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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.

i want to round up values like 124.33 to 124.50 and 124.57 to 125.00 in C# [closed]

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if doublevalue = 124.75 then it shoud be rounded to 125.00
if doublevalue = 124.25 then it shoud be rounded to 124.50
in short after decimal point digit greater than 50 must rounded to 100 and less than 50 should rounded to 50
please help me get this type of code

Simply do the following:
Math.Ceiling(YourValue * 2)/2
Explanation
Suppose your number can be written as X + Y, where X is the integer part and Y is the fractional part. Multiplying it with 2 will make it 2X + 2Y, where 2X will be an even number, double of the integer part. For Y, there are two cases:
If Y >= 0.5, 2Y will be equal to 1 + Z (where 0 <= Z < 1), thus the entire number will be 2X + 1 + Z.
If Y < 0.5, 2Y will be equal to Z (where 0 <= Z < 1), thus the entire number will be 2X + Z.
Doing Math.Ceiling() in the first case will return 2X + 2 and dividing it by 2 will return X + 1, which is the nearest higher integer, (3.7 will become 4).
Doing Math.Ceiling() in the second case will return 2X + 1 and dividing it by 2 will return X + 0.5, or X.5 , (3.3 will become 3.5).

your conditions are not clear from your question, but the principle is to isolate the digits after the decimal point and that decide what to do:
private static double GetFixedNumber(double n)
{
var x = n - Math.Floor(n);
double result = 0;
if (x >= 0.75)
{
result = Math.Floor(n) + 1;
}
else if (x >= 0.25)
{
result = Math.Floor(n) + 0.5;
}
else
{
result = Math.Floor(n);
}
return result;
}

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text.RegularExpressions;
namespace Rextester
{
public class Program
{
public static void Main(string[] args)
{
//Your code goes here
Console.WriteLine( RoundUp(124.25, 2));
}
public static double RoundUp(double input, int places)
{
double multiplier = input*2;
double numberround = Math.Round(multiplier, MidpointRounding.AwayFromZero)/2;
return numberround;
}
}
}

Or compute it yourself:
// flipped the decision around - else it is kindof the same.
static double WorksAsWell(double input)
{
var floored = Math.Floor(input);
if (floored == input) // special case so exact 127. stays that way
return floored;
if (input - floored >= 0.5)
return floored+1;
return floored+0.5;
}
static double RoundToHalfesOrFulls(double input)
{
int asInt = (int)Math.Floor(input);
double remainder = input-asInt;
if(remainder < 0.5)
return asInt+0.5;
return asInt+1.0;
}
public static void Main()
{
for(int i= 12700; i < 12800; i+=5)
System.Console.WriteLine(string.Format("{0} = {1}",i/100.0, RoundToHalfesOrFulls(i/100.0)));
}
}
Output: (RoundToHalfesOrFulls)
127 = 127
127.05 = 127
127.1 = 127
127.15 = 127
127.2 = 127
127.25 = 127
127.3 = 127
127.35 = 127
127.4 = 127
127.45 = 127
127.5 = 127.5
127.55 = 127.5
127.6 = 127.5
127.65 = 127.5
127.7 = 127.5
127.75 = 127.5
127.8 = 127.5
127.85 = 127.5
127.9 = 127.5
127.95 = 127.5

List of Non-Negative Integers - concatenate into largest possible nr - e.g. [1,2,5] = 521 [closed]

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I'm following this blog asking
Five programming problems every Software Engineer should be able to solve in less than 1 hour
I am absolutely stumped by question 4 (5 is a different story too)
Write a function that given a list of non negative integers, arranges them such that they form the largest possible number. For example, given [50, 2, 1, 9], the largest formed number is 95021.
Now the author posted a answer, and I saw an python attempt too:
import math
numbers = [50,2,1,9,10,100,52]
def arrange(lst):
for i in xrange(0, len(lst)):
for j in xrange(0, len(lst)):
if i != j:
comparison = compare(lst[i], lst[j])
if lst[i] == comparison[0]:
temp = lst[j]
lst[j] = lst[i]
lst[i] = temp
return lst
def compare(num1, num2):
pow10_1 = math.floor(math.log10(num1))
pow10_2 = math.floor(math.log10(num2))
temp1 = num1
temp2 = num2
if pow10_1 > pow10_2:
temp2 = (temp2 / math.pow(10, pow10_2)) * math.pow(10, pow10_1)
elif pow10_2 > pow10_1:
temp1 = (temp1 / math.pow(10, pow10_1)) * math.pow(10, pow10_2)
print "Starting", num1, num2
print "Comparing", temp1, temp2
if temp1 > temp2:
return [num1, num2]
elif temp2 > temp1:
return [num2, num1]
else:
if num1 < num2:
return [num1, num2]
else:
return [num2, num1]
print arrange(numbers)
but I'm not going to learn these languages soon. Is anyone willing to share how they would sort the numbers in C# to form the largest number please?
I've also tried straight conversion to C# in VaryCode but when the IComparer gets involved, then it causes erroneous conversions.
The python attempt uses a bubble sort it seems.
Is bubble sort a starting point? What else would be used?

