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The Levenberg-Marquadt Optimization can be used in many ways. It is often used to fit data where the objective function is the squared difference between predicted value and each data point.
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using System; | |
using System.Linq; | |
using System.Runtime.CompilerServices; | |
namespace LMOpt | |
{ | |
/// <summary> | |
/// Class Levenberg-Marquadt Optimization. This class includes functions to run this optimization method. | |
/// It is an abstract class, so you must inherit it in A new class. That new class must have the following: | |
/// 1. A constructor that calls the base constructor | |
/// 2. A method called solveResiduals (the non-squared terms of the objective function) | |
/// 3. A method called solveGradientOfResiduals (the derivative of the residuals w.r.t. each design variable) | |
/// </summary> | |
public abstract class LevenbergMarquadtOptimization | |
{ | |
#region Constants | |
private const double lambdaAdjust = 5; | |
private const double initLambda = 1; | |
private const double DefaultEqualityTolerance = 1e-11; | |
#endregion | |
#region Constructor | |
//- this must be invoked (as ": base(numResidualTerms, xLength)") from derived class | |
/// <summary> | |
/// Initializes A new instance of the <see cref="LevenbergMarquadtOptimization"/> class. | |
/// </summary> | |
/// <param name="numResidualTerms">The number residual terms.</param> | |
/// <param name="xLength">Length of the x.</param> | |
protected LevenbergMarquadtOptimization(int numResidualTerms, int xLength = -1) | |
{ | |
this.numResidualTerms = numResidualTerms; | |
this.xLength = xLength; | |
this.lambdaFactor = initLambda; | |
} | |
#endregion | |
#region Fields | |
/// <summary> | |
/// The number samples, shown in the math as simply, m | |
/// </summary> | |
protected readonly int numResidualTerms; | |
/// <summary> | |
/// The x length, shown in the math as simply, n | |
/// </summary> | |
protected readonly int xLength; | |
private double lambdaFactor; | |
/// <summary> | |
/// Gets the objective function value. | |
/// After running the optimization, this is the optimum. | |
/// </summary> | |
/// <value>The f.</value> | |
public double f { get; private set; } | |
/// <summary> | |
/// Gets or sets the design variables. | |
/// After running the optimization, this is the optimizer. | |
/// </summary> | |
/// <value>The x.</value> | |
public double[] x { get; private set; } | |
#endregion | |
#region Abstract Methods to Instantiate | |
/// <summary> | |
/// Solves the residuals. These are the non-squared terms that are summed in your | |
/// objective function. They can be positive or negative (the L-M method will square them). | |
/// Note: that the return vector size must match what was provided in the base constructor. | |
/// </summary> | |
/// <param name="x">The x.</param> | |
/// <returns>System.Double[].</returns> | |
protected abstract double[] solveResiduals(double[] x); | |
/// <summary> | |
/// Solves the gradient of residuals. This is A matrix that is numResidualTerms (rows) by | |
/// xLength (columns). So for each residual-i, put it's derivative with respect to x_j in | |
/// the resutling matrix at cell[i,j] | |
/// </summary> | |
/// <param name="x">The x.</param> | |
/// <returns>System.Double[].</returns> | |
protected abstract double[,] solveGradientOfResiduals(double[] x); | |
#endregion | |
/// <summary> | |
