Algorithms Chapter 4 Recurrences

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "Algorithms Chapter 4 Recurrences"

Transcription

1 Algorithms Chpter 4 Recurrences Outline The substitution method The recursion tree method The mster method Instructor: Ching Chi Lin 林清池助理教授 Deprtment of Computer Science nd Engineering Ntionl Tiwn Ocen University The purpose of this chpter When n lgorithm contins recursive cll to itself, its running time often be described by recurrence A recurrence is n eqution or inequlity tht describes function in terms of its vlue on smll inputs For exmple: the worst cse running time of Merge Sort The solution is T(n=(nlgn Three methods for solving recurrences the substitution method the recursion tree method the mster method (1 T ( n T ( n / ( n if n 1, if n 1 Techniclities 1/ The running time T(n is only defined when n is n integer, since the size of the input is lwys n integer for most lgorithms For exmple: the running time of Merge Sort is relly Typiclly, we ignore the boundry conditions Since the running time of n lgorithm on constnt sized input is constnt, we hve T(n = (1 for sufficiently smll n Thus, we rewrite the recurrence s 4 (1 T ( n T ( n / T ( n / ( n T(n =T(n/+ (n if n 1, if n 1

2 Techniclities / When we stte nd solve recurrences, we often omit floors, ceilings, nd boundry conditions We forge hed without these detils nd lter determine whether or not they mtter They usully don't, but it is importnt to know when they do The substitution method 1/ The substitution method entils two steps: Guess the form of the solution Use mthemticl induction to find the constnts nd show tht the solution works This method is powerful, but it obviously be pplied only in cses when it is esy to guess the form of the nswer 5 6 The substitution method / For exmple: determine n upper bound on the recurrence guess tht the solution is T(n=O(nlgn prove tht there exist positive constnts c>0 nd n 0 such tht T(nlgn for ll nn 0 Bsis step: T(n =T( n/ + n, where T(1=1 when n=1, T(1 c 1 1lg 1=0, which is odds with T(1=1 since the recurrence does not depend directly on T(1, we replce T(1 by T(=4 nd T(=5 s the bse cses T ( c 1 lg nd T(c 1 lg for ny choice of c 1 thus, we choose c 1 = nd n 0 = The substitution method / Induction step: ssume T( n/ c n/ lg( n/ for n/ then, T(n (c n/ lg( n/ +n c nlg(n/+ n = c nlgn-c nlg+ n = c nlgn-c n + n c nlgn the lst step holds s long s c 1 There exist positive constnts c = mx{,1} nd n 0 = such tht T(n lgn for ll n n 0 7 8

3 Mking good guess Experience: T(n=T( n/ +17+n when n is lrge, the difference between T ( n/ nd T ( n/ +17 is not tht lrge: both cut n nerly evenly in hlf we mke the guess tht T (n = O(nlgn Loose upper nd lower bounds: T(n=T( n/ +n prove lower bound of T (n = (n nd n upper bound of T(n = O(n grdully lower the upper bound nd rise the lower bound until we converge on the tight solution of T (n = (nlgn Recursion trees: will be introduced lter Subtleties Consider the recurrence: T(n = T( n/ +T ( n/ +1 guess tht the solution is O(n, ie, T(n then, T(n c n/ +(c n/+1 = +1 c does not exist new guess T(n-b then, T(n (c n/ -b+(c n/-b+1 = -b+1 -b for b1 lso, the constnt c must be chosen lrge enough to hndle the boundry conditions 9 10 Avoiding pitflls It is esy to err in the use of symptotic nottion For exmple: T(n =T( n/ + n guess tht the solution is O(n, ie, T(n then, T(n (c n/ +n +n = O(n wrong the error is tht we hven't proved the exct form of the inductive hypothesis, tht is, tht T(n Chnging vribles Sometimes, little lgebric mnipultion mke n unknown recurrence similr to one you hve seen before For exmple, consider the recurrence T(n =T( n 1/ +lgn Renming m=lgn yields T( m =T( m/ + m Renming S(m=T( m produces S(m =S(m/ + m S(m = O(mlgm Chnging bck S(mtoT(m, we obtin T(n = T( m = S(m = O(mlgm = O(lgnlglgn 11 1

4 Outline The substitution method The recursion tree method The mster method The recursion tree method A recursion tree is best used to generte good guess, which is then verified by the substitution method Tolerting smll mount of sloppiness, we could use recursion tree to generte good guess One lso use recursion tree s direct proof of solution to recurrence Ides: in recursion tree, ech node represents the cost of single subproblem sum the costs within ech level to obtin set of per level costs sum ll the per level costs to determine the totl cost 1 14 An exmple For exmple: T(n =T( n/4 + (n The construction of recursion tree 1/ T(n =T(n/4 + Tolerting the sloppiness: ignore the floor in the recurrence ssume n is n exct power of 4 Rewrite the recurrence s T(n =T(n/4 + T(n ( T(n/4 T(n/4 T(n/4 (b c(n/4 c(n/4 c(n/4 T(n/ T(n/ T(n/ T(n/ T(n/ T(n/ T(n/ T(n/ T(n/ (c 15

5 log 4 n The construction of recursion tree / Determine the cost of the tree 1/ 17 c(n/4 c(n/4 c(n/4 c(n/ c(n/ c(n/ c(n/ c(n/ c(n/ c(n/ c(n/ c(n/ (/ (/ T(1T(1 T(1 T(1 T(1T(1 (n log 4 log4 n = n log 4 (d Totl:O(n The subproblem size for node t depth i is n/4 i Thus, the tree hs log 4 n + 1 levels (0, 1,,, log 4 n Ech node t depth i, hs cost of c(n/4 i for 0 i log 4 n-1 So, the totl cost over ll nodes t depth i is i *c(n/4 i =(/ i log4 log4 The lst level, t log 4 n,hs n n nodes The cost of the entire tree: 18 T ( n log 4 n 1 i0 ( ( i log 4 n ( / 1 ( / 1 ( n log 4 ( n ( log 4 log 4 n1 ( n log 4 Determine the cost of the tree / Tke dvntge of smll mounts of sloppiness, we hve T ( n log 4 n 1 i 0 i i log 4 ( ( n i 0 1 log 4 ( n 1 ( / ( n 1 O ( n ( ( n log 4 Thus, we hve derived guess of T(n=O(n log 4 Verify the correctness of our guess Now we use the substitution method to verify tht our guess is correct We wnt to show tht T(n dn for some constnt d > 0 Using the sme constnt c > 0 s before, we hve T ( n T ( n / 4 d n / 4 d ( n / 4 /dn dn, where the lst step holds for d (/1c 19 0

