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| Both sides previous revision Previous revision Next revision | Previous revision | ||
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lfa:2022:lab03-dfa-regexp [2022/10/23 15:47] mihai.udubasa |
lfa:2022:lab03-dfa-regexp [2022/10/28 22:42] (current) alexandra.udrescu01 |
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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( ($1^n$)* ) |
| </hidden> | </hidden> | ||
| Line 78: | Line 78: | ||
| **3.1.4** | **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 \} $ | $ 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> | </hidden> | ||
| Line 117: | Line 117: | ||
| </hidden> | </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} \} $ | **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> | <hidden> | ||
| Line 131: | Line 133: | ||
| </hidden> | </hidden> | ||
| + | |||
| **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 )) $ | ||
| + | |||
| <hidden> | <hidden> | ||
| {{:lfa:2022:lfa2022_lab_3.2.3.png?300|}} | {{:lfa:2022:lfa2022_lab_3.2.3.png?300|}} | ||
| </hidden> | </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. | **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|}} | {{:lfa:graf1.png?400|}} | ||
| + | |||
| <hidden> | <hidden> | ||
| Line 147: | Line 153: | ||
| 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> | ||
| Line 164: | Line 170: | ||
| </hidden> | </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) | (hint: BCD) | ||
| + | |||
| <hidden> | <hidden> | ||
| Line 195: | Line 203: | ||
| So, this ugly regex encodes BCD numbers that start with exactly 2 odd digits which are followed by 0 or more even digits. | So, this ugly regex encodes BCD numbers that start with exactly 2 odd digits which are followed by 0 or more even digits. | ||
| </hidden> | </hidden> | ||
| + | |||
| ===== 3.3 Regex Equivalence ===== | ===== 3.3 Regex Equivalence ===== | ||
| Line 205: | Line 214: | ||
| \\ | \\ | ||
| $ E2 = (a|\epsilon)(b|\epsilon) $ | $ E2 = (a|\epsilon)(b|\epsilon) $ | ||
| + | |||
| <hidden><note important> | <hidden><note important> | ||
| Line 211: | Line 221: | ||
| 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> | ||
| + | |||
| ** 3.3.2 ** | ** 3.3.2 ** | ||
| Line 217: | Line 228: | ||
| \\ | \\ | ||
| $ E2 = ab(b|c|d|e)|acd|ace $ | $ E2 = ab(b|c|d|e)|acd|ace $ | ||
| + | |||
| <hidden><note important> | <hidden><note important> | ||
| Line 222: | Line 234: | ||
| 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> | ||
| + | |||
| ** 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)^* $ |
| + | |||
| <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. | ||
| Line 235: | Line 250: | ||
| 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> | ||
| + | |||
| + | |||
| ** 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)^* $ |
| <hidden><note important> | <hidden><note important> | ||
| Line 249: | Line 268: | ||
| Plot twist: They are the same. | 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. | + | The purpose of this exercise is to understand how to approach regex equivalence, not how to solve this given comparison per |
| + | se. | ||
| </note></hidden> | </note></hidden> | ||
| + | |||
| ==== Conclusion ==== | ==== Conclusion ==== | ||