Retain weighted predecessors in many-to-many path queries

CHANGELOG.md

@@ -8,6 +8,7 @@ This project adheres to [Semantic Versioning](https://semver.org/) since version
 
 ### Fixed
 
+- Many-to-many `tcod.path.path2d` queries retain weighted predecessors when reconstructing their route.
 - Copying or pickling Perlin and Simplex noise no longer advances a shared random generator while preparing an unused wavelet tile.
 - `tcod.image.Image.rotate90` updates the width and height of rectangular images, keeping NumPy array conversions consistent with the rotated pixels.
 - Adding a root to an already resolved `tcod.path.Pathfinder` resets its predecessor only when its value improves the current distance. This prevents cyclic paths while preserving routes to cheaper roots.

tcod/path.py

@@ -1559,10 +1559,10 @@ def path2d(  # noqa: C901, PLR0912, PLR0913
             out = out[::-1]
         return out
 
-    # Crude Dijkstra implementation until issues with Pathfinder are fixed
+    # Resolve every endpoint before choosing the nearest start.
     pf.resolve(None)
     best_distance, best_ij = min((pf.distance[ij], ij) for ij in start_points)
     if best_distance == np.iinfo(pf.distance.dtype).max:
         return np.zeros((0, 2), dtype=np.intc)  # All endpoints unreachable
 
-    return hillclimb2d(pf.distance, best_ij, cardinal=bool(cardinal), diagonal=bool(diagonal))
+    return pf.path_from(best_ij)

tests/test_path2d_predecessors.py

@@ -0,0 +1,20 @@
+"""Many-to-many paths must retain movement costs during reconstruction."""
+
+import numpy as np
+import pytest
+
+import tcod.path
+
+
+@pytest.mark.parametrize("diagonal", [0, 1, 2, 3])
+def test_path2d_diagonal_predecessors(diagonal: int) -> None:
+    """An expensive diagonal must not replace two cheaper cardinal moves."""
+    cost = np.ones((4, 4), dtype=np.int16)
+    path = tcod.path.path2d(
+        cost, start_points=[(0, 0), (0, 1)], end_points=[(2, 2), (3, 3)], cardinal=1, diagonal=diagonal
+    )
+    total = sum(diagonal if np.all(a != b) else 1 for a, b in zip(path, path[1:]))
+    assert total == (2 if diagonal == 1 else 3)
+    assert tuple(path[0]) in [(0, 0), (0, 1)]
+    assert tuple(path[-1]) in [(2, 2), (3, 3)]
+    np.testing.assert_array_equal(cost, 1)