Discrete maximum-curvature L-curve parameter selection

Input a real matrix $A\in {\mathbb{R}}^{m\times n}$, data $b\in {\mathbb{R}}^m$, and a strictly increasing finite grid $0<{\lambda }_1<\cdots <{\lambda }_q$ with $q\ge 3$. For each grid point, compute the unique Tikhonov minimizer $x_j=(A^{T}A+{\lambda }_{j}I{)}^{-1}{A}^{T}b,$ and the residual $r_j=\mathrm{\parallel }A{x}_j-b\mathrm{\parallel }$ and solution norm $s_j=\mathrm{\parallel }{x}_j\mathrm{\parallel }$. Discard any index with $r_j=0$ or $s_j=0$. If fewer than three indices remain, return failure status insufficient-valid-points.

For consecutive retained indices $i_0<i_1<i_2$, set $p_k=(\log r_{i_k},\log s_{i_k})\in {\mathbb{R}}^2$, $u=p_1-p_0$, and $v=p_2-p_1$. Define their discrete curvature by $\kappa =\{\begin{array}{ll}{\displaystyle \frac{2\mathrm{\mid }\det (u,v)\mathrm{\mid }}{\mathrm{\parallel }u\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }\mathrm{\parallel }u+v\mathrm{\parallel }}},& \mathrm{\parallel }u\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }\mathrm{\parallel }u+v\mathrm{\parallel }>0,\\ 0,& \text{otherwise}\mathrm{.}\end{array}$ Assign this curvature to the middle retained index. Among the resulting $q^{\mathrm{\prime }}-2$ values, return the original grid parameter whose curvature is largest; if there is a tie, return the tied parameter with the smallest original index. No uniform-grid spacing is assumed in log parameter. This is a finite-grid heuristic and carries no universal optimality claim.

The output is either a selected ${\lambda }_j$ with its discrete curvature and index, or the stated failure status. Every step uses a finite loop over the input grid.