Finite-budget safeguarded Newton–bisection for Euler star pressure

Assume exact-real arithmetic in which each arithmetic operation, comparison, square root, real power on a positive argument, scalar residual evaluation, and call to the one-step Newton procedure terminates. Inputs: admissible primitive states $({\rho }_L,u_L,p_L)$ and $({\rho }_R,u_R,p_R)$, gas constant $\gamma >1$, pressure-width tolerance $\tau >0$, and iteration budget $N\in {\mathbb{N}}_0$. Outputs: inadmissible-input, vacuum, pressure-bracket with an interval, or budget-exhausted with an interval. The procedure never treats a nonzero residual tolerance as an exact root.

1. If ${\rho }_L,{\rho }_R,p_L,p_R$ are not all positive or $\gamma \le 1$, return inadmissible-input before evaluating a wave response.
2. Compute $\mathrm{\Phi }(0+)$ from The ideal-gas star-pressure root exists uniquely exactly below vacuum threshold. If $\mathrm{\Phi }(0+)\ge 0$, return vacuum.
3. Otherwise set the bracket $[a,b]=[0,P]$ with $P$ from that theorem. Its endpoint convention is $\mathrm{\Phi }(a):=\mathrm{\Phi }(0+)<0<\mathrm{\Phi }(b)$.
4. For at most $N$ iterations, in this order:
   1. If $b-a\le \tau$, return pressure-bracket with $[a,b]$.
   2. Let $m=(a+b)\mathrm{/}2$. Call Finite-budget Newton for a square vector residual on the scalar residual $F(p)=\mathrm{\Phi }(p)$ over the open domain $(0,\mathrm{\infty })$, starting at $m$, with residual tolerance $0$ and update budget $1$.
   3. If Newton returns $(r,\mathtt{\text{root}})$, return pressure-bracket with $[r,r]$. If it returns iteration-budget-exhausted with last iterate $q$ and $a+(b-a)\mathrm{/}4\le q\le b-(b-a)\mathrm{/}4$, use $q$ as the query. In every other status or when this central-quarter test fails, use $m$ as the query.
   4. Evaluate $\mathrm{\Phi }$ at the query. If it is zero, return pressure-bracket with the singleton interval. If negative, replace $a$ by the query; if positive, replace $b$ by the query.
5. After the budget, return pressure-bracket if the width is at most $\tau$; otherwise return budget-exhausted with the current bracket.

The returned interval uses the zero lower endpoint as a limiting endpoint, not an admissible pressure. This algorithm gives a finite status and bracket; the separate theorem below proves its invariant and width reduction. It makes no claim that an exhausted budget found a root.