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97 lines
2.6 KiB
Python
97 lines
2.6 KiB
Python
#!/usr/bin/env python
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"""
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##---
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# Julien Lengrand-Lambert
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#Created on : 17 - 10 - 2012
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#
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# DESCRIPTION : Solves problem 55 of Project Euler
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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.
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A number that never forms a palindrome through the reverse
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and add process is called a Lychrel number.
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Due to the theoretical nature of these numbers,
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and for the purpose of this problem, we shall assume that a number is Lychrel
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until proven otherwise.
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In addition you are given that for every number below ten-thousand,
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it will either (i) become a palindrome in less than fifty iterations,
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or, (ii) no one, with all the computing power that exists,
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has managed so far to map it to a palindrome.
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In fact, 10677 is the first number to be shown to require over fifty iterations
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before producing a palindrome: 4668731596684224866951378664
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(53 iterations, 28-digits).
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Surprisingly, there are palindromic numbers that are themselves Lychrel
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numbers; the first example is 4994.
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How many Lychrel numbers are there below ten-thousand?
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NOTE: Wording was modified slightly on 24 April 2007 to emphasise
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the theoretical nature of Lychrel numbers.
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##---
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"""
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def is_palindromic(num):
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"""
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Returns True if num is palindromic
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Ex : 121
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"""
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return (num == reverse(num))
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def reverse(num):
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"""
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Returns the reverse number of num
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ex : 123 returns 321
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"""
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str_num = str(num)
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rev = ''
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for i in range(len(str_num)):
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rev += str_num[len(str_num) - i - 1]
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return int(rev)
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def is_lychrel(num, max_it=50):
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"""
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Given a maximu number of iterations,
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returns True if num is a lychrel number
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"""
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cur = num
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it = 1
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while it < max_it:
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cur = cur + reverse(cur)
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if is_palindromic(cur):
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return False
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it += 1
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return True
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def count_lychrel(max_num, max_it=50):
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"""
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Returns the number of lychrel numbers smaller than max_num
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The maximum iteration number defines how many times the process is repeated
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before considreing a number to be a lychrel number
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"""
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lychrels = []
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for i in xrange(1, max_num + 1):
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if (is_lychrel(i, max_it)):
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lychrels.append(i)
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print lychrels
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return len(lychrels)
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if __name__ == '__main__':
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print "Answer : %d " % (count_lychrel(10000))
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