71 lines
1.6 KiB
Python
71 lines
1.6 KiB
Python
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from functools import lru_cache
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@lru_cache
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def x_naiv(s):
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zcount = s.count(0)
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if zcount == 3:
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return 0
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elif zcount == 2:
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return 1
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elif zcount == 1:
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sl = list(s)
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sl.remove(0)
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if sl[0] == sl[1]:
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return 0
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s = list(s)
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for i in range(len(s)):
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for v in range(1, s[i] + 1):
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ns = list(s)
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ns[i] -= v
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if x_naiv(tuple(sorted(ns))) == 0:
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# Other player loses.
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return 1
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return 0
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def x_nim_sum(s):
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""" I wasn't able to deduce this myself quickly and looked it up instead. """
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return 0 if (s[0] ^ s[1] ^ s[2]) == 0 else 1
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def test_nim_functions():
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assert(x_naiv(tuple([0, 0, 0])) == 0)
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assert(x_naiv(tuple([1, 0, 0])) == 1)
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assert(x_naiv(tuple([1, 1, 0])) == 0)
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assert(x_naiv(tuple([1, 1, 1])) == 1)
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assert(x_naiv(tuple([1, 2, 3])) == 0)
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for n1 in range(0, 8):
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for n2 in range(1, n1 + 1):
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for n3 in range(1, n2 + 1):
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s = tuple([n1, n2, n3])
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assert x_naiv(s) == x_nim_sum(s)
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def euler_301():
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test_nim_functions()
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# log(2**30, 10) ~= 9 -> 1 billion easy brute force
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# Probably a more efficient computation is possible by counting the permutations
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# where all binary digits are zero.
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# 0 0 0 -> 0
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# 0 0 1 -> 0
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# 0 1 1 -> 0
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r = 0
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for n in range(1, 2**30 + 1):
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s = tuple([n, 2 * n, 3 * n])
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xr = x_nim_sum(s)
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if xr == 0:
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r += 1
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return r
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if __name__ == "__main__":
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solution = euler_301()
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print("e301.py: " + str(solution))
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# assert(solution == 0)
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