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.
61 lines
1.4 KiB
Python
61 lines
1.4 KiB
Python
#!/usr/bin/python
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# Consider quadratic Diophantine equations of the form:
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#
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# x^2 – Dy^2 = 1
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#
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# For example, when D=13, the minimal solution in x is 649^2 – 13×180^2 = 1.
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#
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# It can be assumed that there are no solutions in positive integers when D is square.
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#
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# By finding minimal solutions in x for D = {2, 3, 5, 6, 7}, we obtain the following:
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#
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# 3^2 – 2×2^2 = 1
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# 2^2 – 3×1^2 = 1
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# 9^2 – 5×4^2 = 1
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# 5^2 – 6×2^2 = 1
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# 8^2 – 7×3^2 = 1
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#
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# Hence, by considering minimal solutions in x for D ≤ 7, the largest x is obtained when D=5.
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#
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# Find the value of D ≤ 1000 in minimal solutions of x for which the largest value of x is obtained.
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from math import sqrt
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from timeit import default_timer
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from projecteuler import pell_eq
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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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max_ = 0
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max_d = -1
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for i in range(2, 1001):
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if not is_square(i):
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# Solve the Diophantine equation x^2-D*y^2=1 (Pell equation)
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x = pell_eq(i)
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if x > max_:
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max_ = x
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max_d = i
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end = default_timer()
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print('Project Euler, Problem 66')
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print('Answer: {}'.format(max_d))
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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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