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CS172: “Computability & Complexity” Wim van Dam Soda 665

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Today Chapter 0: set notation and languages Chapter 1: deterministic finite automata (DFA) nondeterministic FA (NFA)

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Standard Set Theory: Conditional: A = { x | x N, f(x)=0 } Union: A B Intersection: A B Complement: Cartesian Product: A B Power set: P (A)

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Some Examples L <6 = { x | x N, x<6 } L <6 L prime = {2,3,5} = {0,1} = {(0,0), (0,1), (1,0), (1,1)} Formal: A B = { x | x A and x B}

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Powerset Formal: P (A) = { S | S A} Example: A = {x,y} P (A) = { {}, {x}, {y}, {x,y} } Note the different sizes: | P (A)| = 2 |A| |A A| = |A| 2

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Languages Given an alphabet , we can make a word or string by concatenating the letters of . Concatenation of “x” and “y” is “xy” Typical example: ={0,1}, the possible words over are the finite bit strings. A language is a set of words.

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More about Languages The empty string is the unique string with zero length. Concatenation of two langauges: AB = { xy | x A and y B } Typical examples: L = { x | x is a bit string with two zeros } L = { a n b n | n N } L = {1 n | n is prime}

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A Word of Warning Do not confuse the concatenation of languages with the Cartesian product of sets. For example, let A = {0,00} then AA = { 00, 000, 0000 } with |AA|=3, A A = { (0,0), (0,00), (00,0), (00,00) } with |A A|=4

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Recognizing Languages Let L be a language S a machine M recognizes L if M xSxS “accept” “reject” if and only if x L if and only if x L

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Finite Automaton The most simple machine that is not just a finite list of words. “Read once”, “no write” procedure. Typical is its limited memory. Think cell-phone, elevator door, etc.

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A Simple Automaton (0) q1q1 q2q2 q3q3 10 0,1 01 statestransition rules starting state accepting state

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A Simple Automaton (1) q1q1 q2q2 q3q3 10 0,1 01 on input “0110”, the machine goes: q 1 q 1 q 2 q 2 q 3 = “reject” start accept

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A Simple Automaton (2) q 1 q 2 q 3 q 2 = “accept” q1q1 q2q2 q3q3 10 0,1 01 on input “101”, the machine goes:

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A Simple Automaton (3) 010: reject 11: accept : accept : reject : reject q1q1 q2q2 q3q ,1

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Finite Automaton (def.) A deterministic finite automaton (DFA) M is defined by a 5-tuple M=(Q, , ,q 0,F) –Q: finite set of states – : finite alphabet – : transition function :Q Q –q 0 Q: start state –F Q: set of accepting states

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M = (Q, , ,q,F) states Q = {q 1,q 2,q 3 } alphabet = {0,1} start state q 1 accept states F={q 2 } transition function : q1q1 q2q2 q3q ,1

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Recognizing Languages (def) A finite automaton M = (Q, , ,q,F) accepts a string/word w = w 1 …w n if and only if there is a sequence r 0 …r n of states in Q such that: 1) r 0 = q 0 2) (r i,w i+1 ) = r i+1 for all i = 0,…,n–1 3) r n F

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Regular Languages The language recognized by a finite automaton M is denoted by L(M). A regular language is a language for which there exists a recognizing finite automaton.

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Two DFA Questions Given the description of a finite automaton M = (Q, , ,q,F), what is the language L(M) that it recognizes? In general, what kind of languages can be recognized by finite automata? (What are the regular languages?)

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Union of Two Languages Theorem 1.12: If A 1 and A 2 are regular languages, then so is A 1 A 2. (The regular languages are ‘closed’ under the union operation.) Proof idea: A 1 and A 2 are regular, hence there are two DFA M 1 and M 2, with A 1 =L(M 1 ) and A 2 =L(M 2 ). Out of these two DFA, we will make a third automaton M 3 such that L(M 3 ) = A 1 A 2.

