Global alignment. Genome Rearrangements Finding preserved genes. Lecture 18


 Eustacia Austin
 6 months ago
 Views:
Transcription
1 Computt onl Biology Leture 18 Genome Rerrngements Finding preserved genes We hve seen before how to rerrnge genome to obtin nother one bsed on: Reversls Knowledge of preserved bloks (or genes) Now we re onerned in determining the preserved genes, or more generlly, given two strings nd y, determine ll their possible miml mthes Globl lignment One possibility is to perform globl lignment of nd y with speil soring sheme (for instne +1, 0, 0) nd identify the miml positively soring hunks. Tkes O(mn) time Does not give ll ndidte mthes Might not give miml mthes Emple: = bbbbbb y =bbb bbbbbb bbb missed nonmiml, bbb is better
2 A simple wy: kmers Another possibility is to find ommon kmers. Here s one wy: Algorithm: Denote kmer of by (w, 0, i) where w = i i+k1 Denote kmer of y by (z, 1, j) where z = y j y j+k1 Sort them leiogrphilly Dedue ll k long mthes between nd y Emple: kmers = bb, y = bbb s 3mers: (b, 0, 1), (bb, 0, 2), (bb, 0, 3) y s 3mers: (bb, 1, 1), (b, 1, 2), (b, 1, 3) sort them: (bb, 0, 2), (bb, 0, 3), (bb, 1, 1), (b, 1, 2), (b, 0, 1), (b, 1, 3) Identify mthes: (bb, 0, 3), (bb, 1, 1) nd (b, 0, 1), (b, 1, 3) Disdvntges Worst se running time still O(mn) e.g. O(m) kmers in nd O(n) kmers in y re identil. Thought: ren t we supposed to find these nywy nd, therefore, the O(mn) bound is not neessrily bd? No, they might be smll insignifint mthes, we re interested in miml mthes Mking k lrger redues the running time beuse it results in shortest list of mthes, but we might miss signifint mthes
3 A better solution Suffi tree We will use n effiient dt struture lled suffi tree tht stores ll suffies of string nd supports fst lookup Definition: A suffi tree T of string s of length m is rooted tree suh tht: It hs etly m leves numbered 1 to m Eh internl node other thn the root hs t lest two hildren Eh edge is lbeled by substring of s No two edge lbels out of node strt with the sme hrter For ny lef i, the ontention of the edge lbels on the pth from the root to i spells out the suffi i m s = b Emple of suffi tree 3 b b 5 b Eistene Does suffi tree of string s lwys eist? Consider s = b 3 b 6 If suffi is prefi of nother suffi, then the pth for the first suffi would not end t lef! Solution: lwys terminte string with speil hrter $ tht does not our nywhere. b b
4 Properties of suffi tree A suffi tree stisfies the following: E = V 1 (tree) Number of leves = m + 1 (now s = m + 1) Sine eh internl node hs t lest two hildren, the number of edges E = O( leves) = O(m) Any subtree with k leves stisfies E = O(k) Building suffi tree Here s simple lgorithm: given = 1 m insert speil hrter $ t the end of s initilize the tree T to one root for j = 1 to m + 1 find the longest mth of j m in T strting from the root nd following unique pth split the edge where the mth stops, dd new node w dd n edge (w,j) (j is the new lef) nd lbel it with the remining unmthed hrters of j m s = b Emple b$ 7 $ b $ b$ b b $ b$ 4 $ $ $ $ $ $ 5 $ 3 2 $ 6 1
5 Anlysis Running time: O(m 2 ) Eh suffi requires O(m) time to updte the tree But, there eists n O(m) time suffi tree lgorithm Spe: O(m) How? Eh lbel hs O(m) hrters nd we hve E = O(m) lbels! Solution: do not epliitly store lbels, but store the indies [i,j] of lbel Now wht? How n we use the suffi tree dt struture to identify ll miml mthes between two strings nd y? Consider first the following problem: given string, determine ll lotions where nother string y ours. This n be solved effiiently s desribed net. Find ll ourrenes of y in string mthing Algorithm build suffi tree T for Mth the hrters of y long the unique pth in T until (se 1) either y is ehusted or (se 2) no more mthes re possible if (se 2) y does not our in else the k leves in the subtree below the point of the lst mth give the k lotion of y in (trverse the tree in liner time) O(m) O(n) O(1) O(k)
6 Corretness Why is the string mthing lgorithm orret? If y ours in t position i, then the i th suffi of must strt with y Therefore, lef i must be rehed by the pth determined by y Finding miml mthes Given nd y, we would like to find ll miml mthes between nd y i i+l = y j y j+l Cnnot etend i i+l nd y j y j+l nd obtin mth We will find ll mthes strting t y j tht nnot be etended to the right Build suffi tree for (do this only one) Find the pth in T determined by the longest possible prefi of the suffi y j y n (it ould stop in the middle of n edge e in tht se e is prt of the pth) Let v k, k = 1 p be n internl node on this pth nd T k be the subtree rooted t v k tht eludes v k+1 Identify the leves in eh subtree root Illustrtion L 1 v 1 L 2 Lst point of mth v 2 T1 v p1 L p v p m 1 leves T 2 T p1 T p A lef i in T k gives the lotion in of mth between i i+ L1 + + Lk 1 nd y j y j+ L1 + + Lk 1 tht nnot be etended to the right Running time: O(m) {building T} + O(Σ L k + Σm k ) O(n + m)