6 Another exmple Another exmple: T(n = T(n/+T(n/+O(n The recursion tree: c(n/ c(n/ log / n c(n/9 c(n/9 c(n/9 c(4n/9 Determine the cost of the tree The height of tree is log / n log/ log/ The recursion tree hs fewer thn n n leves The totl cost of ll leves would then be ( n log /, which is ( n lg n Also, not ll levels contribute cost of exctly Thus, we derived guess of T(n=O(nlgn Totl:O(nlgn Verify the correctness of our guess We verify the guess by the substitution method We hve T( n T( n / T(n / d( n / lg( n / d(n / lg(n / ( d( n / lgn d( n / lg ( d(n / lgn d(n / lg(/ dnlgn d(( n / lg (n / lg(/ dnlgn d(( n / lg (n / lg (n / lg dnlgn dn(lg / dnlgn for d c/(lg-(/ Outline The substitution method The recursion tree method The mster method 4

7 The mster method 1/ The mster method provides "cookbook" method for solving recurrences of the form T(n = T(n/b + f(n 1 nd b>1 re constnts f(n is n symptoticlly positive function It requires memoriztion of three cses, but then the solution of mny recurrences be determined quite esily The mster method / The recurrence T(n = T(n/b + f(ndescribes the running time of n lgorithm tht divides problem of size n into subproblems, ech of size n/b ech of subproblems is solved recursively in time T(n/b the cost of dividing nd combining the results is f(n For exmple, the recurrence rising from the MERGE SORT procedure hs =, b =, nd f (n = (n Normlly, we omit the floor nd ceiling functions when writing divide nd conquer recurrences of this form 5 6 Mster theorem Mster theorem: Let 1 nd b > 1 be constnts, let f(n be function, nd let T(n be defined on the nonnegtive integers by the recurrence where we interpret n/b to men either n/b or n/b Then, T(n be bounded symptoticlly s follows log 1 If f ( n O( n for some constnt > 0, then T ( n ( n b log If f ( n ( n b, then T ( n ( n lgn If f ( n ( n for some constnt > 0, nd if f(n/b cf(n for some constnt c<1 nd ll sufficiently lrge n, then T ( n ( f ( n 7 T(n = T(n/b + f(n Intuition behind the mster method Intuitively, the solution to the recurrence is determined by compring the two functions f(n nd n log b b Cse 1: if n log is symptoticlly lrger thn f(n by fctor of n for some constnt > 0, then the solution is T ( n ( n b Cse : if n log is symptoticlly equl to f(n, then the solution is T ( n ( n lgn Cse : if is symptoticlly smller then f(nby fctor of n, nd the function f(n stisfies the "regulrity condition tht f(n/b cf(n, then the solution is T ( n ( f ( n The three cses do not cover ll the possibilities for f(n 8 n log b

8 Using the mster method 1/ Exmple 1: T(n =9T(n/ + n For this recurrence, we hve = 9, b =, f(n = n Thus, log b log n n 9 ( n Since ( ( log 9 f n O n,where = 1, we pply cse 1 The solution is T(n = (n Exmple : T(n = T(n/ + 1 For this recurrence, we hve = 1, b = /, f(n = 1 log / 1 0 Thus, n n n 1 Since f ( n O( n (1,we pply cse The solution is T(n = (lgn Using the mster method / Exmple : T(n =T(n/4 + nlgn For this recurrence, we hve =, b = 4, f(n = nlgn Thus, log b log 079 n n 4 O( n For sufficiently lrge n, f(n/b = (n/4 lg(n/4 (/4nlgn = cf(n for c = /4 log Since ( ( 4 f n n with nd the regulrity condition holds for f(n cse pplies The solution is T(n = (nlgn 9 0 Using the mster method / Exmple 4: T(n =T(n/ + nlgn For this recurrence, we hve =, b =, f(n = nlgn The function f(n = nlgn is symptoticlly lrger log thn n b log n n But, it is not polynomilly lrger since the rtio f ( n / n ( n lg n / n lg n is symptoticlly less thn n for ny positive constnt Consequently, the recurrence flls into the gp between cse nd cse If g(n is symptoticlly lrger thn f(n by fctor of n for some constnt >0, then we sid g(n is polynomilly lrger thn f(n 1

Example 27.1 Draw a Venn diagram to show the relationship between counting numbers, whole numbers, integers, and rational numbers.

