Daniele Fucini
3a09f3d4f5
Added solutions for problem 66, 67, 68, 69 and 70 in C and python. Also, minor improvements to the code for a few older problems.
83 lines
2.3 KiB
Python
83 lines
2.3 KiB
Python
#!/usr/bin/python
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# All square roots are periodic when written as continued fractions and can be written in the form:
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#
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# √N=a0+1/(a1+1/(a2+1/(a3+…
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#
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# For example, let us consider √23:
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#
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# √23=4+√23−4=4+1/(1/(√23−4))=4+1/(1+(√23−3)/7)
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#
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# If we continue we would get the following expansion:
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#
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# √23=4+1/(1+1/(3+1/(1+1/(8+…
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#
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# The process can be summarised as follows:
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# a0=4,1/(√23−4)=(√23+4)/7=1+(√23−3)/7
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# a1=1,7/(√23−3)=7(√23+3)/14=3+(√23−3)/2
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# a2=3,2/(√23−3)=2(√23+3)/14=1+(√23−4)/7
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# a3=1,7/(√23−4)=7(√23+4)/7=8+√23−4
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# a4=8,1/(√23−4)=(√23+4)/7=1+(√23−3)/7
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# a5=1,7/(√23−3)=7(√23+3)/14=3+(√23−3)/2
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# a6=3,2/(√23−3)=2(√23+3)/14=1+(√23−4)/7
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# a7=1,7/(√23−4)=7(√23+4)/7=8+√23−4
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#
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# It can be seen that the sequence is repeating. For conciseness, we use the notation √23=[4;(1,3,1,8)], to indicate that the block (1,3,1,8)
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# repeats indefinitely.
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#
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# The first ten continued fraction representations of (irrational) square roots are:
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#
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# √2=[1;(2)], period=1
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# √3=[1;(1,2)], period=2
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# √5=[2;(4)], period=1
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# √6=[2;(2,4)], period=2
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# √7=[2;(1,1,1,4)], period=4
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# √8=[2;(1,4)], period=2
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# √10=[3;(6)], period=1
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# √11=[3;(3,6)], period=2
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# √12=[3;(2,6)], period=2
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# √13=[3;(1,1,1,1,6)], period=5
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#
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# Exactly four continued fractions, for N≤13, have an odd period.
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#
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# How many continued fractions for N≤10000 have an odd period?
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from math import floor, sqrt
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from timeit import default_timer
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from projecteuler import build_sqrt_cont_fraction
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def is_square(n):
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p = sqrt(n)
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m = int(p)
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if p == m:
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return True
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else:
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return False
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def main():
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start = default_timer()
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count = 0
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for i in range(2, 10000):
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# Perfect squares are obviously not represented as continued fractions.
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# For all other numbers, calculate their period and check if it's odd.
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if not is_square(i):
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# period_cf(i) % 2 != 0:
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fraction, period = build_sqrt_cont_fraction(i, 300)
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if period % 2 != 0:
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count = count + 1
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end = default_timer()
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print('Project Euler, Problem 64')
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print('Answer: {}'.format(count))
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print('Elapsed time: {:.9f} seconds'.format(end - start))
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if __name__ == '__main__':
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main()
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