7 Wht bout left? Given lef i, let left(i) be the hrter i1 If left(i) y j1, then i represents miml mth Therefore, we obtin ll miml mthes between nd y in O(mn) time by repeting the previous lgorithm for every suffi of y Algorithm Build suffi tree T for O(m) for j = 1 to n find the pth in T determined by the longest possible prefi of the suffi y j y n (it ould stop in the middle of n edge e in tht se e is prt of the pth) O(n) let v k, k = 1 p be n internl node on this pth nd T k be the subtree rooted t v k tht eludes v k+1 Let l(v k ) = length of mth up to node v k identify ll leves i in eh subtree suh tht left(i) y j1 Suh lef i in subtree T k represents miml mth of length l(v k ) strting t position i in nd position j in y O(m) Generlized suffi tree for set of strings We n build suffi tree for set of strings s 1, s 2,, s n Append different end of string mrker to eh string in the set ontente ll the strings together build suffi tree for the ontented string The resulting suffi tree will hve lef for eh suffi of the ontented string nd is build in time proportionl to the sum of ll lengths The lef numbers n be esily onverted to two numbers, one identifying string s i nd the other strting position in s i
8 Emple s 1 = b, s 2 = bbb s = b$bbb 1,6 $bbb 2,7 bb 2,1 b 2,5 2,3 b $bbb 1,3 2,8 $bbb b b$bbb 2,4 $bbb b $bbb b 1,5 2,2 1,2 1,4 1,1 Fi lbels of lef edges One defet is tht the tree now represents suffies tht spn more thn one originl string Beuse eh string mrker ours only one, the unwnted suffies re removed by fiing the lbel on lef edges 1,6 $bbb 2,7 bb 2,1 b 2,5 2,3 b $bbb 1,3 2,8 $bbb b b$bbb 2,4 $bbb b $bbb b 1,5 2,2 1,2 1,4 1,1 Suffi tree for nd y Therefore, given two strings nd y, we n build suffi tree for both in O(m + n) (i.e. liner) time. Eh lef in the tree represents Either suffi from Or suffi from y Mrk eh internl node v with (y) if there is lef in the subtree of v representing suffi from (y). This n be done in liner time by bottom up trversl of the tree from leves to the root. Note tht if v is mrked (y), ll nestors of v re mrked (y).
9 Common substrings If αp is substring of nd αq is substring of y for p q, then α orresponds to n internl node v mrked with both nd y nd vievers. Proof: α ours in both nd y suh tht the hrter to the right of α in differs from the hrter to the right of α in y. α,y lef for lef for y onversely, every internl node mrked with both nd y hs to stisfy the sitution depited bove, then αp is substring of nd αq is substring of y for p q. Left diverse node An internl node v is left diverse iff it hs two hildren v 1 nd v 2 with lef i for in v 1 s subtree nd lef j for y in v 2 s subtree, suh tht left(i) left(j) (ssume 0 nd y 0 re different nd distint from ny other hrter) If uαp is substring of nd wαq is substring of y for u w nd p q, then α orresponds to left diverse node v nd vievers. Proof: similr to previous proof Cll suh n α miml ommon substring Compt representtion Therefore, we hve only O(m + n) miml ommon substrings for nd y (but eh miml ommon substring might pper in multiple lotions) If we identify left diverse nodes in liner time, we need only O(m + n) time nd spe to ome up with this ompt representtion of ll miml ommon substrings A miml mth n be represented s (p 1, p 2, l) where p 1 nd p 2 re the positions of miml ommon substring of length l in nd y respetively We n obtin ll miml mthes in O(m + n + k) where k is their number (we will not present the lgorithm)
10 Identifying left diverse nodes For eh node the lgorithm reords: the hrter (v): the left hrter of every lef for in v s subtree, or speil hrter ε if no lef for eists in v s subtree, or speil hrter & the hrter b(v): the left hrter of every lef for y in v s subtree, or speil hrter ε if no lef for y eists in v s subtree, or speil Computing (v) nd b(v) n be done in bottom up pproh in liner time Note tht v is left diverse iff it hs two hildren v 1 nd v 2 with: (v 1 ) b(v 2 ), (v 1 ) ε, b(v 2 ) ε or b(v 1 ) (v 2 ), b(v 1 ) ε, (v 2 ) ε It tkes O( Σ 2 ) time (onstnt) to find two suh hildren or none, where Σ is the lphbet (eh node hs t most Σ hildren)
11/3/13. Indexing techniques. Shortread mapping software. Indexing a text (a genome, etc) Some terminologies. Hashing
