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Keir Mierle8ebb0732012-04-30 23:09:08 -07001// Ceres Solver - A fast non-linear least squares minimizer
2// Copyright 2010, 2011, 2012 Google Inc. All rights reserved.
3// http://code.google.com/p/ceres-solver/
4//
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28//
29// Author: sameeragarwal@google.com (Sameer Agarwal)
30//
31// Purpose: See .h file.
32
33#include "ceres/loss_function.h"
34
35#include <cmath>
36#include <cstddef>
37
38namespace ceres {
39
40void TrivialLoss::Evaluate(double s, double rho[3]) const {
41 rho[0] = s;
42 rho[1] = 1;
43 rho[2] = 0;
44}
45
46void HuberLoss::Evaluate(double s, double rho[3]) const {
47 if (s > b_) {
48 // Outlier region.
49 // 'r' is always positive.
50 const double r = sqrt(s);
51 rho[0] = 2 * a_ * r - b_;
52 rho[1] = a_ / r;
53 rho[2] = - rho[1] / (2 * s);
54 } else {
55 // Inlier region.
56 rho[0] = s;
57 rho[1] = 1;
58 rho[2] = 0;
59 }
60}
61
62void SoftLOneLoss::Evaluate(double s, double rho[3]) const {
63 const double sum = 1 + s * c_;
64 const double tmp = sqrt(sum);
65 // 'sum' and 'tmp' are always positive, assuming that 's' is.
66 rho[0] = 2 * b_ * (tmp - 1);
67 rho[1] = 1 / tmp;
68 rho[2] = - (c_ * rho[1]) / (2 * sum);
69}
70
71void CauchyLoss::Evaluate(double s, double rho[3]) const {
72 const double sum = 1 + s * c_;
73 const double inv = 1 / sum;
74 // 'sum' and 'inv' are always positive, assuming that 's' is.
75 rho[0] = b_ * log(sum);
76 rho[1] = inv;
77 rho[2] = - c_ * (inv * inv);
78}
79
Sameer Agarwalad1f7b72012-08-20 11:10:34 -070080void ArctanLoss::Evaluate(double s, double rho[3]) const {
81 const double sum = 1 + s * s * b_;
82 const double inv = 1 / sum;
83 // 'sum' and 'inv' are always positive.
84 rho[0] = a_ * atan2(s, a_);
85 rho[1] = inv;
86 rho[2] = -2 * s * b_ * (inv * inv);
87}
88
89TolerantLoss::TolerantLoss(double a, double b)
90 : a_(a),
91 b_(b),
92 c_(b * log(1.0 + exp(-a / b))) {
93 CHECK_GE(a, 0.0);
94 CHECK_GT(b, 0.0);
95}
96
97void TolerantLoss::Evaluate(double s, double rho[3]) const {
98 const double x = (s - a_) / b_;
99 // The basic equation is rho[0] = b ln(1 + e^x). However, if e^x is too
100 // large, it will overflow. Since numerically 1 + e^x == e^x when the
101 // x is greater than about ln(2^53) for doubles, beyond this threshold
102 // we substitute x for ln(1 + e^x) as a numerically equivalent approximation.
103 static const double kLog2Pow53 = 36.7; // ln(MathLimits<double>::kEpsilon).
104 if (x > kLog2Pow53) {
105 rho[0] = s - a_ - c_;
106 rho[1] = 1.0;
107 rho[2] = 0.0;
108 } else {
109 const double e_x = exp(x);
110 rho[0] = b_ * log(1.0 + e_x) - c_;
111 rho[1] = e_x / (1.0 + e_x);
112 rho[2] = 0.5 / (b_ * (1.0 + cosh(x)));
113 }
114}
115
116ComposedLoss::ComposedLoss(const LossFunction* f, Ownership ownership_f,
117 const LossFunction* g, Ownership ownership_g)
118 : f_(CHECK_NOTNULL(f)),
119 g_(CHECK_NOTNULL(g)),
120 ownership_f_(ownership_f),
121 ownership_g_(ownership_g) {
122}
123
124ComposedLoss::~ComposedLoss() {
125 if (ownership_f_ == DO_NOT_TAKE_OWNERSHIP) {
126 f_.release();
127 }
128 if (ownership_g_ == DO_NOT_TAKE_OWNERSHIP) {
129 g_.release();
130 }
131}
132
133void ComposedLoss::Evaluate(double s, double rho[3]) const {
134 double rho_f[3], rho_g[3];
135 g_->Evaluate(s, rho_g);
136 f_->Evaluate(rho_g[0], rho_f);
137 rho[0] = rho_f[0];
138 // f'(g(s)) * g'(s).
139 rho[1] = rho_f[1] * rho_g[1];
140 // f''(g(s)) * g'(s) * g'(s) + f'(g(s)) * g''(s).
141 rho[2] = rho_f[2] * rho_g[1] * rho_g[1] + rho_f[1] * rho_g[2];
142}
143
Keir Mierle8ebb0732012-04-30 23:09:08 -0700144void ScaledLoss::Evaluate(double s, double rho[3]) const {
145 if (rho_.get() == NULL) {
146 rho[0] = a_ * s;
147 rho[1] = a_;
148 rho[2] = 0.0;
149 } else {
150 rho_->Evaluate(s, rho);
151 rho[0] *= a_;
152 rho[1] *= a_;
153 rho[2] *= a_;
154 }
155}
156
157} // namespace ceres