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1

Non deterministic finite automata

with transitions

Languages

• Recall. - What is a language? - What is a class of languages?

Finite Automata

• Consists of - A set of states (Q) - A start state (q o - A set of accepting states (F ) - Read symbols () - Transition function () • Let's recap

First there was the DFA

• Deterministic Finite Automata - For every state and every alphabet symbol there is exactly one move that the machine can make. -: Q x Q -is a total function: completely defined. I.e. it is defined for all q Q and a

Then, the NFA

• Non-determinism - When machine is in a given state and reads a symbol, the machine will have a choice of 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.

Non-Deterministic Finite Automata

(NFA) • Transition function -is a function from Q x to 2 Q -(q, a) = subset o (possibly empty) 2 • And now... • Introducing... • The newest in the FA family... • The Non deterministic finite automata with transitions ( -NFA)

Nondeterministic Finite Automata with

transitions (-NFA) • For both DFAs and NFAs, you must read a symbol in order for the machine to make a move. • In Nondeterministic Finite Automata with transitions ( -NFA) - Can make move without reading a symbol off the read tape - Such a move is called a transition

Nondeterministic Finite Automata with

transitions (-NFA) • Example: - Machine to accept decimal numbers q 0 q 1 q 2 q 3 q 5 q 4

0,1,..9

0,1,..9 .

.0,1,..90,1,..9

Nondeterministic Finite Automata with

transitions (-NFA) • How does such a machine accept? - A string will be accepted if there is at least one sequence of state transitions on an input (including transitions) that leaves the machine in an accepting state.

Nondeterministic Finite Automata with

transitions (-NFA) • Example: - -3.45 is accepted - .5678 - 37 is rejected q 0 q 1 q 2 q 3 q 5 q 4

0,1,..9

0,1,..9 .

.0,1,..90,1,..9

Nondeterministic Finite Automata with

transitions (-NFA) • A Non-Deterministic Finite Automata with transitions is a 5-tuple (Q, ,q o , , F) where -Qis a finite set (of states) -is a finite alphabet of symbols -q o

Q is the start state

-F Q is the set of accepting states -is a function from Q x ({}) to 2 Q (transition function) 3

Nondeterministic Finite Automata with

transitions (-NFA) • Transition function -is a function from Q x ({}) to 2 Q -(q, a) = subset o (possibly empty) - In our example •(q 1 , 0) = {q 1 , q 4 •(q 1 , .) = {q 1 •(q 1 •(q 0 , ) = {q 1

Nondeterministic Finite Automata with

transitions (-NFA) • Transition function on a string - is a function from Q x to 2 Q - (q, x) = subset o (possibly empty) - Set of all states that the machine can be in, upon following all possible paths on input x. - We'll need to consider all paths that include the use of transitions

H-Closure

•closure - Before defining the transition function on a string ( (q,x)), it is useful to first define what is known as the closure. - Given a set of states S, the closure will give the set of states reachable from each state in S using only transitions.

H-Closure

•closure: Recursive definition -Let M = (Q, ,q o , , F) be a -NFA - Let S be a subset of Q - The closure, denotes

ECLOSE(S) is defined:

• For each state p S, p ECLOSE(S) • For any q

ECLOSE(S), every element of (q, )

ECLOSE(S)

• No other elements of Q are in

ECLOSE(S)

-Closure •-Closure : Algorithm - 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 (q, ) where q T -ECLOSE(S) = T -Closure •Example 4 -Closure •closure: Example - Find ECLOSE({s}) in our example - T = {s} initial step - T = {s, w} add (s, ) - T = {s, w, q 0 } add (w, ) - T = {s, w, q 0 , p,t} add (q 0 -(w, ) = (w, ) = - We are done, • ECLOSE({s}) = T = {s, w, q 0 , p,t}

Nondeterministic Finite Automata with

transitions (-NFA) • Now lets define

1. For any q Q, (q, ) = ECLOSE ({q})

2. For any y

, a , q Q Set of all states obtained by applying to all states in (q,y) and input a and taking the closure of the result yqp apECLOSEyaq

Nondeterministic Finite Automata with

transitions (-NFA) • Accepting a string - A string x is accepted if running the machine on input x, considering all paths, including the use oftransitions, puts the machine into one of the accepting states - Formally: •x is accepted by M if •(q 0 , x) F

Nondeterministic Finite Automata with

transitions (-NFA) • Are the following strings accepted by the NFA below: -aba -ababa - aaabbb

Nondeterministic Finite Automata with

transitions (-NFA) • I bet that you're asking... - Can JFLAP handle -NFAs? - Well, let's check and see!

Nondeterministic Finite Automata with

transitions (-NFA) • Language accepted by M - The language accepted by M • L(M) = { x | x is accepted by M } • If L is a language over , 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. 5

Nondeterministic Finite Automata with

transitions (-NFA) • Why they're a good idea - Given a regular expression, it is far easier to create an -NFAfor the language described by the expression than it is to create a plain old DFA. - It will also be essential when showing the Fas accept the class of Regular Languages. - Questions?

DFA / NFA / -NFAEquivalence

• Surprisingly enough -transitionsto our NDFA does NOT give it any additional language accepting power. - DFAs and NFAs and -NFAare all equivalent • Every language that can be accepted by a -NFAcan also be accepted by an DFA which can also be accepted by a NFA. - Let's show this -NFA-> DFA •Given -NFAfind DFA -Let E = (Q E E , q 0 , F E ) be a -NFAthen • There exists a DFA, D= (Q D D , q D , F D • Such that L(E) = L(D) -NFA-> DFA • Basic idea - Very similar to the subset construction algorithm • Recall that for a -NFA, : Q x 2 Q • Use the states of D to represent subsets of Q. -NFA-> DFA • Formal definition -E = (Q E E , q 0 , F E ) be a -NFA - We define DFA, D= (Q D D ,q D , F D •Q D = 2 QE •q D = ECLOSE (q 0 •F D = sets containing at least one state from F E -NFA-> DFA • Computing D D (S, a) for S Q D , a •Let S = { p 1 , p 2 , ..., p n • Compute the set of all states reachable from states in

S on input a using transitions from E.

D (S, a) will be the union of the closures of the elements of {r 1 , ..., r m n i iEm aprrr 121
m j jD rECLOSEaS 1 6 -NFA-> DFA q 0 q 1 q 2 q 3 q 5 q 4quotesdbs_dbs14.pdfusesText_20