Finite-budget Gauss–Seidel stationary iteration

Input $n\ge 1$, $A\in M_n(\mathbb{C})$, $b,x^{(0)}\in {\mathbb{C}}^n$, and an iteration budget $m\in {\mathbb{N}}_0$. Use the fixed order $1,\dots ,n$. Check dimensions; on a mismatch return invalid-input. Form $D,L,U$ by Diagonal, lower, and upper parts in the stationary-iteration convention. If some diagonal entry of $A$ is zero, return pivot-failure(i) for the least such index $i$.

Otherwise set $M=D-L$ and $N=U$, so $A=M-N$. For each $k=0,\dots ,m-1$, solve the lower-triangular system

$(D-L)x^{(k+1)}=U{x}^{(k)}+b$

by forward substitution in increasing index order. Equivalently, the component updates are

$x_i^{(k+1)}=\frac{b_i+{\sum }_{j<i}{L}_{ij}{x}_j^{(k+1)}+{\sum }_{j>i}{U}_{ij}{x}_j^{(k)}}{a_{ii}},i=1,\dots ,n\mathrm{.}$

Return completed with $x^{(m)}$. The lower-triangular solve is unique after the pivot check by Forward and backward substitution for triangular linear systems and Forward and backward substitution are correct, unique, and quadratic in scalar operations. After the dimension and pivot checks succeed, a zero budget returns completed with $x^{(0)}$. The procedure reports no convergence or accuracy certificate.