In my array, arr3 has 1000 numbers in it. I have to split this array into k subintervals of width differenceofMaxMin . How can I do that? Later I have to count how many times, the values in arr3 have matched to each interval. But I am stuck at creating intervals from array with a given width.
Any kind of help will be really appreciated!
public partial class Form1 : Form
{
public Form1()
{
InitializeComponent();
}
double[] Statistics1 = new double[500];
double[] Statistics2 = new double[500];
double Alpha1;
double Alpha2;
double RV1;
double RV2;
Random random = new Random();
public double RandomDoubleInclusive() //We are using this method because random.NextDouble() method gives random number
//between 0 and 1 where 0 is inclusive and 1 is exclusive.
//Since the value of probability lies between 0 and 1, both inclusive that's why we need
//to use this method.
{
double d = 0.0;
int i = 0;
do
{
d = random.NextDouble();
i = random.Next(2);
}
while (i == 1 && d > 0);
return d + i;
}
private void label3_Click(object sender, EventArgs e)
{
}
int i,j;
private void button1_Click(object sender, EventArgs e)
{
int SampleSize = Convert.ToInt32(textBox3.Text);
for ( i = 0; i<500;)
{
for (j = 0; j < 500;)
{
Alpha1 = RandomDoubleInclusive();
Alpha2 = RandomDoubleInclusive();
double LnPart = Math.Log(Alpha1);
double part1 = (-2) * LnPart;
double part2 = 2 * 3.14159 * Alpha2;
double CosPart = Math.Cos(part2);
double SinPart = Math.Sin(part2);
RV1 = Math.Sqrt(part1) * CosPart;
Statistics1[i] = RV1;
RV2 = Math.Sqrt(part1) * SinPart;
Statistics2[j] = RV2;
i++;
j++;
}
}
var myList = new List<double>();
myList.AddRange(Statistics1);
myList.AddRange(Statistics2);
double[] arr3 = myList.ToArray();
double Max = arr3.Max();
double Min = arr3.Min();
double differenceofMaxMin = Max - Min; //calculating size of width of interval
double k;
k = Math.Log(SampleSize,2) + 1; //calculating number of subintervals
}
}
I'm not sure I fully understand what exactly you're trying to achieve, but I can certainly try to help you out with an example on how to split an array arr3 into k subintervals with (max) number of elements differenceofMaxMin
var arr3 = Enumerable.Range(0, 1000);
// given: the max number of elements
var differenceofMaxMin = 300;
// determine the number of subintervals
// note that the last subinterval may contain less than differenceofMaxMin elements
var k = (int)Math.Ceiling((double)arr3.Count() / differenceofMaxMin);
var arr3_split = Enumerable.Range(0, k)
.Select(i => arr3.Skip(i * differenceofMaxMin).Take(differenceofMaxMin));
Looking at your method to generate a random double from [0, 1], I think it's overkill since the likelihood to actually draw exactly 1.0 is extremely low.
My problem is like this: I need to generate a random number from Gaussin/Norma distribution and create a histogram of a width of 0.1.
class Gaussian
{
public static double Next(Random r, double mu = 0, double sigma = 1)
{
var u1 = r.NextDouble();
var u2 = r.NextDouble();
var rand_std_normal = Math.Sqrt(-2.0 * Math.Log(u1)) *
Math.Sin(2.0 * Math.PI * u2);
var rand_normal = mu + sigma * rand_std_normal;
return rand_normal;
}
}
I am using the above function to generate a Gaussian random value.
Now, in order to create a histogram, I am in need of such a calculation that is able to automatically convert the gaussian value into an array index. Something like the following:
static void Main(string[] args)
{
const int N = 1000;
int time = N;
const double binsDistance = 0.1;
int binsCount = (int)(N * binsDistance);
Random rand = new Random();
int[] array = new int[binsCount];
int side = 0;
for (int i = 0; i < time; i++)
{
double gauss = Gaussian.Next(rand);
int binNo = Normalization.Normalize(0, binsCount - 1, gauss);
array[binNo]++;
}
}
For that, I tried two calculations.
the first one is here.
the second one is here.
The problem with the first one is, it can't handle negative numbers properly.
The problem with the second one is, it is generating too many zero values.
So, I have two questions:
What is the basic difference between #1 and #2?
How can I achieve what I am trying to do?
I have 5 fields, I want them all to have a generated number between 0 and 100. But, the sum of the 5 fields should be 100.
When I want to give a random number for one field I would do the following:
Random rnd = new Random();
int x= rnd.Next(1, 10);
But how should I do that for multiple fields that needs to have a sum of 100 together?
