题目

COMP30026_2025_SM2 2025 sample exam

匹配题

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

选项

A.True - Valid proof.

B.False - Insufficient proof.

C.False - S not a fooling set.

D.False - S is a fooling set, but ALG does not always choose a distinguishing suffix.

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标准答案

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思路分析

We are given a problem about fooling sets for a language L and several attempted proofs that try to show L is non-regular by providing a fooling set S and an algorithm ALG to choose a distinguishing suffix. Option 1 analysis: 'False - S is a fooling set, but ALG does not always choose a distinguishing suffix.' - First, we must assess whether S forms a fooling set for L. The proposed S = {1, 100, 10000, ...} = {102^m | m ∈ ℕ} seems to be designed to pick words that differ in the number of trailing zeros after a leading 1. To be a fooling set, for any two distinct s_i, s_j in S, there must exist a distinguishing suffix z such that exactly one of s_i z and s_j z is in L. The given ALG for this option says: for two elements 102^i and 102^j with i < j, choose suffix 02^i 1. The suitability of this suffix depends on whether appending 0 2^i 1 to both words yields membership differences consistently. However, the structure 102^i 02^i 1 may fail to distinguish certain pairs or may fail to preserve the distinguishing property across all pairs, depending on the exact encoding of 2^i and the alphabet allowed. Without a rigoro......Login to view full explanation

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类似问题

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 ] 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 = 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 .

Which one is correct?

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