Day 7

aoc7-1.py

+import statistics
+
+def main():
+    """
+    Find the position that requires the fewest total moves for each crab sub to get there.
+    This yields to an insight: given a list of numbers, the value that requires the
+    fewest total "moves" to get there is the median.
+    """
+    with open("aoc7-1.txt", "r") as file:
+        positions = [int(el) for el in file.read().strip().split(",")]
+    med = statistics.median(positions)
+    moves = [abs(pos-med) for pos in positions]
+    print(sum(moves))
+
+
+if __name__ == "__main__":
+    main()

aoc7-2.py

+import statistics
+
+def main():
+    with open("aoc7-1.txt", "r") as file:
+        positions = [int(el) for el in file.read().strip().split(",")]
+    # Gonna brute-force this and see what happens.
+    # I bet I'm right and am just doing math wrong at one point.
+    # By definition, the position we want will be between the lowest
+    # and highest values in the list.
+    # positions = [16,1,2,0,4,2,7,1,2,14]
+    # mn, mx = min(positions), max(positions) # the outer bounds
+    # movegroups = []
+    # for pos in positions:
+    #     movegroup = []
+    #     for i in range(mn, mx+1):
+    #         movegroup.append(sum(range(abs(pos-i)+1)))
+    #     movegroups.append(movegroup)
+    # movetotals = []
+    # for i in range((mx-mn)+1):
+    #     mt = []
+    #     for move in movegroups:
+    #         mt.append(move[i])
+    #     movetotals.append(sum(mt))
+    # print(min(movetotals), mn + movetotals.index(min(movetotals)))
+    # print(statistics.mean(positions))
+    # I WAS RIGHT
+    # Leaving the above in for posterity
+    med = int(statistics.mean(positions))
+    moves = [sum(range(abs(pos-med)+1)) for pos in positions]
+    print(sum(moves))
+
+
+if __name__ == "__main__":
+    main()