5 Boolean Decision Trees (February 11)


 Alexina Thomas
 3 years ago
 Views:
Transcription
1 5 Boolea Decisio Trees (February 11) 5.1 Graph Coectivity Suppose we are give a udirected graph G, represeted as a boolea adjacecy matrix = (a ij ), where a ij = 1 if ad oly if vertices i ad j are coected by a edge. How hard is it to decide whether G is coected? Specifically, how may etries i the adjacecy matrix do we have to examie? s usual, we wat the worstcase ruig time of the best possible algorithm, as a fuctio of, the umber of vertices: D() = mi T (, G) G Here I ll use D() istead of T () to emphasize that we are lookig at determiistic algorithms. Clearly D() (, sice the adjacecy matrix is symmetric ad has zeros o the diagoal. Oe way to derive a lower boud for ay decisio problem is to cosider the size of the smallest proof or certificate that verifies the result. To prove that a graph is coected, it is clearly both ecessary ad sufficiet to reveal the edges i a arbitrary spaig tree of G. This tree costitutes a certificate that G is coected, or a 1certificate for short. Coversely, if we wat to prove that a graph is discoected, we eed to demostrate a cut with o crossig edges, that is, a partitio of the vertices ito two disjoit subsets, such that o edge has oe edpoit i each subset. Such a cut is called a 0certificate. Clearly, ay algorithm that tests coectedess must check all the edges i some 1certificate before returig True ad all the edges i a 0certificate before returig False. Let C 0 () ad C 1 () respectively deote the imum size of a 0 or 1certificate i a vertex graph, ad let C() = {C 0 (), C 1 ()}. We immediately have the followig geeral result. Theorem 1. D() C(). I the case of graph coectivity, we have C 0 () = / / = ( ( mod )/4 ad C 1 () = 1, so D() C() = ( ( mod )/4 = Ω( ). I other words, the trivial algorithm check every edge is optimal up to a small costat factor. Surprisigly, this algorithm is actually exactly optimal! Theorem. D() = ( ) Proof: I ll describe a adversary strategy that requires ay algorithm to examie every edge. The adversary maitais two graphs, Y ad M, each with vertices; iitially, Y is empty ad M is a clique. Y ( yes ) cotais the edges that the algorithm has examied ad foud to be preset i the fictioal iput graph. M ( maybe ) cotais ay edge that might be i the fictioal iput graph; i other words, a edge (i, j) is abset from M if the algorithm kows that a ij = 0. Note that Y is a subgraph of M. The adversary uses the followig simple strategy whe the algorithm examies a potetial edge: retur False uless that aswer would force the fictioal iput graph to be discoected. 1
2 Examie(i, j): if (i, j) Y derisively retur True else if (i, j) M mockigly retur False else if M \ (i, j) is discoected add (i, j) to Y grudgigly retur True else remove (i, j) from M sigh ad retur False We easily observe that with this strategy M is always coected ad Y is always acyclic. Moreover, wheever the adversary adds a edge betwee two compoets of Y, every other edge joiig those two compoets of Y has already bee queried ad removed from M. It follows that Y becomes coected oly after ( queries; at that momet, both Y ad M cosist of the same spaig tree, ad the algorithm ca safely retur True. Before the ( th query, M is coected ad Y is discoected. Sice both graphs are cosistet with the adversary s aswers, either could serve as the fictioal iput graph, which meas the algorithm caot possibly determie the correct output. graph property like coectivity that requires lookig at every possible edge to detect is called evasive. We will retur to evasive graph properties i a future lecture. 5. Strig Properties (Boolea fuctios, laguages, whatever... ) Let s look at these ideas i a little more geerality. strig property 1 is ay fuctio of the form F : {0, 1} {0, 1}. Let T (, s) deote the umber of bits i a bit iput strig s that a algorithm examies before correctly returig F (s). The determiistic decisio tree complexity of a strig property F is, as usual, the worstcase ruig time of ay algorithm to compute it, as a fuctio of the iput size : D() = mi T (, s) s = The model of computatio is the same boolea decisio tree that we saw i the very first lecture. For each iput size, we ca model ay algorithm by a rooted biary tree, where each iteral ode stores the idex of the ext bit to examie, ad each leaf stores a output value. I this model, T (, s) is the depth of the path traversed by the iput strig s i tree, ad the determiistic complexity of ay strig property is the miimum depth of ay tree that correctly computes it. strig property is evasive if D() =. We ca defie the certificate complexity of a strig property F as follows. Ituitively, a 1 certificate is a subset of the bits i the iput s that forces F (s) = 1, ad a 0certificate is a subset of bits that forces F (s) = 0. The certificate complexity C(s) of a strig s is the size of the smallest certificate cosistet with s. Fially, the certificate complexity C() of F is the imum certificate complexity of ay bit iput strig. Equivaletly, we have C() = mi T (, s) s = 1 dmittedly, this is a rather bizarre ame, but it is aalogous to graph properties such as coectedess (which we just saw), acyclicity, plaarity, ad the like. strig property is just a boolea fuctio with argumets, or equivaletly, a set of bit strigs, or equivaletly, a laguage.
