#06 (06/18/2025)

Random numbers ( Monte Carlo simulations)

A program to print random numbers 10 times. 


#include <stdio.h>
#include <stdlib.h>

int main()
{
int i;

for (i=0; i <  10; i++)  printf("%d\n", rand());

printf("\nMAX = %d\n", RAND_MAX);

return 0;
}
rand() returns an integer between 0 and RAND_MAX defined in stdlib.h.
Generate random numbers between 0 and 1.0. 


#include <stdio.h>
#include <stdlib.h>
int main()
{
int i;
for (i=0; i< 10; i++)  printf("%f\n", 1.0*rand()/RAND_MAX);
printf("\nMAX = %d\n", RAND_MAX);
return 0;
}
srand() sets its argument as the seed for a new sequence of pseudo-random integers.
If no seed value is provided, the rand() function is automatically seeded with a value of 1. 


#include <stdio.h>
#include <stdlib.h>

int main()
{
int i;
printf("Enter seed integer = ");
scanf("%d", &i);
srand(i);

printf("%d\n", rand());

return 0;
}
time() defined in <time.h> with the argument NULL returns the elapsed time since 00:00:00 GMT, January 1, 1970, measured in seconds. 


#include <stdio.h>
#include <time.h>

int main()
{
int i;

printf("%d\n", time(NULL));
return 0;
}






#include <stdio.h>
#include <stdlib.h>
#include <time.h>

int main()
{

srand(time(NULL));
printf("%d\n", rand());
return 0;
}

Generating random numbers in arbitrary range

  1. rand() returns an integer between 0 and RAND_MAX (=2147483647).
  2. 1.0*rand()/RAND_MAX returns a floating number between 0 and 1.
  3. 5.0*rand()/RAND_MAX returns a floating number between 0 and 5.
  4. 10.0*rand()/RAND_MAX-5 returns a floating number between −5 and 5.
  5. rand()%7 returns an integer of either 0, 1, 2, 3, 4, 5, or 6.
  6. rand()%7+10 returns an integer of either 10, 11, 12, 13, 14, 15, or 16.

Using random numbers for numerical integration (Monte Carlo Methods)


⌠
⌡ 		1

0 
 1
x
 dx = ?
buffon1.gif

⌠
⌡ 		2

1 
 1
x
 dx = ln x|12 = ln 2 − ln 1 = ln 2.
Generate a random number, x, between 1 and 2, and another random number, y, between 0 and 1. The probability that xy < 1 is proportional to ln 2/1. 


#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <time.h>

int main()
{
float x, y;
int i, count=0;
int n;

printf("Enter iteration number = ");scanf("%d", &n);

srand(time(NULL));

for (i=0; i< n; i++)
{
 x=1.0*rand()/RAND_MAX+1.0;
 y=1.0*rand()/RAND_MAX;

 if (x*y < 1.0) count=count+1;
}

printf("True value = %f\n", log(2));
printf("Appx value = %f\n", 1.0*count/n);

return 0;
}
This approach can be also applied to computing π as
buffon2.gif
As seen in the figure above, the probability that
x2+y2 < 1
if 0 < x < 1 and 0 < y < 1 should be π/4.

Solving a cubic equation

We want to find one of the roots of a cubic equation given by
x3 + x − 1 = 0
which is between 0 and 1 by an iterative method. Modify the equation above to
x = 1
1 + x2
Start with an appropriate initial value and do the iteration above until it is convergent. For example,

x0=0.5,     x1 = 1
1+0.52
 =0.8,    x2 = 1
1+0.82
 =0.609,    x3 = 1
1+0.6092
 =0.728, …
The simplest (minimum) code may look like: 
#include <stdio.h>

int main()
{
int i;
float x;

printf("Enter initial guess =");scanf("%f",&x);

for (i=0;i<100;i++)  /* insert your iteration code */

printf("%f\n",x);
return 0;
}

Largest integer in C

An integer in C (int) is represented by 4 bytes (1 byte = 8 bits). What is the largest integer that can be handled by C ?
Hint:

    (a) Conversion from binary to decimal:

    10112
    =
    1 ×23 + 0 ×22 + 1 ×21 + 1 ×20
    =
    8 + 0 + 2 + 1
    =
    1110.

    (b) If one bit is used as the sign, the maximum number using 4 bytes (32 bits) is
    n=1111111 11111111 11111111 111111112. (31 "1"s).
    Insteading of computing
    n = 1×230+1×229+1×228+ …+ 1×21+1×20,
    consider computing n+1 = 10000000 00000000 00000000 000000002,

    n
    =
    1×230+1×229+1×228+ …+ 1×21+1×20
    =
    (n+1) − 1
    =
    1×231−1
    =
    2147483647.

    (c) Exercise: If an integer is represented by 1 byte (8 bits), what is the maximum integer number with a sign ?


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On 17 Jun 2025, 18:06.