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.
If I have a double (234.004223), etc., I would like to round this to x significant digits in C#.
So far I can only find ways to round to x decimal places, but this simply removes the precision if there are any 0s in the number.
For example, 0.086 to one decimal place becomes 0.1, but I would like it to stay at 0.08.
The framework doesn't have a built-in function to round (or truncate, as in your example) to a number of significant digits. One way you can do this, though, is to scale your number so that your first significant digit is right after the decimal point, round (or truncate), then scale back. The following code should do the trick:
static double RoundToSignificantDigits(this double d, int digits){
if(d == 0)
return 0;
double scale = Math.Pow(10, Math.Floor(Math.Log10(Math.Abs(d))) + 1);
return scale * Math.Round(d / scale, digits);
}
If, as in your example, you really want to truncate, then you want:
static double TruncateToSignificantDigits(this double d, int digits){
if(d == 0)
return 0;
double scale = Math.Pow(10, Math.Floor(Math.Log10(Math.Abs(d))) + 1 - digits);
return scale * Math.Truncate(d / scale);
}
I've been using pDaddy's sigfig function for a few months and found a bug in it. You cannot take the Log of a negative number, so if d is negative the results is NaN.
The following corrects the bug:
public static double SetSigFigs(double d, int digits)
{
if(d == 0)
return 0;
decimal scale = (decimal)Math.Pow(10, Math.Floor(Math.Log10(Math.Abs(d))) + 1);
return (double) (scale * Math.Round((decimal)d / scale, digits));
}
It sounds to me like you don't want to round to x decimal places at all - you want to round to x significant digits. So in your example, you want to round 0.086 to one significant digit, not one decimal place.
Now, using a double and rounding to a number of significant digits is problematic to start with, due to the way doubles are stored. For instance, you could round 0.12 to something close to 0.1, but 0.1 isn't exactly representable as a double. Are you sure you shouldn't actually be using a decimal? Alternatively, is this actually for display purposes? If it's for display purposes, I suspect you should actually convert the double directly to a string with the relevant number of significant digits.
If you can answer those points, I can try to come up with some appropriate code. Awful as it sounds, converting to a number of significant digits as a string by converting the number to a "full" string and then finding the first significant digit (and then taking appropriate rounding action after that) may well be the best way to go.
If it is for display purposes (as you state in the comment to Jon Skeet's answer), you should use Gn format specifier. Where n is the number of significant digits - exactly what you are after.
Here is the the example of usage if you want 3 significant digits (printed output is in the comment of each line):
Console.WriteLine(1.2345e-10.ToString("G3"));//1.23E-10
Console.WriteLine(1.2345e-5.ToString("G3")); //1.23E-05
Console.WriteLine(1.2345e-4.ToString("G3")); //0.000123
Console.WriteLine(1.2345e-3.ToString("G3")); //0.00123
Console.WriteLine(1.2345e-2.ToString("G3")); //0.0123
Console.WriteLine(1.2345e-1.ToString("G3")); //0.123
Console.WriteLine(1.2345e2.ToString("G3")); //123
Console.WriteLine(1.2345e3.ToString("G3")); //1.23E+03
Console.WriteLine(1.2345e4.ToString("G3")); //1.23E+04
Console.WriteLine(1.2345e5.ToString("G3")); //1.23E+05
Console.WriteLine(1.2345e10.ToString("G3")); //1.23E+10
I found two bugs in the methods of P Daddy and Eric. This solves for example the precision error that was presented by Andrew Hancox in this Q&A. There was also a problem with round directions. 1050 with two significant figures isn't 1000.0, it's 1100.0. The rounding was fixed with MidpointRounding.AwayFromZero.
static void Main(string[] args) {
double x = RoundToSignificantDigits(1050, 2); // Old = 1000.0, New = 1100.0
double y = RoundToSignificantDigits(5084611353.0, 4); // Old = 5084999999.999999, New = 5085000000.0
double z = RoundToSignificantDigits(50.846, 4); // Old = 50.849999999999994, New = 50.85
}
static double RoundToSignificantDigits(double d, int digits) {
if (d == 0.0) {
return 0.0;
}
else {
double leftSideNumbers = Math.Floor(Math.Log10(Math.Abs(d))) + 1;
double scale = Math.Pow(10, leftSideNumbers);
double result = scale * Math.Round(d / scale, digits, MidpointRounding.AwayFromZero);
// Clean possible precision error.
