CSE 460: Computabilty and Formal Languages. S. Pramanik

Transcription

1 CSE 46: Computabilty and Formal Languages NF A Λ S. Pramanik

2 NF A Λ Concatanation of languages and () : () 2 O,, O * FA () * FA Λ 2 2 * ()* NFA * ()* NFA- Figure : NF A Λ is an NFA but it also allows Λ-transitions. Advantage of NFA plus flexibility of Λ-transitions. 2

3 Definition: A nondeterministic finite automata with Λ transitions (abbreviated N F A Λ) is a 5-tuple (Q, Σ,, A, δ), where Q and Σ are finite sets, ɛq, A Q, and δ : Q (Σ {Λ}) 2 Q Note: Λ is not part of the alphabet. It is defined for any NF A Λ. Λ 2 * ()* NFA- Figure 2: For the above NFA-Λ Lambda {} {} {2} 2 {} Example: δ(, Λ) = { } δ(, ) = { } δ(, ) = { 2 } δ( 2, ) = { } δ for an NFA Λ 3

4 For xɛσ and pɛq, p is the set of all states ɛq such that there is a seuence of transitions corresponding to x by which the NF A Λ moves from p to. For the above NFA-Λ: x= δ (, ) = {, } δ (, ) = {,, 2 } δ (, ) = not defined Nondeterminism due to Λ-transitions (δ (, ) = {, }). accept x=?, x=?, x=? 2 3 (+) * Figure 3: Language= (+) same as (+) or (+) (+ )( + ) 4

5 GOAL: Find Euivalent NFA. NFA-Λ can make transitions on Λ 2. How can the euivalent NFA simulate these Λ-Transitions. 3. Basic Idea: 4. Λ-transition is a type of nondeterminism. If the transitions of NFA-Λ is as follows: p Λ r then from state p, the input symbol allows us to go to either or r. Thus, we can eliminate the NFA-Λ without changing the states, by simply adding the transitions from p to r on input. Thus, we have the following: For each state of NFA-Λ and each character α of Σ, figure out which states are reachable from taking any number of Λ-transitions and exactly one transition on that character α. In the euivalent NFA, directly connect to each of these states using an arc labeled with α. 5

6 Example: Process state : b NFAb b a b b Process state 2 Process state 3 b b b a a a Similarly process states 4 and 5, NFA: Combine the 5 NFA s b 6

7 Λ-Closure of a Set of States Lambda Closure, Λ(S), for a set of states S is defined as:. Every element of S is an element of Λ(S). 2. For any ɛλ(s), every element of δ(, Λ) is in Λ(S). 3. No other elements of Q are in Λ(S). All states that can be reached through Λ transitions only. p r s w t u v Figure 4: Λ({s}) = {s, w,, p, t} 7

8 NF A Λ to NFA Input NF A Λ = (Q, Σ,, δ, A) Output NF A = (Q, Σ,, δ, A ) Q = Q Σ = Σ = What is A? Computing δ (, a): δ (, a)= the set of states reachable from state in the NFA-Λ taking or more Λ-transitions and exactly one transition on the character a= Λ(δ(Λ(), a) Break this down into three steps: First compute all states reachable from using o or more Λ-transitions - We call this set of states Λ() Next, compute all states reachable from any element of Λ() using the character a We can denote this states as δ(λ(), a) Finally, compute all states reachable from states in δ(λ(), a) using or more Λ-transitions. - We denote these states as Λ(δ(Λ(), a) - This is the desired answer 8

9 Example of Computing δ δ (, b) = {3, 4, 5}. Compute Λ(), all states reachable from state using or more Λ-transitions Λ() = {, 2} 2. Compute δ(λ(), b), all states reachable from any element of Λ() using the character b δ(λ(), b) = δ({, 2}, b) = δ(, b) δ(2, b) = {} {3} = {3} 3. Compute Λ(δ(Λ(), b)), all states reachable from states in δ(λ(), b) using or more Λ-transitions. Λ(δ(Λ(), b)) = Λ(3) = {3, 4, 5} 9

