My Advent of Code solutions in Python.
kevinyap.ca/2019/12/going-fast-in-advent-of-code/
advent-of-code
python
Update solution for 2023/10
Use the polygon area formula that was pivotal to Day 18 to solve Day 10
Part 2. The points "enclosed" by the loop that can't escape is equal to
the internal area of the polygon after removing all points along the
perimeter, so that we don't double-count the loop itself.
+20
-2
+20
-2
2023/day10.py
+20
-2
2023/day10.py
···
59
59
return None
60
60
61
61
node += direction
62
-
loop.append(node)
63
62
64
63
if node == start:
65
64
return loop
65
+
66
+
loop.append(node)
66
67
67
68
69
+
def discrete_polygon_area(points):
70
+
# Use shoelace formula to compute internal area.
71
+
area = 0
72
+
73
+
for a, b in zip(points, points[1:] + [points[0]]):
74
+
area += (b.x + a.x) * (b.y - a.y)
75
+
76
+
area = int(abs(area / 2.0))
77
+
78
+
# Calculate perimeter.
79
+
perimeter = sum(a.dist_manhattan(b) for a, b in zip(points, points[1:] + [points[0]]))
80
+
81
+
# Account for outer perimeter strip in final area computation.
82
+
return area + (perimeter // 2) + 1
83
+
68
84
69
85
board = {}
70
86
for y, line in enumerate(fileinput.input()):
···
82
98
break
83
99
84
100
85
-
# TODO: Solve part 2 programmatically.
101
+
# Solve part 2 by solving for the polygon area and
102
+
# subtracting off all perimeter points.
103
+
print("Part 2:", discrete_polygon_area(loop) - len(loop))