Disjoint Sets { 9} { 1} { 11} Disjoint Sets (cont) Operations. Disjoint Sets (cont) Disjoint Sets (cont) n elements
|
|
- Phebe Simmons
- 6 years ago
- Views:
Transcription
1 Disjoit Sets elemets { x, x, } X =, K Opeatios x Patitioed ito k sets (disjoit sets S, S,, K Fid-Set(x - etu set cotaiig x Uio(x,y - make a ew set by combiig the sets cotaiig x ad y (destoyig them S k MakeSet(x - add a ew elemet x ad make it ito a sigleto set Uio(, Fid-Set( Each set is idetified by its epesetative elemet x {,,5,,,0, } {,,,, } { } { } { } { 0,,5,,, } Fid-Set( Fid-Set( MakeSet(
2 Vesio Each set is a liked list (with poite to tail List head is the epesetative Each elemet has a poite to the head {,,5,,,0, } {,,,, } { } { } { } { 0,,5,,, } 5 0 Vesio A sequece of m opeatios ca take Θ( m time fo i to m/ MakeSet(i edfo fo i to m/ Uio(i-,i edfo i- i- i- poite updates i m / i = Θ( m i i- i- Vesio Fid-Set(x - tavese head poite O( MakeSet(x - ceate x O( Vesio (weighted-uio heuistic Modify Vesio by keepig tack of the size of each set ad appedig the smalle to the lage Uio(x,y - Apped S x to S y ad update head poites O( S x 5 0 Each time x s head poite is updated the size of x s set doubles A head poite has bee updated at most lg times if thee ae a total of elemets i the sets A sequece of m istuctios i which thee ae MakeSets takes time O(m + lg
3 Vesio Each set is a tee with paet poites oly The oot of the tee is the epesetative {,,5,,,0, } {,,,, } { } { } { } { 0,,5,,, } Vesio FidSet(x - Tavese paet poites to each oot (till x=paet(x Wost-case is Θ( 0 5 Fid-Set( 5 0 Vesio MakeSet(x - ceate x O( Uio(x,y - Lik(FidSet(x,FidSet(y O( + time fo FidSets 5 Uio(, 0 Vesio (weighted-uio ad path compessio MakeSet(x Uio(x,y p(x x FidSet(x ak(x 0 FidSet(y Lik(x,y if Lik(, if ak(x>ak(y FidSet(x the p(y x if x p(x else p(x y the p(x FidSet(p(x if ak(x=ak(y etu(p(x the ak(y++ m MakeSets, FidSets, Uios at most m MakeSets, FidSets, Liks
4 Vesio FidSet(x Tavese paet poites to each oot (till x=paet(x 0 Fid-Set( 5 Let size(x be the umbe of odes i the subtee ooted at x ak(x Lemma If x is a oot, the size(x Poof: By iductio o the umbe of Lik opeatios Base: 0 Liks All odes ae oots with size ad ak 0 Step: Assume popety holds afte i Liks ad coside the Lik i+, Lik(x,y Case ak(x>ak(y (o vice vesa The ak(x does ot chage ad size(x iceases so iequality still holds Case ak(x=ak(y= The befoe the lik size(x ad size(y + Afte lik ak(y=+ ad size(y + = = ak ( y QED 5 Lemma If x p(x the ak[x]<ak[p(x] Poof: By iductio o the umbe of opeatios Base: 0 opeatios Tivially tue sice o elemets Step: Assume popety holds afte i opeatios ad coside opeatio i+ MakeSet No chage except fo ew ode with x=p(x FidSet If a ode gets a ew paet, it is the oot of its tee ad must have eve highe ak tha the pevious paet Lik The ode that gets a ew paet, gets oe with highe ak QED Lemma Fo ay 0, the # of odes of ak is at most Poof: A ode gets ak as the esult of a Lik of two tees of ak - This ode will have size at least at that time No ode ca cotibute moe tha oce to a ak chage fom - to Sice thee ae odes, at most odes ca attai ak QED Coollay Fo ay ode x, ak( x lg
5 Fist ty at boudig cost of m istuctios x Maximum cost of a FidSetis O(lg A sequece of m opeatios costs at most O(m lg Ove estimate sice FidSets ae destuctive, educig height of tees Caot fid sequece with Ω(m FidSets each with cost Ω(lg Idea: Boud umbe of paet updates (as i Vesio Befoe FidSet(x Idea: Combie fist ad secod appoaches Thid ty at boudig cost of m istuctios x Afte FidSet(x Chage some paet updates to the ode ad the est to the opeatio Secod ty at boudig cost of m istuctios How may times ca ode x of ak get a ew paet? At most lg - times (Its paets icease i ak fom + to lg The cost of FidSets is at most O(m + # of paet updates lg lg lg # paet updates (lg lg = 0 = 0 = 0 lg + lg lg lg lg lg + lg + = Θ( lg Cost of m opeatios is O(m + lg Ove estimate! Not may baches with cosecutive aks Caot fid Ω( odes with Ω(lg paet updates ak f (k = lg lg - f (i f (i-+ 5 f (0 = 0 f (- = - goup k i 0 goup i has aks i the age [f (i-+, f (i] f ( to be detemied 0
6 Amotized Aalysis usig Aggegate Method If FidSettaveses odes x, x, K, x p the the cost fo tavesig x is chaged to the ode if x ad p(x ae i the same ak goup (befoe ad p(x is ot the oot Path Chage othewise to the opeatio Block Chage Total cost = total path chages + total block chages Maximum block chages to sigle FidSetis f (lg Maximum block chages to a sequece of m opeatios is ( + m f (lg O m f ( i f ( i = ( 0 = 0 0 f f ( i = N i times f ( j = lg j + f ( i f i = f f ( i = = f ( i f ( (lg = lg (lg + = lg Total cost = total path chages + total block chages ( + m f (lg + f (lg = O( m f (lg O m ( lg O m (lg Amotized Aalysis usig Aggegate Method To boud path chages: If ode x has a paet i a diffeet goup, this will emai the case x i goup i ca be chaged at most f (i-(f (i-+ path chages Total path chages f ( i 0 = f ( i + f ( i f ( i = f ( i + f ( i f ( i ( f ( i ( f ( i + f ( i Time to choose f ( f ( i f ( i
= 5! 3! 2! = 5! 3! (5 3)!. In general, the number of different groups of r items out of n items (when the order is ignored) is given by n!
