f(x) = 0.5 * (x0^2 + x1^2), gradient = x.

g(x) = x0 - target, gradient = [1, 0]. Parameterises a simple linear equality constraint `x0 == target`.

---- 1-D scalar function (for penalty-helper tests) ----------------------- `f(x) = a + b * x[0]` with fixed a, b so tests can drive the constraint va

LinearFunction: supports first-order information.

c(x) = 0 for all x. Trivially-feasible equality. Used to pin feasible- start and penalty-schedule tests. We do not multiply by zero inside an expre

Circle: c(x) = x[0]^2 + x[1]^2 This function will be used to form both an equality constraint (forcing the solution to lie on the circle) and an inequ

Second-order 2-D quadratic `f(x) = 0.5 * x^T A x` with `A = diag(4, 2)`. Gradient = A x; Hessian = A. Used only by the mode-downgrade test: we want

EqualityConstraint2: g(x) = x[0] - 0.5 = 0 This forces x[0] to be exactly 0.5. Thus, the optimal solution must have x[0] = 0.5.

Triangle edge g2(x) = 6 - x0 - sqrt(3) * x1 >= 0.

Triangle edge g1(x) = x0 + sqrt(3) * x1 >= 0. Redundant inside the first quadrant but part of the original problem.

Triangle edge g0(x) = x0 / sqrt(3) - x1 >= 0.

---- D.1: Cubic-in-one-axis objective with triangular constraints ------- f(x) = ((x0 - 3)^2 - 9) * x1^3 / (27 * sqrt(3)). Unconstrained, f is unbou

Inequality c(x) = 48 - x0^2 - 2 x1^2 >= 0.

InequalityConstraint3: h(x) = 2 - (x[0] + x[1]) >= 0 This requires x[0] + x[1] <= 2. With x[0]=0.5 from the equality constraint, we have x[1] <= 1.5.

---- D.2: HS029-style product objective with quadratic constraint ------ Hock-Schittkowski 29 (simplified to 2D, f = -x0 * x1 on the ellipse 48 - x0^

f(x) = (x0 - 1)^2 + (x1 - 2)^2, gradient = 2 * (x - [1,2]).

f(x) = 0.5 * ((x0 - 2)^2 + x1^2). Unconstrained optimum (2, 0). Used in the inequality-active KKT test.

QuadraticFunction: supports second-order information.

QuadraticObjective2: f(x) = (x[0]-1)^2 + (x[1]-2)^2 The unconstrained optimum is (1,2) with f(x)=0. However, the constraints (below) force a differen

SumObjective: f(x) = x[0] + x[1] (to be minimized)

Inequality 2 - (x0 + x1) >= 0 (i.e. x0 + x1 <= 2). Gradient = [-1, -1].

Equality constraint `c(alpha) = sum_i alpha_i y_i = 0`. Gradient is the constant label vector `y`.

Dual SVM objective -- same as in `svm_dual_lbfgsb.cc` but restated here so the example stands alone. Precomputes `Q = (y y^T) o (X X^T)`.

Margin constraint for sample `i`: c_i(w, b, xi) = y_i * (w . x_i + b) - 1 + xi_i >= 0. Gradient: dc/dw = y_i * x_i, dc/db = y_i, dc/dxi_j = delta_{i

Primal objective `f(w, b, xi) = 0.5 ||w||^2 + C * sum(xi)`.

Slack non-negativity constraint for sample `i`: c_i(w, b, xi) = xi_i >= 0. Gradient: delta on the `xi_i` entry.

h(x) = bound - x0 >= 0 (i.e. x0 <= bound). Gradient = [-1, 0].

ConstantFunction: supports no derivative information.

`f(x) = x^3 - 3x + 2`, with `f'(0) = -3`. Local minimum at `x = 1` where `f(1) = 0` and `f'(1) = 0`.

----------------------------------------------------------------- Differentiability enum.

`f(x) = 1e-8 * x + x^4` -- nearly flat near 0, a tiny negative slope `-1e-8` at the origin, quartic growth away from 0. The HZ curvature condition `|

f(x,y) = 0.5 * (x^2 - y^2). Indefinite quadratic: the Hessian has eigenvalues (1, -1), so Newton would take an uphill step in y. We use this to test

Dataset bundle: feature matrix (n x d) and label vector (n). Labels are strict {-1, +1} doubles so the SVM margin `y * (w^T x + b)` is a signed scalar

CRTP scalar function: `f(x) = a*x^2 + b*x + c` with `x` a 1-vector. The quadratic case 1 sets `a=1, b=-2, c=0` giving the classical `phi(a) = a^2 - 2a

f(x) = (x^2 - 2)^2 on R^1 -- zero Hessian at x=0, so a pure Newton step is undefined there. TR-Newton must shrink the radius in response to poor mode

Rosenbrock: f(x,y) = 100*(y - x^2)^2 + (1 - x)^2. Unique minimiser at (1, 1), f* = 0. Non-convex on a large portion of R^2 (Hessian becomes indefini

The Rosenbrock objective, second-order. We hand-code the Hessian because TR-Newton requires `DifferentiabilityMode::Second`. The 400-per-coordinate

A minimal first-order function instance just to satisfy the template parameter of `MoreThuente<FunctionType, Ord>`. `cstep` itself is purely scalar a

Status of the solver state.

f(x,y) = 3 x^2 + 10 y^2, strictly convex. Unique minimiser at (0,0), f* = 0. Hessian is constant and diagonal (6, 20), condition number 10/3.