Differences

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--- lfa:2024:lab03 [2024/10/20 13:24]
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+++ lfa:2024:lab03 [2024/10/22 08:13] (current)
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@@ Line 126: / Line 126: @@
 </hidden>
-** 3.2.5 **
-$ E1 = (a\mid b)^*aa^* \mid \epsilon $
-\\
-$ E2 = (a\mid ba)^*(b\mid ba)^* $
-<hidden 3.2.5><note important>
-Both E1 and E2 have an infinite language, so comparing them is not an option.
-We can see that for example E2 accepts b, while E1 does not accept it, so the expressions are not equivalent.
-Fun fact: E1 was proposed by a student as a solution for 3.2.2 last year, while E2 is the actual solution.
-</note></hidden>
-** 3.2.6 **
-$ E1 = ((ab^*a)^+b)^* $
-\\
-$ E2 = (a(b\mid aa)^*ab)^* $
-<hidden 3.2.6><note important>
-Both E1 and E2 have an infinite language, so comparing them is not an option.
-We can try looking for words that are accepted by one and not by the others, but we can't easily find such words. ! This does not mean they are equivalent !
-We should transform each expression into its min DFA and check if they are the same (number of states, transitions, alphabet, initial/final states) (renaming of states might be needed).
-Plot twist: They are the same.
-The purpose of this exercise is to understand how to approach regex equivalence, not how to solve this given comparison per
-se.
-</note></hidden>
@@ Line 207: / Line 171: @@
 We can see that for example E2 accepts b, while E1 does not accept it, so the expressions are not equivalent.
-Fun fact: E1 was proposed by a student as a solution for 3.2.2 last year, while E2 is the actual solution.
+Fun fact: E1 was proposed by a student as a solution for 3.4.1 some year ago, while E2 is the actual solution.
 </note></hidden>