### Prime number factorization

Hey everyone!

I was wondering if someone could help me fix two of my algorithms. I am trying to factorize semi-prime numbers by using a brute force method and a square root method.

The brute force method should start by multiplying all odd numbers between 1 and the semi-prime number. Meaning that 1 should multiply 3 then 5 then 7, etc. Then three should multiply 5 then 7 then 9 and so forth. The first number would never multiply a number equal to or less than itself.

Any help at all would be greatly appreciated!

Here is what I have:

 ``123456789101112`` ``````bruteForce(int x){ f1 = 1; f2 = f1 + 2; while (f1 < x){ for (f2 = x; f2 <= x; f2-2){ f1 * f2; } if (x%f1 == f2 || x%f2 == f1) printFactors(f1,f2); } f1 = f1 + 2; }; ``````

For my other algorithm, I want it to get the square root of the semi-prime and then start multiplying odd numbers around it. I made it so that the square root would have to be set to the next even number (in case it was a factor). If the product of the two odd numbers (a * b) is greater than the original semi-prime, I would multiply b by the next smaller number than a. Conversely, if the product of a and b is less than the original semi-prime, it will multiply a by the next odd number larger than b. This will repeat until the two correct factors of the original semi-prime are found.

Algorithm:

 ``123456789101112131415161718`` ``````SquareRootFactorize(int x){ int N = x; sqrtN = sqrt(N); if (isOdd(sqrtN) == true) sqrtN = sqrtN + 1; f1 = sqrtN - 1; f2 = sqrtN + 1; cout << sqrtN; while (N != n){ n = multFacts(f1,f2); if (N < (f1 * f2)) f1 = f1 + 2; else if (N > (f1 * f2)) f2 = f2 + 2; else if (N = n) printFactors(f1,f2); } }``````
The fun about algorithms is that you sometimes think your brain is bleeding trying to get a glance of what the hell is going on

Giving you the answer wouldn't help you at all (even if you were to understand what's happening)
The key of learning how-to is to come up with a solution yourself and learn the process how to think to get there

Maybe I can point you in a direction and show you that whatever you think the code below is doing isn't what it acutally does.

 ``1234567891011121314`` ``````bruteForce(int x){ f1 = 1; f2 = f1 + 2; // here you set f2 to 3 while (f1 < x){ // looks okay for (f2 = x; f2 <= x; f2-2){ // and here you set f2 to x which overwrites the 3 f1 * f2; // I think your heart is in the right place (is x an odd number?) // if not then you are about to multiply all even numbers from x down to infinity // you have to save the result of f1 * f2 somewhere here so you use the result for your next iteration // On top of all that it looks like an infinite loop (since your f2 is getting smaller it will always be <= x ( btw you wrote f2-2 it should be f2 -= 2) } if (x%f1 == f2 || x%f2 == f1) // I don't get that printFactors(f1,f2); } f1 = f1 + 2; // this seems okay };``````

If you figured this out come back for the second one but don't try to multitask while both Tasks include algorithms ;)

Well I hope this helps you somehow
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