Example 27.1 Draw a Venn diagram to show the relationship between counting numbers, whole numbers, integers, and rational numbers. 2 Rtionl Numbers Integers such s 5 were importnt when solving the eqution x+5 = 0. In similr wy, frctions re importnt for solving equtions like 2x = 1. Wht bout equtions like 2x + 1 = 0? Equtions of this

More information

Polynomial Functions. Polynomial functions in one variable can be written in expanded form as ( )

Polynomial Functions. Polynomial functions in one variable can be written in expanded form as ( ) Polynomil Functions Polynomil functions in one vrible cn be written in expnded form s n n 1 n 2 2 f x = x + x + x + + x + x+ n n 1 n 2 2 1 0 Exmples of polynomils in expnded form re nd 3 8 7 4 = 5 4 +

More information

Factoring Polynomials

Factoring Polynomials Fctoring Polynomils Some definitions (not necessrily ll for secondry school mthemtics): A polynomil is the sum of one or more terms, in which ech term consists of product of constnt nd one or more vribles

More information

Math 135 Circles and Completing the Square Examples

Math 135 Circles and Completing the Square Examples Mth 135 Circles nd Completing the Squre Exmples A perfect squre is number such tht = b 2 for some rel number b. Some exmples of perfect squres re 4 = 2 2, 16 = 4 2, 169 = 13 2. We wish to hve method for

More information

PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY

PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY MAT 0630 INTERNET RESOURCES, REVIEW OF CONCEPTS AND COMMON MISTAKES PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY Contents 1. ACT Compss Prctice Tests 1 2. Common Mistkes 2 3. Distributive

More information

Integration by Substitution

Integration by Substitution Integrtion by Substitution Dr. Philippe B. Lvl Kennesw Stte University August, 8 Abstrct This hndout contins mteril on very importnt integrtion method clled integrtion by substitution. Substitution is

More information

Operations with Polynomials

Operations with Polynomials 38 Chpter P Prerequisites P.4 Opertions with Polynomils Wht you should lern: Write polynomils in stndrd form nd identify the leding coefficients nd degrees of polynomils Add nd subtrct polynomils Multiply

More information

Curve Sketching. 96 Chapter 5 Curve Sketching

Curve Sketching. 96 Chapter 5 Curve Sketching 96 Chpter 5 Curve Sketching 5 Curve Sketching A B A B A Figure 51 Some locl mximum points (A) nd minimum points (B) If (x, f(x)) is point where f(x) reches locl mximum or minimum, nd if the derivtive of

More information

Mathematics Higher Level

Mathematics Higher Level Mthemtics Higher Level Higher Mthemtics Exmintion Section : The Exmintion Mthemtics Higher Level. Structure of the exmintion pper The Higher Mthemtics Exmintion is divided into two ppers s detiled below:

More information

AREA OF A SURFACE OF REVOLUTION

AREA OF A SURFACE OF REVOLUTION AREA OF A SURFACE OF REVOLUTION h cut r πr h A surfce of revolution is formed when curve is rotted bout line. Such surfce is the lterl boundr of solid of revolution of the tpe discussed in Sections 7.

More information

Recurrences and the Complexity of Divide and Conquer Algorithms

Recurrences and the Complexity of Divide and Conquer Algorithms 2006 Antonio Crznig Recurrences nd the Complexity of Divide nd Conquer Algorithms Antonio Crznig Fculty of Informtics Università dell Svizzer itlin October 12, 2010 2006 Antonio Crznig Outline Anlysis

More information

Binary Representation of Numbers Autar Kaw

Binary Representation of Numbers Autar Kaw Binry Representtion of Numbers Autr Kw After reding this chpter, you should be ble to: 1. convert bse- rel number to its binry representtion,. convert binry number to n equivlent bse- number. In everydy

More information

Formal Languages and Automata Exam

Formal Languages and Automata Exam Forml Lnguges nd Automt Exm Fculty of Computers & Informtion Deprtment: Computer Science Grde: Third Course code: CSC 34 Totl Mrk: 8 Dte: 23//2 Time: 3 hours Answer the following questions: ) Consider

More information

Example A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding

Example A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding 1 Exmple A rectngulr box without lid is to be mde from squre crdbord of sides 18 cm by cutting equl squres from ech corner nd then folding up the sides. 1 Exmple A rectngulr box without lid is to be mde

More information

Regular Sets and Expressions

Regular Sets and Expressions Regulr Sets nd Expressions Finite utomt re importnt in science, mthemtics, nd engineering. Engineers like them ecuse they re super models for circuits (And, since the dvent of VLSI systems sometimes finite

More information

MATH 150 HOMEWORK 4 SOLUTIONS

MATH 150 HOMEWORK 4 SOLUTIONS MATH 150 HOMEWORK 4 SOLUTIONS Section 1.8 Show tht the product of two of the numbers 65 1000 8 2001 + 3 177, 79 1212 9 2399 + 2 2001, nd 24 4493 5 8192 + 7 1777 is nonnegtive. Is your proof constructive

More information

RIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS

RIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS RIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS Known for over 500 yers is the fct tht the sum of the squres of the legs of right tringle equls the squre of the hypotenuse. Tht is +b c. A simple proof is

More information

Basic Analysis of Autarky and Free Trade Models

Basic Analysis of Autarky and Free Trade Models Bsic Anlysis of Autrky nd Free Trde Models AUTARKY Autrky condition in prticulr commodity mrket refers to sitution in which country does not engge in ny trde in tht commodity with other countries. Consequently

More information

4.11 Inner Product Spaces

4.11 Inner Product Spaces 314 CHAPTER 4 Vector Spces 9. A mtrix of the form 0 0 b c 0 d 0 0 e 0 f g 0 h 0 cnnot be invertible. 10. A mtrix of the form bc d e f ghi such tht e bd = 0 cnnot be invertible. 4.11 Inner Product Spces

More information

and thus, they are similar. If k = 3 then the Jordan form of both matrices is

and thus, they are similar. If k = 3 then the Jordan form of both matrices is Homework ssignment 11 Section 7. pp. 249-25 Exercise 1. Let N 1 nd N 2 be nilpotent mtrices over the field F. Prove tht N 1 nd N 2 re similr if nd only if they hve the sme miniml polynomil. Solution: If

More information

CHAPTER 11 Numerical Differentiation and Integration

CHAPTER 11 Numerical Differentiation and Integration CHAPTER 11 Numericl Differentition nd Integrtion Differentition nd integrtion re bsic mthemticl opertions with wide rnge of pplictions in mny res of science. It is therefore importnt to hve good methods

More information

A new algorithm for generating Pythagorean triples

A new algorithm for generating Pythagorean triples A new lgorithm for generting Pythgoren triples RH Dye 1 nd RWD Nicklls 2 The Mthemticl Gzette (1998); 82 (Mrch, No. 493), p. 86 91 (JSTOR rchive) http://www.nicklls.org/dick/ppers/mths/pythgtriples1998.pdf

More information

Homework 3 Solutions

Homework 3 Solutions CS 341: Foundtions of Computer Science II Prof. Mrvin Nkym Homework 3 Solutions 1. Give NFAs with the specified numer of sttes recognizing ech of the following lnguges. In ll cses, the lphet is Σ = {,1}.

