regular set? Also write R.E for the following and explain,

Add the following list to your definitions: (define my-list (list 3 6 12 24 48))   Finally, use Racket’s built-in fold function, a lambda of your own creation, a base, and the above my-list to produce the exact same output as your circles function. In this image of code and output, the question marks and “.…” indicate what you should figure out:

Remains code.  Implement regular expression matching with support for '.' and '*'. '.' Matches any single character.'*' Matches zero or more of the preceding element. The matching should cover the entire input string (not partial). The function prototype should be:bool is_match(const char *s, const char *p) Some examples:is_match("aa","a") → falseis_match("aa","aa") → trueis_match("aaa","aa") → falseis_match("aa", "a*") → trueis_match("aa", ".*") → trueis_match("ab", ".*") → trueis_match("aab", "c*a*b") → true""" def is_match(str_a, str_b):    """Finds if `str_a` matches `str_b`     Keyword arguments:    str_a -- string    str_b -- string    """    len_a, len_b = len(str_a) + 1, len(str_b) + 1    matches = [[False] * len_b  for _ in range(len_a)]     # Match empty string with empty pattern    matches[0][0] = True     # Match empty string with .*    for i, element in enumerate(str_b[1:], 2):        matches[0][i] = matches[0][i - 2] and element == '*'     for i, char_a in…

Can someone help me with this question in C++ and not any other programming language No library functions that leverage hash and maps can be used except to make it all lowercase  List sentence(s) with the maximum no. of occurrences of the word "the" in the entire file and also list the corresponding frequency. List sentence(s) with the maximum no. of occurrences of the word "of" in the entire file and also list the corresponding frequency. List sentence(s) with the maximum no. of occurrences of the word "was" in the entire file and also list the corresponding frequency. here is a link too the file its called tiny1 https://drive.google.com/file/d/1ariNN_rW9c3vijK1hvjObcQJafYQ2gzm/view?usp=sharing

please do j, k, and l thanks in advanced

In Kotlin, Write a recursive function called myZip, with an expression body. MyZip takes two Lists and returns a List of Pairs in which each value may be any type (I suggest your start by thinking out how to represent this data type). The pairs consist of corresponding elements in the two lists (the first element of the first list and the first element of the second list, etc). The base case should be that either (or both) of the original lists has length 1, so that, if the lists have different length, the zipping stops when the first list runs out of values.

Q1. Implement regular expression matching with support for '.' and '*'. '.' Matches any single character.'*' Matches zero or more of the preceding element. The matching should cover the entire input string (not partial). The function prototype should be:bool is_match(const char *s, const char *p) Some examples:is_match("aa","a") → falseis_match("aa","aa") → trueis_match("aaa","aa") → falseis_match("aa", "a*") → trueis_match("aa", ".*") → trueis_match("ab", ".*") → trueis_match("aab", "c*a*b") → true""" def is_match(str_a, str_b):    """Finds if `str_a` matches `str_b`     Keyword arguments:    str_a -- string    str_b -- string    """    len_a, len_b = len(str_a) + 1, len(str_b) + 1    matches = [[False] * len_b  for _ in range(len_a)]     # Match empty string with empty pattern    matches[0][0] = True     # Match empty string with .*    for i, element in enumerate(str_b[1:], 2):        matches[0][i] = matches[0][i - 2] and element == '*'     for i, char_a in enumerate(str_a, 1):…

Write a function filter_codes(list_codes) that receives a list of codes and returns another list containing those codes that:    Only have numeric characters; and Are at least 2 characters long; and At most 6 characters long Hint: remember that Python has a built-in function for determining if a string only contains numeric characters. solve using python

I need correct answer unique answer for this. Don't copy from anywhere otherwise I'll report your account for sure. Last time I received wrong answer for this. This time i need right answer. Do not send wrong and plagiarism response