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Proof Union-Theorem (1) M 1 =(Q 1, , 1,q 1,F 1 ) and M 2 =(Q 2, , 2,q 2,F 2 ) Define M 3 = (Q 3, , 3,q 3,F 3 ) by: Q 3 = Q 1 Q 2 = {(r 1,r 2 ) | r 1 Q 1 and r 2 Q 2 } 3 ((r 1,r 2 ),a) = ( 1 (r 1,a), 2 (r 2,a)) q 3 = (q 1,q 2 ) F 3 = {(r 1,r 2 ) | r 1 F 1 or r 2 F 2 }

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Proof Union-Theorem (2) The automaton M 3 = (Q 3, , 3,q 3,F 3 ) runs M 1 and M 2 in ‘parallel’ on a string w. In the end, the final state (r 1,r 2 ) ‘knows’ if w L 1 (via r 1 F 1 ?) and if w L 2 (via r 2 F 2 ?) The accepting states F 3 of M 3 are such that w L(M 3 ) if and only if w L 1 or w L 2, for: F 3 = {(r 1,r 2 ) | r 1 F 1 or r 2 F 2 }.

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Concatenation of L 1 and L 2 Definition: L 1 L 2 = { xy | x L 1 and y L 2 } Example: {a,b} {0,11} = {a0,a11,b0,b11} Theorem 1.13: If L 1 and L 2 are regular langues, then so is L 1 L 2. (The regular languages are ‘closed’ under concatenation.)

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Proving Concatenation Thm. Consider the concatenation: {1,01,11,001,011,…} {0,000,00000,…} (That is: the bit strings that end with a “1”, followed by an odd number of 0’s.) Problem is: given a string w, how does the automaton know where the L 1 part stops and the L 2 substring starts? We need an M with ‘lucky guesses’.

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Nondeterminism Nondeterministic machines are capable of being lucky, no matter how small the probability. A nondeterministic finite automaton has transition rules/possibilities like q1q1 q2q2 q1q1 q2q2 1 q3q3 1

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A Nondeterministic Automaton q1q1 q2q2 q3q3 1 0, 0,1 This automaton accepts “0110”, because there is a possible path that leads to an accepting state, namely: q 1 q 1 q 2 q 3 q 4 q 4 q4q4 1 0,1

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A Nondeterministic Automaton q1q1 q2q2 q3q3 1 0, 0,1 The string 1 gets rejected: on “1” the automaton can only reach: {q 1,q 2,q 3 }. q4q4 1 0,1

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Nondeterminism ~ Parallelism For any (sub)string w, the nondeterministic automaton can be in a set of possible states. If the final set contains an accepting state, then the automaton accepts the string. “The automaton processes the input in a parallel fashion. Its computational path is no longer a line, but a tree.” (Fig. 1.16)

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Nondeterministic FA (def.) A nondeterministic finite automaton (NFA) M is defined by a 5-tuple M=(Q, , ,q 0,F), with –Q: finite set of states – : finite alphabet – : transition function :Q P (Q) –q 0 Q: start state –F Q: set of accepting states

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Nondeterministic :Q P (Q) The function :Q P (Q) is the crucial difference. It means: “When reading symbol “a” while in state q, one can go to one of the states in (q,a) Q.” The in = { } takes care of the empty string transitions.

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Recognizing Languages (def) A nondeterministic FA M = (Q, , ,q,F) accepts a string w = w 1 …w n if and only if we can rewrite w as y 1 …y m with y i and there is a sequence r 0 …r m of states in Q such that: 1) r 0 =q 0 2) r i+1 (r i,y i+1 ) for all i=0,…,m–1 3) r m F

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Exercises (1:3) [Sipser 0.3]: Let A = {x,y,z} and B = {x,y}, answer: 1.Is A a subset of B? 2.Is B a subset of A? 3.What is A B? 4.What is A B? 5.What is A B? 6.What is P (Q)?

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Exercises (2:3) [Sipser 1.5]: Give NFAs with the specified number of states that recognize the following languages over the alphabet ={0,1}: 1.{ w | w ends with 00}, three states 2.{0}; two states 3.{ w | w contains even number of 0s, or exactly two 1s}, six states 4.{0 n | n N }, one state

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Exercises (3:3) Proof the following result: “If L 1 and L 2 are regular languages, then is a regular language too.” Describe the language that is recognized by this nondeterministic automaton: q1q1 q2q2 q3q3 1 0, 1 q4q4 1 0,1

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Additional Practice Problems [Sipser 0.2]: Write the formal descriptions of the sets containing 1.… the numbers 1,10 and … all integers greater than 5 3.… all natural numbers less than 5 4.… the string “aba” 5.… the empty string 6.… nothing Give a formal description of this NFA: Give DFA state diagrams for the following languages over ={0,1}: 1.{ w | w begins with 1 and ends with 0} 2.{ w | w does not contain substring 110} 3.{} 4.all strings except the empty string q1q1 q2q2 q3q3 1 0, 1

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