1. Nondeterministic Finite Automata
CS310
Nondeterministic Finite Automata
Sections:1.2 page 47
September 12, 2008

2. Example (1.30)
Accept string of at least length three that contains a 1 in the third from end
What makes this difficult for a DFA?
Equivalent DFA takes 8 states. Why 8?
= 0,1 ; ∗ 1 0∪1 0∪1
Lets look at the language first. It is clear what this specifies? What does E* mean? U means?

3. Formal Definition of NFA
5 tuple ( Q,Σ, δ , q0,F)
Σ ɛ =Σ∪ 𝑒
δ:𝑄× Σ ɛ →𝑃 𝑄
1, ε 1 q4 q1 q2 q3
0,1 1 ε q1
{q1.q2} ø q2
{q3} q3
{q4}
Same as DFA.
Don’t add e to the Alphabet.
Just add e to the Alphabet for the transition table.
Remember Q x E -> P(Q) means the RANGE of the function must come from P(Q)
The transition table is now made up of sets, some of which are empty!
And contains the e transitions.
What happens with 10, 11, 0, 01

4. Formal Definition of Computing for NFA
Given a machine M= (Q, ∑ , δ,q0, F) and a string w=w1w2…wn over ∑, then M accepts w if there exists a sequence of states r0,r1…rn in Q such that:
r0 = q0
δ (ri, wi+1) = ri+1 ,i=0,…,n-1
rn є F

5. Practice
Construct a NFA with three states that recognizes {w | w ends with two 0s}
Σ = {0,1}

6. Practice Construct a NFA with six states
{w | w even # 0s OR exactly two 1s}
Σ = {0,1}

7. Practice
Construct a NFA with three states
0*1*0*0
Σ = {0,1}