 Solution

Adapt one machine update into one forward iterator element, then apply the skipping iterator of part (a) with $n=\mathrm{step}-1$ and the bidirectional iterator of part (b). Require a positive step and reset once on construction.After $j$ forward results at the unexplored frontier, the returned value is the machine's value after exactly $j\mathrm{step}$ updates. A partial final block contributes no element, and update is never called after shouldStop becomes true. For the printed Add example, the first three observed states areBecause the state machine interface exposes no copy or rollback operation, determining whether enough updates remain generally requires executing them. Thus hasNext may advance the physical machine and cache the next observation. Backward traversal replays stored observations; it does not undo machine effects. This is a trace iterator, with the ordinary cursor semantics of part (b). Assume exclusive use of the machine while this iterator is active, and that returned values remain valid observations rather than mutable views overwritten by later updates.

Each physical update is executed at most once. Previously observed values can be revisited without rerunning the machine, and repeated hasNext calls do not perform duplicate updates once the next result is prepared.