If L1 and L2 are regular languages, pick which of the following operations produce a regular language:多项选择题
A
L1 ^R - L2 ^R
B
(L1 - L2)^R U L1 ^R
C
None is correct.
D
L2 L1 L2
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类似问题
A language L is regular if: (Pick all true statements)
Both the Deterministic and Non-Deterministic finite automata are equivalent in expressive power to the language of regular expressions.
Pick all non-regular languages: Notes Assume they are defined over the appropriate Sigma. Spaces are added to the patterns for readability purpose.
Pick all true statements about formal languages:
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 .
Which one is correct?
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