if ((int)leftSideNumbers >= digits) {
return Math.Round(result, 0, MidpointRounding.AwayFromZero);
}
else {
return Math.Round(result, digits - (int)leftSideNumbers, MidpointRounding.AwayFromZero);
}
}
}
As Jon Skeet mentions: better handle this in the textual domain. As a rule: for display purposes, don't try to round / change your floating point values, it never quite works 100%. Display is a secondary concern and you should handle any special formatting requirements like these working with strings.
My solution below I implemented several years ago and has proven very reliable. It has been thoroughly tested and it performs quite well also. About 5 times longer in execution time than P Daddy / Eric's solution.
Examples of input + output given below in code.
using System;
using System.Text;
namespace KZ.SigDig
{
public static class SignificantDigits
{
public static string DecimalSeparator;
static SignificantDigits()
{
System.Globalization.CultureInfo ci = System.Threading.Thread.CurrentThread.CurrentCulture;
DecimalSeparator = ci.NumberFormat.NumberDecimalSeparator;
}
/// <summary>
/// Format a double to a given number of significant digits.
/// </summary>
/// <example>
/// 0.086 -> "0.09" (digits = 1)
/// 0.00030908 -> "0.00031" (digits = 2)
/// 1239451.0 -> "1240000" (digits = 3)
/// 5084611353.0 -> "5085000000" (digits = 4)
/// 0.00000000000000000846113537656557 -> "0.00000000000000000846114" (digits = 6)
/// 50.8437 -> "50.84" (digits = 4)
/// 50.846 -> "50.85" (digits = 4)
/// 990.0 -> "1000" (digits = 1)
/// -5488.0 -> "-5000" (digits = 1)
/// -990.0 -> "-1000" (digits = 1)
/// 0.0000789 -> "0.000079" (digits = 2)
/// </example>
public static string Format(double number, int digits, bool showTrailingZeros = true, bool alwaysShowDecimalSeparator = false)
{
if (Double.IsNaN(number) ||
Double.IsInfinity(number))
{
return number.ToString();
}
string sSign = "";
string sBefore = "0"; // Before the decimal separator
string sAfter = ""; // After the decimal separator
if (number != 0d)
{
if (digits < 1)
{
throw new ArgumentException("The digits parameter must be greater than zero.");
}
if (number < 0d)
{
sSign = "-";
number = Math.Abs(number);
}
// Use scientific formatting as an intermediate step
string sFormatString = "{0:" + new String('#', digits) + "E0}";
string sScientific = String.Format(sFormatString, number);
string sSignificand = sScientific.Substring(0, digits);
int exponent = Int32.Parse(sScientific.Substring(digits + 1));
// (the significand now already contains the requested number of digits with no decimal separator in it)
StringBuilder sFractionalBreakup = new StringBuilder(sSignificand);
if (!showTrailingZeros)
{
while (sFractionalBreakup[sFractionalBreakup.Length - 1] == '0')
{
sFractionalBreakup.Length--;
exponent++;
}
}
// Place decimal separator (insert zeros if necessary)
int separatorPosition = 0;
if ((sFractionalBreakup.Length + exponent) < 1)
{
sFractionalBreakup.Insert(0, "0", 1 - sFractionalBreakup.Length - exponent);
separatorPosition = 1;
}
else if (exponent > 0)
{
sFractionalBreakup.Append('0', exponent);
separatorPosition = sFractionalBreakup.Length;
}
else
{
separatorPosition = sFractionalBreakup.Length + exponent;
}
sBefore = sFractionalBreakup.ToString();
if (separatorPosition < sBefore.Length)
{
sAfter = sBefore.Substring(separatorPosition);
sBefore = sBefore.Remove(separatorPosition);
}
}
