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Divide-and-Conquer. Matrix multiplication and Strassen’s algorithm. [. [. [. a. r. e. f. s. b. [. [. [. =. t. g. c. d. h. u. Matrix Multiplication. How many operations are needed to multiply two 2 by 2 matrices?. [. [. [. a. r. e. f. s. b. [. [. [. =. t. g. c.
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Divide-and-Conquer • Matrix multiplication and Strassen’s algorithm
[ [ [ a r e f s b [ [ [ = t g c d h u Matrix Multiplication • How many operations are needed to multiply two 2 by 2 matrices?
[ [ [ a r e f s b [ [ [ = t g c d h u Traditional Approach • r = ae + bg • s = af + bh • t = ce + dg • u = cf + dh • 8 multiplications and 4 additions
Each letter represents an n/2 by n/2 matrix We can use the breakdown to form a divide and conquer algorithm [ [ [ A R E F B S [ [ [ = C T G H U D Extending to n by n matrices • R = AE + BG • S = AF + BH • T = CE + DG • U = CF + DH • 8 multiplications of n/2 by n/2 matrices • T(n) = 8 T(n/2) + Q(n2) • T(n) = Q(n3)
R = AE + BG S = AF + BH T = CE + DG U = CF + DH [ [ 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 Example [ [ [ [ 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 = • What are A, B, …, H? • What happens at this level of the D&C multiplication?
p1 = a(f – h) p2 = (a+b)h p3 = (c+d)e p4 = d(g-e) p5 = (a+d)(e + h) p6 = (b-d)(g+h) p7 = (a-c)(e+f) r = p5 + p4 - p2 + p6 s = p1 + p2 t = p3 + p4 u = p5 + p1 – p3 – p7 7 multiplications 18 additions [ [ [ r e a f b s [ [ [ = c t g d h u Strassen’s Approach
Each letter represents an n/2 by n/2 matrix We can use the breakdown to form a divide and conquer algorithm [ [ [ A R E F B S [ [ [ = C T G H U D Extending to n by n matrices • 7 multiplications of n/2 by n/2 matrices • 18 additions of n/2 by n/2 matrices • T(n) = 7 T(n/2) + Q(n2) • T(n) = Q(nlg 7)
p1 = a(f – h) p2 = (a+b)h p3 = (c+d)e p4 = d(g-e) p5 = (a+d)(e + h) p6 = (b-d)(g+h) p7 = (a-c)(e+f) r = p5 + p4 - p2 + p6 s = p1 + p2 t = p3 + p4 u = p5 + p1 – p3 – p7 [ [ [ Example [ [ [ 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 = • What are A, B, …, H? • What happens at this level of the D&C multiplication?
Observations • Comparison: n= 70 • Direct multiplication: 703 = 343,000 • Strassen: 70lg 7 is approximately 150,000 • Crossover point typically around n = 20 • Hopcroft and Kerr have shown 7 multiplications are necessary to multiply 2 by 2 matrices • But we can do better with larger matrices • Current best is O(n2.376) by Coppersmith and Winograd, but it is not practical • Best lower bound is W(n2) (since there are n2 entries) • Matrix multiplication can be used in some graph algorithms as a fundamental step, so theoretical improvements in the efficiency of this operation can lead to theoretical improvements in those algorithms