I am trying to calculate the delta of a number using a formula of bhaskara and delta but it is giving NaN in the message box
//Variables declaration
int a = 800, b = 500, c = 350;
double delta, a1, a2;
//Formulas
delta = (b*b) - (4*a*c);
a1 = (-b + Math.Sqrt(delta)) / (2 * a);
a2 = (-b - Math.Sqrt(delta)) / (2 * a);
//Output
MessageBox.Show(a1.ToString());
MessageBox.Show(a2.ToString());
The value of delta is -870000.
That's not a value you can take the square root of (and get a real number anyway).
From the Math.Sqrt docs:
As Broots said Math.sqrt of negative number will return NaN. There is a Complex struct that should do what you need:
Complex c = Complex.Sqrt(-25); // has a value of approximately 0 + 5i
So I wrote a Quadratic formula program in C#, how do I take the quadratic formula program and modify it so that the program correctly displays the number of solutions.
if there are two solutions,
(x - x1)(x - x2) = 0
if there is only one solution,
(x - x0)^2 = 0
if there are no solutions,
No Solution.
This is the program, if someone could show the solution to this for me that would be wonderful, I'm really stuck on how to do it.
using System;
namespace quadraticequation
{
class MainClass
{
public static void Main(string[] args)
{
Console.WriteLine("Enter a number for a"); //ask the user for information
double a = double.Parse(Console.ReadLine()); //Gets a from the user
Console.WriteLine("Enter a number for b"); //asks the user for information
double b = double.Parse(Console.ReadLine()); //Gets b from the user
Console.WriteLine("Enter a number for c"); //asks the user for information
double c = double.Parse(Console.ReadLine()); //Gets c from the user
//double.Parse --> is used to convert a number or string to a double.
//Console.ReadLine() --> is used to take the input from the user.
//We call a function here
Quadratic(a, b, c);
}
//We need to create a new function
public static void Quadratic(double a, double b, double c)
{
double deltaRoot = Math.Sqrt(b * b - 4 * a * c); //Math.Sqrt takes the square root of the number
if (deltaRoot >= 0) // we use an if statement here to handle information
{
double x1 = (-b + deltaRoot) / 2 * a; //We write the information for x1 here
double x2 = (-b - deltaRoot) / 2 * a; //We write the information for x2 here
Console.WriteLine("x1 = " + x1 + " x2 = " + x2); //we use this to write the roots
}
else // we use an else statement so that we dont return an error when there are no roots
{
Console.WriteLine("There are no roots");
}
}
}
}
I think you have to review your second degree formula solution-skills. You write:
double deltaRoot = Math.Sqrt(b * b - 4 * a * c);
But the test is actually whether b2-4×a×c is larger than or equal to zero: indeed that is actually why we check it: because we cannot take the square root of a negative number (yeah there exist complex numbers that can take the square root of a negative number, but let's ignore that for now).
So the solution is to write it like:
public static void Quadratic(double a, double b, double c) {
double delta = b*b-4*a*c; //only delta
if (delta > 0) {
double deltaRoot = Math.Sqrt(delta);
double x1 = (-b + deltaRoot) / (2 * a); //We write the information for x1 here
double x2 = (-b - deltaRoot) / (2 * a); //We write the information for x2 here
Console.WriteLine("x1 = " + x1 + " x2 = " + x2); //we use this to write the roots
} else if(delta == 0) {
double x1 = -b/(2*a);
Console.WriteLine("x1 = " + x1); //we use this to write the roots
} else {
Console.WriteLine("There are no roots");
}
}
You also have to write (-b + deltaRoot) / (2*a) (with (2*a)), otherwise you will multiply (-b + deltaRoot) / 2 with a instead.
A final note is that equality comparisons with floating points is very tricky so delta == 0 will often fail since the result can be something 1e-20-ish, which is simply an error when performing floating point arithmetic. So it is perhaps better to use a small range of values.