string sReturnValue = sSign + sBefore;
if (sAfter == "")
{
if (alwaysShowDecimalSeparator)
{
sReturnValue += DecimalSeparator + "0";
}
}
else
{
sReturnValue += DecimalSeparator + sAfter;
}
return sReturnValue;
}
}
}
Math.Round() on doubles is flawed (see Notes to Callers in its documentation). The later step of multiplying the rounded number back up by its decimal exponent will introduce further floating point errors in the trailing digits. Using another Round() as #Rowanto does won't reliably help and suffers from other problems. However if you're willing to go via decimal then Math.Round() is reliable, as is multiplying and dividing by powers of 10:
static ClassName()
{
powersOf10 = new decimal[28 + 1 + 28];
powersOf10[28] = 1;
decimal pup = 1, pdown = 1;
for (int i = 1; i < 29; i++) {
pup *= 10;
powersOf10[i + 28] = pup;
pdown /= 10;
powersOf10[28 - i] = pdown;
}
}
/// <summary>Powers of 10 indexed by power+28. These are all the powers
/// of 10 that can be represented using decimal.</summary>
static decimal[] powersOf10;
static double RoundToSignificantDigits(double v, int digits)
{
if (v == 0.0 || Double.IsNaN(v) || Double.IsInfinity(v)) {
return v;
} else {
int decimal_exponent = (int)Math.Floor(Math.Log10(Math.Abs(v))) + 1;
if (decimal_exponent < -28 + digits || decimal_exponent > 28 - digits) {
// Decimals won't help outside their range of representation.
// Insert flawed Double solutions here if you like.
return v;
} else {
decimal d = (decimal)v;
decimal scale = powersOf10[decimal_exponent + 28];
return (double)(scale * Math.Round(d / scale, digits, MidpointRounding.AwayFromZero));
}
}
}
I agree with the spirit of Jon's assessment:
Awful as it sounds, converting to a number of significant digits as a string by converting the number to a "full" string and then finding the first significant digit (and then taking appropriate rounding action after that) may well be the best way to go.
I needed significant-digit rounding for approximate and non-performance-critical computational purposes, and the format-parse round-trip through "G" format is good enough:
public static double RoundToSignificantDigits(this double value, int numberOfSignificantDigits)
{
return double.Parse(value.ToString("G" + numberOfSignificantDigits));
}
This question is similiar to the one you're asking:
Formatting numbers with significant figures in C#
Thus you could do the following:
double Input2 = 234.004223;
string Result2 = Math.Floor(Input2) + Convert.ToDouble(String.Format("{0:G1}", Input2 - Math.Floor(Input2))).ToString("R6");
Rounded to 1 significant digit.
Let inputNumber be input that needs to be converted with significantDigitsRequired after decimal point, then significantDigitsResult is the answer to the following pseudo code.
integerPortion = Math.truncate(**inputNumber**)
decimalPortion = myNumber-IntegerPortion
if( decimalPortion <> 0 )
{
significantDigitsStartFrom = Math.Ceil(-log10(decimalPortion))
scaleRequiredForTruncation= Math.Pow(10,significantDigitsStartFrom-1+**significantDigitsRequired**)
**siginficantDigitsResult** = integerPortion + ( Math.Truncate (decimalPortion*scaleRequiredForTruncation))/scaleRequiredForTruncation
}
else
{
**siginficantDigitsResult** = integerPortion
}
Tested on .NET 6.0
In my opinion, the rounded results are inconsistent due to the defects of the framework and the error of the floating point. Therefore, be careful about use.