I9 Introdution to Bioinformtis, 0 Indeing tehniques Yuzhen Ye (yye@indin.edu) Shool of Informtis & Computing, IUB Contents We hve seen indeing tehnique used in BLAST Applitions tht rely on n effiient indeing
More informationPrefixFree RegularExpression Matching
PrefixFree RegulrExpression Mthing YoSu Hn, Yjun Wng nd Derik Wood Deprtment of Computer Siene HKUST PrefixFree RegulrExpression Mthing p.1/15 Pttern Mthing Given pttern P nd text T, find ll sustrings
More informationData Structures LECTURE 10. Huffman coding. Example. Coding: problem definition
Dt Strutures, Spring 24 L. Joskowiz Dt Strutures LEURE Humn oing Motivtion Uniquel eipherle oes Prei oes Humn oe onstrution Etensions n pplitions hpter 6.3 pp 385 392 in tetook Motivtion Suppose we wnt
More informationAutomata and Languages
Automt nd Lnguges Prof. Mohmed Hmd Softwre Engineering Lb. The University of Aizu Jpn Grmmr Regulr Grmmr Contextfree Grmmr Contextsensitive Grmmr Regulr Lnguges Context Free Lnguges Context Sensitive
More informationOnLine Construction. of Suffix Trees. Overview. Suffix Trees. Notations. goo. Suffix tries
OnLine Cnstrutin Overview Suffix tries f Suffix Trees E. Ukknen Online nstrutin f suffix tries in qudrti time Suffix trees Online nstrutin f suffix trees in liner time Applitins 1 2 Suffix Trees A suffix
More informationThe RiemannStieltjes Integral
Chpter 6 The RiemnnStieltjes Integrl 6.1. Definition nd Eistene of the Integrl Definition 6.1. Let, b R nd < b. ( A prtition P of intervl [, b] is finite set of points P = { 0, 1,..., n } suh tht = 0
More informationNONDETERMINISTIC FSA
Tw o types of nondeterminism: NONDETERMINISTIC FS () Multiple strtsttes; strtsttes S Q. The lnguge L(M) ={x:x tkes M from some strtstte to some finlstte nd ll of x is proessed}. The string x = is
More informationFinite Automata Theory and Formal Languages TMV027/DIT321 LP4 2018
Finite Automt Theory nd Forml Lnguges TMV027/DIT321 LP4 2018 Lecture 10 An Bove April 23rd 2018 Recp: Regulr Lnguges We cn convert between FA nd RE; Hence both FA nd RE ccept/generte regulr lnguges; More
More information6.5 Improper integrals
Eerpt from "Clulus" 3 AoPS In. www.rtofprolemsolving.om 6.5. IMPROPER INTEGRALS 6.5 Improper integrls As we ve seen, we use the definite integrl R f to ompute the re of the region under the grph of y =
More informationArrow s Impossibility Theorem
Rep Voting Prdoxes Properties Arrow s Theorem Arrow s Impossiility Theorem Leture 12 Arrow s Impossiility Theorem Leture 12, Slide 1 Rep Voting Prdoxes Properties Arrow s Theorem Leture Overview 1 Rep
More informationPreview 11/1/2017. Greedy Algorithms. Coin Change. Coin Change. Coin Change. Coin Change. Greedy algorithms. Greedy Algorithms
Preview Greed Algorithms Greed Algorithms Coin Chnge Huffmn Code Greed lgorithms end to e simple nd strightforwrd. Are often used to solve optimiztion prolems. Alws mke the choice tht looks est t the moment,
More informationChapter 4 StateSpace Planning
Leture slides for Automted Plnning: Theory nd Prtie Chpter 4 StteSpe Plnning Dn S. Nu CMSC 722, AI Plnning University of Mrylnd, Spring 2008 1 Motivtion Nerly ll plnning proedures re serh proedures Different
More information(a) A partition P of [a, b] is a finite subset of [a, b] containing a and b. If Q is another partition and P Q, then Q is a refinement of P.
Chpter 7: The Riemnn Integrl When the derivtive is introdued, it is not hrd to see tht the it of the differene quotient should be equl to the slope of the tngent line, or when the horizontl xis is time
More informationIntroduction to Olympiad Inequalities
Introdution to Olympid Inequlities Edutionl Studies Progrm HSSP Msshusetts Institute of Tehnology Snj Simonovikj Spring 207 Contents Wrm up nd AmGm inequlity 2. Elementry inequlities......................
More informationSolutions to Assignment 1
MTHE 237 Fll 2015 Solutions to Assignment 1 Problem 1 Find the order of the differentil eqution: t d3 y dt 3 +t2 y = os(t. Is the differentil eqution liner? Is the eqution homogeneous? b Repet the bove
More informationConnectivity in Graphs. CS311H: Discrete Mathematics. Graph Theory II. Example. Paths. Connectedness. Example
Connetiit in Grphs CSH: Disrete Mthemtis Grph Theor II Instrtor: Işıl Dillig Tpil qestion: Is it possile to get from some noe to nother noe? Emple: Trin netork if there is pth from to, possile to tke trin
More informationQUADRATIC EQUATION. Contents
QUADRATIC EQUATION Contents Topi Pge No. Theory 004 Exerise  0509 Exerise  093 Exerise  3 45 Exerise  4 6 Answer Key 78 Syllus Qudrti equtions with rel oeffiients, reltions etween roots nd oeffiients,
More informationThe First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).
