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lfa:2022:lab03-dfa-regexp [2022/10/27 14:21]
pdmatei [3.3 Regex Equivalence] 		lfa:2022:lab03-dfa-regexp [2022/10/28 22:42] (current)
alexandra.udrescu01
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 $ AB = ? $ $ AB = ? $
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 **3.1.2.** **3.1.2.**
  
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 $ AB = ? $ \\ $ BA = ? $ $ AB = ? $ \\ $ BA = ? $
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 **3.1.3.** **3.1.3.**
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 $ B^* = ? $ $ B^* = ? $
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 A* = {ε} (epsilon is always part of Kleene star) A* = {ε} (epsilon is always part of Kleene star)
  
-B*= {ε} (epsilon is always part of Kleene star) U {1^n} U {(1^(2n)} U {1^(3n)} U ... +B*= {ε} (epsilon is always part of Kleene star) U {$ 1^n $} U {$ 1^{2n} $} U {$ 1^{3n} $} U ... 
-So basically B = L( (1^n)* )+So basically B = L( ($1^n$)* )
  
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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 =====
  
 **3.2.1.** Write a regular expression for the language of arithmetic expressions containing +, * and numbers. **3.2.1.** Write a regular expression for the language of arithmetic expressions containing +, * and numbers.
 **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|}}
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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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 Let's see what words are accepted: Let's see what words are accepted:
   * ab*ab (when we don't loop on state 1)   * ab*ab (when we don't loop on state 1)
-  * ab*(<a way to leave state 2 and return back to it>)*ab => ab*(aab*)ab+  * ab*(<a way to leave state 2 and return back to it>)*ab => ab*(aab*)*ab
   * anything that repeats the previous expression several times   * anything that repeats the previous expression several times
   * ε (the initial state is also a final state)   * ε (the initial state is also a final state)
   * from the previous 2 observations => we can use Kleene star   * from the previous 2 observations => we can use Kleene star
  So, the regex is:  So, the regex is:
-> ( ab*(aab*)ab )*+> ( ab*(aab*)*ab )*
  
 <​note>​ <​note>​
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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.** 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)*] ​
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 (hint: BCD) (hint: BCD)
  
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 0000, 0010, 0100, 0110, 1000. 0000, 0010, 0100, 0110, 1000.
  
-So, this ugly regex encodes BCD numbers that start with exactly 2 odd digits which are followed by 0 or more eve +So, this ugly regex encodes BCD numbers that start with exactly 2 odd digits which are followed by 0 or more even digits.
-n digits.+
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 ===== 3.3 Regex Equivalence ===== ===== 3.3 Regex Equivalence =====
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 $ E2 = (a|\epsilon)(b|\epsilon) $ $ E2 = (a|\epsilon)(b|\epsilon) $
  
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 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.
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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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 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.
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 ** 3.3.3 ** ** 3.3.3 **
  
-$ E1 = (a|b)*aa*| {ε} $+$ E1 = (a\mid b)^*aa^* \mid \epsilon ​$
 \\ \\
-$ E2 = (a|ba)*(b|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.
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 ** 3.3.4 ** ** 3.3.4 **
 +
 $ E1 = ((ab^*a)^+b)^* $ $ E1 = ((ab^*a)^+b)^* $
 \\ \\
-$ E2 = (a(b|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 ====