Develop Reference

Writing an UnHash Method Based on The Following Hash Method

Some time ago I went through an interview process and got stumped on a technical question.
The problem to solve was given the hash method below, write a corresponding unhash method.
static Int64 hashThis (String s)
{
Int64 h = 7;
String letters = "acdegiklmnoprsuw";
for (Int32 i = 0; i < s.Length; i++)
{
h = (h * 37 + letters.IndexOf(s[i]));
}
return h;
}
My failed progress thus far is this:
static String unhashThis(Int64 integer)
{
Int64 h = 7;
String letters = "acdegiklmnoprsuw";
String unhashed = "";
// as long as h < than integer passed in
// this is so we iterate the correct number of times
for (Int32 i = 0; h < integer; i++)
{
h = (h * 37 + letters.IndexOf(letters[i]));
Int64 rem = h % 7;
Int64 rem2 = 7 % h;
//unhashed += letters[(int)h];
//if (h > letters.Length)
//{
// // loop through the letters to find the character
// for (int j = 0; j < h; j++)
// {
// if (j == letters.Length)
// {
// j = 0;
// }
// else
// {
// unhashed += letters[j];
// }
// }
//}
}
return unhashed;
Just to reiterate I am revisiting this as I am trying to get better at problem solving so that I can do better on future interviews.

this works:
static String unHashThis(Int64 hash)
{
String letters = "acdegiklmnoprsuw";
String phrase = "";
while (hash >= 7 * 37)
{
int i = (int)(hash % 37);
phrase = letters[i]+phrase;
hash = (hash - hash % 37) / 37;
}
return phrase;
}
Notes:
First if the string you use to hash and unhash this it's has characters that are not in the "letters" string, it's going to fail
Try to figure why it works, if you can't, tell me and I'll explain
This is closer to encrypton than to Hashing I believe, Hash is not meant to UN-Hash

Nice question. Here is a Python version that also handles characters missing in letters. It uses the fact that find returns -1 when the string is not found (like IndexOf in C#):
C = 37 # must be > len(letters)
letters = "cdategiorkswlm"
def hash_this(s):
h = 7
for ch in s:
h = h * C + letters.find(ch)
return h;
def unhash_this(v):
h = 7
s = ''
while v > h:
j = int(v % C)
if j >= len(letters):
j = -1
v = (v - j) // C
s += '_' if j == -1 else letters[j]
return s[::-1]
hash_this('stackoverflow')
# 402116075536062288302
unhash_this(hash_this('stackoverflow'))
# 'stacko_er_low'

How to convert a string of bits to a HEX [duplicate]

I'm looking for a way to convert a long string of binary to a hex string.
the binary string looks something like this "0110011010010111001001110101011100110100001101101000011001010110001101101011"
I've tried using
hex = String.Format("{0:X2}", Convert.ToUInt64(hex, 2));
but that only works if the binary string fits into a Uint64 which if the string is long enough it won't.
is there another way to convert a string of binary into hex?
Thanks

I just knocked this up. Maybe you can use as a starting point...
public static string BinaryStringToHexString(string binary)
{
if (string.IsNullOrEmpty(binary))
return binary;
StringBuilder result = new StringBuilder(binary.Length / 8 + 1);
// TODO: check all 1's or 0's... throw otherwise
int mod4Len = binary.Length % 8;
if (mod4Len != 0)
{
// pad to length multiple of 8
binary = binary.PadLeft(((binary.Length / 8) + 1) * 8, '0');
}
for (int i = 0; i < binary.Length; i += 8)
{
string eightBits = binary.Substring(i, 8);
result.AppendFormat("{0:X2}", Convert.ToByte(eightBits, 2));
}
return result.ToString();
}

This might help you:
string HexConverted(string strBinary)
{
string strHex = Convert.ToInt32(strBinary,2).ToString("X");
return strHex;
}

Convert.ToInt32("1011", 2).ToString("X");
For string longer than this, you can simply break it into multiple bytes:
var binary = "0110011010010111001001110101011100110100001101101000011001010110001101101011";
var hex = string.Join(" ",
Enumerable.Range(0, binary.Length / 8)
.Select(i => Convert.ToByte(binary.Substring(i * 8, 8), 2).ToString("X2")));

