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- Regular expression to NFA to DFA conversion: Describes the process of taking a unique regular expression, converting that regular expression to an NFA, then converting the NFA into a DFA. Your regular expression must have at minimum two union, two concatenation, and two Kleene star operations. As followed, concatenations of single charaters can be condensed. Your regular expression cannot be a solved problem from any book. You should describe the regular expression with both processes to convert the regular expression to an NFA and the conversion of that NFA into a DFA.Implement the quadratic_formula() function. The function takes 3 arguments, a, b, and c, and computes the two results of the quadratic formula: x1=−b+b2−4ac2a x2=−b−b2−4ac2a The quadratic_formula() function returns the tuple (x1, x2). Ex: When a = 1, b = -5, and c = 6, quadratic_formula() returns (3, 2). Code provided in main.py reads a single input line containing values for a, b, and c, separated by spaces. Each input is converted to a float and passed to the quadratic_formula() function. Ex: If the input is: 2 -3 -77 the output is: Solutions to 2x^2 + -3x + -77 = 0 x1 = 7 x2 = -5.50 code: # TODO: Import math module def quadratic_formula(a, b, c):# TODO: Compute the quadratic formula results in variables x1 and x2return (x1, x2) def print_number(number, prefix_str):if float(int(number)) == number:print("{}{:.0f}".format(prefix_str, number))else:print("{}{:.2f}".format(prefix_str, number)) if __name__ == "__main__":input_line = input()split_line = input_line.split(" ")a =…Complete the function ratio_similarity, which computes the ratio of "similarity" between two non-empty strings. The ratio is computed as follows: 2 (the number of matches of characters) / (total number of characters in both strings) The matches are determined in pairwise fashion. For example, consider the strings abcdef and abdce. The three characters at indexes 0, 1 and 4 match. Therefore, the similarity ratio is 2 * 3/(6 + 5) = 0.5454 .... Examples: Function Call string_similarity('abcdef', 'abdce') string_similarity('dragon', 'flagon') string_similarity('stony', 'stony') string_similarity('Stony Brook University', 'WolfieNet Secure') string_similarity('yabba dabba doo', 'bippity boppity boo') string_similarity('Wolfie the Seawolf', "What's...a Seawolf?") [ ] Return Value 3 4 5 # Test cases 6 print (string_similarity('abcdef', 'abdce')) 7 print (string_similarity('dragon' 1 'flagon')) 'stony')) 0.5454545454545454 0.6666666666666666 1.0 0.0 0.058823529411764705 0.4864864864864865 1…
- function result = result = ''; 1 tokens_to_str_code(token_mat, time_unit) 3 for i =1:size(token_mat,1) 4 5 if(token_mat (i,1) if(token_mat(i,2) > 4*time_unit) result = strcat(result,"-"); elseif((token_mat(i,2) > time_unit) && (token_mat(i,2) 8*time_unit) result = strcat(result,"/"); elseif((token_mat(i,2) 11 == 0) 12 13 14 4*time_unit) && (token_mat(i,2) < 8*time_unit)) 15 result strcat(result," "); %3D 16 end 17 end 18 end 19 end 20 21 Check Test Expected Got tokens %3D [1 4; е 3; 1 3; е 1; 1 3; е 1; 1 1; ө 1; 1 1 ]; time_unit = .5; disp( tokens_to_str_code( tokens, time_unit )) tokens = [ 0 4; 1 1457; 0 463; 1 1457; 0 463; 1 497; 0 1423; 1 1457; ---/..--.. --.---/..--.. о 463; 1 1457; 0 463; 1 1457; ө 3343; 1 497; ө 463; 1 497; 0 463; 1 1457; 0 1423; 1 1457; 0 463; 1 497; 0 463; 1 497; e 379 ]; time_unit = 240; disp( tokens_to_str_code( tokens, time_unit ))Create a function which counts how many lone 1's appear in a given number. Lone means the number doesn't appear twice or more in a row. Examples count LoneOnes (101) ➡2 count LoneOnes (1191) → 1 count LoneOnes (1111) count LoneOnes (462) 0Ex: Let A1 ={x, y}, A2 ={1, 2}, and A3 ={a, b}, Find A1 × A2, (A1 × A2) × A3, A1 × A2 × A3.
- Write a function in C programming language that gets two ints x and y, and returns the sum of all ints between x and y, including them. int sum_interval(int x, int y); For example: - sum_interval(1, 4)returns 1+2+3+4=10. - sum_interval(10, 3)returns 3+4+5+6+7+8+9+10=52. - sum_interval(2, -1)returns (-1)+0+1+2=2. - sum_interval(-1, -1)returns -1. Note that we may have x>=y or x<=y. You may assume that the sum will be within the range of int.Up for the count def counting_series (n): The Champernowme word 1234567891011121314151617181920212223., also known as the counting series, is an infinitely long string of digits made up of all positive integers written out in ascending order without any separators. This function should return the digit at position n of the Champernowne word. Position counting again starts from zero for us budding computer scientists. Of course, the automated tester will throw at your function values of n huge enough that those who construct the Champernowne word as an explicit string will run out of both time and space long before receiving the answer. Instead, note how the fractal structure of this infinite sequence starts with 9 single-digit numbers, followed by 90 two-digit numbers, followed by 900 three-digit numbers, and so on. This observation gifts you a pair of seven league boots that allow their wearer skip prefixes of this series in exponential leaps and bounds, instead of having to crawl…1. Determine |A|, where: {z} b. A = {{z}} _d. A = {z, {z}, {z, {z}}} _e. A = P({z}) _f. A = P({Ø,z}) a. A = _c. A = {z, {z}}
- 2.a Σ : {c,A,G,T}, L = { w : w = CAG™T™C, m = j + n }. For example, CAGTTC E L; CTAGTC ¢ L because the symbols are not in the order specified by the characteristic function; CAGTT ¢ L because it does not end with c; and CAGGTTC € L because the number of T's do not equal the number of A's plus the number of G's. Prove that L¢ RLs using the RL pumping theorem.Please design another function called: multiply_up (int n), which will multiple up from 1 to n, like 1*2*3*4*5....*n, and return the result.Write a generator function named count_seq that doesn't require any arguments and generates a sequence that starts like this: 2, 12, 1112, 3112, 132112, 1113122112, 311311222112, 13211321322112, ...To get a term of the sequence, count how many there are of each digit (in a row) in the **previous** term. For example, the first term is "one 2", which gives us the second term "12". That term is "one 1" followed by "one 2", which gives us the third term "1112". That term is "three 1" followed by "one 2", or 3112. Etc.Your generator function won't just go up to some limit - it will keep going indefinitely. It may need to treat the first one or two terms as special cases, which is fine. It should yield the terms of the sequence as **strings**, rather than numeric values, for example "1112" instead of 1112.