3 where (as above) the mi is take over all algorithms that correctly compute F for all iputs. Origially, certificate complexity was kow as odetermiistic decisio tree complexity, sice it correspods to a odetermiistic variat of the boolea decisio tree model. The best way to thik of a odetermiistic decisio tree is as a family of determiistic decisio trees, where for each iput, we use the best decisio tree i the set for that iput. We ve see oe example of these defiitios already. For the fuctio F : {0, 1} ( {0, 1} that idicates the coectivity of a vertex graph, we have D( ( ( ) = ( ad C( ) ) = mod. s aother example, let F : {0, 1} {0, 1} deote the exactly half fuctio: F (s) = 1 if ad oly if s cotais exactly 1s ad exactly 0s. For this fuctio, we easily observe that D() = C() =. 5.3 Blum s Theorem Oe of the most geeral results about decisio tree complexity was proved by Mauel Blum. Theorem 3 (Blum). For ay strig property, C() D() C(). Proof: The first iequality C() D() is almost trivial, sice ay algorithm must examie all the bits i a certificate before returig its output. lterately, the iequality x mi y f(x, y) mi y x f(x, y) is easy to prove for ay fuctio f(x, y). To prove the other iequality, we describe a algorithm that examies C() bits. Let π s deote the smallest certificate for ay iput strig s. Let S 0 = {π s F (s) = 0} ad S 1 = {π s F (s) = 1} be the sets of all miimal 0 ad 1certificates, respectively. To simplify otatio, let f = C(). Our algorithm works i k phases, examiig at most k bits i each phase. t each phase, we keep oly the certificates that are cosistet with the bits we ve examied so far. Let S0 i deote the subset of S 0 cosistet with the bits see i the first i phases, ad defie S1 i similarly. I particular, we have S0 0 = S 0 ad S1 0 = S 1. However, sice it is poitless to carry aroud ay bit whose value we already kow, the algorithm oly maitais the bits i each certificate that have ot yet bee queried. Thus, after each iput bit x j is queried, ay certificate π that is defied i that bit positio is either removed (if π j x j ) or its legth is decreased by oe (if π j = x j ). I the ith phase, the algorithm simply chooses a arbitrary 0certificate π S i 1 0 ad queries all its (previously uqueried) bits. If all the queries agree with π, the algorithm correctly returs False; otherwise, it cotiues with the ext phase. Now I claim that each phase reduces the legth of every survivig 1certificate by at least 1. Let σ be a arbitrary 1certificate i S i 1 1. Sice every iput strig cotais either a 0certificate or a 1certificate, but ot both, there must be at least oe bit positio that appears i both σ ad the chose 0certificate π. Moreover, this bit positio was ot queried i ay earlier phase, because otherwise, at most oe of them would have survived. If the iput strig agrees with π at that commo bit positio, σ is discarded; otherwise, the legth of σ decreases, as claimed. It follows that after at most k phases, we are left with either o 1certificates, i which case the output must be False, or a sigle empty 1certificate (i.e., a 1certificate that matches the iput i every bit positio), i which case the output must be True. Sice each phase examies at most k bits, the algorithm examies at most k bits altogether. Nodetermiism meas ever havig to admit you re wrog. 3
4 We have already see examples where the first iequality is tight. The secod iequality ca be tight as well. Cosider the fuctio x 1 if k = 0 F k (x 1, x,..., x k) = F k 1 (x 1, x,..., x k 1) F k 1 (x k 1 +1, x k 1 +,..., x k) if k is eve F k 1 (x 1, x,..., x k 1) F k 1 (x k 1 +1, x k 1 +,..., x k) if k is odd This fuctio F k models a dor tree of depth k: a boolea circuit i the form of a complete biary tree, where the leaves are iputs, gates alterate betwee d ad Or at each level, ad the value at the root is the output. (Do t cofuse the dor tree with the decisio tree that evaluates it!) x 1 x 1 x x 1 x x 3 x 4 x 1 x x 3 x 4 x 5 x 6 x 7 x 8 dor trees of depth 0 through 3 It s easy to show that this fuctio is evasive, but that it has much smaller certificate complexity: D(F k ) = k C(F k ) = k/ I fact, it s so easy that I ll leave it as a homework exercise. 5.4 Radomized complexity Recall that a radomized decisio tree is just a probability distributio over a set of determiistic decisio trees, ad that the radomized complexity of a strig property is the worstcase expected ruig time of the fastest radomized decisio tree: R() = mi P s Pr P [] T (, s) gai, I m usig R() istead of the earlier otatio T () to emphasize the radomizatio. We ve already see examples where radomess helps a little bit, but a more extreme example might be more helpful. Let M : {0, 1} 3 {0, 1} deote the boolea media (or majority) fuctio x + y + z M(x, y, z) =. The we ca defie the iterated media fuctio M k : {0, 1} 3k {0, 1} as M 0 (x 1 ) = x 1 ad M k (x 1,..., x 3 k) = M(M k 1 (x 1,..., x 3 k 1), M k 1 (x 3 k 1 +1,..., x 3 k 1), M k 1 (x 3 k 1 +1,..., x 3 k)) for all k Douglas Hofstadter defied a threeplayer game called Hruska based o iterated medias. I a level0 Hruska game, each player chooses a iteger betwee 0 ad 5; the player with the media choice wis the game, ad adds his chose umber to his level1 score. For ay i 1, a leveli game cosists of six level(i 1) games, startig with leveli scores of zero; the wier is the player with the media leveli score at the ed of the game, which is the added to that player s score at level i + 1. I order to make the game fair (ad welldefied), each roud uses a differet permutatio of the players to break ties. I oce wo a level3 Hruska game by havig the media level3 score, by havig the lowest level score, by havig the media level1 score, by choosig the umber 5 o my 16th level0 tur. I do t recommed playig a level4 game, uless you wat your brai to explode. 4
