Anders Englöf Ytterström
a90269f7f9
WALLOFTEXT for part 2, took me 90 minutes to find this important text: > Each step begins with a sequence of letters that > indicate the label of the lens on which the step > operates. The result of running the HASH algorithm > on the label indicates the correct box for that > step. It also clarifies how part 2 and part 1 relates.
63 lines
1.2 KiB
Python
63 lines
1.2 KiB
Python
from collections import OrderedDict, defaultdict
|
|
|
|
from output import answer
|
|
|
|
n = 15
|
|
title = "Lens Library"
|
|
|
|
|
|
@answer(1, "Sum of HASH algorithm results: {}")
|
|
def part_1(presolved):
|
|
return presolved[0]
|
|
|
|
|
|
@answer(2, "Focusing power of the resulting configuration: {}")
|
|
def part_2(presolved):
|
|
return presolved[1]
|
|
|
|
|
|
def presolve(data):
|
|
def h(s):
|
|
v = 0
|
|
for a in s:
|
|
if a == "\n":
|
|
continue
|
|
v += ord(a)
|
|
v *= 17
|
|
v = v % 256
|
|
return v
|
|
|
|
p1 = sum(h(c) for c in data.split(","))
|
|
|
|
b = defaultdict(OrderedDict)
|
|
for lr in data.split(","):
|
|
if "=" in lr:
|
|
l, r = lr.split("=")
|
|
if r == "":
|
|
continue
|
|
k = h(l)
|
|
b[k][l] = r
|
|
if "-" in lr:
|
|
l, _r = lr.split("-")
|
|
k = h(l)
|
|
if l in b[k]:
|
|
del b[k][l]
|
|
p2 = 0
|
|
for i, c in b.items():
|
|
for j, f in enumerate(b[i].values(), 1):
|
|
p2 += (i + 1) * j * int(f)
|
|
|
|
return p1, p2
|
|
|
|
|
|
if __name__ == "__main__":
|
|
with open("./input/16.txt", "r") as f:
|
|
inp = f.read().strip()
|
|
|
|
inp = presolve(inp)
|
|
|
|
a = part_1(inp)
|
|
b = part_2(inp)
|
|
|
|
assert a == 509784
|
|
assert b == 230197
|