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COMP30026_2025_SM2 2025 sample exam - Requires Respondus LockDown Browser

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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 2n 0s where n ∈ ℕ\{0} } examples: "11", "10001", "0110" are in L "1001", "100001" 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? Select the most specific answer from the drop-downs below corresponding to the correctness each of the following proofs. Attempt #1: S = { 11, 1001, 100001, ...} = { 102m1 | m ∈ ℕ } ALG = " Given two elements from S 102i and 102j, where i < j, choose suffix 02i1 " Attempt #2: S = { 1, 110, 11100, ...} = { 1m0m-1 | m ∈ ℕ } ALG = " Given two elements from S 1i0i-1 and 1j0j-1, where i < j, choose suffix 0i1 " Attempt #3: S = { 1, 110, 11100, ...} = { 1m0m-1 | m ∈ ℕ } ALG = " Given two elements from S 1i0i-1 and 1j0j-1, where i < j, choose suffix 02j1 " Attempt #4: S = { 100, 110000, 11100000000, ...} = { 1m02^m | m ∈ ℕ\{0} } ALG = " Given two elements from S 1i02^i and 1j02^j, where i < j, choose suffix 02^i " 1: Attempt #1 2: Attempt #2 3: Attempt #3 4: Attempt #4

Options

A.False - Insufficient proof.

B.True - Valid proof.

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

D.False - S not a fooling set.

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Step-by-Step Analysis

We are evaluating four attempts to construct a fooling set S and a corresponding algorithm (ALG) that picks a distinguishing suffix for pairs of elements in S, to prove that L is non-regular. Option 1: False - Insufficient proof. - The claim tries to use Attempt #1 with S = {11, 1001, 100001, ...} and ALG that, for two elements 102i and 102j with i < j, uses the suffix 02i1. To be a valid fooling-set proof, we need two things: (a) for every i, x_i y_i ∈ L (the concatenation of the pair member with its wildcard counterpart lies in L), and (b) for i ≠ j, x_i y_j ∉ L (the cross-concatenations fail membership). Merely providing a suffix strategy does not verify these two properties for all i, j. Moreover, the strings in S are described in a way that mixes digits (102m1) that are not clearly aligned with the binary alphabet {0,1} in a way that makes checking the L-membership of x_i y_i and x_i y_j nontrivial and potentially incorrect. As a re......Login to view full explanation

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