Develop Reference

How to perform addition of 2 very large (over 50 digits) binary string values in C#

I have been thinking of adding binary numbers where binary numbers are in a form of string and we add those two binary numbers to get a resultant binary number (in string).
So far I have been using this:
long first = Convert.ToInt64(a, 2);
long second = Convert.ToInt64(b, 2);
long addresult = first + second;
string result = Convert.ToString(addresult, 2);
return result;
Courtesy of Stackoverflow: Binary addition of 2 values represented as strings
But, now I want to add numbers which are far greater than the scope of a long data type.
For Example, a Binary value whose decimel result is a BigInteger, i.e., incredibly huge integers as shown below:
111101101000010111101000101010001010010010010110011010100001000010110110110000110001101 which equals to149014059082024308625334669
1111001101011000001011000111100011101011110100101010010001110101011101010100101000001101000010000110001110100010011101011111111110110101100101110001010101011110001010000010111110011011 which equals to23307765732196437339985049250988196614799400063289798555
At least I think it does.
Courtesy of Stackoverflow:
C# Convert large binary string to decimal system
BigInteger to Hex/Decimal/Octal/Binary strings?
I have used logic provided in above links which are more or less perfect.
But, is there a more compact way to add the given two binary strings?
Kindly let me know as this question is rattling in my mind for some time now.

You can exploit the same scheme you used before but with BigInteger:
using System.Linq;
using System.Numerics;
...
BigInteger first = a.Aggregate(BigInteger.Zero, (s, item) => s * 2 + item - '0');
BigInteger second = b.Aggregate(BigInteger.Zero, (s, item) => s * 2 + item - '0');
StringBuilder sb = new StringBuilder();
for (BigInteger addresult = first + second; addresult > 0; addresult /= 2)
sb.Append(addresult % 2);
if (sb.Length <= 0)
sb.Append('0');
string result = string.Concat(sb.ToString().Reverse());

This question was a nostalgic one - thanks. Note that my code is explanatory and inefficient with little to no validation, but it works for your example. You definitely do not want to use anything like this in a real world solution, this is just to illustrate binary addition in principle.
BinaryString#1
111101101000010111101000101010001010010010010110011010100001000010110110110000110001101
decimal:149014059082024308625334669
BinaryString#2
1111001101011000001011000111100011101011110100101010010001110101011101010100101000001101000010000110001110100010011101011111111110110101100101110001010101011110001010000010111110011011
decimal:23307765732196437339985049250988196614799400063289798555
Calculated Sum
1111001101011000001011000111100011101011110100101010010001110101011101010100101000001101000010001101111011100101011010100101010000000111111000100100101001100110100000111001000100101000
decimal:23307765732196437339985049251137210673881424371915133224
Check
23307765732196437339985049251137210673881424371915133224
decimal:23307765732196437339985049251137210673881424371915133224
Here's the code
using System;
using System.Linq;
using System.Numerics;
namespace ConsoleApp3
{
class Program
{
// return 0 for '0' and 1 for '1' (C# chars promotion to ints)
static int CharAsInt(char c) { return c - '0'; }
// and vice-versa
static char IntAsChar(int bit) { return (char)('0' + bit); }
static string BinaryStringAdd(string x, string y)
{
// get rid of spaces
x = x.Trim();
y = y.Trim();
// check if valid binaries
if (x.Any(c => c != '0' && c != '1') || y.Any(c => c != '0' && c != '1'))
throw new ArgumentException("binary representation may contain only '0' and '1'");
// align on right-most bit
if (x.Length < y.Length)
x = x.PadLeft(y.Length, '0');
else
y = y.PadLeft(x.Length, '0');
// NNB: the result may require one more bit than the longer of the two input strings (carry at the end), let's not forget this
var result = new char[x.Length];
// add from least significant to most significant (right to left)
var i = result.Length;
var carry = '0';
while (--i >= 0)
{
// to add x[i], y[i] and carry
// - if 2 or 3 bits are set then we carry '1' again (otherwise '0')
// - if the number of set bits is odd the sum bit is '1' otherwise '0'
var countSetBits = CharAsInt(x[i]) + CharAsInt(y[i]) + CharAsInt(carry);
carry = countSetBits > 1 ? '1' : '0';
result[i] = countSetBits == 1 || countSetBits == 3 ? '1' : '0';
}
// now to make this byte[] a string
var ret = new string(result);
// remember that final carry?
return carry == '1' ? carry + ret : ret;
}
static BigInteger BigIntegerFromBinaryString(string s)
{
var biRet = new BigInteger(0);
foreach (var t in s)
{
biRet = biRet << 1;
if (t == '1')
biRet += 1;
}
return biRet;
}
static void Main(string[] args)
{
var s1 = "111101101000010111101000101010001010010010010110011010100001000010110110110000110001101";
var s2 = "1111001101011000001011000111100011101011110100101010010001110101011101010100101000001101000010000110001110100010011101011111111110110101100101110001010101011110001010000010111110011011";
var sum = BinaryStringAdd(s1, s2);
var bi1 = BigIntegerFromBinaryString(s1);
var bi2 = BigIntegerFromBinaryString(s2);
var bi3 = bi1 + bi2;
Console.WriteLine($"BinaryString#1\n {s1}\n decimal:{bi1}");
Console.WriteLine($"BinaryString#2\n {s2}\n decimal:{bi2}");
Console.WriteLine($"Calculated Sum\n {sum}\n decimal:{BigIntegerFromBinaryString(sum)}");
Console.WriteLine($"Check\n {bi3}\n decimal:{bi3}");
Console.ReadKey();
}
}
}

