ANSWER;
To fall in third case of the master theorem , we need to have a 16. In that case ,the algorithm will be
T (n) = (n 2). for the second case, with a= 16, T(n) = (n 2 log n ) . In the first case of the ,master theorem, to be faster than strassen, we need log 4 a
log2 7, which is a 72=49 . thus, the largest integer value will be 48
4.5-2 Professor Caesar wishes to develop a matrix-multiplication algorithm that is asymptotically faster than Strassen’s algorithm....
Question #4 (15 points) In class, we discussed a divide-and-conquer algorithm for matrix multiplication that involved solving eight subproblems, each half the size of the original, and performing a constant number of e(n) addition operations. Strassen's Algo- rithm, which we did not cover, reduces the number of (half-sized) subproblems to seven, with a constant number of e(n) addition and subtraction operations. Provide clear, concise answers to each of the following related questions. • (7 points). Express the runtime of Strassen's...
I already solved part A and I just need help with part B 1. Matrix Multiplication The product of two n xn matrices X and Y is a third n x n matrix 2 = XY, with entries 2 - 21; = xixYk x k=1 There are n’ entries to compute, each one at a cost of O(n). The formula implies an algorithm with O(nº) running time. For a long time this was widely believed to be the best running...