Solve the following recurrence relations using master theorem: 1. T(n) = 2T(n/4)+ n^0.57 2. T(n) =...

Transcribed Image Text:

Solve the following recurrence relations using master theorem: 1. T(n) = 2T(n/4)+ n^0.57 2. T(n) = 3T(n/3)+ n^0.5 3. T(n) = 2T(n/4)+ n^0.5 • Master theorem. Consider the recurrence relation: T(n) = aT() + f(n), where a> 1 and b > 1 are constants. • Casel. If f(n) • Case2. If f(n) • Case3. If f(n) = = = = (nº) (nº) where c (n) where c < log₂a logħa where с > loga = then T (n) then T(n) then T(n) = 0 (n¹og₂a) = 0 (nº logn) = 0 (f(n)) = another way to write the cases: • Case 1: Iff(n) • Case 2: If nlogba = O(f(n)), then T(n) = O(f(n)logn). • Case 3: If f(n) = N(n¹⁰a+) for some constant > 0 and af(n/b) ≤ cf(n) for some constant c < 1, then, T(n) = O(f(n)). O(n¹⁰a-) for some constant > 0, then T(n) = O(nlogba). Solve the following recurrence relations using master theorem: 1. T(n) = 2T(n/4)+ n^0.57 2. T(n) = 3T(n/3)+ n^0.5 3. T(n) = 2T(n/4)+ n^0.5 • Master theorem. Consider the recurrence relation: T(n) = aT() + f(n), where a> 1 and b > 1 are constants. • Casel. If f(n) • Case2. If f(n) • Case3. If f(n) = = = = (nº) (nº) where c (n) where c < log₂a logħa where с > loga = then T (n) then T(n) then T(n) = 0 (n¹og₂a) = 0 (nº logn) = 0 (f(n)) = another way to write the cases: • Case 1: Iff(n) • Case 2: If nlogba = O(f(n)), then T(n) = O(f(n)logn). • Case 3: If f(n) = N(n¹⁰a+) for some constant > 0 and af(n/b) ≤ cf(n) for some constant c < 1, then, T(n) = O(f(n)). O(n¹⁰a-) for some constant > 0, then T(n) = O(nlogba). Solve the following recurrence relations using master theorem: 1. T(n) = 2T(n/4)+ n^0.57 2. T(n) = 3T(n/3)+ n^0.5 3. T(n) = 2T(n/4)+ n^0.5 • Master theorem. Consider the recurrence relation: T(n) = aT() + f(n), where a> 1 and b > 1 are constants. • Casel. If f(n) • Case2. If f(n) • Case3. If f(n) = = = = (nº) (nº) where c (n) where c < log₂a logħa where с > loga = then T (n) then T(n) then T(n) = 0 (n¹og₂a) = 0 (nº logn) = 0 (f(n)) = another way to write the cases: • Case 1: Iff(n) • Case 2: If nlogba = O(f(n)), then T(n) = O(f(n)logn). • Case 3: If f(n) = N(n¹⁰a+) for some constant > 0 and af(n/b) ≤ cf(n) for some constant c < 1, then, T(n) = O(f(n)). O(n¹⁰a-) for some constant > 0, then T(n) = O(nlogba). Solve the following recurrence relations using master theorem: 1. T(n) = 2T(n/4)+ n^0.57 2. T(n) = 3T(n/3)+ n^0.5 3. T(n) = 2T(n/4)+ n^0.5 • Master theorem. Consider the recurrence relation: T(n) = aT() + f(n), where a> 1 and b > 1 are constants. • Casel. If f(n) • Case2. If f(n) • Case3. If f(n) = = = = (nº) (nº) where c (n) where c < log₂a logħa where с > loga = then T (n) then T(n) then T(n) = 0 (n¹og₂a) = 0 (nº logn) = 0 (f(n)) = another way to write the cases: • Case 1: Iff(n) • Case 2: If nlogba = O(f(n)), then T(n) = O(f(n)logn). • Case 3: If f(n) = N(n¹⁰a+) for some constant > 0 and af(n/b) ≤ cf(n) for some constant c < 1, then, T(n) = O(f(n)). O(n¹⁰a-) for some constant > 0, then T(n) = O(nlogba).

Answer rating: 100% (QA)

Lets solve each of the recurrence relations using the Master Theorem which provides a way to find th... View the full answer