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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 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

Options

A.False - S not a fooling set.

B.False - Insufficient proof.

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

D.True - Valid proof.

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Question restatement: - We are given a language L over {0,1} where w ∈ L if and only if w has no occurrence of two 1s that are separated by exactly 2n zeros for any n ∈ ℕ \ {0}. In other words, we must avoid patterns 1 0^{2n} 1 for any n ≥ 1 within w. The task presents three attempts to form a fooling set S and an accompanying algorithm (ALG) to choose a distinguishing suffix for a pair of elements in S, and asks, for each attempt, to select the most specific matching description from the provided drop-down options. The drop-down options are: - False - S not a fooling set. - False - Insufficient proof. - False - S is a fooling set, but ALG does not always choose a distinguishing suffix. - True - Valid proof. Attempt #1: S = { 1, 100, 10000, ...} = { 1 0^{2m} | m ∈ ℕ } ALG = " Given two elements from S 102i and 102j, where i < j, choose suffix 02i1. " Analysis for Attempt #1: - The set S here consists of strings that start with 1 followed by an even number (2m) of zeros. This construction aims to create a family of prefixes that interact with the no-1s-separated-by-2n-zeros constraint in a way that allows a distinguishing continuation. - A fooling set requires that for any two distinct members x and y of S, there exists a suffix z (which may depend on the chosen pair (x,y)) such that exactly one of xz or yz is in L. The proposed ALG selects a fixed-form suffix depending on i and j (specifically 02^i 1 in the notation given). To be valid, this suffix must distinguish every pair (1 0^{2i}, 1 0^{2j}) with i ≠ j. - Potential issues: (a) The suffix 02i1 must be interpreted precisely; if the intended suffix is 0^{2i} 1 or 0 2^i 1, the meaning changes. Either interpretation must be checked against the actual property of L. (b) It is not immediately obvious that appending 0^{2i} 1 to both 1 0^{2i} and 1 0^{2j} yields one string in L and the other not ......Login to view full explanation

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