This gives:
csharp> MainClass.Quadratic(1,1,1);
There are no roots
csharp> MainClass.Quadratic(1,1,0);
x1 = 0 x2 = -1
csharp> MainClass.Quadratic(1,0,0);
x1 = 0
I am doing 2D trilateration using the MathNet for the matrices and vectors. This i my code:
public static double[] trilaterate2DLinear(double[] pA, double[] pB, double[] pC, double rA, double rB, double rC) {
//Convert doubles to vectors for processing
Vector<double> vA = Vector<double>.Build.Dense(pA);
Vector<double> vB = Vector<double>.Build.Dense(pB);
Vector<double> vC = Vector<double>.Build.Dense(pC);
//Declare elements of b vector
//bBA = 1/2 * (rA^2 - rB^2 + dBA^2)
double[] b = {0, 0};
b[0] = 0.5 * (Math.Pow(rA, 2) - Math.Pow(rB, 2) + Math.Pow(getDistance(pB, pA), 2));
b[1] = 0.5 * (Math.Pow(rA, 2) - Math.Pow(rC, 2) + Math.Pow(getDistance(pC, pA), 2));
//Convert b array to vector form
Vector<double> vb = Vector<double>.Build.Dense(b);
//Build A array
//A = {x2 -x1, y2 - y1}
// {x3 - x1, y3 - y1}
double[,] A = { { pB[0] - pA[0], pB[1] - pA[1] }, { pC[0] - pA[0], pC[1] - pA[1] } };
//Convert A to Matrix form
Matrix<double> mA = Matrix<double>.Build.DenseOfArray(A);
//Declare Transpose of A matrix;
Matrix<double> mAT = mA.Transpose();
//Declare solution vector x to 0
Vector<double> x = Vector<double>.Build.Dense(2);
//Check if A*AT is non-singular (non 0 determinant)
if (mA.Multiply(mAT).Determinant() == 0)
{
//x = ((AT * A)^-1)*AT*b
x = (((mA.Multiply(mAT)).Inverse()).Multiply(mAT)).Multiply(vb);
}
else
{
//TODO case for A*AT to be singular
x = (((mA.Multiply(mAT)).Inverse()).Multiply(mAT)).Multiply(vb);
}
//final position is x + vA
//return as double so as not
return (x.Add(vA)).ToArray();
}
//Gets the Euclidean distance between two points
private static double getDistance(double[] p1, double[] p2)
{
//d^2 = (p1[0] - p2[0])^2 + (p1[1] - p2[1]);
double distSquared = Math.Pow((p1[0] - p2[0]),2) + Math.Pow((p1[1] - p2[1]),2);
return Math.Sqrt(distSquared);
}
pA, pB & pC are the coordinates of the the Beacons and rA, rB & rC are the distances from the each beacon to the user. Is there anything obvious I am doing wrong? Maybe the order of Matrix multiplications need to change but I am not familiar enough with the Linear Least Squares to be able to track the Matrices and tell.
Solved. The if statement condition and calculations inside the if statement where wrong.
Correction:
public static double[] trilaterate2DLinear(double[] pA, double[] pB, double[] pC, double rA, double rB, double rC) {
//Convert doubles to vectors for processing
Vector<double> vA = Vector<double>.Build.Dense(pA);
Vector<double> vB = Vector<double>.Build.Dense(pB);
Vector<double> vC = Vector<double>.Build.Dense(pC);
//Declare elements of b vector
//bBA = 1/2 * (rA^2 - rB^2 + dBA^2)
double[] b = {0, 0};
b[0] = 0.5 * (Math.Pow(rA, 2) - Math.Pow(rB, 2) + Math.Pow(getDistance(pB, pA), 2));
b[1] = 0.5 * (Math.Pow(rA, 2) - Math.Pow(rC, 2) + Math.Pow(getDistance(pC, pA), 2));
//Convert b array to vector form
Vector<double> vb = Vector<double>.Build.Dense(b);
//Build A array
//A = {x2 -x1, y2 - y1}
// {x3 - x1, y3 - y1}
double[,] A = { { pB[0] - pA[0], pB[1] - pA[1] }, { pC[0] - pA[0], pC[1] - pA[1] } };
//Convert A to Matrix form
Matrix<double> mA = Matrix<double>.Build.DenseOfArray(A);
//Declare Transpose of A matrix;
Matrix<double> mAT = mA.Transpose();
//Declare solution vector x to 0
Vector<double> x = Vector<double>.Build.Dense(2);
//Check if A*AT is non-singular (non 0 determinant)
double det = mA.Multiply(mAT).Determinant();
if (mA.Multiply(mAT).Determinant() > 0.1)
{
//x = ((AT * A)^-1)*AT*b
// x = (((mA.Multiply(mAT)).Inverse()).Multiply(mAT)).Multiply(vb);
x = (mA.Transpose() * mA).Inverse() * (mA.Transpose() * vb);
}
else
{
//TODO case for A*AT to be singular
x = (((mA.Multiply(mAT)).Inverse()).Multiply(mAT)).Multiply(vb);
}
//final position is x + vA
//return as double so as not
return (x.Add(vA)).ToArray();
}
You are not calculating the B vector correctly.
