Download presentation

1
**The Growth of Functions: Selected Exercises**

Goals Introduce big-O & big-Omega Show how to estimate the size of functions using this notation.

2
**Copyright © Peter Cappello**

Preface You may use without proof that: The functions below increase asymptotically from top to bottom: f( n ) = k, for some constant k. f( n ) = logk n, for all k N & any constant log base ≥ 2. f( n ) = nq, for all q Q+ f( n ) = kn < (k+1)n, for all k N f( n ) = n! Copyright © Peter Cappello

3
**Copyright © Peter Cappello**

Preface continued The book says that f(n) is O( g( n ) ) when k c n ( n > k | f( n ) | c | g( n ) | ) In computational complexity, we deal exclusively with functions whose domains & ranges are positive. We thus may simplify the definition of O( ) as follows: k c n > k f( n ) c g( n ). You are not responsible for knowing o(). This is different from O(). Copyright © Peter Cappello

4
**Copyright © Peter Cappello**

Exercise 10 Defn: f( n ) is O( g( n ) ) if k c n > k f( n ) cg( n ). Show that: 1) n3 is O( n4 ) 2) n4 is not O( n3 ). Copyright © Peter Cappello

5
**Copyright © Peter Cappello**

Exercise10: Solution Defn: f( n ) is O( g( n ) ) when k c n > k f( n ) cg( n ). Show that: 1) n3 is O( n4 ) 2) n4 is not O( n3 ). For n ≥ 1 & c = 1: n3 1n4 1 n. Proof (by contradiction) Assume k c n > k n4 cn3. k c n > k n4 cn3 k c n > k n c ), which is false. ( Divide both sides by n3 ) 3) Therefore, n4 is not O( n3 ). Copyright © Peter Cappello

6
**Copyright © Peter Cappello**

Theorems You Can Use Thm 1: Let f(x) = anxn + an-1xn-1 + … + a1x + a0, where the ai are real. Then, f(x) is O( xn ). Let f1(x) is O (g1(x) ) & f2(x) is O( g2(x) ). Thm 2: (f1 + f2)(x) is O( max( g1(x), g2(x) ) ) Thm 3: (f1 f2)(x) is O( (g1g2)(x) ). Copyright © Peter Cappello

7
**Copyright © Peter Cappello**

Exercise 20 Give a big-O estimate for the functions: (Use a simple g of smallest order.) a) f( n ) = ( n3 + n2logn )( logn + 1 ) + ( 17logn + 19 )( n3 + 2 ). Copyright © Peter Cappello

8
**Copyright © Peter Cappello**

Exercise 20 a) Solution Give a big-O estimate for the functions: (Use a simple g of smallest order.) a) f( n ) = ( n3 + n2logn )( logn + 1 ) + ( 17logn + 19 )( n3 + 2 ). Using our theorems, ( n3 + n2logn )( logn + 1 ) + ( 17logn + 19 )( n3 + 2 ) Is O( ( n3 )( logn ) + ( 17logn )( n3 ) ) Is O( ( n3 logn ) + ( n3 17logn ) Is O( ( n3 logn ). Copyright © Peter Cappello

9
**Copyright © Peter Cappello**

Exercise 20 b) b) f( n ) = ( 2n + n2 )( n3 + 3n ). Copyright © Peter Cappello

10
**Copyright © Peter Cappello**

Exercise 20 b) Solution b) f( n ) = ( 2n + n2 )( n3 + 3n ). Using our theorems, f( n ) = ( 2n + n2 )( n3 + 3n ) is O( ( 2n )( 3n ) ) which is O( 2n3n ) which is O( 6n ). Copyright © Peter Cappello

11
**Copyright © Peter Cappello**

Exercise 20 c) c) f( n ) = ( nn + n2n + 5n )( n! + 5n ) Copyright © Peter Cappello

12
**Copyright © Peter Cappello**

Exercise 20 c) Solution Defn: f( n ) is O( g( n ) ) when k c n > k f( n ) cg( n ). c) f( n ) = ( nn + n2n + 5n )( n! + 5n ) Using our theorems, f( n ) is O( ( nn + n2n )( n! ) ) In nn + n2n, which is the fastest growing term? Claim: n2n is O ( nn ) : n ≥ 2 2n-1 nn-1. n ≥ 2 n2n 2nn. (Multiply both sides of 1. by 2n.) Thus, f( n ) is O( nnn! ). Copyright © Peter Cappello

