3. †(n) = 6+ (n-1) + 4+ (n-2)+4(3")
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- 4.4-1 For each of the following recurrences, sketch its recursion tree, and guess a good asymptotic upper bound on its solution. Then use the substitution method to verify your answer.A(n) __________ contains 8 __________.sum= 0; for (int i = 0; i 1) { sum++; i= 1/2; } = 2*log2 (n) We denote by Ta(n), Tb (n), Te(n) the running time of the three fragments. 1. Give evaluations for Ta(n), Tb (n), Te(n). 2. Is T(n) = O(Ta(n)) ? Answer YES or NO and justify your answer. 3. Is Te(n) = (Ta(n)) ? Answer YES or NO and justify your answer.
- multiplier 3 make_multiplier (3) print (multiplier_2(5)) # Output: 10 print (multiplier_3(5)) 10 15 In [4]: def sum_of_digits (n): if n < 10: else: return n I return n % 10 + sum_of_digits (n // 10) # Example usage: result sum_of_digits (123456222) print("Sum of digits:", result) # Output: 10 In []: Sum of digits: 27 How did it become 27 I want to explain in detail and give different examples to understand the material11 - The code segment below has time complexity? for (int i=0; i* Is (A – B) N (C – B) = (A N C) - B true FalseQ3: Perform the following operation ( choose FIVE ):- 4- (527 + 321 )aint X[900]; int Y[600]; int sum, sum1, sum2, sum3; //parallelism : dividing outer loop in three parts //i = 1 to 300 for(i=1;i<=300;i++) { for(j=1;j<600;j++) { sum1 = X[i] + Y[j]; } } //i = 301 to 600 for(i=301;i<=600;i++) { for(j=1;j<600;j++) { sum1 = X[i] + Y[j]; } } //i = 601 to 900 for(i=601;i<=900;i++) { for(j=1;j<600;j++) { sum1 = X[i] + Y[j]; } } sum = sum1 + sum2 + sum3;} another way to solve the question that send in the picIn questions 4-10 estimate the Big O value by analyzing the code. Note the algorithms are written in English. Hint: you are interested in the number of operations for each algorithm.Python quesle.com/forms/d/e/1FAlpQLSc6PlhZGOLJ4LOHo5cCGEf9HDChfQ-tT1bES-BKgkKu44eEnw/formResponse The following iterative sequence is defined for the set of positive integers: Sn/2 3n +1 ifn is odd if n is even Un = Using the rule above and starting with 13, we generate the following sequence: 13 u13 = 40 u40 =20 u20 = 10→ u10 =5 u5 = 16 u16 = 8 ug = 4 → Us =2 u2 =1. It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. The below function takes as input an integer n and returns the number of terms generated by the sequence starting at n. function i-Seq (n) u=n; i=%3; while u =1 if statement 1 u=u/2; else statement 2 end i=i+1; end statement 1 and statement 2 should be replaced by: None of the choices statement 1 is "mod(u,2)=D%3D0" and statement 2 is "u = 3*u+1;" statement 1 is "u%2" and statement 2 is "u = 3*u+1;" O statement 1 is "mod(n,2)=30" and statement 2 is "u = 3*n+1;"Evaluate the DFT the sequence x[n]= [2 0 0 2], n20SEE MORE QUESTIONS