I came up with this method. I am new to programming and C# but I hope you will appreciate it:
static string BinToHex(string bin)
{
StringBuilder binary = new StringBuilder(bin);
bool isNegative = false;
if (binary[0] == '-')
{
isNegative = true;
binary.Remove(0, 1);
}
for (int i = 0, length = binary.Length; i < (4 - length % 4) % 4; i++) //padding leading zeros
{
binary.Insert(0, '0');
}
StringBuilder hexadecimal = new StringBuilder();
StringBuilder word = new StringBuilder("0000");
for (int i = 0; i < binary.Length; i += 4)
{
for (int j = i; j < i + 4; j++)
{
word[j % 4] = binary[j];
}
switch (word.ToString())
{
case "0000": hexadecimal.Append('0'); break;
case "0001": hexadecimal.Append('1'); break;
case "0010": hexadecimal.Append('2'); break;
case "0011": hexadecimal.Append('3'); break;
case "0100": hexadecimal.Append('4'); break;
case "0101": hexadecimal.Append('5'); break;
case "0110": hexadecimal.Append('6'); break;
case "0111": hexadecimal.Append('7'); break;
case "1000": hexadecimal.Append('8'); break;
case "1001": hexadecimal.Append('9'); break;
case "1010": hexadecimal.Append('A'); break;
case "1011": hexadecimal.Append('B'); break;
case "1100": hexadecimal.Append('C'); break;
case "1101": hexadecimal.Append('D'); break;
case "1110": hexadecimal.Append('E'); break;
case "1111": hexadecimal.Append('F'); break;
default:
return "Invalid number";
}
}
if (isNegative)
{
hexadecimal.Insert(0, '-');
}
return hexadecimal.ToString();
}

Considering four bits can be expressed by one hex value, you can simply go by groups of four and convert them seperately, the value won't change that way.
string bin = "11110110";
int rest = bin.Length % 4;
if(rest != 0)
bin = new string('0', 4-rest) + bin; //pad the length out to by divideable by 4
string output = "";
for(int i = 0; i <= bin.Length - 4; i +=4)
{
output += string.Format("{0:X}", Convert.ToByte(bin.Substring(i, 4), 2));
}

If you want to iterate over the hexadecimal representation of each byte in the string, you could use the following extension. I've combined Mitch's answer with this.
static class StringExtensions
{
public static IEnumerable<string> ToHex(this String s) {
if (s == null)
throw new ArgumentNullException("s");
int mod4Len = s.Length % 8;
if (mod4Len != 0)
{
// pad to length multiple of 8
s = s.PadLeft(((s.Length / 8) + 1) * 8, '0');
}
int numBitsInByte = 8;
for (var i = 0; i < s.Length; i += numBitsInByte)
{
string eightBits = s.Substring(i, numBitsInByte);
yield return string.Format("{0:X2}", Convert.ToByte(eightBits, 2));
}
}
}
Example:
string test = "0110011010010111001001110101011100110100001101101000011001010110001101101011";
foreach (var hexVal in test.ToHex())
{
Console.WriteLine(hexVal);
}
Prints
06
69
72
75
73
43
68
65
63
6B

If you're using .NET 4.0 or later and if you're willing to use System.Numerics.dll (for BigInteger class), the following solution works fine:
public static string ConvertBigBinaryToHex(string bigBinary)
{
BigInteger bigInt = BigInteger.Zero;
int exponent = 0;
for (int i = bigBinary.Length - 1; i >= 0; i--, exponent++)
{
if (bigBinary[i] == '1')
bigInt += BigInteger.Pow(2, exponent);
}
return bigInt.ToString("X");
}

