Prime Factors In C#

Question

I want to create a program in C# 2005 which calculates prime factors of a given input. i want to use the basic and simplest things, no need to create a method for it nor array things etc. just simple modulus. is there any code which fulfills what i desire?

here is the code for finding simple factors, i need this code to be modified to calculate prime factors

class Program
{
    static void Main(string[] args)
    {
        int a, b;
        Console.WriteLine("Please enter your integer: ");
        a = int.Parse(Console.ReadLine());
        for (b = 1; b <= a; b++)
        {
            if (a % b == 0)
            {
                Console.WriteLine(b + " is a factor of " + a);
            }
        }
        Console.ReadLine();



    }
}

Wurth

It should be simple enough to write yourself. Which bit are you stuck on - what do you need help with? – Recursion

i tried so many codes but none of them working, suppose if 10 prime factors are 2 and 5 then its shown in my program as 25...now what is this. – Wurth

can you post some code that you've tried. What do you have so far? – Autopilot

Theres a good overview of algorithms on this question here. – Horseman

Showing 25 not 2 and 5 - are you just using Console.Write(factor);? You might want to write a space between the numbers Console.Write(' ');, or use Console.WriteLine, or something else. – Recursion

the 25 value is stored in factor, and i cant seperate it – Wurth

Uh, that shouldn't happen. You'd have to jump through hoops to make that happen. You'll need to edit the question and add in your code to show us what you're doing. – Recursion

now just see my code which i posted above for finding simple factors, i need that code to modified in such a way that it calculates prime factors. – Wurth

OK, so how are you going to test for primality? You can either generate primes up to your integer first, or for each new number b you can try dividing it by all primes you've discovered so far to see if it's prime itself or - if you don't want to keep a list and are happy with an inefficient solution - you can try dividing it by all of 2 to b/2 to see if it's prime first. – Recursion

Is this a homework, by the way? – Pylle

Answer 1

int a, b;
Console.WriteLine("Please enter your integer: ");
a = int.Parse(Console.ReadLine());

for (b = 2; a > 1; b++)
    if (a % b == 0)
    {
        int x = 0;
        while (a % b == 0)
        {
            a /= b;
            x++;
        }
        Console.WriteLine($"{b} is a prime factor {x} times!");
    }
Console.WriteLine("Th-Th-Th-Th-Th-... That's all, folks!");

Pylle

You'll need to tell us what the error is. (This will eliminate repeated factors, BTW, not filter it down to primes - and you might want to show when a prime factor is repeated?) Actually your code shows 1 as a factor - is the error that it's stuck in an infinite loop trying to factor out 1? 1 will never be a prime factor so it might be easier to start your loop at 2, but you do need to do the prime filtering too. – Recursion

super upvote for works on my machine badget, I'm stealing this. – Borer

@Chris Marisic - it's an old thing that never grows old. :) – Pylle

@ChrisMarisic upvote to his post and your comment. That's genius ! – Similitude

Can't even lie, poking around 8 years later and this one brings me joy – Robedechambre

This is too slow for: 2145423127 – Maverick

@Maverick Quite possibly. It's not a very optimized code. But it is simple and does the job. – Pylle

Answer 2

public static List<int> Generate(int number)
{
    var primes = new List<int>();

    for (int div = 2; div <= number; div++)
        while (number % div == 0)
        {
            primes.Add(div);
            number = number / div;
        }
    
    return primes;
}

If you want to learn steps of development you can watch video here.

Persaud

The divisor can never be larger than number / 2, so you could optimise by writing: for(int div = 2; div<=number / 2; div++){ – Septivalent

Answer 3

You can go one better as the divisor can never be larger than the square root of the number.

 for(int div = 2; div <= Math.Sqrt(number); div++)

Polard

but after end of for you must add number to primes ... if (number > 1) primes.Add(number); – Aprilette

Answer 4

Try this code (I incorporated various solutions in this code). Although, interpolation is not in 2005 (I think so....)

