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8/27/2009 Sofya Raskhodnikova Intro to Theory of Computation L ECTURE 2 Theory of Computation Finite Automata Operations on languages Nondeterminism L2.1 Sofya Raskhodnikova; based on slides by Nick Hopper Homework 0 due Homework 1 handed out

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JFLAP L2.2 8/27/2009 Sofya Raskhodnikova; based on slides by Rakesh Verma aaa aaaaaa

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Formal definition of FA L2.3 8/27/2009 Sofya Raskhodnikova; based on slides by Nick Hopper Q = {q 0, q 1, q 2, q 3 } Σ = {0,1} : Q Σ → Q transition function * q 0 Q is start state F = {q 1, q 2 } Q accept states M = (Q, Σ, , q 0, F) where M q2q2 0 0, q0q0 q1q1 q3q3 01 q0q0 q0q0 q1q1 q1q1 q2q2 q2q2 q2q2 q3q3 q2q2 q3q3 q0q0 q2q2 *

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A language is regular if it is recognized by a finite automaton L = { w | w contains 001} is regular L = { w | w has an even number of 1s} is regular The language of M, L(M), is the set of strings that M accepts Language of FA 8/27/2009 L1.4Sofya Raskhodnikova; based on slides by Nick Hopper

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8/27/2009 Sofya Raskhodnikova; based on slides by Nick Hopper L1.5 Regular languages Many interesting programs accept regular languages NETWORK PROTOCOLS COMPILERS GENETIC TESTING ARITHMETIC

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INTERNET TRANSMISSION CONTROL PROTOCOL Let TCPS = { w | w is a complete TCP Session} Theorem. TCPS is regular 8/27/2009 Sofya Raskhodnikova; based on slides by Nick Hopper L1.6

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COMMENTS : Are delimited by /* */ Cannot have NESTED /* */ Must be CLOSED by */ */ is ILLEGAL outside a comment COMMENTS = {strings with legal comments} Theorem. COMMENTS is regular. Compilers 8/27/2009 Sofya Raskhodnikova; based on slides by Nick Hopper L1.7

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DNA SEQUENCES are strings over the alphabet {A,C,G,T} GENES are special substrings A GENETIC TEST searches a DNA SEQUENCE for a GENE Theorem. Every GENETIC TEST is regular. Genetic testing 8/27/2009 Sofya Raskhodnikova; based on slides by Nick Hopper L1.8

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LET 3 = A string over 3 has three ROWS Each ROW b 0 b 1 b 2 …b N represents the integer b 0 + 2b 1 + … + 2 N b N. Let ADD = {S | ROW 1 + ROW 2 = ROW 3 } Theorem. ADD is regular. { [ ],[ ],[ ],[ ], [ ],[ ],[ ],[ ]} Arithmetic 8/27/2009 Sofya Raskhodnikova; based on slides by Nick Hopper L1.9

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Operations on languages L2.10 8/27/2009 Sofya Raskhodnikova; based on slides by Nick Hopper Union: A B = { w | w A or w B } Intersection: A B = { w | w A and w B } Complement: A = { w | w A } Reverse: A R = { w 1 …w k | w k …w 1 A } Concatenation: A B = { vw | v A and w B } Star: A* = { w 1 …w k | k ≥ 0 and each w i A } Regular Union: A B = { w | w A or w B } Concatenation: A B = { vw | v A and w B } Star: A* = { w 1 …w k | k ≥ 0 and each w i A }

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Closure properties of the class of regular languages L2.11 8/27/2009 Sofya Raskhodnikova; based on slides by Nick Hopper THEOREM. The class of regular languages is closed under all 6 operations. If A and B are regular, applying any of these operation yields a regular language.

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Closure properties L2.12 8/27/2009 Sofya Raskhodnikova; based on slides by Nick Hopper Union: A B = { w | w A or w B } Complement: A = { w | w A }

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Closure under union L2.13 8/27/2009 Sofya Raskhodnikova; based on slides by Nick Hopper Theorem. The union of two regular languages is also a regular language Proof: Let M 1 = (Q 1, Σ, 1, q 0, F 1 ) be finite automaton for L 1 and M 2 = (Q 2, Σ, 2, q 0, F 2 ) be finite automaton for L 2 Construct a finite automaton M = (Q, Σ, , q 0, F) that recognizes L = L 1 L 2 1 2

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Example L2.14 8/27/2009 Sofya Raskhodnikova; based on slides by Nick Hopper q0q0 q1q p0p0 p1p M1M1 M2M2 M = ?

