# Homework Solution - Set 5 Due: Friday 10/03/08

Save this PDF as:

Size: px
Start display at page: ## Transcription

1 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 2 Add new strt stte nd finl stte. Mke originl finl stte non-finl. 1 2 No unions necessry, so eliminte stte 1. 2 Perform union on loops of stte 2. 2 Eliminte stte 2. No finl unions necessry. ( ) 1

2 (), 1 2 Add new strt stte nd finl stte. Mke originl finl sttes non-finl. 1 2, Perform union on edge from stte 1 to stte Eliminte stte 1. 2 ( ) Perform unions on edges from stte to stte 2 nd from stte to the finl stte. 2

3 2 ( ) Eliminte stte 2. ( ) (( ) ) No unions necessry, so eliminte stte. ( ) ((( ) ) ) ( ) Perform union on remining edges. ( ) ((( ) ) ) ( ) 2. Textook, Pge 88, Exercise () A 1 = {0 n 1 n 2 n n 0} Condensed proof:

4 Proof. uppose A 1 is regulr. Let p e the pumping length given y the pumping lemm. Choose s = 0 p 1 p 2 p. By the lemm, xy p nd y > 0 therefore p 0 so s A 1. Clerly, s p thus s = xyz for some x, y nd z. ince xy p, xy cnnot extend eyond the first p symols of s, mening xy =0 k where 1 k p. Let us write x = 0, y = 0, z = 0 c 1 p 2 p. The numer of 0 s, 1 s nd 2 s in s re given y + + c = p. Leti =0suchthts = xy i z = xz. Thenumerof1 s in s is p wheres the numer of 0 s in s is + c. ors A, thenumerof0 s in s must equl the numer of 1 s in s,nmely + c = p. ustituting for p, wehve + c = + + c with equlity holding when = 0. Becuse y > 0nd y =, >0, thus s / A, contrdiction. Therefore A 1 is non-regulr. Detiled proof: Proof. We wnt to prove the lnguge A 1 is non-regulr. In order to use the pumping lemm, we must ssume A 1 is regulr, since the lemm only pplies to regulr lnguges. The gol is to show our ssumption leds to contrdiction, mening the ssumption is flse nd therefore the opposite must e true. ince our ssumption is tht A 1 is regulr, the opposite of this ssumption is A 1 is non-regulr, which is precisely wht we wnt to show. Once we ssume A 1 is regulr, the lemm provides us with the pumping length, p. We re now free to choose word s which elongs to A 1 nd hs length p. If we choose s ppropritely, we should e le to pump up the size of s in the mnner descried y the pumping lemm nd show the resulting word, s,does not elong to A 1. ince the lemm sttes ll such words should lso elong to the lnguge, this would e contrdiction, leding us to our conclusion tht A 1 is non-regulr. or s to e word in A 1, it must follow the form given y the definition ove. Due to the mnner in which A 1 is defined, to otin unique word, we must fix vlue for n. Given some creful thought, we cn gretly reduce the numer of cses we need to consider sed on the vlue we choose for n. To understnd the ville choices, let us oserve the effects pumping will hve on the word s. If we select s to e lrge enough word from the lnguge, the pumping lemm sttes s cn e divided into three prts s = xyz. rom this division, the lemm descries n infinite set of words of the form s = xy i z where s must lso elong to the lnguge for ny i 0. Depending on the size of y nd likewise where y flls within the word s, we will hve one of the following representtive forms for y (where 1 k p): 4