0 Combiatoial Aalysis Copyight by Deiz Kalı 4 Combiatios Questio 4 What is the diffeece betwee the followig questio i How may 3-lette wods ca you wite usig the lettes A, B, C, D, E ii How may 3-elemet
More informationDisjoint set (Union-Find)
CS124 Lecture 7 Fall 2018 Disjoit set (Uio-Fid) For Kruskal s algorithm for the miimum spaig tree problem, we foud that we eeded a data structure for maitaiig a collectio of disjoit sets. That is, we eed
More informationAuchmuty High School Mathematics Department Sequences & Series Notes Teacher Version
equeces ad eies Auchmuty High chool Mathematics Depatmet equeces & eies Notes Teache Vesio A sequece takes the fom,,7,0,, while 7 0 is a seies. Thee ae two types of sequece/seies aithmetic ad geometic.
More informationMATH Midterm Solutions
MATH 2113 - Midtem Solutios Febuay 18 1. A bag of mables cotais 4 which ae ed, 4 which ae blue ad 4 which ae gee. a How may mables must be chose fom the bag to guaatee that thee ae the same colou? We ca
More informationa) The average (mean) of the two fractions is halfway between them: b) The answer is yes. Assume without loss of generality that p < r.
Solutios to MAML Olympiad Level 00. Factioated a) The aveage (mea) of the two factios is halfway betwee them: p ps+ q ps+ q + q s qs qs b) The aswe is yes. Assume without loss of geeality that p
More informationCHAPTER 5 : SERIES. 5.2 The Sum of a Series Sum of Power of n Positive Integers Sum of Series of Partial Fraction Difference Method
CHAPTER 5 : SERIES 5.1 Seies 5. The Sum of a Seies 5..1 Sum of Powe of Positive Iteges 5.. Sum of Seies of Patial Factio 5..3 Diffeece Method 5.3 Test of covegece 5.3.1 Divegece Test 5.3. Itegal Test 5.3.3
More informationBy the end of this section you will be able to prove the Chinese Remainder Theorem apply this theorem to solve simultaneous linear congruences
Chapte : Theoy of Modula Aithmetic 8 Sectio D Chiese Remaide Theoem By the ed of this sectio you will be able to pove the Chiese Remaide Theoem apply this theoem to solve simultaeous liea cogueces The
More informationUsing Counting Techniques to Determine Probabilities
Kowledge ticle: obability ad Statistics Usig outig Techiques to Detemie obabilities Tee Diagams ad the Fudametal outig iciple impotat aspect of pobability theoy is the ability to detemie the total umbe
More informationThe Pigeonhole Principle 3.4 Binomial Coefficients
Discete M athematic Chapte 3: Coutig 3. The Pigeohole Piciple 3.4 Biomial Coefficiets D Patic Cha School of Compute Sciece ad Egieeig South Chia Uivesity of Techology Ageda Ch 3. The Pigeohole Piciple
More informationLower Bounds for Cover-Free Families
Loe Bouds fo Cove-Fee Families Ali Z. Abdi Covet of Nazaeth High School Gade, Abas 7, Haifa Nade H. Bshouty Dept. of Compute Sciece Techio, Haifa, 3000 Apil, 05 Abstact Let F be a set of blocks of a t-set
More informationBINOMIAL THEOREM An expression consisting of two terms, connected by + or sign is called a
BINOMIAL THEOREM hapte 8 8. Oveview: 8.. A epessio cosistig of two tems, coected by + o sig is called a biomial epessio. Fo eample, + a, y,,7 4 5y, etc., ae all biomial epessios. 8.. Biomial theoem If
More informationCh 3.4 Binomial Coefficients. Pascal's Identit y and Triangle. Chapter 3.2 & 3.4. South China University of Technology
