Question

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 Algorithm as a recurrence relation . (8 points). Using the substitution (induction) method, prove that this recurrence relation is bounded above by O(nº). Carefully show all of your steps, making it clear what you are assuming, and what you are trying to prove. • (5 points extra credit). Can you identify a tighter upper-bound? If so, prove it.

Answer 1

well the divide and conquer approach for matrix multiplication is as-

• Let X and Y be nxn matrices 211 212 ... in 221 222 .. in X = 3 31 32 . Tin nl xn2 .. Inn . We want to compute Z = XY - Zij = EX=1 Xik. Ykj • Naive method uses →na.n= O(n) operations • Divide-and-conquer solution: SA BUJE FUS (A. E+B.G) (A. F+BH) 2= C D G H = (CE+D.G) (C.F+DH) - The above naturally leads to divide-and-conquer solution: * Divide X and Y into 8 sub-matrices A, B, C, and D. * Do 8 matrix multiplications recursively. * Compute z by combining results (doing 4 matrix additions). - Lets assume n = 2€ for some constant c and let A, B, C and D be n/2 x n/2 matrices * Running time of algorithm is T(n) = 8T(n/2) + (na) = T(n) = (na)

the strassen algo and its complexity is as-

Strassen observed the following: SA B E F S 2=CD) (GH) where (S1 + S2 - S4 +) (S&+S4) (54 +55) (S2 + S3 + S5 - S-) ) S2 = (B-D). (G + H) (A + D) (E+H) (A-C).(E +F) (A + B). H A. (F - H) D.G - E) (C+D). E S5 = = = S - Lets test that S6 + S7 is really C.E+D.G S6 + S = D.G - E) + C + D = DG - DE + CE + DE = DG + CE E

(n) This leads to a divide-and-conquer algorithm with running time T(n) = 7T (n/2) + - We only need to perform 7 multiplications recursively. - Division/Combination can still be performed in e(na) time. Lets solve the recurrence using the iteration method T(n) = 7T(n/2) + n2 +7(716) +63) = n²+(5)n? +71(1) = n²+()n? +72(71 )+2) + = x + x + y + z^T = n2+ (z)n? +(7)2n +(3) Br?.... + (5) lossn-1m2 + plos na 2 (23*n° + plop na = nº. 01 (37)log.n-1) + zbogin = np. obywaten) + plon = 12.0(7") + plosn e(7log")

- Now we have the following: zlogn = 710872 (7log, n) (1/log, 2) = n(1/log, 2) log27 = = n log22 plog 7 - Or in general: glogkn = nlogk a So the solution is T(n) = O(nlog 7) = O(n2.8...)