The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples
More informationPeriodic string comparison
Periodi string omprison Alexnder Tiskin Deprtment of Computer Siene University of Wrwik http://www.ds.wrwik..uk/~tiskin Alexnder Tiskin (Wrwik) Periodi string omprison 1 / 51 1 Introdution 2 Semilol string
More informationUniversitaireWiskundeCompetitie. Problem 2005/4A We have k=1. Show that for every q Q satisfying 0 < q < 1, there exists a finite subset K N so that
Problemen/UWC NAW 5/7 nr juni 006 47 Problemen/UWC UniversitireWiskundeCompetitie Edition 005/4 For Session 005/4 we received submissions from Peter Vndendriessche, Vldislv Frnk, Arne Smeets, Jn vn de
More informationLooking for All Palindromes in a String
Looking or All Plindromes in String Shih Jng Pn nd R C T Lee Deprtment o Computer Science nd Inormtion Engineering, Ntionl ChiNn University, Puli, Nntou Hsien,, Tiwn, ROC sjpn@lgdoccsiencnuedutw, rctlee@ncnuedutw
More informationBIFURCATIONS IN ONEDIMENSIONAL DISCRETE SYSTEMS
BIFRCATIONS IN ONEDIMENSIONAL DISCRETE SYSTEMS FRANCESCA AICARDI In this lesson we will study the simplest dynmicl systems. We will see, however, tht even in this cse the scenrio of different possible
More informationWhere did dynamic programming come from?
Where did dynmic progrmming come from? String lgorithms Dvid Kuchk cs302 Spring 2012 Richrd ellmn On the irth of Dynmic Progrmming Sturt Dreyfus http://www.eng.tu.c.il/~mi/cd/ or50/15265463200250010048.pdf
More informationNew Expansion and Infinite Series
Interntionl Mthemticl Forum, Vol. 9, 204, no. 22, 06073 HIKARI Ltd, www.mhikri.com http://dx.doi.org/0.2988/imf.204.4502 New Expnsion nd Infinite Series Diyun Zhng College of Computer Nnjing University
More informationLecture 1  Introduction and Basic Facts about PDEs
* 18.15  Introdution to PDEs, Fll 004 Prof. Gigliol Stffilni Leture 1  Introdution nd Bsi Fts bout PDEs The Content of the Course Definition of Prtil Differentil Eqution (PDE) Liner PDEs VVVVVVVVVVVVVVVVVVVV
More information20 MATHEMATICS POLYNOMIALS
0 MATHEMATICS POLYNOMIALS.1 Introduction In Clss IX, you hve studied polynomils in one vrible nd their degrees. Recll tht if p(x) is polynomil in x, the highest power of x in p(x) is clled the degree of
More information1 From NFA to regular expression
Note 1: How to convert DFA/NFA to regulr expression Version: 1.0 S/EE 374, Fll 2017 Septemer 11, 2017 In this note, we show tht ny DFA cn e converted into regulr expression. Our construction would work
More informationDIRECT CURRENT CIRCUITS
DRECT CURRENT CUTS ELECTRC POWER Consider the circuit shown in the Figure where bttery is connected to resistor R. A positive chrge dq will gin potentil energy s it moves from point to point b through
More informationTypes of Finite Automata. CMSC 330: Organization of Programming Languages. Comparing DFAs and NFAs. NFA for (a b)*abb.
CMSC 330: Orgniztion of Progrmming Lnguges Finite Automt 2 Types of Finite Automt Deterministic Finite Automt () Exctly one sequence of steps for ech string All exmples so fr Nondeterministic Finite Automt
More informationNondeterministic Finite Automata
Nondeterministi Finite utomt The Power of Guessing Tuesdy, Otoer 4, 2 Reding: Sipser.2 (first prt); Stoughton 3.3 3.5 S235 Lnguges nd utomt eprtment of omputer Siene Wellesley ollege Finite utomton (F)
More informationapproaches as n becomes larger and larger. Since e > 1, the graph of the natural exponential function is as below
. Eponentil nd rithmic functions.1 Eponentil Functions A function of the form f() =, > 0, 1 is clled n eponentil function. Its domin is the set of ll rel f ( 1) numbers. For n eponentil function f we hve.
