The recursion base case is when the submatrices degenerate into 1x1 matrices (single numbers) at which point you multiply 2 numbers and return them. Addition and Subtraction of two matrices takes O(N²) time. So the recurrence relation can be written as: T(N) 7T(N/2) + = O(N²) Explain in your solution documentation (in the Java file header) or if you prefer in a separate .txt (text) or .doc/.docx (Word) file: - - Which case of the Master's Theorem applies (CLRS case 1, 2 or 3). - What is the time complexity of the above method in terms of an upper bound O(.)? Do you expect this to run faster than the naive method, which has cubic complexity? This question asks you to implement Strassen's square matrix multiplication method in Java. This is a naive way to multiply two matrices, with a cubic time complexity of O(n³): void multiply(int A☐] [N], int B[] [N], int C[] [N]) { for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { C[i][j] = 0; for (int k = 0; k < N; k++) { C[i][j] += A[i][k]*B[k][j]; } } } } Strassen's matrix multiplication method makes 7 recursive calls. Strassen's method is a divide and conquer method in the sense that this method divides square matrices into sub-matrices of size N/2 x N/2 (similar to the way MergeSort divides arrays as we saw in class). In Strassen's method, the four sub-matrices of the result are calculated as follows. As you can see, there are 7 submatrices with 7 multiplications done (one for each Mk). M1 := (A1,1 A2,2)(B1,1 + B2,2) M2 = (A2,1 + A2,2)B1,1 A1,1 (B1,2 B2,2) - M3 := M₁ == A2,2 (B2,1 B1,1) - M5 M6 := := M7 := = (A1,1 + A1,2) B2,2 (A2,1 A1,1)(B1,1 + B1,2) - (A1,2 A2,2)(B2,1 + B2,2) Now express Ci,j in terms of Mk and join them in the overall matrix C: C1,1 = M1 C1,2 M4 - M5 + M7 M3 + M5 C2,1 M2 M4 = C2.2 M₁ M2 + M3 + M6 =

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