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Advent_of_code/src/P13.py

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Python

# --- Day 13: Shuttle Search ---
# Your ferry can make it safely to a nearby port, but it won't get much
# further. When you call to book another ship, you discover that no ships
# embark from that port to your vacation island. You'll need to get from the
# port to the nearest airport.
# Fortunately, a shuttle bus service is available to bring you from the sea
# port to the airport! Each bus has an ID number that also indicates how often
# the bus leaves for the airport.
# Bus schedules are defined based on a timestamp that measures the number of
# minutes since some fixed reference point in the past. At timestamp 0, every
# bus simultaneously departed from the sea port. After that, each bus travels
# to the airport, then various other locations, and finally returns to the sea
# port to repeat its journey forever.
# The time this loop takes a particular bus is also its ID number: the bus with
# ID 5 departs from the sea port at timestamps 0, 5, 10, 15, and so on. The bus
# with ID 11 departs at 0, 11, 22, 33, and so on. If you are there when the bus
# departs, you can ride that bus to the airport!
# Your notes (your puzzle input) consist of two lines. The first line is your
# estimate of the earliest timestamp you could depart on a bus. The second line
# lists the bus IDs that are in service according to the shuttle company;
# entries that show x must be out of service, so you decide to ignore them.
# To save time once you arrive, your goal is to figure out the earliest bus you
# can take to the airport. (There will be exactly one such bus.)
# For example, suppose you have the following notes:
# 939
# 7,13,x,x,59,x,31,19
# Here, the earliest timestamp you could depart is 939, and the bus IDs in
# service are 7, 13, 59, 31, and 19. Near timestamp 939, these bus IDs depart
# at the times marked D:
# time bus 7 bus 13 bus 59 bus 31 bus 19
# 929 . . . . .
# 930 . . . D .
# 931 D . . . D
# 932 . . . . .
# 933 . . . . .
# 934 . . . . .
# 935 . . . . .
# 936 . D . . .
# 937 . . . . .
# 938 D . . . .
# 939 . . . . .
# 940 . . . . .
# 941 . . . . .
# 942 . . . . .
# 943 . . . . .
# 944 . . D . .
# 945 D . . . .
# 946 . . . . .
# 947 . . . . .
# 948 . . . . .
# 949 . D . . .
# The earliest bus you could take is bus ID 59. It doesn't depart until
# timestamp 944, so you would need to wait 944 - 939 = 5 minutes before it
# departs. Multiplying the bus ID by the number of minutes you'd need to wait
# gives 295.
# What is the ID of the earliest bus you can take to the airport multiplied by
# the number of minutes you'll need to wait for that bus?
from copy import copy
from math import gcd
with open("files/P13.txt", "r") as f:
start = int(f.readline())
buses_x = [bus for bus in f.readline().strip().split(",")]
buses = [int(bus) for bus in buses_x if bus != "x"]
def part_1() -> None:
wait = copy(start)
found = False
while not found:
for bus in buses:
if wait % bus == 0:
print(f"Solution to part 1 is {str((wait - start) * bus)}")
found = True
break
wait += 1
# --- Part Two ---
# The shuttle company is running a contest: one gold coin for anyone that can
# find the earliest timestamp such that the first bus ID departs at that time
# and each subsequent listed bus ID departs at that subsequent minute. (The
# first line in your input is no longer relevant.)
# For example, suppose you have the same list of bus IDs as above:
# 7,13,x,x,59,x,31,19
# An x in the schedule means there are no constraints on what bus IDs must
# depart at that time.
# This means you are looking for the earliest timestamp (called t) such that:
# Bus ID 7 departs at timestamp t.
# Bus ID 13 departs one minute after timestamp t.
# There are no requirements or restrictions on departures at two or three
# minutes after timestamp t.
# Bus ID 59 departs four minutes after timestamp t.
# There are no requirements or restrictions on departures at five minutes
# after timestamp t.
# Bus ID 31 departs six minutes after timestamp t.
# Bus ID 19 departs seven minutes after timestamp t.
# The only bus departures that matter are the listed bus IDs at their specific
# offsets from t. Those bus IDs can depart at other times, and other bus IDs
# can depart at those times. For example, in the list above, because bus ID 19
# must depart seven minutes after the timestamp at which bus ID 7 departs, bus
# ID 7 will always also be departing with bus ID 19 at seven minutes after
# timestamp t.
# In this example, the earliest timestamp at which this occurs is 1068781:
# time bus 7 bus 13 bus 59 bus 31 bus 19
# 1068773 . . . . .
# 1068774 D . . . .
# 1068775 . . . . .
# 1068776 . . . . .
# 1068777 . . . . .
# 1068778 . . . . .
# 1068779 . . . . .
# 1068780 . . . . .
# 1068781 D . . . .
# 1068782 . D . . .
# 1068783 . . . . .
# 1068784 . . . . .
# 1068785 . . D . .
# 1068786 . . . . .
# 1068787 . . . D .
# 1068788 D . . . D
# 1068789 . . . . .
# 1068790 . . . . .
# 1068791 . . . . .
# 1068792 . . . . .
# 1068793 . . . . .
# 1068794 . . . . .
# 1068795 D D . . .
# 1068796 . . . . .
# 1068797 . . . . .
# In the above example, bus ID 7 departs at timestamp 1068788 (seven minutes
# after t). This is fine; the only requirement on that minute is that bus ID 19
# departs then, and it does.
# Here are some other examples:
# The earliest timestamp that matches the list 17,x,13,19 is 3417.
# 67,7,59,61 first occurs at timestamp 754018.
# 67,x,7,59,61 first occurs at timestamp 779210.
# 67,7,x,59,61 first occurs at timestamp 1261476.
# 1789,37,47,1889 first occurs at timestamp 1202161486.
# However, with so many bus IDs in your list, surely the actual earliest
# timestamp will be larger than 100000000000000!
# What is the earliest timestamp such that all of the listed bus IDs depart at
# offsets matching their positions in the list?
# adapted from https://stackoverflow.com/questions/32728526/lcm-using-recursive
def lcm(my_list: list[list[int]]) -> int:
lcm = my_list[0][0]
for pair in my_list[1:]:
lcm = lcm * pair[0] // gcd(lcm, pair[0])
return lcm
def part_2() -> None:
my_buses = []
for idx, bus in enumerate(buses_x):
if bus != "x":
my_buses.append([int(bus), idx])
value = 0
add_to = 1
for i in range(1, len(my_buses)):
good = False
while not good:
good = True
for j in range(i + 1):
if (value + my_buses[j][1]) % my_buses[j][0] != 0:
good = False
break
if good is True:
# find the offset between buses
add_to = lcm(my_buses[:j])
break
else:
value += add_to
print(f"Solution to part 2 is {value}")
if __name__ == "__main__":
part_1()
part_2()