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匹配题

A

False - Insufficient proof.

B

True - Valid proof.

C

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

D

False - S not a fooling set.

登录即可查看完整答案

我们收录了全球超50000道真实原题与详细解析,现在登录,立即获得答案。

类似问题

更多留学生实用工具

加入我们,立即解锁 海量真题独家解析,让复习快人一步!