The square-root Kalman procedure terminates with a finite status and operation-count bound

Every matrix or vector product in [F1] is evaluated by finitely many left-to-right dot products because all dimensions are finite. Each scalar operation, including conversion of a supplied entry at its read, either returns a finite value in the declared arithmetic model or the first range-failure status. Concatenation, zero padding, and block partitioning use finitely many entries.

Each QR call processes a finite number of columns and, within a column, a finite number of active entries. By [F2], each reflector step terminates with an identity, reflector, or range-failure result; hence a QR call either returns its reduced factor or a finite failure status. The lower-triangular innovation solve has finitely many forward-substitution row updates by [F3]; its diagonal is nonzero because [F1] returns failure before the solve otherwise.

A reduced Householder QR call on an $M\times N$ array ($M\ge N\ge 1$) has $N$ passes. Pass $j$ forms a reflector using $O(M-j+1)$ scalar operations and applies it to at most $N-j$ trailing columns using $O((M-j+1)(N-j))$ operations; scans and status checks have no larger order. Summing over $j$ gives $O(M{N}^2)$. Hence the prediction and measurement QR calls at step $k$ cost $O(q_k{d}_k^2)$ and $O(({\bar{r}}_k+d_k)(p_k+d_k{)}^2)$, respectively. A failure can only shorten these loops.

The outer loop has at most $T$ steps, and for $T=0$ the algorithm returns its initial pair immediately. By steps 1.1 and 1.2, every entered step terminates; therefore the whole procedure returns after finitely many operations with a declared status. No success or accuracy guarantee is asserted by this termination result.

The products $F_k{\widehat{m}}_{k-1\mathrm{\mid }k-1}$ and $F_k{L}_{k-1\mathrm{\mid }k-1}$ cost $O(d_k{d}_{k-1})$ and $O(d_k{d}_{k-1}{\rho }_{k-1})$. The product $H_k{L}_{k\mathrm{\mid }k-1}$ costs $O(p_k{d}_k^2)$; the innovation product, triangular solve, factor sign changes, and all block copying/padding cost at most $O(({\bar{r}}_k+d_k)(p_k+d_k{)}^2+q_k{d}_k^2)$ because $p_k,d_k\ge 1$, $q_k\ge r_{Q,k}$, and ${\bar{r}}_k\ge r_{R,k}$. Input conversions add at most one scalar rounding per supplied operand read or copied entry, so they do not change these orders. Together with step 1.3, summing over $k$ gives the stated count. For $T=0$ there is only the constant-time return, with no entry conversions.