5 It is ot hard to prove that D(M k ) = 3 k ad C(M k ) = k ; this is a good warmup for the previous homework exercise. The radomized complexity fits betwee these two bouds. Cosider the followig radomized algorithm for computig M(x, y, z): Query two of x, y, z at radom. If they are equal, retur their commo value. Otherwise, query the remaiig variable ad retur its value. The probability of two variables that agree is at least 1/3, so the expected ruig time of this algorithm is at most 8/3. We ca use this idea recursively to evaluate M k quickly as well. Radomly choose two of the three istaces of M k 1 ad evaluate them recursively. If they retur the same value, we re doe; otherwise (with probability at most /3) we have to evaluate the third istace. We have the recurrece R(M k ) 8 3 R(M k 1), which has the obvious solutio R(M k ) (8/3) k. s it turs out, this algorithm is optimal so R(M k ) = (8/3) k but we do t have the tools to prove this yet. Soo. Promise. Notice that i this case, the radomized complexity falls strictly betwee the certificate complexity ad the determiistic complexity. This should ot be a surprise; eve radomized algorithms have to ru log eough to certify their aswers. I geeral, we have the followig trivial extesio of Blum s theorem for ay boolea fuctio. C() R() T () C() For the dor tree fuctio F k, the radomized complexity is ( ) k k The upper boud follows from a careful radomized algorithm, which I will leave as the third part of the homework exercise. The matchig lower boud is due to Saks ad Widgerso 4. I fact, Saks ad Widgerso cojecture that this is a lower boud o the radomized complexity of ay evasive boolea fuctio. s far as I kow, this cojecture is still ope. 4 Proc. STOC
Divide and Conquer, Solving Recurrences, Integer Multiplication Scribe: Juliana Cook (2015), V. Williams Date: April 6, 2016
CS 6, Lecture 3 Divide ad Coquer, Solvig Recurreces, Iteger Multiplicatio Scribe: Juliaa Cook (05, V Williams Date: April 6, 06 Itroductio Today we will cotiue to talk about divide ad coquer, ad go ito
More informationAsymptotic Growth of Functions
CMPS Itroductio to Aalysis of Algorithms Fall 3 Asymptotic Growth of Fuctios We itroduce several types of asymptotic otatio which are used to compare the performace ad efficiecy of algorithms As we ll
More informationSection IV.5: Recurrence Relations from Algorithms
Sectio IV.5: Recurrece Relatios from Algorithms Give a recursive algorithm with iput size, we wish to fid a Θ (best big O) estimate for its ru time T() either by obtaiig a explicit formula for T() or by
More information4.1 Sigma Notation and Riemann Sums
0 the itegral. Sigma Notatio ad Riema Sums Oe strategy for calculatig the area of a regio is to cut the regio ito simple shapes, calculate the area of each simple shape, ad the add these smaller areas
More informationG Cryptographby and Imperfect Randomness February 21, Lecture 6
G22.322000 Cryptographby ad Imperfect Radomess February 2, 2006 Lecturer: Yevgeiy Dodis Lecture 6 Scribe: Esther Iserovich ad Daa Glaser We bega by fiishig the discussio of multiroud coi flippig ad the
More informationPage 2 of 14 = T(2) + 2 = [ T(3)+1 ] + 2 Substitute T(3)+1 for T(2) = T(3) + 3 = [ T(4)+1 ] + 3 Substitute T(4)+1 for T(3) = T(4) + 4 After i
Page 1 of 14 Search C455 Chapter 4  Recursio Tree Documet last modified: 02/09/2012 18:42:34 Uses: Use recursio tree to determie a good asymptotic boud o the recurrece T() = Sum the costs withi each level
More informationIn nite Sequences. Dr. Philippe B. Laval Kennesaw State University. October 9, 2008
I ite Sequeces Dr. Philippe B. Laval Keesaw State Uiversity October 9, 2008 Abstract This had out is a itroductio to i ite sequeces. mai de itios ad presets some elemetary results. It gives the I ite Sequeces
More information{{1}, {2, 4}, {3}} {{1, 3, 4}, {2}} {{1}, {2}, {3, 4}} 5.4 Stirling Numbers
. Stirlig Numbers Whe coutig various types of fuctios from., we quicly discovered that eumeratig the umber of oto fuctios was a difficult problem. For a domai of five elemets ad a rage of four elemets,
More informationInfinite Sequences and Series
CHAPTER 4 Ifiite Sequeces ad Series 4.1. Sequeces A sequece is a ifiite ordered list of umbers, for example the sequece of odd positive itegers: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29...
More information1 The Binomial Theorem: Another Approach
The Biomial Theorem: Aother Approach Pascal s Triagle I class (ad i our text we saw that, for iteger, the biomial theorem ca be stated (a + b = c a + c a b + c a b + + c ab + c b, where the coefficiets
More informationSoving Recurrence Relations
Sovig Recurrece Relatios Part 1. Homogeeous liear 2d degree relatios with costat coefficiets. Cosider the recurrece relatio ( ) T () + at ( 1) + bt ( 2) = 0 This is called a homogeeous liear 2d degree
More informationMath Discrete Math Combinatorics MULTIPLICATION PRINCIPLE:
Math 355  Discrete Math 4.14.4 Combiatorics Notes MULTIPLICATION PRINCIPLE: If there m ways to do somethig ad ways to do aother thig the there are m ways to do both. I the laguage of set theory: Let
More informationLecture 2: Karger s Min Cut Algorithm
priceto uiv. F 3 cos 5: Advaced Algorithm Desig Lecture : Karger s Mi Cut Algorithm Lecturer: Sajeev Arora Scribe:Sajeev Today s topic is simple but gorgeous: Karger s mi cut algorithm ad its extesio.
