C++ prime testing

Whovian

Alright. I know how incredibly inefficient this algorithm is, but I felt like giving this a whirl.

#include <iostream>
#include <cmath>
using namespace std;

bool prime(int x)
{
bool j = true;
for (double i = 2;i == x;i++)
{
if (x/i == floor(x/i))
{
j = false;
}
}
return j;
}

int main()
{
cout << prime(4) << endl;
return 0;
}

It returns "1" (true.) Why is this? Obviously, an error on my part.

Whovian

This is one of the times I wish I could delete my own threads. I used = instead of ==. Still not working. Edited the original post.

Hobin

Quote by Whovian

This is one of the times I wish I could delete my own threads. I used = instead of ==.

Don't worry about it. It's a classic mistake.

Whovian

I managed to figure the new problem out, I used == when I should've used !=. These =, ==, and != are killing me! :)

jreelawg

You can write it like this if you want.

Code:

for (double i = 2;i != x;i++)
    if (x/i == floor(x/i))
        j = false;

And you could do do it this way instead also

Code:

for (int i = 2; i != x; i++)
    if ((x % i)==0)  
        j = false;

Whovian

Now, for fun, I decided to see if I could make a perfect number tester. What's wrong with this function?

Code:

bool perfect (int x)
{
	int y = 0;
	for (double i = 2;i != x;i++)
	{
		if (x/i == floor(x/i))
		{
			y = y + i;
			}
		}
	return (y == x);
	}

Whatever I input, it returns 0.

EDIT: Figured it out. *bashes head against wall* I started counting factors at 2, not 1.

EDIT: And it's still looking for 33550336.

jreelawg

Quote by Whovian

EDIT: And it's still looking for 33550336.

I don't know the specifics, maybe someone else can help me out about this, but I would think that a comparison between a double and an integer as a loop condition might be problematic.

Integers are exact values while doubles are approximate values.

It seams possible that as you add up i's as doubles, you might end up with a loss of precision enough that i != x when you would expect it to and you will get stuck in an infinite loop.

I just did a test and found that

2.0000000000000001==2 , but
2.000000000000001 !=2

jreelawg

I just ran a test and found this out.

Code:

int x=2;
double y=2;

for ( int i=0; i < 29; ++i){
    y+=y;  //equivalent to y=y+y;
    x+=x;
}

if (x==y) 
    cout << "yes";

Turns out that after 29 additions

Code:

x==y

, but after 30,

Code:

x!=y

If you use the modulus operator then you will have no problems

Code:

(x % i) == //The integer remainder of x/i, which is zero if i is a factor of x.

Hobin

a/b == floor(a/b) is ugly to begin with. Those modulus operators don't exist for no reason.

AlephZero

Quote by jreelawg

I just ran a test and found this out.

Code:

int x=2;
double y=2;

for ( int i=0; i < 29; ++i){
    y+=y;  //equivalent to y=y+y;
    x+=x;
}

if (x==y) 
    cout << "yes";

Turns out that after 29 additions

Code:

x==y

, but after 30,

Code:

x!=y

If you print out the value of x and y at each iteration, you mgiht be surprised as to which of x and y is accurate and which is not...
... but you shouldn't really be surprised that if x is a 32-bit signed integer, it can't hold a number bigger than 2³¹.

On the other hand a double can hold integers up to about 2⁵² exactly, so sometimes storing integer values in doubles wins over storing them in integers.

Try the same code with "long x = 2" (if your implementation uses 64-bit long integers) to see when that falls over. Actually, it might work beyond 52 iterations, because a double can hold integer powers of 2 exactly right up to the maximum value for a double, which is bigger than 10³⁰⁰. But C++ it won't necessarily print those values accurately to 300 digits precision, of course.

TylerH

"long long" is always at least 64 bits wide on any x86 arch, if not 128. Use long long if you need more range, but stay away from floating point numbers.

Use % to check for remainder. For example, if i % p == 0, then p divides i and i isn't prime if p is any number but i or 1.

Also, if you need guaranteed width, look at cstdint.

Hobin

Quote by TylerH

"long long" is always at least 64 bits wide on any x86 arch, if not 128. Use long long if you need more range, but stay away from floating point numbers.

In my experience, it's 64 bits on most compilers: http://gcc.gnu.org/ml/gcc/2004-03/msg01251.html.

"64 bits ought to be anything for anyone." - Me.

Adyssa

When comparing doubles (floating point numbers in general), you should compare to within an error bound to account for small variances in precision.

Example:

Code:

double x, y;
epsilon = 0.000000005 // error bound

// some code assigning values to x and y

if(abs(x - y) < epsilon) // absolute value keeps the result of x - y positive for comparison with the error bound
{
  // they are 'equal', or close enough
}
else
{
  // they are not equal
}

Whovian

Quote by Hobin

a/b == floor(a/b) is ugly to begin with. Those modulus operators don't exist for no reason.

Oh! Modulus! Why didn't I think of that?

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Whovian	This is one of the times I wish I could delete my own threads. I used = instead of ==. Still not working. Edited the original post.

Whovian	C++ prime testing I managed to figure the new problem out, I used == when I should've used !=. These =, ==, and != are killing me! :)

jreelawg	You can write it like this if you want. Code: for (double i = 2;i != x;i++) if (x/i == floor(x/i)) j = false; And you could do do it this way instead also Code: for (int i = 2; i != x; i++) if ((x % i)==0) j = false;

jreelawg	I just ran a test and found this out. Code: int x=2; double y=2; for ( int i=0; i < 29; ++i){ y+=y; //equivalent to y=y+y; x+=x; } if (x==y) cout << "yes"; Turns out that after 29 additions Code: x==y , but after 30, Code: x!=y If you use the modulus operator then you will have no problems Code: (x % i) == //The integer remainder of x/i, which is zero if i is a factor of x.

Hobin	a/b == floor(a/b) is ugly to begin with. Those modulus operators don't exist for no reason.

TylerH	"long long" is always at least 64 bits wide on any x86 arch, if not 128. Use long long if you need more range, but stay away from floating point numbers. Use % to check for remainder. For example, if i % p == 0, then p divides i and i isn't prime if p is any number but i or 1. Also, if you need guaranteed width, look at cstdint.