CSE 460: Computabilty and Formal Languages. Minimization of FA. S. Pramanik

Transcription

1 CSE 46: Computabilty and Formal Languages Minimization of FA S. Pramanik

2 Example of Minimum-State FA. Minimum-State FA for a regular language is an FA for the language that requires the minimum number of states. 2. Example: An FA and a Minimized FA for the language L: (+) C E A B D F FA WITH REDUNDANT STATES FOR C, D and E, F A B C,D E,F MINIMIZED FA 3. Pairwise distinguishable strings for the above language L: {Λ,,,,} Minimum number of states is 5 (including state not shown) 2

3 4. Take a string from L q for each state q above: Note:L q = {x δ(q,x) = q} For Not Minimized FA:{Λ,,,,,} For Minimized FA: {Λ,,,,} 5. See the relationship between the two sets. Note: L C L D = L C,D 3

4 Constructing Minimum-State FA from Equivalence Classes of Strings. Definition: Let L be any language in Σ. Relation I L on Σ is defined as follows: xi L y if, for any zǫσ either both xz and yz are in L or both are not in L. 2. Lemma: I L is an equivalence relation. proof: It is obvious that I L is reflexive and symmetric. For transitivity, suppose xi L y and y I L z. We show that xi L z. Let z be any stringi in Σ a. If xzǫ, then yzǫ L, and therefore wzǫ l, since y I L w (for any z I L holds. (I L partitions Σ into disjoint subsets (equivalence classes). If L is a regular language, the number of equivalence classes is equal to the minimum number of states for the FA.) 3. Lemma: I L is right invariant with respect concatanation. In other words, for any x,yǫσ and any aǫσ, if xi L y, thenxai L ya. Equivalently,if[x]=[y],then[xa]=[ya]. 4

5 (Transition from one equivalent class to another on an input.) [ ] [] [] [] [], 4. Equivalence Classes of strings for I L where L:(+ ) {Λ},{},{,},{,}{},{{,} {,} } {{,} } {{,}{,} } 5. Notation: [x] an equivalence class where x is an element of the class. 6. Equivalence Classes for I L above: [Λ][], [], [], [] 5

6 Constructing Minimum-State FA Construction of minimal-state FA from the equivalence classes: Theorem: Let L Σ, and let Q L be the set of equivalence classes of the relation I L on Σ. If Q L is a finite set, then M L = (Q L,Σ,q,A L,δ) is a finite automaton accepting L, where q = [Λ], A L = {qǫq L q L }, and δ : Q L Σ Q L is defined by the formula δ([x],a) = [xa]. Furthermore, M L has the fewest states of any FA accepting L. [ ] [] [] [] [], 6

7 Algorithm for Constructing Minimum-State FA from a Given FA. Distinguishable strings: Two strings are distinguishable with respect to a language L if there exists a z such that only one of xz and yz is in L. 2. Example: L: (+), x=, y=, z=λ 3. Two states p and q are distinguishable (p q) if xǫl p and yǫl q and x, y are distinguishable with respect to L. Relationship to ([x],[y])? 4. If p is an acceptance state and q is a non-acceptance state, are (p,q) distinguishable states? xǫl p,yǫl q and z = Λ 5. Are p and q distinguishabe for the following? Exists a z. p z q z 7

8 How to Find Distinguishable States. pǫa, q ǫa then p q 2. δ(r,a) = p, δ(s,a) = q p q (r s) 3. Justification: r a p z s a q z (a) p and q are distinguishable because p goes to acceptance state and q goes to non-acceptance state for the same input string z. (b) randsaredistinguishablebecauseforsameinput a, r goes to p and s goes to q. xǫl r,yǫl s and z = az 8

9 Finding All Pairs of Distinguishable States. Note that Null state has to be included. 2. Create a table with states as shown: 3. Mark the table entries to for acceptable-nonacceptable pairs of states. 4. Use each of these marked pairs (p,q) find (r,s) such that they go to p.q on the same input, mark them Now usethose entries with 2 sas (p,q) and find (r,s), markthemas3 s. Continueuntilnomorenew(r,s) s can be created. 6. Entries in the table not yet marked are indistinguishable ,

10 δ(,) = 3,δ(6,) = 6, therefore, δ(2,) = 4,δ(6,) = 6, therefore,

11 Create a Minimum-State FA. Create the states for a minimum-state FA: StartwithstateofthegivenFAandgosequentially with increasing order of states as follows: ForstateoftheFA,createanewstatepofminimumstate FA. p state 2 is not distinguishable from. 2 p state 3 is distinguishable from 2; create a new state p2 for minimum-state FA 3 p2 state 4 is not distinguishable from 3 4 p2 state 5 is not distinguishable from 4 5 p2 state 6 is not distinguishable from 5; create a new state p3 for minimum-state FA. 2. Now we have the following three states of minimumstate FA: {,2} {3,4,5} {6} p p2 p3

12 Complete the Minimum-State FA. Complete the transitions in the new machine by following the transitions for the states,2,3,4,5,6 in the old machine. For example, δ(,) = 3 gives δ(p,) = p ,, {,2} {3,4,5} {6} p p2 p3 2

13 . Theorem 5.2a: Pumping Lemma If L is regular Language then there is an integer n so that for any xǫl with x n, there are strings u, v, and w so that (a) x = uvw (b) uv n (c) v > (d) for any m, uv m wǫl 2. NotethatPumpinglemmacannotproveifalanguage is regular. The theorem is (if then). 3. Pumping lemma does not work for all non-regular languages. It does not work for the language L defined below: L = i=( + ) i (2 + ) i 2 L 2 = {x xǫ{,,2} and x contains at least one or 2 or 22 as a substring} L = L L 2 3

14 Justification: Language L is not regular. If it was regular, then L as given below will be regular (regular languages are closed under intersection and complementation). L = L L Pumping lemma works for L because if we take a string in L and pump on s only th. time the string will be in L2. Pumping more than th. time will make it a member of L. If we take a string in L 2 then pumping lemma holds if we take any string in L 2 of length 3. 4

15 Application of Pumping Lemma. Using Pumping Lemma, show that { i i i } is not regular. Let x = n n u and v must contain all s because uv n Since v >, say v = j and j > For m=2 we have a contradiction because uv 2 w = n+j n ǫl. 5