5 y = 0 k y = 1 k y = 2 k y = 0 k 1 k y = 1 k 2 k y = 0 k 1 n 2 k This mens we hve t lest six different cses to consider if we llow the size nd position of y to e ritrry (with the exception y > 0). However, we re not forced to llow this much vrition in the structure of y. In fct, using the third condition of the pumping lemm, xy p, we effectively limit oth the size nd position of y within the word s. urthermore, depending on the word we choose for s (the vlue we choose for n), we cn lso limit the symols which my pper in y nd hence the relevnce of ech form of y in the ove list. Given the form of the words in A 1, setting n llows us to control the numer of 0 s in the prefix of s. Ifwesetn p, the entire string xy must consist entirely of 0 s since xy consists of no more thn the first p symols of s, ll of which re now 0. Hence, y tkes on single form, nmely y = 0 k where 1 k p. o,yusing condition three nd choosing n ppropritely (specificlly, we will let n = p), we hve nrrowed the numer of cses we need to consider to single cse! Once we select the word s = 0 p 1 p 2 p, for the word to e useful in the context ofthelemm,itmusteevidentthts A 1 nd s p. We hve nerly proved tht s A 1. Wht remins to e shown is n 0. This follows from the conditions y > 0nd xy p. We hve n = p xy y > 0, thus n>0 nd, trivilly, n 0. Lstly, it is esy to show tht s p. ince s =p>p, s p. Now tht it is cler s A 1 nd s p, the lemm llows us to divide s into s = xyz for some x, y nd z. We cn write the representtive forms of x, y nd z s follows: x = 0 y = 0 z = 0 c 1 p 2 p We hve lredy indicted tht x nd y consist entirely of 0 s. As for the form of z, since xy p nd there re p leding 0 s in s, if xy <pthere will e some leftover 0 s which crry over into z, hence the 0 c.(therestofz is just the reminder of s.) urthermore, since the 0 s distriuted cross x, y nd z re from the p leding 0 s of s, + + c must sum to p. At this point, we focus our ttention on the new word, s = xy i z,sprovidedy the lemm. The difference etween s nd s is the numer of times the sustring y 5

6 is llowed to repet. or s, y simply ppers once ut for s, y is llowed to repet ny numer of times. A prticulr instnce of s cn e chosen y fixing the vlue for i. Preferly, we would like vlue for i other thn 1, s i =1wouldmke s = s nd we re trying to construct word which is not in A 1 (recll s needed to e word in A 1 ). The simplest vlue we cn choose for i is i =0. Inthiscse, s = xz. We must now show xz / A 1 in order to form contrdiction with the pumping lemm. (The first condition of the lemm sttes xy i z A 1 for ll i 0.) Recll the representtive forms of x nd z. x = 0 nd z = 0 c 1 p 2 p. One wy to show xz / A 1 is y showing the numer of 0 s in xz does not equl the numer of 1 s in xz since the definition of A 1 requires these quntities to e equl. By the forms of x nd z, it is pprent the numer of 0 s in xz is given y + c nd the numer of 1 s is given y p. Rememering tht + + c = p, we cn determine when the two quntities re equl. + c = p + c = + + c 0= If we cn show tht cnnot possily e 0, then our proof is complete. ortuntely, we cn demonstrte this fct using the condition y > 0. Replcing y with its representtive form, we otin 0 > 0. More or less y definition, 0 = nd therefore >0. o, it follows cnnot e 0, mening the numer of 0 s in xz cnnot equl the numer of 1 s. Thus, xz / A 1 nd since this forms contrdiction with the clims of the pumping lemm, our supposition tht A 1 is regulr must e incorrect. Hence, we conclude A 1 is non-regulr. () A 2 = {www w {, } } Proof. uppose A 2 is regulr. Let p e the pumping length given y the pumping lemm. Choose s = www where w = p. Clerly, s A 2 nd s p, thuss = xyz for some x, y nd z. ince xy p, xy cnnot extend eyond the first p symols of s, mening xy = k where 1 k p. Letuswritex =, y =, z = c p p. The numer of s in ech w is given y ++c = p. Leti =0nds = xy i z = xz. or reference, let xz = w 1 w 2 w where w 1 = c nd w 2 = w = p.thenumer of s in w 2 nd w re ech p wheres the numer of s in w 1 is + c. or s A, thenumerof s in w 1 must equl the numer of s in w 2 nd w,nmely + c = p. ustituting for p, wehve + c = + + c with equlity holding when = 0. Becuse y > 0nd y =, >0, thus s / A, contrdiction. Therefore A 2 is non-regulr. 6

7 (c) A = { 2n n 0} NOTE: n is ssumed to e n integer. Proof. uppose A is regulr. Let p e the pumping length given y the pumping lemm. Choose s = 2n. By the lemm, xy p nd y > 0 therefore p 0nd s A. Clerly, s =2 p p, thuss = xyz for some x, y nd z. Let us write x =, y =, z = c. The numer of s in s is + + c =2 p. Let i =2nd s = xy i z = xyyz. Thenumerof s in s, denoted # (s ), is +2 + c =2 p +. ince y > 0nd y =, >0. rom 2 p = + + c<+2 + c, we conclude 2 p < # (s ). ustituting for on the right-hnd side of +2 + c =2 p +, we find +2 + c =2 p +2 p c. ince xy p, c = xyz xy 2 p p>0, we hve +2 + c<2 p+1,thus# (s ) < 2 p+1. Becuse 2 p < # (s ) < 2 p+1,# (s )isnot n even power of 2 nd s / A, contrdiction. Therefore A is non-regulr. 7