You can use the following approach:
generate 4 random integers in [0, 100]
sort them, let's denote the sorted values as 0 ≤ x1 ≤ x2 ≤ x3 ≤ x4 ≤ 100
use the following 5 values as the random numbers with sum 100:
N1 = x1
N2 = x2 - x1
N3 = x3 - x2
N4 = x4 - x3
N5 = 100 - x4
It basically corresponds to randomly choosing 4 sectioning points on the [0, 100] interval, and using the lengths of the 5 resulting intervals as the random numbers:
const int k = 5;
const int sum = 100;
Random rnd = new Random();
int[] x = new int[k + 1];
// the endpoints of the interval
x[0] = 0;
x[k] = sum;
// generate the k - 1 random sectioning points
for (int i = 1; i < k; i++) {
x[i] = rnd.Next(0, sum + 1);
}
// sort the sectioning points
Array.Sort(x);
// obtain the k numbers with sum s
int[] N = new int[k];
for (int i = 0; i < k; i++) {
N[i] = x[i + 1] - x[i];
}
In order to make your distribution uniform, you could try the following aproach:
Generate some random numbers.
Normalize them.
Correct the last field to get exactly the expected sum, if needed.
The code:
const int ExpectedSum = 100;
Random rnd = new Random();
int[] fields = new int[5];
// Generate 4 random values and get their sum
int sum = 0;
for (int i = 0; i < fields.Length - 1; i++)
{
fields[i] = rnd.Next(ExpectedSum);
sum += fields[i];
}
// Adjust the sum as if there were 5 random values
int actualSum = sum * fields.Length / (fields.Length - 1);
// Normalize 4 random values and get their sum
sum = 0;
for (int i = 0; i < fields.Length - 1; i++)
{
fields[i] = fields[i] * ExpectedSum / actualSum;
sum += fields[i];
}
// Set the last value
fields[fields.Length - 1] = ExpectedSum - sum;
Live example: https://dotnetfiddle.net/5yXwOP
To achieve a truly random distribution, with every element having the chance to be 100 with a total sum of 100, you can use the following solution:
public static int[] GetRandomDistribution(int sum, int amountOfNumbers)
{
int[] numbers = new int[amountOfNumbers];
var random = new Random();
for (int i = 0; i < sum; i++)
{
numbers[random.Next(0, amountOfNumbers)]++;
}
return numbers;
}
static void Main(string[] args)
{
var result = GetRandomDistribution(100, 5);
}
It increases a random number by one until the sum is reached. This should fulfill all your criterias.
After thinking about it, I prefer the following solution, because it's less likely to generate an equal distribution:
public static int[] GetRandomDistribution2(int sum, int amountOfNumbers)
{
int[] numbers = new int[amountOfNumbers];
var random = new Random();
for (int i = 0; i < amountOfNumbers; i++)
{
numbers[i] = random.Next(sum);
}
var compeleteSum = numbers.Sum();
// Scale the numbers down to 0 -> sum
for (int i = 0; i < amountOfNumbers; i++)
{
numbers[i] = (int)(((double)numbers[i] / compeleteSum) * sum);
}
// Due to rounding the number will most likely be below sum
var resultSum = numbers.Sum();
// Add +1 until we reach "sum"
for (int i = 0; i < sum - resultSum; i++)
{
numbers[random.Next(0, amountOfNumbers)]++;
}
return numbers;
}
For Example.
int sum=100;
int i = 5;
Random rnd = new Random();
while (true)
{
int cur;
--i;
if (i == 0) {
Console.WriteLine(sum + " ");
break;
} else
cur=rnd.Next(1, sum);
sum -= cur;
Console.WriteLine(cur + " ");
}
Live Example: https://dotnetfiddle.net/ltIK40
or
Random rnd = new Random();
int x= rnd.Next(1, 10);
int y= rnd.Next(x,x+10);
int y2=rnd.Next(y,y+10);
int y3=rnd.Next(y2,y2+10);
int y4=100-(x+y+y2+y3);
My approach is this:
var rnd = new Random();
var numbers = Enumerable.Range(0, 5).Select(x => rnd.Next(0, 101)).ToArray().OrderBy(x => x).ToArray();
numbers = numbers.Zip(numbers.Skip(1), (n0, n1) => n1 - n0).ToArray();
numbers = numbers.Concat(new[] { 100 - numbers.Sum() }).ToArray();
This is as uniform as I think is possible.
Create your first random number. After that you take the difference between the value of num1 and 100 as the max def of rnd. But to guarantee that their sum is 100, you have to check at the last num if the sum of all nums is 100. If not the value of your last num is the difference that their sum and 100.
