**π‘ Problem Formulation:** Imagine we have a two-dimensional matrix where each row is a sequence of numbers. We need to determine if each row is a circular rotation of the first row. A circular rotation means that shifting elements around will result in the same sequence as another row. For instance, if the input matrix is `[[1, 2, 3], [3, 1, 2], [2, 3, 1]]`

, the desired output is `True`

because each row is a circular rotation of the others.

## Method 1: Iterative Comparison with Rotations

This method compares each row in the matrix with the first row by iteratively rotating it and checking for equality. If any rotation of a row matches the first, the method continues to the next row; otherwise, it returns false.

Here’s an example:

def is_circular_rotation(matrix): # A function to rotate a list def rotate(l, n): return l[n:] + l[:n] first_row = matrix[0] for row in matrix[1:]: if not any(row == rotate(first_row, n) for n in range(len(row))): return False return True # Example matrix matrix = [[1, 2, 3], [3, 1, 2], [2, 3, 1]] print(is_circular_rotation(matrix))

The output of this code snippet is:

True

This function iterates over each row after the first, generating all possible rotations and checking if any rotation matches the first row. This method is easy to understand and implement but can be inefficient for large matrices since it checks all rotations.

## Method 2: Using String Representation

This approach converts rows into strings and checks if the string representation of the first row is a substring of the doubled string of any other row.

Here’s an example:

def has_all_circular_rows(matrix): first_row_str = ''.join(map(str, matrix[0])) for row in matrix[1:]: row_str = ''.join(map(str, row)) if not (first_row_str in row_str + row_str): return False return True matrix = [[1, 2, 3], [3, 1, 2], [2, 3, 1]] print(has_all_circular_rows(matrix))

The output of this code snippet is:

True

This technique takes advantage of string manipulation. Doubling a row string creates a scenario where all circular rotations can be seen as substrings. This method is usually faster than the first for large sizes but can be prone to errors when dealing with multi-digit numbers.

## Method 3: Leveraging Python’s Collections Module

The collections module in Python provides a deque object which has an efficient rotate() method suitable for this problem. This function applies deque’s rotate to compare each row with the first row.

Here’s an example:

from collections import deque def are_circular_rotations(matrix): first_row_deque = deque(matrix[0]) for row in matrix[1:]: row_deque = deque(row) if not any(row_deque == first_row_deque.rotate(n) for n in range(len(row))): return False return True matrix = [[1, 2, 3], [3, 1, 2], [2, 3, 1]] print(are_circular_rotations(matrix))

The output of this code snippet is:

True

This function utilizes the deque data type which is designed for efficient removing and adding elements from either end. Its rotate method simplifies the comparison process. This method can be more efficient than the first but requires additional knowledge of the collections module.

## Method 4: Zip and Set

A more Pythonic and concise way to check for circular rotations uses zip to transpose the matrix and the set data structure to check for unique rotations relative to the first row.

Here’s an example:

def circular_rotation_check(matrix): first_row = matrix[0] for rotation in zip(*matrix): if set(rotation) != set(first_row): return False return True matrix = [[1, 2, 3], [3, 1, 2], [2, 3, 1]] print(circular_rotation_check(matrix))

The output of this code snippet is:

True

By transposing the matrix with zip, this function verifies that each row has the same set of elements as the first one. Although succinct and elegant, it fails in cases where duplicate elements are involved since sets ignore repetitions.

## Bonus One-Liner Method 5: Comprehension with Sets and Zip

This compact one-liner uses list comprehension along with the set and zip functions to verify that all rows are rotations of the first row.

Here’s an example:

matrix = [[1, 2, 3], [3, 1, 2], [2, 3, 1]] all_circular = all(set(r) == set(matrix[0]) for r in zip(*matrix)) print(all_circular)

The output of this code snippet is:

True

In this one-liner, the expression packs the logic of transposing and comparing element sets efficiently. While compact and fast for matrices without duplicates, it suffers from the same limitations as method 4 when dealing with repeated elements.

## Summary/Discussion

**Method 1:** Iterative Comparison with Rotations. Easy to understand. Can be slow for larger matrices.

**Method 2:** Using String Representation. Fast for large matrices. Can misinterpret multi-digit numbers.

**Method 3:** Leveraging Python’s Collections Module. Efficient rotation method. Requires familiarity with the collections module.

**Method 4:** Zip and Set. Pythonic and concise. Not suitable for matrices with duplicate elements.

**Method 5:** Comprehension with Sets and Zip. Elegant one-liner. Same limitations as method 4 regarding duplicate elements.

Emily Rosemary Collins is a tech enthusiast with a strong background in computer science, always staying up-to-date with the latest trends and innovations. Apart from her love for technology, Emily enjoys exploring the great outdoors, participating in local community events, and dedicating her free time to painting and photography. Her interests and passion for personal growth make her an engaging conversationalist and a reliable source of knowledge in the ever-evolving world of technology.