Disc ete Mathem atic Chapte 3: Coutig 3. The Pigeohole Piciple 3.4 Biomial Coefficiets D Patic Cha School of Compute Sciece ad Egieeig South Chia Uivesity of Techology Pigeohole Piciple Suppose that a
More informationBINOMIAL THEOREM NCERT An expression consisting of two terms, connected by + or sign is called a
8. Oveview: 8.. A epessio cosistig of two tems, coected by + o sig is called a biomial epessio. Fo eample, + a, y,,7 4, etc., ae all biomial 5y epessios. 8.. Biomial theoem BINOMIAL THEOREM If a ad b ae
More informationOn a Problem of Littlewood
Ž. JOURAL OF MATHEMATICAL AALYSIS AD APPLICATIOS 199, 403 408 1996 ARTICLE O. 0149 O a Poblem of Littlewood Host Alze Mosbache Stasse 10, 51545 Waldbol, Gemay Submitted by J. L. Bee Received May 19, 1995
More informationTHE ANALYTIC LARGE SIEVE
THE ANALYTIC LAGE SIEVE 1. The aalytic lage sieve I the last lectue we saw how to apply the aalytic lage sieve to deive a aithmetic fomulatio of the lage sieve, which we applied to the poblem of boudig
More informationFinite q-identities related to well-known theorems of Euler and Gauss. Johann Cigler
Fiite -idetities elated to well-ow theoems of Eule ad Gauss Joha Cigle Faultät fü Mathemati Uivesität Wie A-9 Wie, Nodbegstaße 5 email: oha.cigle@uivie.ac.at Abstact We give geealizatios of a fiite vesio
More information4/9/13. Fibonacci Heaps. H.min. H.min. Priority Queues Performance Cost Summary. COMP 160 Algorithms - Tufts University
4/9/ Priority Queues Performace Cost Summary Fiboacci Heas Oeratio Liked List Biary Hea Biomial Hea Fiboacci Hea Relaed Hea make-hea COMP 60 Algorithms - Tufts Uiversity is-emty isert Origial Slides from
More informationCS 270 Algorithms. Oliver Kullmann. Growth of Functions. Divide-and- Conquer Min-Max- Problem. Tutorial. Reading from CLRS for week 2
Geeral remarks Week 2 1 Divide ad First we cosider a importat tool for the aalysis of algorithms: Big-Oh. The we itroduce a importat algorithmic paradigm:. We coclude by presetig ad aalysig two examples.
More informationOn randomly generated non-trivially intersecting hypergraphs
O adomly geeated o-tivially itesectig hypegaphs Balázs Patkós Submitted: May 5, 009; Accepted: Feb, 010; Published: Feb 8, 010 Mathematics Subject Classificatio: 05C65, 05D05, 05D40 Abstact We popose two
More information2012 GCE A Level H2 Maths Solution Paper Let x,
GCE A Level H Maths Solutio Pape. Let, y ad z be the cost of a ticet fo ude yeas, betwee ad 5 yeas, ad ove 5 yeas categoies espectively. 9 + y + 4z =. 7 + 5y + z = 8. + 4y + 5z = 58.5 Fo ude, ticet costs
More informationCounting Functions and Subsets
CHAPTER 1 Coutig Fuctios ad Subsets This chapte of the otes is based o Chapte 12 of PJE See PJE p144 Hee ad below, the efeeces to the PJEccles book ae give as PJE The goal of this shot chapte is to itoduce
More informationConsider unordered sample of size r. This sample can be used to make r! Ordered samples (r! permutations). unordered sample
Uodeed Samples without Replacemet oside populatio of elemets a a... a. y uodeed aagemet of elemets is called a uodeed sample of size. Two uodeed samples ae diffeet oly if oe cotais a elemet ot cotaied
More information9.1 The multiplicative group of a finite field. Theorem 9.1. The multiplicative group F of a finite field is cyclic.
