Thread: Printing Prime Numbers from 1 to 300.

  1. #1
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    Printing Prime Numbers from 1 to 300.

    Code:
    #include<stdio.h>
    main()
    {
    	int number, i;
    
    	for (number=2;number<=300;number++)
    	{
    		for (i=2; i < number ; i++)
    		{
    			if (number%i==0)
    				break;
    			else 
    				printf("\n%d", number);		
    			
    		}
    		
    		continue;
    						
    				
    	}
    
    }
    Trying to print the first 300 Prime Numbers.

    Here's the program that I made..

    Now for some reason unknown to me..it is not giving out the desired output..

    Can anybody point out the mistake?

  2. #2
    Code Goddess Prelude's Avatar
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    >it is not giving out the desired output..
    Let me guess. You're getting lots of duplicates.

    >Can anybody point out the mistake?
    Do a paper run of the program. You'll easily be able to find the problem if you work through your program step by step. I could point out the mistake, but you wouldn't learn anything about debugging if I did.

    >main()
    int main ( void ). You also need to return a value at the end. 0 is the normal success value.

    >continue;
    This is redundant. The loop will continue anyway.
    My best code is written with the delete key.

  3. #3
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    I at least took care of the duplicates..but there's one more problem!

    Every number divisible by 5 is also there in the list!

    Also, how do I take care of the number 2?
    It is also not there on the list!

    Code:
    #include<stdio.h>
    main()
    {
    	int number, i;
    
    	for (number=2;number<=300;number++)
    	{
    		for (i=2; i <number; i++)
    		{
    			if (number%i == 0)
    			{
    				break;
    			}
    			else
    			{
    				printf ("\n%d", number);
    				break;
    			}
    
    		}
    		
    	}
    }

  4. #4
    Cached User mako's Avatar
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    (number=2;number<=300;number++)
    for (i=2; i <number; i++)
    (number%i == 0)

    if you start with number on 2, and with i on 2. Then it should be obvious that number%i is 0...

    and you don't need the second break...

  5. #5
    Code Goddess Prelude's Avatar
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    >Every number divisible by 5 is also there in the list!
    May I suggest an is_prime function? A simple function that takes a number and returns either 0 or 1 depending on whether it's prime or not will save you a lot of headaches. The biggest problem you have is coordinating the two loops to do what you want. If you can do the following, it would be much easier, ne?
    Code:
    int main ( void )
    {
      int i;
    
      for ( i = 2; i < 300; i++ ) {
        if ( is_prime ( i ) )
          printf ( "%d\n", i );
      }
    
      return 0;
    }
    My best code is written with the delete key.

  6. #6
    and the hat of int overfl Salem's Avatar
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    Look up the definition of Prime (again).
    In fact, look at Eratosthenes sieve (google for it).
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.

  7. #7
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    Okay..this is my new program..as lousy as it is..but it works!!!

    Code:
    #include<stdio.h>
    main()
    {
    	int number, i;
    
    	for (number=1;number<=300;number++)
    	{
    		for (i=2; (i<number) || (i==number) ; i++)
    		{
    			if (i==number)
    			{
    				printf ("\n%d", number);
    			}
    
    			if (number%5==0)
    			{
    				break;
    			}
    
    
    			if (number%i == 0 )
    			{
    				break;
    			}
    			else
    			{
    				printf ("\n%d", number);
    				break;
    			}
    
    		}
    		
    	}
    
    }
    @mako
    if you start with number on 2, and with i on 2. Then it should be obvious that number%i is 0...

    and you don't need the second break...
    Yeah man..I could figure that out.

    If I don't put the second break...it gives me a hell lotta duplicates...


    @Prelude

    Not yet done with the functions man...I suppose they will make things easier...hehe


    @Salem

    Look up the definition of Prime (again).
    In fact, look at Eratosthenes sieve (google for it).
    Yeah man..I know the definition of Prime..
    I put int quite a lotta effort into this so called "easy" program..but still did not get the wanted result, so I seeked help.

    Well..the Eratosthenes sieve thing was great!


    Anybody suggesting an improvement in the program ??

  8. #8
    The Richness... Richie T's Avatar
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    >>Anybody suggesting an improvement in the program ??

    Yes - get it to work properly!

    Here is the output of your program for numbers less than 50:

    2
    3
    7
    9
    11
    13
    17
    19
    21
    23
    27
    29
    31
    33
    37
    39
    41
    43
    47
    49
    Now, last I checked, the following numbers are NOT primes:

    9, 21, 27, 33, 39, 49.

    Generally, if your code has to look especially at numbers divisible by 5, there's
    a design flaw in your approach.

