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    <p><strong>There's a new, radically faster version as <a href="https://stackoverflow.com/questions/1269795/unusual-speed-difference-between-python-and-c/1275074#1275074">a separate answer</a>, so this answer is deprecated.</strong></p> <hr> <p>I rewrote your algorithm by making it cache whenever it finds the number to be happy or unhappy. I also tried to make it as pythonic as I could, for example by creating separate functions <code>digits()</code> and <code>happy()</code>. Sorry for using Python 3, but I get to show off a couple a useful things from it as well.</p> <p>This version is <strong>much faster</strong>. It runs at <strong>1.7s</strong> which is <strong>10 times</strong> faster than your original program that takes <strong>18s</strong> (well, my MacBook is quite old and slow :) )</p> <pre><code>#!/usr/bin/env python3 from timeit import Timer from itertools import count print_numbers = False upperBound = 10**5 # Default value, can be overidden by user. def digits(x:'nonnegative number') -&gt; "yields number's digits": if not (x &gt;= 0): raise ValueError('Number should be nonnegative') while x: yield x % 10 x //= 10 def happy(number, known = {1}, happies = {1}) -&gt; 'True/None': '''This function tells if the number is happy or not, caching results. It uses two static variables, parameters known and happies; the first one contains known happy and unhappy numbers; the second contains only happy ones. If you want, you can pass your own known and happies arguments. If you do, you should keep the assumption commented out on the 1 line. ''' # assert 1 in known and happies &lt;= known # &lt;= is expensive if number in known: return number in happies history = set() while True: history.add(number) number = sum(x**2 for x in digits(number)) if number in known or number in history: break known.update(history) if number in happies: happies.update(history) return True def calcMain(): happies = {x for x in range(upperBound) if happy(x) } if print_numbers: print(happies) if __name__ == '__main__': upperBound = eval( input("Pick an upper bound [default {0}]: " .format(upperBound)).strip() or repr(upperBound)) result = Timer(calcMain).timeit(1) print ('This computation took {0} seconds'.format(result)) </code></pre>
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    1. COJust benchmarked your solution. On the first run, it is indeed about 10 times faster. On the second run, the code has already been compiled, and it jumps to 200+ times faster. Nice job! It's almost on par with the C++ code I proposed (which runs 600+ times faster) than the original Python implementation. Your code seems correct but, just for nitpicking, you are missing the upper bound, off-by-one error probably. For instance, if I input 1000, 1000 should be among the happy numbers, but the result lists stops at 998.
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    2. COIt depends on how you define upperBound :) In Python most functions that accept `stop` parameter usually return elements up to `stop-1`, e.g. `range(N)` or `a[:N]`.
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    3. CONote that you don't need the history list. Following Wikipedia's happy numbers page, there is only one repeating sequence of numbers besides 1,1,1,1 and 0,0,0. Therefore, you can test number <= 4 and end the loop, and it's a happy number if number == 1.
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