Let L be the set of strings (N, w) where N is an NFA that accepts the string w. Pick all that apply. L is decidable L recognisable. None of the above.
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- def reverse_sentence(s: str) -> str:"""Given a sentence <s>, we define a word within <s> to be a continuoussequence of characters in <s> that starts with a capital letter andends before the next capital letter in the string or at the end ofthe string, whichever comes first. A word can include a mixture ofpunctuation and spaces.This means that in the string 'ATest string!', there are in fact only twowords: 'A' and 'Test string!'. Again, keep in mind that words start with acapital letter and continue until the next capital letter or the end of thestring, which is why we consider 'Test string!' as one word.This function will reverse each word found in the string, and return a newstring with the reversed words, as illustrated in the doctest below.>>> reverse_sentence('ATest string!')'A!gnirts tseT' """ do it on python, and do not use any list, import or split, or append or join.def reverse_sentence(s: str) -> str:"""Given a sentence <s>, we define a word within <s> to be a continuoussequence of characters in <s> that starts with a capital letter andends before the next capital letter in the string or at the end ofthe string, whichever comes first. A word can include a mixture ofpunctuation and spaces.This means that in the string 'ATest string!', there are in fact only twowords: 'A' and 'Test string!'. Again, keep in mind that words start with acapital letter and continue until the next capital letter or the end of thestring, which is why we consider 'Test string!' as one word.This function will reverse each word found in the string, and return a newstring with the reversed words, as illustrated in the doctest below.>>> reverse_sentence('watchNow')'watchwoN' >>> reverse_sentence('hot')'hot' """ do it on python, and do not use any list, import or split, or append or join.def reverse_sentence(s: str) -> str:"""Given a sentence <s>, we define a word within <s> to be a continuoussequence of characters in <s> that starts with a capital letter andends before the next capital letter in the string or at the end ofthe string, whichever comes first. A word can include a mixture ofpunctuation and spaces.This means that in the string 'ATest string!', there are in fact only twowords: 'A' and 'Test string!'. Again, keep in mind that words start with acapital letter and continue until the next capital letter or the end of thestring, which is why we consider 'Test string!' as one word.This function will reverse each word found in the string, and return a newstring with the reversed words, as illustrated in the doctest below.>>> reverse_sentence('ATest string!')'A!gnirts tseT'""" pass
- def reverse_sentence(s: str) -> str:"""Given a sentence <s>, we define a word within <s> to be a continuoussequence of characters in <s> that starts with a capital letter andends before the next capital letter in the string or at the end ofthe string, whichever comes first. A word can include a mixture ofpunctuation and spaces.This means that in the string 'ATest string!', there are in fact only twowords: 'A' and 'Test string!'. Again, keep in mind that words start with acapital letter and continue until the next capital letter or the end of thestring, which is why we consider 'Test string!' as one word.This function will reverse each word found in the string, and return a newstring with the reversed words, as illustrated in the doctest below.>>> reverse_sentence('ATest string!')'A!gnirts tseT' """Let A be the set of all strings of decimal digits of length 5. For example 00312 and 19483 are two strings in A . You pick a string from A at random. What is the probability that the string has no 4 in it?Let the universal set be the set ? of all real numbers and let ? = {x ∈ r |0 < x ≤ 2},? = {x ∈ r |1 ≤ x < 4}, ? = {x ∈ R |3 ≤ x < 9}. 4 what is A U C
- The reverse of a string x denoted by rev(x) and is defined by the following recursiverule: •rev(ε) = ε•rev(xa) = a.rev(x) Prove that if A is a regular set then so is the following set:rev(A) = {rev(x) : x ∈A}Given a string S, find the longest palindromic substring in S. Substring of string S: S[ i . . . . j ] where 0 ≤ i ≤ j < len(S). Palindrome string: A string which reads the same backwards. More formally, S is palindrome if reverse(S) = S. Incase of conflict, return the substring which occurs first ( with the least starting index). Example 1: Input: S = "aaaabbaa" Output: aabbaa Explanation: The longest Palindromic substring is "aabbaa".With simply the at, length, and substr string methods and the + (concatenate) operator, construct a function that accepts a string s, a start position p, and a length l and returns a subset of s without the characters starting at position p. Strings begin at 0. (“abcdefghijk”, 2, 4) yields “abghijk”.
- Suppose that B is the infinite set of strings made from one or more a’s and b’s. B = { "a", "b", "aa", "ab", "ba", "bb", "aaa", "aab", "aba" ... } Prove that B is countable. Your proof must work for all strings in B. Hints: make sure it works for strings like "a", "aa", "aaa", etc. Also make sure it works for strings like "ab", "aab", "aaab", etc.Implement a method String[] dedup(String[] a) that returns the objects ina[] in sorted order, with duplicates removed.For this problem, please kindly show your work Let ?L be a set of all finite binary strings that representing a binary number divisible by 2 (or 102102). What is a REGEX of ?L? Design an FA ?a such that ?a accepts ?L, i.e., ?=?(?)