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

Slides:

Advertisements

Similar presentations

Introduction to Algorithms

Advertisements

Introduction to Algorithms 6.046J/18.401J

Introduction to Algorithms 6.046J/18.401J/SMA5503 Lecture 2 Prof. Erik Demaine.

Recurrences : 1 Chapter 3. Growth of function Chapter 4. Recurrences.

Introduction to Algorithms 6.046J

5/5/20151 Analysis of Algorithms Lecture 6&7: Master theorem and substitution method.

Nattee Niparnan. Recall What is the measurement of algorithm? How to compare two algorithms? Definition of Asymptotic Notation.

Comp 122, Spring 2004 Divide and Conquer (Merge Sort)

September 12, Algorithms and Data Structures Lecture III Simonas Šaltenis Nykredit Center for Database Research Aalborg University

Divide-and-Conquer Recursive in structure –Divide the problem into several smaller sub-problems that are similar to the original but smaller in size –Conquer.

Algorithm : Design & Analysis [4]

11 Computer Algorithms Lecture 6 Recurrence Ch. 4 (till Master Theorem) Some of these slides are courtesy of D. Plaisted et al, UNC and M. Nicolescu, UNR.

ADA: 4. Divide/Conquer1 Objective o look at several divide and conquer examples (merge sort, binary search), and 3 approaches for calculating their.

8/29/06CS 6463: AT Computational Geometry1 CS 6463: AT Computational Geometry Spring 2006 Convex Hulls Carola Wenk.

Master Theorem Section 7.3 of Rosen Spring 2010 CSCE 235 Introduction to Discrete Structures Course web-page: cse.unl.edu/~cse235 Questions:

Data Structures, Spring 2004 © L. Joskowicz 1 Data Structures – LECTURE 3 Recurrence equations Formulating recurrence equations Solving recurrence equations.

Data Structures, Spring 2006 © L. Joskowicz 1 Data Structures – LECTURE 3 Recurrence equations Formulating recurrence equations Solving recurrence equations.

CS 253: Algorithms Chapter 4 Divide-and-Conquer Recurrences Master Theorem Credit: Dr. George Bebis.

Updates HW#1 has been delayed until next MONDAY. There were two errors in the assignment Merge sort runs in Θ(n log n). Insertion sort runs in Θ(n2).

Recurrences Part 3. Recursive Algorithms Recurrences are useful for analyzing recursive algorithms Recurrence – an equation or inequality that describes.

Recurrence Examples.

Recurrences The expression: is a recurrence. –Recurrence: an equation that describes a function in terms of its value on smaller functions Analysis of.

CS2336: Computer Science II

Chapter 4: Solution of recurrence relationships

Recurrence Relations Connection to recursive algorithms Techniques for solving them.

Analysis of Algorithms CS 477/677 Recurrences Instructor: George Bebis (Appendix A, Chapter 4)

Recurrences The expression: is a recurrence. –Recurrence: an equation that describes a function in terms of its value on smaller functions BIL741: Advanced.

October 1, Algorithms and Data Structures Lecture III Simonas Šaltenis Nykredit Center for Database Research Aalborg University

Introduction to Algorithms Jiafen Liu Sept

Analysis of Algorithms

Analysis of Algorithms CS 477/677

MCA 202: Discrete Mathematics Instructor Neelima Gupta

1 Computer Algorithms Lecture 7 Master Theorem Some of these slides are courtesy of D. Plaisted, UNC and M. Nicolescu, UNR.

9/10/10 A. Smith; based on slides by E. Demaine, C. Leiserson, S. Raskhodnikova, K. Wayne Adam Smith Algorithm Design and Analysis L ECTURES Divide.

DR. NAVEED AHMAD DEPARTMENT OF COMPUTER SCIENCE UNIVERSITY OF PESHAWAR LECTURE-5 Advance Algorithm Analysis.

Rational Root and Complex Conjugates Theorem. Rational Root Theorem Used to find possible rational roots (solutions) to a polynomial Possible Roots :

10/25/20151 CS 3343: Analysis of Algorithms Lecture 6&7: Master theorem and substitution method.

Recursion Algorithm : Design & Analysis [3]. In the last class… Asymptotic growth rate The Sets ,  and  Complexity Class An Example: Maximum Subsequence.

Midterm Review 1. Midterm Exam Thursday, October 15 in classroom 75 minutes Exam structure: –TRUE/FALSE questions –short questions on the topics discussed.

Recurrences – II. Comp 122, Spring 2004.

Solving Recurrences with the Substitution Method.

Lecture 5 Today, how to solve recurrences We learned “guess and proved by induction” We also learned “substitution” method Today, we learn the “master.

Divide and Conquer. Recall Divide the problem into a number of sub-problems that are smaller instances of the same problem. Conquer the sub-problems by.