It should be:
//dBA = 0.5 * (rA^2 - rB^2 - length_vA^2 + length_vB^2)
b[0] = 0.5 * (Math.Pow(rA, 2) - Math.Pow(rB, 2) - Math.Pow(getDistance(pA,{0,0}), 2) + Math.Pow(getDistance(pB,{0,0}), 2));
b[1] = 0.5 * (Math.Pow(rA, 2) - Math.Pow(rC, 2) - Math.Pow(getDistance(pA,{0,0}), 2) + Math.Pow(getDistance(pC,{0,0}), 2));
I looked through the internet, and in terms of Bicubic Interpolation, I can't find a simple equation for it. Wikipedia's page on the subject wasn't very helpful, so is there any easy method to learning how Bicubic Interpolation works and how to implement it? I'm using it to generate Perlin Noise, but using bilinear interpolation is way to choppy for my needs (I already tried it).
If anyone can point me in the right direction by either a good website or just an answer, I would greatly appreciate it. (I'm using C# by the way)
Using this (Thanks to Ahmet Kakıcı who found this), I figured out how to add Bicubic Interpolation. For those also looking for the answer, here is what I used:
private float CubicPolate( float v0, float v1, float v2, float v3, float fracy ) {
float A = (v3-v2)-(v0-v1);
float B = (v0-v1)-A;
float C = v2-v0;
float D = v1;
return A*Mathf.Pow(fracy,3)+B*Mathf.Pow(fracy,2)+C*fracy+D;
}
In order to get 2D Interpolation, I first got the x, then interpolated the y. Eg.
float x1 = CubicPolate( ndata[0,0], ndata[1,0], ndata[2,0], ndata[3,0], fracx );
float x2 = CubicPolate( ndata[0,1], ndata[1,1], ndata[2,1], ndata[3,1], fracx );
float x3 = CubicPolate( ndata[0,2], ndata[1,2], ndata[2,2], ndata[3,2], fracx );
float x4 = CubicPolate( ndata[0,3], ndata[1,3], ndata[2,3], ndata[3,3], fracx );
float y1 = CubicPolate( x1, x2, x3, x4, fracy );
Where ndata is defined as:
float[,] ndata = new float[4,4];
for( int X = 0; X < 4; X++ )
for( int Y = 0; Y < 4; Y++ )
//Smoothing done by averaging the general area around the coords.
ndata[X,Y] = SmoothedNoise( intx+(X-1), inty+(Y-1) );
(intx and inty are the floored values of the requested coordinates. fracx and fracy are the fractional parts of the inputted coordinates, to be x-intx, and y-inty, respectively)
Took Eske Rahn answer and made a single call (note, the code below uses matrix dimensions convention of (j, i) rather than image of (x, y) but that shouldn't matter for interpolation sake):
/// <summary>
/// Holds extension methods.
/// </summary>
public static class Extension
{
/// <summary>
/// Performs a bicubic interpolation over the given matrix to produce a
/// [<paramref name="outHeight"/>, <paramref name="outWidth"/>] matrix.
/// </summary>
/// <param name="data">
/// The matrix to interpolate over.
/// </param>
/// <param name="outWidth">
/// The width of the output matrix.
/// </param>
/// <param name="outHeight">
/// The height of the output matrix.
/// </param>
/// <returns>
/// The interpolated matrix.
/// </returns>
/// <remarks>
/// Note, dimensions of the input and output matrices are in
/// conventional matrix order, like [matrix_height, matrix_width],
/// not typical image order, like [image_width, image_height]. This
/// shouldn't effect the interpolation but you must be aware of it
/// if you are working with imagery.
