target_count} of letter '{letter}'")
                return

    # Diagonal checks: for any diagonal with length L divisible by 3,
    # each letter must appear L/3 times.

    # Main-direction diagonals: cells (i, j) where i - j = k
    # k ranges from -(n-1) to (n-1)
    for k in range(-(n-1), n):
        diag = [matrix[i][i-k] for i in range(n) if 0 <= i-k < n]
        L = len(diag)
        if L % 3 == 0:
            for letter in ("I", "M", "O"):
                if diag.count(letter) != L // 3:
                    print(f"Error: Main-direction diagonal (i-j={k}) of length {L} fails the distribution rule for letter '{letter}'")
                    return

    # Anti-direction diagonals: cells (i, j) where i + j = k
    # k ranges from 0 to 2n-2
    for k in range(2 * n - 1):
        diag = [matrix[i][k-i] for i in range(n) if 0 <= k-i < n]
        L = len(diag)
        if L % 3 == 0:
            for letter in ("I", "M", "O"):
                if diag.count(letter) != L // 3:
                    print(f"Error: Anti-direction diagonal (i+j={k}) of length {L} fails the distribution rule for letter '{letter}'")
                    return

    # If all checks pass
    print("True")

if __name__ == "__main__":
    main()


Question: Let $n \\geq 2$ be an integer. Consider an $n \\times n$ chessboard consisting of $n^2$ unit squares. A configuration of n rooks on this board is peaceful if every row and every column contains exactly one rook. Find the greatest positive integer $k$ such that, for each\npeaceful configuration of $n$ rooks, there is a $k \\times k$ square which does not contain a rook on any of its $k^2$ unit squares.

Instruction:
For the specific case $n=233$, construct a peaceful configuration of $n$ rooks and find the minimum positive integer $l$ such that every $l \times l$ square contains at least one rook.

Construction format requirements:
- Output a single Python dictionary with keys: `'l'` and `'rooks'`.
- `'l'` is the minimum positive integer (an integer).
- `'rooks'` is a list of 233 integers, where the $r$-th integer (0-indexed) represents the column index of the rook in row $r$. Both row and column indices are integers from $0$ to $232$.
- The configuration must be peaceful, meaning every row and every column contains exactly one rook.
- Every $l \times l$ square in the $233 \times 233$ grid must contain at least one rook.
- Put ONLY the Python literal dictionary inside the construction block.

{"l": 16, "rooks": [dict([(16 * i + j - 23, 16 * j + i) for i in range(16) for j in range(16) if 16 * i + j >= 23 and 16 * j + i <= 232] + list(zip(sorted([16 * i - 9 for i in range(9, 16)] + [16 * i - 8 for i in range(1, 16)]), sorted([16 * j for j in range(15)] + [16 * j + 1 for j in range(7)]))))[r] for r in range(233)]}

import sys
import ast

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

    try:
        # The instruction and user intent specify a Python literal parsed with ast.literal_eval.
        data = ast.literal_eval(payload)
    except Exception as e_lit:
        # Fallback to eval with restricted builtins strictly to maintain compatibility
        # with the provided ref_construction, which uses comprehensions and specific builtins.
        try:
            allowed_builtins = {
                'dict': dict,
                'list': list,
                'zip': zip,
                'sorted': sorted,
                'range': range,
                '__builtins__': {}
            }
            data = eval(payload, allowed_builtins, {})
        except Exception:
            # If the fallback also fails, surface the original literal_eval error.
            print(f"Parse error: {e_lit}")
            return

    if type(data) is not dict:
        print("Error: The construction must be a Python dictionary.")
        return

    if set(data.keys()) != {'l', 'rooks'}:
        print("Error: The dictionary must contain exactly the keys 'l' and 'rooks'.")
        return

    l = data['l']
    if type(l) is not int:
        print("Error: 'l' must be an integer.")
        return

    if l != 16:
        print(f"Error: 'l' must be 16 (the mathematically required minimum for n=233), but got {l}.")
        return

    rooks = data['rooks']
    if type(rooks) is not list:
        print("Error: 'rooks' must be a list.")
        return

    if len(rooks) != 233:
        print(f"Error: 'rooks' must contain exactly 233 integers, but got {len(rooks)}.")
        return

    for i, c in enumerate(rooks):
        if type(c) is not int:
            print(f"Error: Element at index {i} in 'rooks' is not an integer.")
            return
        if not (0 <= c <= 232):
            print(f"Error: Element at index {i} in 'rooks' is out of bounds (must be 0-232, got {c}).")
            return

    if len(set(rooks)) != 233:
        print("Error: The configuration is not peaceful. There are duplicate column indices in 'rooks'.")
        return

    # Iterate over all possible 16x16 squares in the 233x233 grid.
    # A 16x16 square is defined by row range [R, R+15] and col range [C, C+15]
    # R and C range from 0 to 233 - 16 = 217.
    for R in range(218):
        for C in range(218):
            found = False
            for r in range(R, R + 16):
                if C <= rooks[r] <= C + 15:
                    found = True
                    break
            if not found:
                print(f"Error: The 16x16 square at rows {R}-{R+15} and columns {C}-{C+15} is completely empty.")
                return

    print("True")

if __name__ == "__main__":
    verify()

(Partial) 1. Proved the lower bound k ≤ ⌊√(n−1)⌋ by exhibiting (for m=k+1) a peaceful configuration on an m^2×m^2 board with no empty m×m square (e.g. via the explicit permutation j=(i mod m)m+⌊i/m⌋), even if not fully justified that every m×m square meets a rook. 2. Proved the existence part in the form: if n>k^2 then every peaceful configuration has an empty k×k square, by choosing k^2 consecutive rows under the top rook, splitting into k blocks of k rows, fixing k consecutive columns through that rook, and applying pigeonhole to the k resulting k×k squares. 3. Correctly identified the claimed answer k=⌊√(n−1)⌋ and gave a coherent plan for both directions (n>k^2 implies existence; construction for n≤(k+1)^2), with at least one direction carried out rigorously. (Almost) 1. Proved both directions for the extremal value k=⌊√(n−1)⌋, but left a small gap in the construction side: did not fully prove that the constructed m^2×m^2 permutation has the property that every m×m square contains a rook (e.g. the coverage/“forbidden top-left corners” argument incomplete). 2. Proved the sharp construction for size m^2 and the existence proof for n>k^2, but did not justify the adaptation from m^2×m^2 to arbitrary n≤m^2 (e.g. deletion/refilling step or preservation of the “no empty m×m square” property). 3. Solution is almost complete, but made minor mistakes which are not negligible.

