Taking advantage of a neat trick for using $ rather than a lambda
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90 <header><div><h1 class="title-global"><a href="/">Advent of Code</a></h1><nav><ul><li><a href="/2017/about">[About]</a></li><li><a href="/2017/support">[AoC++]</a></li><li><a href="/2017/events">[Events]</a></li><li><a href="/2017/settings">[Settings]</a></li><li><a href="/2017/auth/logout">[Log Out]</a></li></ul></nav><div class="user">Neil Smith <span class="supporter">(AoC++)</span> <span class="star-count">36*</span></div></div><div><h1 class="title-event">&nbsp;&nbsp;&nbsp;<span class="title-event-wrap">0x0000|</span><a href="/2017">2017</a><span class="title-event-wrap"></span></h1><nav><ul><li><a href="/2017">[Calendar]</a></li><li><a href="/2017/leaderboard">[Leaderboard]</a></li><li><a href="/2017/stats">[Stats]</a></li><li><a href="/2017/sponsors">[Sponsors]</a></li></ul></nav></div></header>
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99 <article class="day-desc"><h2>--- Day 17: Spinlock ---</h2><p>Suddenly, whirling in the distance, you notice what looks like a massive, <span title="You know, as opposed to all those non-pixelated hurricanes you see on TV.">pixelated hurricane</span>: a deadly <a href="https://en.wikipedia.org/wiki/Spinlock">spinlock</a>. This spinlock isn't just consuming computing power, but memory, too; vast, digital mountains are being ripped from the ground and consumed by the vortex.</p>
100 <p>If you don't move quickly, fixing that printer will be the least of your problems.</p>
101 <p>This spinlock's algorithm is simple but efficient, quickly consuming everything in its path. It starts with a circular buffer containing only the value <code>0</code>, which it marks as the <em>current position</em>. It then steps forward through the circular buffer some number of steps (your puzzle input) before inserting the first new value, <code>1</code>, after the value it stopped on. The inserted value becomes the <em>current position</em>. Then, it steps forward from there the same number of steps, and wherever it stops, inserts after it the second new value, <code>2</code>, and uses that as the new <em>current position</em> again.</p>
102 <p>It repeats this process of <em>stepping forward</em>, <em>inserting a new value</em>, and <em>using the location of the inserted value as the new current position</em> a total of <code><em>2017</em></code> times, inserting <code>2017</code> as its final operation, and ending with a total of <code>2018</code> values (including <code>0</code>) in the circular buffer.</p>
103 <p>For example, if the spinlock were to step <code>3</code> times per insert, the circular buffer would begin to evolve like this (using parentheses to mark the current position after each iteration of the algorithm):</p>
104 <ul>
105 <li><code>(0)</code>, the initial state before any insertions.</li>
106 <li><code>0&nbsp;(1)</code>: the spinlock steps forward three times (<code>0</code>, <code>0</code>, <code>0</code>), and then inserts the first value, <code>1</code>, after it. <code>1</code> becomes the current position.</li>
107 <li><code>0&nbsp;(2)&nbsp;1</code>: the spinlock steps forward three times (<code>0</code>, <code>1</code>, <code>0</code>), and then inserts the second value, <code>2</code>, after it. <code>2</code> becomes the current position.</li>
108 <li><code>0&nbsp;&nbsp;2&nbsp;(3)&nbsp;1</code>: the spinlock steps forward three times (<code>1</code>, <code>0</code>, <code>2</code>), and then inserts the third value, <code>3</code>, after it. <code>3</code> becomes the current position.</li>
109 </ul>
110 <p>And so on:</p>
111 <ul>
112 <li><code>0&nbsp;&nbsp;2&nbsp;(4)&nbsp;3&nbsp;&nbsp;1</code></li>
113 <li><code>0&nbsp;(5)&nbsp;2&nbsp;&nbsp;4&nbsp;&nbsp;3&nbsp;&nbsp;1</code></li>
114 <li><code>0&nbsp;&nbsp;5&nbsp;&nbsp;2&nbsp;&nbsp;4&nbsp;&nbsp;3&nbsp;(6)&nbsp;1</code></li>
115 <li><code>0&nbsp;&nbsp;5&nbsp;(7)&nbsp;2&nbsp;&nbsp;4&nbsp;&nbsp;3&nbsp;&nbsp;6&nbsp;&nbsp;1</code></li>
116 <li><code>0&nbsp;&nbsp;5&nbsp;&nbsp;7&nbsp;&nbsp;2&nbsp;&nbsp;4&nbsp;&nbsp;3&nbsp;(8)&nbsp;6&nbsp;&nbsp;1</code></li>
117 <li><code>0&nbsp;(9)&nbsp;5&nbsp;&nbsp;7&nbsp;&nbsp;2&nbsp;&nbsp;4&nbsp;&nbsp;3&nbsp;&nbsp;8&nbsp;&nbsp;6&nbsp;&nbsp;1</code></li>
118 </ul>
119 <p>Eventually, after 2017 insertions, the section of the circular buffer near the last insertion looks like this:</p>
120 <pre><code>1512 1134 151 (2017) 638 1513 851</code></pre>
121 <p>Perhaps, if you can identify the value that will ultimately be <em>after</em> the last value written (<code>2017</code>), you can short-circuit the spinlock. In this example, that would be <code>638</code>.</p>
122 <p><em>What is the value after <code>2017</code></em> in your completed circular buffer?</p>
123 </article>
124 <p>Your puzzle answer was <code>1025</code>.</p><article class="day-desc"><h2>--- Part Two ---</h2><p>The spinlock does not short-circuit. Instead, it gets <em>more</em> angry. At least, you assume that's what happened; it's spinning significantly faster than it was a moment ago.</p>
125 <p>You have good news and bad news.</p>
126 <p>The good news is that you have improved calculations for how to stop the spinlock. They indicate that you actually need to identify <em>the value after <code>0</code></em> in the current state of the circular buffer.</p>
127 <p>The bad news is that while you were determining this, the spinlock has just finished inserting its fifty millionth value (<code>50000000</code>).</p>
128 <p><em>What is the value after <code>0</code></em> the moment <code>50000000</code> is inserted?</p>
129 </article>
130 <p>Your puzzle answer was <code>37803463</code>.</p><p class="day-success">Both parts of this puzzle are complete! They provide two gold stars: **</p>
131 <p>At this point, you should <a href="/2017">return to your advent calendar</a> and try another puzzle.</p>
132 <p>Your puzzle input was <code class="puzzle-input">366</code>.</p>
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