### CS 311 Homework 3 due 16:30, Thursday, 14 th October 2010 CS 311 Homework 3 due 16:30, Thursdy, 14 th Octoer 2010 Homework must e sumitted on pper, in clss. Question 1. [15 pts.; 5 pts. ech] Drw stte digrms for NFAs recognizing the following lnguges:. L = {w

### Homework 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)

### 5. (±±) Λ = fw j w is string of even lengthg [ 00 = f11,00g 7. (11 [ 00)± Λ = fw j w egins with either 11 or 00g 8. (0 [ ffl)1 Λ = 01 Λ [ 1 Λ 9. Regulr Expressions, Pumping Lemm, Right Liner Grmmrs Ling 106 Mrch 25, 2002 1 Regulr Expressions A regulr expression descries or genertes lnguge: it is kind of shorthnd for listing the memers of lnguge. p-adic Egyptin Frctions Contents 1 Introduction 1 2 Trditionl Egyptin Frctions nd Greedy Algorithm 2 3 Set-up 3 4 p-greedy Algorithm 5 5 p-egyptin Trditionl 10 6 Conclusion 1 Introduction An Egyptin frction

### First Midterm Examination 24-25 Fll Semester First Midterm Exmintion ) Give the stte digrm of DFA tht recognizes the lnguge A over lphet Σ = {, } where A = {w w contins or } 2) The following DFA recognizes the lnguge B over lphet

### Name Ima Sample ASU ID Nme Im Smple ASU ID 2468024680 CSE 355 Test 1, Fll 2016 30 Septemer 2016, 8:35-9:25.m., LSA 191 Regrding of Midterms If you elieve tht your grde hs not een dded up correctly, return the entire pper to p-adic Egyptin Frctions Tony Mrtino My 7, 20 Theorem 9 negtiveorder Theorem 11 clss2 Contents 1 Introduction 1 2 Trditionl Egyptin Frctions nd Greedy Algorithm 2 3 Set-up 3 4 p-greedy Algorithm 5 5 p-egyptin

### For convenience, we rewrite m2 s m2 = m m m ; where m is repeted m times. Since xyz = m m m nd jxyj»m, we hve tht the string y is substring of the fir CSCI 2400 Models of Computtion, Section 3 Solutions to Homework 4 Problem 1. ll the solutions below refer to the Pumping Lemm of Theorem 4.8, pge 119. () L = f n b l k : k n + lg Let's ssume for contrdiction

### 1 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

### Regular Language. Nonregular Languages The Pumping Lemma. The pumping lemma. Regular Language. The pumping lemma. Infinitely long words 3/17/15 Regulr Lnguge Nonregulr Lnguges The Pumping Lemm Models of Comput=on Chpter 10 Recll, tht ny lnguge tht cn e descried y regulr expression is clled regulr lnguge In this lecture we will prove tht not ll

### First Midterm Examination Çnky University Deprtment of Computer Engineering 203-204 Fll Semester First Midterm Exmintion ) Design DFA for ll strings over the lphet Σ = {,, c} in which there is no, no nd no cc. 2) Wht lnguge does

### Parse trees, ambiguity, and Chomsky normal form Prse trees, miguity, nd Chomsky norml form In this lecture we will discuss few importnt notions connected with contextfree grmmrs, including prse trees, miguity, nd specil form for context-free grmmrs

### Grammar. Languages. Content 5/10/16. Automata and Languages. Regular Languages. Regular Languages 5//6 Grmmr Automt nd Lnguges Regulr Grmmr Context-free Grmmr Context-sensitive Grmmr Prof. Mohmed Hmd Softwre Engineering L. The University of Aizu Jpn Regulr Lnguges Context Free Lnguges Context Sensitive

### Formal 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, 14:10-18:30 Techer: Dniel Hedin, phone 021-107052 The exm

### Minimal DFA. minimal DFA for L starting from any other Miniml DFA Among the mny DFAs ccepting the sme regulr lnguge L, there is exctly one (up to renming of sttes) which hs the smllest possile numer of sttes. Moreover, it is possile to otin tht miniml DFA