So, anyways, try this:

// C# Program to print all prime factors 
using System; 

namespace prime
{
    public class Prime
    { 

        public static void PrimeFactors(int n)
        {
            Console.Write($"Prime Factors of {n} are:  ");

            // Prints all the numbers of 2  
            while (n % 2 == 0)
            {
                Console.Write("2 ");
                n /= 2;
            }

            // As no 2 can be further divided, this probably means that n
            // is now an odd number
            for(int i = 3; i <= Math.Sqrt(n); i += 2)
            {
                while (n % i == 0)
                {
                    Console.Write($"{i} ");
                    n /= i;
                }
            }

            // This is for case if n is greater than 2
            if (n > 2)
            {
                Console.Write($"{n} ");
            }

        }

        // Prompts user to give Input to number and passes it on 
        // the PrimeFactors function after checking its format
        public static void RunPrimeFactors()
        {
            Console.Write("Enter a number: ");
            if (int.TryParse(Console.ReadLine(), out int n))
            {
                PrimeFactors(n);
            }
            else
            {
                Console.WriteLine("You entered the wrong format");
            }

        }
        // Driver Code 
        public static void Main()
        {
            RunPrimeFactors();
        }

    }
}

Lithic

The performance of this solution is decent, a little bit slower than EnumeratePrimeFactors in most cases. – Maverick

Answer 5

The most efficient way is to use an optimized implementation of wheel factorization. Here is an example that demonstrates a 2, 3, 5 wheel that only needs to consider factors of the form (30k ± {1, 7, 11, 13}).

public static IEnumerable<T> EnumeratePrimeFactors<T>(this T value) where T : IBinaryInteger<T>, IUnsignedNumber<T> {
    if (BinaryIntegerConstants<T>.Four > value) { yield break; }
    if (BinaryIntegerConstants<T>.Five == value) { yield break; }
    if (BinaryIntegerConstants<T>.Seven == value) { yield break; }
    if (BinaryIntegerConstants<T>.Eleven == value) { yield break; }
    if (BinaryIntegerConstants<T>.Thirteen == value) { yield break; }

    var index = value;

    while (T.Zero == (index & T.One)/* enumerate factors of 2 */) {
        yield return BinaryIntegerConstants<T>.Two;

        index >>= 1;
    }
    while (T.Zero == (index % BinaryIntegerConstants<T>.Three)) { // enumerate factors of 3
        yield return BinaryIntegerConstants<T>.Three;

        index /= BinaryIntegerConstants<T>.Three;
    }
    while (T.Zero == (index % BinaryIntegerConstants<T>.Five)/* enumerate factors of 5 */) {
        yield return BinaryIntegerConstants<T>.Five;

        index /= BinaryIntegerConstants<T>.Five;
    }
    while (T.Zero == (index % BinaryIntegerConstants<T>.Seven)/* enumerate factors of 7 */) {
        yield return BinaryIntegerConstants<T>.Seven;

        index /= BinaryIntegerConstants<T>.Seven;
    }
    while (T.Zero == (index % BinaryIntegerConstants<T>.Eleven)/* enumerate factors of 11 */) {
        yield return BinaryIntegerConstants<T>.Eleven;

        index /= BinaryIntegerConstants<T>.Eleven;
    }
    while (T.Zero == (index % BinaryIntegerConstants<T>.Thirteen)/* enumerate factors of 13 */) {
        yield return BinaryIntegerConstants<T>.Thirteen;

        index /= BinaryIntegerConstants<T>.Thirteen;
    }

    var factor = BinaryIntegerConstants<T>.Seventeen;
    var limit = index.SquareRoot();

    if (factor <= limit) {
        do {
            while (T.Zero == (index % factor)/* enumerate factors of (30k - 13) */) {
                yield return factor;

                index /= factor;
            }

            factor += BinaryIntegerConstants<T>.Two;

            while (T.Zero == (index % factor)/* enumerate factors of (30k - 11) */) {
                yield return factor;