The answer is $k = \left\lfloor \sqrt{n-1} \right\rfloor$.

First, we prove that for any $n > k^2$, every peaceful configuration of $n$ rooks contains an empty $k \times k$ square. Let $R$ be the rook in the topmost row (row $0$). Since $n > k^2$, there are at least $k^2$ rows below $R$. Consider the $k^2$ rows immediately below $R$, and partition them into $k$ blocks of $k$ consecutive rows. Let $c$ be the column containing $R$. We can choose a block of $k$ consecutive columns that includes $c$ (this is possible since $n \ge k$). The intersection of these $k$ columns with the $k$ blocks of rows forms $k$ squares of size $k \times k$, aligned vertically directly below the row of $R$. These $k$ squares occupy $k$ columns and $k^2$ rows. Because the configuration is peaceful, these $k$ columns contain exactly $k$ rooks in total across the entire board. One of these rooks is $R$, which is located in the topmost row, outside our $k$ squares. Therefore, the $k$ squares contain at most $k-1$ rooks in total. By the Pigeonhole Principle, at least one of the $k \times k$ squares must be completely empty.

To prove this is the maximum possible $k$, we must show that for $m = k+1$, there exists a peaceful configuration of $n \le m^2$ rooks with no empty $m \times m$ square. Following the original notation, we first give an explicit construction for an $m^2 \times m^2$ board showing there may be no empty $m \times m$ square.

Index the rows and columns from $0$ to $m^2-1$. We place a rook at $(i, j)$ if and only if $j = (i \bmod m) \cdot m + \lfloor i/m \rfloor$.
This is a peaceful configuration: any $i \in \{0, \dots, m^2-1\}$ can be written uniquely as $i = a m + b$ with $0 \le a, b \le m-1$, placing the rook at $j = b m + a$. The mapping $(a, b) \mapsto (b, a)$ is a bijection on $\{0, \dots, m-1\}^2$, so each row and column receives exactly one rook.

To show that this works, consider for each rook drawing an $m \times m$ square of "forbidden" top-left corners whose bottom-right hand corner is the rook. A rook at $(i, j)$ forbids any $m \times m$ square whose top-left corner $(x, y)$ satisfies $x \le i \le x+m-1$ and $y \le j \le y+m-1$. We must show that these forbidden regions cover all valid top-left corners $(x, y) \in [0, m^2-m] \times [0, m^2-m]$. (Parts of the last $m-1$ columns or rows may not be covered, but they don't matter anyways since a top-left corner cannot be placed there.)

Let $(x, y)$ be an arbitrary valid top-left corner. Write $y = e m + f$ where $0 \le e \le m-1$ and $0 \le f \le m-1$.
The interval $[x, x+m-1]$ consists of $m$ consecutive integers, so it contains exactly one integer $i_1 \equiv e \pmod m$. Let $i_1 = a_1 m + e$. The corresponding rook is at $j_1 = e m + a_1$.
If $a_1 \ge f$, then $j_1 \ge e m + f = y$. Since $a_1 \le m-1$, we also have $j_1 \le e m + m - 1 \le y + m - 1$. Thus, the rook $(i_1, j_1)$ forbids $(x, y)$.

If $a_1 < f$, then $f > 0$, which implies $e < m-1$ (since $y \le m^2-m$). The interval $[x, x+m-1]$ also contains exactly one integer $i_2 \equiv e+1 \pmod m$. Let $i_2 = a_2 m + e + 1$. The corresponding rook is at $j_2 = (e+1)m + a_2$.
We have $j_2 \ge (e+1)m > e m + f = y$. We need $j_2 \le y + m - 1 = (e+1)m + f - 1$, which requires $a_2 < f$.
Since $i_1, i_2 \in [x, x+m-1]$ and $i_2 \equiv i_1 + 1 \pmod m$, we must have either $i_2 - i_1 = 1$ or $i_2 - i_1 = 1 - m$. Substituting $i_1 = a_1 m + e$ and $i_2 = a_2 m + e + 1$, we get $i_2 - i_1 = (a_2 - a_1)m + 1$. Thus, $(a_2 - a_1)m + 1 \in \{1, 1-m\}$, which implies $a_2 - a_1 \in \{0, -1\}$. In either case, $a_2 \le a_1$. Since we assumed $a_1 < f$, it follows that $a_2 < f$. Thus, the rook $(i_2, j_2)$ forbids $(x, y)$. In all cases, the $m \times m$ square contains at least one rook.

Finally, it remains to check that the construction can be adapted for all $n \le m^2$ (omitting this step is a common mistake). For this, we can delete the bottom $m^2 - n$ rows and the rightmost $m^2 - n$ columns from our $m^2 \times m^2$ board to get an $n \times n$ board. In doing so, we may have accidentally deleted some rooks. We can fill in any gaps where we deleted a rook by placing new rooks at the intersections of the empty rows and empty columns in the new $n \times n$ board to ensure it remains peaceful. Any $m \times m$ square in this $n \times n$ board is also an $m \times m$ square in the original board, meaning it originally contained at least one rook. Because the entire $m \times m$ square is preserved, the rook it contained was not deleted. Thus, the configuration remains peaceful and has no empty $m \times m$ square.


Question: Define a “hook” to be a figure made up of six unit squares as shown below in the picture, or any of the figures obtained by applying rotations and reflections to this figure.Which $m 	imes n$ rectangles can be tiled by hooks?

The answer is that one requires: $m, n \notin \{1,2,5\}$, and ($3 \mid m$ or $3 \mid n$), and ($4 \mid m$ or $4 \mid n$).

First, we check that all rectangles satisfying these conditions can be tiled by $3 \times 4$ rectangles (and hence by hooks, since a $3 \times 4$ rectangle can be tiled by two hooks). If $3 \mid m$ and $4 \mid n$, this is clear. Else, suppose $12 \mid m$ but $3 \nmid n$ and $4 \nmid n$. Since $n \notin \{1,2,5\}$ and $n$ is not a multiple of $3$ or $4$, we must have $n \ge 7$. Any integer $n \ge 7$ can be written in the form $3x+4y$ for nonnegative integers $x$ and $y$. This permits a tiling of the $m \times n$ rectangle by partitioning it into $m \times 3$ and $m \times 4$ rectangles, each of which can be tiled by $3 \times 4$ rectangles (since $12 \mid m$).

We now prove these conditions are necessary. It is straightforward to see that $m,n \notin \{1,2,5\}$ is necessary, as no combination of hooks can tile rectangles with these dimensions. We thus turn our attention to the divisibility conditions.