The solution presented uses the approach to sort the numbers in the array in a special way. For example the value 5 comes before 50 because "505" ("50"+"5") comes before "550" ("5"+"50"). The thinking behind this isn't explained, and I'm not convinced that it actually works...
When looking that the problem, I came to this solution:
You can do it recursively. Make a method that loops through the numbers in the array and concatenates each number with the largest value that can be formed by the remaining numbers, to see which one of those is largest:
public static int GetLargest(int[] numbers) {
if (numbers.Length == 1) {
return numbers[0];
} else {
int largest = 0;
for (int i = 0; i < numbers.Length; i++) {
int[] other = numbers.Take(i).Concat(numbers.Skip(i + 1)).ToArray();
int n = Int32.Parse(numbers[i].ToString() + GetLargest(other).ToString());
if (i == 0 || n > largest) {
largest = n;
}
}
return largest;
}
}

If you want to do it by bubble sort, try this:
public static void Main()
{
bool swapped = true;
while (swapped)
{
swapped = false;
for (int i = 0; i < VALUES.Length - 1; i++)
{
if (Compare(VALUES[i], VALUES[i + 1]) > 0)
{
int temp = VALUES[i];
VALUES[i] = VALUES[i + 1];
VALUES[i + 1] = temp;
swapped = true;
}
}
}
String result = "";
foreach (int integer in VALUES)
{
result += integer.ToString();
}
Console.WriteLine(result);
}
public static int Compare(int lhs, int rhs)
{
String v1 = lhs.ToString();
String v2 = rhs.ToString();
return (v1 + v2).CompareTo(v2 + v1) * -1;
}
The Compare method compares the order of two numbers that would create the largest number. When it returns value larger than 0, that means you need to swap.
This is the sorting part:
while (swapped)
{
swapped = false;
for (int i = 0; i < VALUES.Length - 1; i++)
{
if (Compare(VALUES[i], VALUES[i + 1]) > 0)
{
int temp = VALUES[i];
VALUES[i] = VALUES[i + 1];
VALUES[i + 1] = temp;
swapped = true;
}
}
}
It checks to consecutive values in the array and swaps if necessary. After one iteration without swapping, the sort is finished.
Finally, you concatenate the values in the array and print it out.

How to compute the average for every n number in a list [closed]

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Having a List of numbers, I want to take every n(e.g. 5 , 10 , etc.) element, compute their average and put the average in a new list. As an example lets assume we have the following list:
1 , 2 , 3 , 4 , 5 , 6 , 7 , 8
Now we compute the average every 2 elements and we will get the following list as output:
1.5 , 3.5 , 5.5 , 7.5
How can I do that?

You could use a for-loop and Enumerable.Average:
var averages = new List<double>();
for (int i = 0; i < ints.Length; i += 2)
{
int thisInt = ints[i];
int nextInt = i == ints.Length - 1 ? thisInt : ints[i + 1];
averages.Add(new[] { thisInt, nextInt }.Average());
}
Here's a dynamic approach that works with any length:
int take = 2;
for (int i = 0; i < ints.Length; i += take)
{
if(i + take >= ints.Length)
take = ints.Length - i;
int[] subArray = new int[take];
Array.Copy(ints, i, subArray, 0, take);
averages.Add(subArray.Average());
}

This problem is just testing your use of iteration and the modulus operator. Modulus gives you the remainder of division, you can use it to check whether or not the current number should be included in the average as you iterate the array. Here is a sample method;
public float nthsAverage(int n, int[] numbers)
{
// quick check to avoid a divide by 0 error
if (numbers.Length == 0)
return 0;
int sum = 0;
int count = 0;
for (int i = 0; i < numbers.Length; i++)
{
// might want i+1 here instead to compensate for array being 0 indexed, ie 9th number is at the 8th index
if (i % n == 0)
{
sum = sum + numbers[i];
count++;
}
}
return (float)sum / count;
}

public List<double> Average(List<double> number, int nElement)
{
var currentElement = 0;
var currentSum = 0.0;
var newList = new List<double>();
foreach (var item in number)
{
currentSum += item;
currentElement++;
if(currentElement == nElement)
{
newList.Add(currentSum / nElement);
currentElement = 0;
currentSum = 0.0;
}
}
// Maybe the array element count is not the same to the asked, so average the last sum. You can remove this condition if you want
if(currentElement > 0)
{
newList.Add(currentSum / currentElement);
}
return newList;
}