I'll add an alternative solution alongside AlanK's just as an example of how you might go about this without converting the numbers to some form of integer before adding them.
static string BinaryStringAdd(string b1, string b2)
{
char[] c = new char[Math.Max(b1.Length, b2.Length) + 1];
int carry = 0;
for (int i = 1; i <= c.Length; i++)
{
int d1 = i <= b1.Length ? b1[^i] : 48;
int d2 = i <= b2.Length ? b2[^i] : 48;
int sum = carry + (d1-48) + (d2-48);
if (sum == 3)
{
sum = 1;
carry = 1;
}
else if (sum == 2)
{
sum = 0;
carry = 1;
}
else
{
carry = 0;
}
c[^i] = (char) (sum+48);
}
return c[0] == '0' ? String.Join("", c)[1..] : String.Join("", c);
}
Note that this solution is ~10% slower than Alan's solution (at least for this test case), and assumes the strings arrive to the method formatted correctly.

Validate user's input against a range of acceptable values

I have a situation where users can assign locations to items.
locations range from A1 to GG17
I guess it is like a row / column format.. Columns Range from A - GG. After Z will be STRICTLY AA, BB, CC, DD, EE, FF, GG.
And rows 1-17
i.e. users can assign location A1, A2, A3, B4, AA1, BB2... GG17
I want to validate the text the user enters when they assigning a location to stop them from adding HH1 or A20 for example..
I am able to achieve this from A - Z, 1-17, but am stumbling when the validation has to go past Z (AA, BB, CC...)
this is what I have for A-Z and 1-17 so far that works
List<char> allowedColumns = new List<char>();
List<int> allowedRows = new List<int>();
char column ='A';
int row = 1;
if (txtAssignLocation.Text != "")
{
while (column <= 'Z')
{
allowedColumns.Add(column);
column++;
}
while (row <= 17)
{
allowedRows.Add(row);
row++;
}
string enteredText = txtAssignLocation.Text;
string enteredColumn = enteredText.Substring(0, 1);
string enteredRow = enteredText.Substring(1);
if (!allowedColumns.Contains(Convert.ToChar(enteredColumn)) || !allowedRows.Contains(Convert.ToInt32(enteredRow)))
{
lblValidationError.Text = "Entered Location does not exist";
}
}
I am at a bit of a loss since Char cannot be more than one character and it and ++ cannot be applied to string

new Regex("^[A-Z][A-G]?[1-9][0-7]?$") validates most of the combinations. Preventing the numbers from 20 to 97 is a bit more tricky. But play around with regular expressions or just split the string and to an int.TryParse and make sure that the number part is <= 17.
EDIT
Yes, it was a bit quick'n'dirty. This one should do the trick:
#"^([A-Z]|([A-G])\2)([1-9]|1[0-7])$"
Example usage:
Regex regex = new Regex(#"^([A-Z]|([A-G])\2)([1-9]|1[0-7])$");
string userInput = "AG20";
bool ok = regex.Match(userInput).Success;

