lect6-DecisionPropertiesOfRL

lect6-DecisionPropertiesOfRL - Decision Properties of...

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Unformatted text preview: Decision Properties of Regular Languages General Discussion of “Properties” The Pumping Lemma Membership, Emptiness, Etc. 1 Properties of Language Classes x A language class is a set of languages.   We have one example: the regular languages. We’ll see many more in this class. x Language classes have two important kinds of properties: 1. Decision properties. 2. Closure properties. 2 Representation of Languages x Representations can be formal or informal. x Example (formal): represent a language by a RE or DFA defining it. x Example: (informal): a logical or prose statement about its strings:  {0n1n | n is a nonnegative integer}  “The set of strings consisting of some number of 0’s followed by the same number of 1’s.” 3 Decision Properties x A decision property for a class of languages is an algorithm that takes a formal description of a language (e.g., a DFA) and tells whether or not some property holds. x Example: Is language L empty? 4 Subtle Point: Representation Matters x You might imagine that the language is described informally, so if my description is “the empty language” then yes, otherwise no. x But the representation is a DFA (or a RE that you will convert to a DFA). x Can you tell if L(A) = ∅ for DFA A? 5 Why Decision Properties? x When we talked about protocols represented as DFA’s, we noted that important properties of a good protocol were related to the language of the DFA. x Example: “Does the protocol terminate?” = “Is the language finite?” x Example: “Can the protocol fail?” = “Is the language nonempty?” 6 Why Decision Properties – (2) x We might want a “smallest” representation for a language, e.g., a minimum­state DFA or a shortest RE. x If you can’t decide “Are these two languages the same?”  I.e., do two DFA’s define the same language? You can’t find a “smallest.” 7 Closure Properties x A closure property of a language class says that given languages in the class, an operator (e.g., union) produces another language in the same class. x Example: the regular languages are obviously closed under union, concatenation, and (Kleene) closure.  Use the RE representation of languages. 8 Why Closure Properties? 1. Helps construct representations. 2. Helps show (informally described) languages not to be in the class. 9 Example: Use of Closure Property x We can easily prove L1 = {0n1n | n > 0} is not a regular language. x L2 = the set of strings with an = number of 0’s and 1’s isn’t either, but that fact is trickier to prove. x Regular languages are closed under ∩. x If L2 were regular, then L2 ∩L(0*1*) = L1 would be, but it isn’t. 10 The Membership Question x Our first decision property is the question: “is string w in regular language L?” x Assume L is represented by a DFA A. x Simulate the action of A on the sequence of input symbols forming w. 11 Example: Testing Membership 0 1 0 1 1 Next symbol 0 A Start 0,1 1 B 1 C 0 Current state 12 Example: Testing Membership 0 1 0 1 1 Next symbol 0 A Start 1 0,1 B 1 C 0 Current state 13 Example: Testing Membership 0 1 0 1 1 Next symbol 0 A Start 1 0,1 B 1 C 0 Current state 14 Example: Testing Membership 0 1 0 1 1 0 Next symbol A Start 0,1 1 B 1 C 0 Current state 15 Example: Testing Membership 0 1 0 1 1 0 A Start Next symbol 0,1 1 B 1 C 0 Current state 16 Example: Testing Membership 0 1 0 1 1 Next symbol 0 A Start 0,1 1 B 1 C 0 Current state17 What if the Regular Language Is not Represented by a DFA? x There is a circle of conversions from one form to another: RE ε­NFA DFA NFA 18 The Emptiness Problem x Given a regular language, does the language contain any string at all. x Assume representation is DFA. x Construct the transition graph. x Compute the set of states reachable from the start state. x If any final state is reachable, then yes, else no. 19 The Infiniteness Problem x Is a given regular language infinite? x Start with a DFA for the language. x Key idea: if the DFA has n states, and the language contains any string of length n or more, then the language is infinite. x Otherwise, the language is surely finite.  