1. Lecture 6 NFA Subset Construction & Epsilon Transitions
CSCE 355 Foundations of Computation
Lecture 6 NFA Subset Construction & Epsilon Transitions
Topics:
Examples Subset Construction
Author’s Website again
Epsilon transitions
Ruby - dfa1.rb
Sept 15, 2008

2. New: Readings section 2.2.3-2.4
Last Time: Readings 2.3
HW Hints and Induction example
Review
Induction
DFA for Union, revisit Example 2.4
Pop Quiz
Uses of Finite automata
NFA
Delta-hat, a string accepted by an NFA, the language accepted
Subset construction converting NFA  equivalent DFA
TEST 1 – Monday
New: Readings section
Examples Subset construction converting NFA equivalent DFA
Author’s Website Solutions Online

3. NFA example Figure 2.9 Page 56 Transition Table (NFA) x
What does the p mean?
What does the *r mean?
What is δ(s, x) informally? 1 p q r
0,1
State\Input 1 p
{ p, q } q *r

4. Subset Construction example Figure 2.9
Page 56 of text 1 p q r
Equivalent DFA Table
State\Input 1
{ p }
{ p, q}
{ p }
{ p, r}
*{ p, r}
0,1

5. Subset Construction example Figure 2.9
Page 56 of text 1 p q r
Equivalent DFA Table
State\Input 1
{ p }
{ p, q}
{ p }
{ p, r}
*{ p, r}
0,1

6. Subset Construction Significance
Constructing an equivalent DFA from and NFA
What does equivalent mean?
Does equivalent mean have the same number of states?
Equivalent means 
Why convert?
What is better about an NFA?
What is better about a DFA?
We are interested in the power of these models?
Can an NFA recognize a language that a DFA can’t?
Can a DFA recognize a language that an NFA can’t?

7. Exercise 2.2.9 Solutions Online
Author’s Website for Text
HW Solutions for starred (*) problems
2.2.9 page 54 Prove If δ(q0, a) = δ(qf, a) for all a in Σ then for all w != ε we have δ(q0, w) = δ(qf, w) by induction on the length of w.
Basis
Assume
Then we need to show that
Dr. Ullman’s (Jeff’s) slides from CS 154

8. Homework and Test 1 HW 2 Extra Credit HW 3 HW 4 Pop Quiz
HW 3 4a 4b 5 6
HW 4
Pop Quiz

13. Subset Example from Author’s website slides2.pdf

14. Mutual Induction Proof
Write up on back of Lecture Overview

15. Consider our old friend from HW 2. 2. 5b: L = {w ε {0,1}
Consider our old friend from HW 2.2.5b: L = {w ε {0,1}* | the tenth symbol from the right end of w is a ‘1’ }
0,1
0,1
0,1 1
0,1 3 4 … 10 1 2
0,1
If we convert an NFA with n states to a DFA using the subset construction what is the max number states in the DFA?
Can we do better?
Subset construction an example of “lazy evaluation” – i.e. consider only states we can get to from q0
DFA minimization is a topic for later

16. Ruby: Strings and DFAs (dfa1.rb)
# DFA1.rb on Handouts page Now consider how to generate all strings in Σ* of length 6 # Idea: generate them from a list of the strings of length n-1 # by concatenating onto each string w of length n-1 each a ε Σ (a recursive definition) # lists of strings of length n-1 and n strnm1 = Array.new(); strn = Array.new(); strnm1 = ["a", "b"] print "strnm1 = #{strnm1}\n“

17. [2,3,4,5, 6, 7, 8]. each { |len| numstrings = 0 strnm1
[2,3,4,5, 6, 7, 8].each { |len| numstrings = 0 strnm1.each { |str| alphabet.each { |chr| x = chr + str strn[numstrings] = x numstrings = numstrings + 1 } print "Strings of length #{len}:\n" strn.each { |str| print "#{str}\n" } strnm1 = strn strn = Array.new

18. Theorem 2.11 For NFA there is Eq. DFA

19. Theorem 2.12 L is accepted by DFA if and only if L is accepted by NFA

20. Epsilon (ε)-Transitions
Keyword Searching Example : for, format, font

21. Epsilon Closure

22. Equivalent NFA (without ε) for an NFA with ε
Convert NFA with ε to an equivalent NFA without ε
Compute transitive closure of ε arcs
If p can reach state q by ε arcs and δ(r, a) contains p (there is a transition from r to q on input a) then add q to δ(r, a) i.e. add a transition from r to q on input a xx

23. References and Homework
Ruby pickaxe book Online
Author’s Website for Text
Slides, HW, Exams