We use cookies to ensure you have the best browsing experience on our website. In this post, I will try to explain the concept of Strassen’s 4×4 matrix multiplication with an example. $M_{1} = (A_{11} + A_{22})(B_{11} + B_{22})$, $M_{6} = (A_{21} - A_{11}) (B_{11} + B_{12})$, $M_{7} = (A_{12} - A_{22}) (B_{21} + B_{22})$. By using our site, you It is applicable to the matrices of order (n*n). \\ To solve & multiply these matrices we need to follow some steps. Step 2: Write this matrix in the form of four 2×2 matrices and name them accordingly like A1, A2, A3, A4, B1, B2, B3, B4, Step 3: Finally we need to find the multiplications of these matrices one by one in such a way that, So for C1, take consider the matrix A1 & B1, Now for C2, take consider the matrix A2 & B2. Change ), You are commenting using your Twitter account. Simple Divide and Conquer also leads to O(N3), can there be a better way? Before jumping to Strassen's algorithm, it is necessary that you should be familiar with matrix multiplication using the Divide and Conquer method. Submitted by Prerana Jain, on June 22, 2018 . Consider two matrices A and B with 4x4 dimension each as shown below. And $c_{11}$, $c_{12}$, $c_{21}$ and $c_{22}$ are equal to equations 1, 2, 3 and 4 respectively. 2) Calculate following values recursively. So if you observe, I can conclude the following, $A_{11}*B_{11} + A_{12}*B_{21} =    \begin{bmatrix} Strassen’s algorithm makes use of the same divide and conquer approach as above, but instead uses only 7 recursive calls rather than 8 as shown in the equations below. Suppose we have two matrices A & B then the applicable formula is. Fill in your details below or click an icon to log in: You are commenting using your WordPress.com account. & . $c_{11} = a_{11}*b_{11} + a_{12}*b_{21}+a_{13}*b_{31}+a_{14}*b_{41} \qquad(1)$ If we have a 4×4 matrix, then we need to divide it into 4 equal parts. In the above method, we do 8 multiplications for matrices of size N/2 x N/2 and 4 additions.     c_{11} & c_{12} & . \end{cases}$$. Consider a matrix which is of order 4×4, we need to find its multiplication. 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. 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. Backtracking - Explanation and N queens problem, CSS3 Moving Cloud Animation With Airplane, * M1 = (A11 + A22)(B11 + B22) M2 = (A21 + A22) B11 M3 = A11 (B12 - B22) M4 =, * A22 (B21 - B11) M5 = (A11 + A12) B22 M6 = (A21 - A11) (B11 + B12) M7 = (A12 -, * C11 = M1 + M4 - M5 + M7 C12 = M3 + M5 C21 = M2 + M4 C22 = M1 - M2 + M3 + M6, /** join 4 halves into one result matrix **/, /** Function to split parent matrix into child matrices **/, /** Function to join child matrices into parent matrix **/, https://stackoverflow.com/questions/21496538/square-matrix-multiply-recursive-in-java-using-divide-and-conquer/30338712#30338712, https://www.sanfoundry.com/java-program-strassen-algorithm/, C++ : Linked lists in C++ (Singly linked list), Inserting a new node to a linked list in C++. \Theta(1),  & \text{if $n=1$ } \\[2ex] From the Case 1 of Master's Theorem, the time complexity of the above approach is $O(n^{\log_28})$ or $O(n^{3})$ which is the same as the naive method of matrix multiplication. edit ( Log Out /  Introduction. \Theta(1),  & \text{if $n=1$ } \\[2ex] 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.

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