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C'est la Z

A Teacher Looks at Advent of Code 2020 - Days 21 and 24

Day 24 didn't take that much time so I had a chance to go back and finish day 21.

As usual, all my code is up here.

Day 21

Problem Link

It took me a while to get my head around this example. You have a set of recipes. Each recipe has a list of ingredients and a list of allergens. Your job for part 1 was basically to determine which ingredients don't contain any allergens. Part 2 had you determining which ingredient had which allergen.

Part of what I think made this hard was that ingredients were just random strings of letters as opposed to actual ingredients or at least pronounceable words or letter strings.

The key part of the description as that each allergen was present in exactly one ingredient and each ingredient had at most one allergen.

So, for part 1, the tact was for each allergen, take all the recipes that had it. In the example, the first and second recipes had dairy (one indexing). Since only one ingredient can have the dairy allergen the that ingredient must be in both lists.

If you take all the recipes that have the dairy allergen and take each of their lists of ingredients as a set and find all their intersections, you'll be left with all the ingredients that can have the dairy allergen.

Do this with all the allergens and you'll have all the ingredients that could contain them. The other igredients are safe and lead to the answer to part 1.

Part 2 had you identifying each ingredient / allergen pair. Fortunately, the data set we ended up with for part 1 - each allergen and it's list of possible ingredients was made to order. One of them had only one ingredient per allergen. We could identify that one and then remove that ingredient from the remaining allergen's lists. Then repeat to find the next one and then the next. This is similar to the solution to day 16. Of course this isn't a general solution but fortunately the data was all set up for us.

Day 24

Problem Link

Back to Cellular Automata!!!!!! The catch this time is the world is a hex grid. This means that each cell has six neighbors so using a 2D array or similar representation seems like an even worse idea than it was for the 3D and 4D problems earlier this month.

Even before representing the world parsing was an issue. Since a given cell can have neighbors to the east, west, northeast, northwest, southeast or southwest, instructions are given with a line of "moves" describing a tile to flip starting from 0,0.

For example, the line "EEE" would move three east from the origin and flip that tile while "ESEE" would move one east, one southest and then one further east and flipt the tile there.

Fortunately, this wasn't too bad. As we traverse down the line, if we see an e or a w then the instruction is one character so we add the instruction to a list of steps an then continue. Otherwise the instruction is 2 characters so we take 2 and add the instruction adn then proceed.

Next was the data representation. It seemed that a list of live cells would again be the best solution. I figured on using this mapping:

E (-2,0)
W (2,0)
NE (-1,-1)
NW (1,-1)
SE (-1,1)
SW (1,1)

I later found out that this is called "double coordinates." Since we're not storing the full hex grid there's no waste anyway and I wasn't planning on making a visualization so I didn't worry about how this would map to an actual screen.

Now, finding the coordinate of a tile was pretty easy. In Clojure it's:

;; assume deltas is a lookup dictionary of the above mapping

  (defn get-tile-location [steps]
    (reduce (fn [loc step]
              (map + loc (deltas  step))
              ) [0 0] steps))

In Python it would look more like this:

  # assume deltas is a dictionary with the above mapping
  # and steps is a list of instructions ["e","w","se", etc]
  for step in steps:
      loc = [sum(x) for x in zip(loc, deltas[step])]

We can now convert any input line of instructions to a coordinate. If we write a routine to flip a tile we can now set up our world and answer part 1.

Part 2 involved basically turning the part 1 world into a cellular automaton. Fortunately, this was easy given the solution to earlier CA problems. It as just a matter of updating the get-neighbors routine and the rule to go from one generation to the next and we're done.

I enjoyed this - it was a nice twist on the earlier CA questions. First we had a simple CA, then higher dimensions and now changing the layout of the world but still within 2 dimensions. Once we had a way of representing a tile and mapping from a tile to its neighbors we had already solved the problem.

One more day to go but it's been a nice run of problems.

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