1. Theory of Computation CS3102 – Spring 2014 A tale of computers, math, problem solving, life, love and tragic death
Nathan Brunelle Department of Computer Science University of Virginia

2. Midterm Take home? Vote now. Date:
During the week before Spring break (out March 3, due March 6)?
Don’t have to worry over the break
After Spring break (4 days within March 16-23)?
1 more week of content but 2 more weeks to solve problems
I will hold Skype/Google Hangout office hours over the break
Think about it, vote on Thursday.

3. Pumping Lemma
Goal: Give a sufficient condition for showing non-regularity
Consider that L is an infinite regular language
Let M be a DFA, L(M)=L
Let’s say M has p states
There must be some
String w in L s.t. |w|>p
By the pigeon-hole principle:
Some state was visited at least twice!
Taking that loop another time must give
another string in the language. M x2 3 8 p 12 1 20

4. Pumping Lemma
If L is a regular language then there is some number p (called pumping length) where if w is a string in L s.t. |w|>p then w can be divided into 3 pieces: w=xyz which satisfy:
For each i≥0, L
|y|>0
|xy|≤p
Example:
Consider:
By condition 3 we know:
Thus for i=2
we have more a’s than b’s,
so L cannot be regular x M y 3 8 p ℕ 12 1 20 z

5. Pumping Lemma For each i≥0, L |y|>0 |xy|≤p Show: is not regular
Consider
By condition 3 we know y , this leaves 4 cases: y
Problem: in all of these I only change the first half!
I needed to “remember” the first half of the string.

6. Pumping Lemma For each i≥0, L |y|>0 |xy|≤p Show: is not regular
Consider where q is a prime greater than p
y may have only a’s, let y= where m≤p
For i=p+1 we have
Clearly, p(m+1) is not prime
For let then So

7. Pumping Lemma For each i≥0, L |y|>0 |xy|≤p Consider:
This language is pumpable but not regular
3 cases:
If then let x=ε, y=a
If then let x=ε, y=c
If then let x=ε, y=c
Thus every string is pumpable
Nonregular: Then use pumping Lemma!

8. Myhill-Nerode Theorem
Gives a necessary and sufficient condition for regularity!
Idea: If two strings terminate in the same state in a DFA then their membership must be equivalent for any suffix.
If and meet at
state q then for any string
, ends in the same
state as
We say and have
no distinguishing extensions M q

9. Myhill-Nerode Theorem
Recall Equivalence Relation:
A relation ~ is called an equivalence relation if:
Reflexive:
Symmetric:
Transitive:
The relation if and have no distinguishing extensions in language L forms an equivalence relation.
Recall Equivalence Class:
For equivalence relation ~, equivalence class
Myhill Nerode Theorem: L is regular iff has a finite number of equivalence classes under relation

10. Myhill-Nerode Theorem
Myhill Nerode Theorem: L is regular iff has a finite number of equivalence classes under relation
Given a finite set of equivalence classes, construct a DFA:
Each equivalence class becomes a state:
If character takes a string from equivalence class to then add a transition from the state for to the state for
Accept states are those for equivalence classes of strings in the language.

11. Myhill-Nerode Theorem
Extra Credit: Use the Myhill-Nerode Theorem to show the following languages are non-regular: ℕ

12. Language Reversal
Theorem: The regular languages are closed under reversal.
Proof: Construction.
Given a regular language L, show that the language
is also regular.
Let M be a DFA for L, construct M’ to be a NFA for L: M M’ ε q0 ε

13. Half Let Show that HALF preserves regularity.
e.g. if “RingoStarr” is in L, then “Ringo” is in HALF(L)
Let M be a DFA for language L
Intuition: follow the transitions of M for string v. “Check” that there is a path from v to an accept state that consumes |v| characters.
How do we do this “check”? M

14. Half Let Show that HALF preserves regularity.
Intuition: follow the transitions of M for string v. “Check” that there is a path from v to an accept state that consumes |v| characters.
How do we do this “check”? Use the machine for LR M M’ Σ ε Σ Σ ε Σ
F={(q,q)} Σ

15. Double Let Show that DOUBLE preserves regularity.
e.g. if “BamBam” is in L, then “Bam” is in DOUBLE(L)
Let M be a DFA for language L
Intuition: Run L on w, in parallel non-deterministically “guess” the end state in machine M on w and check if starting from that guess puts the machine in an accept state M M
“Guess”

16. Double
Let
Intuition: Run L on w, in parallel non-deterministically “guess” the end state in machine M on w and check if starting from that guess puts the machine in an accept state ε ε M ε M M
Accept if:
M ends in state
M ends in an accept state M 1 M 2 M …