Q1: Write a java program to sum (1 + 2 + . . . + n ) for any positive integer n?

Question

Q1:

Write a java program to sum (1 + 2 + . . . + n ) for any positive integer n?

Q2:

Algorthem A: sum = 0

for i = 1 to n

sum = sum + i.

Algorthem B:

sum = 0

for i = 1 to n

{

for j = 1 to i

sum = sum +1

}

Algorthem C:

sum = n * (n + 1) / 2

Write a Java program that implements the three algorithms and times them for various values of n

.Test your code with n=1000, n=10000, n=100000, n=1000000, ...

A: 1+2+3+...+n

B: 1+(1+1)+(1+1+1)+...

C: Hint:

Option1: long

startMilliseconds = System.currentTimeMillis ();

Option2: Date

current = new Date();longstartTime=current.getTime();

Q3:

Using Big Oh notation, what is the order of the following computation’s time requirement?

for i = 1 to n

{

for j = 1 to 5

Sum = sum + 1

}

Explanation / Answer

Que1: Create a file named code.java and paste following code into it! Flename must be code.java

code.java

public class code
{
public static void main(String[] args)
{
int n = 10;
int sum = 0;
for(int i = 1; i<= n;i++)
{
sum = sum + i;
}
System.out.println("Sum of first " + n + " numbers = " + sum);
}
}

Sample Output:

Sum of first 10 numbers = 55

Que2: Create a file named code2A.java and paste following code into it! Flename must be code2A.java

code2A.java

public class code2A
{
public static void main(String[] args)
{
int n = 1000000;
int sum = 0;
long startMilliseconds = System.currentTimeMillis ();
for(int i = 1; i<= n;i++)
{
sum = sum + i;
}
long endMilliseconds = System.currentTimeMillis ();
long time = endMilliseconds - startMilliseconds;
System.out.println("Time to run the program when n = " + n + ": " + time + " millisecs");
}
}

Sample Output:

Time to run the program when n = 1000000: 4 millisecs

Que2: Create a file named code2B.java and paste following code into it! Flename must be code2B.java

code2B.java

public class code2B
{
public static void main(String[] args)
{
int n = 10000;
int sum = 0;
long startMilliseconds = System.currentTimeMillis ();
for(int i = 1; i<= n;i++)
{
for(int j = 1; j<=i ; j++)
{
sum = sum + 1;
}
}
long endMilliseconds = System.currentTimeMillis ();
long time = endMilliseconds - startMilliseconds;
System.out.println("Time to run the program when n = " + n + ": " + time + " millisecs");
}
}

Sample Output:

Time to run the program when n = 10000: 29 millisecs

Que2: Create a file named code2C.java and paste following code into it! Flename must be code2C.java

code2C.java

public class code2C
{
public static void main(String[] args)
{
int n = 1000;
int sum = 0;
long startMilliseconds = System.currentTimeMillis ();
sum = n*(n + 1)/ 2;
long endMilliseconds = System.currentTimeMillis ();
long time = endMilliseconds - startMilliseconds;
System.out.println("Time to run the program when n = " + n + ": " + time + " millisecs");
}
}

Sample Output:

Time to run the program when n = 1000: 0 millisecs

Que3:

Outer loop runs n times and inner loop runs 5 times.

So, number of operations = 5*n

So time complexity = O(5n) = O(n)

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