The Growth of Functions: Selected Exercises Goals • Introduce big-O & big-Omega • Show how to estimate the size of functions using this notation. 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 2 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 3 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 4 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 ). Copyright © Peter Cappello 5 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 6 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 7 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 8 Exercise 20 b) b) f( n ) = ( 2n + n2 )( n3 + 3n ). Copyright © Peter Cappello 9 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 10 Exercise 20 c) c) f( n ) = ( nn + n2n + 5n )( n! + 5n ) Copyright © Peter Cappello 11 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 ) : 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! ). Copyright © Peter Cappello 12 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 13 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 14 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 15 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 16 End 3.2 Copyright © Peter Cappello 17 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 18 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 ) ). 3. k1, C1, x > k1 f1( x ) C1g1( x ) 4. f2( x ) is O( g2( x ) ). 5. k2, C2, x > k2 f2( x ) C2g2( x ) (1. and defn of Θ) (2.,Defn of O) (1. and defn of Θ) (4., Defn of O) 6. x > max{ k1, k2 } f1 f2( x ) C1C2g1g2( x ) 7. f1 f2( x ) is O( g1g2( x ) ). Copyright © Peter Cappello 2011 (6., Defn of O) 19 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) Copyright © Peter Cappello 2011 20 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). Copyright © Peter Cappello 2011 (Defn of Ω) 21

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