More information

Algebra Review. How well do you remember your algebra?

Algebra Review. How well do you remember your algebra? Algebr Review How well do you remember your lgebr? 1 The Order of Opertions Wht do we men when we write + 4? If we multiply we get 6 nd dding 4 gives 10. But, if we dd + 4 = 7 first, then multiply by then

More information

MODULE 3. 0, y = 0 for all y

MODULE 3. 0, y = 0 for all y Topics: Inner products MOULE 3 The inner product of two vectors: The inner product of two vectors x, y V, denoted by x, y is (in generl) complex vlued function which hs the following four properties: i)

More information

Graphs on Logarithmic and Semilogarithmic Paper

Graphs on Logarithmic and Semilogarithmic Paper 0CH_PHClter_TMSETE_ 3//00 :3 PM Pge Grphs on Logrithmic nd Semilogrithmic Pper OBJECTIVES When ou hve completed this chpter, ou should be ble to: Mke grphs on logrithmic nd semilogrithmic pper. Grph empiricl

More information

LINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES

LINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES LINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES DAVID WEBB CONTENTS Liner trnsformtions 2 The representing mtrix of liner trnsformtion 3 3 An ppliction: reflections in the plne 6 4 The lgebr of

More information

2005-06 Second Term MAT2060B 1. Supplementary Notes 3 Interchange of Differentiation and Integration

2005-06 Second Term MAT2060B 1. Supplementary Notes 3 Interchange of Differentiation and Integration Source: http://www.mth.cuhk.edu.hk/~mt26/mt26b/notes/notes3.pdf 25-6 Second Term MAT26B 1 Supplementry Notes 3 Interchnge of Differentition nd Integrtion The theme of this course is bout vrious limiting

More information

1 Numerical Solution to Quadratic Equations

1 Numerical Solution to Quadratic Equations cs42: introduction to numericl nlysis 09/4/0 Lecture 2: Introduction Prt II nd Solving Equtions Instructor: Professor Amos Ron Scribes: Yunpeng Li, Mrk Cowlishw Numericl Solution to Qudrtic Equtions Recll

More information

Babylonian Method of Computing the Square Root: Justifications Based on Fuzzy Techniques and on Computational Complexity

Babylonian Method of Computing the Square Root: Justifications Based on Fuzzy Techniques and on Computational Complexity Bbylonin Method of Computing the Squre Root: Justifictions Bsed on Fuzzy Techniques nd on Computtionl Complexity Olg Koshelev Deprtment of Mthemtics Eduction University of Texs t El Pso 500 W. University

More information

How fast can we sort? Sorting. Decision-tree model. Decision-tree for insertion sort Sort a 1, a 2, a 3. CS 3343 -- Spring 2009

How fast can we sort? Sorting. Decision-tree model. Decision-tree for insertion sort Sort a 1, a 2, a 3. CS 3343 -- Spring 2009 CS 4 -- Spring 2009 Sorting Crol Wenk Slides courtesy of Chrles Leiserson with smll chnges by Crol Wenk CS 4 Anlysis of Algorithms 1 How fst cn we sort? All the sorting lgorithms we hve seen so fr re comprison

More information

Lecture 3 Gaussian Probability Distribution

Lecture 3 Gaussian Probability Distribution Lecture 3 Gussin Probbility Distribution Introduction l Gussin probbility distribution is perhps the most used distribution in ll of science. u lso clled bell shped curve or norml distribution l Unlike

More information

Use Geometry Expressions to create a more complex locus of points. Find evidence for equivalence using Geometry Expressions.

Use Geometry Expressions to create a more complex locus of points. Find evidence for equivalence using Geometry Expressions. Lerning Objectives Loci nd Conics Lesson 3: The Ellipse Level: Preclculus Time required: 120 minutes In this lesson, students will generlize their knowledge of the circle to the ellipse. The prmetric nd

More information

5.2. LINE INTEGRALS 265. Let us quickly review the kind of integrals we have studied so far before we introduce a new one.

5.2. LINE INTEGRALS 265. Let us quickly review the kind of integrals we have studied so far before we introduce a new one. 5.2. LINE INTEGRALS 265 5.2 Line Integrls 5.2.1 Introduction Let us quickly review the kind of integrls we hve studied so fr before we introduce new one. 1. Definite integrl. Given continuous rel-vlued

More information

Treatment Spring Late Summer Fall 0.10 5.56 3.85 0.61 6.97 3.01 1.91 3.01 2.13 2.99 5.33 2.50 1.06 3.53 6.10 Mean = 1.33 Mean = 4.88 Mean = 3.

Treatment Spring Late Summer Fall 0.10 5.56 3.85 0.61 6.97 3.01 1.91 3.01 2.13 2.99 5.33 2.50 1.06 3.53 6.10 Mean = 1.33 Mean = 4.88 Mean = 3. The nlysis of vrince (ANOVA) Although the t-test is one of the most commonly used sttisticl hypothesis tests, it hs limittions. The mjor limittion is tht the t-test cn be used to compre the mens of only

More information

Or more simply put, when adding or subtracting quantities, their uncertainties add.