More informationMatrices SCHOOL OF ENGINEERING & BUILT ENVIRONMENT. Mathematics (c) 1. Definition of a Matrix
tries Definition of tri mtri is regulr rry of numers enlosed inside rkets SCHOOL OF ENGINEERING & UIL ENVIRONEN Emple he following re ll mtries: ), ) 9, themtis ), d) tries Definition of tri Size of tri
More informationBest Approximation. Chapter The General Case
Chpter 4 Best Approximtion 4.1 The Generl Cse In the previous chpter, we hve seen how n interpolting polynomil cn be used s n pproximtion to given function. We now wnt to find the best pproximtion to given
More informationwhere the box contains a finite number of gates from the given collection. Examples of gates that are commonly used are the following: a b
CS 2942 9/11/04 Quntum Ciruit Model, SolovyKitev Theorem, BQP Fll 2004 Leture 4 1 Quntum Ciruit Model 1.1 Clssil Ciruits  Universl Gte Sets A lssil iruit implements multioutput oolen funtion f : {0,1}
More informationHomework 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 informationA 6Approximation Algorithm for Computing Smallest Common AoNsupertree With Application to the Reconstruction of Glycan Trees
A 6Approximtion Algorithm for Computing Smllest Common AoNsupertree With Applition to the Reonstrution of Glyn Trees Kiyoko F. AokiKinoshit, Minoru Knehis, MingYng Ko, XingYng Li, nd Weizho Wng Abstrt.
More informationSpeech Recognition Lecture 2: Finite Automata and FiniteState Transducers
Speech Recognition Lecture 2: Finite Automt nd FiniteStte Trnsducers Eugene Weinstein Google, NYU Cournt Institute eugenew@cs.nyu.edu Slide Credit: Mehryr Mohri Preliminries Finite lphet, empty string.
More informationThis lecture covers Chapter 8 of HMU: Properties of CFLs
This lecture covers Chpter 8 of HMU: Properties of CFLs Turing Mchine Extensions of Turing Mchines Restrictions of Turing Mchines Additionl Reding: Chpter 8 of HMU. Turing Mchine: Informl Definition B
More information13.3 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS
33 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS As simple ppliction of the results we hve obtined on lgebric extensions, nd in prticulr on the multiplictivity of extension degrees, we cn nswer (in
More information1.3 Regular Expressions
56 1.3 Regulr xpressions These hve n importnt role in describing ptterns in serching for strings in mny pplictions (e.g. wk, grep, Perl,...) All regulr expressions of lphbet re 1.Ønd re regulr expressions,
More informationAlgorithm Design and Analysis
Algorithm Design nd Anlysis LECTURE 8 Mx. lteness ont d Optiml Ching Adm Smith 9/12/2008 A. Smith; sed on slides y E. Demine, C. Leiserson, S. Rskhodnikov, K. Wyne Sheduling to Minimizing Lteness Minimizing
More informationComputing the Optimal Global Alignment Value. B = n. Score of = 1 Score of = a a c g a c g a. A = n. Classical Dynamic Programming: O(n )
Alignment Grph Alignment Mtrix Computing the Optiml Globl Alignment Vlue An Introduction to Bioinformtics Algorithms A = n c t 2 3 c c 4 g 5 g 6 7 8 9 B = n 0 c g c g 2 3 4 5 6 7 8 t 9 0 2 3 4 5 6 7 8
More information5.5 The Substitution Rule
5.5 The Substitution Rule Given the usefulness of the Fundmentl Theorem, we wnt some helpful methods for finding ntiderivtives. At the moment, if n ntiderivtive is not esily recognizble, then we re in
More informationWhat else can you do?
Wht else cn you do? ngle sums The size of specil ngle types lernt erlier cn e used to find unknown ngles. tht form stright line dd to 180c. lculte the size of + M, if L is stright line M + L = 180c( stright
More informationLexical Analysis Finite Automate
Lexicl Anlysis Finite Automte CMPSC 470 Lecture 04 Topics: Deterministic Finite Automt (DFA) Nondeterministic Finite Automt (NFA) Regulr Expression NFA DFA A. Finite Automt (FA) FA re grph, like trnsition
More informationNUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by.
NUMERICAL INTEGRATION 1 Introduction The inverse process to differentition in clculus is integrtion. Mthemticlly, integrtion is represented by f(x) dx which stnds for the integrl of the function f(x) with
More informationGrammar. Languages. Content 5/10/16. Automata and Languages. Regular Languages. Regular Languages
5//6 Grmmr Automt nd Lnguges Regulr Grmmr Contextfree Grmmr Contextsensitive Grmmr Prof. Mohmed Hmd Softwre Engineering L. The University of Aizu Jpn Regulr Lnguges Context Free Lnguges Context Sensitive
More informationMATH FIELD DAY Contestants Insructions Team Essay. 1. Your team has forty minutes to answer this set of questions.
MATH FIELD DAY 2012 Contestnts Insructions Tem Essy 1. Your tem hs forty minutes to nswer this set of questions. 2. All nswers must be justified with complete explntions. Your nswers should be cler, grmmticlly
More informationMath 360: A primitive integral and elementary functions
Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:
More informationT b a(f) [f ] +. P b a(f) = Conclude that if f is in AC then it is the difference of two monotone absolutely continuous functions.