decimal.Parse(doubleValue.ToString("E"), NumberStyles.Float);
example:
using System.Diagnostics;
using System.Globalization;
List<double> doubleList = new();
doubleList.Add( 0.012345);
doubleList.Add( 0.12345 );
doubleList.Add( 1.2345 );
doubleList.Add( 12.345 );
doubleList.Add( 123.45 );
doubleList.Add( 1234.5 );
doubleList.Add(12345 );
doubleList.Add(10 );
doubleList.Add( 0 );
doubleList.Add( 1 );
doubleList.Add(-1 );
doubleList.Add( 0.1);
Debug.WriteLine("");
foreach (var item in doubleList)
{
Debug.WriteLine(decimal.Parse(item.ToString("E2"), NumberStyles.Float));
// 0.0123
// 0.123
// 1.23
// 12.3
// 123
// 1230
// 12300
// 10.0
// 0.00
// 1.00
// -1.00
// 0.100
}
Debug.WriteLine("");
foreach (var item in doubleList)
{
Debug.WriteLine(decimal.Parse(item.ToString("E3"), NumberStyles.Float));
// 0.01235
// 0.1235
// 1.234
// 12.35
// 123.5
// 1234
// 12340
// 10.00
// 0.000
// 1.000
// -1.000
// 0.1000
}
As pointed out by #Oliver Bock is that Math.Round() on doubles is flawed (see Notes to Callers in its documentation). The later step of multiplying the rounded number back up by its decimal exponent will introduce further floating point errors in the trailing digits. Generally, any multiplication by or division by a power of ten gives a non-exact result, since floating-point is typically represented in binary, not in decimal.
Using the following function will avoid floating point errors in the trailing digits:
static double RoundToSignificantDigits(double d, int digits)
{
if (d == 0.0 || Double.IsNaN(d) || Double.IsInfinity(d))
{
return d;
}
// Compute shift of the decimal point.
int shift = digits - 1 - (int)Math.Floor(Math.Log10(Math.Abs(d)));
// Return if rounding to the same or higher precision.
int decimalPlaces = 0;
for (long pow = 1; Math.Floor(d * pow) != (d * pow); pow *= 10) decimalPlaces++;
if (shift >= decimalPlaces)
return d;
// Round to sf-1 fractional digits of normalized mantissa x.dddd
double scale = Math.Pow(10, Math.Abs(shift));
return shift > 0 ?
Math.Round(d * scale, MidpointRounding.AwayFromZero) / scale :
Math.Round(d / scale, MidpointRounding.AwayFromZero) * scale;
}
However if you're willing to go via decimal then Math.Round() is reliable, as is multiplying and dividing by powers of 10:
static double RoundToSignificantDigits(double d, int digits)
{
if (d == 0.0 || Double.IsNaN(d) || Double.IsInfinity(d))
{
return d;
}
decimal scale = (decimal)Math.Pow(10, Math.Floor(Math.Log10(Math.Abs(d))) + 1);
return (double)(scale * Math.Round((decimal)d / scale, digits, MidpointRounding.AwayFromZero));
}
Console.WriteLine("{0:G17}", RoundToSignificantDigits(5.015 * 100, 15)); // 501.5
for me, this one works pretty fine and is also valid for negative numbers:
public static double RoundToSignificantDigits(double number, int digits)
{
int sign = Math.Sign(number);
if (sign < 0)
number *= -1;
if (number == 0)
return 0;
double scale = Math.Pow(10, Math.Floor(Math.Log10(Math.Abs(number))) + 1);
return sign * scale * Math.Round(number / scale, digits);
}
My solution may be helpful in some cases, I use it to display crypto prices which vary greatly in magnitude - it always gives me a specified number of significant figures but unlike ToString("G[number of digits]") it doesn't show small values in scientific notation (don't know a way to avoid this with ToString(), if there is then please let me know!)
const int MIN_SIG_FIGS = 6; //will be one more for < 0
int numZeros = (int)Math.Floor(Math.Log10(Math.Abs(price))); //get number of zeros before first digit, will be negative for price > 0
int decPlaces = numZeros < MIN_SIG_FIGS
? MIN_SIG_FIGS - numZeros < 0
? 0
: MIN_SIG_FIGS - numZeros
: 0; //dec. places: set to MIN_SIG_FIGS + number of zeros, unless numZeros greater than sig figs then no decimal places
return price.ToString($"F{decPlaces}");
Here's a version inspired by Peter Mortensen that adds a couple of safeguards for edge cases such as value being NaN, Inf or very small:
public static double RoundToSignificantDigits(this double value, int digits)
{
if (double.IsNaN(value) || double.IsInfinity(value))
return value;
if (value == 0.0)
return 0.0;
double leftSideNumbers = Math.Floor(Math.Log10(Math.Abs(value))) + 1;
int places = digits - (int)leftSideNumbers;
if (places > 15)
return 0.0;
double scale = Math.Pow(10, leftSideNumbers);
double result = scale * Math.Round(value / scale, digits, MidpointRounding.AwayFromZero);
if (places < 0)
places = 0;
return Math.Round(result, places, MidpointRounding.AwayFromZero);
}
I just did:
int integer1 = Math.Round(double you want to round,
significant figures you want to round to)
Here is something I did in C++
/*
I had this same problem I was writing a design sheet and
the standard values were rounded. So not to give my
values an advantage in a later comparison I need the
number rounded, so I wrote this bit of code.