More informationDepartment of Computer Science, University of Otago
Departmet of Computer Sciece, Uiversity of Otago Techical Report OUCS200609 Permutatios Cotaiig May Patters Authors: M.H. Albert Departmet of Computer Sciece, Uiversity of Otago Micah Colema, Rya Fly
More informationx(x 1)(x 2)... (x k + 1) = [x] k n+m 1
1 Coutig mappigs For every real x ad positive iteger k, let [x] k deote the fallig factorial ad x(x 1)(x 2)... (x k + 1) ( ) x = [x] k k k!, ( ) k = 1. 0 I the sequel, X = {x 1,..., x m }, Y = {y 1,...,
More informationYour organization has a Class B IP address of 166.144.0.0 Before you implement subnetting, the Network ID and Host ID are divided as follows:
Subettig Subettig is used to subdivide a sigle class of etwork i to multiple smaller etworks. Example: Your orgaizatio has a Class B IP address of 166.144.0.0 Before you implemet subettig, the Network
More informationModule 4: Mathematical Induction
Module 4: Mathematical Iductio Theme 1: Priciple of Mathematical Iductio Mathematical iductio is used to prove statemets about atural umbers. As studets may remember, we ca write such a statemet as a predicate
More information2.7 Sequences, Sequences of Sets
2.7. SEQUENCES, SEQUENCES OF SETS 67 2.7 Sequeces, Sequeces of Sets 2.7.1 Sequeces Defiitio 190 (sequece Let S be some set. 1. A sequece i S is a fuctio f : K S where K = { N : 0 for some 0 N}. 2. For
More informationRiemann Sums y = f (x)
Riema Sums Recall that we have previously discussed the area problem I its simplest form we ca state it this way: The Area Problem Let f be a cotiuous, oegative fuctio o the closed iterval [a, b] Fid
More informationCOMP 251 Assignment 2 Solutions
COMP 251 Assigmet 2 Solutios Questio 1 Exercise 8.34 Treat the umbers as 2digit umbers i radix. Each digit rages from 0 to 1. Sort these 2digit umbers ith the RADIXSORT algorithm preseted i Sectio
More informationLecture Notes CMSC 251
We have this messy summatio to solve though First observe that the value remais costat throughout the sum, ad so we ca pull it out frot Also ote that we ca write 3 i / i ad (3/) i T () = log 3 (log ) 1
More information6 Algorithm analysis
6 Algorithm aalysis Geerally, a algorithm has three cases Best case Average case Worse case. To demostrate, let us cosider the a really simple search algorithm which searches for k i the set A{a 1 a...
More informationORDERS OF GROWTH KEITH CONRAD
ORDERS OF GROWTH KEITH CONRAD Itroductio Gaiig a ituitive feel for the relative growth of fuctios is importat if you really wat to uderstad their behavior It also helps you better grasp topics i calculus
More informationChapter 6: Variance, the law of large numbers and the MonteCarlo method
Chapter 6: Variace, the law of large umbers ad the MoteCarlo method Expected value, variace, ad Chebyshev iequality. If X is a radom variable recall that the expected value of X, E[X] is the average value
More informationIncremental calculation of weighted mean and variance
Icremetal calculatio of weighted mea ad variace Toy Fich faf@cam.ac.uk dot@dotat.at Uiversity of Cambridge Computig Service February 009 Abstract I these otes I eplai how to derive formulae for umerically
More informationReview for College Algebra Final Exam
Review for College Algebra Fial Exam (Please remember that half of the fial exam will cover chapters 14. This review sheet covers oly the ew material, from chapters 5 ad 7.) 5.1 Systems of equatios i
More informationFourier Series and the Wave Equation Part 2
Fourier Series ad the Wave Equatio Part There are two big ideas i our work this week. The first is the use of liearity to break complicated problems ito simple pieces. The secod is the use of the symmetries
More informationwhen n = 1, 2, 3, 4, 5, 6, This list represents the amount of dollars you have after n days. Note: The use of is read as and so on.
Geometric eries Before we defie what is meat by a series, we eed to itroduce a related topic, that of sequeces. Formally, a sequece is a fuctio that computes a ordered list. uppose that o day 1, you have
More informationWeek 3 Conditional probabilities, Bayes formula, WEEK 3 page 1 Expected value of a random variable
Week 3 Coditioal probabilities, Bayes formula, WEEK 3 page 1 Expected value of a radom variable We recall our discussio of 5 card poker hads. Example 13 : a) What is the probability of evet A that a 5
More informationLesson 12. Sequences and Series
Retur to List of Lessos Lesso. Sequeces ad Series A ifiite sequece { a, a, a,... a,...} ca be thought of as a list of umbers writte i defiite order ad certai patter. It is usually deoted by { a } =, or
More informationNotes on Hypothesis Testing
Probability & Statistics Grishpa Notes o Hypothesis Testig A radom sample X = X 1,..., X is observed, with joit pmf/pdf f θ x 1,..., x. The values x = x 1,..., x of X lie i some sample space X. The parameter
More informationProperties of MLE: consistency, asymptotic normality. Fisher information.