### Finite 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

### 1 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

### Talen en Automaten Test 1, Mon 7 th Dec, h45 17h30 Tlen en Automten Test 1, Mon 7 th Dec, 2015 15h45 17h30 This test consists of four exercises over 5 pges. Explin your pproch, nd write your nswer to ech exercise on seprte pge. You cn score mximum of 100

### Lecture 08: Feb. 08, 2019 4CS4-6:Theory of Computtion(Closure on Reg. Lngs., regex to NDFA, DFA to regex) Prof. K.R. Chowdhry Lecture 08: Fe. 08, 2019 : Professor of CS Disclimer: These notes hve not een sujected to the usul scrutiny

### MATH 573 FINAL EXAM. May 30, 2007 MATH 573 FINAL EXAM My 30, 007 NAME: Solutions 1. This exm is due Wednesdy, June 6 efore the 1:30 pm. After 1:30 pm I will NOT ccept the exm.. This exm hs 1 pges including this cover. There re 10 prolems.

### Harvard University Computer Science 121 Midterm October 23, 2012 Hrvrd University Computer Science 121 Midterm Octoer 23, 2012 This is closed-ook exmintion. You my use ny result from lecture, Sipser, prolem sets, or section, s long s you quote it clerly. The lphet is

### Theory of Computation Regular Languages. (NTU EE) Regular Languages Fall / 38 Theory of Computtion Regulr Lnguges (NTU EE) Regulr Lnguges Fll 2017 1 / 38 Schemtic of Finite Automt control 0 0 1 0 1 1 1 0 Figure: Schemtic of Finite Automt A finite utomton hs finite set of control

### 1. For each of the following theorems, give a two or three sentence sketch of how the proof goes or why it is not true. York University CSE 2 Unit 3. DFA Clsses Converting etween DFA, NFA, Regulr Expressions, nd Extended Regulr Expressions Instructor: Jeff Edmonds Don t chet y looking t these nswers premturely.. For ech

### I1 = I2 I1 = I2 + I3 I1 + I2 = I3 + I4 I 3 2 The Prllel Circuit Electric Circuits: Figure 2- elow show ttery nd multiple resistors rrnged in prllel. Ech resistor receives portion of the current from the ttery sed on its resistnce. The split is

### List all of the possible rational roots of each equation. Then find all solutions (both real and imaginary) of the equation. 1. Mth Anlysis CP WS 4.X- Section 4.-4.4 Review Complete ech question without the use of grphing clcultor.. Compre the mening of the words: roots, zeros nd fctors.. Determine whether - is root of 0. Show

### 1. For each of the following theorems, give a two or three sentence sketch of how the proof goes or why it is not true. York University CSE 2 Unit 3. DFA Clsses Converting etween DFA, NFA, Regulr Expressions, nd Extended Regulr Expressions Instructor: Jeff Edmonds Don t chet y looking t these nswers premturely.. For ech

### Finite Automata-cont d Automt Theory nd Forml Lnguges Professor Leslie Lnder Lecture # 6 Finite Automt-cont d The Pumping Lemm WEB SITE: http://ingwe.inghmton.edu/ ~lnder/cs573.html Septemer 18, 2000 Exmple 1 Consider L = {ww

### Automata and Languages Automt nd Lnguges Prof. Mohmed Hmd Softwre Engineering Lb. The University of Aizu Jpn Grmmr Regulr Grmmr Context-free Grmmr Context-sensitive Grmmr Regulr Lnguges Context Free Lnguges Context Sensitive

### Improper 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

### CS103B Handout 18 Winter 2007 February 28, 2007 Finite Automata CS103B ndout 18 Winter 2007 Ferury 28, 2007 Finite Automt Initil text y Mggie Johnson. Introduction Severl childrens gmes fit the following description: Pieces re set up on plying ord; dice re thrown or

### CS 275 Automata and Formal Language Theory CS 275 utomt nd Forml Lnguge Theory Course Notes Prt II: The Recognition Prolem (II) Chpter II.5.: Properties of Context Free Grmmrs (14) nton Setzer (Bsed on ook drft y J. V. Tucker nd K. Stephenson)

### CS 330 Formal Methods and Models CS 330 Forml Methods nd Models Dn Richrds, George Mson University, Spring 2017 Quiz Solutions Quiz 1, Propositionl Logic Dte: Ferury 2 1. Prove ((( p q) q) p) is tutology () (3pts) y truth tle. p q p q

