Case Studies: Solving simple scientific and statistical problems

Case Studies

(Solving simple scientific and statistical problems)

1. Write a program to check, whether the given Prime or not.

A prime number is an integer which has no proper divisions. For example, 13 is a prime number, because 13 is not divisible by any number other than 1 and 13.

If 'n' has a factor greater than √n, it also has a factor less than √n. So, it is enough to check whether 'n' is divisible by any number between 2 and √n. If divisible, 'n' is not prime, otherwise it is prime.

/* Program to check the given number is prime or not */

#include <stdio.h>

main()

{

     /* Local definitions */

     int n,i,r;

     printf("Enter the positive integer value.....");

     scanf("%d", &n);

     i=2;

     Step1:

          if(i<=sqrt(n))

          {

               r=n%i;

               if(r==0)

               {

                    printf("%d is not a prime",n);

                    goto end;

               } /* if */

          } /* if */

          else

          {

               i++;

               goto stepl;

          } /* if else */

     printf("%d is prime",n);

end:

     printf(" ");

} * main */

OUTPUT

Enter the positive integer value.....3

3 is prime

EXPLANATION: Read n then checks whether the i<=sort(n), if the condition is true if block is executed and prints the entered value is not a prime. Otherwise else part is executed, then the variable i is increased by 1 then goto step1, after completion of else part finally the entered value is printed as prime.

2. Write a program to find the Greatest Common Division (GCD) (Highest Common Factor) of two given numbers.

Let the numbers be I and J, and I be the bigger number and J be the smaller number. Suppose, we want to find the GCD(HCF) of two numbers 36 and 54. The bigger number 54 must assigned to I and 36 to J.

 I=54, J=36

Divide I by J, the remainder when 54 is divided by 36 is 18, So, I=18, J=36

Divide I by J and find the remainder, when 36 is divided by 18, the remainder is 0, so, I=0, J=18.

Interchanging I and J we get I=18, J=0. Here J=0, hence I is the GCD, so, GCD=18. The Greatest Common Division of 54 and 36 is 18.

/* Program to find ged of two numbers */

#include <stdio.h>

main()

{ /* Local definitions */

    int i,j,k;

    printf("Enter Two numbers.....");

    scanf("%d %d", &i, &j);

    if(j>i)

    {

        k=i;

        i=j;

        j=k;

    } /* if */

step1:

    if(j==0)

        goto step2;

    else

    {

        k=i%j;

        i=j;

        j=k;

    } /* if else */

step2:

    printf("The GCD is..... %d",j);

} /* main */

OUTPUT

Enter Two numbers.....54 36

The GCD is.. 18

EXPLANATION: Read the values i and j, the if statement checks the condition j>i, if condition is true, then the value of i and j is interchanged using third variable k. Otherwise take modular division between i, j. If k becomes zero then prints the value of GCD through the variable j as a output.

3. Program to print Table of Binomial Coefficients.

The Binomial coefficients are used to study the binomial distributions.

It is given by B(m, x) =  , m >= x

A table of binomial coefficients is required to determine the binomial coefficient for any set of 'm' and 'x'. The binomial coefficient can be recursively calculated as follows:

B(m, o)=1, B(m, x)=B(m,x‒1) [(m‒x+1) / x ] , ,x=1, 2, 3, ..., m, B(0,0)=1

That is, the binomial coefficient is one, when either 'x' is zero or 'm' is zero.

/*Program to print table of binomial coefficients */

#include <stdio.h>

main()

{

   int m,x, bin;

   printf("mx");

   for(m=0;m<=5;m++)

      printf("%4d",m);

      printf("\n‒‒‒‒‒‒‒‒‒‒‒‒\n");

   m=0;

   do

   { printf("%2d", m);

      x=0;

      bin=1;

      while(x<=m)

      {

         if(m==0||x==0)

               printf("%4d", bin);

         else

         {

            bin=bin*(m‒x+1)/x;

            printf("%4d", bin);

      } /* if else */

   x=x+1;

   }/* while */

   printf("\n");

   m=m+1;

} /* do */

   while(m<=5);

      printf("\n‒‒‒‒‒‒‒‒‒‒‒‒\n");

} /* main

OUTPUT

mx 0 1 2 3 4 5

‒ ‒ ‒ ‒ ‒ ‒ ‒ ‒ ‒

0  1

1  1

2  1  1

3  1  2  1

4  1  3  3     1

5  1  4  6     4   1

1  1  5  10  10   5   1

‒ ‒ ‒ ‒ ‒ ‒ ‒ ‒ ‒