# Find the sum of all the prime numbers

Discussion in 'C' started by pradeep, Nov 29, 2006.

Joined:
Apr 4, 2005
Messages:
1,645
87
Trophy Points:
0
Occupation:
Programmer
Location:
Kolkata, India
A friend of mine who is learning to program in C, was having diffculty in writing a program to find the sum of all the prime numbers between a specified limit (in her case 3-60). I helped her out with the program, and I thought I'll post it here which might help others.

Code:
``` /*
** Program to find the sum of all numbers between 3 and 60
** @date : 11/29/2006
*/

#include <stdio.h>

void main(void)
{
unsigned int i,j,s=0,is_prime;

for(i=3;i<=60;i++)
{
is_prime = 1; // Assuming that current value of i is prime

// Checking for prime
for(j=2;j<=(i/2);j++)
{
if(i%j==0) // Not prime, set is_prime to 0 and break
{
is_prime = 0;
break;
}
}

if(is_prime) // If the number is prime, sum it up
{
s += i;
// optionally you can print the prime numbers too
printf("%d ",i);
}
}

printf("\n\nThe sum of the prime numbers = %d",s);
}```

2. ### friendsfornirajNew Member

Joined:
Nov 24, 2006
Messages:
40
0
Trophy Points:
0
Occupation:
studying
well this code is good but it ll take more time(for computational work)
please check this code the 2nd for loop

Code:
``` /* ** Program to find the sum of all numbers between 3 and 60 ** @author : Pradeep ** @date : 11/29/2006 */
#include <stdio.h>
#include<math.h>
void main(void)
{
printf("\n please enter the upper limit");
int n;
scanf("%d",&n);\\now n can be given by user
unsigned int i,j,s=0,is_prime;
for(i=3;i<=n;i++)
{
is_prime = 1; // Assuming that current value of i is prime
// Checking for prime
for(j=2;j<=sqrt(i);j++)
{
if(i%j==0) // Not prime, set is_prime to 0 and break
{
is_prime = 0;
break;
}
}
if(is_prime) // If the number is prime, sum it up
{  s += i;             // optionally you can print the prime numbers too
printf("%d ",i);
}
}
printf("\n\nThe sum of the prime numbers = %d",s);
} ```
//see though your program will work proper but if we have to sum up prime nos. between 3 to n
//where n verey big(though within the limits of int )
// then the checkin condition from to n/2 will take lot of time
// suppose ur n==100 then it will check from n=2 to n=50
//so the no. of checking loops required for 100 will be 48
//where as if we go from n=2 to n= 10 we only require 8 loops so it will be much faster
// now think for the case of n=1000 wouldnt it will take lot of time

Last edited by a moderator: Nov 29, 2006

Joined:
Apr 4, 2005
Messages:
1,645
87
Trophy Points:
0
Occupation:
Programmer
Location:
Kolkata, India
I understood your point, but I see you have used sqrt() ! How does that help?? And we need to check whether it got a factor execpt for itself and 1 or not.

Joined:
Jul 12, 2004
Messages:
15,329
377
Trophy Points:
83
I would use
t = sqrt(i);
in
for(j=2;j<=sqrt(i);j++)
e.g.
for(j=2;j<=t;j++)
So it does not need to calc square root all the time it loops through j

5. ### friendsfornirajNew Member

Joined:
Nov 24, 2006
Messages:
40
0
Trophy Points:
0
Occupation:
studying
yes i got you!
the thing is we dont need to go more than the sqrt of the no. which to be checked as prime if it is not divisibe by any no. upto sqrt(n) than there wont be any factor above that you can check it for any no.

Joined:
Apr 4, 2005
Messages:
1,645
87
Trophy Points:
0
Occupation:
Programmer
Location:
Kolkata, India
Its not using square root anywhere, and neither does it require too. Could you please tell me why does it need to use square root??

7. ### friendsfornirajNew Member

Joined:
Nov 24, 2006
Messages:
40
0
Trophy Points:
0
Occupation:
studying
good point shabbir

Joined:
Jul 12, 2004
Messages:
15,329
377
Trophy Points:
83
He edited the code to find a no is prime or not to loop till the sqrt(n) and not till n/2

Joined:
Apr 4, 2005
Messages:
1,645
87
Trophy Points:
0
Occupation:
Programmer
Location:
Kolkata, India
For example, sqrt(60) = 7.46, that means we will loop 7 times, but 60 is divisible by 30 also.
The program works fine with sqrt, but what is the logic?

Joined:
Jul 12, 2004
Messages:
15,329
377
Trophy Points:
83

Joined:
Jul 12, 2004
Messages:
15,329
377
Trophy Points:
83
There is other method also which does not require sqrt.

Joined:
Apr 4, 2005
Messages:
1,645
87
Trophy Points:
0
Occupation:
Programmer
Location:
Kolkata, India
Interesting, so we will find the factor of n within any of the number in sqrt(n) if not then its a prime number.

13. ### friendsfornirajNew Member

Joined:
Nov 24, 2006
Messages:
40
0
Trophy Points:
0
Occupation:
studying
let me tell you the logic now
it is very simple
look if k is any no.
then k can be represented as k=i*j
where i and are k's factor
now i varies from 1to k
and j varies from k to 1
but for prime no. we check for i=2 to k/2
acordingly j will varies from (k/2) to 2
now if 2 is a factor of k
then automatically k/2 is also a factor and so on
that is when i increases, j decreases
so we continue like this then there is condition that both i and j becomes equal
so i*i=k
so i=sqrt(k)
but if k is not a square of any no. then i increases upto a integer just less than sqrt iof k
thats all what i did
now try this problem
for a given no. print all its prime factors that is
if input is 12 out put should be 2*2*3
using the hint that the lowest factor of any no. other than one is always prime
try it

Joined:
Jul 12, 2004
Messages:
15,329
377
Trophy Points:
83
friendsforniraj thats a good logic explained.

15. ### friendsfornirajNew Member

Joined:
Nov 24, 2006
Messages:
40
0
Trophy Points:
0
Occupation:
studying
thank u sir
but ma name is niraj

Joined:
Jul 12, 2004
Messages:
15,329
377
Trophy Points:
83
I will try putting that from now on.

17. ### rahul.mca2001New Member

Joined:
Feb 13, 2008
Messages:
103
0
Trophy Points:
0
why do we need a square root

Joined:
Jan 9, 2008
Messages:
356
14
Trophy Points:
0
Occupation:
Developer
Location:
NOIDA
Excellent explaination. Now any one can understand either he is new or experience

19. ### oleberNew Member

Joined:
Apr 23, 2007
Messages:
37
2
Trophy Points:
0
Occupation:
Software Developer (Perl, C/C++ and Java)
Location:
Hamburg, Germany
for(j=2;j<=sqrt(i);j++) :cryin:

is similar to

for(j=2;j*j <= i;j++) :p

so, no sqrt in here since it is really slow. At least if was some years ago.

Joined:
Apr 23, 2007
Messages:
37