Considering four bits can be expressed by one hex value, you can simply go by groups of four and convert them seperately, the value won't change that way.
string bin = "11110110";
int rest = bin.Length % 4;
bin = bin.PadLeft(rest, '0'); //pad the length out to by divideable by 4
string output = "";
for(int i = 0; i <= bin.Length - 4; i +=4)
{
output += string.Format("{0:X}", Convert.ToByte(bin.Substring(i, 4), 2));
}

static string BinToHex(string bin)
{
if (bin == null)
throw new ArgumentNullException("bin");
if (bin.Length % 8 != 0)
throw new ArgumentException("The length must be a multiple of 8", "bin");
var hex = Enumerable.Range(0, bin.Length / 8)
.Select(i => bin.Substring(8 * i, 8))
.Select(s => Convert.ToByte(s, 2))
.Select(b => b.ToString("x2"));
return String.Join(null, hex);
}

Using LINQ
string BinaryToHex(string binaryString)
{
var offset = 0;
StringBuilder sb = new();
while (offset < binaryString.Length)
{
var nibble = binaryString
.Skip(offset)
.Take(4);
sb.Append($"{Convert.ToUInt32(nibble.toString()), 2):X}");
offset += 4;
}
return sb.ToString();
}

You can take the input number four digit at a time. Convert this digit to ex ( as you did is ok ) then concat the string all together. So you obtain a string representing the number in hex, independetly from the size. Depending on where start MSB on your input string, may be the output string you obtain the way i described must be reversed.

Packet dump / hex view format for byte[] to text file

I want to convert a packet (byte[]) into a hex dump view on a text file but it got messy.
Here is a picture.
I was looping on the main byte[] until I reach mod of 16 (excluding 0) but then if the last line isn't a factor of 16 then ill end up with ASCII representation shifted to the left
I did try several solutions but non is really decent, so I considered it a blind spot and thought I could use some help :)
Also I could use some help on how to check if that character has an ascii representation or should I just replace it with '.'
using (var sw = File.AppendText(Path))
{
sw.WriteLine("[Packet Logged at {0} with direction {1}] with length {2} and type {3}",
DateTime.Now.ToLongTimeString().Replace("PM", string.Empty).Replace("AM", string.Empty).Replace(" ", string.Empty),
ReadWrite.ReadString(Packet,Packet.Length-8,8) == "removed for privacy" ? "Server->Client" : "Client -> Server",
BitConverter.ToUInt16(Packet, 0), BitConverter.ToUInt16(Packet, 2));
for (int i = 0; i < Packet.Length; i++)
{
if (i%16 == 0 && i != 0)
{
}
else
{
sw.Write(Packet[i]);
}
}
}

Taken from http://www.codeproject.com/Articles/36747/Quick-and-Dirty-HexDump-of-a-Byte-Array:
using System.Text;
namespace HexDump
{
class Utils
{
public static string HexDump(byte[] bytes, int bytesPerLine = 16)
{
if (bytes == null) return "<null>";
int bytesLength = bytes.Length;
char[] HexChars = "0123456789ABCDEF".ToCharArray();
int firstHexColumn =
8 // 8 characters for the address
+ 3; // 3 spaces
int firstCharColumn = firstHexColumn
+ bytesPerLine * 3 // - 2 digit for the hexadecimal value and 1 space
+ (bytesPerLine - 1) / 8 // - 1 extra space every 8 characters from the 9th
+ 2; // 2 spaces
int lineLength = firstCharColumn
+ bytesPerLine // - characters to show the ascii value
+ Environment.NewLine.Length; // Carriage return and line feed (should normally be 2)
char[] line = (new String(' ', lineLength - 2) + Environment.NewLine).ToCharArray();
int expectedLines = (bytesLength + bytesPerLine - 1) / bytesPerLine;
StringBuilder result = new StringBuilder(expectedLines * lineLength);
for (int i = 0; i < bytesLength; i += bytesPerLine)
{
line[0] = HexChars[(i >> 28) & 0xF];
line[1] = HexChars[(i >> 24) & 0xF];
line[2] = HexChars[(i >> 20) & 0xF];
line[3] = HexChars[(i >> 16) & 0xF];
line[4] = HexChars[(i >> 12) & 0xF];
line[5] = HexChars[(i >> 8) & 0xF];
line[6] = HexChars[(i >> 4) & 0xF];
line[7] = HexChars[(i >> 0) & 0xF];
int hexColumn = firstHexColumn;
int charColumn = firstCharColumn;
for (int j = 0; j < bytesPerLine; j++)
{
if (j > 0 && (j & 7) == 0) hexColumn++;
if (i + j >= bytesLength)
{
line[hexColumn] = ' ';
line[hexColumn + 1] = ' ';
line[charColumn] = ' ';
}
else
{
byte b = bytes[i + j];
line[hexColumn] = HexChars[(b >> 4) & 0xF];
line[hexColumn + 1] = HexChars[b & 0xF];
line[charColumn] = asciiSymbol( b );
}
hexColumn += 3;
charColumn++;
}
result.Append(line);
}
return result.ToString();
}
static char asciiSymbol( byte val )
{
if( val < 32 ) return '.'; // Non-printable ASCII
if( val < 127 ) return (char)val; // Normal ASCII
// Handle the hole in Latin-1
if( val == 127 ) return '.';
if( val < 0x90 ) return "€.‚ƒ„…†‡ˆ‰Š‹Œ.Ž."[ val & 0xF ];
if( val < 0xA0 ) return ".‘’“”•–—˜™š›œ.žŸ"[ val & 0xF ];
if( val == 0xAD ) return '.'; // Soft hyphen: this symbol is zero-width even in monospace fonts
return (char)val; // Normal Latin-1
}
}
}

