docs: improve Strassen's algorithm docstrings with complexity analysis

- Add O(n^2.807) time complexity explanation to actual_strassen()
- Explain why 7 multiplications beats naive 8 multiplications
- Add docstring to strassen() with args, returns, and raises
- Add doctest to default_matrix_multiplication()
- Add one-line docstrings to helper functions

Ref #14084

Author: Pawansingh3889

divide_and_conquer/strassen_matrix_multiplication.py

@@ -5,7 +5,15 @@
 
 def default_matrix_multiplication(a: list, b: list) -> list:
     """
-    Multiplication only for 2x2 matrices
+    Standard multiplication for 2x2 matrices (base case).
+
+    Used as the base case for Strassen's algorithm when the matrix
+    cannot be subdivided further. Uses 8 multiplications.
+
+    Time complexity: O(1) — fixed size input.
+
+    >>> default_matrix_multiplication([[1, 2], [3, 4]], [[5, 6], [7, 8]])
+    [[19, 22], [43, 50]]
     """
     if len(a) != 2 or len(a[0]) != 2 or len(b) != 2 or len(b[0]) != 2:
         raise Exception("Matrices are not 2x2")
@@ -17,13 +25,15 @@ def default_matrix_multiplication(a: list, b: list) -> list:
 
 
 def matrix_addition(matrix_a: list, matrix_b: list):
+    """Element-wise addition of two matrices of equal dimensions."""
     return [
         [matrix_a[row][col] + matrix_b[row][col] for col in range(len(matrix_a[row]))]
         for row in range(len(matrix_a))
     ]
 
 
 def matrix_subtraction(matrix_a: list, matrix_b: list):
+    """Element-wise subtraction of two matrices of equal dimensions."""
     return [
         [matrix_a[row][col] - matrix_b[row][col] for col in range(len(matrix_a[row]))]
         for row in range(len(matrix_a))
@@ -64,6 +74,7 @@ def split_matrix(a: list) -> tuple[list, list, list, list]:
 
 
 def matrix_dimensions(matrix: list) -> tuple[int, int]:
+    """Return (rows, columns) of a matrix."""
     return len(matrix), len(matrix[0])
 
 
@@ -73,8 +84,22 @@ def print_matrix(matrix: list) -> None:
 
 def actual_strassen(matrix_a: list, matrix_b: list) -> list:
     """
-    Recursive function to calculate the product of two matrices, using the Strassen
-    Algorithm. It only supports square matrices of any size that is a power of 2.
+    Recursive function to calculate the product of two matrices using Strassen's
+    algorithm. Only supports square matrices whose dimensions are a power of 2.
+
+    Strassen's algorithm reduces matrix multiplication from 8 recursive
+    multiplications (naive divide-and-conquer) to 7, at the cost of more
+    additions and subtractions. This gives a better asymptotic complexity:
+
+    - Naive matrix multiplication: O(n^3)
+    - Naive divide-and-conquer:    O(n^3)     — 8 multiplications of n/2 size
+    - Strassen's algorithm:        O(n^2.807) — 7 multiplications of n/2 size
+
+    The 7 intermediate products (t1-t7) are combined to form the four
+    quadrants of the result matrix using only additions and subtractions.
+
+    Reference: Strassen, V. (1969). Gaussian elimination is not optimal.
+    Numerische Mathematik, 13(4), 354-356.
     """
     if matrix_dimensions(matrix_a) == (2, 2):
         return default_matrix_multiplication(matrix_a, matrix_b)
@@ -106,6 +131,26 @@ def actual_strassen(matrix_a: list, matrix_b: list) -> list:
 
 def strassen(matrix1: list, matrix2: list) -> list:
     """
+    Multiply two matrices of arbitrary dimensions using Strassen's algorithm.
+
+    Handles non-square and non-power-of-2 matrices by padding with zeros
+    to the next power of 2, running Strassen's algorithm, then removing
+    the padding from the result.
+
+    Time complexity: O(n^2.807) where n is the padded dimension.
+    Space complexity: O(n^2) for the padded matrices.
+
+    Args:
+        matrix1: First matrix (m x n).
+        matrix2: Second matrix (n x p). Number of columns in matrix1
+                 must equal number of rows in matrix2.
+
+    Returns:
+        Result matrix (m x p).
+
+    Raises:
+        Exception: If matrix dimensions are incompatible for multiplication.
+
     >>> strassen([[2,1,3],[3,4,6],[1,4,2],[7,6,7]], [[4,2,3,4],[2,1,1,1],[8,6,4,2]])
     [[34, 23, 19, 15], [68, 46, 37, 28], [28, 18, 15, 12], [96, 62, 55, 48]]
     >>> strassen([[3,7,5,6,9],[1,5,3,7,8],[1,4,4,5,7]], [[2,4],[5,2],[1,7],[5,5],[7,8]])