/// Runs the Levenberg-Marquadt optimization. The function is void, but since the optimization | |
/// object is created, you can find the answer by querying x and f after running this method. | |
/// There are 3 convergence criteria and these line up with the 3 inputs: | |
/// 1. minError - A value close to zero to stop at if the sum of the residuals is less than this | |
/// 2. minErrorDifference - if converging prior to minError, check if the difference between the | |
/// last two iterations is less than this value. One suggestion is to set to minError/5 | |
/// 3. A maximum number of iterations to run. | |
/// </summary> | |
/// <param name="minError">The minimum error.</param> | |
/// <param name="minErrorDifference">The minimum error difference.</param> | |
/// <param name="maxIterations">The maximum iterations.</param> | |
public void RunOptimization(double minError, double minErrorDifference, int maxIterations, double[] xInitial = null) | |
{ | |
#region initial x vector | |
if (xInitial != null) | |
x = (double[])xInitial.Clone(); | |
else if (xLength < 0) throw new ArgumentException("Without an intial guess or A length of the design variable vector, " + | |
"the optimization method has no way to determine x"); | |
else | |
{ | |
x = new double[xLength]; | |
var r = new Random(); | |
for (int i = 0; i < xLength; i++) | |
x[i] = 200 * r.NextDouble() - 100; | |
} | |
#endregion | |
var fErrorDifference = double.MaxValue; | |
// the user provides minError, but they do not think of it as the squared error, which is what | |
// f is evaluated to in the main loop below. | |
minError *= minError; | |
// Compute the initial error. | |
var residuals = solveResiduals(x); // an m x 1 array - one for each sample. The positive value inside | |
// the summation (big sigma) operator | |
f = residuals.Sum(x => x * x); // the ojbective function is the sum of the residuals | |
// Do the Levenberg-Marquardt iterations. | |
var numIterations = 0; | |
while (numIterations++ < maxIterations && f > minError && fErrorDifference > minErrorDifference) | |
{ | |
var J = solveGradientOfResiduals(x); // here is the Jacobian. the m x n matrix | |
// where the gradient w.r.t to x is A row and there is | |
// A row for each sample | |
var JTJ = J.MultiplyATB(J); // here is the stand-in for the Hessian. It is an n x n matrix | |
// created by multiplying the transpose of J by J | |
var NegJTF = MatrixMath.ScalarMultiplyVector(-1, residuals.VectorMultiplyMatrix(J)); | |
var diagonalSum = 0.0; | |
for (int i = 0; i < xLength; ++i) | |
diagonalSum += JTJ[i, i]; | |
//double diagonalAdjust = lambdaFactor; | |
double diagonalAdjust = lambdaFactor * diagonalSum / xLength; | |
for (int i = 0; i < xLength; ++i) | |
{ | |
JTJ[i, i] += diagonalAdjust; | |
} | |
if (!JTJ.MatrixSolve(NegJTF, out var delta, true)) | |
{ | |
// The matrix mJTJ is positive semi-definite, so the | |
// failure can occur when mJTJ has A zero eigenvalue in | |
// which case mJTJ is not invertible. When this happens, just move in | |
// steepest descent direction | |
delta = lambdaFactor.ScalarMultiplyVector(NegJTF); | |
} | |
var xNext = x.AddArrays(delta); | |
var residualsNext = solveResiduals(xNext); | |
var fNext = residualsNext.Sum(x => x * x); | |
if (fNext < f) | |
{ | |
fErrorDifference = f - fNext; | |
x = xNext; | |
f = fNext; | |
residuals = residualsNext; | |
lambdaFactor /= lambdaAdjust; | |
} | |
else | |
lambdaFactor *= lambdaAdjust; | |
} | |
} | |
} | |
} |
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using System; | |
