fix(rules): a multi-cell creature moving under mixed terrain pays the difficult rate for its whole footprint - SRD Movement, Position and Distance: 'If you occupy squares with different kinds of terrain, you can move only as fast as the most difficult terrain you occupy will allow'; the charge/run/5-foot-step already gated on the whole footprint (_footprint_touches_difficult) but the move/withdraw DESCENT priced every step by the anchor cell alone (grid.step_cost), so a 2x2 creature straddling rubble crossed at full speed; the descent now prices a step by the most difficult square under the footprint via _footprint_step_cost -> a new grid.step_cost_with_terrain (the 5-10-5 + elevation formula stays centralized in grid.py, step_cost delegates to it); 1x1 creatures are bit-identical (footprint == anchor cell); the direction Dijkstra field remains anchor-priced (making it creature-aware rewrites grid.reachable wholesale - the documented deferral); golden untouched (open-floor arena, no difficult terrain); +7 tests (trailing-square orthogonal/diagonal cost, 1x1 bit-identical regression, full-lane move budget 3-vs-6 squares, withdraw hampering)

src/pf1e_simulator/grid.py

@@ -108,14 +108,29 @@ class Grid:
         if cost is None:
             msg = f"cannot step into impassable square {to}"
             raise ValueError(msg)
+        return self.step_cost_with_terrain(frm, to, diagonals_used, cost)
+
+    def step_cost_with_terrain(
+        self, frm: Pos, to: Pos, diagonals_used: int, cost: int
+    ) -> int:
+        """Cost of one step given an explicit destination terrain ``cost``.
+
+        The same 5-10-5 + elevation rule as ``step_cost`` (the single home of
+        that formula), but the caller supplies the terrain cost instead of
+        reading it from the anchor cell. This lets a multi-cell creature price
+        a step by the MOST DIFFICULT square its whole footprint would occupy
+        (SRD "Movement, Position, and Distance": "If you occupy squares with
+        different kinds of terrain, you can move only as fast as the most
+        difficult terrain you occupy will allow") rather than by the anchor
+        square alone.
+        """
         is_diagonal = frm[0] != to[0] and frm[1] != to[1]
+        climb = max(0, self.elevation_at(to) - self.elevation_at(frm))
         if not is_diagonal:
-            return cost + max(0, self.elevation_at(to) - self.elevation_at(frm))
+            return cost + climb
         if cost >= _DIFFICULT_TERRAIN_COST:
-            return 3 + max(0, self.elevation_at(to) - self.elevation_at(frm))
-        return 1 + (diagonals_used % 2) + max(
-            0, self.elevation_at(to) - self.elevation_at(frm)
-        )
+            return 3 + climb
+        return 1 + (diagonals_used % 2) + climb
 
     def diagonal_allowed(self, frm: Pos, to: Pos) -> bool:
         flank_a = (frm[0], to[1])