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Proof (continued) L2.15 8/27/2009 Sofya Raskhodnikova; based on slides by Nick Hopper Idea: Run both M 1 and M 2 at the same time! Q = pairs of states, one from M 1 and one from M 2 = { (q 1, q 2 ) | q 1 Q 1 and q 2 Q 2 } = Q 1 Q 2 q 0 = (q 0, q 0 ) 1 2 F = { (q 1, q 2 ) | q 1 F 1 or q 2 F 2 } ( (q 1,q 2 ), ) = ( 1 (q 1, ), 2 (q 2, ))

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Example (continued) L2.16 8/27/2009 Sofya Raskhodnikova; based on slides by Nick Hopper q0q0 q1q p0p0 p1p M1M1 M2M2 q 0,p 0 q 1,p 0 q 0,p 1 q 1,p M unionintersection

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Closure properties L2.17 8/27/2009 Sofya Raskhodnikova; based on slides by Nick Hopper Union: A B = { w | w A or w B } Intersection: A B = { w | w A and w B } Complement: A = { w | w A } Reverse: A R = { w 1 …w k | w k …w 1 A }

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Closure under reverse L2.18 8/27/2009 Sofya Raskhodnikova; based on slides by Nick Hopper Theorem. The reverse of a regular language is also regular language Proof: Let L be a regular language and M be a finite automaton that recognizes it. Construct a finite automaton M R recognizing L R Idea: Define M R as M with the arrows reversed Swap start and accept states

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Closure under reverse L2.19 8/27/2009 Sofya Raskhodnikova; based on slides by Nick Hopper M R IS NOT ALWAYS A DFA! It may have many start states Some states may have too many outgoing edges, or none

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Example L2.20 8/27/2009 Sofya Raskhodnikova; based on slides by Nick Hopper ,1 0

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L2.21 8/27/2009 Sofya Raskhodnikova; based on slides by Nick Hopper NON DETERMINISM ,1 0 What happens with 100? Nondeterministic Finite Automaton (NFA) accepts if there is a way to make it reach an accept state

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Nondeterminism Ways to think about nondeterminism parallel computation tree of possible computations guessing the “right” choice L2.22 Deterministic Computation Nondeterministic Computation accept or reject accept reject 8/27/2009 Sofya Raskhodnikova; based on slides by Nick Hopper

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Example L2.23 8/27/2009 Sofya Raskhodnikova; based on slides by Nick Hopper L(M)={1,00} ε ε

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Example L2.24 8/27/2009 Sofya Raskhodnikova; based on slides by Nick Hopper 0,1 0,ε1 0,1 1 L(M)={w | w contains 101 or 11}

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Formal Definition 8/25/2009 Q is the set of states Σ is the alphabet q 0 Q is the start state F Q is the set of accept states An NFA is a 5-tuple M = (Q, Σ, , q 0, F) L1.25 Sofya Raskhodnikova; based on slides by Nick Hopper : Q Σ ε → P(Q) is the transition function P(Q) is the set of subsets of Q and Σ ε = Σ {ε}

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Example L2.26 8/27/2009 Sofya Raskhodnikova; based on slides by Nick Hopper (q 3,1) = N = (Q, Σ, , Q 0, F) Q = {q 0, q 1, q 2, q 3, q 4 } Σ = {0,1} F = {q 4 } (q 2,1) = {q 4 } ε ε q1q1 q2q2 q3q3 q4q4 q0q0

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Example L2.27 8/27/2009 Sofya Raskhodnikova; based on slides by Nick Hopper 0,1 0,ε1 0,1 1 (q 1,ε) = N = (Q, Σ, , q 0, F) Q = {q 0, q 1, q 2, q 3 } Σ = {0,1} F = {q 3 } (q 0,1) = {q 0, q 1 } {q 1,q 2 } (q 2,0) = q1q1 q2q2 q3q3 q0q0

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