Project Euler Problem 3: Finding the Largest Prime Factor

  • Thread starter Thread starter SELFMADE
  • Start date Start date
  • Tags Tags
    Euler Project
AI Thread Summary
The discussion focuses on finding the largest prime factor of the number 600851475143 using programming. Initial attempts at coding the solution reveal issues with data types, specifically the need for "unsigned long long" to handle large numbers. Participants point out inefficiencies in the approach, suggesting that starting the search for factors from the square root of the number would be more effective. Additionally, the code structure is criticized for being overly complicated and not utilizing best practices. The conversation emphasizes the importance of optimizing algorithms for large numerical computations.
SELFMADE
Messages
80
Reaction score
0
What is the largest prime factor of the number 600851475143 ?

My attempt:
Code:
#include <stdafx.h>

int main() 
{
	int num1, num2, num3, num4, lrgprime, sum;
	lrgprime=num2=num4=sum=0;
	for (num1=2; num1<300425737572; num1++) {
		num2=600851475143%num1; 

		if (num2==0) {
			for (num3=2; num3<=num1; num3++) {
				num4=num1%num3;
				if (num4==0 && num1==num3 && num1>=lrgprime)
					lrgprime=num1; 
			}		
		}	
	}
	printf("This is the largest prime factor: %d", lrgprime);
	return 0;
}

The machine is just not running this code

What did I do wrong?
 
Physics news on Phys.org
check with some smaller numbers first. see whether you get anything
 
Two problems. First, 600851475143 is too large to store in an int (which is usually 32 bits and only goes up to about 2 billion). Try using long longs instead of ints. Second,
Code:
		if (num2==0) {
			for (num3=2; num3<=num1; num3++) {
				num4=num1%num3;
				if (num4==0 && num1==num3 && num1>=lrgprime)
					lrgprime=num1; 
			}		
		}
doesn't do what you want. It eventually always sets lrgprime to num1, even when num1 has divisors. You need to use a break statement to break out of the loop if num4 == 0 and num1 != num3.
 
SELFMADE said:
What is the largest prime factor of the number 600851475143 ?

My attempt:
Code:
#include <stdafx.h>

int main() 
{
	int num1, num2, num3, num4, lrgprime, sum;
	lrgprime=num2=num4=sum=0;
	for (num1=2; num1<300425737572; num1++) {
		num2=600851475143%num1; 

		if (num2==0) {
			for (num3=2; num3<=num1; num3++) {
				num4=num1%num3;
				if (num4==0 && num1==num3 && num1>=lrgprime)
					lrgprime=num1; 
			}		
		}	
	}
	printf("This is the largest prime factor: %d", lrgprime);
	return 0;
}

The machine is just not running this code

What did I do wrong?

1) stdafx.h is not a standard header
2) The maximum value of a 32-bit signed int is 2147483647. Of course, there is no reason to use integers for this task. The largest basic type you could use would be "unsigned long long", which has a maximum value of 18446744073709551615
3) Your code is a mess...you have tons of useless auxiliary variables and have split things into multiple lines just making it more confusing than it should be.
4) This is not an efficient approach, and will take too long on a big number.

Code:
#include <iostream>
#include <limits>

typedef unsigned long long BigInt;
using namespace std;

BigInt largest_prime_factor( const BigInt &x )
{
	//find factors
	for(BigInt y = x/2; y>2; --y)
	{
		//check factors
		if( x%y == 0 )
		{
			//check if factor is prime
			for(BigInt z=2; z<y; ++z)
				if( y%z == 0 )
					goto not_prime;
			return y;
		}
not_prime: ;
	}
	return x;
}

int main() 
{

	cout << "Max value = " << numeric_limits<BigInt>::max() << "\n";
	cout << largest_prime_factor( 600851475143 );
}
 
Last edited:
Code:
for(BigInt y = x/2; y>2; --y)
Just a suggestion: since the largest prime factor of x cannot not exceed x0=(int)sqrt(x), it pays to start at x0, especially for large numbers.
 
Thank you so much, I am going to save this page up, go home and try it. I have been struggling on this problem for 3 days.
 
Thread 'Have I solved this structural engineering equation correctly?'
Hi all, I have a structural engineering book from 1979. I am trying to follow it as best as I can. I have come to a formula that calculates the rotations in radians at the rigid joint that requires an iterative procedure. This equation comes in the form of: $$ x_i = \frac {Q_ih_i + Q_{i+1}h_{i+1}}{4K} + \frac {C}{K}x_{i-1} + \frac {C}{K}x_{i+1} $$ Where: ## Q ## is the horizontal storey shear ## h ## is the storey height ## K = (6G_i + C_i + C_{i+1}) ## ## G = \frac {I_g}{h} ## ## C...

Similar threads

Replies
2
Views
2K
Replies
6
Views
4K
Replies
7
Views
6K
Replies
8
Views
3K
Replies
2
Views
3K
Replies
39
Views
4K
Back
Top