Rel Vribles, Fll 2014 Problem set 5 Solution suggestions Exerise 1. Let f be bsolutely ontinuous on [, b] Show tht nd T b (f) P b (f) f (x) dx [f ] +. Conlude tht if f is in AC then it is the differene
More informationalong the vector 5 a) Find the plane s coordinate after 1 hour. b) Find the plane s coordinate after 2 hours. c) Find the plane s coordinate
L8 VECTOR EQUATIONS OF LINES HL Mth  Sntowski Vector eqution of line 1 A plne strts journey t the point (4,1) moves ech hour long the vector. ) Find the plne s coordinte fter 1 hour. b) Find the plne
More information5.2 Volumes: Disks and Washers
4 pplictions of definite integrls 5. Volumes: Disks nd Wshers In the previous section, we computed volumes of solids for which we could determine the re of crosssection or slice. In this section, we restrict
More informationLinear Systems with Constant Coefficients
Liner Systems with Constnt Coefficients 4305 Here is system of n differentil equtions in n unknowns: x x + + n x n, x x + + n x n, x n n x + + nn x n This is constnt coefficient liner homogeneous system
More informationMore Properties of the Riemann Integral
More Properties of the Riemnn Integrl Jmes K. Peterson Deprtment of Biologil Sienes nd Deprtment of Mthemtil Sienes Clemson University Februry 15, 2018 Outline More Riemnn Integrl Properties The Fundmentl
More informationFormal languages, automata, and theory of computation
Mälrdlen University TEN1 DVA337 2015 School of Innovtion, Design nd Engineering Forml lnguges, utomt, nd theory of computtion Thursdy, Novemer 5, 14:1018:30 Techer: Dniel Hedin, phone 021107052 The exm
More informationSolutions for HW9. Bipartite: put the red vertices in V 1 and the black in V 2. Not bipartite!
Solutions for HW9 Exerise 28. () Drw C 6, W 6 K 6, n K 5,3. C 6 : W 6 : K 6 : K 5,3 : () Whih of the following re iprtite? Justify your nswer. Biprtite: put the re verties in V 1 n the lk in V 2. Biprtite:
More informationRegular Expressions and NFAs without εtransitions
Regulr Expressions nd NFAs without εtrnsitions Georg chnitger Institut für Informtik, Johnn Wolfgng GoetheUniversität, Robert Myer trße 11 15, 60054 Frnkfurt m Min, Germny georg@thi.informtik.unifrnkfurt.de
More informationDistanceJoin: Pattern Match Query In a Large Graph Database
DistneJoin: Pttern Mth Query In Lrge Grph Dtbse Lei Zou Huzhong University of Siene nd Tehnology Wuhn, Chin zoulei@mil.hust.edu.n Lei Chen Hong Kong University of Siene nd Tehnology Hong Kong leihen@se.ust.hk
More information1 Nondeterministic Finite Automata
1 Nondeterministic Finite Automt Suppose in life, whenever you hd choice, you could try oth possiilities nd live your life. At the end, you would go ck nd choose the one tht worked out the est. Then you
More informationPolynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230
Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given
More informationHandout: Natural deduction for first order logic
MATH 457 Introduction to Mthemticl Logic Spring 2016 Dr Json Rute Hndout: Nturl deduction for first order logic We will extend our nturl deduction rules for sententil logic to first order logic These notes
More informationCS S12 Turing Machine Modifications 1. When we added a stack to NFA to get a PDA, we increased computational power
CS4112015S12 Turing Mchine Modifictions 1 120: Extending Turing Mchines When we dded stck to NFA to get PDA, we incresed computtionl power Cn we do the sme thing for Turing Mchines? Tht is, cn we dd
More information3.1 Review of Sine, Cosine and Tangent for Right Angles
Foundtions of Mth 11 Section 3.1 Review of Sine, osine nd Tngent for Right Tringles 125 3.1 Review of Sine, osine nd Tngent for Right ngles The word trigonometry is derived from the Greek words trigon,
More informationApplied Physics Introduction to Vibrations and Waves (with a focus on elastic waves) Course Outline
Applied Physics Introduction to Vibrtions nd Wves (with focus on elstic wves) Course Outline Simple Hrmonic Motion && + ω 0 ω k /m k elstic property of the oscilltor Elstic properties of terils Stretching,
More informationImproper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:
Improper Integrls The First Fundmentl Theorem of Clculus, s we ve discussed in clss, goes s follows: If f is continuous on the intervl [, ] nd F is function for which F t = ft, then ftdt = F F. An integrl
More informationNumbers and indices. 1.1 Fractions. GCSE C Example 1. Handy hint. Key point
GCSE C Emple 7 Work out 9 Give your nswer in its simplest form Numers n inies Reiprote mens invert or turn upsie own The reiprol of is 9 9 Mke sure you only invert the frtion you re iviing y 7 You multiply
More informationChapter Gauss Quadrature Rule of Integration
Chpter 7. Guss Qudrture Rule o Integrtion Ater reding this hpter, you should e le to:. derive the Guss qudrture method or integrtion nd e le to use it to solve prolems, nd. use Guss qudrture method to
