HW1_sol_2 - Fall Semester 2009 CS300 Algorithms Solution #1...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Fall Semester 2009 CS300 Algorithms Solution #1 1. O ( g 4 ) O ( g 3 ) O ( g 5 ) O ( g 1 ) O ( g 2 ) O ( g 7 ) O ( g 6 ) 2. For case 1, we have f ( n ) = O ( n log b a - ± ), implying that f ( n/b j ) = O (( n/b j ) log b a - ± ). Sub- stituting into g ( n ) yields g ( n ) = O ( log b n - 1 j =0 a j ( n/b j ) log b a - ± ) . We bound the summation within the O -notation by factoring out terms and simplifying, which leaves an increasing geometric series: log b n - 1 X j =0 a j ( n/b j ) log b a - ± = n log b a - ± log b n - 1 X j =0 ( ab ± /b log b a ) j = n log b a - ± log b n - 1 X j =0 ( b ± ) j = n log b a - ± ± b ± log b n - 1 b ± - 1 ² = n log b a - ± ± n ± - 1 b ± - 1 ² Since b and ± are constants, the last expression reduces to n log b a - ± O ( n ± ) = O ( n log b a ). Substituting this expression for the summation in equation (1) yields g ( n ) = O ( n log b a ) , (1) and case 1 is proved. Under the assumption that f ( n ) = Θ( n log b a ) for case 2, we have that f ( n/b j ) = Θ(( n/b j ) log b a ). Substituting into
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 02/04/2010 for the course COMPUTER S cs300 taught by Professor Unkown during the Spring '08 term at Korea Advanced Institute of Science and Technology.

Page1 / 2

HW1_sol_2 - Fall Semester 2009 CS300 Algorithms Solution #1...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online