It will round any double to a given number of significant
figures. But I have a limited range written into the
subroutine. This is to save time as my numbers were not
very large or very small. But you can easily change that
to the full double range, but it will take more time.
Ross Mckinstray
rmckinstray01#gmail.com
*/
#include <iostream>
#include <fstream>
#include <string>
#include <math.h>
#include <cmath>
#include <iomanip>
#using namespace std;
double round_off(double input, int places) {
double roundA;
double range = pow(10, 10); // This limits the range of the rounder to 10/10^10 - 10*10^10 if you want more change range;
for (double j = 10/range; j< 10*range;) {
if (input >= j && input < j*10){
double figures = pow(10, places)/10;
roundA = roundf(input/(j/figures))*(j/figures);
}
j = j*10;
}
cout << "\n in sub after loop";
if (input <= 10/(10*10) && input >= 10*10) {
roundA = input;
cout << "\nDID NOT ROUND change range";
}
return roundA;
}
int main() {
double number, sig_fig;
do {
cout << "\nEnter number ";
cin >> number;
cout << "\nEnter sig_fig ";
cin >> sig_fig;
double output = round_off(number, sig_fig);
cout << setprecision(10);
cout << "\n I= " << number;
cout << "\n r= " <<output;
cout << "\nEnter 0 as number to exit loop";
}
while (number != 0);
return 0;
}
Hopefully I did not change anything formatting it.
This question already has answers here:
How to determine if a decimal/double is an integer?
(17 answers)
Get the decimal part from a double
(18 answers)
Closed 4 years ago.
I have a variable from:
double result = myList.Count / mySeptum;
I want to do the following:
if( result == int ) {
//Do Something...
}
else{
//Do another thing...
}
How can I do this?
I also tried this, but it didn't work:
if ( result%10 == 0 ){
...
}
In an example:
private void button2_Click(object sender, EventArgs e)
{
int r = 10;
int l = 2;
double d = r / l;
if (d % 10 == 0)
{
Console.WriteLine("INTEGER");
}
else
{
Console.WriteLine("DOUBLE");
}
}
For example:
double d = 1.0;
bool isInt = d == (int)d;
modulo:
double d = 1.0;
bool isInt = d % 1 == 0;
In general a floating point number on a computer can not represent every real number but only some discrete values. Thus, only for a few integers it will be possible that a double can be mathematically identical to an integer value. For most integers the closest double will be off by a small amount. So if you are looking for exact matches this will not work.
However, what you could do is to convert your double into an integer an check if the difference between the double and the integer is small enough:
double d = 1.5;
int i = (int) d;
double diff = d - i;
if (diff < 1.0e-6)
{
std::cout << "number is close to integer" << std::endl;
}
How to check my double variable is an integer or not?
From a C point of view (as post was originally tagged):
(I am certain C# has equivalent functions.)
To determine if a double is a whole number, use modf() to return the fractional part.
#include <math.h>
double x = ....;
double ipart;
if (isfinite(x) && modf(x, &ipart) == 0.0) {
// value is a whole number
....
To further test if it is in int range
if (ipart >= INT_MIN && ipart <= INT_MAX) {
int i = (int) ipart;
To check for wider integer types, we need some trickery to insure to no round-off error when forming the limits. Code takes advantage that INT..._MAX are Mersenne numbers
#define INT64_MAX_P1 ((INT64_MAX/2 + 1)*2.0)
if (ipart >= INT64_MIN && ipart < INT64_MAX_P1) {
int64_t i64 = (int64_t) ipart;
Try with typeOf:
if (myInt.GetType() == typeof(int))
How to shift all the whole numbers in a double to the right of the point ?
