Languages Recall. Non deterministic finite automata What is a - PDF document

Languages • Recall. Non deterministic finite automata – What is a language? with ε transitions – What is a class of languages? Finite Automata First there was the DFA • Consists of • Deterministic Finite Automata – A set of states (Q) – For every state and every alphabet symbol there – A start state ( q o ) is exactly one move that the machine can make. – A set of accepting states ( F ) – δ : Q x Σ → Q – Read symbols ( Σ ) – δ is a total function: completely defined. I.e. it – Transition function ( δ ) is defined for all q ∈ Q and a ∈ Σ • Let’s recap Non-Deterministic Finite Automata Then, the NFA (NFA) • Non-determinism • Transition function – δ is a function from Q x Σ to 2 Q – When machine is in a given state and reads a symbol, the machine will have a choice of – δ (q, a) = subset of Q (possibly empty) where to move to next. – There may be states where, after reading a given symbol, the machine has nowhere to go. – Applying the transition function will give, not 1 state, but 0 or more states. 1

Nondeterministic Finite Automata with ε transitions ( ε -NFA) • And now… • For both DFAs and NFAs, you must read a symbol in order for the machine to make a move. • Introducing… • In Nondeterministic Finite Automata with ε transitions ( ε -NFA ) • The newest in the FA family… – Can make move without reading a symbol off • The Non deterministic finite automata with ε the read tape transitions ( ε -NFA ) – Such a move is called a ε transition Nondeterministic Finite Automata with ε Nondeterministic Finite Automata with ε transitions ( ε -NFA) transitions ( ε -NFA) • Example: • How does such a machine accept? – Machine to accept decimal numbers – A string will be accepted if there is at least one 0,1,..9 0,1,..9 sequence of state transitions on an input ε 0,1,..9 (including ε transitions) that leaves the machine ε + - . q 0 q 1 q 2 q 3 q 5 in an accepting state. 0,1,..9 . q 4 Nondeterministic Finite Automata with ε Nondeterministic Finite Automata with ε transitions ( ε -NFA) transitions ( ε -NFA) • A Non-Deterministic Finite Automata with ε • Example: transitions is a 5-tuple ( Q , Σ , q o , δ , F ) where – -3.45 is accepted – Q is a finite set (of states) – .5678 – Σ is a finite alphabet of symbols – 37 is rejected 0,1,..9 – q o ∈ Q is the start state 0,1,..9 – F ⊆ Q is the set of accepting states ε 0,1,..9 ε + - . – δ is a function from Q x ( Σ ∪ { ε }) to 2 Q (transition q 0 q 1 q 2 q 3 q 5 function) 0,1,..9 . q 4 2

Nondeterministic Finite Automata with ε Nondeterministic Finite Automata with ε transitions ( ε -NFA) transitions ( ε -NFA) • Transition function • Transition function on a string – δ is a function from Q x ( Σ ∪ { ε }) to 2 Q is a function from Q x Σ * to 2 Q – δ ^ – δ (q, a) = subset of Q (possibly empty) – (q, x) = subset of Q (possibly empty) δ ^ – In our example – Set of all states that the machine can be in, • δ (q 1 , 0) = {q 1 , q 4 } upon following all possible paths on input x. • δ (q 1 , .) = {q 1 } – We’ll need to consider all paths that include the • δ (q 1 , +) = ∅ use of ε transitions • δ (q 0 , ε ) = {q 1 } ε -Closure ε -Closure • ε closure • ε closure: Recursive definition – Let M = ( Q , Σ , q o , δ , F ) be a ε -NFA – Before defining the transition function on a string ( (q,x)), it is useful to first define what ^ δ – Let S be a subset of Q is known as the ε closure. – The ε closure, denotes ECLOSE (S) is defined: – Given a set of states S, the ε closure will give • For each state p ∈ S, p ∈ ECLOSE (S) the set of states reachable from each state in S • For any q ∈ ECLOSE (S), every element of δ (q, ε ) ∈ using only ε transitions. ECLOSE (S) • No other elements of Q are in ECLOSE (S) ε -Closure ε -Closure • ε -Closure : Algorithm • Example – Since we know that ECLOSE (S) is finite, we can convert the recursive definition to an algorithm. – To find ECLOSE (S) where S is a subset of Q – Let T = S – While (T does not change) do • Add all elements of δ (q, ε ) where q ∈ T – ECLOSE (S) = T 3

