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Euler has design of a homepage and there is a link back to the overview for each solution.

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Felix Martin
2018-06-14 20:11:26 -04:00
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<title>EulerProblem012</title><script src="https://cdnjs.cloudflare.com/ajax/libs/require.js/2.1.10/require.min.js"></script>
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<h1 id="Euler-Problem-12">Euler Problem 12<a class="anchor-link" href="#Euler-Problem-12">&#182;</a></h1><p>The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be $1 + 2 + 3 + 4 + 5 + 6 + 7 = 28$. The first ten terms would be:</p>
<h1 id="Euler-Problem-12">Euler Problem 12<a class="anchor-link" href="#Euler-Problem-12"></a></h1><p><a href="/euler">Back to overview.</a></p><p>The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be $1 + 2 + 3 + 4 + 5 + 6 + 7 = 28$. The first ten terms would be:</p>
<p>$1, 3, 6, 10, 15, 21, 28, 36, 45, 55, \dots$</p>
<p>Let us list the factors of the first seven triangle numbers:</p>
<pre><code> 1: 1
3: 1,3
6: 1,2,3,6
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28: 1,2,4,7,14,28</code></pre>
<p>We can see that 28 is the first triangle number to have over five divisors.</p>
<p>What is the value of the first triangle number to have over five hundred divisors?</p>
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<p>Let's try brute forcing.</p>
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<div class=" highlight hl-ipython3"><pre><span></span><span class="k">def</span> <span class="nf">get_divisors</span><span class="p">(</span><span class="n">n</span><span class="p">):</span>
<span class="k">return</span> <span class="p">[</span><span class="n">i</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="nb">int</span><span class="p">(</span><span class="n">n</span><span class="o">/</span><span class="mi">2</span><span class="p">)</span> <span class="o">+</span> <span class="mi">1</span><span class="p">)</span> <span class="k">if</span> <span class="n">n</span> <span class="o">%</span> <span class="n">i</span> <span class="o">==</span> <span class="mi">0</span><span class="p">]</span> <span class="o">+</span> <span class="p">[</span><span class="n">n</span><span class="p">]</span>
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<span class="c1"># print(t)</span>
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<p>That failed miserably. We have to come up with something more efficient. I looked at my old Haskell solution which looks like this:</p>
<pre><code>divisor_count' x = product . map (succ . length) . group $ prim_factors x</code></pre>
<p>Add first I did not understand what it does at all. But the algorithm is as follows:</p>
<pre><code># get prime factors, for example for 28
2 * 2 * 7
# group the primes
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6</code></pre>
<p>Honestly, I have no idea why this gives as the number of divisors. I will implement the solution and then try to come up with an explanation.</p>
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<div class=" highlight hl-ipython3"><pre><span></span><span class="k">def</span> <span class="nf">is_prime</span><span class="p">(</span><span class="n">n</span><span class="p">,</span> <span class="n">smaller_primes</span><span class="p">):</span>
<span class="k">for</span> <span class="n">s</span> <span class="ow">in</span> <span class="n">smaller_primes</span><span class="p">:</span>
<span class="k">if</span> <span class="n">n</span> <span class="o">%</span> <span class="n">s</span> <span class="o">==</span> <span class="mi">0</span><span class="p">:</span>
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<span class="n">factors</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">remainder</span><span class="p">)</span>
<span class="k">return</span> <span class="n">factors</span>
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<p>This are the prime numbers related functions we already now. Now we implement a group function, the product function we already know, and based on that the algorithm to get the number of divisors mentioned above.</p>
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<div class=" highlight hl-ipython3"><pre><span></span><span class="k">def</span> <span class="nf">group</span><span class="p">(</span><span class="n">xs</span><span class="p">):</span>
<span class="kn">from</span> <span class="nn">functools</span> <span class="k">import</span> <span class="n">reduce</span>
<span class="k">def</span> <span class="nf">f</span><span class="p">(</span><span class="n">xss</span><span class="p">,</span> <span class="n">x</span><span class="p">):</span>
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<span class="k">assert</span><span class="p">(</span><span class="n">get_number_of_divisors</span><span class="p">(</span><span class="mi">28</span><span class="p">)</span> <span class="o">==</span> <span class="mi">6</span><span class="p">)</span>
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<p>Now we are ready to do another brute force attempt.</p>
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<div class=" highlight hl-ipython3"><pre><span></span><span class="n">ts</span> <span class="o">=</span> <span class="n">triangle_number_generator_function</span><span class="p">()</span>
<span class="k">for</span> <span class="n">t</span> <span class="ow">in</span> <span class="n">ts</span><span class="p">:</span>
<span class="k">if</span> <span class="n">get_number_of_divisors</span><span class="p">(</span><span class="n">t</span><span class="p">)</span> <span class="o">&gt;</span> <span class="mi">500</span><span class="p">:</span>
<span class="nb">print</span><span class="p">(</span><span class="n">t</span><span class="p">)</span>
<span class="k">break</span>
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<pre>76576500
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<p>Now the only question is why this crazy algorithm works. Okay, I got it with the help of <a href="https://www.math.upenn.edu/~deturck/m170/wk2/numdivisors.html">this</a> page. The problem is actually an instance of the multiplication principle for counting things. Each prime can be used to calculate (n + 1) other divisors. The incrementation by one is required because we can also choose to not use a certain prime. Of course, to get the potential combinations we have to get the product of the potential divisors for all primes. The web page explains it better. The only question is whether I came up with this algorithm myself back then.</p>
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