1. Lecture 11 Minimization Topics: Strings distinguishing states Equivalence relation October 6, 2008 CSCE 355 Foundations of Computation

2. – 2 – CSCE 355 Fall 2008 Last Time: Readings section 3.1,3.2(skip DFA  RE), 3.3 Questions onTEST 1 – September 29 th Post-class make-up for problem 1 DFA  RegExpr  NFA  DFA Algebraic Laws for Regular Expressions Languages are sets so inherit some properties Associativity, commutivity of + Identity, Annihilator Checking for Equality of Two Regular Expressions r,s Closure properties of regular languages: Intersection, complementNew: ReversalReadings section 3.1,3.2(skip DFA  RE), 3.3 Ruby: Matching in files

3. – 3 – CSCE 355 Fall 2008 Homework  P 124 3.5 Which are true which are false. If false provide a counterexample  R(S+T) = RS +RT  (R*)* = R*  (R*S*)* = (R+S)*  (R+S)* = R*+S*  S(RS+S)*R = RR*S(RR*S)*  (RS+R)*R = R(SR+R) 2.4.4.2

4. – 4 – CSCE 355 Fall 2008 Ruby Matching in Files ### Pattern match from pickaxe book f = File.open("testfile") f.each do |line| puts line puts line if line =~ /Perl|Python/ if line =~ /Perl|Python/ puts "Scripting language mentioned: #{line}" puts "Scripting language mentioned: #{line}" end endendf.close

5. – 5 – CSCE 355 Fall 2008 To follow the construction precisely or to wing it  Construction shortcuts (simplifications less states).  Now Consider a* b*

6. – 6 – CSCE 355 Fall 2008 Induction Proofs in text

7. – 7 – CSCE 355 Fall 2008 Review Relations Again   Reflexive   Irreflexive   Symmetric   Asymmetric   Anti-symmetric   Transitive   Total   Injection   Surjection   Function   Equivalence relation

8. – 8 – CSCE 355 Fall 2008 Strings distinguishing states   A string x in Σ * distinguishes two states q 1 and q 2 if one of δ( q 1, x) and δ( q 2,x) is a final state and one if not.   Examples What string distinguishes q 1 and q 2 in the DFA? What string distinguishes q 3 and q 2 ?   Graphically

9. – 9 – CSCE 355 Fall 2008 Indistinguishable states  Two states are indistinguishable if there is no string that distinguishes them  What if we change the DFA to δ ( q 1,0) = q 3 δ ( q 1,1) = q 1  Then and are not distinguished by any string  Then q 1 and q 2 are not distinguished by any string

10. – 10 – CSCE 355 Fall 2008 Indistinguishable is an Equivalence relation   A relation is an Equivalence relation if 1.. 2.. 3.. Indistinguishable is an Equivalence relation Prove Indistinguishable is an Equivalence relation Proof: Let

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12. – 12 – CSCE 355 Fall 2008 Figure 4.8 DFA page 156   Consider the DFA   Table of inequivalences B C D E F G H ABCDEFG

13. – 13 – CSCE 355 Fall 2008 Partitions Equivalence class of x denoted [x] Π = {X 1, X 2, … X n } 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

14. – 14 – CSCE 355 Fall 2008 Refinement of a partition

15. – 15 – CSCE 355 Fall 2008 Minimization of DFA 1.First eliminate states not reachable from the start state 2.Start with initial partition Π = { F, Q-F } 3.For each X in Π refine X by Considering a in Σ such that two states are in the same subgroup

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17. – 17 – CSCE 355 Fall 2008 Strings distinguishable

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20. – 20 – CSCE 355 Fall 2008 Not all Languages are regular

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22. – 22 – CSCE 355 Fall 2008 Pumping Lemma

23. – 23 – CSCE 355 Fall 2008 Every finite language is Regular