Finite-budget residual-merit damped Newton

Let $U\subseteq {\mathbb{R}}^n$ be open, $n\ge 1$, and let $F:U\to {\mathbb{R}}^n$ be $C^1$ (Continuously differentiable maps, local inverses, and local diffeomorphisms) with Jacobian $J_F$ as in The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case. The matrix of $DF(x)$ in the standard basis is $J_F(x)$ by A total derivative computes every directional derivative, and its matrix is the Jacobian. Define the residual merit $\mathrm{\Phi }(x):=\frac{1}{2}\mathrm{\parallel }F(x){\mathrm{\parallel }}_2^2$. Inputs: $x_0\in U$, residual tolerance $\epsilon \ge 0$, update budget $N\in {\mathbb{N}}_0$, sufficient-decrease parameter $\alpha \in (0,1)$, reduction factor $\beta \in (0,1)$, and finite backtrack budget $B\in {\mathbb{N}}_0$. Output: the last valid iterate and one of root, singular-jacobian, invalid-trial, line-search-exhausted, or iteration-budget-exhausted.

At iteration $k=0,1,\dots ,N$, apply these instructions in order:

1. Evaluate $F(x_k)$. If $\mathrm{\parallel }F(x_k){\mathrm{\parallel }}_2\le \epsilon$, return $(x_k,\mathtt{\text{root}})$. This status records the residual test only.
2. If $k=N$, return $(x_k,\mathtt{\text{iteration-budget-exhausted}})$.
3. Call Newton correction for a square vector residual with $A=J_F(x_k)$ and $r=F(x_k)$. If it returns singular-jacobian, return $(x_k,\mathtt{\text{singular-jacobian}})$. Otherwise let $s_k$ be its correction.
4. For $j=0,1,\dots ,B$, set $t={\beta }^j$ and trial $y=x_k+t{s}_k$. If $y\mathrm{\notin }U$, reject the trial and continue. If $y\in U$, evaluate $\mathrm{\Phi }(y)$ and accept the first trial satisfying $\mathrm{\Phi }(y)\le \mathrm{\Phi }(x_k)-\alpha t\mathrm{\parallel }F(x_k){\mathrm{\parallel }}_2^2\mathrm{.}$ On acceptance set $x_{k+1}=y$ and continue at iteration $k+1$.
5. If no trial was accepted, return $(x_k,\mathtt{\text{invalid-trial}})$ if every tested point was outside $U$; otherwise return $(x_k,\mathtt{\text{line-search-exhausted}})$.

The residual-merit test is defined directly for a vector root problem. A finite backtrack budget can return failure; no global convergence or quadratic-rate claim is made.