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								<p>I tried to solve the problem in a mathematical way, like a Fibonacci sequence. I believe there’s a better way than mine in math.</p>
<p><a href="https://github.com/qhwa/AdventOfCode/blob/master/lib/aoc/2021/day06.exs" rel="noopener nofollow ugc">solution</a></p> 
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								<p>Do you have your <code>DynamicSupervisor</code> solution somewhere? Thanks</p> 
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									<span>Creator of Cure</span>
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								<p><a href="https://github.com/am-kantox/aoc_2021/blob/master/lib/lanternfish.ex" class="onebox" target="_blank" rel="noopener nofollow ugc">https://github.com/am-kantox/aoc_2021/blob/master/lib/lanternfish.ex</a></p> 
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								<p>Cool. We tried to build a DynamicSupervisor solution today at work. Our solution was similar to yours. We quickly stumbled into the process limit of the VM. We could increase it enough for part 1 to work, but the maximum amount was still not enough for part 2 without going multi-node. Maybe we can explore it next time <img src="https://forum.elixirforum.com/images/emoji/apple/slight_smile.png?v=15" title=":slight_smile:" class="emoji" alt=":slight_smile:" loading="lazy" width="20" height="20"></p>
<p>Also we tried casting but then the count of fishes would be wrong.</p> 
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<p>process limit of the VM</p>
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<p>That’s exactly what I meant when I ended my comment with “and boom.”</p>
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								<p>I think there is an elegant mathematical way, using linear algebra: the amount of fish that have specific day can be modeled as a vector, where element 0 is the amount of fish that have 0 days left. Total number of fish at any given day is the sum of elements of that vector.</p>
<p>We can multiply that vector with a matrix <code>K</code>, that produces the next day’s fish counts. Here’s what <code>K</code> could look like (apologies for Octave/Matlab syntax, I used that to verify the approach):</p>
<pre data-code-wrap="elixir"><code class="lang-elixir">K0 = [
  0 0 0 0 0 0 1 0 1;
  1 0 0 0 0 0 0 0 0;
  0 1 0 0 0 0 0 0 0;
  0 0 1 0 0 0 0 0 0;
  0 0 0 1 0 0 0 0 0;
  0 0 0 0 1 0 0 0 0;
  0 0 0 0 0 1 0 0 0;
  0 0 0 0 0 0 1 0 0;
  0 0 0 0 0 0 0 1 0
];

# Fun fact: matrix for computing any initial condition to day 80 is:
K80 = K0 ^ 80;

# Get day 1 from day 0 counts per day:
v1 = v0 * K;

# Get day 80 from day 0 counts per day:
v80 = v0 * K80;
</code></pre>
<p>Full code here for my input: <a href="https://github.com/jarimatti/aoc-2021/blob/main/aoc2021_day6.m" rel="noopener nofollow ugc">aoc2021_day6.m</a></p>
<p>I’ve yet to figure out how to do this in Elixir fluently and it seems it’s time to learn some Nx. <img src="https://forum.elixirforum.com/images/emoji/apple/grin.png?v=15" title=":grin:" class="emoji" alt=":grin:" loading="lazy" width="20" height="20"></p> 
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								<p>OK, that was more fun than expected. (pun not intended)</p>
<p>Nx solution module over here: <a href="https://github.com/jarimatti/aoc-2021/blob/eadf586a84ab557f058cc782d579d3aa258c59eb/lib/aoc2021/day6.ex#L91" rel="noopener nofollow ugc">Aoc2021.Day6.LinAlg</a></p>
<p>Nx multiply and especially <code>Nx.power</code> are done by element, so I had to resort to reduce when multiplying the tensors. Otherwise it pretty much follows the Octave style. I like Nx so far, this was easy to install and play around with.</p> 
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								<p>Thank you for sharing this! This matrix is awesome!</p>
<p>I watched <a class="mention" href="/u/josevalim" rel="nofollow">@josevalim</a> 's <a href="https://www.twitch.tv/videos/1226814298" rel="noopener nofollow ugc">video on day 6</a>, and oh man, it was so amazing and blew my mind.  In the end, he tried the same matrix idea with Nx. It was so delightful to watch him thinking, solving, and explaining.</p> 
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