1. Nondeterministic Finite State Automata (Dr. Torng)
NFA’s

2. Change: d is a relation
For an FSA M, d(q,a) results in one and only one state for all states q and characters a.
That is, d is a function
For an NFA M, d(q,a) can result in a set of states
That is, d is now a relation
Next step is not determined (nondeterministic)

3. Example NFA a b
a,b
Why is this only an NFA and not an FSA? Identify as many reasons as you can.

4. Computing with NFA’s Configurations: same as they are for FSA’s
Computations are different
Initial configuration is identical
However, there may be several next configurations or there may be none.
Computation is no longer a “path” but is now a “graph” (often a tree) rooted at the initial configuration
Definition of halting, accepting, and rejecting configurations is identical
Definition of acceptance must be modified

5. Computation Graph (Tree)b
a,b
Computation Graph (Tree)
Input string aaaaba
(1, aaaaba)
(1, aaaba)
(2, aaaba)
(1, aaba)
(2, aaba)
(3, aaba)
(1, aba)
(2, aba)
(3, aba)
crash
(1, ba)
(2, ba)
(3, ba)
(1, a)
(4, a)
(1, l)
(2, l)
(5, l)

6. Definition of ├ unchanged *b
a,b
Definition of ├ unchanged *
Input string aaaaba
(1, aaaaba)
(1, aaaba)
(2, aaaba)
(1, aaba)
(2, aaba)
(3, aaba)
(1, aba)
(2, aba)
(3, aba)
crash
(1, ba)
(2, ba)
(3, ba)
(1, a)
(4, a)
(1, l)
(2, l)
(5, l)
(1, aaaaba)├ (1, aaaba)
(1, aaaaba)├ (2, aaaba)
(1, aaaaba)├3 (1, aba)
(1, aaaaba)├3 (3, aba)
(1, aaaaba)├* (2, aba)
(1, aaaaba)├* (3, aba)
(1, aaaaba)├* (1, l)
(1, aaaaba)├* (5, l)

7. Acceptance and Rejectionb
a,b
Acceptance and Rejection
Input string aaaaba
M accepts string x if one of the configurations reached is an accepting configuration
(q0, x) ├* (f, l),f in A
M rejects string x if all configurations reached are either not halting configurations or are rejecting configurations
(1, aaaaba)
(1, aaaba)
(2, aaaba)
(1, aaba)
(2, aaba)
(3, aaba)
(1, aba)
(2, aba)
(3, aba)
crash
(1, ba)
(2, ba)
(3, ba)
(1, a)
(4, a)
(1, l)
(2, l)
(5, l)

8. Comparison b a
a,b a a b a b b
FSA a b
a,b
NFA

9. Defining L(M) and LNFA
M accepts string x if one of the configurations reached is an accepting configuration
(q0, x) |-* (f, l),f in A
M rejects string x if all configurations reached are either not halting configurations or are rejecting configurations
L(M)
N(M)
LNFA
Language L is in language class LNFA iff

10. Comparing language classes
LFSA subset of LNFA

11. LFSA subset LNFA Let L be an arbitrary language in LFSA
Let M be the FSA such that L(M) = L
M exists by definition of L in LFSA
Construct an NFA M’ such that L(M’) = L
Argue L(M’) = L
There exists an NFA M’ such that L(M’) = L
L is in LNFA
By definition of L in LNFA

12. Visualization L L LFSA LNFA M M’ FSA’s NFA’s
Let L be an arbitrary language in LFSA
Let M be an FSA such that L(M) = L
M exists by definition of L in LFSA
Construct NFA M’ from FSA M
Argue L(M’) = L
There exists an NFA M’ such that L(M’) = L
L is in LNFA M
LFSA
LNFA
FSA’s
NFA’s M’

13. Construction We need to make M into an NFA M’ such that L(M’) = L(M)
How do we accomplish this?