2015 day 9

adventofcode/2015/day9/day9.py

@@ -0,0 +1,36 @@
+#!/usr/bin/env python3
+import fileinput
+from collections import defaultdict
+from itertools import permutations
+
+
+def parse_locations(inp: list[str]) -> dict[str, dict[str, int]]:
+    "parse list of location pairs into adjacency map"
+    locations = defaultdict(dict)
+    for line in inp:
+        places, dist = line.split(" = ")
+        place_from, place_to = places.split(" to ")
+        locations[place_from][place_to] = int(dist)
+        locations[place_to][place_from] = int(dist)  # store both directions
+    return locations
+
+
+def main(inp):
+    "brute force over permutations of cities, keeping track of min and max cost"
+    locations = parse_locations(inp)
+    min_cost = 99999
+    max_cost = 0
+    for perm in permutations(locations):
+        cost = 0
+        # iterate over pairs of cities and get cost from adjacency map
+        for city_a, city_b in zip(perm, perm[1:]):
+            cost += locations[city_a][city_b]
+        min_cost = min(cost, min_cost)
+        max_cost = max(cost, max_cost)
+    print("Part 1: ", min_cost)
+    print("Part 2: ", max_cost)
+
+
+if __name__ == "__main__":
+    lines = [x.rstrip() for x in fileinput.input()]
+    main(lines)

adventofcode/2024/day18/day18.py

@@ -28,7 +28,10 @@ def can_reach(pos, obstacles, grid_size):
     return False
 
 
-def search(obstacles, start, goal, grid_size):  # could just use bfs?
+def search(obstacles, start, goal, grid_size):
+    """
+    Dijkstra's algorithm. Could as well use BFS since step costs are uniform. (n = Node(new_pos, node.cost + 1))
+    """
     queue = [Node(start)]
     visited = set()
     i = 0