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<div class="section" id="advent-of-code-2017">
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<h1>Advent of Code 2017<a class="headerlink" href="#advent-of-code-2017" title="Permalink to this headline">¶</a></h1>
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<div class="section" id="december-1st">
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<h2>December 1st<a class="headerlink" href="#december-1st" title="Permalink to this headline">¶</a></h2>
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<p>[Given] a sequence of digits (your puzzle input) and find the sum of all
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digits that match the next digit in the list. The list is circular, so
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the digit after the last digit is the first digit in the list.</p>
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<p>For example:</p>
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<ul class="simple">
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<li>1122 produces a sum of 3 (1 + 2) because the first digit (1) matches
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the second digit and the third digit (2) matches the fourth digit.</li>
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<li>1111 produces 4 because each digit (all 1) matches the next.</li>
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<li>1234 produces 0 because no digit matches the next.</li>
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<li>91212129 produces 9 because the only digit that matches the next one
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is the last digit, 9.</li>
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</ul>
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<div class="code ipython2 highlight-default notranslate"><div class="highlight"><pre><span></span><span class="kn">from</span> <span class="nn">notebook_preamble</span> <span class="k">import</span> <span class="n">J</span><span class="p">,</span> <span class="n">V</span><span class="p">,</span> <span class="n">define</span>
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</pre></div>
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</div>
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<p>I’ll assume the input is a Joy sequence of integers (as opposed to a
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string or something else.)</p>
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<p>We might proceed by creating a word that makes a copy of the sequence
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with the first item moved to the last, and zips it with the original to
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make a list of pairs, and a another word that adds (one of) each pair to
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a total if the pair matches.</p>
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<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">AoC2017</span><span class="o">.</span><span class="mi">1</span> <span class="o">==</span> <span class="n">pair_up</span> <span class="n">total_matches</span>
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</pre></div>
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</div>
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<p>Let’s derive <code class="docutils literal notranslate"><span class="pre">pair_up</span></code>:</p>
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<div class="highlight-default notranslate"><div class="highlight"><pre><span></span> <span class="p">[</span><span class="n">a</span> <span class="n">b</span> <span class="n">c</span><span class="p">]</span> <span class="n">pair_up</span>
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<span class="o">-------------------------</span>
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<span class="p">[[</span><span class="n">a</span> <span class="n">b</span><span class="p">]</span> <span class="p">[</span><span class="n">b</span> <span class="n">c</span><span class="p">]</span> <span class="p">[</span><span class="n">c</span> <span class="n">a</span><span class="p">]]</span>
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</pre></div>
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</div>
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<p>Straightforward (although the order of each pair is reversed, due to the
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way <code class="docutils literal notranslate"><span class="pre">zip</span></code> works, but it doesn’t matter for this program):</p>
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<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="p">[</span><span class="n">a</span> <span class="n">b</span> <span class="n">c</span><span class="p">]</span> <span class="n">dup</span>
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<span class="p">[</span><span class="n">a</span> <span class="n">b</span> <span class="n">c</span><span class="p">]</span> <span class="p">[</span><span class="n">a</span> <span class="n">b</span> <span class="n">c</span><span class="p">]</span> <span class="n">uncons</span> <span class="n">swap</span>
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<span class="p">[</span><span class="n">a</span> <span class="n">b</span> <span class="n">c</span><span class="p">]</span> <span class="p">[</span><span class="n">b</span> <span class="n">c</span><span class="p">]</span> <span class="n">a</span> <span class="n">unit</span> <span class="n">concat</span>
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<span class="p">[</span><span class="n">a</span> <span class="n">b</span> <span class="n">c</span><span class="p">]</span> <span class="p">[</span><span class="n">b</span> <span class="n">c</span> <span class="n">a</span><span class="p">]</span> <span class="nb">zip</span>
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<span class="p">[[</span><span class="n">b</span> <span class="n">a</span><span class="p">]</span> <span class="p">[</span><span class="n">c</span> <span class="n">b</span><span class="p">]</span> <span class="p">[</span><span class="n">a</span> <span class="n">c</span><span class="p">]]</span>
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</pre></div>
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</div>
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<div class="code ipython2 highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">define</span><span class="p">(</span><span class="s1">'pair_up == dup uncons swap unit concat zip'</span><span class="p">)</span>
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</pre></div>
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</div>
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<div class="code ipython2 highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">J</span><span class="p">(</span><span class="s1">'[1 2 3] pair_up'</span><span class="p">)</span>