I wrote this solution for dumping binary data:
public static class BinaryVisualiser
{
public static string FormatAsHex(ReadOnlySpan<byte> data)
{
byte ReplaceControlCharacterWithDot(byte character)
=> character < 31 || character >= 127 ? (byte)46 /* dot */ : character;
byte[] ReplaceControlCharactersWithDots(byte[] characters)
=> characters.Select(ReplaceControlCharacterWithDot).ToArray();
var result = new StringBuilder();
const int lineWidth = 16;
for (var pos = 0; pos < data.Length;)
{
var line = data.Slice(pos, Math.Min(lineWidth, data.Length - pos)).ToArray();
var asHex = string.Join(" ", line.Select(v => v.ToString("X2", CultureInfo.InvariantCulture)));
asHex += new string(' ', lineWidth * 3 - 1 - asHex.Length);
var asCharacters = Encoding.ASCII.GetString(ReplaceControlCharactersWithDots(line));
result.Append(FormattableString.Invariant($"{pos:X4} {asHex} {asCharacters}\n"));
pos += line.Length;
}
return result.ToString();
}
}
Note that this implementation leverages Span from C# 7.2 and requires the System.Memory NuGet package from Microsoft.
The output is formatted like this:
0000 EF BB BF 48 65 6C 6C 6F 2C 20 53 74 61 63 6B 20 ...Hello, Stack
0010 4F 76 65 72 66 6C 6F 77 21 Overflow!
Update: configurable byte width.
public static string FormatAsHex(ReadOnlySpan<byte> data, int lineWidth = 16, int byteWidth = 1)
{
byte ReplaceControlCharacterWithDot(byte character) => character < 31 || character >= 127 ? (byte)46 /* dot */ : character;
byte[] ReplaceControlCharactersWithDots(byte[] characters) => characters.Select(ReplaceControlCharacterWithDot).ToArray();
IEnumerable<BigInteger> Chunk(IReadOnlyList<byte> source, int size) => source.Select((item, index) => source.Skip(size * index).Take(size).ToArray()).TakeWhile(bucket => bucket.Any()).Select(e => new BigInteger(e));
var result = new StringBuilder();
for(var pos = 0; pos < data.Length;)
{
var line = data.Slice(pos, Math.Min(lineWidth * byteWidth, data.Length - pos)).ToArray();
var asHex = string.Join(" ", Chunk(line, byteWidth).Select(v => v.ToString("X" + byteWidth * 2, CultureInfo.InvariantCulture)));
asHex += new string(' ', lineWidth * (byteWidth * 2 + 1) - 1 - asHex.Length);
var asCharacters = Encoding.ASCII.GetString(ReplaceControlCharactersWithDots(line));
result.Append(Invariant($"{pos:X4} {asHex} {asCharacters}\n"));
pos += line.Length;
}
return result.ToString();
}
Which allows (encoded as big-endian UTF-16):
0000 4800 6500 6C00 6C00 6F00 2C00 .H.e.l.l.o.,
000C 2000 5300 7400 6100 6300 6B00 . .S.t.a.c.k
0018 2000 4F00 7600 6500 7200 6600 . .O.v.e.r.f
0024 6C00 6F00 7700 .l.o.w

