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Xk+1 ∈ I k+1 . Clearly ω(si , x1 x2 . . xk+1 ) = ω(si , x1 )ω(ν(si , x1 ), x2 . . xk+1 ). (1) Similarly ω(sj , x1 x2 . . xk+1 ) = ω(sj , x1 )ω(ν(sj , x1 ), x2 . . xk+1 ). (2) Since si ∼ =k sj , it follows from (b) that ω(si , x1 ) = ω(sj , x1 ). (3) On the other hand, since ν(si , x1 ) ∼ =k ν(sj , x1 ), it follows that ω(ν(si , x1 ), x2 . . xk+1 ) = ω(ν(sj , x1 ), x2 . . xk+1 ). (4) On combining (1)–(4), we obtain ω(si , x1 x2 . . xk+1 ) = ω(sj , x1 x2 . . xk+1 ). (5) Note now that (5) holds for every x1 x2 .

Describe each of the following languages in A∗ : a) All strings in A containing the digit 2 once. b) All strings in A containing the digit 2 three times. c) All strings in A containing the digit 2. d) All strings in A containing the substring 2112. e) All strings in A in which every block of 2’s has length a multiple of 3. f) All strings in A containing the substring 2112 and in which every block of 2’s has length a multiple of 3. g) All strings in A containing the substring 2112 and in which every block of 2’s has length a multiple of 3 and every block of 1’s has length a multiple of 2.

Suppose again that I = O = {0, 1}, and that we want to design a finite state machine that recognizes the sequence pattern 0101 in the input string x ∈ I ∗ , but only when the last 1 in the sequence pattern 0101 occurs at a position that is a multiple of 3. Consider first of all the situation when the required pattern occurs repeatly and without interruption. Note that the first two inputs do not contribute to any pattern, so we consider the situation when the input string is x1 x2 0101 . . , where x1 , x2 ∈ {0, 1}.

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