56 lines
2.1 KiB
Python
56 lines
2.1 KiB
Python
# -*- encoding: utf-8 -*-
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import sys
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lines = sys.stdin.readlines()
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# the solution relies on replacing rock / paper / scissor by values 0 / 1 / 2 and using
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# modulo-3 arithmetic
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#
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# in modulo-3 arithmetic, the winning move is 1 + the opponent move (e.g., winning move
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# if opponent plays 0 is 1, or 0 if opponent plays 2 (0 = (2 + 1 % 3)))
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#
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# we read the lines in a Nx2 in array with value 0/1/2 instead of A/B/C or X/Y/Z for
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# easier manipulation
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values = [(ord(row[0]) - ord("A"), ord(row[2]) - ord("X")) for row in lines]
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def score_1(ux: int, vx: int) -> int:
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# here ux and vx are both moves: 0 = rock, 1 = paper, 2 = scissor
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#
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# 1. to get the score of the move/shape, we simply add 1 -> vx + 1
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# 2. to get the score of the outcome (loss/draw/win), we use the fact that the
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# winning hand is always the opponent hand (ux) + 1 in modulo-3 arithmetic:
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# - (ux - vx) % 3 gives us 0 for a draw, 1 for a loss and 2 for a win
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# - 1 - ((ux - vx) % 3) gives us -1 for a win, 0 for a loss and 1 for a draw
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# - (1 - ((ux - vx) % 3)) gives us 0 / 1 / 2 for loss / draw / win
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# - the above can be rewritten as ((1 - (ux - vx)) % 3)
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# we can then simply multiply this by 3 to get the outcome score
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#
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return (vx + 1) + ((1 - (ux - vx)) % 3) * 3
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def score_2(ux: int, vx: int) -> int:
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# here ux is the opponent move (0 = rock, 1 = paper, 2 = scissor) and vx is the
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# outcome (0 = loss, 1 = draw, 2 = win)
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#
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# 1. to get the score to the move/shape, we need to find it (as 0, 1 or 2) and then
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# add 1 to it
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# - (vx - 1) gives the offset from the opponent shape (-1 for a loss, 0 for a
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# draw and 1 for a win)
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# - from the offset, we can retrieve the shape by adding the opponent shape and
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# using modulo-3 arithmetic -> (ux + vx - 1) % 3
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# - we then add 1 to get the final shape score
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# 2. to get the score of the outcome, we can simply multiply vx by 3 -> vx * 3
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return (ux + vx - 1) % 3 + 1 + vx * 3
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# part 1 - 13526
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print(f"score 1 is {sum(score_1(*v) for v in values)}")
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# part 2 - 14204
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print(f"score 2 is {sum(score_2(*v) for v in values)}")
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