the growth of functions: selected exercises goals introduce big-o & big-omega show how to...

The Growth of Functions: Selected Exercises

Goals • Introduce big-O & big-Omega• Show how to estimate the size of functions

using this notation.

Copyright © Peter Cappello 2

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!


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().


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 ).


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 ).

1) For n ≥ 1 & c = 1: n3 1n4 1 n.

2) Proof (by contradiction)

1) Assume k c n > k n4 cn3.

2) 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 ).


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) ).


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 ).


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 ).


Exercise 20 b)

b) f( n ) = ( 2n + n2 )( n3 + 3n ).


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 ).


Exercise 20 c)

c) f( n ) = ( nn + n2n + 5n )( n! + 5n )


Exercise 20 c) Solution


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 ) :

1. n ≥ 2 2n-1 nn-1.

2. n ≥ 2 n2n 2nn. (Multiply both sides of 1. by 2n.)

Thus, f( n ) is O( nnn! ).


Exercise 30


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 ).


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?


Generalizing the definitions


What is a good definition of

f( n, m ) is O( g( n, m ) )?


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( ? )


End 3.2

Copyright © Peter Cappello 2011 18

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 ) ).


40 Proof

1. 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 ) ).

2. f1( x ) is O( g1( x ) ). (1. and defn of Θ)

3. k1, C1, x > k1 f1( x ) C1g1( x ) (2.,Defn of O)

4. f2( x ) is O( g2( x ) ). (1. and defn of Θ)

5. k2, C2, x > k2 f2( x ) C2g2( x ) (4., Defn of O)

6. x > max k1, k2 f1 f2( x ) C1C2g1g2( x )

7. f1 f2( x ) is O( g1g2( x ) ). (6., Defn of O)


40 Proof continued

1. f1( x ) is Ω( g1( x ) ). (Previous 1. & defn of Θ)

2. k’1, C’1, x > k’1 f1( x ) ≥ C’1g1( x ) (1. & Defn of Ω)

3. f2( x ) is Ω( g2( x ) ). (Previous 1. & defn of Θ)

4. k’2, C’2, x > k’2 f2( x ) ≥ C’2g2( x ) (3. & Defn of Ω)

5. x > max k’1, k’2 f1 f2( x ) ≥ C’1C’2g1g2( x )

6. f1 f2( x ) is Ω( g1g2( x ) ). (5. & Defn of Ω)

7. f1 f2( x ) is Θ( g1g2( x ) ). (6., previous 7., defn Θ)


50

Show that xy is Ω(xy).

Proof:

1. xy ≥ xy. (Defn of ceiling)

2. Let c = 1.

3. For x > 0, y > 0, xy ≥ cxy.

4. Therefore, xy is Ω(xy). (Defn of Ω)

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