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day02.py
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day02.py
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import io
# --- Day 2: Bathroom Security ---
#
# You arrive at Easter Bunny Headquarters under cover of
# darkness. However, you left in such a rush that you forgot to use the
# bathroom! Fancy office buildings like this one usually have keypad
# locks on their bathrooms, so you search the front desk for the code.
#
# "In order to improve security," the document you find says, "bathroom
# codes will no longer be written down. Instead, please memorize and
# follow the procedure below to access the bathrooms."
#
# The document goes on to explain that each button to be pressed can be
# found by starting on the previous button and moving to adjacent
# buttons on the keypad: U moves up, D moves down, L moves left, and R
# moves right. Each line of instructions corresponds to one button,
# starting at the previous button (or, for the first line, the "5"
# button); press whatever button you're on at the end of each line. If a
# move doesn't lead to a button, ignore it.
#
# You can't hold it much longer, so you decide to figure out the code as
# you walk to the bathroom. You picture a keypad like this:
#
# 1 2 3
# 4 5 6
# 7 8 9
#
# Suppose your instructions are:
#
# ULL
# RRDDD
# LURDL
# UUUUD
#
# You start at "5" and move up (to "2"), left (to "1"), and left
# (you can't, and stay on "1"), so the first button is 1.
#
# Starting from the previous button ("1"), you move right twice (to
# "3") and then down three times (stopping at "9" after two moves
# and ignoring the third), ending up with 9.
#
# Continuing from "9", you move left, up, right, down, and left,
# ending with 8.
#
# Finally, you move up four times (stopping at "2"), then down once,
# ending with 5.
#
# So, in this example, the bathroom code is 1985.
#
# Your puzzle input is the instructions from the document you found at
# the front desk. What is the bathroom code?
# KEYPAD STRUCTURE
#
# j 0 1 2
# i
# 0 1 2 3
# 1 4 5 6
# 2 7 8 9
#
KEYPAD = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
i, j = 1, 1
MAX_I = len(KEYPAD) - 1
MAX_J = len(KEYPAD[0]) - 1
print('PART 1')
with io.open('inputs/day02.txt','r') as f:
line_nb = 0
for instructions in f:
for instruction in instructions:
if instruction == 'U':
i = max(0, i-1)
elif instruction == 'L':
j = max(0, j-1)
elif instruction == 'D':
i = min(MAX_I, i + 1)
elif instruction == 'R':
j = min(MAX_J, j + 1)
print('Code for line {}: {}'.format(line_nb, KEYPAD[i][j]))
line_nb += 1
# --- Part Two ---
#
# You finally arrive at the bathroom (it's a several minute walk from
# the lobby so visitors can behold the many fancy conference rooms and
# water coolers on this floor) and go to punch in the code. Much to your
# bladder's dismay, the keypad is not at all like you imagined
# it. Instead, you are confronted with the result of hundreds of
# man-hours of bathroom-keypad-design meetings:
#
# 1
# 2 3 4
# 5 6 7 8 9
# A B C
# D
#
# You still start at "5" and stop when you're at an edge, but given the
# same instructions as above, the outcome is very different:
#
# You start at "5" and don't move at all (up and left are both
# edges), ending at 5.
#
# Continuing from "5", you move right twice and down three times
# (through "6", "7", "B", "D", "D"), ending at D.
#
# Then, from "D", you move five more times (through "D", "B", "C",
# "C", "B"), ending at B.
#
# Finally, after five more moves, you end at 3.
#
# So, given the actual keypad layout, the code would be 5DB3.
#
# Using the same instructions in your puzzle input, what is the correct
# bathroom code?
#
EMPTY_KEY = 0
KEYPAD = [[EMPTY_KEY, EMPTY_KEY, 1, EMPTY_KEY, EMPTY_KEY],
[EMPTY_KEY, 2, 3, 4, EMPTY_KEY],
[5, 6, 7, 8, 9],
[EMPTY_KEY, 'A', 'B', 'C', EMPTY_KEY],
[EMPTY_KEY, EMPTY_KEY, 'D', EMPTY_KEY, EMPTY_KEY]]
i, j = 2, 0
MAX_I = len(KEYPAD) - 1
MAX_J = len(KEYPAD[0]) - 1
print('')
print('PART 2')
with io.open('inputs/day02.txt','r') as f:
line_nb = 0
for instructions in f:
for instruction in instructions:
if instruction == 'U':
new_i = max(0, i-1)
if KEYPAD[new_i][j] != EMPTY_KEY:
i = new_i
elif instruction == 'L':
new_j = max(0, j-1)
if KEYPAD[i][new_j] != EMPTY_KEY:
j = new_j
elif instruction == 'D':
new_i = min(MAX_I, i + 1)
if KEYPAD[new_i][j] != EMPTY_KEY:
i = new_i
elif instruction == 'R':
new_j = min(MAX_J, j + 1)
if KEYPAD[i][new_j] != EMPTY_KEY:
j = new_j
print('Code for line {}: {}'.format(line_nb, KEYPAD[i][j]))
line_nb += 1