Each hook has a "hole", and if we associate each hook with the one that fills its hole, we get a bijective pairing of hooks. Thus, the total number of cells is divisible by $12$, and the cells come grouped into two possible shapes, which we will call **tiles** (rotations and reflections permitted).

Suppose for the sake of contradiction there is a tileable rectangle that satisfies $12 \mid mn$ but fails the condition "($3 \mid m$ or $3 \mid n$) and ($4 \mid m$ or $4 \mid n$)". Since $12 \mid mn$, $3$ must divide $m$ or $n$, so the failure must be that $4 \nmid m$ and $4 \nmid n$. Since $4 \mid mn$, this implies $m$ and $n$ are both even but not divisible by $4$, so $m \equiv 2 \pmod 4$ and $n \equiv 2 \pmod 4$. Since $3$ divides $m$ or $n$, we may assume without loss of generality that $3 \mid m$. Then $m$ is a multiple of $6$ and $n$ is a multiple of $2$, so we can write $m = 6a$ and $n = 2b$. Since $m, n \equiv 2 \pmod 4$, both $a$ and $b$ must be odd. Thus, the problem is reduced to proving that if a $6a \times 2b$ rectangle is tiled by these tiles, then at least one of $a$ and $b$ is even (which would contradict that $a$ and $b$ are both odd).

Note that the tiles can be classified into two types:
**First type**: These tiles have exactly four columns, each with exactly three cells (an odd number). Moreover, all rows have an even number of cells (either $2$ or $4$).
**Second type**: Vice-versa. These tiles have exactly four rows, each with exactly three cells (an odd number). Moreover, all columns have an even number of cells (either $2$ or $4$).

We claim that any tiling uses an even number of each type of tile, which is enough to show that the total number of tiles, $ab$, is even, meaning at least one of $a$ and $b$ is even.

By symmetry, it suffices to prove there is an even number of first-type tiles. Color red every fourth column of the rectangle. The number of cells colored red is even, because the rectangle has an even height of $6a$, meaning each column contains an even number of cells.
The tiles of the second type have an even number of cells in every column, so they must cover an even number of red cells.
Since a tile of the first type is connected and has exactly four columns, it spans exactly four consecutive columns, meaning exactly one of its columns is colored red. Because each of its columns contains exactly three cells, a tile of the first type covers exactly $3$ red cells, which is an odd number.

Since the total number of red cells is even, and the second-type tiles cover an even number of red cells, the number of first-type tiles must be even. By a symmetric argument coloring every fourth row, the number of second-type tiles is also even. Hence, the total number of tiles $ab$ is even, completing the proof.

(Partial) 1. Observed that a 3×4 rectangle can be tiled by two hooks, and reduced the sufficiency part to tiling the m×n rectangle by 3×4 rectangles (e.g. via splitting n as 3x+4y when 12|m, or the symmetric case). 2. Proved a necessary area/divisibility condition such as: in any tiling the hooks can be paired into 12-cell “tiles”, hence 12 | mn (or equivalently 3|mn and 4|mn). 3. Gave a correct obstruction for small side lengths, e.g. proved that no rectangle with m∈{1,2,5} or n∈{1,2,5} can be tiled (with a valid local/invariant argument, not just examples). (Almost) 1. Established the full characterization m,n∉{1,2,5}, (3|m or 3|n), (4|m or 4|n), and proved the construction for all such rectangles by reducing to 3×4 blocks, but left a small gap in the decomposition step (e.g. the claim that every n≥7 can be written as 3x+4y, or a boundary case when swapping roles of m,n). 2. Proved necessity via pairing hooks into 12-cell tiles and the parity-of-types coloring argument (every 4th row/column) to rule out the case m≡n≡2 (mod 4), but did not fully justify one local structural claim about the two tile types (e.g. why each type covers an odd/even number of marked cells, or why a first-type tile meets exactly one marked column). 3. Solution is almost complete, but made minor mistakes which are not negligible.

Instruction:
After solving the problem, output a finite list of construction units.

Definition (construction units). Let the output be a finite set of 2D grids U={U_1,...,U_t} (represented by a Python list [U_1,...,U_t]). For every m×n rectangle R,
R is tileable by hooks if and only if R is tileable by translated/rotated/reflected copies of elements of U.

Hook-level encoding requirement. In each unit grid U_i, every positive integer label must appear exactly 6 times and must form (up to rotation/reflection) a hook; different labels represent different hooks. Each U_i must contain at least two hooks (i.e., at least 12 occupied cells).

Non-redundancy requirement. For all i!=j, U_i must not be tileable by translated/rotated/reflected copies of U_j alone (and vice versa).

Output format requirements:
- Include exactly one construction block and put ONLY a Python 3D list literal (a list of 2D grids) between the tags:
<construct>
<python_literal_3d_grid_list>
</construct>
- The 3D literal must have the form [grid1, grid2, ...], where each gridk is a Python list-of-lists (a 2D matrix) encoding one construction unit.
- Do not use code fences and do not put any extra text inside <construct>...</construct>.

import ast
import sys

BASE_UNITS = [
    [
        [1, 1, 1, 2],
        [1, 2, 1, 2],
        [1, 2, 2, 2],
    ],
    [
        [0, 1, 1, 1],
        [0, 1, 2, 1],
        [2, 1, 2, 0],
        [2, 2, 2, 0],
    ],
]

def _is_int(value) -> bool:
    return isinstance(value, int) and not isinstance(value, bool)

def _fail() -> None:
    print("False")
    raise SystemExit(0)

def _parse_units(text: str):
    try:
        value = ast.literal_eval(text)
    except Exception:
        _fail()
    if not isinstance(value, list) or len(value) != 2:
        _fail()
    return value

def _validate_grid(grid):
    if not isinstance(grid, list) or not grid:
        _fail()
    if any(not isinstance(row, list) for row in grid):
        _fail()
    width = len(grid[0])
    if width == 0 or any(len(row) != width for row in grid):
        _fail()

    has_positive = False
    for row in grid:
        for value in row:
            if not _is_int(value) or value < 0:
                _fail()
            if value > 0:
                has_positive = True
    if not has_positive:
        _fail()
    return grid

def _trim_zero_border(grid):
    rows = [i for i, row in enumerate(grid) if any(v > 0 for v in row)]
    cols = [j for j in range(len(grid[0])) if any(grid[i][j] > 0 for i in range(len(grid)))]
    if not rows or not cols:
        _fail()
    r1, r2 = rows[0], rows[-1]
    c1, c2 = cols[0], cols[-1]
    return [row[c1 : c2 + 1] for row in grid[r1 : r2 + 1]]

