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--- lfa:2022:lab03-dfa-regexp [2022/10/07 20:06]
alexandra.udrescu01
+++ lfa:2022:lab03-dfa-regexp [2022/10/28 22:42] (current)
alexandra.udrescu01
@@ Line 30: / Line 30: @@
 $ B = \{ 1^n \mid n \geq 1 \} $
 \\
-$ AB = ? $ \\ $ BA = ?
+$ AB = ? $ \\ $ BA = ? $
 <hidden>
@@ Line 70: / Line 70: @@
 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( ($1^n$)* )
 </hidden>
@@ Line 78: / Line 78: @@
 **3.1.4**
-$math[A = \{ 0^n 1^n 2^m \mid m \geq n  \geq 1 \}]
+$math[A = \{ 0^n 1^n 0^m \mid m \geq n  \geq 1 \}]
 \\
 $ B = \{ 0^n \mid n \geq 1 \} $
@@ Line 87: / Line 87: @@
-$math[AB = \{ 0^n 1^n 2^m 0^k \mid m \geq n  \geq 1, k \geq 1 \}]
+$math[AB = \{ 0^n 1^n 0^{m+k} \mid m \geq n  \geq 1, k \geq 1 \}]. Deci $math[AB = A].
-Note that the n in the definition of language A **is different** from the n in in the definition of B, they are **independent** when used in defining diffrent sets/languages. However, when n is used several times in the definition of one language, such as the 2 times it appears in langauge A, it is **the same** value.
+Note that the n in the definition of language A **is different** from the n in in the definition of B, they are **independent** when used in defining different sets/languages. However, when n is used several times in the definition of one language, such as the 2 times it appears in language A, it is **the same** value.
-$math[BA = \{ 0^{(n+k)} 1^n 2^m \mid m \geq n  \geq 1, k \geq 1 \}]
+$math[BA = \{ 0^{(n+k)} 1^n 0^m \mid m \geq n  \geq 1, k \geq 1 \}]. Equivalently: $math[BA = \{0^x 1^y 0^z \mid x \geq y\geq 1 \text{ and } z \geq y \geq 1 \}]
 </hidden>
@@ Line 117: / Line 117: @@
 </hidden>
 **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} \} $
 <hidden>
@@ Line 131: / Line 133: @@
 </hidden>
 **3.2.3.** Write a DFA for $ L(( 10 \cup 0) ^* ( 1 \cup \epsilon )) $
 <hidden>
 {{:lfa:2022:lfa2022_lab_3.2.3.png?300|}}
 </hidden>
 **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.
 {{:lfa:graf1.png?400|}}
 <hidden>
@@ Line 147: / Line 153: @@
 Let's see what words are accepted:
   * 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
   * ε (the initial state is also a final state)
   * from the previous 2 observations => we can use Kleene star
  So, the regex is:
-> ( ab*(aab*)ab )*
+> ( ab*(aab*)*ab )*
 <note>
@@ Line 164: / Line 170: @@
 </hidden>
-**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 $math[((0(1 \cup 0)(1 \cup 0)) \cup 100)1((0(1 \cup 0)(1 \cup 0)) \cup 100)1(((0(1 \cup 0)(1 \cup 0)) \cup 100)0)*]
 (hint: BCD)
 <hidden>
@@ Line 176: / Line 184: @@
 Ugly, yes. But let's see the language of this little regex:
+, 0011, 0101, 0111, 1001
+What are these in BCD?
+, 3, 5, 7, 9.
+So, we got so far BCD numbers that start with an odd digit, BUT this pattern repeats twice. So, we have BCD numbers starting with 2 odd digits.
+NEXT:
-What are these in BCD?
+( ( 0 ( 1 U 0 ) ( 1 U 0 ) ) U 100 ) 0 )*
+We have a Kleene star, so something repeats any number of times.
+And, inside this Kleene star, there is ( 0 (1 U 0) (1 U 0) ) U 100 )0 that accepts the language:
+, 0010, 0100, 0110, 1000.
+So, this ugly regex encodes BCD numbers that start with exactly 2 odd digits which are followed by 0 or more even digits.
+</hidden>
-.
+===== 3.3 Regex Equivalence =====
-So, we got so far BCD numbers that start with an odd digit, BUT this pattern repeats twice. So BCD numbers starting with 2 odd digits.
+Are the following regex pair equivalent?
-NEXT:
+** 3.3.1 **
-( ( 0 ( 1 U 0 ) ( 1 U 0 ) ) U 100 ) 0 )*
+$ E1 = ab|a|b $
+\\
+$ E2 = (a|\epsilon)(b|\epsilon) $
-We have a Kleene star, so something repeats any number of times.
-And, inside this Kleene star, there is ( 0 (1 U 0) (1 U 0) ) U 100 )0 that accepts the language:
+<hidden><note important>
+We can observe that E2 accepts ε, while E1 does not so they are not equivalent.
+\\
+Another approach is to compute the language of each expression (since they are finite) and check if they are equivalent.
+</note></hidden>
+** 3.3.2 **
+$ E1 = a(b|c)(d|e)|abb|abc $
+\\
+$ E2 = ab(b|c|d|e)|acd|ace $
-.
-So this ugly regex encodes BCD numbers that start with exactly 2 odd digits which are follwed by 0 or more even digits.
+<hidden><note important>
-</hidden>
+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.
+</note></hidden>
+** 3.3.3 **
+$ E1 = (a\mid b)^*aa^* \mid \epsilon $
+\\
+$ E2 = (a\mid ba)^*(b\mid ba)^* $
+<hidden><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>
+** 3.3.4 **
+$ E1 = ((ab^*a)^+b)^* $
+\\
+$ E2 = (a(b\mid aa)^*ab)^* $
+<hidden><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>
-===== Extra HARD - click here if you want nightmares =====
+==== Conclusion ====
 <hidden><note important>
 What have we learned today?