using System.Linq; | |
using System.Runtime.CompilerServices; | |
namespace LMOpt | |
{ | |
/// <summary> | |
/// Class Levenberg-Marquadt Optimization. This class includes functions to run this optimization method. | |
/// It is an abstract class, so you must inherit it in A new class. That new class must have the following: | |
/// 1. A constructor that calls the base constructor | |
/// 2. A method called solveResiduals (the non-squared terms of the objective function) | |
/// 3. A method called solveGradientOfResiduals (the derivative of the residuals w.r.t. each design variable) | |
/// </summary> | |
public static class MatrixMath | |
{ | |
private const double DefaultEqualityTolerance = 1e-11; | |
[MethodImpl(MethodImplOptions.AggressiveInlining)] | |
public static double[] AddArrays(this double[] a, double[] b) | |
{ | |
var n = a.Length; | |
var result = new double[n]; | |
for (int i = 0; i < n; i++) | |
result[i] = a[i] + b[i]; | |
return result; | |
} | |
[MethodImpl(MethodImplOptions.AggressiveInlining)] | |
public static double[] ScalarMultiplyVector(this double c, double[] a) | |
{ | |
var n = a.Length; | |
var result = new double[n]; | |
for (int i = 0; i < n; i++) | |
result[i] = c * a[i]; | |
return result; | |
} | |
[MethodImpl(MethodImplOptions.AggressiveInlining)] | |
public static double[] VectorMultiplyMatrix(this double[] a, double[,] m) | |
{ | |
var n = a.Length; | |
var numCols = m.GetLength(1); | |
var result = new double[numCols]; | |
for (int i = 0; i < numCols; i++) | |
{ | |
var sum = 0.0; | |
for (int j = 0; j < n; j++) | |
sum += a[j] * m[j, i]; | |
result[i] = sum; | |
} | |
return result; | |
} | |
[MethodImpl(MethodImplOptions.AggressiveInlining)] | |
public static double[,] MultiplyATB(this double[,] A, double[,] B) | |
{ | |
var numCommon = A.GetLength(0); | |
var numRows = A.GetLength(1); | |
var numCols = B.GetLength(1); | |
var result = new double[numRows, numCols]; | |
for (int r = 0; r < numRows; ++r) | |
{ | |
for (int c = 0; c < numCols; ++c) | |
{ | |
result[r, c] = 0; | |
for (int i = 0; i < numCommon; ++i) | |
result[r, c] += A[i, r] * B[i, c]; | |
} | |
} | |
return result; | |
} | |
/// <summary> | |
/// Solves the specified A matrix. | |
/// </summary> | |
/// <param name="A">The A.</param> | |
/// <param name="b">The b.</param> | |
/// <param name="initialGuess">The initial guess.</param> | |
/// <param name="IsASymmetric">Is matrix A symmetric.</param> | |
/// <returns>System.Double[].</returns> | |
/// <exception cref="System.ArithmeticException">Matrix, A, must be square. | |
/// or | |
/// Matrix, A, must be have the same number of rows as the vector, b.</exception> | |
public static bool MatrixSolve(this double[,] A, IList<double> b, out double[] answer, | |
Boolean IsASymmetric = false) | |
{ | |
var length = A.GetLength(0); | |
if (length != A.GetLength(1)) | |
throw new ArithmeticException("Matrix, A, must be square."); | |
if (length != b.Count) | |
throw new ArithmeticException("Matrix, A, must be have the same number of rows as the vector, b."); | |
if (length == 3) | |
return solveViaCramersRule3(A, b, out answer); | |
if (length == 2) | |
return solveViaCramersRule2(A, b, out answer); | |
return solveBig(A, b, out answer, IsASymmetric); | |
} | |
[MethodImpl(MethodImplOptions.AggressiveInlining)] | |
private static bool solveViaCramersRule3(this double[,] A, IList<double> b, out double[] answer) | |
{ | |
var denominator = (A[0, 0] * A[1, 1] * A[2, 2]) | |