13
**Copyright © Peter Cappello**

Exercise 30 Defn: f( n ) is O( g( n ) ) when k c n > k f( n ) cg( n ). Defn. f( n ) is Ω( g( n ) ) when k c > 0 n > k f( n ) ≥ cg( n ). What does it mean for f( n ) to be Ω( 1 )? Hint: graph f( n ). Copyright © Peter Cappello

14
**Copyright © Peter Cappello**

Exercise 30 Solution Defn. f( n ) is Ω( g( n ) ) when k c > 0 n > k f( n ) ≥ c g( n ). What does it mean for a function to be Ω( 1 )? From the definition, f( n ) is Ω( 1 ) when k c > 0 n > k f( n ) ≥ c. f( n ) ≥ c > 0, for sufficiently large n. f( n ) = 1/n is Ω( 1 ). True or false? Copyright © Peter Cappello

15
**Generalizing the definitions**

Defn: f( n ) is O( g( n ) ) when k c n > k f( n ) cg( n ). What is a good definition of f( n, m ) is O( g( n, m ) )? Copyright © Peter Cappello

16
**Time Complexity of Bubble Sort**

void bubblesort( int[] a ) { for ( int i = 0; i < a.length – 1; i++ ) for ( int j = 0; j < a.length – 1 – i; j++ ) if ( a[ j ] < a[ j + 1 ] ) int temp = a[ j ]; a[ j ] = a[ j + 1 ]; a[ j + 1 ] = temp; } Let a.length = n. What is the total # of comparisons as a function of n? This number is O( ? ) Copyright © Peter Cappello

17
**Copyright © Peter Cappello**

End 3.2 Copyright © Peter Cappello

18
**Copyright © Peter Cappello 2011**

40 Defn. f( n ) is Θ( g( n ) ) when f( n ) is O( g ( n ) ) and f( n ) is Ω( g( n ) ). Show that: if f1( x ) & f2( x ) are functions from Z+ to R and f1( x ) is Θ( g1( x ) ) and f2( x ) is Θ( g2( x ) ), then f1f2 ( x ) is Θ( g1g2( x ) ). Copyright © Peter Cappello 2011

19
**Copyright © Peter Cappello 2011**

40 Proof Assume f1( x ) & f2( x ) are functions from Z+ to R and f1( x ) is Θ( g1( x ) ) and f2( x ) is Θ( g2( x ) ). f1( x ) is O( g1( x ) ) (1. and defn of Θ) k1, C1, x > k1 f1( x ) C1g1( x ) (2.,Defn of O) f2( x ) is O( g2( x ) ) (1. and defn of Θ) k2, C2, x > k2 f2( x ) C2g2( x ) (4., Defn of O) x > max{ k1, k2 } f1 f2( x ) C1C2g1g2( x ) f1 f2( x ) is O( g1g2( x ) ) (6., Defn of O) Copyright © Peter Cappello 2011

20
**Copyright © Peter Cappello 2011**

40 Proof continued f1( x ) is Ω( g1( x ) ) (Previous 1. & defn of Θ) k’1, C’1, x > k’1 f1( x ) ≥ C’1g1( x ) (1. & Defn of Ω) f2( x ) is Ω( g2( x ) ) (Previous 1. & defn of Θ) k’2, C’2, x > k’2 f2( x ) ≥ C’2g2( x ) (3. & Defn of Ω) x > max{ k’1, k’2 } f1 f2( x ) ≥ C’1C’2g1g2( x ) f1 f2( x ) is Ω( g1g2( x ) ) (5. & Defn of Ω) f1 f2( x ) is Θ( g1g2( x ) ) (6., previous 7., defn Θ) Copyright © Peter Cappello 2011

21
**Copyright © Peter Cappello 2011**

50 Show that ┌ xy ┐ is Ω(xy). Proof: ┌ xy ┐ ≥ xy (Defn of ceiling) Let c = 1. For x > 0, y > 0, ┌ xy ┐ ≥ cxy. Therefore,┌ xy ┐ is Ω(xy) (Defn of Ω) Copyright © Peter Cappello 2011

Similar presentations

OK

Chapter 2 Fundamentals of the Analysis of Algorithm Efficiency Copyright © 2007 Pearson Addison-Wesley. All rights reserved.

Chapter 2 Fundamentals of the Analysis of Algorithm Efficiency Copyright © 2007 Pearson Addison-Wesley. All rights reserved.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on water our lifeline assistance Ppt on tcp ip protocol suite 3rd Ppt on sap r 3 architecture Download small ppt on global warming Download ppt on pulse code modulation encoder Download ppt on query processing and optimization ppt Ppt on standing order meaning Urinary tract anatomy and physiology ppt on cells Ppt on business etiquettes ppt Ppt on forward rate agreement hedging