Or more simply put, when adding or subtracting quantities, their uncertainties add. Propgtion of Uncertint through Mthemticl Opertions Since the untit of interest in n eperiment is rrel otined mesuring tht untit directl, we must understnd how error propgtes when mthemticl opertions re

More information

4: RIEMANN SUMS, RIEMANN INTEGRALS, FUNDAMENTAL THEOREM OF CALCULUS

4: RIEMANN SUMS, RIEMANN INTEGRALS, FUNDAMENTAL THEOREM OF CALCULUS 4: RIEMA SUMS, RIEMA ITEGRALS, FUDAMETAL THEOREM OF CALCULUS STEVE HEILMA Contents 1. Review 1 2. Riemnn Sums 2 3. Riemnn Integrl 3 4. Fundmentl Theorem of Clculus 7 5. Appendix: ottion 10 1. Review Theorem

More information

Rotating DC Motors Part II

Rotating DC Motors Part II Rotting Motors rt II II.1 Motor Equivlent Circuit The next step in our consiertion of motors is to evelop n equivlent circuit which cn be use to better unerstn motor opertion. The rmtures in rel motors

More information

Module 2. Analysis of Statically Indeterminate Structures by the Matrix Force Method. Version 2 CE IIT, Kharagpur

Module 2. Analysis of Statically Indeterminate Structures by the Matrix Force Method. Version 2 CE IIT, Kharagpur Module Anlysis of Stticlly Indeterminte Structures by the Mtrix Force Method Version CE IIT, Khrgpur esson 9 The Force Method of Anlysis: Bems (Continued) Version CE IIT, Khrgpur Instructionl Objectives

More information

Mathematics. Vectors. hsn.uk.net. Higher. Contents. Vectors 128 HSN23100

Mathematics. Vectors. hsn.uk.net. Higher. Contents. Vectors 128 HSN23100 hsn.uk.net Higher Mthemtics UNIT 3 OUTCOME 1 Vectors Contents Vectors 18 1 Vectors nd Sclrs 18 Components 18 3 Mgnitude 130 4 Equl Vectors 131 5 Addition nd Subtrction of Vectors 13 6 Multipliction by

More information

Physics 43 Homework Set 9 Chapter 40 Key

Physics 43 Homework Set 9 Chapter 40 Key Physics 43 Homework Set 9 Chpter 4 Key. The wve function for n electron tht is confined to x nm is. Find the normliztion constnt. b. Wht is the probbility of finding the electron in. nm-wide region t x

More information

Integration. 148 Chapter 7 Integration

Integration. 148 Chapter 7 Integration 48 Chpter 7 Integrtion 7 Integrtion t ech, by supposing tht during ech tenth of second the object is going t constnt speed Since the object initilly hs speed, we gin suppose it mintins this speed, but

More information

Bayesian Updating with Continuous Priors Class 13, 18.05, Spring 2014 Jeremy Orloff and Jonathan Bloom

Bayesian Updating with Continuous Priors Class 13, 18.05, Spring 2014 Jeremy Orloff and Jonathan Bloom Byesin Updting with Continuous Priors Clss 3, 8.05, Spring 04 Jeremy Orloff nd Jonthn Bloom Lerning Gols. Understnd prmeterized fmily of distriutions s representing continuous rnge of hypotheses for the

More information

19. The Fermat-Euler Prime Number Theorem

19. The Fermat-Euler Prime Number Theorem 19. The Fermt-Euler Prime Number Theorem Every prime number of the form 4n 1 cn be written s sum of two squres in only one wy (side from the order of the summnds). This fmous theorem ws discovered bout

More information

Finite Automata. Informatics 2A: Lecture 3. John Longley. 25 September School of Informatics University of Edinburgh

Finite Automata. Informatics 2A: Lecture 3. John Longley. 25 September School of Informatics University of Edinburgh Lnguges nd Automt Finite Automt Informtics 2A: Lecture 3 John Longley School of Informtics University of Edinburgh jrl@inf.ed.c.uk 25 September 2015 1 / 30 Lnguges nd Automt 1 Lnguges nd Automt Wht is

More information

Assuming all values are initially zero, what are the values of A and B after executing this Verilog code inside an always block? C=1; A <= C; B = C;

Assuming all values are initially zero, what are the values of A and B after executing this Verilog code inside an always block? C=1; A <= C; B = C; B-26 Appendix B The Bsics of Logic Design Check Yourself ALU n [Arthritic Logic Unit or (rre) Arithmetic Logic Unit] A rndom-numer genertor supplied s stndrd with ll computer systems Stn Kelly-Bootle,

More information

Econ 4721 Money and Banking Problem Set 2 Answer Key

Econ 4721 Money and Banking Problem Set 2 Answer Key Econ 472 Money nd Bnking Problem Set 2 Answer Key Problem (35 points) Consider n overlpping genertions model in which consumers live for two periods. The number of people born in ech genertion grows in

More information

Double Integrals over General Regions

Double Integrals over General Regions Double Integrls over Generl egions. Let be the region in the plne bounded b the lines, x, nd x. Evlute the double integrl x dx d. Solution. We cn either slice the region verticll or horizontll. ( x x Slicing

More information

Math Review 1. , where α (alpha) is a constant between 0 and 1, is one specific functional form for the general production function.

Math Review 1. , where α (alpha) is a constant between 0 and 1, is one specific functional form for the general production function. Mth Review Vribles, Constnts nd Functions A vrible is mthemticl bbrevition for concept For emple in economics, the vrible Y usully represents the level of output of firm or the GDP of n economy, while

More information

1. In the Bohr model, compare the magnitudes of the electron s kinetic and potential energies in orbit. What does this imply?

