Question

Code that needs to be modified:

#include

#include

#include

int main()

{

int i,N,x;

unsigned int seed;

double R;

printf("\nEnter number of iterations and seed");

printf("\n");

scanf("%i %u", &N,&seed);

srand(seed);

  

for(i=0;i

{

R=(double)rand()/RAND_MAX;

if (R<0.5)

x=x+1;

else

x=x-1;

}

printf("Final location is ");

printf("%d",x);

printf("\n");

}

Question: Write a code that generates N pairs of random numbers. Call the first member of each pair x and the second member y. Count how many of the N pairs obey x^2+y^2<1. Call that number M, and have your code print out M/N. What do you get for M/N when N=10^2? How about N=10^4, N=10^6 and N=10^8? Explain your result. The language used is C.

Hint: It might be useful to multiply the value you get for M/N by four.

Warning: Be careful how you define N and M. If you have variables N,M that are integers and compute N/M the computer will give you an integer which is the number of times M foes into N. Thus in N=5 an M=3 then N/M=5/3=1. Similarly if N=13 and M=2 then N/M=13/2=6. Finally, if N=1 and M=2 then N/M=1/2=0. if you want 13/2=6.5 you need to tell the computer 13 is a double (or float) by declaring it as a double(or float). This is an important point to remember in coding. If you ever write a line of C code like y=(1/2)*x; the computer will set y=0 because it will compute 1/2=0 before multiplying by x.

******* Please make sure you answer all the questions asked and please do it 100%correctly. Please write neat and very clear that way i can be able to read and understand what is being said. The language used is C.

Answer 1

CODE:

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#include<time.h>
double calculate(int n) //Function to calculate m
{
double x,y;
double sum;
int m=0,i;
for(i=0;i<n;i++)
{
x=(double)rand()/RAND_MAX;
y=(double)rand()/RAND_MAX;
double sum=(x*x)+(y*y);
if(sum<1)
m++;
}
double finite=(double)m/n;
return finite;
}
int main()
{
int i,N;
int m=0;
srand(time(NULL)); // Generate different number different time
double result100=calculate(100); //calling function for 10^2
printf("For 10^2 iterations%f\n",result100);
double result10000=calculate(10000); //calling function for 10^4
printf("For 10^4 iterations%f\n",result10000);
double result1000000=calculate(1000000); //calling function for 10^6
printf("For 100^6 iterations%f\n",result1000000);
double result100000000=calculate(100000000); // calling function for 10^8
printf("For 100^8 iterations%f\n",result100000000);
}

Output:

For 10^2 iterations0.890000
For 10^4 iterations0.786700
For 100^6 iterations0.785321
For 100^8 iterations0.785413