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

### CMPSCI 250: Introduction to Computation. Lecture #31: What DFA s Can and Can t Do David Mix Barrington 9 April 2014 CMPSCI 250: Introduction to Computtion Lecture #31: Wht DFA s Cn nd Cn t Do Dvid Mix Brrington 9 April 2014 Wht DFA s Cn nd Cn t Do Deterministic Finite Automt Forml Definition of DFA s Exmples of DFA

### Theoretical foundations of Gaussian quadrature Theoreticl foundtions of Gussin qudrture 1 Inner product vector spce Definition 1. A vector spce (or liner spce) is set V = {u, v, w,...} in which the following two opertions re defined: (A) Addition of

### Intermediate Math Circles Wednesday, November 14, 2018 Finite Automata II. Nickolas Rollick a b b. a b 4 Intermedite Mth Circles Wednesdy, Novemer 14, 2018 Finite Automt II Nickols Rollick nrollick@uwterloo.c Regulr Lnguges Lst time, we were introduced to the ide of DFA (deterministic finite utomton), one

### 1.4 Nonregular Languages 74 1.4 Nonregulr Lnguges The number of forml lnguges over ny lphbet (= decision/recognition problems) is uncountble On the other hnd, the number of regulr expressions (= strings) is countble Hence, ll

### Theory of Computation Regular Languages Theory of Computtion Regulr Lnguges Bow-Yw Wng Acdemi Sinic Spring 2012 Bow-Yw Wng (Acdemi Sinic) Regulr Lnguges Spring 2012 1 / 38 Schemtic of Finite Automt control 0 0 1 0 1 1 1 0 Figure: Schemtic of

### Coalgebra, Lecture 15: Equations for Deterministic Automata Colger, Lecture 15: Equtions for Deterministic Automt Julin Slmnc (nd Jurrin Rot) Decemer 19, 2016 In this lecture, we will study the concept of equtions for deterministic utomt. The notes re self contined

### Compiler Design. Fall Lexical Analysis. Sample Exercises and Solutions. Prof. Pedro C. Diniz University of Southern Cliforni Computer Science Deprtment Compiler Design Fll Lexicl Anlysis Smple Exercises nd Solutions Prof. Pedro C. Diniz USC / Informtion Sciences Institute 4676 Admirlty Wy, Suite

### CS 301. Lecture 04 Regular Expressions. Stephen Checkoway. January 29, 2018 CS 301 Lecture 04 Regulr Expressions Stephen Checkowy Jnury 29, 2018 1 / 35 Review from lst time NFA N = (Q, Σ, δ, q 0, F ) where δ Q Σ P (Q) mps stte nd n lphet symol (or ) to set of sttes We run n NFA

### FORMAL LANGUAGES, AUTOMATA AND COMPUTABILITY. FLAC (15-453) - Spring L. Blum 15-453 FORMAL LANGUAGES, AUTOMATA AND COMPUTABILITY THE PUMPING LEMMA FOR REGULAR LANGUAGES nd REGULAR EXPRESSIONS TUESDAY Jn 21 WHICH OF THESE ARE REGULAR? B = {0 n 1 n n 0} C = { w w hs equl numer of

### State Minimization for DFAs Stte Minimiztion for DFAs Red K & S 2.7 Do Homework 10. Consider: Stte Minimiztion 4 5 Is this miniml mchine? Step (1): Get rid of unrechle sttes. Stte Minimiztion 6, Stte is unrechle. Step (2): Get rid

### CSE396 Prelim I Answer Key Spring 2017 Nme nd St.ID#: CSE96 Prelim I Answer Key Spring 2017 (1) (24 pts.) Define A to e the lnguge of strings x {, } such tht x either egins with or ends with, ut not oth. Design DFA M such tht L(M) = A. A node-rc

### 1 Structural induction Discrete Structures Prelim 2 smple questions Solutions CS2800 Questions selected for Spring 2018 1 Structurl induction 1. We define set S of functions from Z to Z inductively s follows: Rule 1. For ny

### Lecture 3: Equivalence Relations Mthcmp Crsh Course Instructor: Pdric Brtlett Lecture 3: Equivlence Reltions Week 1 Mthcmp 2014 In our lst three tlks of this clss, we shift the focus of our tlks from proof techniques to proof concepts

### Converting Regular Expressions to Discrete Finite Automata: A Tutorial Converting Regulr Expressions to Discrete Finite Automt: A Tutoril Dvid Christinsen 2013-01-03 This is tutoril on how to convert regulr expressions to nondeterministic finite utomt (NFA) nd how to convert

