1. CSCI 2670 Introduction to Theory of Computing
August 30, 2005

2. Announcements Homework due next Tues, Sept 6
1.7b, 1.8a, 1.10a, 1.15, 1.17a&b, 1.31, 1.32
Old Text 1.5b, 1.6a, 1.8a, 1.24, 1.25 (1.15 & 1.17 are not in old text)
Office hours with TA Ryan Foster
Monday 2:30 – 3:30
Friday 10:00 – 11:00
Room 301 B (knock if locked)
Change in syllabus
First midterm exam Tues, September 27

3. Agenda Last class This week Combined regular languages
Introduced non-determinism
This week
Section 1.2
Start Section 1.3

4. Find M such that L(M) = L(M1)  L(M2)q1 q2 q3
0,1 1 M1 1
q1’
q2’
q3’ M2
Example
Find M such that L(M) = L(M1)  L(M2) q1 q2 q3
0,1 1
q1’
q2’
q3’ ε
Can jump to q1’ non-deterministically

5. NFA vs. DFA
DFA
NFA
Exactly one arrow exiting each node for each symbol of the alphabet
Number of exiting arrows can be anything from 0 to the number of nodes
Each arrow is labeled only with characters of the alphabet
Characters may be labeled with an empty string, ε
Each string reaches exactly one state at the end of the input
A string may reach several different states states at the end of the input

6. Computing on an NFA
For each symbol of the string, keep track of all possible transitions in parallel
When input ends, there may be several possible ending states
If at least one of the possibilities is an accepting state, then the NFA accepts the string

7. Example Does this NFA accept the string 001? From state(s) Input
0,1 1 q1 q2 ε q3 1 1
Does this NFA accept the string 001?
From state(s)
Input
To state(s) q1
q2, q3 1
q1, q3
Yes
q1, q3

8. Example Any string containing two 1’s separated by at most one 0 0,1 1ε
Any string containing two 1’s separated by at most one 0

9. Example Zero or more concatenations of the strings aba and ab   b a

10. Example ε
a, b, c
a, b, d
a, c, d
b, c, d
Strings of any length containing at most three of the symbols in  = {a, b, c, d}

11. The power of non-determinism
Strings of any length containing at most two of the symbols in  = {a, b, c} c a b
a, b, c
a, c
b,c
a, b
DFA
a, b
a, c
b, c ε
NFA

12. Non-deterministic finite automaton
A non-deterministic finite automaton is a
5-tuple (Q,,,q0,F), where
Q is a finite set of states
 is a (finite) alphabet
 : Q × ε  P(Q) is the transition function
 maps to sets of states
q0 is the start state, and
F  Q is the set of accept states

13. Equivalence of DFAs and NFAs
Theorem: Every non-deterministic finite automaton has an equivalent deterministic finite automaton
Both FAs accept the same language
Proof method
Construction
Similar to method used for calculating strings
Follow all paths in parallel where states represent parallel paths

14. Proof idea
Given NFA M1={Q,,,q0,F} construct DFA M2={Q’,,’,q0’,F’} with L(M1)=L(M2)
Intuition
Recall :Q×εP(Q)
Our DFA’s transition function will generate paths within P(Q)
’: P(Q)×P(Q)

15. Defining M2 Determine Q’, q0’, and F’
Q’ = P(Q)
q0’ = {q0}
F’ = {R  Q’ | R  F  }
i.e., R contains at least one of M1’s accept states
Defining ’ (for now ignore ε jumps)