DivideConquer Algorithms Recurrence Relations Selected Exercises Exercise 10

Divide-&-Conquer Algorithms & Recurrence Relations: Selected Exercises

Exercise 10 Find f( n ) when n = 2 k, where f satisfies the recurrence relation f( n ) = f( n/2 ) + 1, with f( 1 ) = 1. Copyright © Peter Cappello 2

Exercise 10 Solution We are asked for the value of f. We are given that f( n ) = f( n / 2 ) + 1, with f( 1 ) = 1, where n = 2 k. f( 2 k ) = f( 2 k - 1 ) + 1, with f( 20 ) = 1. Answer: f( 2 k ) = k + 1. (Proof would be inductive. ) Copyright © Peter Cappello 3

Master Theorem Let f be an increasing function that such that f( n ) = a. f( n / b ) + c. nd whenever n = bk, where a, b, k Z+ & c > 0 & d ≥ 0: If ( a < bd ) then f( n ) is O( nd ) If ( a = bd ) then f( n ) is O( ndlogn ) If ( a > bd ) then f( n ) is O( nlogba ). Copyright © Peter Cappello 4

Exercise 10 Solution (2) • If we are interested only in the asymptotic growth rate, use the Master Theorem. • For f( n ) = f( n / 2 ) + 1 = 1 f( n / 2 ) + 1 n 0 a=1 b=2 d=0 So, a = bd So, f(n) is O( ndlog n ) = O( log n ). Copyright © Peter Cappello 5

Merge Sort Procedure mergesort( List a ) { assert a != null && !a. is. Empty(); int n = a. size(); If ( n == 1 ) return a; List L 1 = new List containing a( 0 ), . . . , a( n/2 ); List L 2 = new List containing a( n/2 + 1 ), …, a( n ); return merge( mergesort( L 1 ), mergesort( L 2 ); } Copyright © Peter Cappello 6

Give a recurrence relation, f( n ), the time to run merge sort on a list of size n. Copyright © Peter Cappello 7

Solution A Divide&Conquer Recurrence Relation f( n ) = 2 f( n / 2 ) + cn 1, where c is some constant. Give a big-O estimate of the time to run merge sort on a List of size n. Copyright © Peter Cappello 8

Give a big-O estimate of the time to run merge sort on a List of size n. 1. f( n ) = 2 f( n / 2 ) + cn 1. 2. a = 2; b = 2; d = 1. 3. Since 2 = 21, f( n ) is O( n 1 log n ). Copyright © Peter Cappello 9

Strassen’s Matrix Product Algorithm • Let C = A x B, where A, B, & C are n x n matrices. • Do this recursively, in terms of n/2 x n/2 matrices. Done conventionally, this results in: 8 n/2 x n/2 matrix products and 4 n/2 x n/2 matrix additions. (Show on board) • The time to perform the arithmetic is modeled by the following recurrence equation: T( n ) = 8 T( n/2 ) + 4( n/2 )2 (explain) • Applying the Master Theorem, where a = 8, b = 2, d = 2, we obtain a big-O estimate: O( nlogb(a) ) = O( nlog 2(8) ) = O( n 3 ). Copyright © Peter Cappello 10

• Strassen showed how to do this with: 7 n/2 x n/2 matrix products and 18 n/2 x n/2 matrix additions. • The time to do the arithmetic is modeled by recurrence: T( n ) = 7 T( n/2 ) + 18( n/2 )2 • Apply the Master Theorem, where a = 7, b = 2, d = 2, to obtain a big-O estimate: O( nlogb(a) ) = O( nlog 2(7) ) ≈ O( n 2. 81). Copyright © Peter Cappello 11

End Copyright © Peter Cappello 12

Exercise 20 (a) Set up a divide-&-conquer recurrence relation for the # of modular multiplies (MM) required to compute an mod m, where a, m, n Z+, using the recursive algorithm from Example 3 in Section 4. 4. Let b = a mod m. The algorithm then is: – an mod m = ( an/2 mod m )2 mod m, for even n (1 MM) – an mod m = ([(a(n-1)/2 mod m)2 mod m]. b) mod n, for odd n (2 MMs) Copyright © Peter Cappello 13

Exercise 20 (a) Solution The algorithm is: an mod m = ( an/2 mod m )2 mod m, for even n (1 MM) an mod m = ([(a(n-1)/2 mod m)2 mod m]. b) mod n, for odd n (2 MMs) Let f( n ) denote the # of multiplies. For even n, f( n/2 ) + 1 MM is used. For odd n, f( n/2 ) + 2 MMs are used. Worst case, f( n ) = f( n/2 ) + 2 (The recurrence is used in a big-O estimate of run time. It suffices to know that the additive term is a constant. ) Copyright © Peter Cappello 14

Exercise 20 (b) Using f( n ) = f( n/2 ) + 2, construct a big-O estimate for the # of modular multiplies used to compute an mod m using the recursive algorithm. Copyright © Peter Cappello 15

Exercise 20 (b) Solution Use the Master Theorem: f( n ) = a. f( n/b ) + c. nd f( n ) = f( n/2 ) + 2 = 1 f( n/2 ) + 2 n 0 a = 1, b = 2, d = 0. So, a = bd So, f( n ) is O( ndlog n ) = O( log n ). Copyright © Peter Cappello 16

Exercise 30 assumes that f satisfies the conditions of the Master Theorem. Use Exercise 29 to show that, for f(n) = a. f(n/b) + c. nd if a = bd, then f(n) is O( nd log n ). Exercise 29 gives us: If a = bd & n = bk then f(n) = f(1)nd + cnd logbn. Copyright © Peter Cappello 17

Exercise 30 Solution To show: f(n) = f(1)nd + cnd logbn is O( nd log n ). So: – f(n) growth clearly is dominated by cnd logbn. – c is a constant. – logbn differs from log 2 n by only a constant factor. ( log 2 b ). logbn = log 2( b logbn ) = log 2 n. – cnd logbn therefore is O( nd log n ). Copyright © Peter Cappello 18

Theorem 1 Let f be an increasing function such that f(n) = a. f(n/b) + c whenever b | n, where a ≥ 1, b >1 are integers, & c > 0 is real. If ( a > 1 ) then f(n) is O( nlogba ) If ( a = 1 ) then f(n) is O( logn ). When n = bk where k > 0 is integer, and a > 1, f(n) = C 1 nlogba + C 2, where C 1 = f(1) + c/(a – 1) and C 2 = – c/(a – 1). Copyright © Peter Cappello 19