Example 1.11 I Fig 1.12 September 13, 2020 1 / 12
Example 1.11 II s a b q 1 a r 1 b a a b b q 2 r 2 a b September 13, 2020 2 / 12
Example 1.11 III L ( M ) =? a . . . a , b . . . b where “ . . . ” can be any string of a and b September 13, 2020 3 / 12
Example 1.13 I Figure 1.14 September 13, 2020 4 / 12
Example 1.13 II 0 q 1 � t e s e 1 r � , 2 1 2 2 q 0 q 2 0 1 , � reset � 0 , � reset � September 13, 2020 5 / 12
Example 1.13 III Σ = {� reset � , 0 , 1 , 2 } L ( M ) = . . . . . . � reset � . . . � reset � . . . = { sum of the last segment mod 3 = 0 } Example: 10 � reset � 22 � reset � 012 September 13, 2020 6 / 12
Example 1.13 IV Running this string � reset � 1 0 2 2 q 0 − → q 1 − → q 1 − − − → q 0 − → q 2 − → q 1 � reset � 0 1 2 − − − → q 0 − → q 0 − → q 1 − → q 0 Accepted. September 13, 2020 7 / 12
Formal Definition of Computation I M accepts w = w 1 · · · w n if ∃ states r 0 · · · r n such that r 0 = q 0 1 δ ( r i , w i +1 ) = r i +1 , i = 0 , . . . , n − 1 2 r n ∈ F 3 Definition: a language is regular if recognized by some automata September 13, 2020 8 / 12
Designing Automata I Given a language, how do we construct a machine to recognize it? Basically we need to get a state diagram (where the number of states is finite) An automaton recognizing { 0 , 1 } strings with odd # of 1’s Fig 1.20 September 13, 2020 9 / 12
Designing Automata II 0 0 1 q e q o 1 Examples 01 0 1 − → q e − → q o q e 010101 0 1 0 1 0 1 − → q e → q o − − → q o − → q e − → q e − → q o q e September 13, 2020 10 / 12
Designing Automata III Two ways to think about the design After the first 1, we go to q o . Subsequently, every 1 , . . . , 1 pair is cancelled out by 1 1 − → q e → · · · → q e → q o − q o q e , q o respectively remember whether the number of 1’s so far is even or odd Example 1.21 strings contain 001 Fig 1.22 September 13, 2020 11 / 12
Designing Automata IV 1 0 0 0 1 q q 0 q 00 q 001 1 0 , 1 q 0 , q 00 indicate that before the current input character, we have 0 and 00, respectively September 13, 2020 12 / 12
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