implementation

CF 958A1 - Death Stars (easy)

Rating: 1400
Tags: implementation
Model: gpt-5-3-mini
Solve time: 1m 10s
Verified: yes (1/1 samples)

Solution

Problem Understanding

We are given two square grids of size $N \times N$, where each cell is either a star or empty space. The task is to determine whether these two grids can represent the same pattern if we are allowed to transform one of them using rigid symmetries of a square.

The allowed transformations correspond to physically rotating the grid by 90, 180, or 270 degrees, or flipping it horizontally or vertically. After applying any sequence of these operations, if one grid can be made identical to the other cell by cell, then they are considered the same configuration.

From a computational perspective, each map is just a matrix of characters. The question becomes whether two matrices are equivalent under a small finite set of geometric transformations.

The constraint $N \le 10$ is extremely small. Even operations that are cubic or involve trying all transformations explicitly are trivial at this scale. This immediately tells us we do not need hashing tricks, search pruning, or asymptotically optimal matching techniques. Direct simulation of all candidate transformations is sufficient.

A subtle point is that transformations are discrete but can overlap. For example, a horizontal flip followed by a vertical flip is equivalent to a 180-degree rotation. If one attempts to implement only a subset of transformations without understanding closure, it is easy to miss a valid match.

A few edge situations are worth keeping in mind. If both grids are already identical, no transformation is needed and the answer should be Yes. If the pattern has high symmetry, such as all cells being identical, every transformation preserves equality, which can incorrectly mask bugs in an implementation that does not actually apply transformations correctly. Another failure case is attempting to compare only rotations without flips; for example, a pattern and its mirror image are not related by rotation alone.

Approaches

The brute-force idea is straightforward: enumerate every transformation allowed on the first grid, apply it, and compare the result with the second grid. Since the grid is at most $10 \times 10$, each comparison costs $O(N^2)$, which is negligible. The only question is how many transformations must be checked.

A naive mistake would be to try arbitrary sequences of flips and rotations, which could generate an unbounded search space. However, the structure of a square’s symmetries is closed and finite. Any sequence of allowed rotations and flips reduces to one of exactly eight distinct transformations: four rotations (0, 90, 180, 270 degrees) and their mirrored counterparts. This is the dihedral group of the square.

The key observation is that instead of exploring sequences, we can directly construct each of these eight resulting grids and compare them. This turns what could have been a combinatorial search into a constant-factor enumeration.

Approach Time Complexity Space Complexity Verdict
Brute Force (all transformation sequences) O(∞ conceptual, effectively unbounded) O(N^2) Too slow / ill-defined
Try all 8 symmetries explicitly O(8 · N^2) O(N^2) Accepted

Algorithm Walkthrough

We explicitly generate all distinct transformed versions of the first grid and check whether any matches the second grid.

  1. Read the two grids and store them as lists of strings.
  2. Define a function that rotates a grid 90 degrees clockwise by mapping cell $(i, j)$ to $(j, N-1-i)$. This is the fundamental building block, since all rotations can be derived from repeated application.
  3. Construct the four rotations of the grid: 0 degrees (original), 90 degrees, 180 degrees, and 270 degrees. Each rotation is obtained from the previous one using the same mapping rule.
  4. For each rotated grid, also construct its horizontal reflection. A horizontal flip reverses each row.
  5. Each rotation and its flipped version gives a total of 8 candidate grids.
  6. Compare each candidate grid with the target grid. If any match is identical cell-by-cell, output Yes immediately.
  7. If no transformation matches after checking all 8 possibilities, output No.

The reason we do not mix arbitrary sequences is that any combination of allowed operations always collapses into one of these eight canonical forms. Enumerating them directly avoids redundancy and guarantees completeness.

Why it works

The transformations of a square form a closed group of size 8. Every allowed operation is an element of this group, and compositions never produce anything outside it. By generating one representative for each group element, we ensure that every possible transformed version of the first grid is considered exactly once. This guarantees that if an equivalence exists under any allowed sequence of operations, it will appear in our enumeration.

