 Solution

Interpret the source's $m$ as the maximum finite shortest-path distance among reachable pairs, taking $m=0$ when only length-zero graph paths are needed. This is the length bound required for reflexive transitive closure. A directed graph cycle permits arbitrarily long walks, so the maximum length of all possible walks would not be a finite termination parameter.^[ref]

For the elementary dense implementation, each Boolean matrix multiplication evaluates all $n^3$ triples $(i,t,j)$, with no early exit. Each equality test costs $O(n^2)$. Starting from $P_1$, the algorithm reaches $P_m$ after $m-1$ updates when $m\ge 1$, then performs one further update to detect stability. Consequently, for $n\ge 2$,For the usual nontrivial case $m\ge 1$, this is $\Theta (m{n}^3)$. Since $m\le n-1$, the overall worst case is $\Theta (n^4)$. This bound is tight for this implementation: the directed chain $0\to 1\to \cdots \to n-1$ has $m=n-1$ and gains previously unknown pairs at every length until then, while every update scans all triples. This tightness does not claim an inherent lower bound for the reflexive transitive closure problem; it concerns the specified repeated-extension algorithm.