### Revision Sheet. (a) Give a regular expression for each of the following languages: Theoreticl Computer Science (Bridging Course) Dr. G. D. Tipldi F. Bonirdi Winter Semester 2014/2015 Revision Sheet University of Freiurg Deprtment of Computer Science Question 1 (Finite Automt, 8 + 6 points)

### Bases for Vector Spaces Bses for Vector Spces 2-26-25 A set is independent if, roughly speking, there is no redundncy in the set: You cn t uild ny vector in the set s liner comintion of the others A set spns if you cn uild everything

### Math 4310 Solutions to homework 1 Due 9/1/16 Mth 4310 Solutions to homework 1 Due 9/1/16 1. Use the Eucliden lgorithm to find the following gretest common divisors. () gcd(252, 180) = 36 (b) gcd(513, 187) = 1 (c) gcd(7684, 4148) = 68 252 = 180 1

### CHAPTER 1 Regular Languages. Contents Finite Automt (FA or DFA) CHAPTE 1 egulr Lnguges Contents definitions, exmples, designing, regulr opertions Non-deterministic Finite Automt (NFA) definitions, euivlence of NFAs nd DFAs, closure under regulr

### W. We shall do so one by one, starting with I 1, and we shall do it greedily, trying Vitli covers 1 Definition. A Vitli cover of set E R is set V of closed intervls with positive length so tht, for every δ > 0 nd every x E, there is some I V with λ(i ) < δ nd x I. 2 Lemm (Vitli covering)

### Improper Integrals, and Differential Equations Improper Integrls, nd Differentil Equtions October 22, 204 5.3 Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted

### NON-DETERMINISTIC FSA Tw o types of non-determinism: NON-DETERMINISTIC FS () Multiple strt-sttes; strt-sttes S Q. The lnguge L(M) ={x:x tkes M from some strt-stte to some finl-stte nd ll of x is proessed}. The string x = is

### CS 373, Spring Solutions to Mock midterm 1 (Based on first midterm in CS 273, Fall 2008.) CS 373, Spring 29. Solutions to Mock midterm (sed on first midterm in CS 273, Fll 28.) Prolem : Short nswer (8 points) The nswers to these prolems should e short nd not complicted. () If n NF M ccepts

### CS103 Handout 32 Fall 2016 November 11, 2016 Problem Set 7 CS103 Hndout 32 Fll 2016 Novemer 11, 2016 Prolem Set 7 Wht cn you do with regulr expressions? Wht re the limits of regulr lnguges? On this prolem set, you'll find out! As lwys, plese feel free to drop

### Farey Fractions. Rickard Fernström. U.U.D.M. Project Report 2017:24. Department of Mathematics Uppsala University U.U.D.M. Project Report 07:4 Frey Frctions Rickrd Fernström Exmensrete i mtemtik, 5 hp Hledre: Andres Strömergsson Exmintor: Jörgen Östensson Juni 07 Deprtment of Mthemtics Uppsl University Frey Frctions

### USA 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 1-8, one digit in ech circle with no digit repeted, so tht no two circles tht re connected by line segment contin consecutive digits.

### Assignment 1 Automata, Languages, and Computability. 1 Finite State Automata and Regular Languages Deprtment of Computer Science, Austrlin Ntionl University COMP2600 Forml Methods for Softwre Engineering Semester 2, 206 Assignment Automt, Lnguges, nd Computility Smple Solutions Finite Stte Automt nd

### Polynomial 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

### Convert the NFA into DFA Convert the NF into F For ech NF we cn find F ccepting the sme lnguge. The numer of sttes of the F could e exponentil in the numer of sttes of the NF, ut in prctice this worst cse occurs rrely. lgorithm:

### 1.3 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,

### QUADRATIC RESIDUES MATH 372. FALL INSTRUCTOR: PROFESSOR AITKEN QUADRATIC RESIDUES MATH 37 FALL 005 INSTRUCTOR: PROFESSOR AITKEN When is n integer sure modulo? When does udrtic eution hve roots modulo? These re the uestions tht will concern us in this hndout 1 The

### MTH 505: Number Theory Spring 2017 MTH 505: Numer Theory Spring 207 Homework 2 Drew Armstrong The Froenius Coin Prolem. Consider the eqution x ` y c where,, c, x, y re nturl numers. We cn think of \$ nd \$ s two denomintions of coins nd \$c