Lecture 3 Properties of MLE: cosistecy, asymptotic ormality. Fisher iformatio. I this sectio we will try to uderstad why MLEs are good. Let us recall two facts from probability that we be used ofte throughout
More informationA Gentle Introduction to Algorithms: Part II
A Getle Itroductio to Algorithms: Part II Cotets of Part I:. Merge: (to merge two sorted lists ito a sigle sorted list.) 2. Bubble Sort 3. Merge Sort: 4. The BigO, BigΘ, BigΩ otatios: asymptotic bouds
More informationCS103X: Discrete Structures Homework 4 Solutions
CS103X: Discrete Structures Homewor 4 Solutios Due February 22, 2008 Exercise 1 10 poits. Silico Valley questios: a How may possible sixfigure salaries i whole dollar amouts are there that cotai at least
More informationProject Deliverables. CS 361, Lecture 28. Outline. Project Deliverables. Administrative. Project Comments
Project Deliverables CS 361, Lecture 28 Jared Saia Uiversity of New Mexico Each Group should tur i oe group project cosistig of: About 612 pages of text (ca be loger with appedix) 612 figures (please
More informationTHE ABRACADABRA PROBLEM
THE ABRACADABRA PROBLEM FRANCESCO CARAVENNA Abstract. We preset a detailed solutio of Exercise E0.6 i [Wil9]: i a radom sequece of letters, draw idepedetly ad uiformly from the Eglish alphabet, the expected
More information1 Computing the Standard Deviation of Sample Means
Computig the Stadard Deviatio of Sample Meas Quality cotrol charts are based o sample meas ot o idividual values withi a sample. A sample is a group of items, which are cosidered all together for our aalysis.
More informationExample 2 Find the square root of 0. The only square root of 0 is 0 (since 0 is not positive or negative, so those choices don t exist here).
BEGINNING ALGEBRA Roots ad Radicals (revised summer, 00 Olso) Packet to Supplemet the Curret Textbook  Part Review of Square Roots & Irratioals (This portio ca be ay time before Part ad should mostly
More informationMeasures of Spread and Boxplots Discrete Math, Section 9.4
Measures of Spread ad Boxplots Discrete Math, Sectio 9.4 We start with a example: Example 1: Comparig Mea ad Media Compute the mea ad media of each data set: S 1 = {4, 6, 8, 10, 1, 14, 16} S = {4, 7, 9,
More informationLesson 15 ANOVA (analysis of variance)
Outlie Variability betwee group variability withi group variability total variability Fratio Computatio sums of squares (betwee/withi/total degrees of freedom (betwee/withi/total mea square (betwee/withi
More informationHandout: How to calculate time complexity? CSE 101 Winter 2014
Hadout: How to calculate time complexity? CSE 101 Witer 014 Recipe (a) Kow algorithm If you are usig a modied versio of a kow algorithm, you ca piggyback your aalysis o the complexity of the origial algorithm
More informationRecursion and Recurrences
Chapter 5 Recursio ad Recurreces 5.1 Growth Rates of Solutios to Recurreces Divide ad Coquer Algorithms Oe of the most basic ad powerful algorithmic techiques is divide ad coquer. Cosider, for example,
More informationDiscrete Mathematics and Probability Theory Spring 2014 Anant Sahai Note 13
EECS 70 Discrete Mathematics ad Probability Theory Sprig 2014 Aat Sahai Note 13 Itroductio At this poit, we have see eough examples that it is worth just takig stock of our model of probability ad may
More informationRunning Time ( 3.1) Analysis of Algorithms. Experimental Studies ( 3.1.1) Limitations of Experiments. Pseudocode ( 3.1.2) Theoretical Analysis
Ruig Time ( 3.) Aalysis of Algorithms Iput Algorithm Output A algorithm is a stepbystep procedure for solvig a problem i a fiite amout of time. Most algorithms trasform iput objects ito output objects.
More information4. Trees. 4.1 Basics. Definition: A graph having no cycles is said to be acyclic. A forest is an acyclic graph.