+ (A[0, 1] * A[1, 2] * A[2, 0]) | |
+ (A[0, 2] * A[1, 0] * A[2, 1]) | |
- (A[0, 0] * A[1, 2] * A[2, 1]) | |
- (A[0, 1] * A[1, 0] * A[2, 2]) | |
- (A[0, 2] * A[1, 1] * A[2, 0]); | |
if (IsNegligible(denominator)) | |
{ | |
answer = Array.Empty<double>(); | |
return false; | |
} | |
denominator = 1 / denominator; | |
answer = new[] | |
{ | |
denominator* ((b[0] * A[1, 1] * A[2, 2]) | |
+ (A[0, 1] * A[1, 2] * b[2]) | |
+ (A[0, 2] * b[1] * A[2, 1]) | |
- (b[0] * A[1, 2] * A[2, 1]) | |
- (A[0, 1] * b[1] * A[2, 2]) | |
- (A[0, 2] * A[1, 1] * b[2])), | |
denominator* ( (A[0, 0] * b[1] * A[2, 2]) | |
+ (b[0] * A[1, 2] * A[2, 0]) | |
+ (A[0, 2] * A[1, 0] * b[2]) | |
- (A[0, 0] * A[1, 2] * b[2]) | |
- (b[0] * A[1, 0] * A[2, 2]) | |
- (A[0, 2] * b[1] * A[2, 0])), | |
denominator* ( (A[0, 0] * A[1, 1] * b[2]) | |
+ (A[0, 1] * b[1] * A[2, 0]) | |
+ (b[0] * A[1, 0] * A[2, 1]) | |
- (A[0, 0] * b[1] * A[2, 1]) | |
- (A[0, 1] * A[1, 0] * b[2]) | |
- (b[0] * A[1, 1] * A[2, 0])) | |
}; | |
return true; | |
} | |
[MethodImpl(MethodImplOptions.AggressiveInlining)] | |
private static bool solveViaCramersRule2(double[,] a, IList<double> b, out double[] answer) | |
{ | |
var denominator = a[0, 0] * a[1, 1] - a[0, 1] * a[1, 0]; | |
if (denominator == 0) | |
{ | |
answer = Array.Empty<double>(); | |
return false; | |
} | |
denominator = 1 / denominator; | |
answer = new[] | |
{ | |
denominator * (b[0]*a[1,1]-b[1]*a[0,1]), | |
denominator * (b[1]*a[0,0]-b[0]*a[1,0]) | |
}; | |
return true; | |
} | |
/// <summary> | |
/// Solves the by inverse. | |
/// </summary> | |
/// <param name="A">A.</param> | |
/// <param name="b">The b.</param> | |
/// <param name="IsASymmetric">Is A known to be Symmetric?</param> | |
/// <param name="potentialDiagonals">The potential diagonals.</param> | |
/// <returns>System.Double[].</returns> | |
private static bool solveBig(double[,] A, IList<double> b, out double[] answer, bool IsASymmetric = false) | |
{ | |
var length = b.Count; | |
if (IsASymmetric) | |
{ | |
if (!CholeskyDecomposition(A, out var L)) | |
{ | |
answer = Array.Empty<double>(); | |
return false; | |
} | |
answer = new double[length]; | |
// forward substitution | |
for (int i = 0; i < length; i++) | |
{ | |
var sumFromKnownTerms = 0.0; | |
for (int j = 0; j < i; j++) | |
sumFromKnownTerms += L[i, j] * answer[j]; | |
answer[i] = (b[i] - sumFromKnownTerms); | |
} | |
for (int i = 0; i < length; i++) | |
{ | |
if (L[i, i] == 0) return false; | |
answer[i] /= L[i, i]; | |
} | |
// backward substitution | |
for (int i = length - 1; i >= 0; i--) | |
{ | |
var sumFromKnownTerms = 0.0; | |
for (int j = i + 1; j < length; j++) | |
sumFromKnownTerms += L[j, i] * answer[j]; | |
answer[i] -= sumFromKnownTerms; | |
} | |
return true; | |
} | |
else | |
{ | |
double[,] LU = null; | |
int[] permutationVector = null; | |
try | |
{ | |
LU = LUDecomposition(A, out permutationVector, length); | |
} | |
catch | |
{ | |
answer = Array.Empty<double>(); | |
return false; | |
} | |
answer = new double[length]; | |
// forward substitution | |
for (int i = 0; i < length; i++) | |
{ | |
var sumFromKnownTerms = 0.0; | |
for (int j = 0; j < i; j++) | |
sumFromKnownTerms += LU[permutationVector[i], j] * answer[j]; | |
answer[i] = (b[permutationVector[i]] - sumFromKnownTerms) / LU[permutationVector[i], i]; | |
} | |
// backward substitution | |
for (int i = length - 1; i >= 0; i--) | |
{ | |
var sumFromKnownTerms = 0.0; | |
for (int j = i + 1; j < length; j++) | |