/// </remarks>
public static float[,] BicubicInterpolation(
this float[,] data,
int outWidth,
int outHeight)
{
if (outWidth < 1 || outHeight < 1)
{
throw new ArgumentException(
"BicubicInterpolation: Expected output size to be " +
$"[1, 1] or greater, got [{outHeight}, {outWidth}].");
}
// props to https://stackoverflow.com/a/20924576/240845 for getting me started
float InterpolateCubic(float v0, float v1, float v2, float v3, float fraction)
{
float p = (v3 - v2) - (v0 - v1);
float q = (v0 - v1) - p;
float r = v2 - v0;
return (fraction * ((fraction * ((fraction * p) + q)) + r)) + v1;
}
// around 6000 gives fastest results on my computer.
int rowsPerChunk = 6000 / outWidth;
if (rowsPerChunk == 0)
{
rowsPerChunk = 1;
}
int chunkCount = (outHeight / rowsPerChunk)
+ (outHeight % rowsPerChunk != 0 ? 1 : 0);
var width = data.GetLength(1);
var height = data.GetLength(0);
var ret = new float[outHeight, outWidth];
Parallel.For(0, chunkCount, (chunkNumber) =>
{
int jStart = chunkNumber * rowsPerChunk;
int jStop = jStart + rowsPerChunk;
if (jStop > outHeight)
{
jStop = outHeight;
}
for (int j = jStart; j < jStop; ++j)
{
float jLocationFraction = j / (float)outHeight;
var jFloatPosition = height * jLocationFraction;
var j2 = (int)jFloatPosition;
var jFraction = jFloatPosition - j2;
var j1 = j2 > 0 ? j2 - 1 : j2;
var j3 = j2 < height - 1 ? j2 + 1 : j2;
var j4 = j3 < height - 1 ? j3 + 1 : j3;
for (int i = 0; i < outWidth; ++i)
{
float iLocationFraction = i / (float)outWidth;
var iFloatPosition = width * iLocationFraction;
var i2 = (int)iFloatPosition;
var iFraction = iFloatPosition - i2;
var i1 = i2 > 0 ? i2 - 1 : i2;
var i3 = i2 < width - 1 ? i2 + 1 : i2;
var i4 = i3 < width - 1 ? i3 + 1 : i3;
float jValue1 = InterpolateCubic(
data[j1, i1], data[j1, i2], data[j1, i3], data[j1, i4], iFraction);
float jValue2 = InterpolateCubic(
data[j2, i1], data[j2, i2], data[j2, i3], data[j2, i4], iFraction);
float jValue3 = InterpolateCubic(
data[j3, i1], data[j3, i2], data[j3, i3], data[j3, i4], iFraction);
float jValue4 = InterpolateCubic(
data[j4, i1], data[j4, i2], data[j4, i3], data[j4, i4], iFraction);
ret[j, i] = InterpolateCubic(
jValue1, jValue2, jValue3, jValue4, jFraction);
}
}
});
return ret;
}
}
I'm a bit confused on the third degree polynomial used.
Yes it gives the correct values in 0 and 1, but the derivatives of neighbouring cells does not fit, as far as I can calculate. If the grid-data is linear, it does not even return a line....
And it is not point symmetric in x=0.5
The polynomial that fits in 0 and 1 AND also have the same derivatives for neighbouring cells, and thus is smooth, is (almost) as easy to calculate.
(and it reduces to linear form if that fits the data)
In the labelling p and m is a short hand for plus and minus e.g. vm1 is v(-1)
//Bicubic convolution algorithm, cubic Hermite spline
static double CubicPolateConv
(double vm1, double v0, double vp1, double vp2, double frac) {
//The polynomial of degree 3 where P(x)=f(x) for x in {0,1}
//and P'(1) in one cell matches P'(0) in the next, gives a continuous smooth curve.