1. In the Bohr model, compare the magnitudes of the electron s kinetic and potential energies in orbit. What does this imply? Assignment 3: Bohr s model nd lser fundmentls 1. In the Bohr model, compre the mgnitudes of the electron s kinetic nd potentil energies in orit. Wht does this imply? When n electron moves in n orit, the

More information

2.4 Circular Waveguide

2.4 Circular Waveguide .4 Circulr Wveguide y x Figure.5: A circulr wveguide of rdius. For circulr wveguide of rdius (Fig..5, we cn perform the sme sequence of steps in cylindricl coordintes s we did in rectngulr coordintes to

More information

Section 5-4 Trigonometric Functions

Section 5-4 Trigonometric Functions 5- Trigonometric Functions Section 5- Trigonometric Functions Definition of the Trigonometric Functions Clcultor Evlution of Trigonometric Functions Definition of the Trigonometric Functions Alternte Form

More information

The Chain Rule. rf dx. t t lim " (x) dt " (0) dx. df dt = df. dt dt. f (r) = rf v (1) df dx

The Chain Rule. rf dx. t t lim  (x) dt  (0) dx. df dt = df. dt dt. f (r) = rf v (1) df dx The Chin Rule The Chin Rule In this section, we generlize the chin rule to functions of more thn one vrible. In prticulr, we will show tht the product in the single-vrible chin rule extends to n inner

More information

Warm-up for Differential Calculus

Warm-up for Differential Calculus Summer Assignment Wrm-up for Differentil Clculus Who should complete this pcket? Students who hve completed Functions or Honors Functions nd will be tking Differentil Clculus in the fll of 015. Due Dte:

More information

Lecture 15 - Curve Fitting Techniques

Lecture 15 - Curve Fitting Techniques Lecture 15 - Curve Fitting Techniques Topics curve fitting motivtion liner regression Curve fitting - motivtion For root finding, we used given function to identify where it crossed zero where does fx

More information

Lectures 8 and 9 1 Rectangular waveguides

Lectures 8 and 9 1 Rectangular waveguides 1 Lectures 8 nd 9 1 Rectngulr wveguides y b x z Consider rectngulr wveguide with 0 < x b. There re two types of wves in hollow wveguide with only one conductor; Trnsverse electric wves

More information

Cypress Creek High School IB Physics SL/AP Physics B 2012 2013 MP2 Test 1 Newton s Laws. Name: SOLUTIONS Date: Period:

Cypress Creek High School IB Physics SL/AP Physics B 2012 2013 MP2 Test 1 Newton s Laws. Name: SOLUTIONS Date: Period: Nme: SOLUTIONS Dte: Period: Directions: Solve ny 5 problems. You my ttempt dditionl problems for extr credit. 1. Two blocks re sliding to the right cross horizontl surfce, s the drwing shows. In Cse A

More information

Square Roots Teacher Notes

Square Roots Teacher Notes Henri Picciotto Squre Roots Techer Notes This unit is intended to help students develop n understnding of squre roots from visul / geometric point of view, nd lso to develop their numer sense round this

More information

DlNBVRGH + Sickness Absence Monitoring Report. Executive of the Council. Purpose of report

DlNBVRGH + Sickness Absence Monitoring Report. Executive of the Council. Purpose of report DlNBVRGH + + THE CITY OF EDINBURGH COUNCIL Sickness Absence Monitoring Report Executive of the Council 8fh My 4 I.I...3 Purpose of report This report quntifies the mount of working time lost s result of

More information

SPECIAL PRODUCTS AND FACTORIZATION

SPECIAL PRODUCTS AND FACTORIZATION MODULE - Specil Products nd Fctoriztion 4 SPECIAL PRODUCTS AND FACTORIZATION In n erlier lesson you hve lernt multipliction of lgebric epressions, prticulrly polynomils. In the study of lgebr, we come

More information

Experiment 6: Friction

Experiment 6: Friction Experiment 6: Friction In previous lbs we studied Newton s lws in n idel setting, tht is, one where friction nd ir resistnce were ignored. However, from our everydy experience with motion, we know tht

More information

1.2 The Integers and Rational Numbers

1.2 The Integers and Rational Numbers .2. THE INTEGERS AND RATIONAL NUMBERS.2 The Integers n Rtionl Numers The elements of the set of integers: consist of three types of numers: Z {..., 5, 4, 3, 2,, 0,, 2, 3, 4, 5,...} I. The (positive) nturl

More information

Section A-4 Rational Expressions: Basic Operations

Section A-4 Rational Expressions: Basic Operations A- Appendi A A BASIC ALGEBRA REVIEW 7. Construction. A rectngulr open-topped bo is to be constructed out of 9- by 6-inch sheets of thin crdbord by cutting -inch squres out of ech corner nd bending the

More information

Novel Methods of Generating Self-Invertible Matrix for Hill Cipher Algorithm

Novel Methods of Generating Self-Invertible Matrix for Hill Cipher Algorithm Bibhudendr chry, Girij Snkr Rth, Srt Kumr Ptr, nd Sroj Kumr Pnigrhy Novel Methods of Generting Self-Invertible Mtrix for Hill Cipher lgorithm Bibhudendr chry Deprtment of Electronics & Communiction Engineering

More information

CS473 - Algorithms I

CS473 - Algorithms I CS473 - Algorithms I Lecture 4 The Divide-and-Conquer Design Paradigm View in slide-show mode 1 Reminder: Merge Sort Input array A sort this half sort this half Divide Conquer merge two sorted halves Combine

More information

Many algorithms, particularly divide and conquer algorithms, have time complexities which are naturally

Many algorithms, particularly divide and conquer algorithms, have time complexities which are naturally Recurrence Relations Many algorithms, particularly divide and conquer algorithms, have time complexities which are naturally modeled by recurrence relations. A recurrence relation is an equation which

More information

Lecture 5. Inner Product

Lecture 5. Inner Product Lecture 5 Inner Product Let us strt with the following problem. Given point P R nd line L R, how cn we find the point on the line closest to P? Answer: Drw line segment from P meeting the line in right

More information

Linear Equations in Two Variables

Linear Equations in Two Variables Liner Equtions in Two Vribles In this chpter, we ll use the geometry of lines to help us solve equtions. Liner equtions in two vribles If, b, ndr re rel numbers (nd if nd b re not both equl to 0) then

More information

PHY 140A: Solid State Physics. Solution to Homework #2

PHY 140A: Solid State Physics. Solution to Homework #2 PHY 140A: Solid Stte Physics Solution to Homework # TA: Xun Ji 1 October 14, 006 1 Emil: jixun@physics.ucl.edu Problem #1 Prove tht the reciprocl lttice for the reciprocl lttice is the originl lttice.