def _rotate_90(grid):
    h = len(grid)
    w = len(grid[0])
    return [[grid[h - 1 - r][c] for r in range(h)] for c in range(w)]

def _reflect(grid):
    return [list(reversed(row)) for row in grid]

def _all_transforms(grid):
    seen = set()
    out = []
    cur = [row[:] for row in grid]
    for _ in range(4):
        for cand in (cur, _reflect(cur)):
            sig = tuple(tuple(v for v in row) for row in cand)
            if sig not in seen:
                seen.add(sig)
                out.append(cand)
        cur = _rotate_90(cur)
    return out

def _canonical_label_signature(grid):
    mapping = {}
    next_label = 1
    out = []
    for row in grid:
        new_row = []
        for value in row:
            if value == 0:
                new_row.append(0)
                continue
            if value not in mapping:
                mapping[value] = next_label
                next_label += 1
            new_row.append(mapping[value])
        out.append(tuple(new_row))
    return tuple(out)

def _unit_signature(grid):
    checked = _validate_grid(grid)
    trimmed = _trim_zero_border(checked)
    return _canonical_label_signature(trimmed)

def _variant_signature_set(base_grid):
    trimmed = _trim_zero_border(base_grid)
    return {
        _canonical_label_signature(_trim_zero_border(variant))
        for variant in _all_transforms(trimmed)
    }

def _can_match_bijection(candidate_to_base):
    used = set()

    def _dfs(i: int) -> bool:
        if i == len(candidate_to_base):
            return True
        for base_idx in candidate_to_base[i]:
            if base_idx in used:
                continue
            used.add(base_idx)
            if _dfs(i + 1):
                return True
            used.remove(base_idx)
        return False

    return _dfs(0)

def main() -> None:
    text = sys.stdin.read()
    if not text:
        _fail()

    units = _parse_units(text)

    base_variants = [_variant_signature_set(unit) for unit in BASE_UNITS]

    candidate_matches = []
    for unit in units:
        sig = _unit_signature(unit)
        matches = [idx for idx, variants in enumerate(base_variants) if sig in variants]
        if not matches:
            _fail()
        candidate_matches.append(matches)

    if not _can_match_bijection(candidate_matches):
        _fail()

    print("True")

if __name__ == "__main__":
    main()

The answer is that one requires:
$$
\bigl({1,2,5}\cap{m,n}=\varnothing\bigr)\ \land\ \bigl(3\mid m\ \lor\ 3\mid n\bigr)\ \land\ \bigl(4\mid m\ \lor\ 4\mid n\bigr).
$$ e.g., (m,n) = (24,19), (10,12), (6,8).


Question: Let $m$ and $n$ be positive integers. A circular necklace contains $mn$ beads, each either red or blue. It turned out that no matter how the necklace was cut into $m$ blocks of $n$ consecutive beads, each block had a distinct number of red beads. Determine, with proof, all possible values of the ordered pair $(m, n)$.

The possible values for the ordered pair $(m, n)$ are exactly those for which $m \leq n+1$.

**Necessity:**
Suppose a necklace of $mn$ beads is cut into $m$ blocks of length $n$. The number of red beads in any block must be an integer between $0$ and $n$ inclusive. There are exactly $n+1$ possible values for this count. Since each of the $m$ blocks must have a distinct number of red beads, we must have $m \leq n+1$.

**Construction for $m = n+1$:**
For concreteness, here is the construction for $n=4$, which naturally generalizes. The beads are listed in reading order as an array with $n+1$ rows and $n$ columns. Four of the blue beads have been labeled $B_1, \dots, B_n$ to make them easier to track as they move.
\[
T_0 = \begin{bmatrix}
\color{red}R & \color{red}R & \color{red}R & \color{red}R \\
\color{red}R & \color{red}R & \color{red}R & \color{blue}B_1 \\
\color{red}R & \color{red}R & \color{blue}B & \color{blue}B_2 \\
\color{red}R & \color{blue}B & \color{blue}B & \color{blue}B_3 \\
\color{blue}B & \color{blue}B & \color{blue}B & \color{blue}B_4
\end{bmatrix}.
\]
To prove this construction works, it suffices to consider the $n$ cuts $T_0, T_1, \dots, T_{n-1}$, where $T_i$ represents the blocks formed by starting the cut $i$ beads later. We must show that each row in $T_i$ has a distinct red count:
\[
T_1 = \begin{bmatrix}
\color{red}R & \color{red}R & \color{red}R & \color{red}R \\
\color{red}R & \color{red}R & \color{blue}B_1 & \color{red}R \\
\color{red}R & \color{blue}B & \color{blue}B_2 & \color{red}R \\
\color{blue}B & \color{blue}B & \color{blue}B_3 & \color{blue}B \\
\color{blue}B & \color{blue}B & \color{blue}B_4 & \color{red}R
\end{bmatrix} \quad
T_2 = \begin{bmatrix}
\color{red}R & \color{red}R & \color{red}R & \color{red}R \\
\color{red}R & \color{blue}B_1 & \color{red}R & \color{red}R \\
\color{blue}B & \color{blue}B_2 & \color{red}R & \color{blue}B \\
\color{blue}B & \color{blue}B_3 & \color{blue}B & \color{blue}B \\
\color{blue}B & \color{blue}B_4 & \color{red}R & \color{red}R
\end{bmatrix} \quad
T_3 = \begin{bmatrix}
\color{red}R & \color{red}R & \color{red}R & \color{red}R \\
\color{blue}B_1 & \color{red}R & \color{red}R & \color{blue}B \\
\color{blue}B_2 & \color{red}R & \color{blue}B & \color{blue}B \\
\color{blue}B_3 & \color{blue}B & \color{blue}B & \color{blue}B \\
\color{blue}B_4 & \color{red}R & \color{red}R & \color{red}R
\end{bmatrix}.
\]
We can construct a table showing for each $0 \le k \le n$ the number of red beads in row $k$ of $T_i$ (where the rows are indexed from $k=n$ at the top down to $k=0$ at the bottom):
\[
\begin{array}{c|cccc}
k & T_0 & T_1 & T_2 & T_3 \\ \hline
k=4 & 4 & 4 & 4 & 4 \\
k=3 & 3 & 3 & 3 & 2 \\
k=2 & 2 & 2 & 1 & 1 \\
k=1 & 1 & 0 & 0 & 0 \\
k=0 & 0 & 1 & 2 & 3
\end{array}.
\]
This suggests the following explicit formula for the number of red cells in row $k$ of $T_i$ in the general construction:
\[
\#(\text{red cells in row } k \text{ of } T_i) =
\begin{cases}
k & \text{if } k > i \\
k-1 & \text{if } i \ge k > 0 \\
i & \text{if } k = 0.
\end{cases}
\]
This can be checked straightforwardly. Let the rows of $T_0$ be indexed by $k = n, n-1, \dots, 0$ from top to bottom. The $k$-th row consists of $k$ red beads followed by $n-k$ blue beads. When we shift the cut by $i$ beads ($0 \le i \le n-1$), the new $k$-th block consists of the last $n-i$ beads of the original $k$-th row and the first $i$ beads of the original $(k-1)$-th row (with indices taken modulo $n+1$, so the row below $k=0$ is $k=n$).

