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lfa:2023:lab06 [2023/11/13 19:24] mihai.udubasa created |
lfa:2023:lab06 [2023/11/23 23:34] (current) alexandra.udrescu01 |
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| </hidden> | </hidden> | ||
| - | /* | + | |
| <hidden DFA2> | <hidden DFA2> | ||
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| </hidden> | </hidden> | ||
| - | */ | + | |
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| * $math[q_2 = A^*(Bq_1\cup \epsilon)] | * $math[q_2 = A^*(Bq_1\cup \epsilon)] | ||
| * We replace $math[q_2] into the first equation: | * We replace $math[q_2] into the first equation: | ||
| - | * $math[q_1 = A q_1 \cup BA^*(B q_1 \cup \epsilon) = (A \cup BA^*) q_1 \cup BA^*] | + | * $math[q_1 = A q_1 \cup BA^*(B q_1 \cup \epsilon) = (A \cup BA^*B^*) q_1 \cup BA^*] |
| - | * and we apply Arden's lemma again: $math[q_1 = (A \cup BA^*)^*BA^*] | + | * and we apply Arden's lemma again: $math[q_1 = (A \cup BA^*B)^*BA^*] |
| === Exercise === | === Exercise === | ||
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| - | /* | + | |
| <hidden Solution> | <hidden Solution> | ||
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| => q2 = B*A(A+B+A | B+A)*A* | => q2 = B*A(A+B+A | B+A)*A* | ||
| </hidden> | </hidden> | ||
| - | */ | + | |