1. CSC 4170
Theory of Computation
Nondeterminism
Section 1.2

2. An NFA without -transitionsq1 1 q2
0,1 q3
0,1
What language does this NFA recognize?

3. An NFA with -transitions
1.2.b 1 b  a a 3
a,b 2
b a b a b
Does this NFA accept:
 ?
a ?
b ?
aaa…a ?

4. Formal definition of a nondeterministic finite automaton
Let  =   {}
A NFA is a 5-tuple (Q, , , s, F), where:
Q is a finite set called the states,
 is a finite set called the alphabet,
 is a function of the type Q  P(Q) called the transition function,
s is an element of Q called the start state,
F is a subset of Q called the set of accept states.

5. Our automaton formalized1 Q: : : s: F: b  a a
a b  1 2 3 3
a,b 2
A = (Q, , , s, F)

6. Formal definition of accepting by NFA
M = (Q, , , s, F) 1 b  a a 3
a,b 2
M accepts the string x iff x can be written as
u1 u2 … un
where each ui is in , and there is a sequence
r1, r2, …, rn, rn+1
of states such that:
r1= s
ri+1 (ri,ui), for each i with 1 i  n
rn+1  F
Example: aa
u u2 … un
aa =
r1, r2, …, rn, rn+1

7. What language does this NFA recognize? 

8. What language does this DFA recognize?1 2 3 5 4

9. Equivalence of NFAs and DFAs
1.2.h
Two machines are said to be equivalent, if they recognize the same language.
Theorem Every NFA has an equivalent DFA.
Proof. Consider an NFA
N = (Q, , , s, F)
We need construct an equivalent DFA
D = (Q’, , ’, s’, F’)
using a procedure called the subset construction (next slide).
Notation:
For RQ, let
R = {q | q can be reached from R by traveling
along 0 or more -arrows}
Explain the idea: keep track of what set of states is possible a a
For RQ and a, let
(R,a) = {q | q can be reached from R by traveling
along an a-arrow} a

10. The Subset Construction
Constructing DFA D = (Q’, , ’, s’, F’) from NFA N = (Q, , , s, F)
Q’ = P (Q)
’(R,a) = (R,a)
s’ = {s}
F’= {R | R is a subset of Q containing an accept state of N}
D obviously works correctly:
at every step in the computation, it clearly enters a state that
corresponds to the subset of states that N could be in at that point. a a   a

11. Applying the subset construction to our NFA
1.2.j
Q’: :
’:
s’:
F’:
N = (Q, , , s, F) 1
a b 
{1}
{2}
{3}
{1,2}
{1,3}
{2,3}
{1,2,3} b  a a 3
a,b 2
Q’ = P (Q)
’(R,a) = (R,a)
s’ = {s}
F’= {R | R is a subset of Q containing an
accept state of N}

12. D {,{1},{2},{3}, {{1},{1,2},{1,3},{1,2,3}} The resulting DFA
1.2.k
{,{1},{2},{3},
{1,2},{1,3},{2,3},{1,2,3}} D
Q’: :
’:
s’:
F’:
a,b
{3} b 
{a,b} a a
a b 
{1}
{2}
{3}
{1,2}
{1,3}
{2,3}
{1,2,3} b
 
{1,3}
{1}
 {2} b
{2,3} {3} a b b
{1,3} 
{2,3} {2,3}
{1,3} {2}
{2,3} a
{2}
{1,2,3} {3}
{1,2,3} {2,3} a b
a,b
{1,2,3}
{1,2}
{1,3}
{{1},{1,2},{1,3},{1,2,3}} a

13. Removing unreachable states
{,{1},{2},{3},
{1,2},{1,3},{2,3},{1,2,3}} D
Q’: :
’:
s’:
F’:
a,b
{3} b 
{a,b} a
a b 
{1}
{2}
{3}
{1,2}
{1,3}
{2,3}
{1,2,3} b
 
{1,3}
 {2} b
{2,3} {3} a b
{1,3} 
{2,3} {2,3}
{1,3} {2}
{2,3} a
{2}
{1,2,3} {3}
{1,2,3} {2,3} a b
{1,2,3}
{1,3}
{{1},{1,2},{1,3},{1,2,3}} a

14. N D b a a Testing in work 1 b  a a 3 a,b 2 a,b {3} b  a b {1,3} b a
1.2.m
N D
a,b
{3} b  1 a b  a b
{1,3} b a a b 3
a,b 2
{2,3} a
{2} a b
b a a
{1,2,3} a