More information

Review guide for the final exam in Math 233

Review guide for the final exam in Math 233 Review guide for the finl exm in Mth 33 1 Bsic mteril. This review includes the reminder of the mteril for mth 33. The finl exm will be cumultive exm with mny of the problems coming from the mteril covered

More information

Vectors 2. 1. Recap of vectors

Vectors 2. 1. Recap of vectors Vectors 2. Recp of vectors Vectors re directed line segments - they cn be represented in component form or by direction nd mgnitude. We cn use trigonometry nd Pythgors theorem to switch between the forms

More information

COMPONENTS: COMBINED LOADING

COMPONENTS: COMBINED LOADING LECTURE COMPONENTS: COMBINED LOADING Third Edition A. J. Clrk School of Engineering Deprtment of Civil nd Environmentl Engineering 24 Chpter 8.4 by Dr. Ibrhim A. Asskkf SPRING 2003 ENES 220 Mechnics of

More information

Plotting and Graphing

Plotting and Graphing Plotting nd Grphing Much of the dt nd informtion used by engineers is presented in the form of grphs. The vlues to be plotted cn come from theoreticl or empiricl (observed) reltionships, or from mesured

More information

FUNCTIONS AND EQUATIONS. xεs. The simplest way to represent a set is by listing its members. We use the notation

FUNCTIONS AND EQUATIONS. xεs. The simplest way to represent a set is by listing its members. We use the notation FUNCTIONS AND EQUATIONS. SETS AND SUBSETS.. Definition of set. A set is ny collection of objects which re clled its elements. If x is n element of the set S, we sy tht x belongs to S nd write If y does

More information

All pay auctions with certain and uncertain prizes a comment

All pay auctions with certain and uncertain prizes a comment CENTER FOR RESEARC IN ECONOMICS AND MANAGEMENT CREAM Publiction No. 1-2015 All py uctions with certin nd uncertin prizes comment Christin Riis All py uctions with certin nd uncertin prizes comment Christin

More information

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics. W02D3_0 Group Problem: Pulleys and Ropes Constraint Conditions

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics. W02D3_0 Group Problem: Pulleys and Ropes Constraint Conditions MSSCHUSES INSIUE OF ECHNOLOGY Deprtment of hysics 8.0 W02D3_0 Group roblem: ulleys nd Ropes Constrint Conditions Consider the rrngement of pulleys nd blocks shown in the figure. he pulleys re ssumed mssless

More information

Reasoning to Solve Equations and Inequalities

Reasoning to Solve Equations and Inequalities Lesson4 Resoning to Solve Equtions nd Inequlities In erlier work in this unit, you modeled situtions with severl vriles nd equtions. For exmple, suppose you were given usiness plns for concert showing

More information

MATLAB Workshop 13 - Linear Systems of Equations

MATLAB Workshop 13 - Linear Systems of Equations MATLAB: Workshop - Liner Systems of Equtions pge MATLAB Workshop - Liner Systems of Equtions Objectives: Crete script to solve commonly occurring problem in engineering: liner systems of equtions. MATLAB

More information

Strong acids and bases

Strong acids and bases Monoprotic Acid-Bse Equiliri (CH ) ϒ Chpter monoprotic cids A monoprotic cid cn donte one proton. This chpter includes uffers; wy to fi the ph. ϒ Chpter 11 polyprotic cids A polyprotic cid cn donte multiple

More information

www.mathsbox.org.uk e.g. f(x) = x domain x 0 (cannot find the square root of negative values)

www.mathsbox.org.uk e.g. f(x) = x domain x 0 (cannot find the square root of negative values) www.mthsbo.org.uk CORE SUMMARY NOTES Functions A function is rule which genertes ectl ONE OUTPUT for EVERY INPUT. To be defined full the function hs RULE tells ou how to clculte the output from the input

More information

Distributions. (corresponding to the cumulative distribution function for the discrete case).

Distributions. (corresponding to the cumulative distribution function for the discrete case). Distributions Recll tht n integrble function f : R [,] such tht R f()d = is clled probbility density function (pdf). The distribution function for the pdf is given by F() = (corresponding to the cumultive

More information

Redistributing the Gains from Trade through Non-linear. Lump-sum Transfers

Redistributing the Gains from Trade through Non-linear. Lump-sum Transfers Redistributing the Gins from Trde through Non-liner Lump-sum Trnsfers Ysukzu Ichino Fculty of Economics, Konn University April 21, 214 Abstrct I exmine lump-sum trnsfer rules to redistribute the gins from

More information

6.2 Volumes of Revolution: The Disk Method

6.2 Volumes of Revolution: The Disk Method mth ppliction: volumes of revolution, prt ii Volumes of Revolution: The Disk Method One of the simplest pplictions of integrtion (Theorem ) nd the ccumultion process is to determine so-clled volumes of

More information

Appendix D: Completing the Square and the Quadratic Formula. In Appendix A, two special cases of expanding brackets were considered:

Appendix D: Completing the Square and the Quadratic Formula. In Appendix A, two special cases of expanding brackets were considered: Appendi D: Completing the Squre nd the Qudrtic Formul Fctoring qudrtic epressions such s: + 6 + 8 ws one of the topics introduced in Appendi C. Fctoring qudrtic epressions is useful skill tht cn help you

More information

The remaining two sides of the right triangle are called the legs of the right triangle.

