67 lines
1.7 KiB
Python
67 lines
1.7 KiB
Python
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from lib_misc import is_palindrome
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def euler_004():
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r = 0
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for a in range(999, 99, -1):
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for b in range(a, r // a, -1):
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if is_palindrome(a * b) and a * b > r:
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r = a * b
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return r
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def euler_004_original():
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""" The solution that I came up with originally. """
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r = 0
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for a in range(999, 99, -1):
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if a * a < r:
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break
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for b in range(a, 99, -1):
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c = a * b
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if c > r and is_palindrome(c):
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r = c
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return r
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def euler_004_forum():
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"""
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A solution from etatsui in the project Euler forum.
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It is impressive to see how much fast it is relying on
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mathematical tricks:
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11(9091a + 910b + 100c) = mn;
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Let:
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11 * 10 < m < 11 * 90
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"""
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for a in range(9, 0, -1):
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for b in range(9, -1, -1):
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for c in range(9, -1, -1):
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num = 9091 * a + 910 * b + 100 * c
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for m in range(90, 9, -1):
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if num % m == 0:
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if num / m > 999:
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break
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else:
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result = num * 11
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return result
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assert(euler_004() == 906609)
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print("e004.py: {}".format(euler_004()))
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def time_tests():
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from timeit import timeit
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assert(euler_004() == 906609)
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assert(euler_004_original() == 906609)
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assert(euler_004_forum() == 906609)
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print(timeit(lambda: euler_004(), number=100))
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print(timeit(lambda: euler_004_original(), number=100))
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print(timeit(lambda: euler_004_forum(), number=100))
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# time_tests()
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# 0.5044660240000667
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# 0.7896412069985672
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# 0.04794573199978913
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