RecurrenceAndMasterTheorem.ppt
- 1. Recurrences • The expression: is a recurrence. – Recurrence: an equation that describes a function in terms of its value on smaller functions 1 2 2 1 ) ( n cn n T n c n T
- 2. 2 Master Method/Theorem • Theorem 4.1 (page 73) – for T(n) = aT(n/b)+f(n), n/b may be n/b or n/b. – where a 1, b>1 are positive integers, f(n) be a non- negative function. 1. If f(n)=O(nlogb a-) for some >0, then T(n)= (nlogb a ). 2. If f(n)= (nlogb a ), then T(n)= (nlogb a lg n). 3. If f(n)=(nlogb a+) for some >0, and if af(n/b) cf(n) for some c<1 and all sufficiently large n, then T(n)= (f(n)).
- 3. 3 Implications of Master Theorem • Comparison between f(n) and nlogb a (<,=,>) • Must be asymptotically smaller (or larger) by a polynomial, i.e., n for some >0. • In case 3, the “regularity” must be satisfied, i.e., af(n/b) cf(n) for some c<1 . • There are gaps – between 1 and 2: f(n) is smaller than nlogb a , but not polynomially smaller. – between 2 and 3: f(n) is larger than nlogb a , but not polynomially larger. – in case 3, if the “regularity” fails to hold.
- 4. 4 Application of Master Theorem • T(n) = 9T(n/3)+n; – a=9,b=3, f(n) =n – nlogb a = nlog3 9 = (n2) – f(n)=O(nlog3 9-) for =1 – By case 1, T(n) = (n2). • T(n) = T(2n/3)+1 – a=1,b=3/2, f(n) =1 – nlogb a = nlog3/2 1 = (n0) = (1) – By case 2, T(n)= (lg n).
- 5. 5 Application of Master Theorem • T(n) = 3T(n/4)+nlg n; – a=3,b=4, f(n) =nlg n – nlogb a = nlog4 3 = (n0.793) – f(n)= (nlog4 3+) for 0.2 – Moreover, for large n, the “regularity” holds for c=3/4. • af(n/b) =3(n/4)lg (n/4) (3/4)nlg n = cf(n) – By case 3, T(n) = (f(n))= (nlg n).
- 6. 6 Exception to Master Theorem • T(n) = 2T(n/2)+nlg n; – a=2,b=2, f(n) =nlg n – nlogb a = nlog2 2 = (n) – f(n) is asymptotically larger than nlogb a , but not polynomially larger because – f(n)/nlogb a = lg n, which is asymptotically less than n for any >0. – Therefore,this is a gap between 2 and 3.
- 7. 7 Where Are the Gaps nlogba f(n), case 2: within constant distances c1 c2 n f(n), case 1, at least polynomially smaller Gap between case 1 and 2 n Gap between case 3 and 2 f(n), case 3, at least polynomially larger Note: 1. for case 3, the regularity also must hold. 2. if f(n) is lg n smaller, then fall in gap in 1 and 2 3. if f(n) is lg n larger, then fall in gap in 3 and 2 4. if f(n)=(nlogba lgk n), then T(n)=(nlogba lgk+1 n). (as exercise)