### Problems with my algorithms, any help appreciated

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.

Here is what I have:

 ``1234567891011`` ``````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); } }``````

Any help at all would be a greatly appreciated!
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