4. Trees Oe of the importat classes of graphs is the trees. The importace of trees is evidet from their applicatios i various areas, especially theoretical computer sciece ad molecular evolutio. 4.1 Basics
More informationSequences II. Chapter 3. 3.1 Convergent Sequences
Chapter 3 Sequeces II 3. Coverget Sequeces Plot a graph of the sequece a ) = 2, 3 2, 4 3, 5 + 4,...,,... To what limit do you thik this sequece teds? What ca you say about the sequece a )? For ǫ = 0.,
More information1.3. VERTEX DEGREES & COUNTING
35 Chapter 1: Fudametal Cocepts Sectio 1.3: Vertex Degrees ad Coutig 36 its eighbor o P. Note that P has at least three vertices. If G x v is coected, let y = v. Otherwise, a compoet cut off from P x v
More informationConvexity, Inequalities, and Norms
Covexity, Iequalities, ad Norms Covex Fuctios You are probably familiar with the otio of cocavity of fuctios. Give a twicedifferetiable fuctio ϕ: R R, We say that ϕ is covex (or cocave up) if ϕ (x) 0 for
More informationSECTION 1.5 : SUMMATION NOTATION + WORK WITH SEQUENCES
SECTION 1.5 : SUMMATION NOTATION + WORK WITH SEQUENCES Read Sectio 1.5 (pages 5 9) Overview I Sectio 1.5 we lear to work with summatio otatio ad formulas. We will also itroduce a brief overview of sequeces,
More information.04. This means $1000 is multiplied by 1.02 five times, once for each of the remaining sixmonth
Questio 1: What is a ordiary auity? Let s look at a ordiary auity that is certai ad simple. By this, we mea a auity over a fixed term whose paymet period matches the iterest coversio period. Additioally,
More informationCS85: You Can t Do That (Lower Bounds in Computer Science) Lecture Notes, Spring 2008. Amit Chakrabarti Dartmouth College
CS85: You Ca t Do That () Lecture Notes, Sprig 2008 Amit Chakrabarti Dartmouth College Latest Update: May 9, 2008 Lecture 1 Compariso Trees: Sortig ad Selectio Scribe: William Che 1.1 Sortig Defiitio 1.1.1
More information8.1 Arithmetic Sequences
MCR3U Uit 8: Sequeces & Series Page 1 of 1 8.1 Arithmetic Sequeces Defiitio: A sequece is a comma separated list of ordered terms that follow a patter. Examples: 1, 2, 3, 4, 5 : a sequece of the first
More informationMATH 361 Homework 9. Royden Royden Royden
MATH 61 Homework 9 Royde..9 First, we show that for ay subset E of the real umbers, E c + y = E + y) c traslatig the complemet is equivalet to the complemet of the traslated set). Without loss of geerality,
More informationLecture 4: Cheeger s Inequality
Spectral Graph Theory ad Applicatios WS 0/0 Lecture 4: Cheeger s Iequality Lecturer: Thomas Sauerwald & He Su Statemet of Cheeger s Iequality I this lecture we assume for simplicity that G is a dregular
More informationCS103A Handout 23 Winter 2002 February 22, 2002 Solving Recurrence Relations
CS3A Hadout 3 Witer 00 February, 00 Solvig Recurrece Relatios Itroductio A wide variety of recurrece problems occur i models. Some of these recurrece relatios ca be solved usig iteratio or some other ad
More informationSolving DivideandConquer Recurrences
Solvig DivideadCoquer Recurreces Victor Adamchik A divideadcoquer algorithm cosists of three steps: dividig a problem ito smaller subproblems solvig (recursively) each subproblem the combiig solutios
More informationUnit 20 Hypotheses Testing
Uit 2 Hypotheses Testig Objectives: To uderstad how to formulate a ull hypothesis ad a alterative hypothesis about a populatio proportio, ad how to choose a sigificace level To uderstad how to collect
More informationAlgorithms and Data Structures DV3. Arne Andersson
Algorithms ad Data Structures DV3 Are Adersso Today s lectures Textbook chapters 14, 5 Iformatiostekologi Itroductio Overview of Algorithmic Mathematics Recurreces Itroductio to Recurreces The Substitutio
More informationMath 475, Problem Set #6: Solutions
Math 475, Problem Set #6: Solutios A (a) For each poit (a, b) with a, b oegative itegers satisfyig ab 8, cout the paths from (0,0) to (a, b) where the legal steps from (i, j) are to (i 2, j), (i, j 2),
More informationThe Power of Free Branching in a General Model of Backtracking and Dynamic Programming Algorithms
The Power of Free Brachig i a Geeral Model of Backtrackig ad Dyamic Programmig Algorithms SASHKA DAVIS IDA/Ceter for Computig Scieces Bowie, MD sashka.davis@gmail.com RUSSELL IMPAGLIAZZO Dept. of Computer
More informationCS 253: Algorithms. Chapter 4. DivideandConquer Recurrences Master Theorem. Credit: Dr. George Bebis
CS 5: Algorithms Chapter 4 DivideadCoquer Recurreces Master Theorem Credit: Dr. George Bebis Recurreces ad Ruig Time Recurreces arise whe a algorithm cotais recursive calls to itself Ruig time is represeted
More informationSearching Algorithm Efficiencies
Efficiecy of Liear Search Searchig Algorithm Efficiecies Havig implemeted the liear search algorithm, how would you measure its efficiecy? A useful measure (or metric) should be geeral, applicable to ay
More informationNPTEL STRUCTURAL RELIABILITY
NPTEL Course O STRUCTURAL RELIABILITY Module # 0 Lecture 1 Course Format: Web Istructor: Dr. Aruasis Chakraborty Departmet of Civil Egieerig Idia Istitute of Techology Guwahati 1. Lecture 01: Basic Statistics
More informationLecture 3. denote the orthogonal complement of S k. Then. 1 x S k. n. 2 x T Ax = ( ) λ x. with x = 1, we have. i = λ k x 2 = λ k.
18.409 A Algorithmist s Toolkit September 17, 009 Lecture 3 Lecturer: Joatha Keler Scribe: Adre Wibisoo 1 Outlie Today s lecture covers three mai parts: CouratFischer formula ad Rayleigh quotiets The
More informationCME 302: NUMERICAL LINEAR ALGEBRA FALL 2005/06 LECTURE 8
CME 30: NUMERICAL LINEAR ALGEBRA FALL 005/06 LECTURE 8 GENE H GOLUB 1 Positive Defiite Matrices A matrix A is positive defiite if x Ax > 0 for all ozero x A positive defiite matrix has real ad positive
More information0.7 0.6 0.2 0 0 96 96.5 97 97.5 98 98.5 99 99.5 100 100.5 96.5 97 97.5 98 98.5 99 99.5 100 100.5
Sectio 13 KolmogorovSmirov test. Suppose that we have a i.i.d. sample X 1,..., X with some ukow distributio P ad we would like to test the hypothesis that P is equal to a particular distributio P 0, i.e.