//And we also wants it to reduce nicely to a line, if that matches the data
//P(x)=Dx³+Cx²+Bx+A=((Dx+C)x+B)x+A
//P'(x)=3Dx²+2Cx+B
//P(0)=A =v0
//P(1)=D+C+B+A =Vp1
//P'(0)=B =(vp1-vm1)/2
//P'(1)=3D+2C+B=(vp2-v0 )/2
//Subtracting expressions for A and B from D+C+B+A
//D+C =vp1-B-A = (vp1+vm1)/2 - v0
//Subtracting that twice and a B from the P'(1)
//D=(vp2-v0)/2 - 2(D+C) -B =(vp2-v0)/2 - (Vp1+vm1-2v0) - (vp1-vm1)/2
// = 3(v0-vp1)/2 + (vp2-vm1)/2
//C=(D+C)-D = (vp1+vm1)/2 - v0 - (3(v0-vp1)/2 + (vp2-vm1)/2)
// = vm1 + 2vp1 - (5v0+vp2)/2;
//It is quite easy to calculate P(½)
//P(½)=D/8+C/4+B/2+A = (9*(v0+vp1)-(vm1+vp2))/16
//i.e. symmetric in its uses, and a mean of closest adjusted by mean of next ones
double B = (vp1 - vm1) / 2;
double DpC =(vp1 -v0) -B; //D+C+B+A - A - B
double D = (vp2 - v0) / 2 - 2 * DpC - B;
double C = DpC - D;
//double C = vm1 + 2 * vp1 - (5 * v0 + vp2) / 2;
//double D = (3*(v0 - vp1) + (vp2 - vm1)) / 2;
return ((D * frac + C) * frac + B) * frac + A;
}
Inspired by the comment of ManuTOO below I made it as fourth order too, with an optional parameter, you can set/calculate as you please, without breaking the smoothness of the curve.
It is basically the same with an added term all the way in the calculations. And If you set E to Zero, it will be identical to the above.
(Obviously if the data is actually on a line your calculation of E must assure that E is zero to get a linear output)
//The polynomial of degree 4 where P(x)=f(x) for x in {0,1}
//and P'(1) in one cell matches P'(0) in the next, gives a continuous smooth curve.
//And we also wants the it to reduce nicely to a line, if that matches the data
//With high order quotient weight of your choice....
//P(x)=Ex⁴+Dx³+Cx²+Bx+A=((((Ex+D)x+C)x+B)x+A
//P'(x)=4Ex³+3Dx²+2Cx+B
//P(0)=A =v0
//P(1)=E+D+C+B+A =Vp1
//P'(0)=B =(vp1-vm1)/2
//P'(1)=4E+3D+2C+B=(vp2-v0 )/2
//Subtracting Expressions for A, B and E from the E+D+C+B+A
//D+C =vp1-B-A -E = (vp1+vm1)/2 - v0 -E
//Subtracting that twice, a B and 4E from the P'(1)
//D=(vp2-v0)/2 - 2(D+C) -B -4E =(vp2-v0)/2 - (Vp1+vm1-2v0-2E) - (vp1-vm1)/2 -4E
// = 3(v0-vp1)/2 + (vp2-vm1)/2 -2E
//C=(D+C)-(D) = (vp1+vm1)/2 - v0 -E - (3(v0-vp1)/2 + (vp2-vm1)/2 -2E)
// = vm1 + 2vp1 - (5v0+vp2)/2 +E;
double E = ????????; //Your feed.... If Zero, cubic, see below
double B = (vp1 - vm1) / 2;
double DpC =(vp1 -v0) -B -E; //E+D+C+B+A - A - B -E
double D = (vp2 - v0) / 2 - 2 * DpC - B - 4 * E;
double C = DpC - D;
return (((E * frac + D) * frac + C) * frac + B) * frac + v0;
ADD:
Quoted from suggestion by #ManTOO below:
double E = (v0 - vm1 + vp1 - vp2) * m_BicubicSharpness;
With m_BicubicSharpness at 1.5, it's very close of Photoshop's Bicubic Sharper ; personally, I set it to 1.75 for a bit of extra kick.
Note that if data is on a line, this suggestion reduces nicely to zero
Note: The Bicubic formula given by Nicholas above (as the answer) has an error in it. by interpolating a sinusoid, I was able to see it was not correct.
