Anders Englöf Ytterström
39e09dd36e
Learned a lot about the Josephus' Problem today! Solved part 1 by using a dict, but eventually ended up just adding the mathematical shortcut and rewriting both parts to use deque() for performance. Part 2 was tricky, since k (the elf to remove after all presents were stolen from them) is a index that changes over time. No tries for a solution that was performant enough using lists and dicts were succesfull, so by inspiration from the subreddit the final solution code is based on 2 deque() that pops and appends between them. There are 2 part 1 solutions. - A correct implementation of the Josephus' Problem, using deque(). Recursion would have worked as well, but Python do not like recursions. - The mathematical superior version, with a link to the Youtube video were it is introduced.
52 lines
1.1 KiB
Python
52 lines
1.1 KiB
Python
from collections import deque
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def solve(data):
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elfes = int(data)
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p1 = left_adjacent_rule(elfes)
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p1b = mathematical_superiority(elfes)
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p2 = opposite_side_rule(elfes)
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assert p1 == p1b
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return p1, p2
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def mathematical_superiority(num_elfes):
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# Josephus' problem, quick method:
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# https://www.youtube.com/watch?v=uCsD3ZGzMgE
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b = format(num_elfes, "b")
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return int(b[1:] + b[0], 2)
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def left_adjacent_rule(num_elfes):
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# https://en.wikipedia.org/wiki/Josephus_problem
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q = deque(list(range(1, num_elfes + 1)))
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while len(q) > 1:
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q.rotate(-1)
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q.popleft()
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return q.pop()
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def opposite_side_rule(num_elfes):
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elfes = list(range(1, num_elfes + 1))
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separator = num_elfes // 2
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L, R = deque(elfes[:separator]), deque(elfes[separator:])
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while L and R:
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R.popleft()
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l2r = L.popleft()
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R.append(l2r)
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if len(R) - len(L) != 1:
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r2l = R.popleft()
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L.append(r2l)
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return R.pop()
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if __name__ == "__main__":
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with open("./input/19.txt", "r") as f:
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inp = f.read().strip()
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p1, p2 = solve(inp)
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print(p1)
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print(p2)
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