Q 2. (a) Consider the following piece of code and answer the following questions:for (i 0;i

for (i = 0; i< n; i++) for(j = 0; j < log2n; j++) print(j) Q. What…

Answered: Q 2. (a) Consider the following piece…

Computer Networking: A Top-Down Approach (7th Edition)

7th Edition

ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

Chapter1: Computer Networks And The Internet

Section: Chapter Questions

Problem R1RQ: What is the difference between a host and an end system? List several different types of end...

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a) You are given two different codes for finding the n-th Fibonacci number. Find the time complexity of both implementations and compare the two. Implementation 1 def fibonacci_1 (n) : if n < 0: print ("Invalid input!") elif n <= 1: return n else: return fibonacci_1 (n-1) +fibonacci_1 (n-2) n = int (input ("Enter a number: ")) nth_fib= fibonacci_1 (n) print("The {}th fibonacci number is {}.". format (n, nth_fib)) Implementation 2 def fibonacci_2 (n) : if n<0: return "Invalid Input" if n<=1: return n fib [0] (n+1) fib[0] = 0 fib[1] = 1 for i in range (2, n+1): fib[i] = fib[i-1]+ fib[i-2] return fib[n] n = int(input ("Enter a number: ")) nth_fib = fibonacci_2 (n) print("The {} th fibonacci number is {}.". format (n, nth_fib))
Trace the following code and write what it does. void Finals(X, Y, m, n) { int i=1, j-1, k=13; while (i <=m && j<=n) { if (X[i]
please include full solutions and provide explanation. will give thumbs up for complete answer
loop 2 is infinite !? : how calculation Time-complexity and worst-case complexity
List two ways to rewrite the following code with code optimization methods for(i=0;i
The time complexity of the following code is O(n^2). In C++, write a code to confirm the time complexity of the following pseudocode: int j = 2 while (j<n) { int k = j while (k<n) { sum += a[k]=b[k] k += n^1/3 log n } j = j*sqrt(5) }
void- find and replace (char A, char *oldc, char newc) for (int i-0; i<5; i++) for (int j%3D0; j<3; j++) == oldc[jl) A[i] = newc[j]: if (A[i] The procedure is tested with the following four test cases. 1. oldc 2. oldc "abc" , newc = "cde", newc = "dab" "bcd" 3. oldc 4. oldc = "abc", newc = "bca", newc= "cda" = "bac"
No plagarism please! Correct and detailed answer will be Upvoted else downvoted. Thank you!
Find the worst-case running time of the following code in "big-Oh" notation: for(i=0; i < n; i++) for(j=0; j<n; j++) for(k = 0; k <m; k++) statement;
Determine how many addition assignment statements are done in the worst case scenario of the following code. Then determine how many multiplication assignment statements are done in the worst case scenario. Assume that all variables are properly declared. y = 1; for (i=1; i<=n;i++) { for (j = 1; j<=n;j++) { x = i + j; x = x * y; } x = y + x; }
*JAVA*The transpose of matrix M is a matrix T such that T[i , j] = M[j , i], for all i and j. a.) Write a parallel program using shared variables that computes the transpose of an n*n matrix M. Use P worker processes. For simplicity, assume that n is a multiple of P. b.) Modify your programs to handle the case when n is not a multiple of P.
Bottom: Given b, set the low-order b bits of x to 1; the others to 0. For example, if b is 3, x should be 7. Pay special attention to the edge cases: if b is 32 x should be −1; if b is 0 x should be 0. Do not use - in your solution. Write code in Java. Provided input(s): b Permitted: 40 operations (may use !, ~, +, <<, >>, &, ^, |) Hint: The obvious solution ~(0xFFFFFFFF << b) won’t work. Bit shifts always do a modulo on their right-hand operand, so a << b does the same thing as a << (b % (8*sizeof(a)). Thus, << 32 and << 0 do the same thing.

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Q 2. (a) Consider the following piece of code and answer the following questions: for (i 0; i