Limited to strings of length n or less. 20 Proof of Key Idea x If an n­state DFA accepts a string w of length n or more, then there must be a state that appears twice on the path labeled w from the start state to a final state. x Because there are at least n+1 states along the path. 21 Proof – (2) w = xyz x y z Then xyiz is in the language for all i > 0. Since y is not ε, we see an infinite number of strings in L. 22 Infiniteness – Continued x We do not yet have an algorithm. x There are an infinite number of strings of length > n, and we can’t test them all. x Second key idea: if there is a string of length > n (= number of states) in L, then there is a string of length between n and 2n­1. 23 Proof of 2nd Key Idea y x Remember: x z x We can choose y to be the first cycle on the path. x So |xy| < n; in particular, 1 < |y| < n. x Thus, if w is of length 2n or more, there is a shorter string in L that is still of length at least n. x Keep shortening to reach [n, 2n­1]. 24 Completion of Infiniteness Algorithm x Test for membership all strings of length between n and 2n­1.  If any are accepted, then infinite, else finite. x A terrible algorithm. x Better: find cycles between the start state and a final state. 25 Finding Cycles 1. Eliminate states not reachable from the start state. 2. Eliminate states that do not reach a final state. 3. Test if the remaining transition graph has any cycles. 26 The Pumping Lemma x We have, almost accidentally, proved a statement that is quite useful for showing certain languages are not regular. x Called the pumping lemma for regular languages. 27 Statement of the Pumping Lemma Number of states of DFA for L For every regular language L There is an integer n, such that For every string w in L of length > n We can write w = xyz such that: 1. |xy| < n. Labels along 2. |y| > 0. first cycle on path labeled w 3. For all i > 0, xyiz is in L. 28 Example: Use of Pumping Lemma x We have claimed {0k1k | k > 1} is not a regular language. x Suppose it were. Then there would be an associated n for the pumping lemma. x Let w = 0n1n. We can write w = xyz, where x and y consist of 0’s, and y ≠ ε. x But then xyyz would be in L, and this string has more 0’s than 1’s. 29 Decision Property: Equivalence x Given regular languages L and M, is L = M? x Algorithm involves constructing the product DFA from DFA’s for L and M. x Let these DFA’s have sets of states Q and R, respectively. x Product DFA has set of states Q × R.  I.e., pairs [q, r] with q in Q, r in R. 30 Product DFA – Continued x Start state = [q0, r0] (the start states of the DFA’s for L, M). x Transitions: δ([q,r], a) = [δL(q,a), δM(r,a)]  δL, δM are the transition functions for the DFA’s of L, M.  That is, we simulate the two DFA’s in the two state components of the product DFA. 31 Example: Product DFA 0 A 1 B [A,C] 0, 1 1 1 C 0 0 D 1 1 0 0 [A,D] 1 0 [B,C] 0 [B,D] 1 32 Equivalence Algorithm x Make the final states of the product DFA be those states [q, r] such that exactly one of q and r is a final state of its own DFA. x Thus, the product accepts w iff w is in exactly one of L and M. 33 Example: Equivalence 0 A 1 B [A,C] 0, 1 1 1 C 0 0 D 1 1 0 0 [A,D] 1 0 [B,C] 0 [B,D] 1 34 Equivalence Algorithm – (2) x The product DFA’s language is empty iff L = M. x But we already have an algorithm to test whether the language of a DFA is empty. 35 Decision Property: Containment x Given regular languages L and M, is L ⊆ M? x Algorithm also uses the product automaton. x How do you define the final states [q, r] of the product so its language is empty iff L ⊆ M? Answer: q is final; r is not. 36 Example: Containment 0 A 1 B [A,C] 0, 1 1 1 C 0 1 0 D 1 1 0 0 [A,D] 1 0 0 [B,C] [B,D] Note: the only final state is unreachable, so containment holds. 37 The Minimum­State DFA for a Regular Language x In principle, since we can test for equivalence of DFA’s we can, given a DFA A find the DFA with the fewest states accepting L(A). x Test all smaller DFA’s for equivalence with A. x But that’s a terrible algorithm. 38 Efficient State Minimization x Construct a table with all pairs of states. x If you find a string that distinguishes two states (takes exactly one to an accepting state), mark that pair. x Algorithm is a recursion on the length of the shortest distinguishing string. 39 State Minimization – (2) x Basis: Mark a pair if exactly one is a final state. x Induction: mark [q, r] if there is some input symbol a such that [δ(q,a), δ(r,a)] is marked. x After no more marks are possible, the unmarked pairs are equivalent and can be merged into one state. 