Let be their Manhattan distance. Each joint action can reduce by at most two, so use the simultaneous rendezvous distance
This heuristic function is nonzero whenever the robots are apart. For a transition , the triangle inequality gives
Applying the ceiling function yields , and . Thus is a consistent heuristic, hence an admissible heuristic.
In the unobstructed grid this estimate is exact. Along a shortest route of length , both robots approach a middle room, taking at most steps each; when is an odd integer, the robot with the shorter route waits once. That final wait prevents a crossing from being mistaken for a meeting. With obstacles the same formula remains a relaxation heuristic, though it need not be exact.
Figure 1.
An original nine-edge route needs five simultaneous steps: the last wait produces a meeting rather than a crossing
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