Based on what David said in the comments why not simply have a "hard coded" list of acceptable values since there aren't that many and simply check any entered text against this list?
// This will be a list of all valid locations from 'A1' to 'GG17'
var locations = new HashSet<string>();
// Add all values from 'A1' to 'Z17'
for (char c = 'A'; c <= 'Z'; c++)
{
for (int i = 1; i <= 17; i++)
{
locations.Add($"{c}{i}");
}
}
// Add the values for 'AA1' to 'GG17'
for (char c = 'A'; c <= 'G'; c++)
{
for (int i = 1; i <= 17; i++)
{
locations.Add($"{c}{c}{i}");
}
}
Now you can simply check against this list for validation:
locations.Contains("A1"); // true
locations.Contains("BB10"); // true
locations.Contains("AF7"); // false
locations.Contains("GA10"); // false

You can do this with just providing the maxColumn and maxRow in your method (or passed in as variables) instead of an array or other hardcoding of logic. You could enhance this with LINQ, but providing answer as a loop for clarity:
public static bool ValidateInput(string input)
{
//set maxColumn and maxRow sizes
const string maxColumn = "BG";
const int maxRow = 155;
//initialize input parse variables
string inputColumn = "";
int inputRow = int.MaxValue;
//use only upper-case (to aid in comparison)
input = input.ToUpper();
//parse input into inputColumn and inputRow
for (int i = 0; i < input.Length; i++)
{
if (char.IsDigit(input[i]))
{
inputRow = int.Parse(input.Substring(i));
break;
}
inputColumn += input[i];
}
//make sure the length is at least as long as maxColumn (for comparing single letter column to double-letter column, for example)
inputColumn.PadLeft(maxColumn.Length, ' ');
//return comparison result (inclusive to maxColum and maxRow)
return string.Compare(inputColumn, maxColumn) <= 0 && inputRow <= maxRow;
}
UPDATE: If you are interested in the LINQ version, here it is:
public static bool ValidateInput(string input)
{
//set maxColumn and maxRow sizes
const string maxColumn = "BG";
const int maxRow = 155;
//parse input into inputColumn and inputRow
string inputColumn = new string(input.TakeWhile(char.IsLetter).Select(char.ToUpper).ToArray());
int inputRow = int.Parse(new string(input.Skip(inputColumn.Length).ToArray()));
//make sure the length is at least as long as maxColumn (for comparing single letter column to double-letter column, for example)
inputColumn.PadLeft(maxColumn.Length, ' ');
//return comparison result (inclusive to maxColum and maxRow)
return string.Compare(inputColumn, maxColumn) <= 0 && inputRow <= maxRow;
}

Find all possible combinations of word with and without hyphens

For a string that may have zero or more hyphens in it, I need to extract all the different possibilities with and without hyphens.
For example, the string "A-B" would result in "A-B" and "AB" (two possibilities).
The string "A-B-C" would result in "A-B-C", "AB-C", "A-BC" and "ABC" (four possibilities).
The string "A-B-C-D" would result in "A-B-C-D", "AB-C-D", "A-BC-D", "A-B-CD", "AB-CD", "ABC-D", "A-BCD" and "ABCD" (eight possibilities).
...etc, etc.
I've experimented with some nested loops but haven't been able to get anywhere near the desired result. I suspect I need something recursive unless there is some simple solution I am overlooking.
NB. This is to build a SQL query (shame that SQL Server does't have MySQL's REGEXP pattern matching).
Here is one attempt I was working on. This might work if I do this recursively.
string keyword = "A-B-C-D";
List<int> hyphens = new List<int>();
int pos = keyword.IndexOf('-');
while (pos != -1)
{
hyphens.Add(pos);
pos = keyword.IndexOf('-', pos + 1);
}
for (int i = 0; i < hyphens.Count(); i++)
{
string result = keyword.Substring(0, hyphens[i]) + keyword.Substring(hyphens[i] + 1);
Response.Write("<p>" + result);
}
A B C D are words of varying length.

Take a look at your sample cases. Have you noticed a pattern?
With 1 hyphen there are 2 possibilities.
With 2 hyphens there are 4 possibilities.
With 3 hyphens there are 8 possibilities.
The number of possibilities is 2n.
This is literally exponential growth, so if there are too many hyphens in the string, it will quickly become infeasible to print them all. (With just 30 hyphens there are over a billion combinations!)
That said, for smaller numbers of hyphens it might be interesting to generate a list. To do this, you can think of each hyphen as a bit in a binary number. If the bit is 1, the hyphen is present, otherwise it is not. So this suggests a fairly straightforward solution:
Split the original string on the hyphens
Let n = the number of hyphens
Count from 2n - 1 down to 0. Treat this counter as a bitmask.
For each count begin building a string starting with the first part.
Concatenate each of the remaining parts to the string in order, preceded by a hyphen only if the corresponding bit in the bitmask is set.
Add the resulting string to the output and continue until the counter is exhausted.
Translated to code we have:
public static IEnumerable<string> EnumerateHyphenatedStrings(string s)
{
string[] parts = s.Split('-');
int n = parts.Length - 1;
if (n > 30) throw new Exception("too many hyphens");
for (int m = (1 << n) - 1; m >= 0; m--)
{
StringBuilder sb = new StringBuilder(parts[0]);
for (int i = 1; i <= n; i++)
{
if ((m & (1 << (i - 1))) > 0) sb.Append('-');
sb.Append(parts[i]);
}
yield return sb.ToString();
}
}
Fiddle: https://dotnetfiddle.net/ne3N8f

