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MYHILL NERODE THEOREM

MYHILL NERODE THEOREM. By Anusha Tilkam. Myhill Nerode Theorem:. The following three statements are equivalent The set L є ∑* is accepted by a FSA L is the union of some of the equivalence classes of a right invariant equivalence relation of finite index.

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MYHILL NERODE THEOREM

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  1. MYHILL NERODE THEOREM By AnushaTilkam

  2. MyhillNerode Theorem: The following three statements are equivalent • The set L є ∑* is accepted by a FSA • L is the union of some of the equivalence classes of a right invariant equivalence relation of finite index. • Let equivalence relation RL be defined by : xRLy iff for all z in ∑* xz is in L exactly when yz is in L. Then RL is of finite index.

  3. Theorem Proof: • There are three conditions: • Condition (i) implies condition (ii) • Condition (ii) implies condition (iii) • Condition (iii) implies condition (i)

  4. Equivalence Relation A binary relation ̴ over a set X is an equivalence relation if it satisfies • Reflexivity • Symmetry • Transitivity

  5. Condition (i) implies condition (ii) Proof: Let L be a regular language accepted by a DFSA M = (Q,∑,δ,q0,F). Define RMon ∑* xRMy if δ(q0, x) = δ(q0 , y) In order to show that its an equivalence relation it has to satisfy three properties.

  6. δ(q0 , x) = δ(q0 , x) --- Reflexive • If δ(q0 , x) = δ(q0 , y) then δ(q0 , y) = δ(q0 , x) --- Symmetry • If δ(q0 , x) = δ(q0 , y) δ(q0 , y) = δ(q0 , z) then δ(q0 , x) = δ(q0 , z) --- Transitive

  7. Index of an Equivalence relation: There are N states If This RMis an Equivalence Relation, Then the index of RM is at most the number of States of M q0 q1 q2 qn-1

  8. Right invariant If x RM y Then xzRMyz for any z є ∑* Then we say RMis Right invariant Proof: δ(q0 , x) = δ(q0, y) δ(q0 , xz) = δ( δ(q0 , x), z ) = δ( δ(q0 , y), z ) = δ(q0 , yz) Therefore RMis right invariant

  9. L is the union of sum of the equivalence classes of that relation. If the Equivalence Relation RMhas n states. S0 , S1 , S2, ……, Si ,…….. , Sn-1 | | | | | q0 , q1 , q2 ,….., qi ,…..…, qn-1

  10. Condition (ii) implies condition (iii) : Proof: Let E be an equivalence relation as defined in (ii). We have to prove that E is a Refinement of RL. What is Refinement?

  11. x E y | x,y є to same equivalence class of E xz E yz | xz is related to yz for any z є ∑* L is the union of sum of the equivalence classes of E. If L contains this equivalence class then xz and yz are in L or it may not be in L. Then we can say that x RL y Hence it is proved that every equivalence class in E is an Equivalence class in RL Then we can say that E is a Refinement of RL E is of finite index Index of RL <= index of E therefore RL is of Finite index.

  12. b b b a a q0 q2 q1 • Example : DFA L ={ w | w contains a stings having atleast one a ,no sequence of b} ∑* is partioned into three equivalence class J0,J1,J2 a

  13. J0 – strings which do not contain an a J1 – strings which contain odd number of a’s J2 - strings which contain even number of a’s L = J1 U J2

  14. Condition (iii) implies condition (i) Proof: RL is right invariant x RL y if xzє L yzє L Therefore if z = wz then xwzє L ywzє L for any w and z Then xwz RLywz Hence RL is Right invariant Define an FSA M’ = (Q’, ∑,δ’,q0 ’,F’) as follows: For each equivalence class of RL ,we have a state in Q’. |Q’| = index of RL

  15. If x є ∑* denote the Equivalence class of RL to which x є to [x] q0’ = [є] belongs to initial state / one equivalence class. For symbol a є ∑ δ’([x],a) = [xa] This definition is consistent because RL is right invariant. If xRLy then δ([x],a) = [ya] Because x,y belong to same class and Right invariant. Therefore we can say that L is accepted by a FSA.

  16. Example : J0 and J1 U J2 are the two equivalence classes in RL a,b b J0 a J1 , J2

  17. To show that a given language is not Regular: • L = {anbn |n>=1} Assume that L is Regular Then by MyhillNerode theorem we can say that L is the union of sum of the Equivalence classes and etc a, aa,aaa,aaaa,…….. Each of this cannot be in different equivalence classes. an ~ am for m ≠ n By Right invariance anbn ~ am bn for m ≠ n Hence contradiction The L cannot be regular.

  18. Conclusion • Shown how the MyhillNerode theorem helps in minimizing the number of states in a DFA. • How it shows that the language is not regular.

  19. References • Languages and MachinesThomas A. Sudkamp, Addison Wesley • http://en.wikipedia.org/wiki/Myhill%E2%80%93Nerode_theorem

  20. Thank You

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