### Designing finite automata II Designing finite utomt II Prolem: Design DFA A such tht L(A) consists of ll strings of nd which re of length 3n, for n = 0, 1, 2, (1) Determine wht to rememer out the input string Assign stte to ech of

### UNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 UNIFORM CONVERGENCE Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 Suppose f n : Ω R or f n : Ω C is sequence of rel or complex functions, nd f n f s n in some sense. Furthermore,

### Recursively 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

### September 13 Homework Solutions College of Engineering nd Computer Science Mechnicl Engineering Deprtment Mechnicl Engineering 5A Seminr in Engineering Anlysis Fll Ticket: 5966 Instructor: Lrry Cretto Septemer Homework Solutions. Are Qudrtic Forms Recll the Simon & Blume excerpt from n erlier lecture which sid tht the min tsk of clculus is to pproximte nonliner functions with liner functions. It s ctully more ccurte to sy tht we pproximte

### Lecture 3. In this lecture, we will discuss algorithms for solving systems of linear equations. Lecture 3 3 Solving liner equtions In this lecture we will discuss lgorithms for solving systems of liner equtions Multiplictive identity Let us restrict ourselves to considering squre mtrices since one

### Non-Deterministic Finite Automata. Fall 2018 Costas Busch - RPI 1 Non-Deterministic Finite Automt Fll 2018 Costs Busch - RPI 1 Nondeterministic Finite Automton (NFA) Alphbet ={} q q2 1 q 0 q 3 Fll 2018 Costs Busch - RPI 2 Nondeterministic Finite Automton (NFA) Alphbet

### Linear Inequalities. Work Sheet 1 Work Sheet 1 Liner Inequlities Rent--Hep, cr rentl compny,chrges \$ 15 per week plus \$ 0.0 per mile to rent one of their crs. Suppose you re limited y how much money you cn spend for the week : You cn spend

### Solutions Problem Set 2. Problem (a) Let M denote the DFA constructed by swapping the accept and non-accepting state in M. Solution Prolem Set 2 Prolem.4 () Let M denote the DFA contructed y wpping the ccept nd non-ccepting tte in M. For ny tring w B, w will e ccepted y M, tht i, fter conuming the tring w, M will e in n ccepting

### How do we solve these things, especially when they get complicated? How do we know when a system has a solution, and when is it unique? XII. LINEAR ALGEBRA: SOLVING SYSTEMS OF EQUATIONS Tody we re going to tlk out solving systems of liner equtions. These re prolems tht give couple of equtions with couple of unknowns, like: 6= x + x 7=

### CSCI 340: Computational Models. Kleene s Theorem. Department of Computer Science CSCI 340: Computtionl Models Kleene s Theorem Chpter 7 Deprtment of Computer Science Unifiction In 1954, Kleene presented (nd proved) theorem which (in our version) sttes tht if lnguge cn e defined y ny

### CS 330 Formal Methods and Models Dana Richards, George Mason University, Spring 2016 Quiz Solutions CS 330 Forml Methods nd Models Dn Richrds, George Mson University, Spring 2016 Quiz Solutions Quiz 1, Propositionl Logic Dte: Ferury 9 1. (4pts) ((p q) (q r)) (p r), prove tutology using truth tles. p

### Lecture 2 : Propositions DRAFT CS/Mth 240: Introduction to Discrete Mthemtics 1/20/2010 Lecture 2 : Propositions Instructor: Dieter vn Melkeeek Scrie: Dlior Zelený DRAFT Lst time we nlyzed vrious mze solving lgorithms in order to illustrte

### Chapter 4 Regular Grammar and Regular Sets. (Solutions / Hints) C K Ngpl Forml Lnguges nd utomt Theory Chpter 4 Regulr Grmmr nd Regulr ets (olutions / Hints) ol. (),,,,,,,,,,,,,,,,,,,,,,,,,, (),, (c) c c, c c, c, c, c c, c, c, c, c, c, c, c c,c, c, c, c, c, c, c, c,

### CSCI FOUNDATIONS OF COMPUTER SCIENCE 1 CSCI- 2200 FOUNDATIONS OF COMPUTER SCIENCE Spring 2015 My 7, 2015 2 Announcements Homework 9 is due now. Some finl exm review problems will be posted on the web site tody. These re prcqce problems not Mth 348 Fll 7 Lecture 3: Curves in Clculus Disclimer. As we hve textook, this lecture note is for guidnce nd supplement only. It should not e relied on when prepring for exms. In this lecture we set up

