hw02_sol

# hw02_sol - CS170 Spring 2010 HW2 Solutions 1(DPV 2.5 a T(n...

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CS170 - Spring 2010 HW2 - Solutions 1. (DPV 2.5) a) T ( n ) = 2 T ( n/ 3) + 1 = Θ( n log 3 2 ) by the Master theorem. b) T ( n ) = 5 T ( n/ 4) + n = Θ( n log 4 5 ) by the Master theorem. c) T ( n ) = 7 T ( n/ 7) + n = Θ( n log n ) by the Master theorem. d) T ( n ) = 9 T ( n/ 3) + n 2 = Θ( n 2 log n ) by the Master theorem. e) T ( n ) = 8 T ( n/ 2) + n 3 = Θ( n 3 log n ) by the Master theorem. f) T ( n ) = 49 T ( n/ 25) + n 3 / 2 log n = Θ( n 3 / 2 log n ). By using the same reasoning used in the proof of the Master Theorem, we see that at level i of the recursion we have 49 i · ( n (25) i ) 3 / 2 · log ( n 25 i ) = ( 49 25 3 / 2 ) i · n 3 / 2 · (log n - i · log25). Since log 25 n i =0 ( 49 25 3 / 2 ) i = Θ(1) we see our expression is O ( n 3 / 2 log n ). Also, T ( n ) is clearly Ω( n 3 / 2 log n ), since this is a lower bound on the running time of the ﬁrst level of the recursion. g) T ( n ) = T ( n - 1) + 2 = Θ( n ). Since T ( n ) = n i =0 2. h) T ( n ) = T ( n - 1) + n c = n i =0 i c = Θ( n c +1 ) ( c 1 is a constant). Shown in HW1. i) T ( n ) = T ( n - 1) + c n = n i =0 c i = Θ( c n ) ( c > 1 is a constant). Shown in HW1. j) T ( n ) = 2 T ( n - 1) + 1 = n i =0 2 i · Θ(1) = Θ(2 n ). j) T ( n ) = T ( n ) + 1 = Θ(loglog n ). In each level of the recursion we have one process which takes O (1). Therefore T ( n ) + 1 = k i =0 1+ T ( b ) where b is some small enough constant and k is the number of levels of the recursion taken to reach b from n . We need to ﬁnd k , and plug it into the equation k i =0 1+ T ( b ) to compute the running time. At step i of the recursion the input is of size n ( 1 2 i ) . We need to ﬁnd k such that n ( 1 2 k ) = b . Doing the math we get k = loglog n - loglog b log2 = Θ(loglog n ), and we see that T ( n ) = k i =0 1 + T ( b ) = Θ(loglog n ). 1

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2. (DPV 2.11) Let X = ± A B C D ² and Y = ± E F G H ² be two n -by- n matrices, where A,B,C,D,E,F,G and H are n/ 2-by- n/ 2 submatrices. For any i n 2 and j n 2 : ( X · Y ) ij = n X k =1 X ik · Y kj = n/ 2 X k =1 X ik · Y kj + n X k = n 2 +1 X ik · Y kj = n/ 2 X k =1 A ik · E kj + n/ 2 X k =1 B ik · G kj = ( A · E + B · G ) ij A similar proof holds for the values of ( i,j ) in the other sectors of the matrix. 3. (DPV 2.15)
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## This note was uploaded on 04/30/2010 for the course CS 170 taught by Professor Henzinger during the Spring '02 term at Berkeley.

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hw02_sol - CS170 Spring 2010 HW2 Solutions 1(DPV 2.5 a T(n...

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