An algorithm for a number divisible to n

At first user gives a number (n) to program, for example 5.
the program must find the smallest number that can be divided to n (5).
and this number can only consist of digits 0 and 9 not any other digits.
for example if user gives 5 to program.
numbers that can be divided to 5 are:
5, 10, 15, 20, 25, 30, ..., 85, 90, 95, ...
but 90 here is the smallest number that can be divided to 5 and also consist of digits (0 , 9). so answer for 5 must be 90.
and answer for 9 is 9, because it can be divided to 9 and consist of digit (9).
my code
string a = txtNumber.Text;
Int64 x = Convert.ToInt64(a);
Int64 i ,j=1,y=x;
bool t = false;
for (i = x + 1; t == false; i++)
{
if (i % 9 == 0 && i % 10 == 0 && i % x == 0)
{
j = i;
for (; (i /= 10) != 0; )
{
i /= 10;
if (i == 0)
t = true;
continue;
}
}
}
lblAnswer.Text = Convert.ToString(j);

If you're happy to go purely functional then this works:
Func<IEnumerable<long>> generate = () =>
{
Func<long, IEnumerable<long>> extend =
x => new [] { x * 10, x * 10 + 9 };
Func<IEnumerable<long>, IEnumerable<long>> generate2 = null;
generate2 = ns =>
{
var clean = ns.Where(n => n > 0).ToArray();
return clean.Any()
? clean.Concat(generate2(clean.SelectMany(extend)))
: Enumerable.Empty<long>();
};
return generate2(new[] { 9L, });
};
Func<long, long?> f = n =>
generate()
.Where(x => x % n == 0L)
.Cast<long?>()
.FirstOrDefault();
So rather than iterate through all possible values and test for 0 & 9 and divisibility, this just generates only numbers with 0 & 9 and then only tests for visibility. It is much faster this way.
I can call it like this:
var result = f(5L); // 90L
result = f(23L); //990909L
result = f(123L); //99999L
result = f(12321L); //90900999009L
result = f(123212L); //99909990090000900L
result = f(117238L); //990990990099990990L
result = f(1172438L); //null == No answer
These results are super fast. f(117238L) returns a result on my computer in 138ms.

You can try this way :
string a = txtNumber.Text;
Int64 x = Convert.ToInt64(a);
int counter;
for (counter = 1; !isValid(x * counter); counter++)
{
}
lblAnswer.Text = Convert.ToString(counter*x);
code above works by searching multiple of x incrementally until result that satisfy criteria : "consist of only 0 and or 9 digits" found. By searching only multiple of x, it is guaranteed to be divisible by x. So the rest is checking validity of result candidate, in this case using following isValid() function :
private static bool isValid(int number)
{
var lastDigit = number%10;
//last digit is invalid, return false
if (lastDigit != 0 & lastDigit != 9) return false;
//last digit is valid, but there is other digit(s)
if(number/10 >= 1)
{
//check validity of digit(s) before the last
return isValid(number/10);
}
//last digit is valid, and there is no other digit. return true
return true;
}
About strange empty for loop in snippet above, it is just syntactic sugar, to make the code a bit shorter. It is equal to following while loop :
counter = 1;
while(!isValid(input*counter))
{
counter++;
}

Use this simple code
int inputNumber = 5/*Or every other number, you can get this number from input.*/;
int result=1;
for (int i = 1; !IsOk(result,inputNumber); i++)
{
result = i*inputNumber;
}
Print(result);
IsOk method is here:
bool IsOk(int result, int inputNumber)
{
if(result%inputNumber!=0)
return false;
if(result.ToString().Replace("9",string.Empty).Replace("0",string.Empty).Length!=0)
return false;
return true;
}

My first solution has very bad performance, because of converting a number to string and looking for characters '9' and '0'.
New solution:
My new solution has very good performance and is a technical approach since of using Breadth-first search(BFS).
Algorithm of this solution:
For every input number, test 9, if it is answer print it, else add 2 child numbers (90 & 99) to queue, and continue till finding answer.
int inputNumber = 5;/*Or every other number, you can get this number from input.*/
long result;
var q = new Queue<long>();
q.Enqueue(9);
while (true)
{
result = q.Dequeue();
if (result%inputNumber == 0)
{
Print(result);
break;
}
q.Enqueue(result*10);
q.Enqueue(result*10 + 9);
}
Trace of number creation:
9
90,99
900,909,990,999
9000,9009,9090,9099,9900,9909,9990,9999
.
.
.

I wrote this code for console, and i used goto command however it is not prefered but i could not write it with only for.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace main
{
class Program
{
static void Main(string[] args)
{
Console.WriteLine("Enter your number");
Int64 x = Convert.ToInt64(Console.ReadLine());
Int64 y, j, i, k, z = x;
x = x + 5;
loop:
x++;
for (i = 0, y = x; y != 0; i++)
y /= 10;
for (j = x, k = i; k != 0; j /= 10, k--)
{
if (j % 10 != 9)
if (j % 10 != 0)
goto loop;
}
if (x % z != 0)
goto loop;
Console.WriteLine("answer:{0}",x);
Console.ReadKey();
}
}
}