6. For the same processor as problem 4, answer the same questions for the following C code. int N = 1000; int A[N] [N], B[N] int i, j; for (i=0; i< N; i++) { int temp = 0; for (j=0; j < N; j++) { temp = A[i][j] * B[j] } C[i] temp; Instruction class CPI #instructions x10 1 add mul load/store branch 222 Total Cycles x10 Cycles Percentage

6. For the same processor as problem 4, answer the same questions for the following C code. int N = 1000; int A[N] [N], B[N] int i, j; for (i=0; i< N; i++) { int temp = 0; for (j=0; j < N; j++) { temp = A[i][j] * B[j] } C[i] temp; Instruction class CPI #instructions x10 1 add mul load/store branch 222 Total Cycles x10 Cycles Percentage

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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2. The function f is defined for non-negative integers a and b recursively as follows: f (а, b) 3D1 if a = 0 or b = 0 f (a - 1, b f (a f (а, а) 1) + 2a - 1 if a if a > b if a < b b b, b) + f (b, b) + f (b а, а) } Compute f (2, 3) by drawing a recursion tree showing all of the computation required and then use your tree to compute the answer.
If f(n) is the number of times that y=y+A[i]; is executed in the code segment below, which of the following is the correct big-Oh notation for f? y=0;for(j=0;j<n;j++) { for (i=0;i<n;i++) { y=y+A[i]; }} Group of answer choices O(1) O(n2) O(nlog2n) O(n)
c) Estimate the time complexity for the following C++ program: //IIIII// Start //////////// #include void BSort(int A[n][n]) { int temp; for(int j=0 ; j A[i+1]) { const int n=10; int Fact (int); void BSort(int A[n][n]); int main() { int A[n][n]; for(int i=0;i>A[i][j]; temp=A[i]; A[i]=A[i+1]; A[i+1]=temp; } } BSort(A); cout<<"Factorial = "<
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))
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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) }
Evaluate the performance per line
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1. Analyze the time complexity of the code segment and find their Big-O.void myfunction1(int n){for(int i=0; i < n; i++) {for(int j=0; j < n; j++) {for(int k=0; k < n; k++) {for(int m=0; m < n; m++) {printf("Hello!");} } } }}
Q5. Find the time complexity of the following code int b = 0; for (j = 0; jj; i--) { b=b+j+i;