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Fix Tukey loss function

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Since the output of LossFunction::Evaluate is multiplied by 0.5, the
current implementation of the Tukey loss function must be multiplied
by 2.

Change-Id: Ia94753eef1a375fe48cc2e0d2cc61350904c8248
This commit is contained in:
Enrique Fernandez
2020-02-17 15:52:58 +01:00
parent cf4185c4e3
commit 6da364713f
3 changed files with 47 additions and 10 deletions
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include/ceres/loss_function.h
+7 -6
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@@ -186,7 +186,7 @@ class CERES_EXPORT HuberLoss : public LossFunction {
//
// rho(s) = 2 (sqrt(1 + s) - 1).
//
// At s = 0: rho = [0, 1, -1/2].
// At s = 0: rho = [0, 1, -1 / (2 * a^2)].
class CERES_EXPORT SoftLOneLoss : public LossFunction {
public:
explicit SoftLOneLoss(double a) : b_(a * a), c_(1 / b_) {}
@@ -203,7 +203,7 @@ class CERES_EXPORT SoftLOneLoss : public LossFunction {
//
// rho(s) = log(1 + s).
//
// At s = 0: rho = [0, 1, -1].
// At s = 0: rho = [0, 1, -1 / a^2].
class CERES_EXPORT CauchyLoss : public LossFunction {
public:
explicit CauchyLoss(double a) : b_(a * a), c_(1 / b_) {}
@@ -276,12 +276,13 @@ class CERES_EXPORT TolerantLoss : public LossFunction {
// This is the Tukey biweight loss function which aggressively
// attempts to suppress large errors.
//
// The term is computed as:
// The term is computed as follows where the equations are scaled by a
// factor of 2 because the cost function is given by 1/2 rho(s):
//
// rho(s) = a^2 / 6 * (1 - (1 - s / a^2)^3 ) for s <= a^2,
// rho(s) = a^2 / 6 for s > a^2.
// rho(s) = a^2 / 3 * (1 - (1 - s / a^2)^3 ) for s <= a^2,
// rho(s) = a^2 / 3 for s > a^2.
//
// At s = 0: rho = [0, 0.5, -1 / a^2]
// At s = 0: rho = [0, 1, -2 / a^2]
class CERES_EXPORT TukeyLoss : public ceres::LossFunction {
public:
explicit TukeyLoss(double a) : a_squared_(a * a) {}