Formula 1. This recurrence arises for a recursive program that loops through the input to eliminate one item: CN = CN-1+N, for N 2 with C₁ = 1.
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- Question 2 Consider the following algorithm: g1 = 7 g2 = 6 for k in range(3,8): gk = (k-1)·gk-1 + gk-2 What is the last term, g8, of the recursive sequence generated as a result of executing this algorithm? Your Answer: Question 2 options: AnswerExercise 2. Give a recursive definition for the factorial operation k! n! for n ≥ 1. (remember that 1! = 0! = 1) Provide an algorithm in pseudo code to evaluate k! n! as one function Provide an algorithm in pseudo code to evaluate k! n! as three functions Evaluate the complexity of the algorithm at point 2 Evaluate the complexity of the algorithm at point 3Q10. Consider the following algorithm: g1 = 7 g2 = 6 For k starting at 3 and ending with 8: gk = (k-1)·gk-1 + gk-2 What is the last term, g8, of the recursive sequence generated as a result of executing this algorithm?
- A game is played by moving a marker ahead either 2 or 3 steps on a linear path. Let cn be the number of different ways a path of length n can be covered. Given, Cn =Cn-2 + Cn-3, Ci=0, c2=1, c3=1 Write a recursive algorithm to compute Cn.Draw a recursion tree for the following function for myFunc(5), then give the output of the int myFunc(int n){ if (n==1) return 1; else return (myFunc(n-1) + n); }Question 2 Q10. Consider the following algorithm: g1 = 8 g2 = 5 for k in range(3,8): gk = (k-1)·gk-1 + gk-2 What is the last term, g8, of the recursive sequence generated as a result of executing this algorithm?
- Write the pseudocode for a recursive algorithm to compute b3k, where b is a real number and k is a positive integer. Use the fact that b3k+1=(b3k)3.Consider the Sort-and-Count algorithm explained in section 5.3 of our text: "Counting Inversions"Suppose that the initial list is: 92 71 36 91 27 48 14 34 81 26 24 65 78 51 37 22 Sort-and-Count makes two recursive calls. The first recursive call inputs the first half of the initial list: 92 71 36 91 27 48 14 34 and returns the sorted version of the first half, as well as the number of inversions found in the first half (22). The second recursive call inputs the second half of the initial list: 81 26 24 65 78 51 37 22 and returns the sorted version of the second half, as well as the number of inversions found in the second half (19). Sort-and-Count then calls Merge-and-Count. To Merge-and-Count, Sort-and-Count passes the sorted versions of the two halves of the original list: 14 27 34 36 48 71 91 92, and 22 24 26 37 51 65 78 81 Merge-and-Count begins merging the two half-lists together, while counting…Fibonacci sequence is given by the recursive relation:F(0) = 1 and F(1) = 1F(n) = F(n-1) + F(n-2)Ultimately, as discussed in lecture 6, Slide 23, it can be shown that: F(n) =1/√5 ((1+√5)/2)n - 1/√5((1-√5)/2)n The number (1+√5)/2 = 1.618 is called the Golden Ratio Clearly F(n) = O(2n).Write a recursive program like the one described in Lecture 5, Slide 32.Then, perform a numeric asymtotic analysis and try to determine a more precise asymptotic estimate of time needed to calculate F(n) (e.g., use a time function).Esentially, try to estimate/infer the tight bound, (f(n)). More details on Big O and in Lecture 5b.Upload both the program (upload your exact cpp, java, py file - source files) and a pdf file analyzing the output of your program.
- The binomial coefficient C(N,k) can be defined recursively as follows: C(N,0) = 1, C(N,N) = 1, and for 0 < k < N, C(N,k) = C(N-1,k) + C(N - 1,k - 1). Write a function and give an analysis of the running time to compute the binomial coefficients as follows: A. The function is written recursively.g1 = 5 g2 = 6 for k in range(3,8): gk = (k-1)·gk-1 + gk-2 What is the last term, g8, of the recursive sequence generated as a result of executing this algorithm?Assume that for each number I n is not 2. How could the algorithm be modified to handle the situation where n is odd? I have two approaches: one that directly adjusts the recursive method and the other that mixes the iterative and recursive approaches. Just one of the two tasks must be completed (as long as it works and does not increase the BigOh of the running time.)