More informationElementary Theory of Russian Roulette
Elemetary Theory of Russia Roulette iterestig patters of fractios Satoshi Hashiba Daisuke Miematsu Ryohei Miyadera Itroductio. Today we are goig to study mathematical theory of Russia roulette. If some
More informationChapter 5 O A Cojecture Of Erdíos Proceedigs NCUR VIII è1994è, Vol II, pp 794í798 Jeærey F Gold Departmet of Mathematics, Departmet of Physics Uiversity of Utah Do H Tucker Departmet of Mathematics Uiversity
More informationThe second difference is the sequence of differences of the first difference sequence, 2
Differece Equatios I differetial equatios, you look for a fuctio that satisfies ad equatio ivolvig derivatives. I differece equatios, istead of a fuctio of a cotiuous variable (such as time), we look for
More informationLecture 7: Borel Sets and Lebesgue Measure
EE50: Probability Foudatios for Electrical Egieers JulyNovember 205 Lecture 7: Borel Sets ad Lebesgue Measure Lecturer: Dr. Krisha Jagaatha Scribes: Ravi Kolla, Aseem Sharma, Vishakh Hegde I this lecture,
More informationConcept: Types of algorithms
Discrete Math for Bioiformatics WS 10/11:, by A. Bockmayr/K. Reiert, 18. Oktober 2010, 21:22 1001 Cocept: Types of algorithms The expositio is based o the followig sources, which are all required readig:
More informationLecture 13. Lecturer: Jonathan Kelner Scribe: Jonathan Pines (2009)
18.409 A Algorithmist s Toolkit October 27, 2009 Lecture 13 Lecturer: Joatha Keler Scribe: Joatha Pies (2009) 1 Outlie Last time, we proved the BruMikowski iequality for boxes. Today we ll go over the
More informationLearning outcomes. Algorithms and Data Structures. Time Complexity Analysis. Time Complexity Analysis How fast is the algorithm? Prof. Dr.
Algorithms ad Data Structures Algorithm efficiecy Learig outcomes Able to carry out simple asymptotic aalysisof algorithms Prof. Dr. Qi Xi 2 Time Complexity Aalysis How fast is the algorithm? Code the
More informationTHE HEIGHT OF qbinary SEARCH TREES
THE HEIGHT OF qbinary SEARCH TREES MICHAEL DRMOTA AND HELMUT PRODINGER Abstract. q biary search trees are obtaied from words, equipped with the geometric distributio istead of permutatios. The average
More informationMeasurable Functions
Measurable Fuctios Dug Le 1 1 Defiitio It is ecessary to determie the class of fuctios that will be cosidered for the Lebesgue itegratio. We wat to guaratee that the sets which arise whe workig with these
More informationOutput Analysis (2, Chapters 10 &11 Law)
B. Maddah ENMG 6 Simulatio 05/0/07 Output Aalysis (, Chapters 10 &11 Law) Comparig alterative system cofiguratio Sice the output of a simulatio is radom, the comparig differet systems via simulatio should
More informationCHAPTER 3 DIGITAL CODING OF SIGNALS
CHAPTER 3 DIGITAL CODING OF SIGNALS Computers are ofte used to automate the recordig of measuremets. The trasducers ad sigal coditioig circuits produce a voltage sigal that is proportioal to a quatity
More information1. MATHEMATICAL INDUCTION
1. MATHEMATICAL INDUCTION EXAMPLE 1: Prove that for ay iteger 1. Proof: 1 + 2 + 3 +... + ( + 1 2 (1.1 STEP 1: For 1 (1.1 is true, sice 1 1(1 + 1. 2 STEP 2: Suppose (1.1 is true for some k 1, that is 1
More informationFactoring x n 1: cyclotomic and Aurifeuillian polynomials Paul Garrett <garrett@math.umn.edu>
(March 16, 004) Factorig x 1: cyclotomic ad Aurifeuillia polyomials Paul Garrett Polyomials of the form x 1, x 3 1, x 4 1 have at least oe systematic factorizatio x 1 = (x 1)(x 1
More information8.5 Alternating infinite series
65 8.5 Alteratig ifiite series I the previous two sectios we cosidered oly series with positive terms. I this sectio we cosider series with both positive ad egative terms which alterate: positive, egative,
More informationCS100: Introduction to Computer Science
Review: History of Computers CS100: Itroductio to Computer Sciece Maiframes Miicomputers Lecture 2: Data Storage  Bits, their storage ad mai memory Persoal Computers & Workstatios Review: The Role of
More informationTHE ARITHMETIC OF INTEGERS.  multiplication, exponentiation, division, addition, and subtraction
THE ARITHMETIC OF INTEGERS  multiplicatio, expoetiatio, divisio, additio, ad subtractio What to do ad what ot to do. THE INTEGERS Recall that a iteger is oe of the whole umbers, which may be either positive,
More informationarxiv:1012.1336v2 [cs.cc] 8 Dec 2010
Uary SubsetSum is i Logspace arxiv:1012.1336v2 [cs.cc] 8 Dec 2010 1 Itroductio Daiel M. Kae December 9, 2010 I this paper we cosider the Uary SubsetSum problem which is defied as follows: Give itegers
More informationChair for Network Architectures and Services Institute of Informatics TU München Prof. Carle. Network Security. Chapter 2 Basics
Chair for Network Architectures ad Services Istitute of Iformatics TU Müche Prof. Carle Network Security Chapter 2 Basics 2.4 Radom Number Geeratio for Cryptographic Protocols Motivatio It is crucial to