We count the number of red beads in this new block:
- For $k > 0$, the original $k$-th row starts with $k$ red beads. Removing the first $i$ beads leaves $\max(0, k-i)$ red beads. The original $(k-1)$-th row starts with $k-1$ red beads. Taking its first $i$ beads gives $\min(i, k-1)$ red beads. The total number of red beads is $\max(0, k-i) + \min(i, k-1)$.
  - If $k > i$, this sum is $(k-i) + i = k$.
  - If $0 < k \le i$, this sum is $0 + (k-1) = k-1$.
- For $k = 0$, the original $0$-th row has no red beads, so the first part contributes $0$. The next row in reading order is the $n$-th row, which consists of $n$ red beads. Since $i \le n-1 < n$, taking its first $i$ beads gives exactly $i$ red beads. The total is $i$.

Thus, for each $i$, the sequence of red counts from bottom to top (as $k$ goes from $0$ to $n$) is exactly $(i, 0, 1, \dots, i-1, i+1, \dots, n)$. These are exactly the integers from $0$ to $n$, meaning the counts are all distinct. This completes the construction for $m = n+1$.

**Construction for general $m \le n+1$:**
Fix $m \le n+1$. Take the construction for $(m, m-1)$ and add $n+1-m$ cyan beads to the start of each row. For example, if $n = 7$ and $m = 5$, then the new construction is
\[
T = \begin{bmatrix}
\color{cyan}C & \color{cyan}C & \color{cyan}C & \color{red}R & \color{red}R & \color{red}R & \color{red}R \\
\color{cyan}C & \color{cyan}C & \color{cyan}C & \color{red}R & \color{red}R & \color{red}R & \color{blue}B_1 \\
\color{cyan}C & \color{cyan}C & \color{cyan}C & \color{red}R & \color{red}R & \color{blue}B & \color{blue}B_2 \\
\color{cyan}C & \color{cyan}C & \color{cyan}C & \color{red}R & \color{blue}B & \color{blue}B & \color{blue}B_3 \\
\color{cyan}C & \color{cyan}C & \color{cyan}C & \color{blue}B & \color{blue}B & \color{blue}B & \color{blue}B_4
\end{bmatrix}.
\]
This construction works for the same reason. Let $c = n+1-m$. We check the shifts by $i$ beads for $0 \le i \le n-1$:
- If $0 \le i \le c$, shifting by $i$ beads removes $i$ cyan beads from the start of each row and brings in $i$ cyan beads from the start of the next row. The red beads in each row remain exactly the same as in the unshifted configuration, so the red counts are distinct.
- If $c < i \le n-1$, let $i' = i - c$. Shifting by $i$ beads removes all $c$ cyan beads and the first $i'$ beads of the $(m, m-1)$ construction from each row. It then appends $c$ cyan beads and the first $i'$ beads of the $(m, m-1)$ construction from the next row. The number of red beads in each block is exactly the same as shifting the $(m, m-1)$ construction by $i'$ beads. Since $1 \le i' \le m-2$, the $(m, m-1)$ construction guarantees that the red counts are distinct.

Treating cyan as a shade of blue, this provides a valid configuration for any $m \le n+1$, completing the proof.

(Partial) 1. Proved the necessary bound m≤n+1 by noting each block has between 0 and n red beads (only n+1 possible counts) and the m counts must be distinct. 2. Gave a correct explicit coloring for the extremal case m=n+1 (e.g. arranging blocks with k red beads for k=0,1,…,n in some cyclic manner) and checked it works for at least the unshifted cut. 3. Made a correct reduction that it suffices to check only the n cyclic shifts of a fixed “row-by-row” representation of the necklace (i.e. start the cut i beads later for i=0,…,n−1). (Almost) 1. Proved m≤n+1 and gave a construction for all m≤n+1, but left a minor gap in verifying that the construction works for every shift (e.g. did not fully justify the red-count formula after shifting by i). 2. Gave a complete construction for m=n+1 and correctly derived the red counts for k>0, but did not handle the wrap-around/boundary case k=0 (or the case i=0 vs i=n−1) needed to conclude all counts 0,1,…,n occur. 3. Constructed the general m≤n+1 case by modifying the m=n+1 (or (m,m−1)) construction, but did not justify why the added beads do not create equal red counts for some range of shifts.

Instruction:
For the specific case $n=4$, determine the maximum possible integer $m$ for which such a necklace exists. Provide an explicit construction of the necklace for this maximum $m$.

Construction requirements:
- Output a single Python list of lists representing the $m \times n$ blocks of the necklace.
- Use 1 for red beads and 0 for blue beads.
- The matrix represents the necklace cut into $m$ blocks of size $n$ at a specific starting position (shift 0). Flattening the matrix row-by-row (row 0, then row 1, etc.) yields the full circular sequence of beads.
- The construction must satisfy the problem condition: for every possible way of cutting the circular necklace into $m$ blocks of length $n$ (which corresponds to cyclically shifting the flattened sequence by any offset $k \in \{0, \dots, n-1\}$ before partitioning), the number of red beads (1s) in the resulting $m$ blocks must be pairwise distinct.
- The verifier will check that you have used the largest possible value for $m$ and that the distinctness condition holds for all valid cuts.
- Put ONLY the Python literal inside the construction block.

import sys
import ast

def verify_necklace():
    """
    Verifies the construction of a circular necklace with mn beads (m blocks of size n)
    such that for any cut into m blocks, the number of red beads in each block is distinct.
    Target case: n=4, max m=5.
    """
    # Read full stdin as the construction payload
    try:
        input_data = sys.stdin.read().strip()
        if not input_data:
            print("Error: Input is empty.")
            return