You should be able to track each hyphen position, and basically say its either there or not there. Loop through all the combinations, and you got all your strings. I found the easiest way to track it was using a binary, since its easy to add those with Convert.ToInt32
I came up with this:
string keyword = "A-B-C-D";
string[] keywordSplit = keyword.Split('-');
int combinations = Convert.ToInt32(Math.Pow(2.0, keywordSplit.Length - 1.0));
List<string> results = new List<string>();
for (int j = 0; j < combinations; j++)
{
string result = "";
string hyphenAdded = Convert.ToString(j, 2).PadLeft(keywordSplit.Length - 1, '0');
// Generate string
for (int i = 0; i < keywordSplit.Length; i++)
{
result += keywordSplit[i] +
((i < keywordSplit.Length - 1) && (hyphenAdded[i].Equals('1')) ? "-" : "");
}
results.Add(result);
}

This works for me:
Func<IEnumerable<string>, IEnumerable<string>> expand = null;
expand = xs =>
{
if (xs != null && xs.Any())
{
var head = xs.First();
if (xs.Skip(1).Any())
{
return expand(xs.Skip(1)).SelectMany(tail => new []
{
head + tail,
head + "-" + tail
});
}
else
{
return new [] { head };
}
}
else
{
return Enumerable.Empty<string>();
}
};
var keyword = "A-B-C-D";
var parts = keyword.Split('-');
var results = expand(parts);
I get:
ABCD
A-BCD
AB-CD
A-B-CD
ABC-D
A-BC-D
AB-C-D
A-B-C-D

I've tested this code and it is working as specified in the question. I stored the strings in a List<string>.
string str = "AB-C-D-EF-G-HI";
string[] splitted = str.Split('-');
List<string> finalList = new List<string>();
string temp = "";
for (int i = 0; i < splitted.Length; i++)
{
temp += splitted[i];
}
finalList.Add(temp);
temp = "";
for (int diff = 0; diff < splitted.Length-1; diff++)
{
for (int start = 1, limit = start + diff; limit < splitted.Length; start++, limit++)
{
int i = 0;
while (i < start)
{
temp += splitted[i++];
}
while (i <= limit)
{
temp += "-";
temp += splitted[i++];
}
while (i < splitted.Length)
{
temp += splitted[i++];
}
finalList.Add(temp);
temp = "";
}
}

I'm not sure your question is entirely well defined (i.e. could you have something like A-BCD-EF-G-H?). For "fully" hyphenated strings (A-B-C-D-...-Z), something like this should do:
string toParse = "A-B-C-D";
char[] toParseChars = toPase.toCharArray();
string result = "";
string binary;
for(int i = 0; i < (int)Math.pow(2, toParse.Length/2); i++) { // Number of subsets of an n-elt set is 2^n
binary = Convert.ToString(i, 2);
while (binary.Length < toParse.Length/2) {
binary = "0" + binary;
}
char[] binChars = binary.ToCharArray();
for (int k = 0; k < binChars.Length; k++) {
result += toParseChars[k*2].ToString();
if (binChars[k] == '1') {
result += "-";
}
}
result += toParseChars[toParseChars.Length-1];
Console.WriteLine(result);
}
The idea here is that we want to create a binary word for each possible hyphen. So, if we have A-B-C-D (three hyphens), we create binary words 000, 001, 010, 011, 100, 101, 110, and 111. Note that if we have n hyphens, we need 2^n binary words.
Then each word maps to the output you desire by inserting the hyphen where we have a '1' in our word (000 -> ABCD, 001 -> ABC-D, 010 -> AB-CD, etc). I didn't test the code above, but this is at least one way to solve the problem for fully hyphenated words.
Disclaimer: I didn't actually test the code

Reverse binary representation of int (only significant bits)