Chapte 9 Pimitive Roots 9.1 The multiplicative goup of a finite fld Theoem 9.1. The multiplicative goup F of a finite fld is cyclic. Remak: In paticula, if p is a pime then (Z/p) is cyclic. In fact, this
More informationUsing Difference Equations to Generalize Results for Periodic Nested Radicals
Usig Diffeece Equatios to Geealize Results fo Peiodic Nested Radicals Chis Lyd Uivesity of Rhode Islad, Depatmet of Mathematics South Kigsto, Rhode Islad 2 2 2 2 2 2 2 π = + + +... Vieta (593) 2 2 2 =
More informationEDEXCEL NATIONAL CERTIFICATE UNIT 28 FURTHER MATHEMATICS FOR TECHNICIANS OUTCOME 2- ALGEBRAIC TECHNIQUES TUTORIAL 1 - PROGRESSIONS
EDEXCEL NATIONAL CERTIFICATE UNIT 8 FURTHER MATHEMATICS FOR TECHNICIANS OUTCOME - ALGEBRAIC TECHNIQUES TUTORIAL - PROGRESSIONS CONTENTS Be able to apply algebaic techiques Aithmetic pogessio (AP): fist
More informationEXAMPLES. Leader in CBSE Coaching. Solutions of BINOMIAL THEOREM A.V.T.E. by AVTE (avte.in) Class XI
avtei EXAMPLES Solutios of AVTE by AVTE (avtei) lass XI Leade i BSE oachig 1 avtei SHORT ANSWER TYPE 1 Fid the th tem i the epasio of 1 We have T 1 1 1 1 1 1 1 1 1 1 Epad the followig (1 + ) 4 Put 1 y
More informationDANIEL YAQUBI, MADJID MIRZAVAZIRI AND YASIN SAEEDNEZHAD
MIXED -STIRLING NUMERS OF THE SEOND KIND DANIEL YAQUI, MADJID MIRZAVAZIRI AND YASIN SAEEDNEZHAD Abstact The Stilig umbe of the secod id { } couts the umbe of ways to patitio a set of labeled balls ito
More informationICS141: Discrete Mathematics for Computer Science I
Uivesity of Hawaii ICS141: Discete Mathematics fo Compute Sciece I Dept. Ifomatio & Compute Sci., Uivesity of Hawaii Ja Stelovsy based o slides by D. Bae ad D. Still Oigials by D. M. P. Fa ad D. J.L. Goss
More information( ) 1 Comparison Functions. α is strictly increasing since ( r) ( r ) α = for any positive real number c. = 0. It is said to belong to
Compaiso Fuctios I this lesso, we study stability popeties of the oautoomous system = f t, x The difficulty is that ay solutio of this system statig at x( t ) depeds o both t ad t = x Thee ae thee special
More informationMATH /19: problems for supervision in week 08 SOLUTIONS
MATH10101 2018/19: poblems fo supevisio i week 08 Q1. Let A be a set. SOLUTIONS (i Pove that the fuctio c: P(A P(A, defied by c(x A \ X, is bijective. (ii Let ow A be fiite, A. Use (i to show that fo each
More informationConditional Convergence of Infinite Products
Coditioal Covegece of Ifiite Poducts William F. Tech Ameica Mathematical Mothly 106 1999), 646-651 I this aticle we evisit the classical subject of ifiite poducts. Fo stadad defiitios ad theoems o this
More informationCS / MCS 401 Homework 3 grader solutions
CS / MCS 401 Homework 3 grader solutios assigmet due July 6, 016 writte by Jāis Lazovskis maximum poits: 33 Some questios from CLRS. Questios marked with a asterisk were ot graded. 1 Use the defiitio of
More informationChapter 3: Theory of Modular Arithmetic 38
Chapte 3: Theoy of Modula Aithmetic 38 Section D Chinese Remainde Theoem By the end of this section you will be able to pove the Chinese Remainde Theoem apply this theoem to solve simultaneous linea conguences
More informationProgression. CATsyllabus.com. CATsyllabus.com. Sequence & Series. Arithmetic Progression (A.P.) n th term of an A.P.
Pogessio Sequece & Seies A set of umbes whose domai is a eal umbe is called a SEQUENCE ad sum of the sequece is called a SERIES. If a, a, a, a 4,., a, is a sequece, the the expessio a + a + a + a 4 + a
More informationLESSON 15: COMPOUND INTEREST
High School: Expoeial Fuctios LESSON 15: COMPOUND INTEREST 1. You have see this fomula fo compoud ieest. Paamete P is the picipal amou (the moey you stat with). Paamete is the ieest ate pe yea expessed
More informationRecursion. Algorithm : Design & Analysis [3]
Recusio Algoithm : Desig & Aalysis [] I the last class Asymptotic gowth ate he Sets Ο, Ω ad Θ Complexity Class A Example: Maximum Susequece Sum Impovemet of Algoithm Compaiso of Asymptotic Behavio Aothe
More informationSkip Lists. Presentation for use with the textbook, Algorithm Design and Applications, by M. T. Goodrich and R. Tamassia, Wiley, 2015 S 3 S S 1
Presetatio for use with the textbook, Algorithm Desig ad Applicatios, by M. T. Goodrich ad R. Tamassia, Wiley, 2015 Skip Lists S 3 15 15 23 10 15 23 36 Skip Lists 1 What is a Skip List A skip list for
More informationSolution. 1 Solutions of Homework 1. Sangchul Lee. October 27, Problem 1.1
Solutio Sagchul Lee October 7, 017 1 Solutios of Homework 1 Problem 1.1 Let Ω,F,P) be a probability space. Show that if {A : N} F such that A := lim A exists, the PA) = lim PA ). Proof. Usig the cotiuity
More informationLecture 24: Observability and Constructibility
ectue 24: Obsevability ad Costuctibility 7 Obsevability ad Costuctibility Motivatio: State feedback laws deped o a kowledge of the cuet state. I some systems, xt () ca be measued diectly, e.g., positio
More informationr, this equation is graphed in figure 1.