Yes partition 2 is the bootable partition with 80
The size of partition for the image above is 2056237 bytes
dd if=/dev/zero of=/dev/sdx bs=512 skip=2265165 count=1028160
The alignment doesn't matter for the end sector , at least not for performance reasons . Alignment of the start sectors affects in the partition; alignmengt of the last sectore only afftets the last few sectors of the partition , if at all. So sectors are numberes from 0 and the fdisk is suggesting the last sector on your disk which has above results.
hope this was helpful.

Decoding Bitcoin Base58 address to byte array

I'm trying to decode bitcoin address from Base58 string into byte array, and to do that I rewrited original function from Satoshi repository (https://github.com/bitcoin/bitcoin/blob/master/src/base58.cpp), written in c++, to c# (which I'm using).
Original code
static const char* pszBase58 = "123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz";
bool DecodeBase58(const char *psz, std::vector<unsigned char>& vch) {
// Skip leading spaces.
while (*psz && isspace(*psz))
psz++;
// Skip and count leading '1's.
int zeroes = 0;
while (*psz == '1') {
zeroes++;
psz++;
}
// Allocate enough space in big-endian base256 representation.
std::vector<unsigned char> b256(strlen(psz) * 733 / 1000 + 1); // log(58) / log(256), rounded up.
// Process the characters.
while (*psz && !isspace(*psz)) {
// Decode base58 character
const char *ch = strchr(pszBase58, *psz);
if (ch == NULL)
return false;
// Apply "b256 = b256 * 58 + ch".
int carry = ch - pszBase58;
for (std::vector<unsigned char>::reverse_iterator it = b256.rbegin(); it != b256.rend(); it++) {
carry += 58 * (*it);
*it = carry % 256;
carry /= 256;
}
assert(carry == 0);
psz++;
}
// Skip trailing spaces.
while (isspace(*psz))
psz++;
if (*psz != 0)
return false;
// Skip leading zeroes in b256.
std::vector<unsigned char>::iterator it = b256.begin();
while (it != b256.end() && *it == 0)
it++;
// Copy result into output vector.
vch.reserve(zeroes + (b256.end() - it));
vch.assign(zeroes, 0x00);
while (it != b256.end())
vch.push_back(*(it++));
return true;
}
Mine rewrited c# version
private static string Base58characters = "123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz";
public static bool Decode(string source, ref byte[] destination)
{
int i = 0;
while (i < source.Length)
{
if (source[i] == 0 || !Char.IsWhiteSpace(source[i]))
{
break;
}
i++;
}
int zeros = 0;
while (source[i] == '1')
{
zeros++;
i++;
}
byte[] b256 = new byte[(source.Length - i) * 733 / 1000 + 1];
while (i < source.Length && !Char.IsWhiteSpace(source[i]))
{
int ch = Base58characters.IndexOf(source[i]);
if (ch == -1) //null
{
return false;
}
int carry = Base58characters.IndexOf(source[i]);
for (int k = b256.Length - 1; k > 0; k--)
{
carry += 58 * b256[k];
b256[k] = (byte)(carry % 256);
carry /= 256;
}
i++;
}
while (i < source.Length && Char.IsWhiteSpace(source[i]))
{
i++;
}
if (i != source.Length)
{
return false;
}
int j = 0;
while (j < b256.Length && b256[j] == 0)
{
j++;
}
destination = new byte[zeros + (b256.Length - j)];
for (int kk = 0; kk < destination.Length; kk++)
{
if (kk < zeros)
{
destination[kk] = 0x00;
}
else
{
destination[kk] = b256[j++];
}
}
return true;
}
Function that I'm using for converting from byte-array to HexString
public static string ByteArrayToHexString(byte[] source)
{
return BitConverter.ToString(source).Replace("-", "");
}
To test if everything is working correctly I've used test cases found online here (https://github.com/ThePiachu/Bitcoin-Unit-Tests/blob/master/Address/Address%20Generation%20Test%201.txt). Good thing is that 97% of this test are passed correctly but for 3 there is a little error and I do not know where it is coming from. So my ask to you is to point me what could for these test go wrong or where in rewriting I've made an error. Thank you in advance.
The test cases where errors occures are 1, 21 and 25.
1.
Input:
16UwLL9Risc3QfPqBUvKofHmBQ7wMtjvM
Output:
000966776006953D5567439E5E39F86A0D273BEED61967F6
Should be:
00010966776006953D5567439E5E39F86A0D273BEED61967F6
21.
Input:
1v3VUYGogXD7S1E8kipahj7QXgC568dz1
Output:
0008201462985DF5255E4A6C9D493C932FAC98EF791E2F22
Should be:
000A08201462985DF5255E4A6C9D493C932FAC98EF791E2F22
25.
Input:
1axVFjCkMWDFCHjQHf99AsszXTuzxLxxg
Output:
006C0B8995C7464E89F6760900EA6978DF18157388421561
Should be:
00066C0B8995C7464E89F6760900EA6978DF18157388421561

