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MYHILL NERODE THEOREM By Anusha Tilkam Myhill Nerode Theorem: The following three statements are equivalent 1. The set L є ∑* is accepted by a FSA 2. L is the union of some of the equivalence classes of a right invariant equivalence relation of finite index. 3. 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. Theorem Proof: • There are three conditions: 1. Condition (i) implies condition (ii) 2. Condition (ii) implies condition (iii) 3. Condition (iii) implies condition (i) Equivalence Relation A binary relation ̴ over a set X is an equivalence relation if it satisfies • Reflexivity • Symmetry • Transitivity Condition (i) implies condition (ii) Proof: Let L be a regular language accepted by a DFSA M = (Q,∑,δ,q0,F). Define RM on ∑* x RM y if δ(q0 , x) = δ(q0 , y) In order to show that its an equivalence relation it has to satisfy three properties. • 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 • Index of an Equivalence relation: There are N states q0 q1 q2 qn-1 If This RM is an Equivalence Relation, Then the index of RM is at most the number of States of M • Right invariant If x RM y Then xz RM yz for any z є ∑* Then we say RM is 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 RM is right invariant • L is the union of sum of the equivalence classes of that relation. If the Equivalence Relation RM has n states. S0 , S1 , S2, ……, Si ,…….. , Sn-1 | | | | | q0 , q1 , q2 ,….., qi ,…..…, qn-1 • 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? 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. • Example : DFA b b b a q0 a q2 q1 a L ={ w | w contains a stings having atleast one a ,no sequence of b} ∑* is partioned into three equivalence class J0,J1,J2 J0 J1 J2 є a aa b ba aba bb babaa babab …… so on ……so on ……..so on 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 • Condition (iii) implies condition (i) Proof: RL is right invariant x RL y if xz є L yz є L Therefore if z = wz then xwz є L ywz є L for any w and z Then xwz RL ywz 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 • 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 xRL y 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. • Example : J0 and J1 U J2 are the two equivalence classes in RL a,b b a J0 J1 , J 2 To show that a given language is not Regular: • L = {anbn |n>=1} Assume that L is Regular Then by Myhill Nerode 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. Conclusion • Shown how the Myhill Nerode theorem helps in minimizing the number of states in a DFA. • How it shows that the language is not regular. References • Languages and Machines Thomas A. Sudkamp, Addison Wesley • http://en.wikipedia.org/wiki/Myhill%E2%80%9 3Nerode_theorem Thank You