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lfa:2023:lab03 [2023/10/23 16:17] mihai.udubasa |
lfa:2023:lab03 [2023/10/31 11:43] (current) alexandra.udrescu01 |
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| $ AB = ? $ \\ $ BA = ? $ | $ AB = ? $ \\ $ BA = ? $ | ||
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| **3.1.3.** | **3.1.3.** | ||
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| **3.1.4** | **3.1.4** | ||
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| $ AB = ? $ \\ $ BA = ? $ | $ AB = ? $ \\ $ BA = ? $ | ||
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| ===== 3.2. Writing Regular Expressions ===== | ===== 3.2. Writing Regular Expressions ===== | ||
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| **Hint:** you can abbreviate $ 0 \cup 1 \cup ... \cup 9 $ by $ [0-9] $ | **Hint:** you can abbreviate $ 0 \cup 1 \cup ... \cup 9 $ by $ [0-9] $ | ||
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| We start by defining the regex for a number: | We start by defining the regex for a number: | ||
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| **3.2.2.** Write a regular expression for $ L = \{ \omega \text{ in } \text{{0,1}} ^* \text{ | EVERY sequence of two or more consecutive zeros appears before ANY sequence of two or more consecutive ones} \} $ | **3.2.2.** Write a regular expression for $ L = \{ \omega \text{ in } \text{{0,1}} ^* \text{ | EVERY sequence of two or more consecutive zeros appears before ANY sequence of two or more consecutive ones} \} $ | ||
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| **3.2.3.** Write a DFA for $ L(( 10 \cup 0) ^* ( 1 \cup \epsilon )) $ | **3.2.3.** Write a DFA for $ L(( 10 \cup 0) ^* ( 1 \cup \epsilon )) $ | ||
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| {{:lfa:2022:lfa2022_lab_3.2.3.png?300|}} | {{:lfa:2022:lfa2022_lab_3.2.3.png?300|}} | ||
| </hidden> | </hidden> | ||
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| **3.2.4.** Write a regular expression which generates the accepted language of A. Then try to find the most simple and easy to understand way to write it. | **3.2.4.** Write a regular expression which generates the accepted language of A. Then try to find the most simple and easy to understand way to write it. | ||
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| {{:lfa:graf1.png?400|}} | {{:lfa:graf1.png?400|}} | ||
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| - | **3.2.5.** Describe as precisely as possible the language generated by $math[((0(1 \cup 0)(1 \cup 0)) \cup 100)1((0(1 \cup 0)(1 \cup 0)) \cup 100)1(((0(1 \cup 0)(1 \cup 0)) \cup 100)0)*] | + | |
| + | **3.2.5.** | ||
| + | Find a regular expression for the set of all binary strings with the property that none of its prefixes has two more 0's than 1's nor two more 1's than 0's. | ||
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| + | <hidden> | ||
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| + | We the given property implies that the word is composed of repeated sequences of '10' and '01' (so elements at odd positions are different from their left neighbour), possibly followed by a final 1 or 0: | ||
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| + | After any prefix of even length with an equal number of 0's and 1's (including $\epsilon$), you can either finish the word, add a single character and then finish the word (both of these keeping the rule true) or add at least 2 characters. In the last case, if the characters are equal the prefix ending at these two characters breaks the rule. | ||
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| + | $(01\cup10)^*(0\cup1\cup\epsilon)$ | ||
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| + | </hidden> | ||
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| + | /* this is the old exercise | ||
| + | Describe as precisely as possible the language generated by $math[((0(1 \cup 0)(1 \cup 0)) \cup 100)1((0(1 \cup 0)(1 \cup 0)) \cup 100)1(((0(1 \cup 0)(1 \cup 0)) \cup 100)0)*] | ||
| (hint: BCD) | (hint: BCD) | ||
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| <hidden> | <hidden> | ||
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| $ E2 = (a|\epsilon)(b|\epsilon) $ | $ E2 = (a|\epsilon)(b|\epsilon) $ | ||
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| <hidden><note important> | <hidden><note important> | ||
| We can observe that E2 accepts ε, while E1 does not so they are not equivalent. | We can observe that E2 accepts ε, while E1 does not so they are not equivalent. | ||
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| Another approach is to compute the language of each expression (since they are finite) and check if they are equivalent. | Another approach is to compute the language of each expression (since they are finite) and check if they are equivalent. | ||
| </note></hidden> | </note></hidden> | ||
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| ** 3.3.2 ** | ** 3.3.2 ** | ||
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| $ E2 = ab(b|c|d|e)|acd|ace $ | $ E2 = ab(b|c|d|e)|acd|ace $ | ||
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| <hidden><note important> | <hidden><note important> | ||
| Since both E1 and E2 have a finite language, we could just compute the language and check if they are equivalent. | Since both E1 and E2 have a finite language, we could just compute the language and check if they are equivalent. | ||
| Language is L = {abb, abc, abd, abe, acd, ace}, therefore they are equivalent. | Language is L = {abb, abc, abd, abe, acd, ace}, therefore they are equivalent. | ||
| </note></hidden> | </note></hidden> | ||
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| ** 3.3.3 ** | ** 3.3.3 ** | ||
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| $ E2 = (a\mid ba)^*(b\mid ba)^* $ | $ E2 = (a\mid ba)^*(b\mid ba)^* $ | ||
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| <hidden><note important> | <hidden><note important> | ||
| Both E1 and E2 have an infinite language, so comparing them is not an option. | Both E1 and E2 have an infinite language, so comparing them is not an option. | ||
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| 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 3.2.2 last year, while E2 is the actual solution. | ||
| </note></hidden> | </note></hidden> | ||
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| ** 3.3.4 ** | ** 3.3.4 ** | ||
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| $ E2 = (a(b\mid aa)^*ab)^* $ | $ E2 = (a(b\mid aa)^*ab)^* $ | ||
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| <hidden><note important> | <hidden><note important> | ||
| Both E1 and E2 have an infinite language, so comparing them is not an option. | Both E1 and E2 have an infinite language, so comparing them is not an option. | ||
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| se. | se. | ||
| </note></hidden> | </note></hidden> | ||
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| ==== Conclusion ==== | ==== Conclusion ==== | ||