Master theorem
- 1. Master Theorem CSE235 Introduction Pitfalls Examples 4th Condition Master Theorem Slides by Christopher M. Bourke Instructor: Berthe Y. Choueiry Spring 2006 Computer Science & Engineering 235 Introduction to Discrete Mathematics 1 / 25
- 2. Master Theorem CSE235 Introduction Pitfalls Examples 4th Condition Master Theorem I When analyzing algorithms, recall that we only care about the asymptotic behavior. Recursive algorithms are no different. Rather than solve exactly the recurrence relation associated with the cost of an algorithm, it is enough to give an asymptotic characterization. The main tool for doing this is the master theorem. 2 / 25
- 3. Master Theorem CSE235 Introduction Pitfalls Examples 4th Condition Master Theorem II Theorem (Master Theorem) Let T(n) be a monotonically increasing function that satisfies T(n) = aT(n b ) + f(n) T(1) = c where a ≥ 1, b ≥ 2, c > 0. If f(n) ∈ Θ(nd) where d ≥ 0, then T(n) = Θ(nd) if a < bd Θ(nd log n) if a = bd Θ(nlogb a) if a > bd 3 / 25
- 4. Master Theorem CSE235 Introduction Pitfalls Examples 4th Condition Master Theorem Pitfalls You cannot use the Master Theorem if T(n) is not monotone, ex: T(n) = sin n f(n) is not a polynomial, ex: T(n) = 2T(n 2 ) + 2n b cannot be expressed as a constant, ex: T(n) = T( √ n) Note here, that the Master Theorem does not solve a recurrence relation. Does the base case remain a concern? 4 / 25
- 5. Master Theorem CSE235 Introduction Pitfalls Examples 4th Condition Master Theorem Example 1 Let T(n) = T n 2 + 1 2 n2 + n. What are the parameters? a = b = d = Therefore which condition? 5 / 25
- 6. Master Theorem CSE235 Introduction Pitfalls Examples 4th Condition Master Theorem Example 1 Let T(n) = T n 2 + 1 2 n2 + n. What are the parameters? a = 1 b = d = Therefore which condition? 6 / 25
- 7. Master Theorem CSE235 Introduction Pitfalls Examples 4th Condition Master Theorem Example 1 Let T(n) = T n 2 + 1 2 n2 + n. What are the parameters? a = 1 b = 2 d = Therefore which condition? 7 / 25
- 8. Master Theorem CSE235 Introduction Pitfalls Examples 4th Condition Master Theorem Example 1 Let T(n) = T n 2 + 1 2 n2 + n. What are the parameters? a = 1 b = 2 d = 2 Therefore which condition? 8 / 25
- 9. Master Theorem CSE235 Introduction Pitfalls Examples 4th Condition Master Theorem Example 1 Let T(n) = T n 2 + 1 2 n2 + n. What are the parameters? a = 1 b = 2 d = 2 Therefore which condition? Since 1 < 22, case 1 applies. 9 / 25
- 10. Master Theorem CSE235 Introduction Pitfalls Examples 4th Condition Master Theorem Example 1 Let T(n) = T n 2 + 1 2 n2 + n. What are the parameters? a = 1 b = 2 d = 2 Therefore which condition? Since 1 < 22, case 1 applies. Thus we conclude that T(n) ∈ Θ(nd ) = Θ(n2 ) 10 / 25
- 11. Master Theorem CSE235 Introduction Pitfalls Examples 4th Condition Master Theorem Example 2 Let T(n) = 2T n 4 + √ n + 42. What are the parameters? a = b = d = Therefore which condition? 11 / 25
- 12. Master Theorem CSE235 Introduction Pitfalls Examples 4th Condition Master Theorem Example 2 Let T(n) = 2T n 4 + √ n + 42. What are the parameters? a = 2 b = d = Therefore which condition? 12 / 25
- 13. Master Theorem CSE235 Introduction Pitfalls Examples 4th Condition Master Theorem Example 2 Let T(n) = 2T n 4 + √ n + 42. What are the parameters? a = 2 b = 4 d = Therefore which condition? 13 / 25
