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FORMAL LANGUAGES, AUTOMATA, AND COMPUTABILITY

15-453. FORMAL LANGUAGES, AUTOMATA, AND COMPUTABILITY. Q. Σ. q 0. F. M = ({p,q}, {0,1}, , p, {q}). 0. 0. THEOREM: L(M) = {w | w has odd number of 1s }. 1. q. p. 1. Proof: By induction on n , the length of a string.

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FORMAL LANGUAGES, AUTOMATA, AND COMPUTABILITY

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  1. 15-453 FORMAL LANGUAGES, AUTOMATA, AND COMPUTABILITY

  2. Q Σ q0 F M = ({p,q}, {0,1}, , p, {q}) 0 0 THEOREM: L(M) = {w | w has odd number of 1s } 1 q p 1 Proof: By induction on n, the length of a string. Base Case: n=0:ε  L and ε  L(M). Induction Step: Suppose for all w Σ*, |w| = n, w L(M) iffw has odd number of 1s Any string of length n+1 has the form w0 or w1 If w has odd # of 1’s, M is in state q after reading w  M is in state p after reading w1 M rejects w1  M is in state q after reading w0  M accepts w0

  3. UNION THEOREM The union of two regular languages is also a regular language “Regular Languages Are Closed Under Union” INTERSECTION THEOREM The intersection of two regular languages is also a regular language

  4. Complement THEOREM The complement of a regular language is also a regular language In other words, if L is regular than so is L, where L= { w  Σ* | w  L } Proof ?

  5. THE REGULAR OPERATIONS Union: A  B = { w | w  A or w  B } Intersection: A  B = { w | w  A and w  B } Negation:A = { w  Σ* | w  A } Reverse: AR = { w1 …wk | wk …w1 A } Concatenation: A  B = { vw | v  A and w  B } Star: A* = { s1 … sk | k ≥ 0 and each si A }

  6. REVERSE THEOREM The reverse of a regular language is also a regular language ``Regular Languages Are Closed Under Reverse” If a language can be recognized by a DFA that reads strings from right to left, then there is an “normal” DFA that accepts the same language Counterintuitive! DFAs have finite memory…

  7. REVERSING DFAs Assume L is a regular language. Let M be a DFA that recognizes L We’ll build a machine MR that accepts LR If M accepts w, then w describes a directed path in M from start to an accept First Attempt: Try to define MR as M with the arrows reversed, turn start state into a final state, turn final states into starts

  8. MR IS NOT ALWAYS A DFA! It could have many start states Some states may have too many outgoing edges, or none at all!

  9. 0,1 1 0 0 0 1 1

  10. NONDETERMINISM 0,1 1 0 0 0 1 1 What happens with 100? We will say that this machine accepts a string if there is some paththat reaches an accept state from a start state

  11. IBM JOURNAL APRIL 1959 Turing Award winningpaper

  12. NFA EXAMPLES 0,1 0,1 0, ε 0 At each state, we can have any number of out arrows for some letter  Σ, including ε

  13. NFA EXAMPLES 1 1 0,1 ε 0 0 L(M) = {0i1j | i  {0,1}, j ≥ 0} Possibly many start states

  14. NFA EXAMPLES 1 0 0 L(M)={1,00}

  15. A non-deterministic finite automaton (NFA) is a 5-tuple N = (Q, Σ, , Q0, F) Q is the set of states Σ is the alphabet  : Q Σε→ 2Q is the transition function Q0 Q is the set of start states F  Q is the set of accept states 2Q is the set of all possible subsets of Q Σε = Σ {ε}

  16. Let w Σ* and suppose w can be written as • w1... wn where wi Σε(ε = empty string) • Then N acceptsw if there are r0, r1, ..., rn Qsuch that • r0 Q0 • ri+1 (ri, wi+1 ) for i = 0, ..., n-1, and • rn  F L(N) = the language recognized by N = set of all strings machine N accepts A language L is recognized by an NFA N if L = L(N).

