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* Create __init__.py * Add files via upload * Update sol1.py
77 lines
2.3 KiB
Python
77 lines
2.3 KiB
Python
"""
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If we take 47, reverse and add, 47 + 74 = 121, which is palindromic.
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Not all numbers produce palindromes so quickly. For example,
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349 + 943 = 1292,
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1292 + 2921 = 4213
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4213 + 3124 = 7337
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That is, 349 took three iterations to arrive at a palindrome.
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Although no one has proved it yet, it is thought that some numbers, like 196,
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never produce a palindrome. A number that never forms a palindrome through the
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reverse and add process is called a Lychrel number. Due to the theoretical nature
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of these numbers, and for the purpose of this problem, we shall assume that a number
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is Lychrel until proven otherwise. In addition you are given that for every number
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below ten-thousand, it will either (i) become a palindrome in less than fifty
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iterations, or, (ii) no one, with all the computing power that exists, has managed
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so far to map it to a palindrome. In fact, 10677 is the first number to be shown
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to require over fifty iterations before producing a palindrome:
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4668731596684224866951378664 (53 iterations, 28-digits).
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Surprisingly, there are palindromic numbers that are themselves Lychrel numbers;
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the first example is 4994.
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How many Lychrel numbers are there below ten-thousand?
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"""
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def is_palindrome(n: int) -> bool:
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"""
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Returns True if a number is palindrome.
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>>> is_palindrome(12567321)
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False
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>>> is_palindrome(1221)
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True
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>>> is_palindrome(9876789)
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True
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"""
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return str(n) == str(n)[::-1]
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def sum_reverse(n: int) -> int:
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"""
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Returns the sum of n and reverse of n.
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>>> sum_reverse(123)
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444
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>>> sum_reverse(3478)
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12221
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>>> sum_reverse(12)
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33
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"""
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return int(n) + int(str(n)[::-1])
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def compute_lychrel_nums(limit: int) -> int:
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"""
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Returns the count of all lychrel numbers below limit.
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>>> compute_lychrel_nums(10000)
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249
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>>> compute_lychrel_nums(5000)
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76
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>>> compute_lychrel_nums(1000)
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13
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"""
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lychrel_nums = []
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for num in range(1, limit):
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iterations = 0
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a = num
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while iterations < 50:
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num = sum_reverse(num)
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iterations += 1
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if is_palindrome(num):
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break
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else:
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lychrel_nums.append(a)
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return len(lychrel_nums)
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if __name__ == "__main__":
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print(f"{compute_lychrel_nums(10000) = }")
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