In your for-loop:
for (int k = b256.Length - 1; k > 0; k--)
The loop condition should be k >= 0 so that you don't skip the first byte in b256.

Converting long string of binary to hex c#

I'm looking for a way to convert a long string of binary to a hex string.
the binary string looks something like this "0110011010010111001001110101011100110100001101101000011001010110001101101011"
I've tried using
hex = String.Format("{0:X2}", Convert.ToUInt64(hex, 2));
but that only works if the binary string fits into a Uint64 which if the string is long enough it won't.
is there another way to convert a string of binary into hex?
Thanks

I just knocked this up. Maybe you can use as a starting point...
public static string BinaryStringToHexString(string binary)
{
if (string.IsNullOrEmpty(binary))
return binary;
StringBuilder result = new StringBuilder(binary.Length / 8 + 1);
// TODO: check all 1's or 0's... throw otherwise
int mod4Len = binary.Length % 8;
if (mod4Len != 0)
{
// pad to length multiple of 8
binary = binary.PadLeft(((binary.Length / 8) + 1) * 8, '0');
}
for (int i = 0; i < binary.Length; i += 8)
{
string eightBits = binary.Substring(i, 8);
result.AppendFormat("{0:X2}", Convert.ToByte(eightBits, 2));
}
return result.ToString();
}

This might help you:
string HexConverted(string strBinary)
{
string strHex = Convert.ToInt32(strBinary,2).ToString("X");
return strHex;
}

Convert.ToInt32("1011", 2).ToString("X");
For string longer than this, you can simply break it into multiple bytes:
var binary = "0110011010010111001001110101011100110100001101101000011001010110001101101011";
var hex = string.Join(" ",
Enumerable.Range(0, binary.Length / 8)
.Select(i => Convert.ToByte(binary.Substring(i * 8, 8), 2).ToString("X2")));

I came up with this method. I am new to programming and C# but I hope you will appreciate it:
static string BinToHex(string bin)
{
StringBuilder binary = new StringBuilder(bin);
bool isNegative = false;
if (binary[0] == '-')
{
isNegative = true;
binary.Remove(0, 1);
}
for (int i = 0, length = binary.Length; i < (4 - length % 4) % 4; i++) //padding leading zeros
{
binary.Insert(0, '0');
}
StringBuilder hexadecimal = new StringBuilder();
StringBuilder word = new StringBuilder("0000");
for (int i = 0; i < binary.Length; i += 4)
{
for (int j = i; j < i + 4; j++)
{
word[j % 4] = binary[j];
}
switch (word.ToString())
{
case "0000": hexadecimal.Append('0'); break;
case "0001": hexadecimal.Append('1'); break;
case "0010": hexadecimal.Append('2'); break;
case "0011": hexadecimal.Append('3'); break;
case "0100": hexadecimal.Append('4'); break;
case "0101": hexadecimal.Append('5'); break;
case "0110": hexadecimal.Append('6'); break;
case "0111": hexadecimal.Append('7'); break;
case "1000": hexadecimal.Append('8'); break;
case "1001": hexadecimal.Append('9'); break;
case "1010": hexadecimal.Append('A'); break;
case "1011": hexadecimal.Append('B'); break;
case "1100": hexadecimal.Append('C'); break;
case "1101": hexadecimal.Append('D'); break;
case "1110": hexadecimal.Append('E'); break;
case "1111": hexadecimal.Append('F'); break;
default:
return "Invalid number";
}
}
if (isNegative)
{
hexadecimal.Insert(0, '-');
}
return hexadecimal.ToString();
}