40 Transitivity of “Indistinguishable” x If state p is indistinguishable from q, and q is indistinguishable from r, then p is indistinguishable from r. x Proof: The outcome (accept or don’t) of p and q on input w is the same, and the outcome of q and r on w is the same, then likewise the outcome of p and r. 41 Constructing the Minimum­ State DFA x Suppose q1,…,qk are indistinguishable states. x Replace them by one state q. x Then δ(q1, a),…, δ(qk, a) are all indistinguishable states.  Key point: otherwise, we should have marked at least one more pair. x Let δ(q, a) = the representative state for that group. 42 Example: State Minimization {1} {2,4} {5} {2,4,6,8} {1,3,5,7} * {1,3,7,9} * {1,3,5,7,9} r b {2,4} {5} {2,4,6,8} {1,3,5,7} {2,4,6,8} {1,3,7,9} {2,4,6,8} {1,3,5,7,9} {2,4,6,8} {1,3,5,7,9} {2,4,6,8} {5} {2,4,6,8} {1,3,5,7,9} r b A B C B D E C D F D D G E D G * F D C * G D G Remember this DFA? It was constructed for the chessboard NFA by the subset construction. 43 Here it is with more convenient state names Example – Continued r b A B C B D E C D F D D G E D G * F D C * G D G G F E D C B Axx Bxx Cxx Dxx Exx F Start with marks for the pairs with one of the final states F or G. 44 Example – Continued r b A B C B D E C D F D D G E D G * F D C * G D G G F E D C B Axx Bxx Cxx Dxx Exx F Input r gives no help, because the pair [B, D] is not marked. 45 Example – Continued r b A B C B D E C D F D D G E D G * F D C * G D G G F E D C B Axx x x x Bxx x x x Cxx Dxx Exx Fx But input b distinguishes {A,B,F} from {C,D,E,G}. For example, [A, C] gets marked because [C, F] is marked. 46 Example – Continued r b A B C B D E C D F D D G E D G * F D C * G D G G F E D C B Axx x x x Bxx x x x Cxx x x Dxx Exx Fx [C, D] and [C, E] are marked because of transitions on b to marked pair [F, G]. 47 Example – Continued r b A B C B D E C D F D D G E D G * F D C * G D G G F E D C B Axx x x x x Bxx x x x Cxx x x Dxx Exx Fx [A, B] is marked because of transitions on r to marked pair [B, D]. [D, E] can never be marked, because on both inputs they go to the same state. 48 Example – Concluded r b A B C B D E C D F D D G E D G * F D C * G D G r b A B C B H H C H F H H G * F H C * G H G G F E D C B Axx x x x x Bxx x x x Cxx x x Dxx Exx Fx Replace D and E by H. Result is the minimum­state DFA. 49 Eliminating Unreachable States x Unfortunately, combining indistinguishable states could leave us with unreachable states in the “minimum­state” DFA. x Thus, before or after, remove states that are not reachable from the start state. 50 Clincher x We have combined states of the given DFA wherever possible. x Could there be another, completely unrelated DFA with fewer states? x No. The proof involves minimizing the DFA we derived with the hypothetical better DFA. 51 Proof: No Unrelated, Smaller DFA x Let A be our minimized DFA; let B be a smaller equivalent. x Consider an automaton with the states of A and B combined. x Use “distinguishable” in its contrapositive form:  If states q and p are indistinguishable, so are δ(q, a) and δ(p, a). 52 Inferring Indistinguishability q0 Start states of A and B indistinguishable because L(A) = L(B). a q b Must be indistinguishable p0 a p r Must be indistinguishable b s 53 Inductive Hypothesis x Every state q of A is indistinguishable from some state of B. x Induction is on the length of the shortest string taking you from the start state of A to q. 54 Proof – (2) x Basis: Start states of A and B are indistinguishable, because L(A) = L(B). x Induction: Suppose w = xa is a shortest string getting A to state q. x By the IH, x gets A to some state r that is indistinguishable from some state p of B. x Then δ(r, a) = q is indistinguishable from δ(p, a). 55 Proof – (3) x However, two states of A cannot be indistinguishable from the same state of B, or they would be indistinguishable from each other.  Violates transitivity of “indistinguishable.” x Thus, B has at least as many states as A. 56 ...
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This note was uploaded on 03/30/2012 for the course CS 154 taught by Professor Motwani,r during the Spring '08 term at Stanford.

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