More informationLecture 14: Quadrature
Lecture 14: Qudrture This lecture is concerned with the evlution of integrls fx)dx 1) over finite intervl [, b] The integrnd fx) is ssumed to be relvlues nd smooth The pproximtion of n integrl by numericl
More informationLine Integrals and Entire Functions
Line Integrls nd Entire Funtions Defining n Integrl for omplex Vlued Funtions In the following setions, our min gol is to show tht every entire funtion n be represented s n everywhere onvergent power series
More informationProject 6: Minigoals Towards Simplifying and Rewriting Expressions
MAT 51 Wldis Projet 6: Minigols Towrds Simplifying nd Rewriting Expressions The distriutive property nd like terms You hve proly lerned in previous lsses out dding like terms ut one prolem with the wy
More informationLine Integrals. Partitioning the Curve. Estimating the Mass
Line Integrls Suppose we hve curve in the xy plne nd ssocite density δ(p ) = δ(x, y) t ech point on the curve. urves, of course, do not hve density or mss, but it my sometimes be convenient or useful to
More informationChapter 4 Contravariance, Covariance, and Spacetime Diagrams
Chpter 4 Contrvrince, Covrince, nd Spcetime Digrms 4. The Components of Vector in Skewed Coordintes We hve seen in Chpter 3; figure 3.9, tht in order to show inertil motion tht is consistent with the Lorentz
More informationSCHOOL OF ENGINEERING & BUILT ENVIRONMENT. Mathematics
SCHOOL OF ENGINEERING & BUIL ENVIRONMEN Mthemtics An Introduction to Mtrices Definition of Mtri Size of Mtri Rows nd Columns of Mtri Mtri Addition Sclr Multipliction of Mtri Mtri Multipliction 7 rnspose
More informationThe Ellipse. is larger than the other.
The Ellipse Appolonius of Perg (5 B.C.) disovered tht interseting right irulr one ll the w through with plne slnted ut is not perpendiulr to the is, the intersetion provides resulting urve (oni setion)
More informationThermodynamics. Question 1. Question 2. Question 3 3/10/2010. Practice Questions PV TR PV T R
/10/010 Question 1 1 mole of idel gs is rought to finl stte F y one of three proesses tht hve different initil sttes s shown in the figure. Wht is true for the temperture hnge etween initil nd finl sttes?
More informationTries and suffixes trees
Trie: A dtstructure for set of words Tries nd suffixes trees Alon Efrt Comuter Science Dertment University of Arizon All words over the lhet Σ={,,..z}. In the slides, let sy tht the lhet is only {,,c,d}
More informationSymmetrical Components 1
Symmetril Components. Introdution These notes should e red together with Setion. of your text. When performing stedystte nlysis of high voltge trnsmission systems, we mke use of the perphse equivlent
More informationH (2a, a) (u 2a) 2 (E) Show that u v 4a. Explain why this implies that u v 4a, with equality if and only u a if u v 2a.
Chpter Review 89 IGURE ol hord GH of the prol 4. G u v H (, ) (A) Use the distne formul to show tht u. (B) Show tht G nd H lie on the line m, where m ( )/( ). (C) Solve m for nd sustitute in 4, otining
More information1 APL13: Suffix Arrays: more space reduction
1 APL13: Suffix Arrys: more spce reduction In Section??, we sw tht when lphbet size is included in the time nd spce bounds, the suffix tree for string of length m either requires Θ(m Σ ) spce or the minimum
More informationSOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 2014
SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 014 Mrk Scheme: Ech prt of Question 1 is worth four mrks which re wrded solely for the correct nswer.
More informationUSA Mathematical Talent Search Round 1 Solutions Year 21 Academic Year
1/1/21. Fill in the circles in the picture t right with the digits 18, one digit in ech circle with no digit repeted, so tht no two circles tht re connected by line segment contin consecutive digits.
More informationTopic 1 Notes Jeremy Orloff
Topic 1 Notes Jerem Orloff 1 Introduction to differentil equtions 1.1 Gols 1. Know the definition of differentil eqution. 2. Know our first nd second most importnt equtions nd their solutions. 3. Be ble
More informationINTRODUCTION TO LINEAR ALGEBRA
ME Applied Mthemtics for Mechnicl Engineers INTRODUCTION TO INEAR AGEBRA Mtrices nd Vectors Prof. Dr. Bülent E. Pltin Spring Sections & / ME Applied Mthemtics for Mechnicl Engineers INTRODUCTION TO INEAR
More informationCounting intersections of spirals on a torus
Counting intersections of spirls on torus 1 The problem Consider unit squre with opposite sides identified. For emple, if we leve the centre of the squre trveling long line of slope 2 (s shown in the first
More informationPART 1 MULTIPLE CHOICE Circle the appropriate response to each of the questions below. Each question has a value of 1 point.