Considering four bits can be expressed by one hex value, you can simply go by groups of four and convert them seperately, the value won't change that way.
string bin = "11110110";
int rest = bin.Length % 4;
if(rest != 0)
bin = new string('0', 4-rest) + bin; //pad the length out to by divideable by 4
string output = "";
for(int i = 0; i <= bin.Length - 4; i +=4)
{
output += string.Format("{0:X}", Convert.ToByte(bin.Substring(i, 4), 2));
}

If you want to iterate over the hexadecimal representation of each byte in the string, you could use the following extension. I've combined Mitch's answer with this.
static class StringExtensions
{
public static IEnumerable<string> ToHex(this String s) {
if (s == null)
throw new ArgumentNullException("s");
int mod4Len = s.Length % 8;
if (mod4Len != 0)
{
// pad to length multiple of 8
s = s.PadLeft(((s.Length / 8) + 1) * 8, '0');
}
int numBitsInByte = 8;
for (var i = 0; i < s.Length; i += numBitsInByte)
{
string eightBits = s.Substring(i, numBitsInByte);
yield return string.Format("{0:X2}", Convert.ToByte(eightBits, 2));
}
}
}
Example:
string test = "0110011010010111001001110101011100110100001101101000011001010110001101101011";
foreach (var hexVal in test.ToHex())
{
Console.WriteLine(hexVal);
}
Prints
06
69
72
75
73
43
68
65
63
6B

If you're using .NET 4.0 or later and if you're willing to use System.Numerics.dll (for BigInteger class), the following solution works fine:
public static string ConvertBigBinaryToHex(string bigBinary)
{
BigInteger bigInt = BigInteger.Zero;
int exponent = 0;
for (int i = bigBinary.Length - 1; i >= 0; i--, exponent++)
{
if (bigBinary[i] == '1')
bigInt += BigInteger.Pow(2, exponent);
}
return bigInt.ToString("X");
}

Considering four bits can be expressed by one hex value, you can simply go by groups of four and convert them seperately, the value won't change that way.
string bin = "11110110";
int rest = bin.Length % 4;
bin = bin.PadLeft(rest, '0'); //pad the length out to by divideable by 4
string output = "";
for(int i = 0; i <= bin.Length - 4; i +=4)
{
output += string.Format("{0:X}", Convert.ToByte(bin.Substring(i, 4), 2));
}

static string BinToHex(string bin)
{
if (bin == null)
throw new ArgumentNullException("bin");
if (bin.Length % 8 != 0)
throw new ArgumentException("The length must be a multiple of 8", "bin");
var hex = Enumerable.Range(0, bin.Length / 8)
.Select(i => bin.Substring(8 * i, 8))
.Select(s => Convert.ToByte(s, 2))
.Select(b => b.ToString("x2"));
return String.Join(null, hex);
}

Using LINQ
string BinaryToHex(string binaryString)
{
var offset = 0;
StringBuilder sb = new();
while (offset < binaryString.Length)
{
var nibble = binaryString
.Skip(offset)
.Take(4);
sb.Append($"{Convert.ToUInt32(nibble.toString()), 2):X}");
offset += 4;
}
return sb.ToString();
}

You can take the input number four digit at a time. Convert this digit to ex ( as you did is ok ) then concat the string all together. So you obtain a string representing the number in hex, independetly from the size. Depending on where start MSB on your input string, may be the output string you obtain the way i described must be reversed.