1. Use analytical/automatic derivatives.

    This is the single most important piece of advice we can give to you. It is tempting to take the easy way out and use numeric differentiation. This is a bad idea. Numeric differentiation is slow, ill-behaved, hard to get right, and results in poor convergence behaviour.

    Ceres allows the user to define templated functors which will be automatically differentiated. For most situations this is enough and we recommend using this facility. In some cases the derivatives are simple enough or the performance considerations are such that the overhead of automatic differentiation is too much. In such cases, analytic derivatives are recommended.

    The use of numerical derivatives should be a measure of last resort, where it is simply not possible to write a templated implementation of the cost function.

    In many cases it is not possible to do analytic or automatic differentiation of the entire cost function, but it is generally the case that it is possible to decompose the cost function into parts that need to be numerically differentiated and parts that can be automatically or analytically differentiated.

    To this end, Ceres has extensive support for mixing analytic, automatic and numeric differentiation. See CostFunctionToFunctor.

  2. When using Quaternions, consider using QuaternionParameterization.

    Quaternions are a four dimensional parameterization of the space of three dimensional rotations \(SO(3)\). However, the \(SO(3)\) is a three dimensional set, and so is the tangent space of a Quaternion. Therefore, it is sometimes (not always) benefecial to associate a local parameterization with parameter blocks representing a Quaternion. Assuming that the order of entries in your parameter block is \(w,x,y,z\), you can use QuaternionParameterization.


    If you are using Eigen’s Quaternion object, whose layout is \(x,y,z,w\), then you should use EigenQuaternionParameterization.

  3. How do I solve problems with general linear & non-linear inequality constraints with Ceres Solver?

    Currently, Ceres Solver only supports upper and lower bounds constraints on the parameter blocks.

    A crude way of dealing with inequality constraints is have one or more of your cost functions check if the inequalities you are interested in are satisfied, and if not return false instead of true. This will prevent the solver from ever stepping into an infeasible region.

    This requires that the starting point for the optimization be a feasible point. You also risk pre-mature convergence using this method.

  4. How do I solve problems with general linear & non-linear equality constraints with Ceres Solver?

    There is no built in support in ceres for solving problems with equality constraints. Currently, Ceres Solver only supports upper and lower bounds constraints on the parameter blocks.

    The trick described above for dealing with inequality constraints will not work for equality constraints.

  5. How do I set one or more components of a parameter block constant?

    Using SubsetParameterization.

  6. Putting Inverse Function Theorem to use.

    Every now and then we have to deal with functions which cannot be evaluated analytically. Computing the Jacobian in such cases is tricky. A particularly interesting case is where the inverse of the function is easy to compute analytically. An example of such a function is the Coordinate transformation between the ECEF and the WGS84 where the conversion from WGS84 to ECEF is analytic, but the conversion back to WGS84 uses an iterative algorithm. So how do you compute the derivative of the ECEF to WGS84 transformation?

    One obvious approach would be to numerically differentiate the conversion function. This is not a good idea. For one, it will be slow, but it will also be numerically quite bad.

    Turns out you can use the Inverse Function Theorem in this case to compute the derivatives more or less analytically.

    The key result here is. If \(x = f^{-1}(y)\), and \(Df(x)\) is the invertible Jacobian of \(f\) at \(x\). Then the Jacobian \(Df^{-1}(y) = [Df(x)]^{-1}\), i.e., the Jacobian of the \(f^{-1}\) is the inverse of the Jacobian of \(f\).

    Algorithmically this means that given \(y\), compute \(x = f^{-1}(y)\) by whatever means you can. Evaluate the Jacobian of \(f\) at \(x\). If the Jacobian matrix is invertible, then its inverse is the Jacobian of \(f^{-1}(y)\) at \(y\).

    One can put this into practice with the following code fragment.

    Eigen::Vector3d ecef; // Fill some values
    // Iterative computation.
    Eigen::Vector3d lla = ECEFToLLA(ecef);
    // Analytic derivatives
    Eigen::Matrix3d lla_to_ecef_jacobian = LLAToECEFJacobian(lla);
    bool invertible;
    Eigen::Matrix3d ecef_to_lla_jacobian;
    lla_to_ecef_jacobian.computeInverseWithCheck(ecef_to_lla_jacobian, invertible);