CODEWITHSUNDEEP
c++primes
user1318194
user1318194

Reputation:

C++ code for checking for prime numbers not working

I'm having trouble with this C++ code. The integer num is a number that I want to check if it is prime. However this program is always returning false. It's probably something simple but I can't find anything.

for(int i=2;i<num;i++){ //primes are allowed to be divided by 1 so we start at 2
        if(num % i == 0){ //can be divided by a number other than itself or 1 so we trip out
            return false;
        } else if(i == num){ //if we've already done checks as high as possible and not tripped out yet then report success
            return true;
        }
}

Upvotes: 0

Views: 1413

Answers (6)

Will Ness
Will Ness

Reputation: 71070

Here's the proper way to write what you meant:

int i=2;                     // move declaration out
for(/*int i=2*/;i<num;i++){ 
        if(num % i == 0){ 
            return false;
        } // else            // and the final test too
}
if(i == num){                
    return true;
}

But that's not efficient. You only have to check for i's not exceeding of sqrt(num). Plus, if you check num%2, there's no more need to check any other even numbers, so you can use an increment of 2. Or you can even count by 6:

if( num == 2 || num == 3 ) return true;
if( num < 2 || num % 2 == 0 || num % 3 == 0 ) return false;
for( int i=5, j=7, lim=sqrt(num); i<=lim; i+=6, j+=6 ){   
        if( num % i == 0 || num % j == 0 ){ 
            return false;
        } 
}
return true;

(notice: this is more efficient than another answer here, which says it's an "optimization" of an initial version of this answer).

Upvotes: 0

Rajeev
Rajeev

Reputation: 1

bool isprime(int n)
{
    if(n<2) return false;
    if(n==2)return true;
    for(int i=2;i<=sqrt(n);i++)
        if(n%i==0) return false;
    return true;
}

Upvotes: 0

JP_
JP_

Reputation: 1666

bool CheckPrime(int num) {
    bool yayornay = true;
    for(int i = 2; i < num; i++) {
         if(num % i == 0) {
             yayornay = false;
             break;
         }
    }
    return yayornay;
}

Upvotes: 1

Flash
Flash

Reputation: 16703

i == num will never occur, since your loop condition is i<num. Try:

for(int i=2;i<num;i++){ //primes are allowed to be divided by 1 so we start at 2
    if(num % i == 0){ //can be divided by a number other than itself or 1 so we trip out
        return false;
    } else if(i == num-1){ //if we've already done checks as high as possible and not tripped out yet then report success
        return true;
    }
}

As pointed out below, the else condition here is redundant, and you only need to check from 2 to sqrt(num) - since the remaining factors have already been checked.

There are more improvements that can be made depending on how complex you want to make the problem. Most methods in reality use probabilistic algorithms.

Upvotes: 7

Zelter Ady
Zelter Ady

Reputation: 6338

A small optimization for Will Ness's code, just calculate the sqrt of the number outside the for. The condition check executes many times and has no sense to calculate sqrt each time.

if( num == 2 ) return true;
if( num < 2 || num % 2 == 0 ) return false;
int sqrt = sqrt(num);

for( int i=3; i<=sqrt; i+=2 ){   
        if(num % i == 0){ 
            return false;
        } 
}
return true;

So far I think that this is the most efficient way!

Upvotes: 1

Bo Persson
Bo Persson

Reputation: 92211

You don't have to check every number, as a lot of them can easily be ruled out. For example, after checking that num is not divisible by 2, you can skip all other even numbers. That saves you half the tests.

We also definitely know that any other factor must be less than num/2 (or really sqrt(num), but that is harder to compute). That knowledge can save us another half of the tests.

So now we have:

if (num % 2 == 0)
    return false;

for(int i = 3; i < num / 2; i += 2){ 
     if(num % i == 0){ //can be divided by a number other than itself or 1 so we trip out
         return false;
     }
}

// arriving here we have found no factors, so it must be a prime
return true;

Upvotes: 4

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