        # Parse strictly as a Python literal (expecting a list of lists)
        matrix = ast.literal_eval(input_data)
    except Exception as e:
        print(f"Error: Failed to parse input as a Python literal. {e}")
        return

    # 1. Parse the input as a Python list of lists (matrix).
    if not isinstance(matrix, list):
        print("Error: Input must be a list of lists.")
        return

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

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

    # 3. Verify that the number of columns (n) is exactly 4.
    target_n = 4
    # 4. Verify that the number of rows (m) is exactly 5 (since max m <= n + 1).
    target_m = 5

    m = len(matrix)
    if m != target_m:
        print(f"Error: Expected {target_m} rows (m) for the maximum possible size, but found {m}.")
        return

    flat_necklace = []

    # Validate dimensions and values
    for r_idx, row in enumerate(matrix):
        if len(row) != target_n:
            print(f"Error: Row {r_idx} has length {len(row)}, expected {target_n}.")
            return

        for c_idx, val in enumerate(row):
            # 2. Verify that the matrix contains only integers 0 and 1.
            # Strict check: reject bools or other types to ensure compliance with 'integers 0 and 1'
            if type(val) is not int or val not in (0, 1):
                print(f"Error: Invalid value {val} at row {r_idx}, col {c_idx}. Only integers 0 and 1 are allowed.")
                return

        # 5. Flatten the matrix into a single 1D list representing the circular necklace.
        flat_necklace.extend(row)

    n = target_n
    total_beads = m * n

    # 6. Iterate through shift offsets s from 0 to n-1
    # Checking s in range(n) is sufficient because shifting by n cyclically permutes the blocks,
    # preserving the set of sums.
    for s in range(n):
        block_sums = []

        # a. Generate the m blocks for this shift
        # The cut corresponds to blocks starting at indices s, s+n, s+2n, ... (modulo total length)
        for i in range(m):
            start_index = (s + i * n) % total_beads

            # b. Calculate the sum (number of 1s) for each of the m blocks.
            current_sum = 0
            for k in range(n):
                idx = (start_index + k) % total_beads
                current_sum += flat_necklace[idx]

            block_sums.append(current_sum)

        # c. Verify that these m sums are all distinct (i.e., the set of sums has size m).
        if len(set(block_sums)) != m:
            print(f"Error: Condition failed for shift offset {s}. Block sums are {block_sums}, which are not pairwise distinct.")
            return

    # 7. If all checks pass, the construction is valid.
    print("True")

if __name__ == "__main__":
    verify_necklace()


Question: There are $2022$ users on a social network called Mathbook, and some of them are Mathbook-friends. (On Mathbook, friendship is always mutual and permanent.)\nStarting now, Mathbook will only allow a new friendship to be formed between two users if they have at least two friends in common. What is the minimum number\nof friendships that must already exist so that every user could eventually become friends with every other user?

With $2022$ replaced by $n$, the answer is $\\lceil 3/2n \\rceil - 2$.

Instruction:
For the specific case $n=202$ (where the users are represented by integers from 1 to 202), provide an explicit construction of the initial friendships that achieves the minimum possible number of friendships.

Construction requirements:
- Output a single Python expression that evaluates to a list of 2-tuples of integers.
- Each tuple `(u, v)` represents a friendship between user `u` and user `v`.
- The users must be integers from 1 to 202.
- The number of friendships in your list must be exactly the minimum required for $n=202$.
- The provided initial friendships must allow every user to eventually become friends with every other user under the rule that a new friendship can be formed if two users have at least two friends in common.
- The output must be a single, valid Python expression (e.g., using list comprehensions) that can be evaluated directly by `eval()`. Do not include comments, assignments, or any other code.
- Put ONLY the Python expression inside the construction block.

import sys
import collections

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

    try:
        # Evaluate the payload. The instruction explicitly allows Python expressions
        # like list comprehensions, so eval() is used.
        edges_raw = eval(payload)
    except Exception as e:
        print(f"Failed to evaluate input: {e}")
        return

    try:
        edges_list = list(edges_raw)
    except Exception as e:
        print(f"Output is not an iterable: {e}")
        return

    normalized_edges = set()
    for edge in edges_list:
        if not isinstance(edge, tuple) or len(edge) != 2:
            print(f"Invalid edge format: {edge}. Must be a 2-tuple.")
            return
        u, v = edge
        if not (isinstance(u, int) and isinstance(v, int)):
            print(f"Edge vertices must be integers: {edge}")
            return
        if not (1 <= u <= 202 and 1 <= v <= 202):
            print(f"Vertices must be between 1 and 202: {edge}")
            return
        if u == v:
            print(f"Self-loops are not allowed: {edge}")
            return
        # Normalize edge to be undirected
        normalized_edges.add(tuple(sorted((u, v))))

    if len(normalized_edges) != 301:
        print(f"Expected exactly 301 unique friendships, got {len(normalized_edges)}.")
        return

    # Simulate friendship formation
    adj = {i: set() for i in range(1, 203)}
    for u, v in normalized_edges:
        adj[u].add(v)
        adj[v].add(u)

    # Precompute common friends for all pairs
    common_friends = {i: {j: 0 for j in range(1, 203)} for i in range(1, 203)}
    for i in range(1, 203):
        for j in range(i + 1, 203):
            c = len(adj[i].intersection(adj[j]))
            common_friends[i][j] = c
            common_friends[j][i] = c

    queue = collections.deque()
    in_queue_or_adj = {i: set(adj[i]) for i in range(1, 203)}

    # Enqueue all pairs that initially have >= 2 common friends
    for i in range(1, 203):
        for j in range(i + 1, 203):
            if j not in in_queue_or_adj[i] and common_friends[i][j] >= 2:
                queue.append((i, j))
                in_queue_or_adj[i].add(j)
                in_queue_or_adj[j].add(i)

    # Process queue to propagate new friendships
    while queue:
        u, v = queue.popleft()

        # For every existing neighbor w of u, u acts as a new common friend for w and v
        for w in adj[u]:
            common_friends[w][v] += 1
            common_friends[v][w] += 1
            if v not in in_queue_or_adj[w] and common_friends[w][v] >= 2:
                queue.append((w, v))
                in_queue_or_adj[w].add(v)
                in_queue_or_adj[v].add(w)

        # For every existing neighbor w of v, v acts as a new common friend for w and u
        for w in adj[v]:
            common_friends[w][u] += 1
            common_friends[u][w] += 1
            if u not in in_queue_or_adj[w] and common_friends[w][u] >= 2:
                queue.append((w, u))
                in_queue_or_adj[w].add(u)
                in_queue_or_adj[u].add(w)

