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Spline interpolation in C#
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Diagnostics;
namespace Interpolation
{
/// <summary>
/// Spline interpolation class.
/// </summary>
public class SplineInterpolator
{
private readonly double[] _keys;
private readonly double[] _values;
private readonly double[] _h;
private readonly double[] _a;
/// <summary>
/// Class constructor.
/// </summary>
/// <param name="nodes">Collection of known points for further interpolation.
/// Should contain at least two items.</param>
public SplineInterpolator(IDictionary<double, double> nodes)
{
if (nodes == null)
{
throw new ArgumentNullException("nodes");
}
var n = nodes.Count;
if (n < 2)
{
throw new ArgumentException("At least two point required for interpolation.");
}
_keys = nodes.Keys.ToArray();
_values = nodes.Values.ToArray();
_a = new double[n];
_h = new double[n];
for (int i = 1; i < n; i++)
{
_h[i] = _keys[i] - _keys[i - 1];
}
if (n > 2)
{
var sub = new double[n - 1];
var diag = new double[n - 1];
var sup = new double[n - 1];
for (int i = 1; i <= n - 2; i++)
{
diag[i] = (_h[i] + _h[i + 1]) / 3;
sup[i] = _h[i + 1] / 6;
sub[i] = _h[i] / 6;
_a[i] = (_values[i + 1] - _values[i]) / _h[i + 1] - (_values[i] - _values[i - 1]) / _h[i];
}
SolveTridiag(sub, diag, sup, ref _a, n - 2);
}
}
/// <summary>
/// Gets interpolated value for specified argument.
/// </summary>
/// <param name="key">Argument value for interpolation. Must be within
/// the interval bounded by lowest ang highest <see cref="_keys"/> values.</param>
public double GetValue(double key)
{
int gap = 0;
var previous = double.MinValue;
// At the end of this iteration, "gap" will contain the index of the interval
// between two known values, which contains the unknown z, and "previous" will
// contain the biggest z value among the known samples, left of the unknown z
for (int i = 0; i < _keys.Length; i++)
{
if (_keys[i] < key && _keys[i] > previous)
{
previous = _keys[i];
gap = i + 1;
}
}
var x1 = key - previous;
var x2 = _h[gap] - x1;
return ((-_a[gap - 1] / 6 * (x2 + _h[gap]) * x1 + _values[gap - 1]) * x2 +
(-_a[gap] / 6 * (x1 + _h[gap]) * x2 + _values[gap]) * x1) / _h[gap];
}
/// <summary>
/// Solve linear system with tridiagonal n*n matrix "a"
/// using Gaussian elimination without pivoting.
/// </summary>
private static void SolveTridiag(double[] sub, double[] diag, double[] sup, ref double[] b, int n)
{
int i;
for (i = 2; i <= n; i++)
{
sub[i] = sub[i] / diag[i - 1];
diag[i] = diag[i] - sub[i] * sup[i - 1];
b[i] = b[i] - sub[i] * b[i - 1];
}
b[n] = b[n] / diag[n];
for (i = n - 1; i >= 1; i--)
{
b[i] = (b[i] - sup[i] * b[i + 1]) / diag[i];
}
}
}
}
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Diagnostics;
namespace Interpolation
{
class Program
{
static void Main(string[] args)
{
var r = 1000;
var known = new Dictionary<double, double>
{
{ 0.0, 0.0 },
{ 100.0, 0.50 * r },
{ 300.0, 0.75 * r },
{ 500.0, 1.00 * r },
};
foreach (var pair in known)
{
Debug.WriteLine(String.Format("{0:0.000}\t{1:0.000}", pair.Key, pair.Value));
}
var scaler = new SplineInterpolator(known);
var start = known.First().Key;
var end = known.Last().Key;
var step = (end - start) / 50;
for (var x = start; x <= end; x += step)
{
var y = scaler.GetValue(x);
Debug.WriteLine(String.Format("\t\t{0:0.000}\t{1:0.000}", x, y));
}
}
}
}
@swharden
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swharden commented Jan 22, 2022

I recently found a more reliable method for achieving cubic spline interpolation in C#. The code I'm linking to here is simpler, faster, stateless, and doesn't throw exceptions.

https://swharden.com/blog/2022-01-22-spline-interpolation

spline-interpolation

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