    I'd go with Prelude's suggestion to make an isprime function, test it thoroughly
    over a range of values easy to spot errors, and fix the program from there.
    Also, you should allow the user to enter the upper limit on numbers you want
    to test, or even specify the range - allow the user to get primes between any
    two numbers.

    Lastly:

    1. main returns int, and since you are not using command line arguments, the parameter
    list should be (void)

    2. (i<number) || (i==number) is the same as (i <= number)

    3. You can make this program run more efficiently by using (i < ((number/2) + 1)), since
    a number cannot be evenly divided by any number greater than half its value
    No No's:
    fflush (stdin); gets (); void main ();


    Goodies:
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    ASCII stupid question, get a stupid ANSI

  9. #9
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    Since you haven't learned functions yet ...

    Code:
    int is_prime
    for all numbers you want to test {
       is_prime = 1 //start by assuming prime
       for (i=2; i < number ; i++) {  //or a more efficient loop as suggested
          if (number % i == 0 )  {
            //not prime, set flag, break inner loop
          }
       }
       if(is_prime) {
          print number
       }
    }

  10. #10
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    I have tried to do the same thing man...

    Code:
    for (number=1; number<=300; number++)
    {
    	for (i=2; i<=number; i++)
    	{
             if (i==number)
                { 
                   printf("\n%d", number);   /*this basically prints out '2' */
                 }
    
    		if (number%i==0)
    		{
    			break;   /*cuz the number is not prime and no 
    			           further checking would be necessary.*/
    		}
    
    		else 
    		{
    			printf("\n%d");  /*the number is indeed prime, and so needs to be printed out!*/
    
    			break; /*people say that this break is redundant 
    				   but without this i am getting duplicates of numbers 
    				   (and they are not prime too!)*/
    		} 
    	}
    }
    I cannot think any further!

    For checking if a number is prime or not...i know what needs to be done..

    I need to divide the number by a number greater than one and a number less than the number itself..that would be

    for (i=2; i<number; i++) OR for (i=2; i<=number-1; i++)

    The outer loop would be for (i=2; i<=300; i++)
    ...to increment the numbers that are gonna be checked for PRIME...


    Bottomline..i am outta ideas...(
    Last edited by duffmckagan; 07-27-2006 at 12:47 PM.

  11. #11
    ex-DECcie
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    Quote Originally Posted by duffmckagan
    I have tried to do the same thing man...


    I need to divide the number by a number greater than one and a number less than the number itself..that would be

    Actually, you only need to divide the number by a number between 2 on the lower end and number/2 on the upper end. Dividing by anything more than number/2 is just a waste of cycles....
    Mr. Blonde: You ever listen to K-Billy's "Super Sounds of the Seventies" weekend? It's my personal favorite.

  12. #12
    ex-DECcie
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    Because you seem about at wits end, here's something that might work....

    Quote Originally Posted by duffmckagan

    Code:
    
    printf("1\n");
    printf("2\n");
    
    for (number=3; number<=300; number++)
    {
            int isPrime = 0;
    	for (i=2; i<=number/2; i++)
    	{
     		if (number%i==0)
    		{
                           isPrime = 0;
    			break;   /*cuz the number is not prime and no 
    			           further checking would be necessary.*/
    		}
                    isPrime = 1;
    	}
    
            if (isPrime)
                printf("%d\n",number);
           
    }
    I
    I think that should work. I'm winging it while I'm doing real work here today.... It'll get you closer if it doesn't get you all the way there....
    Mr. Blonde: You ever listen to K-Billy's "Super Sounds of the Seventies" weekend? It's my personal favorite.

  13. #13
    and the hat of int overfl Salem's Avatar
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    > for (i=2; i<=number/2; i++)
    1. you only need to check upto sqrt(number)
    2. if you eliminate %2 as a special case, then both the inner and outer loops can increment by 2 each time, not 1
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.

  14. #14
    Registered User r3d.h4tt3r's Avatar
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    i wrote a program like this in c++

    didnt have time to port it over to c but should still be readable

    Code:
    #include <iostream> //header file
    #include <cmath> //header file
    using namespace std;
    
    int main() 
    {
    	int x; 
    	int i;
    	int prime;
    
    	for(x=1; x<=10000; x=x+2) 
    	{
    		prime=1; //sets prime
    		for(i=2; i<=sqrt(x); i++)
    			{
    				if(x % i == 0)	
    				{
    					prime = 0;	
    					i=sqrt(x)+1;	
    				}
    
    			}
    		if(prime == 1)	
    		{
    		       cout << x << " is prime.\n"; //"cout" equal to the "printf" command in c
    		}
    
    	}
    	return 0;
    }

  15. #15
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    sqrt() is a floating point function. Introducing it when working with integers also introduces potential for rounding errors (finite precision of floating point types) to affect end conditions.

    A condition that is more certain of working is "i*i < x"

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