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By Samuel Eilenberg
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Extra resources for Automata, Languages, and Machines
Sample text
A = ( T h e set 0 I p ~5 q } is not recognizable. Indeed if A is recognizable, then so is the set B = {gW I 5p} obtained from A by reversal and interchanging the roles of c and z. Therefore A n B is recognizable. 2. 4. Let E = (0,T } . For every word s E E" let I s lo denote the number of 0's in s. Similarly let I s IT = I s I - I s I n . ) is not recognizable. 2. Similarly the set defined by the condition I s ,I 5 I s Ir is not recognizable. 1 , d = ( Q , I , T ) , and n = card Q. 1. Let A = I Show that A is finite iff I s I < n for all S E A.
3 under the assumption that f: r" 4Z" is a morphism such that 1 = If - l . References See the References list at the end of Chapter 111. 111 CHAPTER Deterministic Automata ‘The notions of a deterministic and of a complete automaton bring the subject of automata closer to abstract algebra. ” Thus, gradually, algebraic concepts, methods, and techniques are introduced and brought to bear upon the theory of automata. 1. 2) For q E Q and CT E L‘ there is at most one edge CT: y + p in d. , A = 2, c = a.
I) => (v). Let d =(0,i, T ) , q E 0 and t E Z*. Since XV' is accessible we have is = q for some s E 1". Since there exists u E L'* such that stir E A , we have istu E T and thus qt = ist f Since A f @ it also follows that i f ~2and thus LY' is complete. (v) = (vi). Obvious. (vi) (vii). Obvious. is complete (vii) 3 (i). Let ,d=(Q, i, 2') and s E P. T h u s su E A I a. 4 hold. If Ld-, is not complete, then s-*A = for some s E 2". T h e completion d,Lc of is then obtained by adjoining a sink state 0.