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</pre></div>
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</div>
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<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="p">[[</span><span class="mi">2</span> <span class="mi">1</span><span class="p">]</span> <span class="p">[</span><span class="mi">3</span> <span class="mi">2</span><span class="p">]</span> <span class="p">[</span><span class="mi">1</span> <span class="mi">3</span><span class="p">]]</span>
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</pre></div>
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</div>
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<div class="code ipython2 highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">J</span><span class="p">(</span><span class="s1">'[1 2 2 3] pair_up'</span><span class="p">)</span>
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</pre></div>
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</div>
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<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="p">[[</span><span class="mi">2</span> <span class="mi">1</span><span class="p">]</span> <span class="p">[</span><span class="mi">2</span> <span class="mi">2</span><span class="p">]</span> <span class="p">[</span><span class="mi">3</span> <span class="mi">2</span><span class="p">]</span> <span class="p">[</span><span class="mi">1</span> <span class="mi">3</span><span class="p">]]</span>
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</pre></div>
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</div>
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<p>Now we need to derive <code class="docutils literal notranslate"><span class="pre">total_matches</span></code>. It will be a <code class="docutils literal notranslate"><span class="pre">step</span></code> function:</p>
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<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">total_matches</span> <span class="o">==</span> <span class="mi">0</span> <span class="n">swap</span> <span class="p">[</span><span class="n">F</span><span class="p">]</span> <span class="n">step</span>
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</pre></div>
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</div>
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<p>Where <code class="docutils literal notranslate"><span class="pre">F</span></code> will have the pair to work with, and it will basically be a
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<code class="docutils literal notranslate"><span class="pre">branch</span></code> or <code class="docutils literal notranslate"><span class="pre">ifte</span></code>.</p>
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<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">total</span> <span class="p">[</span><span class="n">n</span> <span class="n">m</span><span class="p">]</span> <span class="n">F</span>
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</pre></div>
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</div>
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<p>It will probably be easier to write if we dequote the pair:</p>
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<div class="highlight-default notranslate"><div class="highlight"><pre><span></span> total [n m] i F′
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----------------------
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total n m F′
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</pre></div>
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</div>
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<p>Now <code class="docutils literal notranslate"><span class="pre">F′</span></code> becomes just:</p>
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<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">total</span> <span class="n">n</span> <span class="n">m</span> <span class="p">[</span><span class="o">=</span><span class="p">]</span> <span class="p">[</span><span class="n">pop</span> <span class="o">+</span><span class="p">]</span> <span class="p">[</span><span class="n">popop</span><span class="p">]</span> <span class="n">ifte</span>
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</pre></div>
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</div>
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<p>So:</p>
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<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">F</span> <span class="o">==</span> <span class="n">i</span> <span class="p">[</span><span class="o">=</span><span class="p">]</span> <span class="p">[</span><span class="n">pop</span> <span class="o">+</span><span class="p">]</span> <span class="p">[</span><span class="n">popop</span><span class="p">]</span> <span class="n">ifte</span>
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</pre></div>
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</div>
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<p>And thus:</p>
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<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">total_matches</span> <span class="o">==</span> <span class="mi">0</span> <span class="n">swap</span> <span class="p">[</span><span class="n">i</span> <span class="p">[</span><span class="o">=</span><span class="p">]</span> <span class="p">[</span><span class="n">pop</span> <span class="o">+</span><span class="p">]</span> <span class="p">[</span><span class="n">popop</span><span class="p">]</span> <span class="n">ifte</span><span class="p">]</span> <span class="n">step</span>
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</pre></div>
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</div>
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<div class="code ipython2 highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">define</span><span class="p">(</span><span class="s1">'total_matches == 0 swap [i [=] [pop +] [popop] ifte] step'</span><span class="p">)</span>
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</pre></div>
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</div>
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<div class="code ipython2 highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">J</span><span class="p">(</span><span class="s1">'[1 2 3] pair_up total_matches'</span><span class="p">)</span>
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</pre></div>
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</div>
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<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="mi">0</span>
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</pre></div>
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</div>
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<div class="code ipython2 highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">J</span><span class="p">(</span><span class="s1">'[1 2 2 3] pair_up total_matches'</span><span class="p">)</span>
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</pre></div>
