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strassen's matrix multiplication 4x4 example code

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1) Divide matrices A and B in 4 sub-matrices of size N/2 x N/2 as shown in the below diagram. . You can find a tip somewhere at the end of the article on how to generalize this algorithm for any value of n. Source : https://stackoverflow.com/questions/21496538/square-matrix-multiply-recursive-in-java-using-divide-and-conquer/30338712#30338712, For multiplying two matrices of size n x n, we make 8 recursive calls above, each on a matrix/subproblem with size n/2 x n/2. So the time complexity can be written as. \\ From the above equations, the recurrence relation of the Strassen’s approach is, $$T(n) = close, link https://www.youtube.com/watch?v=QXY4RskLQcI, Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. Following is simple Divide and Conquer method to multiply two square matrices. Time Complexity of above method is O(N3). Attention reader! Strassen’s method is similar to above simple divide and conquer method in the sense that this method also divide matrices to sub-matrices of size N/2 x N/2 as shown in the above diagram, but in Strassen’s method, the four sub-matrices of result are calculated using following formulae. ( Log Out /  Writing code in comment? However, $O(n^{2.81})$ is not much improvement though but enough for $n$ having large value as depicted in the graph below. And the matrix multiplication of the two 2x2 matrices A11 and B11 is. Note: Here the dimension n is of the power of 2. Introduction to Algorithms 3rd Edition by Clifford Stein, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest

The matrix multiplication of the above two matrices A and B is Matrix C, where, Strassen algorithm is a recursive method for matrix multiplication where we divide the matrix into 4 sub-matrices of dimensions n/2 x n/2 in each recursive step. Take two submatrices from the above two matrices A and B each as ($A_{11}$ &  $A_{12}$) and ($B_{11}$ & $B_{21}$) as shown below.

Here we save one recursive call, but have several new additions of n/2 x n/2 matrices. \begin{cases} \begin{cases} Please use ide.geeksforgeeks.org, generate link and share the link here. \Theta(1),  & \text{if $n=1$ } \\[2ex]     \end{bmatrix}$. $c_{21} = a_{21}*b_{11} + a_{22}*b_{21}+a_{23}*b_{31}+a_{24}*b_{41} \qquad(3)$ Addition of two matrices takes O(N2) time. & .

8T(\frac{n}{2}) + \Theta(n^2), & \text{if $n>1$} Consider two matrices A and B with 4x4 dimension each as shown below, The matrix multiplication of the above two matrices A and B is Matrix C, 4) Because of the limited precision of computer arithmetic on noninteger values, larger errors accumulate in Strassen’s algorithm than in Naive Method (Source: CLRS Book), Easy way to remember Strassen’s Matrix Equation, References: In this article, we are going to discuss about the strassen matrix multiplication, formula of matrix multiplication and algorithms for strassen matrix multiplication. So, from Case 1 of Master's Theorem, the time complexity of the above approach is $O(n^{\log_27})$ or $O(n^{2.81})$ which beats the divide and conquer approach asymptotically. ( Log Out /  $c_{12} = a_{11}*b_{12} + a_{12}*b_{22}+a_{13}*b_{32}+a_{14}*b_{42} \qquad(2)$ Naive Method \end{cases}$$. Before jumping to Strassen's algorithm, it is necessary that you should be familiar with matrix multiplication using the Divide and Conquer method. & . & .     c_{21} & c_{22} & . $\frac{n^2}{4}$, so each addition takes $\Theta(\frac{n^{2}}{4})$ time.

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