  17. 1 q4 q2 0 N = (Q, Σ, , Q0, F) ε q3 Q = {q1, q2, q3, q4} Σ = {0,1} 0 Q0 = {q1, q2} q1 F = {q4} Q {q4} (q2,1) = 00 L(N)? 01 L(N)? (q3,1) =  { q3} (q1,0) =

  18. Deterministic Computation Non-Deterministic Computation reject accept or reject accept

  19. MULTIPLE START STATES We allow multiple start states for NFAs, Can easily convert NFA with many start states into one with a single start state: ε ε ε

  20. UNION THEOREM FOR NFAs? 0 0 1 0 1

  21. NFAs ARE SIMPLER THAN DFAs An NFA that recognizes the language {1}: 1 A DFA that recognizes the language {1}: 0,1 0 1 0,1

  22. BUT DFAs CAN SIMULATENFAs! Theorem: Every NFA has an equivalent*DFA Corollary: A language is regular iff it is recognized by an NFA Corollary: L is regular iff LR is regular * N is equivalent to M if L(N) = L (M)

  23. FROM NFA TO DFA Input: N = (Q, Σ, , Q0, F) Output: M = (Q, Σ, , q0, F) To learn if NFA accepts, we could do the computation in parallel, maintaining the set of all possible states that can be reached Q = 2Q reject Idea: Q = 2Q accept

  24. (R,) =  ε( (r,) ) rR FROM NFA TO DFA Input: NFA N = (Q, Σ, , Q0, F) Output: DFA M = (Q, Σ, , q0, F) Q = 2Q  : Q Σ→ Q * q0 = ε(Q0) F = { R  Q | f  R for some f  F } * For R  Q, the ε-closure of R,ε(R) = {q that can be reached from some r R by traveling along zero or more ε arrows}

  25. EXAMPLE OF ε-CLOSURE 0,1 0,1 0,ε 0,ε q2 q0 q1 ε({q0}) = {q0 , q1, q2} {q1, q2} ε({q1}) = {q2} ε({q2}) =

  26. Given: NFA N = ( {1,2,3}, {a,b},  , {1}, {1} ) Construct: Equivalent DFA M M = (2{1,2,3}, {a,b}, , {1,3}, …) N a a b 1 {1,3} {3}  a b b a b b a,b ε a , b {2} {2,3} a 2 3 a b a {1}, {1,2} ? ε({1}) = {1,3} {1,2,3}

  27. NFAs CAN MAKE PROOFS MUCH EASIER! Remember this on your Homework!

  28. REGULAR LANGUAGES ARE CLOSED UNDER THE REGULAR OPERATIONS Union: A  B = { w | w  A or w  B } Intersection: A  B = { w | w  A and w  B } Negation:A = { w  Σ* | w  A } Reverse: AR = { w1 …wk | wk …w1 A } Concatenation: A  B = { vw | v  A and w  B } Star: A* = { w1 …wk | k ≥ 0 and each wi A }

  29. SOME LANGUAGES ARE NOT REGULAR B = {0n1n | n ≥ 0} is NOT regular!

  30. WHICH OF THESE ARE REGULAR C = { w | w has equal number of 1s and 0s} NOT REGULAR D = { w | w has equal number of occurrences of 01 and 10 } REGULAR!!!

  31. THE PUMPING LEMMA Let L be a regular language with |L| =  Then there is a positive integer P s.t. if w  L and |w| ≥ P then can write w = xyz, where: 1. |y| > 0 (y isn’t ε) 2. |xy| ≤ P 3. For everyi ≥ 0, xyiz  L Why is it called the pumping lemma? The word w gets PUMPED into something longer…

  32. q|w| Proof: Let M be a DFA that recognizes L Let P be the number of states in M Assume w  L is such that |w| ≥ P 1. |y| > 0 2. |xy| ≤ P 3. xyiz  L for all i ≥ 0 We show: w = xyz y x z … q0 qj qk There must be j and k such that j < k ≤ P, and qj = qk

  33. USING THE PUMPING LEMMA Let’s prove that B = {0n1n | n ≥ 0} is not regular Assume B is regular. Let w = 0P1P If B is regular, can write w = xyz, |y| > 0, |xy| ≤ P, and for any i ≥ 0, xyiz is also in B Contradiction! xyyz has more 0s than 1s If y is all 0s: THIS CASE CAN’T HAPPEN! If y is all 1s: |xy| ≤ P If y has both 1s and 0s: THIS CASE CAN’T HAPPEN!

  34. USING THE PUMPING LEMMA C = { w | w has equal number of 1s and 0s}is not regular Assume C is regular. Let w = 0P1P(w is in C!) If C is regular, can write w = xyz, |y| > 0, |xy| ≤ P, where for any i ≥ 0, xyiz is also in C xyyz has more 0s than 1s If y is all 0s: CAN’T HAPPEN If y is all 1s: If y has both 1s and 0s: CAN’T HAPPEN

  35. FLAC Read Chapters 1.3 and 2.1 of the book for next time

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