And to simply your code and get a clean strcuture, put that code in a loop and instead of single numbers work with an int[5] array.
private int[] CreateRandomNumbersWithSum()
{
int[] nums = new int[5];
int difference = 100;
Random rnd = new Random();
for (int i = 0; i < nums.Length; i++)
{
nums[i] = rnd.Next(0, difference);
difference -= nums[i];
}
int sum = 0;
foreach (var num in nums)
sum += num;
if (sum != 100)
{
nums[4] = 100 - sum;
}
return nums;
}
I think this is a very simple solution:
public void GenerateRandNr(int total)
{
var rnd = new Random();
var nr1 = rnd.Next(0, total);
var nr2 = rnd.Next(0, total - nr1);
var nr3 = rnd.Next(0, total - nr1 - nr2);
var nr4 = rnd.Next(0, total - nr1 - nr2 - nr3);
var nr5 = total - nr1 - nr2 - nr3 - nr4;
}
EDIT:
Just tested it, works fine for me:
The solution is that it's not the numbers that need to be random so much as the distribution needs to be random. The randomness of the numbers will be a side effect of their random distribution.
So you would start with five random numbers in a given range. The exact range doesn't matter as long as the range is the same for all five, although a broader range allows for more variation. I'd use Random.NextDouble() which returns random numbers between 0 and 1.
Each of those individual numbers divided by the sum of those numbers represents a distribution.
For example, say your random numbers are .4, .7, .2, .5, .2. (Using fewer digits for simplicity.)
The total of those numbers is 2. So now the distributions are each of those numbers divided by the total.
.4 / 2 = .20
.7 / 2 = .35
.2 / 2 = .10
.5 / 2 = .25
.2 / 2 = .10
You'll notice that those distributions will equal 100% or really close to it if there are a lot more decimal places.
The output is going to be each of those distributions times the target number, in this case, 100. In other words, each of those numbers represents a piece of 100.
So multiplying each of those distributions times the target, we get 20, 35, 10, 25, and 100, which add up to 100.
The trouble is that because of rounding your numbers won't always perfectly add up to 100. To fix that you might add one to the smallest number if the sum is less than 100, or subtract one from the largest number of the the sum is greater than 100. Or you could choose to add or subtract on one of the numbers at random.
Here's a class to create the distributions. (I'm just playing around so I haven't exactly optimized this to death.)
public class RandomlyDistributesNumbersTotallingTarget
{
public IEnumerable<int> GetTheNumbers(int howManyNumbers, int targetTotal)
{
var random = new Random();
var distributions = new List<double>();
for (var addDistributions = 0; addDistributions < howManyNumbers; addDistributions++)
{
distributions.Add(random.NextDouble());
}
var sumOfDistributions = distributions.Sum();
var output = distributions.Select(
distribution =>
(int)Math.Round(distribution / sumOfDistributions * targetTotal, 0)).ToList();
RoundUpOutput(output, targetTotal);
return output;
}
private void RoundUpOutput(List<int> output, int targetTotal)
{
var difference = targetTotal - output.Sum();
if (difference !=0)
{
var indexToAdjust =
difference > 0 ? output.IndexOf(output.Min()) : output.IndexOf(output.Max());
output[indexToAdjust]+= difference;
}
}
}
And here's a not-perfectly-scientific unit test that tests it many times over and ensures that the results always total 100.
[TestMethod]
public void OutputTotalsTarget()
{
var subject = new RandomlyDistributesNumbersTotallingTarget();
for (var x = 0; x < 10000; x++)
{
var output = subject.GetTheNumbers(5, 100);
Assert.AreEqual(100, output.Sum());
}
}
Some sample outputs:
5, 30, 27, 7, 31
15, 7, 26, 27, 25
10, 11, 23, 2, 54
The numbers are always going to average to 20, so while 96, 1, 1, 1 is a hypothetical possibility they're going to tend to hover closer to 20.
Okay. Having been burned by my previous attempt at this seemingly trivial problem, I decided to have another go. Why not normalise all the numbers after generation? This guarantees randomness, and avoids O(n log n) performance from a sort. It also has the advantage that even with my basic maths, I can work out that the numbers are uniformly distributed.
public static int[] UniformNormalization(this Random r, int valueCount, int valueSum)
{
var ret = new int[valueCount];
long sum = 0;
for (int i = 0; i < valueCount; i++)
{
var next = r.Next(0, valueSum);
ret[i] = next;
sum += next;
}
var actualSum = 0;
for (int i = 0; i < valueCount; i++)
{
actualSum += ret[i] = (int)((ret[i] * valueSum) / sum);
}
//Fix integer rounding errors.
if (valueSum > actualSum)
{
for (int i = 0; i < valueSum - actualSum; i++)
{
ret[r.Next(0, valueCount)]++;
}
}
return ret;
}
This should also be one of the fastest solutions.
how to find the min and max for quadratic equation using c# ??
f(x,y) = x^2 + y^2 + 25 * (sin(x)^2 + sin(y)^2) ,where (x,y) from (-2Pi, 2Pi) ??
in the manual solving I got min is = 0 , max = 8Pi^2 = 78.957 .
I tried to write the code based on liner quadratic code but something goes totally wrong
this code give the min = -4.?? and the max = 96 could you help to know where is my mistake please ??
