Anders Ytterström
c0d352cdf1
I struggled a lot with the direction, but realised way too much later that the mappings are bidirectional. Part 2 was exceptionally hard, but I realised early that the order did not matter and that a simple boolean lock was all that was needed. The challenge was to know when to watch for duplicates. The seen was initially a set, but since new sets must be sent instead of references, a list is used instead to be able to use the seen + [] hack.
75 lines
1.4 KiB
Python
75 lines
1.4 KiB
Python
import unittest
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from solutions.day_12 import Solution
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class Day12TestCase(unittest.TestCase):
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def setUp(self):
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self.solution = Solution()
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self.puzzle_input = self.solution.parse_input(
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"""
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start-A
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start-b
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A-c
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A-b
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b-d
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A-end
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b-end
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"""
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)
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def test_parse_puzzle_input(self):
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data = """
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a-b
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q1w2-e3r4
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"""
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assert self.solution.parse_input(data) == [["a", "b"], ["q1w2", "e3r4"]]
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def test_solve_first_part(self):
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assert self.solution.solve(self.puzzle_input) == 10
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def test_solve_second_part(self):
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assert self.solution.solve_again(self.puzzle_input) == 36
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if __name__ == "__main__":
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unittest.main()
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"""
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start,A,b,A,b,A,c,A,end
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start,A,b,A,b,A,end
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start,A,b,A,b,end
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start,A,b,A,c,A,b,A,end
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start,A,b,A,c,A,b,end
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start,A,b,A,c,A,c,A,end
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start,A,b,A,c,A,end
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start,A,b,A,end
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start,A,b,d,b,A,c,A,end
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start,A,b,d,b,A,end
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start,A,b,d,b,end
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start,A,b,end
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start,A,c,A,b,A,b,A,end
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start,A,c,A,b,A,b,end
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start,A,c,A,b,A,c,A,end
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start,A,c,A,b,A,end
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start,A,c,A,b,d,b,A,end
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start,A,c,A,b,d,b,end
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start,A,c,A,b,end
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start,A,c,A,c,A,b,A,end
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start,A,c,A,c,A,b,end
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start,A,c,A,c,A,end
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start,A,c,A,end
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start,A,end
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start,b,A,b,A,c,A,end
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start,b,A,b,A,end
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start,b,A,b,end
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start,b,A,c,A,b,A,end
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start,b,A,c,A,b,end
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start,b,A,c,A,c,A,end
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start,b,A,c,A,end
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start,b,A,end
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start,b,d,b,A,c,A,end
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start,b,d,b,A,end
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start,b,d,b,end
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start,b,end
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"""
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