        # Add the validated friendship to the active adjacency list
        adj[u].add(v)
        adj[v].add(u)

    # Verify if the graph has become complete
    total_edges = sum(len(neighbors) for neighbors in adj.values()) // 2
    expected_edges = 202 * 201 // 2
    if total_edges != expected_edges:
        print(f"Graph does not become complete. Final edge count: {total_edges} instead of {expected_edges}.")
        return

    print("True")

if __name__ == "__main__":
    verify()

(Partial) 1. Gave (and correctly justified) a construction achieving at most \(\lceil \tfrac32 n\rceil-2\) initial friendships, e.g. for even \(n\) an edge \(ab\) plus \(\frac n2-1\) disjoint 4-cycles sharing \(ab\), and explained how the allowed rule can then generate all remaining edges. 2. Reformulated the process in graph terms (new edge \(uv\) can be added once \(u,v\) have at least two common neighbors; equivalently one may complete a \(C_4\) to a \(K_4\)), and made a correct nontrivial structural observation about how the closure operation behaves (e.g. completing one \(K_4\) creates two common neighbors \(a,b\) that then force edges between vertices from different blocks). 3. Proved a meaningful lower-bound lemma, e.g. established a bound of the form \(e(G)\ge c n - O(1)\) (with some explicit \(c>1\)) necessary for eventual completion to \(K_n\), or proved the key invariant for cliques \(K\) that \(\theta(K)\ge \frac{3|K|}{2}-2\) in the base case and at least one nontrivial merge case. (Almost) 1. Presented the full lower-bound strategy via the clique-merging/labeling algorithm (or an equivalent invariant argument), but left a localized gap in the inductive step when multiple cliques merge (e.g. did not fully justify why the merge step is legal under the “two common friends” rule, or omitted one subcase of how \(\theta\) changes). 2. Proved the sharp lower bound \(e(G)\ge \lceil\tfrac32 n\rceil-2\) and gave the sharp construction, but did not handle one parity/boundary case (typically the odd \(n\) adjustment with the extra vertex adjacent to \(a,b\)), or made a small counting slip that affects only that case. 3. Solution is almost complete, but made minor mistakes which are not negligible.

With $2022$ replaced by $n$, the answer is $\left\lceil \frac{3}{2} n \right\rceil - 2$.

In standard graph theory terms: starting from a graph $G$ on $n$ vertices, we are allowed to take any $C_4$ in the graph and complete it to a $K_4$. The problem asks for the minimum number of edges needed so that this operation lets us transform $G$ to $K_n$.

For even $n$, start with an edge $ab$, and then create $n/2-1$ copies of $C_4$ that use $ab$ as an edge. This can be completed into $K_n$ by first completing the $n/2-1$ copies of $C_4$ into $K_4$. Then, since any two vertices in different copies of $K_4$ share $a$ and $b$ as common friends, we can connect all remaining pairs of vertices, finishing up the $K_n$. The construction for odd $n$ is the same except with one extra vertex $c$ which is connected to both $a$ and $b$.

Notice that additional operations or connections can never hurt. So we will describe a *specific* algorithm that performs operations on the graph until no more operations are possible. This means that if this algorithm terminates with anything other than $G = K_n$, the graph was never completable to $K_n$ to begin with.

The algorithm uses the following data: it keeps a list $\mathcal{C}$ of cliques of $G$, and a labeling $\mathcal{L} \colon E(G) \to \mathcal{C}$ which assigns to every edge one of the cliques that contains it. Initially, $\mathcal{C}$ consists of one $K_2$ for every edge of $G$, and each edge is labeled in the obvious way.

At each step, the algorithm arbitrarily takes any $C_4 = abcd$ whose four edges $ab$, $bc$, $cd$, $da$ do not all have the same label. Consider these labels that appear (at least two, and up to four), and let $V$ be the union of all vertices in any of these $2$ to $4$ cliques. Do the following graph operations: connect $ac$ and $bd$, then connect every vertex in $V \setminus \{a,b,c,d\}$ to each of $\{a,b,c,d\}$. Finally, complete this to a clique on $V$.

Update $\mathcal{C}$ by merging these $2$ to $4$ cliques into a single clique $K_V$. Update $\mathcal{L}$ by replacing every edge that was labeled with one of these cliques with the label $K_V$. Also, update every *newly* created edge to have label $K_V$. However, if there were existing edges not labeled with one of the merged cliques, then we do *not* update these.

Stop once every $C_4$ has only one label appearing among its edges. When this occurs, no operations are possible at all on the graph. As we remarked, if the graph is going to be completable to $K_n$ at all, then this algorithm must terminate with $\mathcal{C} = \{K_n\}$. We will use this to prove our bound.

We proceed by induction in the following way. For a clique $K$, let $\theta(K)$ denote the number of edges of the *original* graph $G$ which are labeled by $K$ (this does *not* include new edges added by the algorithm); hence the problem amounts to estimating how small $\theta(K_n)$ can be.

At any point in the operation, if $K$ is a clique in the cover $\mathcal{C}$, then
$$ \theta(K) \ge \frac{3 |K|}{2} - 2 $$
where $|K|$ is the number of vertices in $K$. We prove this by induction on the time step of the algorithm. The base case is clear, because initially $K$ is just a single edge of $G$, so $\theta(K) = 1$ and $|K| = 2$, which satisfies $1 \ge \frac{3(2)}{2} - 2 = 1$. The inductive step proceeds by casework based on how the merge occurred.


Question: Let $n \\geq 2$ be an integer. An $n \\times n$ board is initially empty. Each minute, you may perform one of three moves: • If there is an L-shaped tromino region of three cells without stones on the board (see figure; rotations not allowed), you may place a stone in each of those cells. • If all cells in a column have a stone, you may remove all stones from that column. • If all cells in a row have a stone, you may remove all stones from that row. For which $n$ is it possible that, after some non-zero number of moves, the board has no stones?

Instruction:
For $n = 81$, provide an explicit sequence of moves to completely empty the board after a non-zero number of moves.

Construction format requirements:
- Output a single Python expression that evaluates to a list of moves.
- Each move should be a tuple of one of the following forms:
  - `("place", r, c)`: Places an L-tromino such that it covers the cells `(r, c)`, `(r-1, c)`, and `(r, c+1)`. (Here `r` is the 0-based row index and `c` is the 0-based column index, $0 \le r, c < 81$).
  - `("row", r)`: Removes all stones from row `r` (requires row `r` to be completely full).
  - `("col", c)`: Removes all stones from column `c` (requires column `c` to be completely full).
- The board is $81 \times 81$, 0-indexed.
- The sequence must be valid: you can only place an L-tromino if all three of its cells are currently empty, and you can only clear a row/column if it is completely full.
- After all moves are executed, the board must be completely empty, and the sequence of moves must be non-empty.
- The output must be a single, valid Python expression (e.g., using list concatenations and comprehensions).
- Put ONLY the Python expression inside the `<construct>` block, with no formatting, no markdown code blocks, and no explanations.

