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    <p>My answer here isn't pretty or elegant, it is still brute force. But, it simplifies the problem space a little and terminates successfully in less than 10 seconds.</p> <p><strong>Getting factors of n:</strong> Like @usethedeathstar mentioned, it is possible to test for factors only up to <code>n/2</code>. However, we can do better by testing only up to the square root of n:</p> <pre><code>let n = 36 =&gt; factors(n) : (1x36, 2x18, 3x12, 4x9, 6x6, 9x4, 12x3, 18x2, 36x1) </code></pre> <p>As you can see, it loops around after 6 (the square root of 36). We also don't need to explicitly return the factors, just find out how many there are... so just count them off with a generator inside of sum():</p> <pre><code>import math def get_factors(n): return sum(2 for i in range(1, round(math.sqrt(n)+1)) if not n % i) </code></pre> <p><strong>Testing the triangular numbers</strong></p> <p>I have used a generator function to yield the triangular numbers:</p> <pre><code>def generate_triangles(limit): l = 1 while l &lt;= limit: yield sum(range(l + 1)) l += 1 </code></pre> <p>And finally, start testing:</p> <pre><code>def test_triangles(): triangles = generate_triangles(100000) for i in triangles: if get_factors(i) &gt; 499: return i </code></pre> <p>Running this with the profiler, it completes in less than 10 seconds:</p> <pre><code>$ python3 -m cProfile euler12.py 361986 function calls in 8.006 seconds </code></pre> <p>The BIGGEST time saving here is <code>get_factors(n)</code> testing only up to the square root of n - this makes it heeeaps quicker and you save heaps of memory overhead by not generating a list of factors.</p> <p>As I said, it still isn't pretty - I am sure there are more elegant solutions. But, it fits the bill of being faster :)</p>
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    1. POOptimise the solution to Project Euler 12 (Python)
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    1. USNick Burns
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    1. POOptimise the solution to Project Euler 12 (Python)
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    1. COwow - when you complete this you get to read Project Euler's overview on the problem. They have some very cool approaches :)
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    2. COHi, thanks for the reply. def get_factors(n): return sum(2 for i in range(1, round(math.sqrt(n)+1)) if not n % i) You said you use a generator within the sum. Could you explain this please
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    3. COMy solution already counted to sqrt(n) instead of n/2?, and the fact of not writing them to memory but just counting doesnt improve much?
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