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Mikael CAPELLE 2022-12-02 15:06:37 +01:00
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2022/day2.py
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@ -2,6 +2,13 @@
from pathlib import Path
# the solution relies on replacing rock / paper / scissor by values 0 / 1 / 2 and using
# modulo-3 arithmetic
#
# in modulo-3 arithmetic, the winning move is 1 + the opponent move (e.g., winning move
# if opponent plays 0 is 1, or 0 if opponent plays 2 (0 = (2 + 1 % 3)))
#
# we read the lines in a Nx2 in array with value 0/1/2 instead of A/B/C or X/Y/Z for
# easier manipulation
with open(Path(__file__).parent.joinpath("inputs", "day2.txt")) as fp:
@ -11,17 +18,34 @@ with open(Path(__file__).parent.joinpath("inputs", "day2.txt")) as fp:
def score_1(ux: int, vx: int) -> int:
# explanation:
# - (1 + vx) is just the score of the shape
# - ((1 - (ux - vx)) % 3) gives 0 for loss, 1 for draw and 2 for win, that we
# can multiply with 3 to get the outcome score
return (1 + vx) + ((1 - (ux - vx)) % 3) * 3
# here ux and vx are both moves: 0 = rock, 1 = paper, 2 = scissor
#
# 1. to get the score of the move/shape, we simply add 1 -> vx + 1
# 2. to get the score of the outcome (loss/draw/win), we use the fact that the
# winning hand is always the opponent hand (ux) + 1 in modulo-3 arithmetic:
# - (ux - vx) % 3 gives us 0 for a draw, 1 for a loss and 2 for a win
# - 1 - ((ux - vx) % 3) gives us -1 for a win, 0 for a loss and 1 for a draw
# - (1 - ((ux - vx) % 3)) gives us 0 / 1 / 2 for loss / draw / win
# - the above can be rewritten as ((1 - (ux - vx)) % 3)
# we can then simply multiply this by 3 to get the outcome score
#
return (vx + 1) + ((1 - (ux - vx)) % 3) * 3
def score_2(ux: int, vx: int) -> int:
# explanation:
# - (ux + vx - 1) % 3 gives the target shape (0, 1, 2), we add one to get the score
# - vx * 3 is simply the outcome score
# here ux is the opponent move (0 = rock, 1 = paper, 2 = scissor) and vx is the
# outcome (0 = loss, 1 = draw, 2 = win)
#
# 1. to get the score to the move/shape, we need to find it (as 0, 1 or 2) and then
# add 1 to it
# - (vx - 1) gives the offset from the opponent shape (-1 for a loss, 0 for a
# draw and 1 for a win)
# - from the offset, we can retrieve the shape by adding the opponent shape and
# using modulo-3 arithmetic -> (ux + vx - 1) % 3
# - we then add 1 to get the final shape score
# 2. to get the score of the outcome, we can simply multiply vx by 3 -> vx * 3
return (ux + vx - 1) % 3 + 1 + vx * 3