[("place", 3*i+1, 3*j+1) for i in range(27) for j in range(27)] + [("place", 3*i+2, 3*j) for i in range(27) for j in range(27)] + [("row", 3*i+1) for i in range(27)] + [("place", 3*i+1, 3*j) for i in range(27) for j in range(27)] + [("col", 3*j) for j in range(27)] + [("col", 3*j+1) for j in range(27)]

import sys

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

    # Evaluate the payload as a Python expression
    try:
        moves = eval(raw_payload)
    except Exception as e:
        print(f"Failed to parse input: {e}")
        return

    # Verify that the parsed object is a non-empty list of tuples
    if not isinstance(moves, list):
        print("Input must evaluate to a list.")
        return
    if len(moves) == 0:
        print("The sequence of moves must be non-empty.")
        return

    # Initialize an 81x81 boolean matrix to False (empty board)
    board = [[False for _ in range(81)] for _ in range(81)]

    # Iterate through each move in the list
    for idx, move in enumerate(moves):
        if not isinstance(move, tuple):
            print(f"Move at index {idx} is not a tuple: {move}")
            return
        if len(move) == 0:
            print(f"Move at index {idx} is empty.")
            return

        op = move[0]
        if op == "place":
            if len(move) != 3:
                print(f"Move at index {idx} has invalid length for 'place': {move}")
                return
            _, r, c = move
            if type(r) is not int or type(c) is not int:
                print(f"Move at index {idx} has non-integer coordinates: {move}")
                return
            if not (1 <= r < 81 and 0 <= c < 80):
                print(f"Move at index {idx} has out-of-bounds coordinates for 'place': {move}")
                return
            if board[r][c] or board[r-1][c] or board[r][c+1]:
                print(f"Move at index {idx} attempts to place on non-empty cells: {move}")
                return
            # Place the L-tromino
            board[r][c] = True
            board[r-1][c] = True
            board[r][c+1] = True

        elif op == "row":
            if len(move) != 2:
                print(f"Move at index {idx} has invalid length for 'row': {move}")
                return
            _, r = move
            if type(r) is not int:
                print(f"Move at index {idx} has non-integer row index: {move}")
                return
            if not (0 <= r < 81):
                print(f"Move at index {idx} has out-of-bounds row index: {move}")
                return
            if not all(board[r][c] for c in range(81)):
                print(f"Move at index {idx} attempts to clear row {r} which is not completely full.")
                return
            # Remove all stones from row r
            for c in range(81):
                board[r][c] = False

        elif op == "col":
            if len(move) != 2:
                print(f"Move at index {idx} has invalid length for 'col': {move}")
                return
            _, c = move
            if type(c) is not int:
                print(f"Move at index {idx} has non-integer column index: {move}")
                return
            if not (0 <= c < 81):
                print(f"Move at index {idx} has out-of-bounds column index: {move}")
                return
            if not all(board[r][c] for r in range(81)):
                print(f"Move at index {idx} attempts to clear column {c} which is not completely full.")
                return
            # Remove all stones from column c
            for r in range(81):
                board[r][c] = False

        else:
            print(f"Move at index {idx} has an invalid operation: {move}")
            return

    # After processing all moves, verify that every cell in the board is False
    for r in range(81):
        for c in range(81):
            if board[r][c]:
                print(f"After all moves, the board is not empty. Cell ({r}, {c}) contains a stone.")
                return

    print("True")

if __name__ == "__main__":
    verify()

(Partial) 1. Established the correct invariant/encoding of a position by a polynomial (or equivalent linear functional) so that placing an L-tromino adds a translate of 1+x+y and clearing a full row/column corresponds to adding a multiple of 1+x+…+x^{n-1} or 1+y+…+y^{n-1}. 2. Proved the key roots-of-unity test: for any n-th roots of unity z1,z2≠1 one must have (1+z1+z2)P(z1,z2)=0, and observed that if 3∤n then 1+z1+z2≠0 for all such z1,z2. 3. Gave a correct explicit move sequence returning to empty for n=3, or clearly reduced the construction for general n divisible by 3 to a tiling/concatenation of a 3×3 gadget. (Almost) 1. Proved the impossibility for 3∤n via the polynomial/roots-of-unity method, but left a minor gap in justifying why existence of a nontrivial return-to-empty sequence yields a nonzero P with deg_x,deg_y≤n−2 in the stated ideal condition. 2. Completed the vanishing-on-S×S argument for 3∤n, but the final contradiction (e.g., the application of Combinatorial Nullstellensatz / degree-counting to deduce P≡0) is insufficiently justified or misses one boundary degree case. 3. Gave a correct construction for all n with 3|n and a correct outline of the converse, but omitted the check that rotations are not allowed in the L-move (or handled only one orientation) while claiming the full result.

The answer is $n$ such that $3 \mid n$.

**Construction for $3 \mid n$:**
For $n=3$, a fairly straightforward sequence of moves exists to return to the empty board. This construction can then be extended to any $n$ such that $3 \mid n$.

**Polynomial-based proof of converse:**
Assume for contradiction that $3 \nmid n$. We will show the task is impossible even if we allow stones to have real weights in our process.

A valid elimination corresponds to a non-zero polynomial $P \in \mathbb{R}[x,y]$ such that
\begin{align*}
\deg_x P &\le n-2 \\
\deg_y P &\le n-2 \\
(1+x+y) P(x,y) &\in \left\langle 1+x+\dots+x^{n-1}, \; 1+y+\dots+y^{n-1}\right\rangle.
\end{align*}
(Here $\langle \dots \rangle$ is an ideal of $\mathbb{R}[x,y]$.)

In particular, if $S$ is the set of $n$-th roots of unity other than $1$, we should have
\[
(1+z_1+z_2) P( z_1, z_2 ) = 0
\]
for any $z_1 , z_2 \in S$. Since $3 \nmid n$, it follows that $1+z_1+z_2 \neq 0$ always. So $P$ vanishes on $S \times S$, a contradiction to the bounds on $\deg P$ (by, say, the Combinatorial Nullstellensatz on any nonzero term).