To observe the importance of algorithm efficiency and the order of growth,…

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To observe the importance of algorithm efficiency and the order of growth, you are required to perform the following experiment. Assume an input of size 100,000 (denoted as N) (a) Linear function (O(N)): Write a function that loops over the input items (a single loop) and print the elapsed time taken by the function in milliseconds (b) Quadratic function (O(N2)): Write a function that loops over the input items with two-levels of nesting and print the elapsed time taken by the function in milliseconds. The function should have two nested loops (outer and inner) where each one starts from 0 to N (c) Cubicfunction (O(N3): Write a function that loops over the input items with three-levels of nesting and print the elapsed time taken by the function in milliseconds. The function should have three nested loops where each one starts from 0 to N Hint: The actual code inside the loop is not important. So make it simple and define a double variable in the beginning of your program, and then increment this variable inside the inner most loop (The example below is for two-level nesting). Double sum = 0; For (j=0, End Loop; j<=N ; j++) End Loop; Deliverables of Question I (1) A single executable program. The program should contain the three functions mentioned above. The program takes one argument with values 1 (calls the linear function), 2 (calls the quadratic function). 3 (calls the cubic function) (2) In your pdf report, plot a graph (e.g., Excel graph), where the x-axis is the passed argument (1, 2, or 3), and the y-axis is the execution time that you measured (can be in log scale if needed)

To observe the importance of algorithm efficiency and the order of growth, you are required to perform the following experiment. Assume an input of size 100,000 (denoted as N) (a) Linear function (O(N)): Write a function that loops over the input items (a single loop) and print the elapsed time taken by the function in milliseconds (b) Quadratic function (O(N2)): Write a function that loops over the input items with two-levels of nesting and print the elapsed time taken by the function in milliseconds. The function should have two nested loops (outer and inner) where each one starts from 0 to N (c) Cubicfunction (O(N3): Write a function that loops over the input items with three-levels of nesting and print the elapsed time taken by the function in milliseconds. The function should have three nested loops where each one starts from 0 to N Hint: The actual code inside the loop is not important. So make it simple and define a double variable in the beginning of your program, and then increment this variable inside the inner most loop (The example below is for two-level nesting). Double sum = 0; For (j=0, End Loop; j

Expert Answer

answers

public class Main

{

static void linearFunction(int n)

{

long start = System.currentTimeMillis();

for(int i = 0; i < n; i++);

long end=System.currentTimeMillis();

long elapsedTime = end – start;

System.out.println(“Elapsed time taken by linear function for “+n+ ” iterations: “+elapsedTime);

}

 

static void quadraticFunction(int n)

{

long start = System.currentTimeMillis();

for(int i = 0 ; i <n;i++)

for(int j = 0; j < n; j++);

long end=System.currentTimeMillis();

long elapsedTime = end – start;

System.out.println(“Elapsed time taken by quadratic function for “+n+” iterations: “+elapsedTime);

}

 

static void cubicFunction(int n)

{

long start= System.currentTimeMillis();

for(int i = 0 ; i <n;i++)

for(int j = 0; j < n; j++)

for(int k = 0; k <n; k++);

 

long end=System.currentTimeMillis();

long elapsedTime = end – start;

System.out.println(“Elapsed time taken by cubic function for “+n+” iterations: “+elapsedTime);

}

 

 

public static void main(String[] args) {

//System.out.println(“Hello World”);

System.out.println(“Calling linear function for 1000 iterations”);

linearFunction(1000);

System.out.println(“Calling quadratic function for 1000 iterations”);

quadraticFunction(1000);

System.out.println(“Calling cubic function for 1000 iterations”);

cubicFunction(1000);

System.out.println(“——————————————–“);

System.out.println(“Calling linear function for 10000 iterations”);

linearFunction(10000);

System.out.println(“Calling quadratic function for 10000 iterations”);

quadraticFunction(10000);

System.out.println(“Calling cubic function for 10000 iterations”);

cubicFunction(10000);

System.out.println(“——————————————–“);

System.out.println(“Calling linear function for 100000 iterations”);

linearFunction(100000);

System.out.println(“Calling quadratic function for 100000 iterations”);

quadraticFunction(100000);

System.out.println(“Calling cubic function for 100000 iterations”);

cubicFunction(100000);

System.out.println(“——————————————–“);

System.out.println(“Calling linear function for 500000 iterations”);

linearFunction(500000);

System.out.println(“Calling quadratic function for 500000 iterations”);

quadraticFunction(500000);

System.out.println(“Calling cubic function for 500000 iterations”);

cubicFunction(500000);

System.out.println(“——————————————–“);

}

}

—————————————————————————–

//output

Calling linear function for 1000 iterations
Elapsed time taken by linear function for 1000 iterations: 0
Calling quadratic function for 1000 iterations
Elapsed time taken by quadratic function for 1000 iterations: 6
Calling cubic function for 1000 iterations
Elapsed time taken by cubic function for 1000 iterations: 11
——————————————–
Calling linear function for 10000 iterations
Elapsed time taken by linear function for 10000 iterations: 0
Calling quadratic function for 10000 iterations
Elapsed time taken by quadratic function for 10000 iterations: 1
Calling cubic function for 10000 iterations
Elapsed time taken by cubic function for 10000 iterations: 50
——————————————–
Calling linear function for 100000 iterations
Elapsed time taken by linear function for 100000 iterations: 2
Calling quadratic function for 100000 iterations
Elapsed time taken by quadratic function for 100000 iterations: 0
Calling cubic function for 100000 iterations
Elapsed time taken by cubic function for 100000 iterations: 0
——————————————–
Calling linear function for 500000 iterations
Elapsed time taken by linear function for 500000 iterations: 3
Calling quadratic function for 500000 iterations
Elapsed time taken by quadratic function for 500000 iterations: 0
Calling cubic function for 500000 iterations
Elapsed time taken by cubic function for 500000 iterations: 0
——————————————–

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