Washigto Uivesity i St Louis Spig 8 Depatmet of Ecoomics Pof James Moley Ecoomics 4 Homewok # 3 Suggested Solutio Note: This is a suggested solutio i the sese that it outlies oe of the may possible aswes
More informationMath 104: Homework 2 solutions
Math 04: Homework solutios. A (0, ): Sice this is a ope iterval, the miimum is udefied, ad sice the set is ot bouded above, the maximum is also udefied. if A 0 ad sup A. B { m + : m, N}: This set does
More informationLecture 6: October 16, 2017
Ifomatio ad Codig Theoy Autum 207 Lectue: Madhu Tulsiai Lectue 6: Octobe 6, 207 The Method of Types Fo this lectue, we will take U to be a fiite uivese U, ad use x (x, x 2,..., x to deote a sequece of
More informationMath 166 Week-in-Review - S. Nite 11/10/2012 Page 1 of 5 WIR #9 = 1+ r eff. , where r. is the effective interest rate, r is the annual
Math 66 Week-i-Review - S. Nite // Page of Week i Review #9 (F-F.4, 4.-4.4,.-.) Simple Iteest I = Pt, whee I is the iteest, P is the picipal, is the iteest ate, ad t is the time i yeas. P( + t), whee A
More informationThis Lecture. Divide and Conquer. Merge Sort: Algorithm. Merge Sort Algorithm. MergeSort (Example) - 1. MergeSort (Example) - 2
This Lecture Divide-ad-coquer techique for algorithm desig. Example the merge sort. Writig ad solvig recurreces Divide ad Coquer Divide-ad-coquer method for algorithm desig: Divide: If the iput size is
More information4. PERMUTATIONS AND COMBINATIONS
4. PERMUTATIONS AND COMBINATIONS PREVIOUS EAMCET BITS 1. The umbe of ways i which 13 gold cois ca be distibuted amog thee pesos such that each oe gets at least two gold cois is [EAMCET-000] 1) 3 ) 4 3)
More informationMetric Space Properties
Metric Space Properties Math 40 Fial Project Preseted by: Michael Brow, Alex Cordova, ad Alyssa Sachez We have already poited out ad will recogize throughout this book the importace of compact sets. All
More informationMA541 : Real Analysis. Tutorial and Practice Problems - 1 Hints and Solutions
MA54 : Real Aalysis Tutorial ad Practice Problems - Hits ad Solutios. Suppose that S is a oempty subset of real umbers that is bouded (i.e. bouded above as well as below). Prove that if S sup S. What ca
More informationA note on random minimum length spanning trees
A ote o adom miimum legth spaig tees Ala Fieze Miklós Ruszikó Lubos Thoma Depatmet of Mathematical Scieces Caegie Mello Uivesity Pittsbugh PA15213, USA ala@adom.math.cmu.edu, usziko@luta.sztaki.hu, thoma@qwes.math.cmu.edu
More information1. By using truth tables prove that, for all statements P and Q, the statement
Author: Satiago Salazar Problems I: Mathematical Statemets ad Proofs. By usig truth tables prove that, for all statemets P ad Q, the statemet P Q ad its cotrapositive ot Q (ot P) are equivalet. I example.2.3
More informationMA131 - Analysis 1. Workbook 9 Series III
MA3 - Aalysis Workbook 9 Series III Autum 004 Cotets 4.4 Series with Positive ad Negative Terms.............. 4.5 Alteratig Series.......................... 4.6 Geeral Series.............................
More informationRandomized Algorithms I, Spring 2018, Department of Computer Science, University of Helsinki Homework 1: Solutions (Discussed January 25, 2018)
Radomized Algorithms I, Sprig 08, Departmet of Computer Sciece, Uiversity of Helsiki Homework : Solutios Discussed Jauary 5, 08). Exercise.: Cosider the followig balls-ad-bi game. We start with oe black
More informationSOME ARITHMETIC PROPERTIES OF OVERPARTITION K -TUPLES
#A17 INTEGERS 9 2009), 181-190 SOME ARITHMETIC PROPERTIES OF OVERPARTITION K -TUPLES Deick M. Keiste Depatmet of Mathematics, Pe State Uivesity, Uivesity Pak, PA 16802 dmk5075@psu.edu James A. Selles Depatmet
More informationMDIV. Multiple divisor functions
MDIV. Multiple divisor fuctios The fuctios τ k For k, defie τ k ( to be the umber of (ordered factorisatios of ito k factors, i other words, the umber of ordered k-tuples (j, j 2,..., j k with j j 2...
