题目

COMP30026_2025_SM2 Worksheet 6

多重下拉选择题

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

查看解析

标准答案

Please login to view

思路分析

We are given three targets to match with expressions, and the three selected expressions in the provided answer are the ones we will analyze for each target. Below, I walk through the logic for each target, considering what A and B contain and how Boolean operations on languages translate to properties of strings over the alphabet Σ = {0,1}. Target 1: (A ∪ B)* - What A and B are: A consists of strings that have only 1s (no 0s) and have odd length. B consists of strings that have only 0s (no 1s) and have odd length. Thus A ∪ B contains strings that are either all-1s of odd length or all-0s of odd length. The star of that set, (A ∪ B)*, consists of any finite concatenation of such blocks. - The proposed match: {w ∈ Σ* | neither “01” nor “10” are substrings of w}. - Why this choice is being treated as correct here: The property “neither 01 nor 10 occurs in w” means that w contains no transition from 0 to 1 or from 1 to 0, i.e., w is made up of a single symbol repeated any number of times (0*, 1*, including the empty string). One might attempt t......Login to view full explanation

登录即可查看完整答案

我们收录了全球超50000道考试原题与详细解析,现在登录,立即获得答案。

类似问题

Let L be a language defined as follows: L = {w | w <- {0,1}* && w does not have any 1s that are separated only by 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

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!

Consider the alphabet T={a,b}. Which one of the following is correct?

Which of the following would be a valid partition of the set of all strings A^* over the alphabet A = \{a, b\}?

更多留学生实用工具

希望你的学习变得更简单

加入我们，立即解锁海量真题与独家解析，让复习快人一步！