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</div>
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<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="mi">2</span>
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</pre></div>
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</div>
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<p>Now we can define our main program and evaluate it on the examples.</p>
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<div class="code ipython2 highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">define</span><span class="p">(</span><span class="s1">'AoC2017.1 == pair_up total_matches'</span><span class="p">)</span>
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</pre></div>
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</div>
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<div class="code ipython2 highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">J</span><span class="p">(</span><span class="s1">'[1 1 2 2] AoC2017.1'</span><span class="p">)</span>
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</pre></div>
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</div>
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<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="mi">3</span>
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</pre></div>
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</div>
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<div class="code ipython2 highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">J</span><span class="p">(</span><span class="s1">'[1 1 1 1] AoC2017.1'</span><span class="p">)</span>
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</pre></div>
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</div>
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<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="mi">4</span>
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</pre></div>
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</div>
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<div class="code ipython2 highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">J</span><span class="p">(</span><span class="s1">'[1 2 3 4] AoC2017.1'</span><span class="p">)</span>
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</pre></div>
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</div>
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<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="mi">0</span>
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</pre></div>
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</div>
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<div class="code ipython2 highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">J</span><span class="p">(</span><span class="s1">'[9 1 2 1 2 1 2 9] AoC2017.1'</span><span class="p">)</span>
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</pre></div>
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</div>
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<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="mi">9</span>
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</pre></div>
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</div>
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<div class="code ipython2 highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">J</span><span class="p">(</span><span class="s1">'[9 1 2 1 2 1 2 9] AoC2017.1'</span><span class="p">)</span>
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</pre></div>
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</div>
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<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="mi">9</span>
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</pre></div>
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</div>
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<div class="highlight-default notranslate"><div class="highlight"><pre><span></span> <span class="n">pair_up</span> <span class="o">==</span> <span class="n">dup</span> <span class="n">uncons</span> <span class="n">swap</span> <span class="n">unit</span> <span class="n">concat</span> <span class="nb">zip</span>
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<span class="n">total_matches</span> <span class="o">==</span> <span class="mi">0</span> <span class="n">swap</span> <span class="p">[</span><span class="n">i</span> <span class="p">[</span><span class="o">=</span><span class="p">]</span> <span class="p">[</span><span class="n">pop</span> <span class="o">+</span><span class="p">]</span> <span class="p">[</span><span class="n">popop</span><span class="p">]</span> <span class="n">ifte</span><span class="p">]</span> <span class="n">step</span>
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<span class="n">AoC2017</span><span class="o">.</span><span class="mi">1</span> <span class="o">==</span> <span class="n">pair_up</span> <span class="n">total_matches</span>
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</pre></div>
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</div>
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<p>Now the paired digit is “halfway” round.</p>
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<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="p">[</span><span class="n">a</span> <span class="n">b</span> <span class="n">c</span> <span class="n">d</span><span class="p">]</span> <span class="n">dup</span> <span class="n">size</span> <span class="mi">2</span> <span class="o">/</span> <span class="p">[</span><span class="n">drop</span><span class="p">]</span> <span class="p">[</span><span class="n">take</span> <span class="n">reverse</span><span class="p">]</span> <span class="n">cleave</span> <span class="n">concat</span> <span class="nb">zip</span>
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</pre></div>
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</div>
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<div class="code ipython2 highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">J</span><span class="p">(</span><span class="s1">'[1 2 3 4] dup size 2 / [drop] [take reverse] cleave concat zip'</span><span class="p">)</span>
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</pre></div>
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</div>
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<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="p">[[</span><span class="mi">3</span> <span class="mi">1</span><span class="p">]</span> <span class="p">[</span><span class="mi">4</span> <span class="mi">2</span><span class="p">]</span> <span class="p">[</span><span class="mi">1</span> <span class="mi">3</span><span class="p">]</span> <span class="p">[</span><span class="mi">2</span> <span class="mi">4</span><span class="p">]]</span>
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</pre></div>
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</div>
