permutations with repeats algorithm with recursion

Question

I'm having some trouble getting this to work using one function, instead of having to use many.

If I want to get permutations with repeats like 2^3. permutations with repeats

to get:

000
001
101
011
100
101
110
111

I can have this function:

   static void Main(string[] args)
    {
        three_permutations(2);
        Console.ReadLine();
    }


    static void three_permutations(int y)
    {

        for (int aa = 0; aa < y; aa++)
        {
            for (int bb = 0; bb < y; bb++)
            {
                for (int cc = 0; cc < y; cc++)
                {
                    Console.Write((aa));
                    Console.Write((bb));
                    Console.Write((cc));
                    Console.WriteLine();
                }
            }
        }

    }

But then to do 4 (like 2^4), the only way I can think is this:

  static void four_permutations(int y)
    {
            for (int aa = 0; aa < y; aa++)
            {
                for (int bb = 0; bb < y; bb++)
                {
                    for (int cc = 0; cc < y; cc++)
                    {
                        for (int dd = 0; dd < y; dd++)
                        {
                            Console.Write((aa));
                            Console.Write((bb));
                            Console.Write((cc));
                            Console.Write((dd));
                            Console.WriteLine();
                        }
                    }
                }
            }
     }

but I'm sure there's a better way using recursion I'm just not sure how to do it. I appreciate any help. Thanks.

Answer 1

void permutations(string text, int numberOfDigits, int numberOfChars)
{
    if (numberOfDigits > 0)
        for (int j = 0; j < numberOfChars; j++)
            permutations(text + j.ToString(), numberOfDigits - 1, numberOfChars);
    else textBox1.Text += text + "\r\n";
}

and call:

permutations("", 3, 2);

Answer 2

Permutations with repetition is essentially counting in another base.

public static void Permutations(int digits, int options)
{
    double maxNumberDouble = Math.Ceiling(Math.Pow(options, digits));
    int maxNumber = (int)maxNumberDouble;
    for (int i = 0; i < maxNumber; i++)
    {
        Console.WriteLine(Convert.ToString(i, options).PadLeft(3, '0'));
    }
}

The example that you've printed is essentially counting from 0 to 8 in base 2.