More informationTaking DCOP to the Real World: Efficient Complete Solutions for Distributed MultiEvent Scheduling
Taig DCOP to the Real World: Efficiet Complete Solutios for Distributed MultiEvet Schedulig Rajiv T. Maheswara, Milid Tambe, Emma Bowrig, Joatha P. Pearce, ad Pradeep araatham Uiversity of Souther Califoria
More information2. Degree Sequences. 2.1 Degree Sequences
2. Degree Sequeces The cocept of degrees i graphs has provided a framewor for the study of various structural properties of graphs ad has therefore attracted the attetio of may graph theorists. Here we
More informationCHAPTER 3 THE TIME VALUE OF MONEY
CHAPTER 3 THE TIME VALUE OF MONEY OVERVIEW A dollar i the had today is worth more tha a dollar to be received i the future because, if you had it ow, you could ivest that dollar ad ear iterest. Of all
More informationSection 73 Estimating a Population. Requirements
Sectio 73 Estimatig a Populatio Mea: σ Kow Key Cocept This sectio presets methods for usig sample data to fid a poit estimate ad cofidece iterval estimate of a populatio mea. A key requiremet i this sectio
More informationLecture 5: Span, linear independence, bases, and dimension
Lecture 5: Spa, liear idepedece, bases, ad dimesio Travis Schedler Thurs, Sep 23, 2010 (versio: 9/21 9:55 PM) 1 Motivatio Motivatio To uderstad what it meas that R has dimesio oe, R 2 dimesio 2, etc.;
More informationSAMPLE QUESTIONS FOR FINAL EXAM. (1) (2) (3) (4) Find the following using the definition of the Riemann integral: (2x + 1)dx
SAMPLE QUESTIONS FOR FINAL EXAM REAL ANALYSIS I FALL 006 3 4 Fid the followig usig the defiitio of the Riema itegral: a 0 x + dx 3 Cosider the partitio P x 0 3, x 3 +, x 3 +,......, x 3 3 + 3 of the iterval
More informationThe Euler Totient, the Möbius and the Divisor Functions
The Euler Totiet, the Möbius ad the Divisor Fuctios Rosica Dieva July 29, 2005 Mout Holyoke College South Hadley, MA 01075 1 Ackowledgemets This work was supported by the Mout Holyoke College fellowship
More information0,1 is an accumulation
Sectio 5.4 1 Accumulatio Poits Sectio 5.4 BolzaoWeierstrass ad HeieBorel Theorems Purpose of Sectio: To itroduce the cocept of a accumulatio poit of a set, ad state ad prove two major theorems of real
More informationHere are a couple of warnings to my students who may be here to get a copy of what happened on a day that you missed.
This documet was writte ad copyrighted by Paul Dawkis. Use of this documet ad its olie versio is govered by the Terms ad Coditios of Use located at http://tutorial.math.lamar.edu/terms.asp. The olie versio
More informationModified Line Search Method for Global Optimization
Modified Lie Search Method for Global Optimizatio Cria Grosa ad Ajith Abraham Ceter of Excellece for Quatifiable Quality of Service Norwegia Uiversity of Sciece ad Techology Trodheim, Norway {cria, ajith}@q2s.tu.o
More information3. Greatest Common Divisor  Least Common Multiple
3 Greatest Commo Divisor  Least Commo Multiple Defiitio 31: The greatest commo divisor of two atural umbers a ad b is the largest atural umber c which divides both a ad b We deote the greatest commo gcd
More informationMath Background: Review & Beyond. Design & Analysis of Algorithms COMP 482 / ELEC 420. Solving for Closed Forms. Obtaining Recurrences
Math Backgroud: Review & Beyod. Asymptotic otatio Desig & Aalysis of Algorithms COMP 48 / ELEC 40 Joh Greier. Math used i asymptotics 3. Recurreces 4. Probabilistic aalysis To do: [CLRS] 4 # T() = O()
More informationApproximating Area under a curve with rectangles. To find the area under a curve we approximate the area using rectangles and then use limits to find
1.8 Approximatig Area uder a curve with rectagles 1.6 To fid the area uder a curve we approximate the area usig rectagles ad the use limits to fid 1.4 the area. Example 1 Suppose we wat to estimate 1.
More information4 n. n 1. You shold think of the Ratio Test as a generalization of the Geometric Series Test. For example, if a n ar n is a geometric sequence then
SECTION 2.6 THE RATIO TEST 79 2.6. THE RATIO TEST We ow kow how to hadle series which we ca itegrate (the Itegral Test), ad series which are similar to geometric or pseries (the Compariso Test), but of
More informationRepeating Decimals are decimal numbers that have number(s) after the decimal point that repeat in a pattern.
5.5 Fractios ad Decimals Steps for Chagig a Fractio to a Decimal. Simplify the fractio, if possible. 2. Divide the umerator by the deomiator. d d Repeatig Decimals Repeatig Decimals are decimal umbers
More informationTHE REGRESSION MODEL IN MATRIX FORM. For simple linear regression, meaning one predictor, the model is. for i = 1, 2, 3,, n
We will cosider the liear regressio model i matrix form. For simple liear regressio, meaig oe predictor, the model is i = + x i + ε i for i =,,,, This model icludes the assumptio that the ε i s are a sample
More information