Questions

COMP30026_2025_SM2 Worksheet 7

Multiple dropdown selections

Complete the following statements about the languages given, over the alphabet Σ={b, e}: Let R = eb(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 .

View Explanation

Verified Answer

Please login to view

Step-by-Step Analysis

We are asked to complete several statements about languages over the alphabet Σ = {b, e} and specific constructions R, B, D, K. I will go option by option, explaining why each statement is right or wrong, using standard automata/regular-language reasoning. First statement set (about A and R): Option 1: "regular, R is a correct regular expression". Here, R = e*b*(ebb*)*(ε ∪ e) is presented as all strings that do not contain the substring bees. The claim is that A = L(R) is regular and that R is a correct regular expression for it. Explanation: The language of all strings over {b,e} that avoid the fixed forbidden substring bee is a standard regular-language set; such avoidance can be captured by a regular expression or by a finite automaton that tracks partial matches of bee and rejects when a full bee is seen. The given pattern e*b*(ebb*)*(ε ∪ e) attempts to encode this avoidance; while a detailed construction would be needed to verify every edge case, in general the class of strings not containing a given substring is regular, and R is a syntactically valid regular expression. Therefore this option is plausible and aligns with regular-language theory. Option 2: "not regular, R is not a correct regular expression". This would claim the language avoiding bee is nonregular, which contradicts standard results: for any fixed forbidden substring, the set of strings avoiding it is regular. Additionally, if R is intended to describe A, R as written is a legitimate regexp form, though one could debate minor formatting details; in any case, the mainstream view is that the avoidance language is regular, so this option is incorrect. Option 3: "not regular, R is a correct regular expression". This claims R is a correct regexp but the language is nonregular. As explained above, the avoidance-of-bee language is regular; thus the combination of a regular language with a statement of nonregularity is inconsistent. This option is therefore incorrect on two counts. Option 4: "regular, R i......Login to view full explanation

Log in for full answers

We've collected over 50,000 authentic exam questions and detailed explanations from around the globe. Log in now and get instant access to the answers!