Nondeterministic Finite Automata with ε ε -Closure transitions ( ε -NFA) • ε closure: Example • Now lets define δ ^ For any q ∈ Q, (q, ε ) = ECLOSE ({q}) – Find ECLOSE({s}) in our example 1. δ ^ For any y ∈ Σ * , a ∈ Σ , q ∈ Q – T = {s} initial step 2. – T = {s, w} add δ (s, ε ) ⎛ ⎞ ⎜ ⎟ δ ˆ = δ ∈ U – T = {s, w, q 0 } add δ (w, ε ) ( q , ya ) ECLOSE ( p , a ) ⎜ ⎟ ⎝ ⎠ – T = {s, w, q 0 , p,t} add δ (q 0 , ε ) δ ˆ p ( q , y ) δ (w, ε ) = δ (w, ε ) = ∅ – Set of all states obtained by applying δ to all states in δ * (q,y) and input a and taking the ε closure of the result – We are done, • ECLOSE({s}) = T = {s, w, q 0 , p,t} Nondeterministic Finite Automata with ε Nondeterministic Finite Automata with ε transitions ( ε -NFA) transitions ( ε -NFA) • Are the following strings accepted by the ε - • Accepting a string NFA below: – A string x is accepted if running the machine on input x, considering all paths, including the use – aba – ababa of ε transitions, puts the machine into one of – aaabbb the accepting states – Formally: • x ∈ Σ * is accepted by M if (q 0 , x) ∩ F ≠ ∅ • δ ^ Nondeterministic Finite Automata with ε Nondeterministic Finite Automata with ε transitions ( ε -NFA) transitions ( ε -NFA) • Language accepted by M • I bet that you’re asking… – Can JFLAP handle ε -NFAs ? – The language accepted by M • L(M) = { x ∈ Σ * | x is accepted by M } – Well, let’s check and see! • If L is a language over Σ , L is accepted by M iff L = L(M). – For all x ∈ L, x is accepted by M. – For all x ∉ L, x is rejected by M. 4

Nondeterministic Finite Automata with ε DFA / NFA / ε -NFA Equivalence transitions ( ε -NFA) • Surprisingly enough • Why they’re a good idea – ε transitions to our NDFA does NOT give it any – Given a regular expression, it is far easier to create an ε -NFA for the language described by additional language accepting power. – DFAs and NFAs and ε -NFA are all equivalent the expression than it is to create a plain old DFA. • Every language that can be accepted by a ε -NFA can also be accepted by an DFA which can also be – It will also be essential when showing the Fas accepted by a NFA. accept the class of Regular Languages. – Questions? – Let’s show this ε -NFA -> DFA ε -NFA -> DFA • Given ε -NFA find DFA • Basic idea – Let E = (Q E , Σ , δ E , q 0 , F E ) be a ε -NFA then – Very similar to the subset construction algorithm • There exists a DFA, D = (Q D , Σ , δ D , q D , F D ) • Recall that for a ε -NFA , δ : Q x Σ → 2 Q • Such that L(E) = L(D) • Use the states of D to represent subsets of Q. ε -NFA -> DFA ε -NFA -> DFA • Computing δ D • Formal definition – δ D (S, a) for S ∈ Q D , a ∈ Σ – E = (Q E , Σ , δ E , q 0 , F E ) be a ε -NFA • Let S = { p 1 , p 2 , …, p n } – We define DFA, D = (Q D , Σ , δ D , q D , F D ) • Compute the set of all states reachable from states in • Q D = 2 QE S on input a using transitions from E. • q D = ECLOSE (q 0 ) n = U δ { , , L , } ( , ) r r r p a 1 2 m E i • F D = sets containing at least one state from F E = 1 i • δ D (S, a) will be the union of the ε closures of the elements of {r 1 , …, r m } m δ = U ( S , a ) ECLOSE ( r ) D j = j 1 5

ε -NFA -> DFA ε -NFA -> DFA ε closure State 0,1,..9 0,1,..9 q 0 {q 0 , q 1 } ε 0,1,..9 ε + - . q 0 q 1 q 2 q 3 q 5 q 1 {q 1 } q 2 {q 2 } 0,1,..9 . q 3 {q 3 , q 5 } q 4 q 4 {q 4 } q 5 {q 5 } ε -NFA -> DFA ε -NFA -> DFA 0,1,…,9 • Now we must show that D accepts the same . q 1 q 4 q 2 q 3 q 5 language as E 0,1,…,9 – Can be shown (using induction) that for all x ∈ 0,1,…,9 Σ * + - q 0 q 1 q 1 0,1,…,9 • δ D ^ (q, x) = ^ δ E (q, x) . . q 2 q 3 q 5 • See Theorem 2.22 0,1,…,9 0,1,…,9 ε -NFA -> DFA DFA -> ε -NFA • Show that D and E recognize the same • The other direction is fairly straighforward. language – For any DFA, there is an equivalent NFA (q 0 , x) ∩ F E ≠ ∅ – An NFA is nothing more than a ε -NFA with no ε – x is accepted by E iff ^ δ E – x is accepted by D iff (q D , x) ∩ F E ≠ ∅ transitions. Thus ^ δ D – Thus, • δ (q, ε ) for all states q = ∅ • x is accepted by D iff x is accepted by E • Questions? 6

What have we shown Questions? • For every DFA, there is an NFA that • Let’s take a break. accepts the same language and visa versa • For every DFA, there is a ε -NFA that accepts the same language, and visa versa • Thus, for every NFA there is a ε -NFA that accepts the same language, and visa versa • DFAs, NFAs, and ε -NFA s are equivalent! 7