I uploaded the code to dropbox if anyone can have look : https://www.dropbox.com/s/p7y6krk2gk29i9e/Program.cs
double[] X, Y, Result; // Range array and result array.
private void BtnRun_Click(object sender, EventArgs e)
{
//Set any Range for the function
X = setRange(-2 * Math.PI, 2 * Math.PI, 10000);
Y = setRange(-2 * Math.PI, 2 * Math.PI, 10000);
Result = getOutput_twoVariablesFunction(X, Y);
int MaxIndex = getMaxIndex(Result);
int MinIndex = getMinIndex(Result);
TxtMin.Text = Result[MinIndex].ToString();
TxtMax.Text = Result[MaxIndex].ToString();
}
private double twoVariablesFunction(double x,double y)
{
double f;
//Set any two variables function
f = Math.Pow(x, 2) + Math.Pow(y, 2) + 25 * (Math.Pow(Math.Sin(x), 2) + Math.Pow(Math.Sin(y), 2));
return f;
}
private double[] setRange(double Start, double End, int Sample)
{
double Step = (End - Start) / Sample;
double CurrentVaue = Start;
double[] Array = new double[Sample];
for (int Index = 0; Index < Sample; Index++)
{
Array[Index] = CurrentVaue;
CurrentVaue += Step;
}
return Array;
}
private double[] getOutput_twoVariablesFunction(double[] X, double[] Y)
{
int Step = X.Length;
double[] Array = new double[Step];
for (int Index = 0; Index < X.Length ; Index++)
{
Array[Index] = twoVariablesFunction(X[Index], Y[Index]);
}
return Array;
}
private int getMaxIndex(double[] ValuesArray)
{
double M = ValuesArray.Max();
int Index = ValuesArray.ToList().IndexOf(M);
return Index;
}
private int getMinIndex(double[] ValuesArray)
{
double M = ValuesArray.Min();
int Index = ValuesArray.ToList().IndexOf(M);
return Index;
}
Do you want to compute (sin(x))^2 or sin(x^2)? In your f(x,y) formula it looks like (sin(x))^2, but in your method twoVariablesFunction like sin(x^2).
I need to get a random number in C#, within (-15, 15) but without generate values between (-10, 10) as float numbers. Like the random should come with in (-15,-10) and (10,15).
Is it possible to get?
This can be done in one line, but I separated it out for clarity.
public double GetRandomNumber()
{
//Between 0 and 1
Random random = new Random();
double randomNumber = random.NextDouble();
//Between -0.5 and 0.5;
randomNumber -= 0.5;
//Between -5.0 and 5.0;
randomNumber *= 10.0;
//Between [-15.0, -10.0] or [10.0, 15.0]
randomNumber += Math.Sign(randomNumber) * 10.0;
return randomNumber;
}
One way would be to use this:
var random = new Random();
var result = random.NextDouble();
if(result < 0.5)
result = -15 + result * 10;
else
result = 5 + result * 10;
Random.NextDouble generates a number between 0.0 and 1.0.
If it is less than 0.5 we treat this as the indicator to create a negative number.
Another approach, assuming that you do want a decimal place in your result (since it isn't entirely clear).
Random rand = new Random();
var intValue = rand.Next(10,15);
var decimalValue = rand.NextDouble();
var sign = rand.Next(0,1);
if (sign == 0) sign = -1;
return (intValue + decimalValue) * sign;
Generate two random numbers. The first random number will choose whether you're in the -15 to -10 range, or the 10 to 15 range, the second - how far along in that range you want to be.
Try:
Random random = new Random();
float value = (float) (Math.Sign((random.NextDouble() - 1)) * (10 + 5 * random.NextDouble()));
static Random r = new Random(2);
static void Main(string[] args)
{
int d = r.Next(-15, 15);
while ((d >= -15 && d <= -10) || (d >= 10 && d <= 15))
Console.WriteLine(d);
Console.ReadLine();
}
If you want integers and you want it to be performant, I would use:
var rand = new Random();
int value = rand.Next(-5, 7);
return value > 0 ? value + 9 : value - 10;
I am assuming you want numbers from the ranges: [-15, -10] and [10, -15]. Adjust hard-coded values above appropriately. To make it more performant, there may be a way to do some bit arithmetic and avoid the comparison but I'll leave that for the comments.
Here is a static method to accomplish that:
public static int RandomRange(int min, int max, bool includeNegatives = false)
{
if( min >= max) throw new Exception("min can't be greater than max");
Random rdm = new Random();
int num = 0;
while (num < min) num = rdm.Next(max + 1);
return num * (rdm.Next() % 2 == 0 ? -1 : 1);
}
To call it:
var TenToFifteenNegatives = RandomRange(10, 15, true);
var TenToFifteenNoNegatives = RandomRange(10, 15);