- 14. Master Theorem CSE235 Introduction Pitfalls Examples 4th Condition Master Theorem Example 2 Let T(n) = 2T n 4 + √ n + 42. What are the parameters? a = 2 b = 4 d = 1 2 Therefore which condition? 14 / 25
- 15. Master Theorem CSE235 Introduction Pitfalls Examples 4th Condition Master Theorem Example 2 Let T(n) = 2T n 4 + √ n + 42. What are the parameters? a = 2 b = 4 d = 1 2 Therefore which condition? Since 2 = 4 1 2 , case 2 applies. 15 / 25
- 16. Master Theorem CSE235 Introduction Pitfalls Examples 4th Condition Master Theorem Example 2 Let T(n) = 2T n 4 + √ n + 42. What are the parameters? a = 2 b = 4 d = 1 2 Therefore which condition? Since 2 = 4 1 2 , case 2 applies. Thus we conclude that T(n) ∈ Θ(nd log n) = Θ( √ n log n) 16 / 25
- 17. Master Theorem CSE235 Introduction Pitfalls Examples 4th Condition Master Theorem Example 3 Let T(n) = 3T n 2 + 3 4 n + 1. What are the parameters? a = b = d = Therefore which condition? 17 / 25
- 18. Master Theorem CSE235 Introduction Pitfalls Examples 4th Condition Master Theorem Example 3 Let T(n) = 3T n 2 + 3 4 n + 1. What are the parameters? a = 3 b = d = Therefore which condition? 18 / 25
- 19. Master Theorem CSE235 Introduction Pitfalls Examples 4th Condition Master Theorem Example 3 Let T(n) = 3T n 2 + 3 4 n + 1. What are the parameters? a = 3 b = 2 d = Therefore which condition? 19 / 25
- 20. Master Theorem CSE235 Introduction Pitfalls Examples 4th Condition Master Theorem Example 3 Let T(n) = 3T n 2 + 3 4 n + 1. What are the parameters? a = 3 b = 2 d = 1 Therefore which condition? 20 / 25
- 21. Master Theorem CSE235 Introduction Pitfalls Examples 4th Condition Master Theorem Example 3 Let T(n) = 3T n 2 + 3 4 n + 1. What are the parameters? a = 3 b = 2 d = 1 Therefore which condition? Since 3 > 21, case 3 applies. 21 / 25
- 22. Master Theorem CSE235 Introduction Pitfalls Examples 4th Condition Master Theorem Example 3 Let T(n) = 3T n 2 + 3 4 n + 1. What are the parameters? a = 3 b = 2 d = 1 Therefore which condition? Since 3 > 21, case 3 applies. Thus we conclude that T(n) ∈ Θ(nlogb a ) = Θ(nlog2 3 ) 22 / 25
- 23. Master Theorem CSE235 Introduction Pitfalls Examples 4th Condition Master Theorem Example 3 Let T(n) = 3T n 2 + 3 4 n + 1. What are the parameters? a = 3 b = 2 d = 1 Therefore which condition? Since 3 > 21, case 3 applies. Thus we conclude that T(n) ∈ Θ(nlogb a ) = Θ(nlog2 3 ) Note that log2 3 ≈ 1.5849 . . .. Can we say that T(n) ∈ Θ(n1.5849) ? 23 / 25
- 24. Master Theorem CSE235 Introduction Pitfalls Examples 4th Condition “Fourth” Condition Recall that we cannot use the Master Theorem if f(n) (the non-recursive cost) is not polynomial. There is a limited 4-th condition of the Master Theorem that allows us to consider polylogarithmic functions. Corollary If f(n) ∈ Θ(nlogb a logk n) for some k ≥ 0 then T(n) ∈ Θ(nlogb a logk+1 n) This final condition is fairly limited and we present it merely for completeness. 24 / 25
- 25. Master Theorem CSE235 Introduction Pitfalls Examples 4th Condition “Fourth” Condition Example Say that we have the following recurrence relation: T(n) = 2T n 2 + n log n Clearly, a = 2, b = 2 but f(n) is not a polynomial. However, f(n) ∈ Θ(n log n) for k = 1, therefore, by the 4-th case of the Master Theorem we can say that T(n) ∈ Θ(n log2 n) 25 / 25