PART MULTIPLE CHOICE Circle the pproprite response to ech of the questions below. Ech question hs vlue of point.. If in sequence the second level difference is constnt, thn the sequence is:. rithmetic
More information1 Online Learning and Regret Minimization
2.997 DecisionMking in LrgeScle Systems My 10 MIT, Spring 2004 Hndout #29 Lecture Note 24 1 Online Lerning nd Regret Minimiztion In this lecture, we consider the problem of sequentil decision mking in
More informationImproper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral.
Improper Integrls Introduction When we defined the definite integrl f d we ssumed tht f ws continuous on [, ] where [, ] ws finite, closed intervl There re t lest two wys this definition cn fil to e stisfied:
More informationCalculus Cheat Sheet. Integrals Definitions. where F( x ) is an antiderivative of f ( x ). Fundamental Theorem of Calculus. dx = f x dx g x dx
Clulus Chet Sheet Integrls Definitions Definite Integrl: Suppose f ( ) is ontinuous AntiDerivtive : An ntiderivtive of f ( ) on [, ]. Divide [, ] into n suintervls of is funtion, F( ), suh tht F = f.
More informationPythagoras Theorem. The area of the square on the hypotenuse is equal to the sum of the squares on the other two sides
Pythgors theorem nd trigonometry Pythgors Theorem The hypotenuse of rightngled tringle is the longest side The hypotenuse is lwys opposite the rightngle 2 = 2 + 2 or 2 = 22 or 2 = 22 The re of the
More informationCS 360 Exam 2 Fall 2014 Name
CS 360 Exm 2 Fll 2014 Nme 1. The lsses shown elow efine singlylinke list n stk. Write three ifferent O(n)time versions of the reverse_print metho s speifie elow. Eh version of the metho shoul output
More informationHomework 4. 0 ε 0. (00) ε 0 ε 0 (00) (11) CS 341: Foundations of Computer Science II Prof. Marvin Nakayama
CS 341: Foundtions of Computer Science II Prof. Mrvin Nkym Homework 4 1. UsetheproceduredescriedinLemm1.55toconverttheregulrexpression(((00) (11)) 01) into n NFA. Answer: 0 0 1 1 00 0 0 11 1 1 01 0 1 (00)
More informationHomework Solution  Set 5 Due: Friday 10/03/08
CE 96 Introduction to the Theory of Computtion ll 2008 Homework olution  et 5 Due: ridy 10/0/08 1. Textook, Pge 86, Exercise 1.21. () 1 2 Add new strt stte nd finl stte. Mke originl finl stte nonfinl.
More informationLecture 3. Limits of Functions and Continuity
Lecture 3 Limits of Functions nd Continuity Audrey Terrs April 26, 21 1 Limits of Functions Notes I m skipping the lst section of Chpter 6 of Lng; the section bout open nd closed sets We cn probbly live
More informationLogic Optimization 1. Logic Optimization. Optimization vs. Tradeoff. Twolevel Logic Optimization. ECE 474A/57A ComputerAided Logic Design
ECE 474A/57A ComputerAided Logic Design Logic Optimition Logic Optimition We now know how to build digitl circuits How cn we build better circuits? Let s consider two importnt design criteri Del the time
More informationRecursively Enumerable and Recursive. Languages
Recursively Enumerble nd Recursive nguges 1 Recll Definition (clss 19.pdf) Definition 10.4, inz, 6 th, pge 279 et S be set of strings. An enumertion procedure for Turing Mchine tht genertes ll strings
More informationQuantum Physics II (8.05) Fall 2013 Assignment 2
Quntum Physics II (8.05) Fll 2013 Assignment 2 Msschusetts Institute of Technology Physics Deprtment Due Fridy September 20, 2013 September 13, 2013 3:00 pm Suggested Reding Continued from lst week: 1.
More information1 Error Analysis of Simple Rules for Numerical Integration
cs41: introduction to numericl nlysis 11/16/10 Lecture 19: Numericl Integrtion II Instructor: Professor Amos Ron Scries: Mrk Cowlishw, Nthnel Fillmore 1 Error Anlysis of Simple Rules for Numericl Integrtion
More informationTransition systems (motivation)
Trnsition systems (motivtion) Course Modelling of Conurrent Systems ( Modellierung neenläufiger Systeme ) Winter Semester 2009/0 University of DuisurgEssen Brr König Tehing ssistnt: Christoph Blume In
More informationa) Read over steps (1) (4) below and sketch the path of the cycle on a P V plot on the graph below. Label all appropriate points.
Prole 3: Crnot Cyle of n Idel Gs In this prole, the strting pressure P nd volue of n idel gs in stte, re given he rtio R = / > of the volues of the sttes nd is given Finlly onstnt γ = 5/3 is given You
More information3 x x 3x x. 3x x x 6 x 3. PAKTURK 8 th National Interschool Maths Olympiad, h h
PAKTURK 8 th Ntionl Interschool Mths Olmpid,.9. Q: Evlute 6.9. 6 6 6... 8 8...... Q: Evlute bc bc. b. c bc.9.9b.9.9bc Q: Find the vlue of h in the eqution h 7 9 7.. bc. bc bc. b. c bc bc bc bc......9 h
More information