More informationInduction: Solutions
Writig Proofs Misha Lavrov Iductio: Solutios Wester PA ARML Practice March 6, 206. Prove that a 2 2 chessboard with ay oe square removed ca always be covered by shaped tiles. Solutio : We iduct o. For
More informationSolutions to Tutorial 3 (Week 4)
The Uiversity of Sydey School of Mathematics ad Statistics Solutios to Tutorial Week 4 MATH2962: Real ad Complex Aalysis Advaced Semester 1, 2017 Web Page: http://www.maths.usyd.edu.au/u/ug/im/math2962/
More informationHomework 3. = k 1. Let S be a set of n elements, and let a, b, c be distinct elements of S. The number of k-subsets of S is
Homewor 3 Chapter 5 pp53: 3 40 45 Chapter 6 p85: 4 6 4 30 Use combiatorial reasoig to prove the idetity 3 3 Proof Let S be a set of elemets ad let a b c be distict elemets of S The umber of -subsets of
More informationEinstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,
MB BINOMIAL THEOREM Biomial Epessio : A algebaic epessio which cotais two dissimila tems is called biomial epessio Fo eample :,,, etc / ( ) Statemet of Biomial theoem : If, R ad N, the : ( + ) = a b +
More informationA Bijective Approach to the Permutational Power of a Priority Queue
A Bijective Appoach to the Pemutational Powe of a Pioity Queue Ia M. Gessel Kuang-Yeh Wang Depatment of Mathematics Bandeis Univesity Waltham, MA 02254-9110 Abstact A pioity queue tansfoms an input pemutation
More informationMath 2784 (or 2794W) University of Connecticut
ORDERS OF GROWTH PAT SMITH Math 2784 (or 2794W) Uiversity of Coecticut Date: Mar. 2, 22. ORDERS OF GROWTH. Itroductio Gaiig a ituitive feel for the relative growth of fuctios is importat if you really
More informationIf a subset E of R contains no open interval, is it of zero measure? For instance, is the set of irrationals in [0, 1] is of measure zero?
2 Lebesgue Measure I Chapter 1 we defied the cocept of a set of measure zero, ad we have observed that every coutable set is of measure zero. Here are some atural questios: If a subset E of R cotais a
More informationLecture 12: November 13, 2018
Mathematical Toolkit Autum 2018 Lecturer: Madhur Tulsiai Lecture 12: November 13, 2018 1 Radomized polyomial idetity testig We will use our kowledge of coditioal probability to prove the followig lemma,
More information2 Banach spaces and Hilbert spaces
2 Baach spaces ad Hilbert spaces Tryig to do aalysis i the ratioal umbers is difficult for example cosider the set {x Q : x 2 2}. This set is o-empty ad bouded above but does ot have a least upper boud
More informationFIXED POINT AND HYERS-ULAM-RASSIAS STABILITY OF A QUADRATIC FUNCTIONAL EQUATION IN BANACH SPACES
IJRRAS 6 () July 0 www.apapess.com/volumes/vol6issue/ijrras_6.pdf FIXED POINT AND HYERS-UAM-RASSIAS STABIITY OF A QUADRATIC FUNCTIONA EQUATION IN BANACH SPACES E. Movahedia Behbaha Khatam Al-Abia Uivesity
More informationLecture Notes for Analysis Class
Lecture Notes for Aalysis Class Topological Spaces A topology for a set X is a collectio T of subsets of X such that: (a) X ad the empty set are i T (b) Uios of elemets of T are i T (c) Fiite itersectios
More informationDifferent kinds of Mathematical Induction
Differet ids of Mathematical Iductio () Mathematical Iductio Give A N, [ A (a A a A)] A N () (First) Priciple of Mathematical Iductio Let P() be a propositio (ope setece), if we put A { : N p() is true}
More informationExam 2 CMSC 203 Fall 2009 Name SOLUTION KEY Show All Work! 1. (16 points) Circle T if the corresponding statement is True or F if it is False.
1 (1 poits) Circle T if the correspodig statemet is True or F if it is False T F For ay positive iteger,, GCD(, 1) = 1 T F Every positive iteger is either prime or composite T F If a b mod p, the (a/p)
More informationThe Boolean Ring of Intervals
MATH 532 Lebesgue Measure Dr. Neal, WKU We ow shall apply the results obtaied about outer measure to the legth measure o the real lie. Throughout, our space X will be the set of real umbers R. Whe ecessary,
More informationChapter 0. Review of set theory. 0.1 Sets
Chapter 0 Review of set theory Set theory plays a cetral role i the theory of probability. Thus, we will ope this course with a quick review of those otios of set theory which will be used repeatedly.
