# Lanternfish and lots of data (AOC 2021 Day 6)

Today we had to model the growth of the lanternfish population (problem here).

Lanternfish spawn new lanternfish every seven days. The trick is that the original starting population consists of fish at different points in the cycle. For instance, if your input data was 3,2,4 then each fish would spawn a new fish in three, two, and four days respectively. The new fish would set their timers to 8 and start counting down to their spawn date on the next day and the original fish would reset it's timer to 6.

Of course, lanternfish never die so the population is going to get quite large.

At first glance this seems to be an easy problem to model. Read the fish into a list or array and start processing them a day at a time. Add fish to the end as needed.

I'm guessing this would work for part 1 but doubt it would work for part 2. Even before I started solving this just seemed like a problem where part 2 would blow away our data set getting much to large much too fast.

So, what to do? It turns out the problem is cyclical. Instead of looking at fish, we can look at days. We have an 8 day cycle. Instead of storing the fish, store the number of fish that will spawn on any given day.

That's the insight.

I started by playing around with mod and trying to figure out how to walk around the list while summing up fish but after a while my brain started to hurt. Instead, I just did things in the most straightforward way I could.

Fist, I made a list with my initial state. The example problem had a data set of `"3,4,3,1 2"` which led to this list:

``[0 1 1 2 1 0 0 0 0]``

Notice that I have 9 not 8 slots (indexed 0 - 8). That extra one at the end is where I move the fish that are spawning today (day 0) so they can start the cycle again.

From there, it's just following the rules:

1. Grab the number of fish to spawn today.
2. Move all the fish down a day - that is, the fish in day 1 go to day 0, day 2 to day 1 etc.
3. Add the new fish to slot 6 (as per the specification).
4. Add the number that spawned today to the end.

Repeat this for the requisite number of days and then add them all up.

The Clojure code for the complete solution is:

``````      (defn solve [data days]
(let [start-state (reduce
(fn [sofar next]
(update sofar next inc))
[0 0 0 0 0 0 0 0 0 ] data )]
(apply + (loop [i 0
gens start-state]
(if (< i days)
(let [last (first gens)
gens  (into []  ( drop 1 gens))
gens  (assoc gens 6 (+ (nth gens 6) last))
gens  (conj gens last)
]
(recur (inc i) gens)
)
gens)))))``````

For those of you not familiar with Clojure, that `reduce` in the let on the second line is how we take the data and build our starting state. For those of you who DO know clojure, I could probably have made it cleaner with a threading macro instead of the three step assignment to gens in the bottom let.

In any event, we walk away with a memory efficient time efficient solution.

There's probably some recurrence that could be set up and solved to do this all with math but since the mod stuff made my head hurt working on a recurrence equation would probably make it explode.

There's usually at least one problem like this in advent of code each year. Something where it looks like you'll have to calculate forever but it turns out you can set up a short repeating cycle to get the job done. Unfortunately, they're all too advanced for my average intro CS students. Sure, I could go over the problem and they'd kindof get it but I don't think the majority would really grok it. Alternatively they could just follow the rules given and led in class to setting up the cycle but I'd love to come up with one where they can really discover the magic.