I'm trying to write a program for reversing numbers in binary. For instance, the binary representation of 13 is 1101, and reversing it gives 1011, which corresponds to number 11 right?
Here's my code:
static void Main(string[] args)
{
Console.WriteLine("Enter a Number");
int numb = int.Parse(Console.ReadLine());
int reverse = 0;
while (numb > 0)
{
int rem = numb % 10;
reverse = (reverse * 10) + rem;
numb = numb / 10;
}
Console.WriteLine("Reverse number={0}", reverse);
Console.ReadLine();
}
By this code I only get the numbers to reverse (13 -> 31)...
The input should contain a single line with an integer N, 1≤N≤1000000000 and I want my output in one line with one integer, the number I want to get by reversing the binary representation of N.

Something like that
// 13 = 1101b
int value = 13;
// 11 = 1011b
int result = Convert.ToInt32(new String(
Convert.ToString(value, 2)
.Reverse()
.ToArray()), 2);
Explanation:
Convert.ToString(value, 2) returns value in binary representation ("1101")
Reverse().ToArray() - reverse the string ('1','0','1','1') as sequence of characters and converts to array char[].
new String(...) constructs string "1011" from array of char
finally, Convert.ToInt32(..., 2) convert binary representation back to int

You can use Convert.ToString and Convert.ToInt32 methods, where 2 means binary:
int numb = int.Parse(Console.ReadLine());
var reversedString = Convert.ToString(numb, 2).ReverseString();
var result = Convert.ToInt32(reversedString, 2);
...
public static string ReverseString(this string s)
{
char[] arr = s.ToCharArray();
Array.Reverse(arr);
return new string(arr);
}

A fun excercise would be doing this without using the string conversion.
I have very little experience with bit twiddling so there is probably a faster and better way of doing this, but this seems to work:
public static IEnumerable<bool> ToBinary(this int n)
{
for (int i = 0; i < 32; i++)
{
yield return (n & (1 << i)) != 0;
}
}
public static int ToInt(this IEnumerable<bool> b)
{
var n = 0;
var counter = 0;
foreach (var i in b.Trim().Take(32))
{
n = n | (i ? 1 : 0) << counter;
counter++
}
return n;
}
private static IEnumerable<bool> Trim(this IEnumerable<bool> list)
{
bool trim = true;
foreach (var i in list)
{
if (i)
{
trim = false;
}
if (!trim)
{
yield return i;
}
}
}
And now you'd call it like this:
var reversed = n.ToBinary().Reverse().ToInt();

Mask out part first 12 characters of string with *?

How can I take the value 123456789012345 or 1234567890123456 and turn it into:
************2345 and ************3456
The difference between the strings above is that one contains 15 digits and the other contains 16.
I have tried the following, but it does not keep the last 4 digits of the 15 digit number and now matter what the length of the string, be it 13, 14, 15, or 16, I want to mask all beginning digits with a *, but keep the last 4. Here is what I have tried:
String.Format("{0}{1}", "************", str.Substring(11, str.Length - 12))

Something like this:
string s = "1234567890123"; // example
string result = s.Substring(s.Length - 4).PadLeft(s.Length, '*');
This will mask all but the last four characters of the string. It assumes that the source string is at least 4 characters long.

using System;
class Program
{
static void Main()
{
var str = "1234567890123456";
if (str.Length > 4)
{
Console.WriteLine(
string.Concat(
"".PadLeft(12, '*'),
str.Substring(str.Length - 4)
)
);
}
else
{
Console.WriteLine(str);
}
}
}

Easiest way: Create an extension method to extract the last four digits. Use that in your String.Format call.
For example:
public static string LastFour(this string value)
{
if (string.IsNullOrEmpty(value) || value.length < 4)
{
return "0000";
}
return value.Substring(value.Length - 4, 4)
}
In your code:
String.Format("{0}{1}", "************", str.LastFour());
In my opinion, this leads to more readable code, and it's reusable.
EDIT: Perhaps not the easiest way, but an alternative way that may produce more maintainable results. <shrug/>

Try this:
var maskSize = ccDigits.Length - 4;
var mask = new string('*', maskSize) + ccDigits.Substring(maskSize);

