题目
单项选择题
Consider the alphabet T={a,b}. Which one of the following is correct?
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标准答案
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思路分析
The question asks us to consider the alphabet T = {a, b} and determine which statement is correct among the given options.
First, I note that the provided data includes an answer label: 'c. For every finite language over T, the complement is infinite.' but the list of answer options is empty, so we cannot directly evaluate each option as presented. Nevertheless, I can analyze the general statement about finite languages over a finite alphabet and t......Login to view full explanation登录即可查看完整答案
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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!
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!
Which of the following would be a valid partition of the set of all strings A^* over the alphabet A = \{a, b\}?
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