"""
Another version, using a generator of Fibonacci "triples"...
Encapsulating a discovery -- not a proof.
By David Koski (Python by K. Urner)
Failure at around 37th power is due to floating point limitations.
"""
import unittest
def fib_triple(a=0, b=1):
"""
>>> from anyproject import fib_triple
>>> gem = fib_triple()
>>> next(gem)
(0, 1, 1)
>>> next(gem)
(1, 1, 2)
>>> next(gem)
(1, 2, 3)
>>> next(gem)
(2, 3, 5)
>>> next(gem)
(3, 5, 8)
>>> gem = fib_triple(13,-8)
>>> next(gem)
(13, -8, 5)
>>> next(gem)
(-8, 5, -3)
>>> next(gem)
(5, -3, 2)
>>> next(gem)
(-3, 2, -1)
>>> next(gem)
(2, -1, 1)
>>> next(gem)
(-1, 1, 0)
"""
c = a + b
while True:
yield a, b, a + b
a, b, c = b, a + b, b + c
series = fib_triple(13,-8) # 13, -8, 5, -3, 2, -1, 1, 0 , 1, 2, 3, 5, 8
RT5 = pow(5, .5) # "square root" of five
PHI = (1 + RT5)/2 # golden proportion
def cool_formula():
"""
Two fib numbers, two apart, plus the one in the middle * sqrt(5)
all over 2, gives PHI to a power. Advancing all three sequences
gives successive powers.
"""
while True:
f0, f1, f2 = next(series)
yield (f0 + f2 + f1 * RT5)/2.0
class TestPhi(unittest.TestCase):
def test_loop(self):
gem = cool_formula()
for e in range(-6, 30): # adjust range (fails about 37th power)
answer = next( gem )
self.assertAlmostEqual( PHI ** e, answer)
if __name__ == "__main__":
unittest.main()
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