More informationElectron states in a periodic potential. Assume the electrons do not interact with each other. Solve the single electron Schrodinger equation: KJ =
Electo states i a peiodic potetial Assume the electos do ot iteact with each othe Solve the sigle electo Schodige equatio: 2 F h 2 + I U ( ) Ψ( ) EΨ( ). 2m HG KJ = whee U(+R)=U(), R is ay Bavais lattice
More informationLecture 2 February 8, 2016
MIT 6.854/8.45: Advaced Algorithms Sprig 206 Prof. Akur Moitra Lecture 2 February 8, 206 Scribe: Calvi Huag, Lih V. Nguye I this lecture, we aalyze the problem of schedulig equal size tasks arrivig olie
More informationSTAT Homework 1 - Solutions
STAT-36700 Homework 1 - Solutios Fall 018 September 11, 018 This cotais solutios for Homework 1. Please ote that we have icluded several additioal commets ad approaches to the problems to give you better
More informationMath 7409 Homework 2 Fall from which we can calculate the cycle index of the action of S 5 on pairs of vertices as
Math 7409 Hoewok 2 Fall 2010 1. Eueate the equivalece classes of siple gaphs o 5 vetices by usig the patte ivetoy as a guide. The cycle idex of S 5 actig o 5 vetices is 1 x 5 120 1 10 x 3 1 x 2 15 x 1
More informationThe number of r element subsets of a set with n r elements
Popositio: is The umbe of elemet subsets of a set with elemets Poof: Each such subset aises whe we pick a fist elemet followed by a secod elemet up to a th elemet The umbe of such choices is P But this
More informationON EUCLID S AND EULER S PROOF THAT THE NUMBER OF PRIMES IS INFINITE AND SOME APPLICATIONS
Joual of Pue ad Alied Mathematics: Advaces ad Alicatios Volume 0 Numbe 03 Pages 5-58 ON EUCLID S AND EULER S PROOF THAT THE NUMBER OF PRIMES IS INFINITE AND SOME APPLICATIONS ALI H HAKAMI Deatmet of Mathematics
More informationREAL ANALYSIS II: PROBLEM SET 1 - SOLUTIONS
REAL ANALYSIS II: PROBLEM SET 1 - SOLUTIONS 18th Feb, 016 Defiitio (Lipschitz fuctio). A fuctio f : R R is said to be Lipschitz if there exists a positive real umber c such that for ay x, y i the domai
More informationKEY. Math 334 Midterm II Fall 2007 section 004 Instructor: Scott Glasgow
KEY Math 334 Midtem II Fall 7 sectio 4 Istucto: Scott Glasgow Please do NOT wite o this exam. No cedit will be give fo such wok. Rathe wite i a blue book, o o you ow pape, pefeably egieeig pape. Wite you
More informationWeek 5-6: The Binomial Coefficients
Wee 5-6: The Biomial Coefficiets March 6, 2018 1 Pascal Formula Theorem 11 (Pascal s Formula For itegers ad such that 1, ( ( ( 1 1 + 1 The umbers ( 2 ( 1 2 ( 2 are triagle umbers, that is, The petago umbers
More informationSVD ( ) Linear Algebra for. A bit of repetition. Lecture: 8. Let s try the factorization. Is there a generalization? = Q2Λ2Q (spectral theorem!
Liea Algeba fo Wieless Commuicatios Lectue: 8 Sigula Value Decompositio SVD Ove Edfos Depatmet of Electical ad Ifomatio echology Lud Uivesity it 00-04-06 Ove Edfos A bit of epetitio A vey useful matix
More informationBertrand s Postulate
Bertrad s Postulate Lola Thompso Ross Program July 3, 2009 Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 1 / 33 Bertrad s Postulate I ve said it oce ad I ll say it agai: There s always a
More informationSome Integral Mean Estimates for Polynomials
Iteatioal Mathematical Foum, Vol. 8, 23, o., 5-5 HIKARI Ltd, www.m-hikai.com Some Itegal Mea Estimates fo Polyomials Abdullah Mi, Bilal Ahmad Da ad Q. M. Dawood Depatmet of Mathematics, Uivesity of Kashmi
More information# fixed points of g. Tree to string. Repeatedly select the leaf with the smallest label, write down the label of its neighbour and remove the leaf.
Combiatorics Graph Theory Coutig labelled ad ulabelled graphs There are 2 ( 2) labelled graphs of order. The ulabelled graphs of order correspod to orbits of the actio of S o the set of labelled graphs.
More informationAssignment 5: Solutions
McGill Uiversity Departmet of Mathematics ad Statistics MATH 54 Aalysis, Fall 05 Assigmet 5: Solutios. Let y be a ubouded sequece of positive umbers satisfyig y + > y for all N. Let x be aother sequece
More informationw (1) ˆx w (1) x (1) /ρ and w (2) ˆx w (2) x (2) /ρ.
2 5. Weighted umber of late jobs 5.1. Release dates ad due dates: maximimizig the weight of o-time jobs Oce we add release dates, miimizig the umber of late jobs becomes a sigificatly harder problem. For
More informationMA131 - Analysis 1. Workbook 3 Sequences II
MA3 - Aalysis Workbook 3 Sequeces II Autum 2004 Cotets 2.8 Coverget Sequeces........................ 2.9 Algebra of Limits......................... 2 2.0 Further Useful Results........................
More informationAnalytic Continuation
Aalytic Cotiuatio The stadard example of this is give by Example Let h (z) = 1 + z + z 2 + z 3 +... kow to coverge oly for z < 1. I fact h (z) = 1/ (1 z) for such z. Yet H (z) = 1/ (1 z) is defied for
More informationHOMEWORK #4 - MA 504
HOMEWORK #4 - MA 504 PAULINHO TCHATCHATCHA Chapter 2, problem 19. (a) If A ad B are disjoit closed sets i some metric space X, prove that they are separated. (b) Prove the same for disjoit ope set. (c)
More informationIt is often useful to approximate complicated functions using simpler ones. We consider the task of approximating a function by a polynomial.