### 3 Regular expressions 3 Regulr expressions Given n lphet Σ lnguge is set of words L Σ. So fr we were le to descrie lnguges either y using set theory (i.e. enumertion or comprehension) or y n utomton. In this section we shll

### set is not closed under matrix [ multiplication, ] and does not form a group. Prolem 2.3: Which of the following collections of 2 2 mtrices with rel entries form groups under [ mtrix ] multipliction? i) Those of the form for which c d 2 Answer: The set of such mtrices is not closed

### Handout: 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

### CSE : Exam 3-ANSWERS, Spring 2011 Time: 50 minutes CSE 260-002: Exm 3-ANSWERS, Spring 20 ime: 50 minutes Nme: his exm hs 4 pges nd 0 prolems totling 00 points. his exm is closed ook nd closed notes.. Wrshll s lgorithm for trnsitive closure computtion is

### (e) if x = y + z and a divides any two of the integers x, y, or z, then a divides the remaining integer Divisibility In this note we introduce the notion of divisibility for two integers nd b then we discuss the division lgorithm. First we give forml definition nd note some properties of the division opertion. Finite Automt Let's strt with n exmple: Here you see leled circles tht re sttes, nd leled rrows tht re trnsitions. One of the sttes is mrked "strt". One of the sttes hs doule circle; this is terminl stte

### Regular expressions, Finite Automata, transition graphs are all the same!! CSI 3104 /Winter 2011: Introduction to Forml Lnguges Chpter 7: Kleene s Theorem Chpter 7: Kleene s Theorem Regulr expressions, Finite Automt, trnsition grphs re ll the sme!! Dr. Neji Zgui CSI3104-W11 1

### Fundamental Theorem of Calculus Fundmentl Theorem of Clculus Recll tht if f is nonnegtive nd continuous on [, ], then the re under its grph etween nd is the definite integrl A= f() d Now, for in the intervl [, ], let A() e the re under

### and that at t = 0 the object is at position 5. Find the position of the object at t = 2. 7.2 The Fundmentl Theorem of Clculus 49 re mny, mny problems tht pper much different on the surfce but tht turn out to be the sme s these problems, in the sense tht when we try to pproimte solutions we

### Non Deterministic Automata. Linz: Nondeterministic Finite Accepters, page 51 Non Deterministic Automt Linz: Nondeterministic Finite Accepters, pge 51 1 Nondeterministic Finite Accepter (NFA) Alphbet ={} q 1 q2 q 0 q 3 2 Nondeterministic Finite Accepter (NFA) Alphbet ={} Two choices

### 2.4 Linear Inequalities and Interval Notation .4 Liner Inequlities nd Intervl Nottion We wnt to solve equtions tht hve n inequlity symol insted of n equl sign. There re four inequlity symols tht we will look t: Less thn , Less thn or

### 1 Structural induction, finite automata, regular expressions Discrete Structures Prelim 2 smple uestions s CS2800 Questions selected for spring 2017 1 Structurl induction, finite utomt, regulr expressions 1. We define set S of functions from Z to Z inductively s

### MAA 4212 Improper Integrals Notes by Dvid Groisser, Copyright c 1995; revised 2002, 2009, 2014 MAA 4212 Improper Integrls The Riemnn integrl, while perfectly well-defined, is too restrictive for mny purposes; there re functions which

### CS 330 Formal Methods and Models CS 330 Forml Methods nd Models Dn Richrds, section 003, George Mson University, Fll 2017 Quiz Solutions Quiz 1, Propositionl Logic Dte: Septemer 7 1. Prove (p q) (p q), () (5pts) using truth tles. p q

### Mathematics Number: Logarithms plce of mind F A C U L T Y O F E D U C A T I O N Deprtment of Curriculum nd Pedgogy Mthemtics Numer: Logrithms Science nd Mthemtics Eduction Reserch Group Supported y UBC Teching nd Lerning Enhncement

### Lecture 09: Myhill-Nerode Theorem CS 373: Theory of Computtion Mdhusudn Prthsrthy Lecture 09: Myhill-Nerode Theorem 16 Ferury 2010 In this lecture, we will see tht every lnguge hs unique miniml DFA We will see this fct from two perspectives 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 Models of Comput:on Lecture #8 Chpter 7 con:nued Any lnguge tht e defined y regulr expression, finite utomton, or trnsi:on grph cn e defined y ll three methods We prove this y showing tht ny lnguge defined