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();



    }
}

It should be simple enough to write yourself. Which bit are you stuck on - what do you need help with? — Rup, May 03 '11 at 16:59
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. — Aliza, May 03 '11 at 17:02
can you post some code that you've tried. What do you have so far? — BZink, May 03 '11 at 17:03
Theres a good overview of algorithms on this question [here](http://stackoverflow.com/questions/23287/prime-factors). — Matt, May 03 '11 at 17:03
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. — Rup, May 03 '11 at 17:05
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. — Rup, May 03 '11 at 17:07
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. — Aliza, May 03 '11 at 17:10
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. — Rup, May 03 '11 at 17:25

score 51 · Answer 1 · edited Aug 22 '22 at 08:56

51

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!");

Works on my machine!

edited Aug 22 '22 at 08:56

Renat Zamaletdinov

1,210
1
21
36

answered May 03 '11 at 17:12

Vilx-

104,512
87
279
422

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. – Rup May 03 '11 at 17:21
35

super upvote for works on my machine badget, I'm stealing this. – Chris Marisic May 03 '11 at 19:33
@Chris Marisic - it's an [old thing](http://www.codinghorror.com/blog/2007/03/the-works-on-my-machine-certification-program.html) that never grows old. :) – Vilx- May 03 '11 at 19:52
@ChrisMarisic upvote to his post and your comment. That's genius ! – gilles emmanuel Sep 03 '14 at 23:07
Can't even lie, poking around 8 years later and this one brings me joy – zgc7009 Oct 02 '22 at 16:32
This is too slow for: 2145423127 – Charles Aug 15 '23 at 13:09
@Charles Quite possibly. It's not a very optimized code. But it is simple and does the job. – Vilx- Aug 15 '23 at 14:23

score 5 · Answer 2 · edited Nov 24 '20 at 21:40

5

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.

edited Nov 24 '20 at 21:40

AustinWBryan

3,249
3
24
42

answered Jan 28 '16 at 09:44

Bayram Üçüncü

433
3
12

3

The divisor can never be larger than number / 2, so you could optimise by writing: for(int div = 2; div<=number / 2; div++){ – stevieg Apr 24 '16 at 06:57

score 4 · Answer 3 · edited Nov 24 '20 at 21:41

4

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++)

edited Nov 24 '20 at 21:41

AustinWBryan

3,249
3
24
42

answered Sep 19 '16 at 10:00

John Conacher

41
1

but after end of `for` you must add `number` to `primes` ... `if (number > 1) primes.Add(number);` – Cyrus Jan 06 '19 at 08:03

score 2 · Answer 4 · edited May 16 '19 at 18:58

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();
        }

    }
}

The performance of this solution is decent, a little bit slower than `EnumeratePrimeFactors` in most cases. — Charles, Aug 15 '23 at 13:13

Kittoes0124 · Answer 5 · 2023-07-15T00:11:43.743

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;
    }
}

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. — Charles, Aug 15 '23 at 13:15

score 0 · Answer 6 · answered Mar 08 '17 at 03:02

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());
        }
    }

score -1 · Answer 7 · answered Apr 13 '20 at 15:31

-1

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;
        }
    }
}

answered Apr 13 '20 at 15:31

Vincent Amoako-Ampofo

9
1

3

Please consider explaining your answer outside of just providing the code necessary to solve the problem. – codewario Apr 13 '20 at 16:13
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.. – Vincent Amoako-Ampofo Apr 14 '20 at 21:59
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. – Bip901 Feb 03 '21 at 12:41

Prime Factors In C#

7 Answers7

Linked

Related