Is it possible to formulate a nested cost function?

[zhexu...@outlook.com]

For the convinience of modelling, I may want to creat a nested cost function. Take rosenbrock cost function for example:  f(x,y) = (1-x)^2 + 100(y - x^2)^2.

Instead of directed calculate residue like rosenbrock.cc, can i have:

function 1: f1 = x^2

function 2: f2 = y - f1

function 3: f3 = 1-x

cost function f(x, y) = (1-x)^2 + 100(y - f1)^2

I tried to do like this but fail to compile. Error was generated at line that call F1. I use MSVC 2022.

struct F1 {
    template <typename T>
    bool operator()(const T* const x, T* residual) const {
        // f1 = x * x;
        residual[0] = x[0] * x[0];
        return true;
    }
};

struct F2 {
    template <typename T>
    bool operator()(const T* const x, const T* const y, T* residual) const {
        // f2 = y - f1
        T* temp;
        F1(x, temp); //error C2440

        //F1<T>(x, temp); //error C2760
        residual = y[0] - temp;
        return true;
    }
};

Can someone help me? Thanks!

[Sameer Agarwal]

This works.

struct F1 {
    template <typename T>
    bool operator()(const T* const x, T* residual) const {
        // f1 = x * x;
        residual[0] = x[0] * x[0];
        return true;
    }
};

struct F2 {
    template <typename T>
    bool operator()(const T* const x, const T* const y, T* residual) const {
        // f2 = y - f1

        T temp;
        F1()(x, &temp);
        residual[0] = y[0] - temp;
        return true;
    }
};

// f(x,y) = (1-x)^2 + 100(y - x^2)^2;
struct Rosenbrock {
  template <typename T>
  bool operator()(const T* parameters, T* cost) const {
    const T x = parameters[0];
    const T y = parameters[1];
    T temp;
    cost[0] = (1.0 - x) * (1.0 - x) + 100.0 * F2()(&x,&y, &temp);
    return true;
  }
};

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[zhexu...@outlook.com]

Thank you for your reply! The code you provided was able to compile, but failed to get the right answer (1.0, 1.0).  Optimization of rosenbrock.cc in example folder was suceessful. I tried to figure out why and modified the line cost[0] = (1.0 - x) * (1.0 - x) + 100.0 * F2()(&x,&y, &temp); because the F2() function seems to return 1 or 0 after conversion. But my code still cannot get (1.0, 1.0), It went to (-1.1056, 1.22376). Followed is my full code. Thanks for your kindly help again!

#include <vector>

#include "ceres/ceres.h"
#include "gflags/gflags.h"
#include "glog/logging.h"

using ceres::AutoDiffCostFunction;
using ceres::CostFunction;
using ceres::Problem;
using ceres::Solve;
using ceres::Solver;

struct F1 {
    template <typename T>
    bool operator()(const T* const x, T* residual) const {
        // f1 = x * x;
        residual[0] = x[0] * x[0];
        return true;
    }
};

struct F2 {
    template <typename T>
    bool operator()(const T* const x, const T* const y, T* residual) const {
        // f2 = y - f1
        T temp;
        F1()(x, &temp);
        residual[0] = y[0] - temp;
        return true;
    }
};

// f(x,y) = (1-x)^2 + 100(y - x^2)^2;
struct Rosenbrock {
    template <typename T>
    bool operator()(const T* parameters, T* cost) const {
        const T x = parameters[0];
        const T y = parameters[1];
        T temp;

        //cost[0] = (1.0 - x) * (1.0 - x) + 100.0 * F2()(&x, &y, &temp);
        F2()(&x, &y, &temp);
        cost[0] = (1.0 - x) * (1.0 - x) + 100.0 * temp * temp;
        return true;
    }
};

int main(int argc, char** argv) {
    GFLAGS_NAMESPACE::ParseCommandLineFlags(&argc, &argv, true);
    google::InitGoogleLogging(argv[0]);

    double x[2] = { -1.2, 1 };

    Problem problem;

    problem.AddResidualBlock(
        new AutoDiffCostFunction<Rosenbrock, 1, 2>(new Rosenbrock), nullptr, x);

    Solver::Options options;

    std::cout << "Initial x = " << x[0]
        << ", y = " << x[1]
        << "\n";

    Solver::Summary summary;
    Solve(options, &problem, &summary);

    std::cout << summary.FullReport() << "\n";

    std::cout << "Final x = " << x[0]
        << ", y = " << x[1]
        << "\n";

    return 0;
}

[zhexu...@outlook.com]

I know what's wrong here... It is solved as a Non-linear Least Squares instead of normal problem, which make it harder to solve. Followed code works. Thanks!

#include <vector>

#include "ceres/ceres.h"
#include "gflags/gflags.h"
#include "glog/logging.h"

using ceres::AutoDiffCostFunction;
using ceres::CostFunction;
using ceres::Problem;
using ceres::Solve;
using ceres::Solver;

struct F1 {
    template <typename T>
    bool operator()(const T* const x, T* residual) const {
        // f1 = x * x;
        residual[0] = x[0] * x[0];
        return true;
    }
};

struct RosenbrockPart2 {

    template <typename T>
    bool operator()(const T* const x, const T* const y, T* residual) const {
        // f2 = y - f1
        T temp;
        F1()(x, &temp);
        residual[0] = y[0] - temp;
        return true;
    }
};

struct RosenbrockPart1 {

    template <typename T>
    bool operator()(const T* const x, T* residual) const {

        residual[0] = 1.0 - x[0];

        return true;
    }
};


int main(int argc, char** argv) {
    GFLAGS_NAMESPACE::ParseCommandLineFlags(&argc, &argv, true);
    google::InitGoogleLogging(argv[0]);

    double x = -1.2;
    double y = 1;

    Problem problem;

    problem.AddResidualBlock(
        new AutoDiffCostFunction<RosenbrockPart1, 1, 1>(new RosenbrockPart1), nullptr, &x);
    problem.AddResidualBlock(
        new AutoDiffCostFunction<RosenbrockPart2, 1, 1, 1>(new RosenbrockPart2), nullptr, &x, &y);

    Solver::Options options;
    options.minimizer_progress_to_stdout = true;
    options.max_num_iterations = 300;

    std::cout << "Initial x = " << x

        << ", y = " << y

        << "\n";

    Solver::Summary summary;
    Solve(options, &problem, &summary);

    std::cout << summary.FullReport() << "\n";

    std::cout << "Final x = " << x

        << ", y = " << y
        << "\n";

    return 0;
}