Example i have 5342, i want the function to return 0.5342. I do not know the number of digits in the double, it's randomly generated. Should be fairly easy but i can't find any answers.
private static void Main(string[] args)
{
Console.WriteLine(MyFunction(5127));
Console.WriteLine(MyFunction(1));
Console.WriteLine(MyFunction(51283271));
Console.WriteLine(MyFunction(-512));
Console.WriteLine(MyFunction(0));
}
public static double MyFunction(double myNumber)
{
return Math.Floor(myNumber) / Math.Pow(10, Math.Abs(myNumber).ToString().Length);
}
This sounds like a pretty bizarre task, to be honest, but you could use:
while (Math.Abs(value) >= 1)
{
value = value / 10;
}
That will go into an infinite loop if the input is infinity though - and you may well lose information as you keep dividing. The latter point is important - if what you're really interested in is the decimal representation, you should consider using decimal instead of double.
You could potentially use a mixture of Math.Log and Math.Pow to do it, but the above is probably what I'd start with.
This will get you most of the way there
public static string test()
{
double dub = 5432;
string dubTxt = dub.ToString();
string text = "0.";
string test = string.Concat(text + dubTxt);
if (1 == 1)
{
MessageBox.Show(test);
return test;
}
}
You will have to develop more if statements to handle the negative numbers.
public static string test()
{
double dub = 5432;
string dubTxt = dub.ToString();
string text = "0.";
string test = string.Concat(text + dubTxt);
if (dub < 0)
{
//Do this code instead
}
}
Good Luck. Please bump me if you choose it!! I need the cred so I can do other junk. :-D
Just divide by 10 until the number is less than 1.
public static double SomeMethod(double n)
{
double d = n;
bool isNegative = (d < 0);
if(isNegative)
d = d * -1;
while(d >= 1)
d = d/10;
if(isNegative)
d = d * -1;
return d;
}
Alternative (and more precise) option:
public static double SomeMethod2(double n)
{
double d = n;
bool isNegative = (d < 0);
if(isNegative)
d = d * -1;
int numberOfDigits = ((int)d).ToString().Length;
int divisor = 1;
for(int i = 0; i < numberOfDigits; i++)
divisor = divisor * 10;
d = d/divisor;
if(isNegative)
d = d * -1;
return d;
}
This is a problem statement.
Consider a number 2345. If you multiply its digits then you get the number 120. Now if you again multiply digits of 120 then you will get number 0 which is a one digit number. If I add digits of 2345 then I will get 14. If I add digits of 14 then I will get 5 which is a one digit number.
Thus any number can be converted into two one digit numbers in some number of steps. You can see 2345 is converted to 0 by using multiplication of digits in 2 steps and it is converted to 5 by using addition of digits in 2 steps. Now consider any number N. Let us say that it can be converted by multiplying digits to a one digit number d1 in n1 steps and by adding digits to one digit number d2 in n2 steps.
Your task is to find smallest number greater than N and less than 1000000000 which can be converted by multiplying its digits to d1 in less than or equal to n1 steps and by adding its digits to d2 in less than or equal to n2 steps.
How to solve it in C#...
I think you're simply approaching / interpreting the problem incorrectly; here's a stab in the dark:
using System;
using System.Diagnostics;
static class Program
{
static void Main()
{
// check our math first!
// You can see 2345 is converted to 0 by using multiplication of digits in 2 steps
int value, steps;
value = MultiplyToOneDigit(2345, out steps);
Debug.Assert(value == 0);
Debug.Assert(steps == 2);
// and it is converted to 5 by using addition of digits in 2 steps
value = SumToOneDigit(2345, out steps);
Debug.Assert(value == 5);
Debug.Assert(steps == 2);
// this bit is any random number
var rand = new Random();
for (int i = 0; i < 10; i++)
{
int N = rand.Next(0, MAX);
int result = Execute(N);
Console.WriteLine("For N={0}, our answer is {1}", N, result);
}
}
const int MAX = 1000000000;
//Now consider any number N.
static int Execute(int N)
{
// Let us say that it can be converted by multiplying digits to a one digit number d1 in n1
// steps and by adding digits to one digit number d2 in n2 steps.
