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source: branches/HeuristicLab.Problems.GrammaticalOptimization/DynamicDataDisplay/Common/BezierBuilder.cs @ 13757

Last change on this file since 13757 was 12503, checked in by aballeit, 10 years ago

#2283 added GUI and charts; fixed MCTS

File size: 3.6 KB
Line 
1using System;
2using System.Collections.Generic;
3using System.Linq;
4using System.Text;
5using System.Windows;
6
7namespace Microsoft.Research.DynamicDataDisplay.Charts.NewLine
8{
9  // todo check a license
10  public static class BezierBuilder
11  {
12    public static IEnumerable<Point> GetBezierPoints(Point[] points)
13    {
14      if (points == null)
15        throw new ArgumentNullException("points");
16
17      Point[] firstControlPoints;
18      Point[] secondControlPoints;
19
20      int n = points.Length - 1;
21      if (n < 1)
22        throw new ArgumentException("At least two knot points required", "points");
23      if (n == 1)
24      { // Special case: Bezier curve should be a straight line.
25        firstControlPoints = new Point[1];
26        // 3P1 = 2P0 + P3
27        firstControlPoints[0].X = (2 * points[0].X + points[1].X) / 3;
28        firstControlPoints[0].Y = (2 * points[0].Y + points[1].Y) / 3;
29
30        secondControlPoints = new Point[1];
31        // P2 = 2P1 – P0
32        secondControlPoints[0].X = 2 *
33          firstControlPoints[0].X - points[0].X;
34        secondControlPoints[0].Y = 2 *
35          firstControlPoints[0].Y - points[0].Y;
36
37        return Join(points, firstControlPoints, secondControlPoints);
38      }
39
40      // Calculate first Bezier control points
41      // Right hand side vector
42      double[] rhs = new double[n];
43
44      // Set right hand side X values
45      for (int i = 1; i < n - 1; ++i)
46        rhs[i] = 4 * points[i].X + 2 * points[i + 1].X;
47      rhs[0] = points[0].X + 2 * points[1].X;
48      rhs[n - 1] = (8 * points[n - 1].X + points[n].X) / 2.0;
49      // Get first control points X-values
50      double[] x = GetFirstControlPoints(rhs);
51
52      // Set right hand side Y values
53      for (int i = 1; i < n - 1; ++i)
54        rhs[i] = 4 * points[i].Y + 2 * points[i + 1].Y;
55      rhs[0] = points[0].Y + 2 * points[1].Y;
56      rhs[n - 1] = (8 * points[n - 1].Y + points[n].Y) / 2.0;
57      // Get first control points Y-values
58      double[] y = GetFirstControlPoints(rhs);
59
60      // Fill output arrays.
61      firstControlPoints = new Point[n];
62      secondControlPoints = new Point[n];
63      for (int i = 0; i < n; ++i)
64      {
65        // First control point
66        firstControlPoints[i] = new Point(x[i], y[i]);
67        // Second control point
68        if (i < n - 1)
69          secondControlPoints[i] = new Point(2 * points
70            [i + 1].X - x[i + 1], 2 *
71            points[i + 1].Y - y[i + 1]);
72        else
73          secondControlPoints[i] = new Point((points
74            [n].X + x[n - 1]) / 2,
75            (points[n].Y + y[n - 1]) / 2);
76      }
77
78      return Join(points, firstControlPoints, secondControlPoints);
79    }
80
81    private static IEnumerable<Point> Join(Point[] points, Point[] firstControlPoints, Point[] secondControlPoints)
82    {
83      var length = firstControlPoints.Length;
84      for (int i = 0; i < length; i++)
85      {
86        yield return points[i];
87        yield return firstControlPoints[i];
88        yield return secondControlPoints[i];
89      }
90
91      yield return points[length];
92    }
93
94    /// <summary>
95    /// Solves a tridiagonal system for one of coordinates (x or y)
96    /// of first Bezier control points.
97    /// </summary>
98    /// <param name="rhs">Right hand side vector.</param>
99    /// <returns>Solution vector.</returns>
100    private static double[] GetFirstControlPoints(double[] rhs)
101    {
102      int n = rhs.Length;
103      double[] x = new double[n]; // Solution vector.
104      double[] tmp = new double[n]; // Temp workspace.
105
106      double b = 2.0;
107      x[0] = rhs[0] / b;
108      for (int i = 1; i < n; i++) // Decomposition and forward substitution.
109      {
110        tmp[i] = 1 / b;
111        b = (i < n - 1 ? 4.0 : 3.5) - tmp[i];
112        x[i] = (rhs[i] - x[i - 1]) / b;
113      }
114      for (int i = 1; i < n; i++)
115        x[n - i - 1] -= tmp[n - i] * x[n - i]; // Backsubstitution.
116
117      return x;
118    }
119  }
120}
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