**π‘ Problem Formulation:** Duplicate rows in a binary matrix can skew data analysis and lead to inaccurate results. This article aims to provide Python solutions to identify and extract these redundant rows. Imagine a binary matrix where each row represents a binary number. The goal is for you to input this matrix and receive an output listing the rows that have duplicates.

## Method 1: Brute Force Comparison

This method involves a nested loop to compare each row with all other rows, identifying duplicates by direct element-wise comparison. Although this method is straightforward, it doesn’t scale well with larger matrices due to its O(n^2) time complexity.

Here’s an example:

def find_duplicate_rows(matrix): duplicates = set() for i in range(len(matrix)): for j in range(i+1, len(matrix)): if matrix[i] == matrix[j] and i not in duplicates: duplicates.add(i) break return duplicates matrix = [ [1, 0, 0], [0, 1, 0], [1, 0, 0] ] print(find_duplicate_rows(matrix))

Output:

{0}

This code snippet defines a function `find_duplicate_rows()`

that takes a binary matrix as input and returns the set of indices of the duplicate rows. The nested for-loops identify row duplicates by comparison, adding them to a set if a duplicate is found.

## Method 2: Hashing with Dictionary

With this method, each row is converted to a tuple and hashed to check for duplicates by storing row indices in a dictionary. This is more efficient than brute force, especially for large datasets, due to its average case time complexity of O(n).

Here’s an example:

def find_duplicates(matrix): seen = {} for idx, row in enumerate(matrix): row_tup = tuple(row) if row_tup in seen: seen[row_tup].append(idx) else: seen[row_tup] = [idx] return [indices for indices in seen.values() if len(indices) > 1] matrix = [ [1, 0, 0], [0, 1, 0], [1, 0, 0] ] print(find_duplicates(matrix))

Output:

[[0, 2]]

The `find_duplicates()`

function processes the binary matrix by hashing each row as a tuple and storing indices in a dictionary. Redundant rows are easily spotted when the stored index list has more than one element.

## Method 3: Sorting the Matrix

By sorting the binary matrix, duplicate rows become adjacent and easier to detect. This technique is quick for matrices with a limited number of rows but can become less efficient as row size increases due to the O(nlogn) time complexity of sorting.

Here’s an example:

def sort_and_find(matrix): sorted_matrix = sorted(matrix) return [sorted_matrix[i] for i in range(len(sorted_matrix) - 1) if sorted_matrix[i] == sorted_matrix[i + 1]] matrix = [ [1, 0, 0], [0, 1, 0], [1, 0, 0] ] print(sort_and_find(matrix))

Output:

[[1, 0, 0]]

After sorting the matrix, the function `sort_and_find()`

checks for equality between adjacent rows and returns a list of duplicate rows. Sorting makes the comparison linear with respect to the number of rows, following the initial sort.

## Method 4: Set Operations

This method includes creating sets from matrix rows and comparing sizes to identify duplicates. This approach is very intuitive and performs well for small to medium-sized matrices.

Here’s an example:

def find_duplicates_set(matrix): unique_rows = set(map(tuple, matrix)) return len(matrix) != len(unique_rows) matrix = [ [1, 0, 0], [0, 1, 0], [1, 0, 0] ] print(find_duplicates_set(matrix))

Output:

True

The function `find_duplicates_set()`

converts the matrix rows to tuples, creating a set to automatically remove duplicate rows. It then compares the length of the set to the original matrix to determine if any rows were duplicates.

## Bonus One-Liner Method 5: Using NumPy Library

For Python developers already using the NumPy library, this one-liner takes advantage of efficient array operations to find duplicate rows. Requires NumPy to be installed.

Here’s an example:

import numpy as np matrix = np.array([ [1, 0, 0], [0, 1, 0], [1, 0, 0] ]) duplicates = np.unique(matrix, axis=0, return_index=True, return_counts=True) print(duplicates[1][duplicates[2] > 1])

Output:

[0]

Using NumPy’s `unique()`

function, the code quickly extracts unique rows along with the original indices and counts of each row. By selecting indices where the count is greater than one, we identify the rows that have duplicates more efficiently.

## Summary/Discussion

**Method 1: Brute Force Comparison.**It’s simple to implement. It’s best for small matrices. Performance degrades with larger matrices.**Method 2: Hashing with Dictionary.**More efficient for larger datasets. Average case time complexity is O(n). Requires additional space for storing indices.**Method 3: Sorting the Matrix.**Simplifies the process of finding duplicates by making them adjacent. Time complexity increases with the number of rows due to sorting.**Method 4: Set Operations.**Straightforward and effective on small to medium matrices. It may not be the best choice for very large datasets due to space constraints of sets.**Bonus One-Liner Method 5: Using NumPy Library.**Fast and succinct. Requires NumPy, which might not be suitable for all environments or for those looking to keep dependencies minimal.