Taylor Polyomials ad Taylor Series It is ofte useful to approximate complicated fuctios usig simpler oes We cosider the task of approximatig a fuctio by a polyomial If f is at least -times differetiable
More informationMa 530 Introduction to Power Series
Ma 530 Itroductio to Power Series Please ote that there is material o power series at Visual Calculus. Some of this material was used as part of the presetatio of the topics that follow. What is a Power
More informationUniversity of Colorado Denver Dept. Math. & Stat. Sciences Applied Analysis Preliminary Exam 13 January 2012, 10:00 am 2:00 pm. Good luck!
Uiversity of Colorado Dever Dept. Math. & Stat. Scieces Applied Aalysis Prelimiary Exam 13 Jauary 01, 10:00 am :00 pm Name: The proctor will let you read the followig coditios before the exam begis, ad
More informationComplex Numbers Solutions
Complex Numbers Solutios Joseph Zoller February 7, 06 Solutios. (009 AIME I Problem ) There is a complex umber with imagiary part 64 ad a positive iteger such that Fid. [Solutio: 697] 4i + + 4i. 4i 4i
More informationThe Discrete Fourier Transform
(7) The Discete Fouie Tasfom The Discete Fouie Tasfom hat is Discete Fouie Tasfom (DFT)? (ote: It s ot DTFT discete-time Fouie tasfom) A liea tasfomatio (mati) Samples of the Fouie tasfom (DTFT) of a apeiodic
More informationSeunghee Ye Ma 8: Week 5 Oct 28
Week 5 Summary I Sectio, we go over the Mea Value Theorem ad its applicatios. I Sectio 2, we will recap what we have covered so far this term. Topics Page Mea Value Theorem. Applicatios of the Mea Value
More informationFind a formula for the exponential function whose graph is given , 1 2,16 1, 6
Math 4 Activity (Due by EOC Apr. ) Graph the followig epoetial fuctios by modifyig the graph of f. Fid the rage of each fuctio.. g. g. g 4. g. g 6. g Fid a formula for the epoetial fuctio whose graph is
More informationMath 451: Euclidean and Non-Euclidean Geometry MWF 3pm, Gasson 204 Homework 3 Solutions
Math 451: Euclidea ad No-Euclidea Geometry MWF 3pm, Gasso 204 Homework 3 Solutios Exercises from 1.4 ad 1.5 of the otes: 4.3, 4.10, 4.12, 4.14, 4.15, 5.3, 5.4, 5.5 Exercise 4.3. Explai why Hp, q) = {x
More informationHomework 1 Solutions. The exercises are from Foundations of Mathematical Analysis by Richard Johnsonbaugh and W.E. Pfaffenberger.
Homewor 1 Solutios Math 171, Sprig 2010 Hery Adams The exercises are from Foudatios of Mathematical Aalysis by Richard Johsobaugh ad W.E. Pfaffeberger. 2.2. Let h : X Y, g : Y Z, ad f : Z W. Prove that
More informationBertrand s Postulate. Theorem (Bertrand s Postulate): For every positive integer n, there is a prime p satisfying n < p 2n.
Bertrad s Postulate Our goal is to prove the followig Theorem Bertrad s Postulate: For every positive iteger, there is a prime p satisfyig < p We remark that Bertrad s Postulate is true by ispectio for,,
More informationDavenport-Schinzel Sequences and their Geometric Applications
Advaced Computatioal Geometry Sprig 2004 Daveport-Schizel Sequeces ad their Geometric Applicatios Prof. Joseph Mitchell Scribe: Mohit Gupta 1 Overview I this lecture, we itroduce the cocept of Daveport-Schizel
More informationHomework 9. (n + 1)! = 1 1
. Chapter : Questio 8 If N, the Homewor 9 Proof. We will prove this by usig iductio o. 2! + 2 3! + 3 4! + + +! +!. Base step: Whe the left had side is. Whe the right had side is 2! 2 +! 2 which proves
More informationEnd-of-Year Contest. ERHS Math Club. May 5, 2009
Ed-of-Year Cotest ERHS Math Club May 5, 009 Problem 1: There are 9 cois. Oe is fake ad weighs a little less tha the others. Fid the fake coi by weighigs. Solutio: Separate the 9 cois ito 3 groups (A, B,
More informationMATHS FOR ENGINEERS ALGEBRA TUTORIAL 8 MATHEMATICAL PROGRESSIONS AND SERIES
MATHS FOR ENGINEERS ALGEBRA TUTORIAL 8 MATHEMATICAL PROGRESSIONS AND SERIES O completio of this ttoial yo shold be able to do the followig. Eplai aithmetical ad geometic pogessios. Eplai factoial otatio
More information