10 Recursive Definition of δ for an NF A Λ Definition of δ. For any ɛq, δ (, Λ) = Λ({}) 2. For any ɛq, yɛσ, and aɛσ, δ (, ya) = Λ( rɛδ (,y)δ(r, a)) A string x is accepted by M if δ (, x) A. Similar to δ of NFA except. Start with Λ infront of the input. 2. After each input symbol do Λ closure. We will use recursive definition of δ to calculate the set δ (, abc). This set is defined in terms of δ (, ab), which is defined interms of δ (, a and which is then defined interms of δ (, Λ) which is Λ( ) (the base case). We, therefore approach the calculation from bottom up, calculating δ (, Λ) first. δ (, abc) = δ (p, Λ/abc) where pɛp = Λ(δ(, Λ)) = δ (r, Λa/bc) where rɛr = Λ((δ(p, a), pɛp ) = δ (s, Λab/c) where sɛs = Λ((δ(r, b), rɛr) = δ (m, Λabc/) where mɛm = Λ((δ(s, c), sɛs)

11 p r s w t u v Figure 5: δ (, ) = δ (r, Λ/), rɛ{, p, t} = δ (s, Λ/), sɛλ(δ(, ) δ(p, ) δ(t, )) = Λ({p, u}) = {p, u, } = δ (m, Λ/), mɛλ(δ(p, ) δ(u, )) = Λ({r}) = {r}

12 Constructing NFA from NF A Λ Theorem: If L is accepted by an NF A Λ, there is an euivalent NFA that also accepts L. δ(, Λ) δ(, ) δ(, ) A {B} {A} B {D} {C} C {B} D {D} δ (A, ) = {A, B, C, D}, δ (A, ) = {????}, δ (B, ) = {C, D} δ (B, ) = {?????}, δ (C, ) = {?????}, δ (C, ) = {?} δ (D, ) = {????}, δ (D, ) = {????} Draw NFA using the above transitions. Show δ for the same inputs as above for NFA are the same. Prove that δ (, x) = δ (, x) for any x except x= Λ. See the difference: δ (A, Λ={A,B,D} but δ (A, Λ) = {A} Justify conditions for acceptance states:. Acceptance state of the NF A Λ becomes the acceptance states of the NFA. 2. If Λ({ }) contains an acceptance state then also becomes an acceptance state of the NFA. Justification: 2

13 C B A D Λ Λ C B A D Figure 6: 3. For the exercise problem: Λ({A}) = {A, B, D} containing the final state D Therefore, Both D and A are going to be final states. 3

14 Theorem : For every language L Σ accepted by an NFA-Λ: M = (Q, Σ,, A, δ), there is an NFA (without Λ transitions): M = (Q, Σ,, A, δ ) that also accepts L. Proof: We define M as Q = Q Σ = Σ = δ (, Λ) = (there is no Λ-transitions in M ɛq and σɛσ δ (, σ) = δ (, σ) and A = A { } if ΛɛL else A = A We want to show L(M ) = L(M) = L First, assume x = This is the base case for the proof by induction later. If Λ is accepted by M then Λ({ }) A which includes. By definition of M, =. Therefore, ɛa by definition of M. Λ is accepted by M If the string Λ ɛl(m), than A = A ; therefore, ɛa, and Λ ɛl(m ) Second, assume x We prove δ (, x) = δ (, x) for x We prove this by induction: 4

