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Lecture Theory of automata - Lecture 24 includes the following content: Regular languages, complement of a language, theorem, proof, example, intersection of two regular languages. | Lecture # 10 Theory Of Automata By Dr. MM Alam 1 1 Lecture 9 at a glance Kleene Theorem Part I and Part II Kleene Theorem Part III (Union) 2 Repeat – Kleene Part III Every Regular Expression can be represented by an FA We already know that a regular expression has a corresponding FA. However, the difficult part is that a regular expression can be combined with other regular expression through union (sum), concatenation and closure of FA. Thus, we need to devise methods for FA1+FA2, FA1FA2, FA1 Closure. 3 Repeat – Kleene Theorem Part III (Union) If r1+r2 represents a regular expression r3, then FA+FA2 represents an FA3 that correspond to r3. Start by taking both FA’s initial state and traversing on each input symbol in the respective FA Since one initial state is allowed in FA, therefore, only one state can be marked as initial state During the process, any state encountered final, the resultant state will be final. This is due to the fact that multiple final states are allowed in FA. | Lecture # 10 Theory Of Automata By Dr. MM Alam 1 1 Lecture 9 at a glance Kleene Theorem Part I and Part II Kleene Theorem Part III (Union) 2 Repeat – Kleene Part III Every Regular Expression can be represented by an FA We already know that a regular expression has a corresponding FA. However, the difficult part is that a regular expression can be combined with other regular expression through union (sum), concatenation and closure of FA. Thus, we need to devise methods for FA1+FA2, FA1FA2, FA1 Closure. 3 Repeat – Kleene Theorem Part III (Union) If r1+r2 represents a regular expression r3, then FA+FA2 represents an FA3 that correspond to r3. Start by taking both FA’s initial state and traversing on each input symbol in the respective FA Since one initial state is allowed in FA, therefore, only one state can be marked as initial state During the process, any state encountered final, the resultant state will be final. This is due to the fact that multiple final states are allowed in FA. 4 Old States Reading at a Reading at b z1-≡(q0,p0) (q1,p1)≡z2 (q1,p1)≡z2 z2+≡(q1,p1) (q3,p1)≡z3 (q2,p1)≡z4 z3+≡(q3,p1) (q3,p1)≡z3 (q3,p1)≡z3 z4+≡(q2,p1) (q2,p1)≡z4 (q2,p1)≡z4 5 6 Kleene Theorem Part III (Concatenation) If r1r2 represents a regular expression r3, then FA1FA2 represents an FA3 that should correspond to r3. Start by taking the first FA’s initial state and traversing on each input symbol in the respective FA. Since one initial state is allowed in FA, therefore, only one state can be marked as initial state During the process, any state encountered final of the second FA only, the resultant state will be final. Further, the second FA will be concatenated through first FA’s initial state. However, if the final state of the second FA is encountered, it will not be combined with the first FA. 7 3 Questions for Concatenation FA1: (a+b)b(a+b)* FA2: (a+b)+ FA3: (aaa+b)+ 8 Question (concatenation) Find FA1FA2 for the following: 9 10 Verification: (a+b)b(a+b)*(a+b)+ bba, 11 .