LINQ:
char maskBy = '*';
string input = "123456789012345";
int count = input.Length <= 4 ? 0 : input.Length - 4;
string output = new string(input.Select((c, i) => i < count ? maskBy : c).ToArray());

static private String MaskInput(String input, int charactersToShowAtEnd)
{
if (input.Length < charactersToShowAtEnd)
{
charactersToShowAtEnd = input.Length;
}
String endCharacters = input.Substring(input.Length - charactersToShowAtEnd);
return String.Format(
"{0}{1}",
"".PadLeft(input.Length - charactersToShowAtEnd, '*'),
endCharacters
);
}
Adjust the function header as required, call with:
MaskInput("yourInputHere", 4);

private string MaskDigits(string input)
{
//take first 6 characters
string firstPart = input.Substring(0, 6);
//take last 4 characters
int len = input.Length;
string lastPart = input.Substring(len - 4, 4);
//take the middle part (****)
int middlePartLenght = len - (firstPart.Length + lastPart.Length);
string middlePart = new String('*', middlePartLenght);
return firstPart + middlePart + lastPart;
}
MaskDigits("1234567890123456");
// output : "123456******3456"

Try the following:
private string MaskString(string s)
{
int NUM_ASTERISKS = 4;
if (s.Length < NUM_ASTERISKS) return s;
int asterisks = s.Length - NUM_ASTERISKS;
string result = new string('*', asterisks);
result += s.Substring(s.Length - NUM_ASTERISKS);
return result;
}

Regex with a match evaluator will do the job
string filterCC(string source) {
var x=new Regex(#"^\d+(?=\d{4}$)");
return x.Replace(source,match => new String('*',match.Value.Length));
}
This will match any number of digits followed by 4 digits and the end (it won't include the 4 digits in the replace). The replace function will replace the match with a string of * of equal length.
This has the additional benefit that you could use it as a validation algorthim too. Change the first + to {11,12} to make it match a total of 15 or 16 chars and then you can use x.IsMatch to determine validity.
EDIT
Alternatively if you always want a 16 char result just use
return x.Replace(source,new String('*',12));

// "123456789".MaskFront results in "****56789"
public static string MaskFront(this string str, int len, char c)
{
var strArray = str.ToCharArray();
for (var i = 0; i < len; i++)
{
if(i < strArray.Length)
{
strArray[i] = c;
}
else
{
break;
}
}
return string.Join("", strArray);
}
// "123456789".MaskBack results in "12345****"
public static string MaskBack(this string str, int len, char c)
{
var strArray = str.ToCharArray();
var tracker = strArray.Length - 1;
for (var i = 0; i < len; i++)
{
if (tracker > -1)
{
strArray[tracker] = c;
tracker--;
}
else
{
break;
}
}
return string.Join("", strArray);
}

Try this out:
static string Mask(string str)
{
if (str.Length <= 4) return str;
Regex rgx = new Regex(#"(.*?)(\d{4})$");
string result = String.Empty;
if (rgx.IsMatch(str))
{
for (int i = 0; i < rgx.Matches(str)[0].Groups[1].Length; i++)
result += "*";
result += rgx.Matches(str)[0].Groups[2];
return result;
}
return str;
}

Mask from start and from end with sending char
public static string Maskwith(this string value, int fromStart, int fromEnd, char ch)
{
return (value?.Length >= fromStart + fromEnd) ?
string.Concat(Enumerable.Repeat(ch, fromStart)) + value.Substring(fromStart, value.Length - (fromStart + fromEnd)) + string.Concat(Enumerable.Repeat(ch, fromEnd))
: "";
} //Console.WriteLine("mytestmask".Maskwith(2,3,'*')); **testm***
show chars from start and from end by passing value and mask the middle
public static string MasktheMiddle(this string value, int visibleCharLength, char ch)
{
if (value?.Length <= (visibleCharLength * 2))
return string.Concat(Enumerable.Repeat(ch,value.Length));
else
return value.Substring(0, visibleCharLength) + string.Concat(Enumerable.Repeat(ch, value.Length - (visibleCharLength * 2))) + value.Substring(value.Length - visibleCharLength);
} //Console.WriteLine("mytestmask".MasktheMiddle(2,'*')); Result: my******sk

How can I take the value 123456789012345 or 1234567890123456 and turn it into:
************2345 and ************3456
one more way to do this:
var result = new string('*',0,value.Length - 4) + new string(value.Skip(value.Length - 4).ToArray())
// or using string.Join

An extension method using C# 8's index and range:
public static string MaskStart(this string input, int showNumChars, char maskChar = '*') =>
input[^Math.Min(input.Length, showNumChars)..]
.PadLeft(input.Length, maskChar);

A simple way
string s = "1234567890123"; // example
int l = s.Length;
s = s.Substring(l - 4);
string r = new string('*', l);
r = r + s;