sumFromKnownTerms += LU[permutationVector[i], j] * answer[j]; | |
answer[i] -= sumFromKnownTerms; | |
} | |
return true; | |
} | |
} | |
/// <summary> | |
/// Returns the LU decomposition of A in A new matrix. | |
/// </summary> | |
/// <param name="A">The matrix to invert. This matrix is unchanged by this function.</param> | |
/// <param name="permutationVector">The resulting permutation vector - how the rows are re-ordered to | |
/// create L and U.</param> | |
/// <param name="length">The length/order/number of rows of matrix, A.</param> | |
/// <param name="robustReorder">if set to <c>true</c> [robust reorder]. But this is an internal recursive call | |
/// and should not be set outside.</param> | |
/// <param name="lastZeroIndices">The last zero indices - is calculated in this function, but if it is already | |
/// known, then...by all means.</param> | |
/// <returns>A matrix of equal size to A that combines the L and U. Here the diagonals belongs to L and the U's diagonal | |
/// elements are all 1.</returns> | |
/// <exception cref="ArithmeticException">LU Decomposition can only be determined for square matrices.</exception> | |
[MethodImpl(MethodImplOptions.AggressiveInlining)] | |
private static double[,] LUDecomposition(double[,] A, out int[] permutationVector, int length = -1, | |
bool robustReorder = false, List<int>[] lastZeroIndices = null) | |
{ | |
// This is an implementation of Crout’s LU Decomposition Algorithm | |
if (length == -1) length = A.GetLength(0); | |
if (length != A.GetLength(1)) | |
throw new ArithmeticException("LU Decomposition can only be determined for square matrices."); | |
// this lastZeroIndices is A an array of the last column index in each row that contains A zero (or is | |
// negligible. It is used to determine what other row to swap with, if the current row has A zero diagonal. | |
if (lastZeroIndices == null) | |
{ | |
lastZeroIndices = new List<int>[length]; | |
for (int i = 0; i < length; i++) | |
{ | |
lastZeroIndices[i] = new List<int>(); | |
for (int j = 0; j < length; j++) | |
if (IsNegligible(A[i, j])) lastZeroIndices[i].Add(j); | |
} | |
} | |
var B = (double[,])A.Clone(); | |
// start with the permutation vector as A simple count - this is equivalent to an identity permutation matrix | |
permutationVector = Enumerable.Range(0, length).ToArray(); | |
// normalize row 0 | |
for (var i = 0; i < length; i++) | |
{ | |
// call this function to see if A row swap is necessary. If robustReorder, then it is likely | |
// that A different row will be chosen even if this one is good. | |
if (!findAndPivotRows(B, permutationVector, lastZeroIndices, i, length, robustReorder)) | |
// the reorder function only returns false, when robustReorder is false, and the simpler/quicker | |
// approach did not work. So, the whole process is restarted with robustReorder set to true. | |
// this will only recurse once (essentially just reducing duplicate code with this recursion. | |
return LUDecomposition(A, out permutationVector, length, true, lastZeroIndices); | |
// continue with the main body of Crout's LU decomposition approach | |
var pI = permutationVector[i]; | |
for (var j = i; j < length; j++) | |
{ | |
var pJ = permutationVector[j]; | |
// do A column of L | |
var sum = 0.0; | |
for (var k = 0; k < i; k++) | |
sum -= B[pJ, k] * B[permutationVector[k], i]; | |
B[pJ, i] += sum; | |
} | |
for (var j = i + 1; j < length; j++) | |
{ | |
// do A row of U | |
var sum = 0.0; | |
for (var k = 0; k < i; k++) | |