int n1, n2;
int d1 = MultiplyToOneDigit(N, out n1),
d2 = SumToOneDigit(N, out n2);
// Your task is to find smallest number greater than N and less than 1000000000
for (int i = N + 1; i < MAX; i++)
{
int value, steps;
// which can be converted by multiplying its digits to d1 in less than or equal to n1 steps
value = MultiplyToOneDigit(i, out steps);
if (value != d1 || steps > n1) continue; // no good
// and by adding its digits to d2 in less than or equal to n2 steps.
value = SumToOneDigit(i, out steps);
if(value != d2 || steps > n2) continue; // no good
return i;
}
return -1; // no answer
}
static int MultiplyToOneDigit(int value, out int steps)
{
steps = 0;
while (value > 10)
{
value = MultiplyDigits(value);
steps++;
}
return value;
}
static int SumToOneDigit(int value, out int steps)
{
steps = 0;
while (value > 10)
{
value = SumDigits(value);
steps++;
}
return value;
}
static int MultiplyDigits(int value)
{
int acc = 1;
while (value > 0)
{
acc *= value % 10;
value /= 10;
}
return acc;
}
static int SumDigits(int value)
{
int total = 0;
while (value > 0)
{
total += value % 10;
value /= 10;
}
return total;
}
}
There are two memory problems I can see; the first is the generation of lots of strings - you might want to approach that something like:
static int SumDigits(int value)
{
int total = 0;
while (value > 0)
{
total += value % 10;
value /= 10;
}
return total;
}
(which is completely untested)
The second problem is the huge list; you don't need to store (in lstString) every value just to find a minimum. Just keep track of the best you've done so far. Or if you need the data for every value, then: don't store them as a string. Indeed, the i can be implied anyway (from the position in the list/array), so all you would really need would be an int[] of the cnt values for every value. And int[1000000000] is 4GB just by itself, so would require the large-array support in recent .NET versions (<gcAllowVeryLargeObjects>). But much better would be: just don't store it.
But it's throwing System.OutOfMemoryException .
That simply mean you're running out of memory. Your limit is 1,000,000,000 or roughly 1G. Times 4 bytes for a string reference that's already too large for a 32 bit system. Even without the actual strings.
You can store your answers more compactly in an int[] array but that would still show the same problem.
So, lower your limit or compile and run on a 64 bit PC.
A for effort :)
Now doing together. You can of course do refactoring.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace _17082903_smallest_greatest_number
{
class Program
{
static void Main(string[] args)
{
int N = 2344;
int n1 = 0;
int n2 = 0;
int d1 = SumDigits(N, ref n1);
int d2 = ProductDigits(N, ref n2);
bool sumFound = false, productFound = false;
for (int i = N + 1; i < 1000000000; i++)
{
if (!sumFound)
{
int stepsForSum = 0;
var res = SumDigits(i, ref stepsForSum);
if (res == d1 && stepsForSum <= n1)
{
Console.WriteLine("the smallest number for sum is: " + i);
Console.WriteLine(string.Format("sum result is {0} in {1} steps only", res, stepsForSum));
sumFound = true;
}
stepsForSum = 0;
}
if (!productFound)
{
int stepsForProduct = 0;
var res2 = ProductDigits(i, ref stepsForProduct);
if (res2 == d2 && stepsForProduct <= n2)
{
Console.WriteLine("the smallest number for product is: " + i);
Console.WriteLine(string.Format("product result is {0} in {1} steps only", res2, stepsForProduct));
productFound = true;
}
stepsForProduct = 0;
}
if (productFound && sumFound)
{
break;
}
}
}
static int SumDigits(int value, ref int numOfSteps)
{
int total = 0;
while (value > 0)
{
total += value % 10;
value /= 10;
}
numOfSteps++;
if (total < 10)
{
return total;
}
else
{
return SumDigits(total, ref numOfSteps);
}
}
static int ProductDigits(int value, ref int numOfSteps)
{
int total = 1;
while (value > 0)
{
total *= value % 10;
value /= 10;
}
numOfSteps++;
if (total < 10)
{
return total;
}
else
{
return ProductDigits(total, ref numOfSteps);
}
}
}
}