Questions

COMP30026_2025_SM2 Worksheet 6

Multiple dropdown selections

Let Σ = {0, 1} and let A = {w ∈ Σ* | w has no 0's and has odd length}, B = {w ∈ Σ* | w has no 1's and has odd length}. Select the correct expressions to make these equations true: (A ∪ B)* = [ Select ] {w ∈ Σ* | w has an even number of 0's, and an odd number of 1's} {w ∈ Σ* | the length of w is odd} Σ* {w ∈ Σ* | neither “01” nor “10” are substrings of w} (A ◦ A) ◦ A = [ Select ] A \ {1} A {w ∈ Σ* | w has no 0's and has length 6n + 3 for some nonnegative integer n} {w ∈ Σ* | w has no 0's} BC ∩ (B ◦ B)C = [ Select ] {ε} {w ∈ Σ* | w = ε or 1 occurs in w} A* {w ∈ Σ* | 1 occurs in w} Hint: set intersection, union, and complement are Boolean operations!

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Step-by-Step Analysis

We start by restating the given problem components so we can reason about each expression piece by piece. - Σ = {0, 1} - A = { w in Σ* | w has no 0's and has odd length } = {1^k | k is odd} - B = { w in Σ* | w has no 1's and has odd length } = {0^k | k is odd} - The three drop-downs to fill are: 1) (A ∪ B)* = [Select] 2) (A ◦ A) ◦ A = [Select] 3) BC ∩ (B ◦ B)C = [Select] (here ◦ is string concatenation and C denotes complement relative to Σ*, so X C means the complement of X) - The provided answer choices for each dropdown are as given in the prompt, and the user-provided selections are the three specific strings in the answer array. Analysis of option 1 for (A ∪ B)* = [Select] - A ∪ B consists of two kinds of blocks: a block of only 1’s of odd length (from A) and a block of only 0’s of odd length (from B). When you take the Kleene star (A ∪ B)*, you are concatenating any finite sequence of such blocks. - If you concatenate two blocks of different symbols (e.g., 1^odd followed by 0^odd), at the boundary you inevitably create the substrings 10 or 01. Hence, many words in (A ∪ B)* will contain 01 or 10 as substrings. - The only words that avoid both 01 and 10 substrings are precisely the words that consist of a single symbol only (all 0’s, all 1’s), including the empty word ε. Equivalently, these are the words w with no occurrences of 01 or 10. - Therefore, the set of words with no 01 or 10 substrings equals {ε} ∪ 0* ∪ 1*. This is exactly the language of words over {0,1} that contain no switch between symbols. - Among the given options ......Login to view full explanation

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