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<p>I realized that each pair is repeated…</p>
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<div class="code ipython2 highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">J</span><span class="p">(</span><span class="s1">'[1 2 3 4] dup size 2 / [drop] [take reverse] cleave zip'</span><span class="p">)</span>
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</pre></div>
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</div>
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<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="p">[</span><span class="mi">1</span> <span class="mi">2</span> <span class="mi">3</span> <span class="mi">4</span><span class="p">]</span> <span class="p">[[</span><span class="mi">1</span> <span class="mi">3</span><span class="p">]</span> <span class="p">[</span><span class="mi">2</span> <span class="mi">4</span><span class="p">]]</span>
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</pre></div>
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</div>
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<div class="code ipython2 highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">define</span><span class="p">(</span><span class="s1">'AoC2017.1.extra == dup size 2 / [drop] [take reverse] cleave zip swap pop total_matches 2 *'</span><span class="p">)</span>
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</pre></div>
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</div>
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<div class="code ipython2 highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">J</span><span class="p">(</span><span class="s1">'[1 2 1 2] AoC2017.1.extra'</span><span class="p">)</span>
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</pre></div>
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</div>
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<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="mi">6</span>
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||
</pre></div>
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</div>
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<div class="code ipython2 highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">J</span><span class="p">(</span><span class="s1">'[1 2 2 1] AoC2017.1.extra'</span><span class="p">)</span>
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</pre></div>
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</div>
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<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="mi">0</span>
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</pre></div>
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</div>
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<div class="code ipython2 highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">J</span><span class="p">(</span><span class="s1">'[1 2 3 4 2 5] AoC2017.1.extra'</span><span class="p">)</span>
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</pre></div>
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</div>
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<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="mi">4</span>
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</pre></div>
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</div>
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</div>
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</div>
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<div class="section" id="refactor-ftw">
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<h1>Refactor FTW<a class="headerlink" href="#refactor-ftw" title="Permalink to this headline">¶</a></h1>
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<p>With Joy a great deal of the heuristics from Forth programming carry
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over nicely. For example, refactoring into small, well-scoped commands
|
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with mnemonic names…</p>
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<div class="highlight-default notranslate"><div class="highlight"><pre><span></span> <span class="n">rotate_seq</span> <span class="o">==</span> <span class="n">uncons</span> <span class="n">swap</span> <span class="n">unit</span> <span class="n">concat</span>
|
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<span class="n">pair_up</span> <span class="o">==</span> <span class="n">dup</span> <span class="n">rotate_seq</span> <span class="nb">zip</span>
|
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<span class="n">add_if_match</span> <span class="o">==</span> <span class="p">[</span><span class="o">=</span><span class="p">]</span> <span class="p">[</span><span class="n">pop</span> <span class="o">+</span><span class="p">]</span> <span class="p">[</span><span class="n">popop</span><span class="p">]</span> <span class="n">ifte</span>
|
||
<span class="n">total_matches</span> <span class="o">==</span> <span class="p">[</span><span class="n">i</span> <span class="n">add_if_match</span><span class="p">]</span> <span class="n">step_zero</span>
|
||
|
||
<span class="n">AoC2017</span><span class="o">.</span><span class="mi">1</span> <span class="o">==</span> <span class="n">pair_up</span> <span class="n">total_matches</span>
|
||
|
||
<span class="n">half_of_size</span> <span class="o">==</span> <span class="n">dup</span> <span class="n">size</span> <span class="mi">2</span> <span class="o">/</span>
|
||
<span class="n">split_at</span> <span class="o">==</span> <span class="p">[</span><span class="n">drop</span><span class="p">]</span> <span class="p">[</span><span class="n">take</span> <span class="n">reverse</span><span class="p">]</span> <span class="n">cleave</span>
|
||
<span class="n">pair_up</span><span class="o">.</span><span class="n">extra</span> <span class="o">==</span> <span class="n">half_of_size</span> <span class="n">split_at</span> <span class="nb">zip</span> <span class="n">swap</span> <span class="n">pop</span>
|
||
|
||
<span class="n">AoC2017</span><span class="o">.</span><span class="mf">1.</span><span class="n">extra</span> <span class="o">==</span> <span class="n">pair_up</span><span class="o">.</span><span class="n">extra</span> <span class="n">total_matches</span> <span class="mi">2</span> <span class="o">*</span>
|
||
</pre></div>
|
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|
||
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<span xmlns:dct="http://purl.org/dc/terms/" property="dct:title">Thun Documentation</span> by <a xmlns:cc="http://creativecommons.org/ns#" href="https://joypy.osdn.io/" property="cc:attributionName" rel="cc:attributionURL">Simon Forman</a> is licensed under a <a rel="license" href="http://creativecommons.org/licenses/by-nc-sa/4.0/">Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License</a>.<br />Based on a work at <a xmlns:dct="http://purl.org/dc/terms/" href="https://osdn.net/projects/joypy/" rel="dct:source">https://osdn.net/projects/joypy/</a>.
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