Algorithms and Computations Complexity Lecture 4 Growth of Functions Instructor : Haya Sammaneh

1. Algorithms and Computations ComplexityLecture 4Growth of FunctionsInstructor : Haya Sammaneh

2. Recurrences 􀂄 Describes the running time for a function contains recursion. 􀂄 Three methods to find the recurrences: 􀂅Substitution method. 􀂅Recursion-tree method. 􀂅The master method which provides the bounds for recurrences of the form T (n) = aT (n/b) + f (n).

3. The substitution method 􀂄 Entails two steps: 􀂅 Guess the form of the solution. 􀂅 Use mathematical induction to find the constants and show that the solution works. 􀂄 Can be used to establish either upper or lower bounds on a recurrence.

4. Example on 􀂄 Suppose T(1)= O(1) 􀂄 We guess that the solution is T (n) = O(n lg n). 􀂄 Prove that T (n) ≤ c nlg n for an appropriate choice of the constant c > 0. 􀂅 T (n/2) ≤ c (n/2)lg (n/2) = cn lg(n/2) + n =cn lg n - cn lg 2 + n because lg(n/2)= lg(n)-lg(2) T(n) =cn lg n - cn + n because lg(2)=1 ≤ cn lg n. 􀂅 where the last step holds as long as c ≥ 1.

5. Substitution Method : Example • Let’s find the running time of the merge sort (let’s assume that n=2b, for some b).

6. Substitution Method (Ex. 2) • (let’s assume that n=2b, for some b).

7. Substitution Method (Ex. 3) • Let’s find the running time of this algorithm for a 2n x2n board.

8. Avoiding pitfalls مخاطرand Changingvariables

9. The recursion-tree method 􀂄 A straightforward way to devise a good guess. 􀂅Each node represents the cost of a single sub-problem somewhere in the set of recursive function invocations. 􀂅We sum the costs within each level of the tree and then we sum all the per-level costs to determine the total cost of all levels of the recursion.

14. Example • If n= 43 (size of input) • T(n)=3 T(n/4)+cn2 • T(43)= 3 T(43/4)+c(43)2 • = 3T(42)+c(43)2 • = 3[3T(41)+c(42)2 ] +c(43)2 • = 9 T(41)+3 c(42)2 +c(43)2 • =9[ 3 T(1)+ c(4)2 ]+ 3 c(42)2 +c(43)2 • = 27 T(1) + 9 c(4)2 + 3 c(42)2 + 1c(43)2 • level =i =3 level=i=2 level=i= 1 level=i=0 33 node each with cost = T(1)3i node each with cost = c (n/4i)2 • # of level in this tree = log4 n= log4(43) = 3 levels

15. The master method The master method applies to recurrences of the form T(n) = aT(n/b) + f (n) , where a ≥ 1, b > 1, and fis asymptotically positive.

16. Three common cases Compare f (n) with nlogba: 1.f (n) =O(n logba – c) for some constant c > 0. •f (n) grows polynomially slower than nlogba (by an ncfactor). Solution:T(n) = Θ(nlogba) . 2.f (n)=Θ(nlogbalgkn) for some constant k ≥ 0. • f (n) andnlogbagrow at similar rates. Solution:T(n) = Θ(nlogbalgk+1n) .

17. Three common cases (cont.) Compare f (n) with nlogba: 3.f (n) = Ω(nlogba+c) for some constant c > 0. • f (n) grows polynomially faster thannlogba(by an nc factor), and f (n) satisfies the regularity conditionthat a f (n/b) ≤ c f (n) for some constant c < 1. Solution:T(n) = Θ( f (n)) .

18. The master method

19. Understanding the master theory

22. Examples Ex.T(n) = 4T(n/2) + n a = 4, b = 2 ⇒ nlogba= n2; f (n) = n. CASE 1: f (n) = O(n2– c) for c = 1. ∴ T(n) = Θ(n2). Ex.T(n) = 4T(n/2) + n2 a = 4, b = 2 ⇒ nlogba= n2; f (n) = n2. CASE 2: f (n) = Θ(n2lg0n), that is, k = 0. ∴ T(n) = Θ(n2lgn).

23. Examples Ex.T(n) = 4T(n/2) + n3 a = 4, b = 2 ⇒ nlogba= n2; f (n) = n3. CASE 3: f (n) = Ω(n2 + c) for c = 1 and 4(n/2) 3 ≤ cn3 (reg. cond.) for c = 1/2. ∴ T(n) = Θ(n3). [ a f (n/b) ≤ c f (n) for some constant c < 1] Ex.T(n) = 4T(n/2) + n2/lgn a = 4, b = 2 ⇒ nlogba= n2; f (n) = n2/lgn. Master method does not apply.