1. Lecture 12 Non-Regular Languages
CSCE 355 Foundations of Computation
Lecture 12 Non-Regular Languages
Topics:
Strings distinguishing states
Equivalence relation
October 8, 2008

2. New: Last Time: Readings section 3.1,3.2(skip DFA RE), 3.3
Ruby: Matching in files (file.rb)
Review Relations Again
Strings distinguishing states
Indistinguishable states
Indistinguishable is an Equivalence relation
Table of inequivalences
Partitions
New:
If L is regular is the Reversal LR regular?
Ruby: Matching in files (fileArgv.rb)
Unix Shell “regular expressions”

3. Homework
1. Modify fileArgv.rb to substitute “xxx-xxxx” for every occurrence of phone number in every file in a directory
4.4.2

4. Unix Shell “regular expressions” Aren’t really regular expressions
Examples
ls *.rb // * is not the closure operator but wildcard
ls [a-d]*
grep “[a-d]*” *

5. Ruby Matching in Files II, fileARGV.rb
ARGV.each do |arg| p arg f = File.open(p) f.each do |line| puts line if line =~ /Perl|Python/ puts "Scripting language mentioned:#{line}" end f.close

6. Command-Line Options -i [extension]
Edits ARGV files in place. For each file named in ARGV, anything you write to standard output will be saved back as the contents of that file.
A backup copy of the file will be made if extension is supplied.
Builtin-Variable $< (or ARGF) - An object that provides access to the concatenation of the contents of all the files given as command-line arguments or $stdin (in the case where there are no arguments).

7. Strings distinguishing states
A string x in Σ* distinguishes two states q1 and q2 if one of δ(q1, x) and δ(q2,x) is a final state and one if not.
Examples
What string distinguishes q1 and q2 in the DFA?
What string distinguishes q3 and q2?
Graphically

8. Indistinguishable states
Two states are indistinguishable if there is no string that distinguishes them
What if we change the DFA to
δ (q1,0) = q3
δ (q1,1) = q1
Then q1 and q2 are not distinguished by any string

9. Indistinguishable is an Equivalence relation
A relation is an Equivalence relation if .
Prove Indistinguishable is an Equivalence relation
Proof: Let

10. Figure 4.8 DFA page 156 Consider the DFA
Now ε distinguishes C from everything else
Table of state inequivalences 1 A B C D E F G H . X 1 A B C D 1 1 1 1 1 E F G H 1

11. Table of state inequivalences1
Table of state inequivalences 1 A B C D 1 1 A B C D E F G H . X 1 1 1 E F G H 1
Consider the string “0”
Find pairs of states (p,r) with
δ(q, 0)is distinguished from δ(r, 0)
since δ(D, 0) = C and δ(X, 0) !=C; D is disting. from X = A, B, E, G, H
since δ(F, 0) = C and δ(X, 0) !=C; F is disting. from X = A, B, E, G, H
Now consider the string “0”

12. Table of state inequivalences1
Table of state inequivalences 1 A B C D 1 1 A B C D E F G H . X 1 1 1 E F G H 1
Consider the string “1”
Find pairs of states (p,r) with
δ(q, 1)is distinguished from δ(r, 1)
Is δ(A, 1)=B distinguished from δ(B, 1)=C?
Is δ(A, 1)=B distinguished from δ(E, 1)=F?
Is δ(A, 1)=B distinguished from δ(F, 1)=G?
Is δ(A, 1)=B distinguished from δ(G, 1)=E? …
Now consider the string “0”

13. Table of state inequivalences1
Table of state inequivalences 1 A B C D 1 1 A B C D E F G H . X 1 1 1 E F G H 1
Do iterations considering both 0 and 1.
If for an iteration there is no change then there never will be. So stop.
Now consider the string “0”

14. Partitions Equivalence class of x denoted [x]
Π = {X1, X2, … Xn} is a partition of a set S if .
If R is an equivalence relation on a set S then R induces a partition on S which means the equivalence classes form a partition

15. Minimization of DFA
First eliminate states not reachable from the start state
Partition the states into sets of indistinguishable states (equivalence classes)
Figure 4.12 Minimum state DFA equivalent to DFA in figure 4.8 (slide 10)

17. Strings distinguishable for a Language
Two strings y and z are distinguished by a language L if there is a string x such that
exactly one of yx and zx is in L
Example: L

18. Consequences for DFA
Suppose a DFA D=(Q, Σ, δ, q0, F) has n states and that
Language L has than n distinguished prefixes
Then L cannot be the language accepted by D, (L(D).
Why?

20. Applications

21. Not all Languages are regular

23. Pumping Lemma for regular languages