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Added 24 and 25 to ipython.

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Felix Martin
2018-02-06 22:22:00 +01:00
parent b7e7314e99
commit ee6e6e6445
8 changed files with 948 additions and 7 deletions
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ipython/EulerProblem023.html
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@@ -188,21 +188,124 @@ div#notebook {
<p>As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24. By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit.</p>
<p>Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.</p>
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<p>We reuse the sum of proper divisors function and use it to tell whether a number is abundant or not.</p>
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<div class=" highlight hl-ipython3"><pre><span></span><span class="k">def</span> <span class="nf">sum_of_proper_divisors</span><span class="p">(</span><span class="n">n</span><span class="p">):</span>
<span class="kn">from</span> <span class="nn">math</span> <span class="k">import</span> <span class="n">sqrt</span>
<span class="k">if</span> <span class="n">n</span> <span class="o">==</span> <span class="mi">1</span><span class="p">:</span>
<span class="k">return</span> <span class="mi">0</span>
<span class="n">s</span> <span class="o">=</span> <span class="mi">1</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">2</span><span class="p">,</span> <span class="nb">int</span><span class="p">(</span><span class="n">sqrt</span><span class="p">(</span><span class="n">n</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="n">s</span> <span class="o">+=</span> <span class="n">i</span>
<span class="n">x</span> <span class="o">=</span> <span class="p">(</span><span class="n">n</span> <span class="o">//</span> <span class="n">i</span><span class="p">)</span>
<span class="k">if</span> <span class="n">x</span> <span class="o">!=</span> <span class="n">i</span><span class="p">:</span>
<span class="n">s</span> <span class="o">+=</span> <span class="n">x</span>
<span class="k">return</span> <span class="n">s</span>
<span class="k">def</span> <span class="nf">is_abundant</span><span class="p">(</span><span class="n">n</span><span class="p">):</span>
<span class="k">return</span> <span class="n">sum_of_proper_divisors</span><span class="p">(</span><span class="n">n</span><span class="p">)</span> <span class="o">&gt;</span> <span class="n">n</span>
<span class="k">assert</span><span class="p">(</span><span class="n">is_abundant</span><span class="p">(</span><span class="mi">7</span><span class="p">)</span> <span class="o">==</span> <span class="kc">False</span><span class="p">)</span>
<span class="k">assert</span><span class="p">(</span><span class="n">is_abundant</span><span class="p">(</span><span class="mi">12</span><span class="p">)</span> <span class="o">==</span> <span class="kc">True</span><span class="p">)</span>
<span class="n">abundant_numbers_smaller_30000</span> <span class="o">=</span> <span class="p">[</span><span class="n">n</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">1</span><span class="p">,</span> <span class="mi">30001</span><span class="p">)</span> <span class="k">if</span> <span class="n">is_abundant</span><span class="p">(</span><span class="n">n</span><span class="p">)]</span>
<span class="n">abundant_numbers_smaller_30000_set</span> <span class="o">=</span> <span class="nb">set</span><span class="p">(</span><span class="n">abundant_numbers_smaller_30000</span><span class="p">)</span>
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<p>Now that we have all abundant numbers in sorted order we can write a function which tests whether a number can be written as a sum of two abundant numbers.</p>
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<div class=" highlight hl-ipython3"><pre><span></span><span class="k">def</span> <span class="nf">is_sum_of_two_abundants</span><span class="p">(</span><span class="n">n</span><span class="p">):</span>
<span class="k">for</span> <span class="n">a</span> <span class="ow">in</span> <span class="n">abundant_numbers_smaller_30000</span><span class="p">:</span>
<span class="k">if</span> <span class="n">a</span> <span class="o">&gt;</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="k">return</span> <span class="kc">False</span>
<span class="n">r</span> <span class="o">=</span> <span class="n">n</span> <span class="o">-</span> <span class="n">a</span>
<span class="k">if</span> <span class="n">r</span> <span class="ow">in</span> <span class="n">abundant_numbers_smaller_30000_set</span><span class="p">:</span>
<span class="k">return</span> <span class="kc">True</span>
<span class="k">return</span> <span class="kc">False</span>
<span class="k">assert</span><span class="p">(</span><span class="n">is_sum_of_two_abundants</span><span class="p">(</span><span class="mi">24</span><span class="p">)</span> <span class="o">==</span> <span class="kc">True</span><span class="p">)</span>
<span class="k">assert</span><span class="p">(</span><span class="n">is_sum_of_two_abundants</span><span class="p">(</span><span class="mi">23</span><span class="p">)</span> <span class="o">==</span> <span class="kc">False</span><span class="p">)</span>
<span class="k">assert</span><span class="p">(</span><span class="n">is_sum_of_two_abundants</span><span class="p">(</span><span class="mi">28123</span><span class="p">)</span> <span class="o">==</span> <span class="kc">True</span><span class="p">)</span>
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<p>Now we try to brute force.</p>
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<div class=" highlight hl-ipython3"><pre><span></span><span class="n">s</span> <span class="o">=</span> <span class="nb">sum</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="mi">30000</span><span class="p">)</span> <span class="k">if</span> <span class="ow">not</span> <span class="n">is_sum_of_two_abundants</span><span class="p">(</span><span class="n">i</span><span class="p">)])</span>
<span class="nb">print</span><span class="p">(</span><span class="n">s</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">4179871</span><span class="p">)</span>
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<pre>4179871
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