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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
parent 894160d1a0
commit c34ebd6181
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<!DOCTYPE html>
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<head><meta charset="utf-8" />
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<title>EulerProblem028</title><script src="https://cdnjs.cloudflare.com/ajax/libs/require.js/2.1.10/require.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/2.0.3/jquery.min.js"></script>
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<h1 id="Euler-Problem-28">Euler Problem 28<a class="anchor-link" href="#Euler-Problem-28">&#182;</a></h1><p>Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:</p>
<h1 id="Euler-Problem-28">Euler Problem 28<a class="anchor-link" href="#Euler-Problem-28"></a></h1><p><a href="/euler">Back to overview.</a></p><p>Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:</p>
<pre><code>21 22 23 24 25
20 7 8 9 10
19 6 1 2 11
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<p>$1 + 3 + 5 + 7 + 9 + 13 + 17 + 21 + 25 = 101$</p>
<p>It can be verified that the sum of the numbers on the diagonals is 101.</p>
<p>What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral formed in the same way?</p>
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<p>When we know the corner of a certain spiral we can calculate it's total like $f_n = 4 c_n + 6 (n - 1)$. We then only have to update the corner value for each spiral.</p>
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<div class=" highlight hl-ipython3"><pre><span></span><span class="n">total</span> <span class="o">=</span> <span class="mi">1</span>
<span class="n">current_corner</span> <span class="o">=</span> <span class="mi">3</span>
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<span class="n">s</span> <span class="o">=</span> <span class="n">total</span>
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<div class=" highlight hl-ipython3"><pre><span></span><span class="k">assert</span><span class="p">(</span><span class="n">s</span> <span class="o">==</span> <span class="mi">669171001</span><span class="p">)</span>
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<pre>669171001</pre>
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<p>$c_n = (n - 1)^2 - (n - 2) = n^2 - 2n + 1 - n + 2 = n^2 - 3n + 3$.</p>
<p>Now, we can insert $c_n$ into $f_n$ which gives as the sum of corners for the nth spiral:</p>
<p>$f_n = 4n^2 - 12n + 12 + 6n - 6 = 4n^2 - 6n + 6$</p>
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<div class=" highlight hl-ipython3"><pre><span></span><span class="n">s</span> <span class="o">=</span> <span class="mi">1</span> <span class="o">+</span> <span class="nb">sum</span><span class="p">([</span><span class="mi">4</span> <span class="o">*</span> <span class="n">n</span> <span class="o">*</span> <span class="n">n</span> <span class="o">-</span> <span class="mi">6</span> <span class="o">*</span> <span class="n">n</span> <span class="o">+</span> <span class="mi">6</span> <span class="k">for</span> <span class="n">n</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">3</span><span class="p">,</span> <span class="mi">1002</span><span class="p">,</span> <span class="mi">2</span><span class="p">)])</span>
<span class="k">assert</span><span class="p">(</span><span class="n">s</span> <span class="o">==</span> <span class="mi">669171001</span><span class="p">)</span>
<span class="n">s</span>
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<pre>669171001</pre>
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