The correct formula is:
float A = 0.5f * (v3 - v0) + 1.5f * (v1 - v2);
float B = 0.5f * (v0 + v2) - v1 - A;
float C = 0.5f * (v2 - v0);
float D = v1;
For the derivation, see https://www.paulinternet.nl/?page=bicubic
I want to use a random number generator that creates random numbers in a gaussian range where I can define the median by myself. I already asked a similar question here and now I'm using this code:
class RandomGaussian
{
private static Random random = new Random();
private static bool haveNextNextGaussian;
private static double nextNextGaussian;
public static double gaussianInRange(double from, double mean, double to)
{
if (!(from < mean && mean < to))
throw new ArgumentOutOfRangeException();
int p = Convert.ToInt32(random.NextDouble() * 100);
double retval;
if (p < (mean * Math.Abs(from - to)))
{
double interval1 = (NextGaussian() * (mean - from));
retval = from + (float)(interval1);
}
else
{
double interval2 = (NextGaussian() * (to - mean));
retval = mean + (float)(interval2);
}
while (retval < from || retval > to)
{
if (retval < from)
retval = (from - retval) + from;
if (retval > to)
retval = to - (retval - to);
}
return retval;
}
private static double NextGaussian()
{
if (haveNextNextGaussian)
{
haveNextNextGaussian = false;
return nextNextGaussian;
}
else
{
double v1, v2, s;
do
{
v1 = 2 * random.NextDouble() - 1;
v2 = 2 * random.NextDouble() - 1;
s = v1 * v1 + v2 * v2;
} while (s >= 1 || s == 0);
double multiplier = Math.Sqrt(-2 * Math.Log(s) / s);
nextNextGaussian = v2 * multiplier;
haveNextNextGaussian = true;
return v1 * multiplier;
}
}
}
Then to verify the results I plotted them with gaussianInRange(0, 0.5, 1) for n=100000000
As one can see the median is really at 0.5 but there isn't really a curve visible. So what I'm doing wrong?
EDIT
What i want is something like this where I can set the highest probability by myself by passing a value.
The simplest way to draw normal deviates conditional on them being in a particular range is with rejection sampling:
do {
retval = NextGaussian() * stdev + mean;
} while (retval < from || to < retval);
The same sort of thing is used when you draw coordinates (v1, v2) in a circle in your unconditional normal generator.
Simply folding in values outside the range doesn't produce the same distribution.
Also, if you have a good implementation of the error function and its inverse, you can calculate the values directly using an inverse CDF. The CDF of a normal distribution is
F(retval) = (1 + erf((retval-mean) / (stdev*sqrt(2)))) / 2
The CDF of a censored distribution is
C(retval) = (F(retval) - F(from)) / (F(to) - F(from)), from ≤ x < to
To draw a random number using a CDF, you draw v from a uniform distribution on [0, 1] and solve C(retval) = v. This gives
double v = random.NextDouble();
double t1 = erf((from - mean) / (stdev*sqrt(2)));
t2 = erf((to - mean) / (stdev*sqrt(2)));
double retval = mean + stdev * sqrt(2) * erf_inv(t1*(1-v) + t2*v);
You can precalculate t1 and t2 for specific parameters. The advantage of this approach is that there is no rejection sampling, so you only need a single NextDouble() per draw. If the [from, to] interval is small this will be faster.
However, it sounds like you might want the binomial distribution instead.
I have similar methods in my Graph generator (had to modify it a bit):
Returns a random floating-point number using a generator function with a specific range:
private double NextFunctional(Func<double, double> func, double from, double to, double height, out double x)
{
double halfWidth = (to - from) / 2;
double distance = halfWidth + from;
x = this.rand.NextDouble() * 2 - 1;// -1 .. 1
double y = func(x);
x = halfWidth * x + distance;
y *= height;
return y;
}
Gaussian function:
private double Gauss(double x)
{
// Graph should look better with double-x scale.
x *= 2;
double σ = 1 / Math.Sqrt(2 * Math.PI);
double variance = Math.Pow(σ, 2);
double exp = -0.5 * Math.Pow(x, 2) / variance;
double y = 1 / Math.Sqrt(2 * Math.PI * variance) * Math.Pow(Math.E, exp);
return y;
}
A method that generates a graph using the random numbers:
private void PlotGraph(Graphics g, Pen p, double from, double to, double height)
{
for (int i = 0; i < 1000; i++)
{
double x;
double y = this.NextFunctional(this.Gauss, from, to, height, out x);
this.DrawPoint(g, p, x, y);
}
}
I would rather used a cosine function - it is much faster and pretty close to the gaussian function for your needs:
double x;
double y = this.NextFunctional(a => Math.Cos(a * Math.PI), from, to, height, out x);
The out double x parameter in the NextFunctional() method is there so you can easily test it on your graphs (I use an iterator in my method).