> For the complete documentation index, see [llms.txt](https://ceres-solver-tutorial-cn.gitbook.io/ceres/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://ceres-solver-tutorial-cn.gitbook.io/ceres/optimization-tutorial/non-linear-least-squares/powells-function.md). # Powell’s Function 考虑一个略微复杂点的例子，最小化 Powell's 函数。 其中 $$x=\left\[x\_{1}, x\_{2}, x\_{3}, x\_{4}\right]$$ 残差项为： $$ \begin{aligned} f\_{1}(x) & =x\_{1}+10 x\_{2} \ f\_{2}(x) & =\sqrt{5}\left(x\_{3}-x\_{4}\right) \ f\_{3}(x) & =\left(x\_{2}-2 x\_{3}\right)^{2} \ f\_{4}(x) & =\sqrt{10}\left(x\_{1}-x\_{4}\right)^{2} \ F(x) & =\left\[f\_{1}(x), f\_{2}(x), f\_{3}(x), f\_{4}(x)\right] \end{aligned} $$ 则在该例子中 $$f\_i(x)$$ 就是 ResidualBlock，$$x$$ 就是整个参数块，例如 $$f\_1(x)$$ 对应的参数块为 $$\[x\_1,x\_2]$$。Cost 的计算方式为 $$x\_1+10x\_2$$。$$F(x)$$ 拥有四个参数，拥有四个残差项，我们想要最小化 $$\frac{1}{2}||F(x)||^2$$。接下来我们使用 Ceres 来解决该问题。 同样，第一步是定义评估目标函数中每个残差项的 functor。评估 $$f\_4(x\_1,x\_4)$$ 的 functor 如下： ```cpp struct F4 { template bool operator()(const T* const x1, const T* const x4, T* residual) const { residual[0] = sqrt(10.0) * (x1[0] - x4[0]) * (x1[0] - x4[0]); return true; } }; ``` 类似的我们可以定义 F1，F2，F3 去评估 $$f\_1(x\_1,x\_2), f\_2(x\_3,x\_4), f\_3(x\_3,x\_2)$$。最终构建优化问题的方式如下： ```cpp double x1 = 3.0; double x2 = -1.0; double x3 = 0.0; double x4 = 1.0; Problem problem; // Add residual terms to the problem using the autodiff // wrapper to get the derivatives automatically. problem.AddResidualBlock( new AutoDiffCostFunction(), nullptr, &x1, &x2); problem.AddResidualBlock( new AutoDiffCostFunction(), nullptr, &x3, &x4); problem.AddResidualBlock( new AutoDiffCostFunction(), nullptr, &x2, &x3); problem.AddResidualBlock( new AutoDiffCostFunction(), nullptr, &x1, &x4); ``` AutoDiffCostFunction\() 的模板参数分别代表 Cost 的计算方式，残差的维度，第一个参数的维度，第二个参数的维度。这里需要注意的是，每个 ResidualBlock 仅依赖于相应残差函数所依赖的两个参数，而不依赖于所有四个参数，例如 $$f\_1(x\_1,x\_2)$$ 仅仅依赖于 $$x\_1,x\_2$$。编译并运行 [examples/powell.cc](https://ceres-solver.googlesource.com/ceres-solver/+/master/examples/powell.cc) 可以得到如下结果： ```shell Initial x1 = 3, x2 = -1, x3 = 0, x4 = 1 iter cost cost_change |gradient| |step| tr_ratio tr_radius ls_iter iter_time total_time 0 1.075000e+02 0.00e+00 1.55e+02 0.00e+00 0.00e+00 1.00e+04 0 2.91e-05 3.40e-04 1 5.036190e+00 1.02e+02 2.00e+01 0.00e+00 9.53e-01 3.00e+04 1 4.98e-05 3.99e-04 2 3.148168e-01 4.72e+00 2.50e+00 6.23e-01 9.37e-01 9.00e+04 1 2.15e-06 4.06e-04 3 1.967760e-02 2.95e-01 3.13e-01 3.08e-01 9.37e-01 2.70e+05 1 9.54e-07 4.10e-04 4 1.229900e-03 1.84e-02 3.91e-02 1.54e-01 9.37e-01 8.10e+05 1 1.91e-06 4.14e-04 5 7.687123e-05 1.15e-03 4.89e-03 7.69e-02 9.37e-01 2.43e+06 1 1.91e-06 4.18e-04 6 4.804625e-06 7.21e-05 6.11e-04 3.85e-02 9.37e-01 7.29e+06 1 1.19e-06 4.21e-04 7 3.003028e-07 4.50e-06 7.64e-05 1.92e-02 9.37e-01 2.19e+07 1 1.91e-06 4.25e-04 8 1.877006e-08 2.82e-07 9.54e-06 9.62e-03 9.37e-01 6.56e+07 1 9.54e-07 4.28e-04 9 1.173223e-09 1.76e-08 1.19e-06 4.81e-03 9.37e-01 1.97e+08 1 9.54e-07 4.32e-04 10 7.333425e-11 1.10e-09 1.49e-07 2.40e-03 9.37e-01 5.90e+08 1 9.54e-07 4.35e-04 11 4.584044e-12 6.88e-11 1.86e-08 1.20e-03 9.37e-01 1.77e+09 1 9.54e-07 4.38e-04 12 2.865573e-13 4.30e-12 2.33e-09 6.02e-04 9.37e-01 5.31e+09 1 2.15e-06 4.42e-04 13 1.791438e-14 2.69e-13 2.91e-10 3.01e-04 9.37e-01 1.59e+10 1 1.91e-06 4.45e-04 14 1.120029e-15 1.68e-14 3.64e-11 1.51e-04 9.37e-01 4.78e+10 1 2.15e-06 4.48e-04 Solver Summary (v 2.2.0-eigen-(3.4.0)-lapack-suitesparse-(7.1.0)-metis-(5.1.0)-acceleratesparse-eigensparse) Original Reduced Parameter blocks 4 4 Parameters 4 4 Residual blocks 4 4 Residuals 4 4 Minimizer TRUST_REGION Dense linear algebra library EIGEN Trust region strategy LEVENBERG_MARQUARDT Given Used Linear solver DENSE_QR DENSE_QR Threads 1 1 Linear solver ordering AUTOMATIC 4 Cost: Initial 1.075000e+02 Final 1.120029e-15 Change 1.075000e+02 Minimizer iterations 15 Successful steps 15 Unsuccessful steps 0 Time (in seconds): Preprocessor 0.000311 Residual only evaluation 0.000002 (14) Jacobian & residual evaluation 0.000023 (15) Linear solver 0.000043 (14) Minimizer 0.000163 Postprocessor 0.000012 Total 0.000486 Termination: CONVERGENCE (Gradient tolerance reached. Gradient max norm: 3.642190e-11 <= 1.000000e-10) Final x1 = 0.000146222, x2 = -1.46222e-05, x3 = 2.40957e-05, x4 = 2.40957e-05 ``` 不难看出，当目标函数值为 0 时，最优解位于 $$x\_{1}=0, x\_{2}=0, x\_{3}=0, x\_{4}=0$$ 处。 经过 15 次迭代，Ceres 得到的 cost 为 $$1.120029e-15$$。 ## Footnotes * [examples/powell.cc](https://ceres-solver.googlesource.com/ceres-solver/+/master/examples/powell.cc).