Fizz Buzz without conditionals or booleans

Re:Fizz Buzz without conditionals or booleans by susam on Wednesday November 19, 2025 @09:49AM (#45980229)

Late to the party but here's a solution I worked out using roots of unity and cosines:

  from math import cos, pi
  for n in range(1, 101):
      print([n, 'Fizz', 'Buzz', 'FizzBuzz'][round((1 + 2 * cos(2 * pi * n / 3)) / 3 + 2 * (1 + 2 * cos(2 * pi * n / 5) + 2 * cos(4 * pi * n / 5)) / 5)])

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Re:Fizz Buzz without conditionals or booleans by nik9000 on Wednesday November 19, 2025 @06:59PM (#45987011)

As it should be.

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