Question

1. (10 points) Write an efficient iterative (i.e., loop-based) function Fibonnaci(n) that returns the nth Fibonnaci number. By definition Fibonnaci(0) is 1, Fibonnaci(1) is 1, Fibonnaci(2) is 2, Fibonnaci(3) is 3, Fibonnaci(4) is 5, and so on. Your function may only use a constant amount of memory (i.e. no auxiliary array). Argue that the running time of the function is Θ(n), i.e. the function is linear in n. 2. (10 points) Order the following functions by growth rate: N, \N, N1.5, N2, N log N, N log logN, N log2 N, N log (N2), 2/N, 2V, 2N/2, 37, N2 log N, N3. Indicate which functions grow at the same rate. 3. (10 points) Compute the running time T(n) of the program fragment below and provide an analysis of the running time (Big-Oh notation will do). For convenience, assume that operations inside for loops take constant time, i.e. Θ(1) sum = 0 for( i-0; i<n; i++) for j0; 3 for (k=0; sum++ k 〈 j, k++) 4. (10 points) An algorithm takes 0.5ms for input size 100. How long will it take for input size 500 if the running time is the following (assume low-order terms are negligible): linear, O(n log n), quadratic, and cubic? 5. (10 points) Give an O(logN) algorithm to determine if there exists an integer i such that Ai-i in an array of integers A1くA·Ag < … < AN. For example, in {-10,-3, 3, 5, 7}, A,-3. In {2, 3, 4, 5, 6, 7), there 1s no such 6. (10 points) Show that X62, can be computed with only eight multiplications. 7. (10 points) Write a recursive version of the binary search function to find if x is in A 8. (10 points) Given two functions T1(n) = 20n and T2 (n) = n2 + 10n-200 and assuming integer n > 0, what is the minimum value of n such that T1 (n) < T2 (n)? 9. (10 points) Is a linear-time algorithm always faster than a quadratic-time algorithm? Why or why not? 10. (10 points) Explain why recursion should never be used as a substitute for a simple for loop?

Answer 1

If you post ore than 1 question, then as HOMEWORKLIB RULES guidelines I just have to answer one question.

1.

function Fibonnaci(n):

{

    // first term of series

    if n == 0

        return 1

   

    if n == 1

        return 2

    a = 1

    b = 2

    for i <- 2 to n

    {

        c = a + b

        a = b

        b = c

    }

    return b

}

The above function computes n th fibonnaci number in

        O(n) Time Complexity

and     O(1) Space Complexity

It is so because the above function checks the two conditions for n == 0 and n == 1 in O(1) time.

The for loop runs for n - 2 times.

Hence, Time Complexity = O(1) + O(1) + O(n - 2)

                       = O(n - 2) (as O(1) is negligible)

                       = O(n) (as (n - 2) = n if n is very large)