How can I efficiently check whether one unit arrangement can become another in a single turn?

Simply comparing each region with its neighbors isn’t enough, because several destinations may all depend on the same nearby supply. For example, two regions might each appear able to receive units, while the only available units can actually satisfy just one of them. The flow formulation catches that shared-supply conflict. If the map is small, you can instead use backtracking or integer programming, but max flow is the clean general solution.

There’s also a useful necessary-and-sufficient way to think about it: for every set of destination regions, the starting units that can reach that set must be at least as numerous as the units required there. Checking all such sets directly is expensive, but a max-flow algorithm performs those checks efficiently. Build the graph with source → starting regions, allowed movement edges, and destination regions → sink; capacities on the outer edges are the starting and ending unit counts. Run max flow and compare it with the total number of units.

Treat it as a transportation or max-flow problem. Make one node for every region in the starting state and another node for every region in the target state. Connect a starting-region node to a target-region node when a unit can travel between them within the allowed number of spaces: distance at most 1 for a one-space move, or at most 2 for a two-space move. Give each starting node supply equal to its unit count and each target node demand equal to its final count. A feasible flow that satisfies every demand means the move is legal; if no such flow exists, some units would have to travel too far or appear from nowhere. This automatically handles units staying in place and avoids trying to match individual units by hand.