15 Assume y = n, δ (, y) = δ (, y) (hypothesis) Then yσ = n + and δ (, ya) = {δ (p, σ) pɛδ (, y)} by definition of δ = {δ (p, σ) pɛδ (, y)} by the induction hypothesis = {δ (p, σ) pɛδ (, y)} by definition of δ. Now if xɛl(m) we want to show xɛl(m ) Because xɛl(m), δ (, x) contains an element of A; therefore since δ (, x) = δ (, x) and A A, xɛl(m ). Now if xɛl(m ) we show xɛl(m) If xɛl(m ) then δ (, x) contains an element of A. If A = A then xɛl(m). However, if A A then A has and A does not. If A contains then ΛɛL(M ) i.e., δ (, x) contains. However, we have shown δ (, x) = δ (, x) Thus, δ (, x) contains and therefore, contains Λ({ )} Thus, even when A A xɛl(m) NOTE: even if x, M can go to (for example string x =, and x is accepted by reaching ). In M, the string Λ does not have to be accepted by reaching ; it is accepted by reaching any of the states in Λ({ }). 5

16 Theorem 4.3. L can be recognized by an FA 2. L can be recognized by an NFA 3. L can be recognized by an NF A Λ 6

17 Kleene s Theorem Examples: 2 UNION Figure 7: Does Not work: 2 UNION? 3, 2 3 Figure 8: Kleene s Theorem 4.4 (Part ): Any regular language can be accepted by a finite automata. Proof: By theorem 4.3 (same L can be recognized by FA, NFA and Λ-NFA) it is sufficient to show that any regular language can be accepted by an NF A Λ. Definition 3. gives the definition of regular languages and we construct NF A Λ based on this definition. NF A Λ for the three basic languages (figure ):, {Λ}, {a}(aɛσ) Now if L and L 2 are two regular languages recognized by two NF A Λ s M:(Q, Σ,, A, δ ) and M2:(Q 2, Σ, 2, A 2, δ 2 ) 7

18 a a Figure 9:, then we give construction for NF A Λs union, concatanation and kleen *, i.e., M u, M c, and M k, recognizing languages L L 2, L.L 2 and L, respectively. We will give construction for M c : (Q c, Σ, c, A c, δ c ) only: By renaming states we can assume Q Q 2 =.. Q c = Q Q 2 2. c = 3. A c = A 2 4. Include all transitions of M and M 2 in M c 5. Add Λ transitions from each of the states in A to 2. Refer to the following figure: UNION OF M AND M2 f M CONCATANATION OF M and M2 u f f 2 = c f 2 f 2 2 M2 f f 2 f 2 KLEENE STAR OF M k f M f Figure : Show that if x is accepted by M and x 2 is accepted by M 2 then x x 2 is accepted by M c Since x is accepted by M and M c has all the states and the transitions of M, M c will reach the state cor- 8

19 responding to an acceptance state of M. It will then follow the Λ transition from this state to the state of M c that corresponds to the start state of M 2. Since x 2 is accepted by M 2 and M c has all the states and the transitions of M 2, M c will reach the state that correspond to the acceptance state of M 2 which is also an acceptance state of M c. Thus M c accepts x x 2. Show if x is accepted by M c then x = x x 2, and x is accepted by M and x 2 is accepted by M 2 : Since x is accepted by M c, it starts in c = and ends in ɛa c = A 2. This can only be done through an intermediate Λ transition as defined above. Since Q Q 2 =, this crossing can happen only once. All transitions before crossing are due to those of machine M and after crossing are due to those of M 2. Thus, x = x x 2, x is accepted by M c and x is used before crossing and is accepted by M, x 2 is used after crossing and is accepted by M 2. Proof for M k : Base case is similar to above that there exists NF A Λ for empty set, etc. We assume there is a NF A Λ for L then there is an NF A Λ for L. If xɛl then xɛl(m k ) and if xɛl(m k ) then xɛl xɛl means that x = x x 2...x m, where each x i ɛl x can be written as (Λx Λ)(Λx 2 Λ)...(Λx m Λ) Where each one of them (Λx m Λ) gets accepted by M and comes back to start state of L(M k ). Therefore x is accepted by L(M k ) 9

20 Construct an N F A Λ for the following regular expression using Kleene s theorem: ( + ) (+) * Figure : 2