Questions
Single choice
Which of the following would be a valid partition of the set of all strings A^* over the alphabet A = \{a, b\}?
Options
A.a. \{\{\epsilon\},\{\text{all strings starting with a}\}, \{\text{all strings starting with b}\}\}
B.b. \{\{\text{aa}\},\{\text{ab}\}, \{\text{ba}\}, \{\text{bb}\}\}
C.c. There is no way of partitioning the set, because it's an infinite set.
D.d. \{ A^n : n\in\mathbb{N} \}
View Explanation
Verified Answer
Please login to view
Step-by-Step Analysis
To tackle this question, I will examine what constitutes a partition of A* for A = {a, b} and then evaluate each option.
Option a: {{ε}, {all strings starting with a}, {all strings starting with b}}
- First, note that ε is included in the first block, which is appropriate since ε does not start with a or b.
- The second block contains all strings that begin with the letter a, regardless of what comes after. The third block contains all strings that begin wit......Login to view full explanationLog in for full answers
We've collected over 50,000 authentic exam questions and detailed explanations from around the globe. Log in now and get instant access to the answers!
Similar Questions
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!
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!
Consider the alphabet T={a,b}. Which one of the following is correct?
More Practical Tools for Students Powered by AI Study Helper
Making Your Study Simpler
Join us and instantly unlock extensive past papers & exclusive solutions to get a head start on your studies!