Python Solution

import sys
input = sys.stdin.readline

def rotate(grid):
    n = len(grid)
    return ["".join(grid[n - 1 - r][c] for r in range(n)) for c in range(n)]

def flip_h(grid):
    return grid[::-1]

def solve():
    n = int(input())
    a = [input().strip() for _ in range(n)]
    b = [input().strip() for _ in range(n)]

    variants = []

    cur = a
    for _ in range(4):
        variants.append(cur)
        variants.append(flip_h(cur))
        cur = rotate(cur)

    for v in variants:
        if v == b:
            print("Yes")
            return

    print("No")

if __name__ == "__main__":
    solve()

The solution relies on two primitive transformations. The rotation function builds a new grid by reading columns of the original grid bottom-to-top, which corresponds to a 90-degree clockwise turn. The flip function reverses row order, producing a vertical reflection; depending on interpretation, this is sufficient because rotations combined with this flip generate all mirror cases.

We generate rotations iteratively so that each new grid is derived from the previous one, avoiding repeated recomputation from the original. This keeps the implementation simple and reduces the chance of indexing mistakes.

The comparison is done by direct string equality of row lists, which is safe because each row fully encodes the grid structure.

Worked Examples

Example 1

Input grids:

A:

XO
OX

B:

OX
XO
Step Transformation Grid
1 Original XO / OX
2 Rotate 90 OX / XO
3 Compare matches B

This shows that a single rotation is sufficient to align the grids.

Example 2

Input grids:

A:

XO
OO

B:

OX
OO
Step Transformation Grid
1 Original XO / OO
2 Horizontal flip OX / OO
3 Compare matches B

This case demonstrates a mirror symmetry rather than rotation, showing why flips are necessary.

Complexity Analysis

Measure Complexity Explanation
Time O(N²) We build at most 8 grids, each requiring a full traversal of the matrix, and compare them once
Space O(N²) We store a constant number of grids of size $N \times N$

Since $N \le 10$, the total number of operations is tiny, well within limits even under strict constraints.

Test Cases

import sys, io

def run(inp: str) -> str:
    old_stdin = sys.stdin
    old_stdout = sys.stdout
    sys.stdin = io.StringIO(inp)
    sys.stdout = io.StringIO()
    try:
        solve()
        return sys.stdout.getvalue().strip()
    finally:
        sys.stdin = old_stdin
        sys.stdout = old_stdout

def solve():
    import sys
    input = sys.stdin.readline

    def rotate(grid):
        n = len(grid)
        return ["".join(grid[n - 1 - r][c] for r in range(n)) for c in range(n)]

    def flip_h(grid):
        return grid[::-1]

    n = int(input())
    a = [input().strip() for _ in range(n)]
    b = [input().strip() for _ in range(n)]

    variants = []
    cur = a
    for _ in range(4):
        variants.append(cur)
        variants.append(flip_h(cur))
        cur = rotate(cur)

    print("Yes" if any(v == b for v in variants) else "No")

# provided sample
assert run("""4
XOOO
XXOO
OOOO
XXXX
XOOO
XOOO
XOXO
XOXX
""") == "Yes"

# identical grids
assert run("""2
XO
OX
XO
OX
""") == "Yes"

# needs rotation
assert run("""2
XO
OO
OO
XO
""") == "Yes"

# needs flip
assert run("""2
XO
OO
OX
OO
""") == "Yes"

# impossible case
assert run("""2
XO
XO
OX
XX
""") == "No"
Test input Expected output What it validates
sample 1 Yes correctness under mixed rotation/flip
identical grids Yes zero transformation case
rotation match Yes rotation handling
flip match Yes mirror symmetry
impossible case No rejecting non-equivalent patterns

Edge Cases

A minimal $1 \times 1$ grid always matches itself regardless of transformation. The algorithm handles this naturally because rotation and flip operations return the same single-cell grid, so the only variant equals the target if the characters match.

Highly symmetric grids, such as those filled entirely with 'X', produce identical variants under all transformations. In such cases, every generated candidate equals the target, so the first comparison succeeds immediately. This does not affect correctness because the algorithm does not assume uniqueness of transformations, only existence.

Another subtle case is when the correct transformation is a combination like flip followed by rotation. The enumeration still captures it because that combined transformation appears as one of the eight canonical variants generated during the rotation-flip cycle.