修改平方和[ceres solver]
modify the squared sum [ceres-solver]
我正在尝试修改ceres的默认行为,即将残差的平方和计算为成本函数。我希望它只计算一个和(残差已经以只能为正的方式计算)
根据文档,我应该使用ConditionedCostFunction
这就是我所做的:我定义了取1个残差和1个参数的调节器
struct Conditioners : ceres::CostFunction
{
public:
Conditioners()
{
set_num_residuals(1);
mutable_parameter_block_sizes()->push_back(1);
}
~Conditioners()
{}
template<typename T>
T operator() (T x)
{
return T(x * x);
}
bool Evaluate(double const* const* parameters, double* residuals, double** jacobians) const
{
return true;
}
};
我把调理剂放在矢量里
std::vector<ceres::CostFunction*> conditioners;
for(int i = 0; i < 1; i++)
conditioners.push_back(new Conditioners());
ceres::ConditionedCostFunction* ccf =
new ceres::ConditionedCostFunction(cost_function, conditioners, ceres::TAKE_OWNERSHIP);
problem.AddResidualBlock(ccf, NULL, &x);
它编译一切。但这并不能解决问题。它甚至还没有启动。上面写着:
Ceres Solver Report: Iterations: 0, Initial cost: 4.512500e+01, Final cost: 4.512500e+01, Termination: CONVERGENCE
x : 0.5 -> 0.5
而不是:
iter cost cost_change |gradient| |step| tr_ratio tr_radius ls_iter iter_time total_time
0 4.512500e+01 0.00e+00 9.50e+00 0.00e+00 0.00e+00 1.00e+04 0 2.99e-04 1.04e-03
1 4.511598e-07 4.51e+01 9.50e-04 9.50e+00 1.00e+00 3.00e+04 1 3.84e-04 9.72e-03
2 5.012552e-16 4.51e-07 3.17e-08 9.50e-04 1.00e+00 9.00e+04 1 2.98e-05 9.92e-03
Ceres Solver Report: Iterations: 2, Initial cost: 4.512500e+01, Final cost: 5.012552e-16, Termination: CONVERGENCE
x : 0.5 -> 10
(如果你想自己尝试,这个例子会修改hellowword的例子)你知道哪里出了问题吗??(ceres报告没有更具体)
我找到了解决方案,它是:
struct Conditioners : ceres::CostFunction
{
public:
Conditioners()
{
set_num_residuals(1);
mutable_parameter_block_sizes()->push_back(1);
}
~Conditioners()
{}
template<typename T>
T operator() (T x)
{
return T(x * x);
}
bool Evaluate(double const* const* parameters, double* residuals, double** jacobians) const
{
residuals[0] = parameters[0][0] * parameters[0][0]
if (jacobians)
jacobians[0][0] = 2.0 * parameters[0][0]
return true;
}
};
我的错误是认为我只需要重新实现()运算符,ceres就会自动找到jacobian。事实并非如此。