John McCardle
767d0d4b0f
Phase A (Python surface): - New mcrfpy.Heuristic IntEnum: EUCLIDEAN, MANHATTAN, CHEBYSHEV, DIAGONAL, ZERO - Grid.find_path() accepts heuristic= and weight= kwargs (weighted A*) - Grid.get_dijkstra_map() accepts roots= (list of positions or DiscreteMap mask) Phase B (FLEE primitives): - DijkstraMap.invert() returns a new map with inverted distance field - DijkstraMap.descent_step(pos) returns steepest-descent neighbor or None DijkstraMap internally switched from the C++ TCODDijkstra wrapper to the C API (TCOD_dijkstra_*) because multi-root compute and invert/get_descent are not exposed on the wrapper. Single-root Dijkstra cache is preserved for backward compatibility; multi-root and mask paths bypass the cache since cache keys would be ill-defined. New tests: heuristic_enum_test, find_path_heuristic_test, multi_root_dijkstra_test, dijkstra_flee_test. Baseline JSONs for dijkstra_bench and gridview_render_bench refreshed against the new implementation. Co-Authored-By: Claude Opus 4.7 (1M context) <noreply@anthropic.com>
61 lines
2.1 KiB
Python
61 lines
2.1 KiB
Python
"""Grid.find_path heuristic/weight kwargs produce valid paths across each built-in."""
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import mcrfpy
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import sys
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def make_open_grid(w, h):
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g = mcrfpy.Grid(grid_size=(w, h))
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for y in range(h):
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for x in range(w):
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c = g.at(x, y)
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c.walkable = True
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c.transparent = True
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return g
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def main():
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g = make_open_grid(30, 30)
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# On an obstacle-free grid every admissible heuristic yields an optimal-length path.
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# libtcod returns steps (excluding origin), so a diagonal-permitting move from (0,0)
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# to (20,20) is 20 steps.
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for h in (mcrfpy.Heuristic.EUCLIDEAN,
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mcrfpy.Heuristic.MANHATTAN,
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mcrfpy.Heuristic.CHEBYSHEV,
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mcrfpy.Heuristic.DIAGONAL,
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mcrfpy.Heuristic.ZERO):
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p = g.find_path((0, 0), (20, 20), heuristic=h)
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assert p is not None, f"no path for {h}"
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steps = list(p)
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assert len(steps) == 20, f"heuristic {h} gave {len(steps)} steps, expected 20"
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# Last step must be the goal.
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assert (int(steps[-1].x), int(steps[-1].y)) == (20, 20), \
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f"heuristic {h} did not end at goal"
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# Weighted A* with weight>=1 must still find a path (not necessarily optimal).
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for w in (1.0, 1.5, 2.0):
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p = g.find_path((0, 0), (20, 20), heuristic=mcrfpy.Heuristic.EUCLIDEAN, weight=w)
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assert p is not None, f"no path for weight={w}"
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steps = list(p)
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assert len(steps) >= 20, f"weight={w} gave impossibly short path"
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# With an obstacle, the path still reaches the goal.
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g2 = make_open_grid(30, 30)
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for y in range(5, 25):
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g2.at(15, y).walkable = False
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p = g2.find_path((0, 0), (29, 0), heuristic=mcrfpy.Heuristic.MANHATTAN)
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assert p is not None
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steps = list(p)
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assert (int(steps[-1].x), int(steps[-1].y)) == (29, 0)
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# No step may land on a blocked cell.
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for s in steps:
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assert not (int(s.x) == 15 and 5 <= int(s.y) < 25), \
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f"path stepped through wall at {s}"
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print("PASS")
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if __name__ == "__main__":
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main()
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sys.exit(0)
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