The remaining two sides of the right triangle are called the legs of the right triangle. 10 MODULE 6. RADICAL EXPRESSIONS 6 Pythgoren Theorem The Pythgoren Theorem An ngle tht mesures 90 degrees is lled right ngle. If one of the ngles of tringle is right ngle, then the tringle is lled right

More information

9 CONTINUOUS DISTRIBUTIONS

9 CONTINUOUS DISTRIBUTIONS 9 CONTINUOUS DISTIBUTIONS A rndom vrible whose vlue my fll nywhere in rnge of vlues is continuous rndom vrible nd will be ssocited with some continuous distribution. Continuous distributions re to discrete

More information

6 Energy Methods And The Energy of Waves MATH 22C

6 Energy Methods And The Energy of Waves MATH 22C 6 Energy Methods And The Energy of Wves MATH 22C. Conservtion of Energy We discuss the principle of conservtion of energy for ODE s, derive the energy ssocited with the hrmonic oscilltor, nd then use this

More information

Mechanics Cycle 1 Chapter 5. Chapter 5

Mechanics Cycle 1 Chapter 5. Chapter 5 Chpter 5 Contct orces: ree Body Digrms nd Idel Ropes Pushes nd Pulls in 1D, nd Newton s Second Lw Neglecting riction ree Body Digrms Tension Along Idel Ropes (i.e., Mssless Ropes) Newton s Third Lw Bodies

More information

Solving BAMO Problems

Solving BAMO Problems Solving BAMO Problems Tom Dvis tomrdvis@erthlink.net http://www.geometer.org/mthcircles Februry 20, 2000 Abstrct Strtegies for solving problems in the BAMO contest (the By Are Mthemticl Olympid). Only

More information

Square & Square Roots

Square & Square Roots Squre & Squre Roots Squre : If nuber is ultiplied by itself then the product is the squre of the nuber. Thus the squre of is x = eg. x x Squre root: The squre root of nuber is one of two equl fctors which

More information

Rational Functions. Rational functions are the ratio of two polynomial functions. Qx bx b x bx b. x x x. ( x) ( ) ( ) ( ) and

Rational Functions. Rational functions are the ratio of two polynomial functions. Qx bx b x bx b. x x x. ( x) ( ) ( ) ( ) and Rtionl Functions Rtionl unctions re the rtio o two polynomil unctions. They cn be written in expnded orm s ( ( P x x + x + + x+ Qx bx b x bx b n n 1 n n 1 1 0 m m 1 m + m 1 + + m + 0 Exmples o rtionl unctions

More information

#A12 INTEGERS 13 (2013) THE DISTRIBUTION OF SOLUTIONS TO XY = N (MOD A) WITH AN APPLICATION TO FACTORING INTEGERS

#A12 INTEGERS 13 (2013) THE DISTRIBUTION OF SOLUTIONS TO XY = N (MOD A) WITH AN APPLICATION TO FACTORING INTEGERS #A1 INTEGERS 13 (013) THE DISTRIBUTION OF SOLUTIONS TO XY = N (MOD A) WITH AN APPLICATION TO FACTORING INTEGERS Michel O. Rubinstein 1 Pure Mthemtics, University of Wterloo, Wterloo, Ontrio, Cnd mrubinst@uwterloo.c

More information

EQUATIONS OF LINES AND PLANES

EQUATIONS OF LINES AND PLANES EQUATIONS OF LINES AND PLANES MATH 195, SECTION 59 (VIPUL NAIK) Corresponding mteril in the ook: Section 12.5. Wht students should definitely get: Prmetric eqution of line given in point-direction nd twopoint

More information

THE RATIONAL NUMBERS CHAPTER

THE RATIONAL NUMBERS CHAPTER CHAPTER THE RATIONAL NUMBERS When divided by b is not n integer, the quotient is frction.the Bbylonins, who used number system bsed on 60, epressed the quotients: 0 8 s 0 60 insted of 8 s 7 60,600 0 insted

More information

One Minute To Learn Programming: Finite Automata

One Minute To Learn Programming: Finite Automata Gret Theoreticl Ides In Computer Science Steven Rudich CS 15-251 Spring 2005 Lecture 9 Fe 8 2005 Crnegie Mellon University One Minute To Lern Progrmming: Finite Automt Let me tech you progrmming lnguge

More information

Geometry 7-1 Geometric Mean and the Pythagorean Theorem

Geometry 7-1 Geometric Mean and the Pythagorean Theorem Geometry 7-1 Geometric Men nd the Pythgoren Theorem. Geometric Men 1. Def: The geometric men etween two positive numers nd is the positive numer x where: = x. x Ex 1: Find the geometric men etween the

More information

Intro. to the Divide-and-Conquer Strategy via Merge Sort CMPSC 465 CLRS Sections 2.3, Intro. to and various parts of Chapter 4

Intro. to the Divide-and-Conquer Strategy via Merge Sort CMPSC 465 CLRS Sections 2.3, Intro. to and various parts of Chapter 4 Intro. to the Divide-and-Conquer Strategy via Merge Sort CMPSC 465 CLRS Sections 2.3, Intro. to and various parts of Chapter 4 I. Algorithm Design and Divide-and-Conquer There are various strategies we

More information

Variable Dry Run (for Python)

Variable Dry Run (for Python) Vrile Dr Run (for Pthon) Age group: Ailities ssumed: Time: Size of group: Focus Vriles Assignment Sequencing Progrmming 7 dult Ver simple progrmming, sic understnding of ssignment nd vriles 20-50 minutes

More information

Homework #6 Chapter 7 Homework Acids and Bases

Homework #6 Chapter 7 Homework Acids and Bases Homework #6 Chpter 7 Homework Acids nd Bses 18. ) H O(l) H 3O (q) OH (q) H 3 O OH Or H O(l) H (q) OH (q) H OH ) HCN(q) H O(l) H 3O (q) CN (q) H 3 O HCN CN Or HCN(q) H (q) CN (q) H CN HCN c) NH 3(q) H O(l)

More information

Thinking out of the Box... Problem It s a richer problem than we ever imagined

Thinking out of the Box... Problem It s a richer problem than we ever imagined From the Mthemtics Techer, Vol. 95, No. 8, pges 568-574 Wlter Dodge (not pictured) nd Steve Viktor Thinking out of the Bo... Problem It s richer problem thn we ever imgined The bo problem hs been stndrd

More information