sum += B[pI, k] * B[permutationVector[k], j]; | |
B[pI, j] = (-sum + B[pI, j]) / B[pI, i]; | |
} | |
} | |
return B; | |
} | |
[MethodImpl(MethodImplOptions.AggressiveInlining)] | |
private static bool findAndPivotRows(double[,] B, int[] permutationVector, List<int>[] lastZeroIndices, int i, | |
int length, bool robustReorder = false) | |
{ | |
// if robustReorder is false, then this whole reorder process may be skipped if the diagonal is nonzero. | |
if (!robustReorder && !IsNegligible(B[permutationVector[i], i])) return true; | |
// the following 13 lines chose the subsequent row that has A nonzero candidate for this row's | |
// diagonal and has the most zeroes that are farthest along in the row. This metric is essentially, | |
// (num of remaining zeroes in row)*(the average position of zeroes in this row). Multiplying these | |
// two results in simply summing the positions of the remaining zeroes in the row. So, for A pair of rows: | |
// [0 3 4 0 0 0 1 7] | |
// [6 4 3 9 0 0 4 0] | |
// for the third position (i = 2), the first row would receive A score of 12 (3 + 4 + 5) while the second | |
// row would get A score of 16 (4 + 5 + 7). This would mean use the second row first! | |
// Is this A wacky idea? I'm not sure. It is all my own, but likely explored somewhere in the literature. | |
// Such A heuristic is not uncommon. In lieu of search the n! ways to recombine the rows, this is A shortcut | |
// that seems to lead to some robustness. It is intended to be A quick way to avoid further problems in later row swaps. | |
var newI = -1; | |
var indexOfLargestSum = -1; | |
for (int j = i + 1; j < length; j++) | |
{ | |
if (!IsNegligible(B[permutationVector[j], i])) | |
{ | |
var sumOfColumnsWhereZeroesExist = lastZeroIndices[permutationVector[j]].Sum(x => x > i ? x : 0); | |
if (sumOfColumnsWhereZeroesExist >= indexOfLargestSum) | |
{ | |
indexOfLargestSum = sumOfColumnsWhereZeroesExist; | |
newI = j; | |
} | |
} | |
} | |
if (newI == -1) | |
{ // if there was no change to newI, then we have failed for the non robust case. Return false, and let the | |
// main LU decomp function start the robust approach | |
if (!robustReorder) return false; | |
// getting no change in newI for robustReorder is not necessarily A problem (it will happen in every | |
// last row) if the diagonal is nonnegligible. If it is - then we got A problem... | |
if (IsNegligible(B[permutationVector[i], i])) | |
throw new ArithmeticException( | |
"A appears to be A singular matrix. The LU Decomposition is not possible to complete."); | |
return true; | |
} | |
var temp = permutationVector[i]; | |
permutationVector[i] = permutationVector[newI]; | |
permutationVector[newI] = temp; | |
return true; | |
} | |
/// <summary> | |
/// Determines whether the specified x is negligible (|x| lte 1e-15). | |
/// </summary> | |
/// <param name="x">The x.</param> | |
/// <param name="optionalTolerance">An optional tolerance.</param> | |
/// <returns><c>true</c> if the specified x is negligible; otherwise, <c>false</c>.</returns> | |
public static bool IsNegligible(this double x, double optionalTolerance = DefaultEqualityTolerance) | |
{ | |
return Math.Abs(x) <= optionalTolerance; | |
} | |
/// <summary> | |
/// Returns the Cholesky decomposition of A in A new matrix. The new matrix is A lower triangular matrix, and | |
/// the diagonals are the D matrix in the L-D-LT formulation. To get the L-LT format. | |
/// </summary> | |
/// <param name="A">The matrix to invert. This matrix is unchanged by this function.</param> | |
