Questions
Languages and Computation (COMP2049 UNNC) (SPC1 24-25) Lab Quiz 4 - Group A
Single choice
Which one is correct?
Options
A.a. If L_1 is a regular language and L_2 is non-regular, then L_1 \cap L_2 is either \emptyset or \{\lambda\}.
B.b. If L_1 \subseteq L_2 and L_2 is a regular language, then L_1 is also regular.
C.c. If L_1 is regular and L_2 is non-regular, then L_1 \cup L_2 is non-regular.
D.d. If L is a non-regular language, then \bar{L} is an infinite language.
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Step-by-Step Analysis
Consider the four statements about language properties and think through their implications carefully.
Option a: 'If L1 is a regular language and L2 is non-regular, then L1 ∩ L2 is either ∅ or {λ}.' In fact, the intersection of a regular language with any language is always regular, because regular languages are closed under intersection with regular languages; here L2 may be non-regular, but the intersection L1 ∩ L2 is a subset of t......Login to view full explanationLog in for full answers
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Let L be a language defined as follows: L = { w | w ∈ {0,1}* && w does not have any 1s that are separated only by 2*n 0's where n ∈ ℕ\{0} } examples: "11", "101", "00", "010" are in L "1001", "10100001" are not in L Which of the following attempts to prove that L is a non-regular language provides a valid fooling set 'S' + algorithm to choose a distinguishing suffix for a pair of elements in S? For each of the following attempts, select the most specific answer from the respective drop-down. Attempt #1: S = { 1, 100, 10000, ...} = { 102m | m ∈ ℕ } ALG = " Given two elements from S 102i and 102j, where i < j, choose suffix 02i1 " Attempt #2: S = { 1, 10, 100, ...} = { 10m | m ∈ ℕ } ALG = " Given two elements from S 10i and 10j, where i < j: IF (i even and j odd || j even and i odd) -> Choose suffix "1" ELSE -> Choose suffix "0i1" " Attempt #3: S = { 0, 10, 100 } ALG = " Choose the suffix according to the following map of element pairings: (0,10) -> choose suffix "01" (0,100) -> choose suffix "1" (10,100) -> choose suffix "1" " 1: Attempt #1 2: Attempt #2 3: Attempt #3
Complete the following statements about the languages given, over the alphabet Σ={b, e}: Let R = e*b*(ebb*)*(ε ∪ e), which is all strings that do not contain “bee”. The language A = L(R) is [ Select ] regular, R is not a correct regular expression not regular, R is not a correct regular expression regular, R is a correct regular expression not regular, R is a correct regular expression . Let B = {bee}. The language B is [ Select ] not regular regular . The language D = {w∈Σ* | w contains the substring bee exactly once} is [ Select ] regular, because D = A◦B◦A and regular languages are closed under concatenation unknown, because D = A◦B◦A, where one of the languages is not regular not regular, because D = A∪B, where one of the languages is not regular regular, because D = A∪B and regular languages are closed under concatenation unknown, because D = A∪B, where one of the languages is not regular not regular, because D = A◦B◦A, where one of the languages is not regular . The language K = {w∈Σ* | w contains n occurrences of the substring bee, where n≥1} is [ Select ] (D◦A◦B)* D* D∪B D*\A DD* , and is therefore [ Select ] unknown because at least one of the languages isn’t regular not regular because regular languages aren't closed under Kleene star not regular because at least one of the languages isn’t regular regular because regular languages are closed under concatenation and Kleene star regular because the regular languages are closed under union not regular because the regular languages aren’t closed under concatenation .
Complete the following statements about the languages given, over the alphabet Σ={b, e}: Let R = e*b*(ebb*)*(ε ∪ e), which is all strings that do not contain “bee”. The language A = L(R) is [ Select ] not regular, R is not a correct regular expression not regular, R is a correct regular expression regular, R is not a correct regular expression regular, R is a correct regular expression . Let B = {bee}. The language B is [ Select ] regular not regular . The language D = {w∈Σ* | w contains the substring bee exactly once} is [ Select ] not regular, because D = A∪B, where one of the languages is not regular unknown, because D = A∪B, where one of the languages is not regular not regular, because D = A◦B◦A, where one of the languages is not regular unknown, because D = A◦B◦A, where one of the languages is not regular regular, because D = A∪B and regular languages are closed under concatenation regular, because D = A◦B◦A and regular languages are closed under concatenation . The language K = {w∈Σ* | w contains n occurrences of the substring bee, where n≥1} is [ Select ] D∪B DD* D* (D◦A◦B)* D*\A , and is therefore [ Select ] regular because regular languages are closed under concatenation and Kleene star not regular because the regular languages aren’t closed under concatenation not regular because regular languages aren't closed under Kleene star not regular because at least one of the languages isn’t regular unknown because at least one of the languages isn’t regular regular because the regular languages are closed under union .
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