                index /= factor;
            }

            factor += BinaryIntegerConstants<T>.Four;

            while (T.Zero == (index % factor)/* enumerate factors of (30k - 7) */) {
                yield return factor;

                index /= factor;
            }

            factor += BinaryIntegerConstants<T>.Six;

            while (T.Zero == (index % factor)/* enumerate factors of (30k - 1) */) {
                yield return factor;

                index /= factor;
            }

            factor += BinaryIntegerConstants<T>.Two;

            while (T.Zero == (index % factor)/* enumerate factors of (30k + 1) */) {
                yield return factor;

                index /= factor;
            }

            factor += BinaryIntegerConstants<T>.Six;

            while (T.Zero == (index % factor)/* enumerate factors of (30k + 7) */) {
                yield return factor;

                index /= factor;
            }

            factor += BinaryIntegerConstants<T>.Four;

            while (T.Zero == (index % factor)/* enumerate factors of (30k + 11) */) {
                yield return factor;

                index /= factor;
            }

            factor += BinaryIntegerConstants<T>.Two;

            while (T.Zero == (index % factor)/* enumerate factors of (30k + 13) */) {
                yield return factor;

                index /= factor;
            }

            factor += BinaryIntegerConstants<T>.Four;
            limit = index.SquareRoot();
        } while (factor <= limit);
    }

    if ((index != T.One) && (index != value)) {
        yield return index;
    }
}

Casias

This is the fastest solution in most cases, but it uses yield, we might not want yield. It uses C# 7 Features that might not be available everywhere. It doesn't return the number if it is a prime itself. – Maverick

Answer 6

This version lists all the factors as an explicit formula:

static void Main(string[] args)
    {
        Console.WriteLine("Please enter your integer (0 to stop): ");

        int a = int.Parse(Console.ReadLine());
        while(a>0)
        {
            List<int> primeFactors = PrimeFactors(a);
            LogFactorList(primeFactors);
            a = int.Parse(Console.ReadLine());
        }
        Console.WriteLine("Goodbye.");
    }


    /// <summary>
    /// Find prime factors
    /// </summary>
    public static List<int> PrimeFactors(int a)
    {
        List<int> retval = new List<int>();
        for (int b = 2; a > 1; b++)
        {               
            while (a % b == 0)
            {
                a /= b;
                retval.Add(b);
            }               
        }
        return retval;
    }

    /// <summary>
    /// Output factor list to console
    /// </summary>
    private static void LogFactorList(List<int> factors)
    {
        if (factors.Count == 1)
        {
            Console.WriteLine("{0} is Prime", factors[0]);
        }
        else
        {
            StringBuilder sb = new StringBuilder();
            for (int i = 0; i < factors.Count; ++i)
            {
                if (i > 0)
                {
                    sb.Append('*');
                }
                sb.AppendFormat("{0}", factors[i]);
            }
            Console.WriteLine(sb.ToString());
        }
    }

Osgood

Answer 7

using static System.Console;

namespace CodeX
{
    public class Program
    {
        public static void Main(string[] args)
        {
            for (int i = 0; i < 20; i++)
            {
                if (IsPrime(i))
                {
                    Write($"{i} ");
                }
            }
        }

        private static bool IsPrime(int number)
        {
            if (number <= 1) return false;  // prime numbers are greater than 1

            for (int i = 2; i < number; i++)
            {
                // only if is not a product of two natural numbers
                if (number % i == 0)       
                    return false;
            }

            return true;
        }
    }
}

Glean

Please consider explaining your answer outside of just providing the code necessary to solve the problem. – Jaala

Found this definition @ wikipedia on prime numbers. Prime numbers starts from 2 and then IsPrime(number) method loops through all numbers starting from 2 but less than the given number to check for a number that can divide the given number without a remainder.. – Glean

This code does not answer the question. It prints all prime numbers from 0 to 19, not the prime factors of a number. It is also very inefficient. – Scaffold