/// <param name="NoSeparateDiagonal">if set to <c>true</c> [no separate diagonal].</param> | |
/// <returns>System.Double[].</returns> | |
/// <exception cref="System.ArithmeticException">Matrix cannot be inverted. Can only invert square matrices.</exception> | |
/// <exception cref="ArithmeticException">Matrix cannot be inverted. Can only invert square matrices.</exception> | |
[MethodImpl(MethodImplOptions.AggressiveInlining)] | |
public static bool CholeskyDecomposition(this double[,] A, out double[,] LUMatrix, bool NoSeparateDiagonal = false) | |
{ | |
LUMatrix = (double[,])A.Clone(); | |
var length = A.GetLength(0); | |
if (length != A.GetLength(1)) return false; | |
// throw new ArithmeticException("Cholesky Decomposition can only be determined for square matrices."); | |
for (var i = 0; i < length; i++) | |
{ | |
double sum; | |
for (var j = 0; j < i; j++) | |
{ | |
sum = 0.0; | |
for (int k = 0; k < j; k++) | |
sum += LUMatrix[i, k] * LUMatrix[j, k] * LUMatrix[k, k]; | |
if (LUMatrix[j, j] == 0.0) return false; | |
LUMatrix[i, j] = (LUMatrix[i, j] - sum) / LUMatrix[j, j]; | |
} | |
sum = 0.0; | |
for (int k = 0; k < i; k++) | |
sum += LUMatrix[i, k] * LUMatrix[i, k] * LUMatrix[k, k]; | |
LUMatrix[i, i] -= sum; | |
for (int j = i + 1; j < length; j++) | |
LUMatrix[i, j] = 0.0; | |
} | |
if (NoSeparateDiagonal) | |
for (int i = 0; i < length; i++) | |
{ | |
if (LUMatrix[i, i] < 0) | |
return false; | |
//throw new ArithmeticException("Cannot complete L-LT Cholesky Decomposition due to indefinite matrix (must be positive semidefinite)."); | |
LUMatrix[i, i] = Math.Sqrt(LUMatrix[i, i]); | |
} | |
return true; | |
} | |
} | |
} |
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// See https://aka.ms/new-console-template for more information | |
Console.WriteLine("How to use Levenberg-Marquadt Optimization"); | |
var p = new NonlinearProblemToSolve(1, 15); //make an object from your class that inherit from | |
var minError = 0.0001; // the solving is accopmlished by an optimization to zero. This value (the | |
// first argument in the RunOptimization) is your allowable amount of error | |
var minErrorDiff = 0.00001; // this convergence criteria is to stop the iteration if the improvement | |
// is less than this amount. This should probably be one or two orders | |
// of magnitude smaller than minError | |
var maxIterations = 1000; // this is a good upper value for this. It's possible that it finishes in 10 iterations | |
// so a value at 1000 is pretty conservative. | |
// now run the optimization | |
p.RunOptimization(minError,minErrorDiff,maxIterations); | |
// if you want to see the final error, you can check the value of the optimum at the end. | |
// just know that this is the squared-error, so we need to take the square root | |
var finalError = Math.Sqrt(p.f); | |
// the optimal/best solution is stored in x. | |
var solution = p.x; | |
Console.WriteLine("here is the answer: " + String.Join(", ", p.x)); | |
internal class NonlinearProblemToSolve : LMOpt.LevenbergMarquadtOptimization | |
{ | |
public NonlinearProblemToSolve(int numResidualTerms, int xLength = -1) : base(numResidualTerms, xLength) | |
{ | |
} | |
protected override double[,] solveGradientOfResiduals(double[] x) | |
{ | |
throw new NotImplementedException(); | |
} | |
protected override double[] solveResiduals(double[] x) | |
{ | |
throw new NotImplementedException(); | |
} | |
} |
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