1Computer Sciences. 2 GROWTH OF FUNCTIONS 3.2 STANDARD NOTATIONS AND COMMON FUNCTIONS.

Master Method Some of the slides are from Prof. Plaisted’s resources at University of North Carolina at Chapel Hill.

Recurrences (in color) It continues…. Recurrences When an algorithm calls itself recursively, its running time is described by a recurrence. When an algorithm.

BY Lecturer: Aisha Dawood. A recurrence is a function is defined in terms of:  one or more base cases, and  itself, with smaller arguments. 2.

Chapter 4: Solution of recurrence relationships Techniques: Substitution: proof by induction Tree analysis: graphical representation Master theorem: Recipe.

Equal costs at all levels

Introduction to Algorithms

Mathematical Foundations (Solving Recurrence)

Introduction to Algorithms: Recurrences

Lecture 11. Master Theorem for analyzing Recursive relations

Chapter 4: Divide and Conquer

Algorithm : Design & Analysis [4]

T(n) = aT(n/b) + cn = a(aT(n/b/b) + cn/b) + cn 2

Introduction to Algorithms 6.046J

Master Theorem Section 7.3 of Rosen Fall 2008

Algorithms and Data Structures Lecture III

Recursion-tree method

CS 3343: Analysis of Algorithms

Master Theorem Section 8.3 of Rosen Spring 2017

Introduction to Algorithms

Using The Master Method Case 1

Master Theorem Section 8.3 of Rosen Spring 2012

Analysis of Algorithms

CS200: Algorithm Analysis

Master Theorem Section 8.3 of Rosen Spring 2013

Master Theorem Section 8.3 of Rosen Spring 2018

Algorithms and Data Structures Lecture III

Presentation transcript:

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

f (n/b) Idea of master theorem f (n/b)   (1) … Recursion tree: … f (n)f (n) a f (n/b 2 ) … a h = log b n f (n)f (n) a f (n/b) a 2 f (n/b 2 ) … #leaves = a h = a log b n = n log b a n log b a   (1)

Three common cases Compare f (n) with n log b a : 1. f (n) = O(n log b a –  ) for some constant  > 0. f (n) grows polynomially slower than n log b a (by an n  factor). Solution: T(n) =  (n log b a ).

f (n/b) Idea of master theorem f (n/b)   (1) … Recursion tree: … f (n)f (n) a f (n/b 2 ) … a h = log b n f (n)f (n) a f (n/b) a 2 f (n/b 2 ) … n log b a   (1) C ASE 1: The weight increases geometrically from the root to the leaves. The leaves hold a constant fraction of the total weight.  (n log b a )

Three common cases Compare f (n) with n log b a : 2. f (n) =  (n log b a ) f (n) and n log b a grow at similar rates. Solution: T(n) =  (n log b a lg n).

f (n/b) Idea of master theorem f (n/b)   (1) … Recursion tree: … f (n)f (n) a f (n/b 2 ) … a h = log b n f (n)f (n) a f (n/b) a 2 f (n/b 2 ) … n log b a   (1) C ASE 2: The weight is approximately the same on each of the log b n levels.  (n log b a lg n)

Three common cases (cont.) Compare f (n) with n log b a : 3. f (n) =  (n log b a +  ) for some constant  > 0. f (n) grows polynomially faster than n log b a (by an n  factor), and f (n) satisfies the regularity condition that a f (n/b)  c f (n) for some constant c < 1. Solution: T(n) =  ( f (n) ).

f (n/b) Idea of master theorem f (n/b)   (1) … Recursion tree: … f (n)f (n) a f (n/b 2 ) … a h = log b n f (n)f (n) a f (n/b) a 2 f (n/b 2 ) … n log b a   (1) C ASE 3: The weight decreases geometrically from the root to the leaves. The root holds a constant fraction of the total weight.  ( f (n))

Examples Ex. T(n) = 4T(n/2) + n a = 4, b = 2  n log b a = n 2 ; f (n) = n. C ASE 1: f (n) = O(n 2 –  ) for  = 1.  T(n) =  (n 2 ). Ex. T(n) = 4T(n/2) + n 2 a = 4, b = 2  n log b a = n 2 ; f (n) = n 2. C ASE 2: f (n) =  (n 2 lg 0 n), that is, k = 0.  T(n) =  (n 2 lg n).

Examples Ex. T(n) = 4T(n/2) + n 3 a = 4, b = 2  n log b a = n 2 ; f (n) = n 3. C ASE 3: f (n) =  (n 2 +  ) for  = 1 and 4(cn/2) 3  cn 3 (reg. cond.) for c = 1/2.  T(n) =  (n 3 ). Ex. T(n) = 4T(n/2) + n 2 /lg n a = 4, b = 2  n log b a = n 2 ; f (n) = n 2 /lg n. Master method does not apply. In particular, for every constant  > 0, we have n   (lg n).