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1. The conventional way of multiplying two n-by-n matrices requires O(n3) time. There are much better ways of multiplying large matrices, bascd on a dividc-and-conquer approach. For example, Strassen's algorithm breaks cach n-by-n matrix into four submatrices of sizc n/2- by-n/2 each, and computes the result by performing 7 recursive multiplications on n/2-by-n/2 matrices plus a constant number of additions of n/2-by-n/2 matrices. Since addition of n-by-n matrices only requires O(n2) time, you get a recurrence for the running time of T(n) = 7T(n/2) + cn2 for some constant c (which, asymptotically, doesn't matter). Hypothetically, we can beat Strassen's approach if we can break our two n-by-n matrices into y2 submatrices of size n/y-by-n/y cach, and perform the multiplication of n-by-n matrices by using x recursive multiplications of n/y-by-n/y matrices plus any constant number of additions of n/y-by-n/y matrices. What is the relationship between x and y that would make this hypothetical algorithm have a lower running time than Strassen's algorithm? Identify one such pair below. Note, because the exact calculation requires the computation of logarithms, you may wish to find and use a scientific calculator of some sort. c) d) x-344: y-8. x=153; y=6