pset1sol-f10 - 18.335 Problem Set 1 Solutions Problem 1:...

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Unformatted text preview: 18.335 Problem Set 1 Solutions Problem 1: Gaussian elimination The inner loop of LU, the loop over rows, subtracts from each row a different multiple of the pivot row. But this is exactly a rank-1 update U U- xy T , where x is the column-vector of multipliers and y T is the pivot row. More explicitly, we can rewrite Gaussian elimination without row swaps (pivoting) as: U = A for k = 1 to m- 1 x = u k +1: m,k /u kk U k +1: m,k : m = U k +1: m,k : m- xu k,k : m Note that I have used Matlab notation n : n to denote ranges (from n to n ) of rows or columns. In particular, note that we only have to do a rank-1 update of a submatrix of U , and that u k,k : m is a row vector of the k-th row of U from column k to column m . Problem 2: Asymptotic notation (a) means both O and . Lets do one at a time. By the definition of O , f ( n ) C 1 F ( n ) for n > N 1 and g ( n ) C 2 G ( n ) for n > N 2 (no absolute values since the functions were given to be nonnegative), for some constants C 1 , 2 and N 1 , 2 . If we let C = max( C 1 ,C 2 ) and N = max( N 1 ,N 2 ) , then f ( n ) + g ( n ) C 1 F ( n ) + C 2 G ( n ) C [ F ( n ) + G ( n )] for n > N , hence f + g is O ( F + G ) . Similarly for , replacing with and max with min , so f + g is ( F + G ) . Hence f + g is ( F + G ) . (b) f ( n ) O [ g ( n )] | f ( n ) | C | g ( n ) | for some C > and n > N | g ( n ) | C- 1 | f ( n ) | for C- 1 > and n > N g ( n ) [ g ( n )] . Q.E.D. (c) f ( n ) O [ F ( n )] | f ( n ) | C | F ( n ) | for n > N . If h ( n ) O [ f ( n ) + cF ( n )] then for n > N we have | h ( n ) | C | f ( n ) + cF ( n ) | C | f ( n ) | + C | c || F ( n ) | for some C > . For n > max( N ,N ) , | h ( n ) | ( C C + C | c | ) | F ( n ) | where C C + C | c | > , and thus h ( n ) O [ F ( n )] . However, the same inference is not true if we replace O with ; as a simple example, consider f ( n ) = n 3 and F ( n ) = n 3- n 2 with c =- 1 : in this case n 3 O ( n 3- n 2 ) , but f ( n ) + cF ( n ) = n 2 and ( n 2 ) is not a subset of ( n 3- n 2 ) = ( n 3 ) . (d) If the running time is O ( n 2 ) , that means that the time is n 2 multiplied by a constant, asymptotically. If it is O ( n 2 ) or worse, that would mean that the time is bounded above by any function n 2 , which is true of every function! Usually, when you hear things like this, what people really mean is ( n 2 ) or worse, or equivalently...
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pset1sol-f10 - 18.335 Problem Set 1 Solutions Problem 1:...

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