How to find all of the options for binary number

Question

I have a string array that contains binary numbers: { "xx0x", "110x", "100x", "010x", "x01x", "1010", "0011", "0111", "1111", "xxxx", "0001", "1010", "0110" };

The 'x' can be 0 or 1. I need to create a new array that contains all of the possibilities that exists.

I thought about creating an array with all the options and than compare it, but I'm not sure how to do the compare.

I have edited your title. Please see, "[Should questions include “tags” in their titles?](http://meta.stackexchange.com/questions/19190/)", where the consensus is "no, they should not". — John Saunders, Dec 08 '14 at 14:13
Sounds like a plan to me. Just don't compare the position when you have an `x` in the mask. Note, though, that `xxxx` would cover all possible outcomes, so was that a typo? — 500 - Internal Server Error, Dec 08 '14 at 14:13
I'm still not sure how to make the compare, how do i ignore part of the chars? Yeah xxxx gives me all the options, but I got extra 4 constant digits at the beginning of the number that I don't need to check and that makes the difference:) — user3976173, Dec 08 '14 at 14:16
Post your code without the ignore part that you can't figure out and we can probably help you from there. — 500 - Internal Server Error, Dec 08 '14 at 14:26
This is the full array: string[] InputsToSOD = new string[] { "0000xx0x", "0000110x", "0000100x", "0000010x", "0000x01x", "00001010", "00000011", "00000111", "00001111", "1000xxxx", "10000001", "10001010", "10000110" }; — user3976173, Dec 08 '14 at 14:46

score 0 · Accepted Answer · answered Dec 08 '14 at 14:57

Here is a code who will enumerate all possible values.

           string[] patterns = { "xx0x", "110x", "100x", "010x", "x01x", "1010", "0011", "0111", "1111", "xxxx", "0001", "1010", "0110" };

        // - Loop for each input pattern
        foreach (string pattern in patterns)
        {
            // - Count the varying characters and memorize their position
            int wildcardsCount = pattern.Count(x => x == 'x');
            int[] positions = new int[wildcardsCount];
            int index = 0;
            int position = 0;
            foreach (char c in pattern)
            {
                if (c == 'x')
                    positions[index++] = position;
                position++;
            }

            // - Loop in all possible values taken by missing bits
            int count = (int)Math.Pow(2, wildcardsCount);
            for (int currentValue = 0; currentValue < count; currentValue++)
            {
                // - Prepare the binary string
                string currentValueStr = Convert.ToString(currentValue, 2).PadLeft(wildcardsCount, '0');

                // - Replace each character with known wildcard positions
                char[] tmpOutput = pattern.ToCharArray();
                int charIndex = 0;
                for (charIndex = 0; charIndex < wildcardsCount; charIndex++)
                    tmpOutput[positions[charIndex]] = currentValueStr[charIndex];
                string output = new string(tmpOutput);

                // - Here we are
                Debug.WriteLine(pattern + " (" + (currentValue + 1).ToString() + "/" + count.ToString() + ") => " + output);
            }
        }

Output :

xx0x (1/8) => 0000
xx0x (2/8) => 0001
xx0x (3/8) => 0100
xx0x (4/8) => 0101
xx0x (5/8) => 1000
xx0x (6/8) => 1001
xx0x (7/8) => 1100
xx0x (8/8) => 1101
110x (1/2) => 1100
110x (2/2) => 1101
100x (1/2) => 1000
100x (2/2) => 1001
010x (1/2) => 0100
010x (2/2) => 0101
x01x (1/4) => 0010
x01x (2/4) => 0011
x01x (3/4) => 1010
x01x (4/4) => 1011
1010 (1/1) => 1010
0011 (1/1) => 0011
0111 (1/1) => 0111
1111 (1/1) => 1111
xxxx (1/16) => 0000
xxxx (2/16) => 0001
xxxx (3/16) => 0010
xxxx (4/16) => 0011
xxxx (5/16) => 0100
xxxx (6/16) => 0101
xxxx (7/16) => 0110
xxxx (8/16) => 0111
xxxx (9/16) => 1000
xxxx (10/16) => 1001
xxxx (11/16) => 1010
xxxx (12/16) => 1011
xxxx (13/16) => 1100
xxxx (14/16) => 1101
xxxx (15/16) => 1110
xxxx (16/16) => 1111
0001 (1/1) => 0001
1010 (1/1) => 1010
0110 (1/1) => 0110

Best regards.

How to find all of the options for binary number

1 Answers1