Day 17 brought back Cellular Automata. It was a nice follow up to day 11. In my writeup I talked about data representation - how a Cellular Automoton like Conway's game of life is a nice 2D array project in a class like APCS-A but multi dimensional arrays are only one way to represent a cellular automaton. Day 17 really drove that home.
The actual rules were pretty simple - if a cell is active and has 2 or 3 active neighbors it stays active. If it's inactive and has three active it becomes active. Otherwise the cell is inactive.
The catch for part 1 is that this CS is in three dimensions. Each and a cell's neighbors are defined as all coordinates that differ by one in any of the three dimensions. So, if a cell is at an (x,y,z) location it's neighbors will be at (x+1,y,z), (x-1,y,z), (x+1,y+1,z), (x+1,y-1,z), etc. for 26 neighbors in all.
You could use a list within a list within a list or a three dimensional array to represent your world but that's tricky and error prone. What's worse, part 2 took the CA into the fourth dimension.
Better is to just keep a list or set of active cells. Then the problem becomes pretty easy. You need to be able to:
Find all of a cell's neighbors - this is pretty easy because you can iterate over all the +1 and -1 possibilities for each of the x, y, and z values.
Find all the potential cells for the next state - this is also pretty easy because it's the set of all cells that are currently active along with all of their neighbors.
Count a given cell's active neighbors - this is easy once you've done the find neighbors routine.
A way to test if a cell is active which is just checking to see if it's in your active cells list or set.
Then, it's pretty easy to run the CA:
# pythonesque pseudocode potential_cells = find_all_neighbors(current_active_cells) new_cells =  for cell in potential_cells: n = count_neighbors(cell) if is_active(cell) and (n==2 or n=3): new_cells.append(cell) elif (not is_active(cell)) and n==3: new_cells.append(cell)
Then you just have to run generate new states until you get the answer.
Part 2 extended the CA to 4 dimensions. If you had a multidimensional array this would get super message but with a list of active cells, the changes are minimal - just add an extra coordinate, update getting the neighbors and you're good to go.
This is a case of where thinking through your data representation can be a big win.
Clojure code here.
Day 18 was all about evaluating math expressions. For part 1 you had parenthesized expressions consisting of numbers * and + that you had to evaluate but you had to do it by first doing parens then left to right - multiplication was not a higher precedence.
This sounds like a parsing first problem but it turns out I was able to exploit some of Clojure's language features. Looking at the subreddit after solving it seems that a bunch of other languages also have features that could be exploited.
Clojure represents data (and programs) as S-Expressions - basically
stuff in parens. As a prefix language, instead of writing 10+20, in
Clojure you'd write
(+ 10 20), that is run the plus function on 10
and 20. If you have something lie
(+ 10 (* 20 3)), Clojure has to
evaluate the inner S-Expression (sexp) before it can add that to +10
so Clojure can do the parsing for us. We can take an input string and
convert it to an sexp using
read-string but if we just try to do
(read-string "1 + 2 + 3") we'd get an error because "1 + 2 + 3"
isn't a valid sexp so we just surround it by parens:
(def equation-sexp (read-string (str "(" "1 + 2 * 3 + (4 * 5 )" ")")))
The above would leave us with the sexp (1 + 2 * 3 + (4 * 5 )).
Next, forgetting the inner parens, we can write a function that will evaluate an sexp of the form (1 + 2 * 3 + …) etc. Basically, this can be done with a reduce. Start with the first value then take the rest of the list two at a time, the first of each pair is an operator and the second is an operand so apply the operand to the other number in the pair and your overall result so far.
In Clojure it looks like this:
(defn part1-eval [f & r] (reduce (fn [ans [op next]] (apply op [ans next] )) f (partition 2 r)))
Next, we insert that function name to the start of each sexp so (1 + 2 * 3 + (4 * 5*)) becomes (part1-eval 1 + 2 * 3 + (part1-eval 4 * 5)). Finally we can do a Clojure eval on this form which will run part1-eval on the rest of the sexp which will first run part1-eval on the 4 * 5, that will return the 20 and then the first part1-eval will finish it's calculations to give you the answer.
Part 2 was similar but there you had to perform addition before multiplication. All that was necessary was write a part2-eval function that would stand in for the part1-eval.
The idea is to take an sexp like (1 + 2 * 3 + 4 * 5) we first split this list around the * this gives us (1 + 2) () (3 + 4) () (5). We then filter this to remove the non numbers which gives (1 2) () (3 4) (5) (). Then we remove the empty lists: (1 2) (3 4) (5). Add the elements of each list: 3 7 5 and then multiply them together.
All the code is here.
I like day 17 a lot or some variant for students to discuss data representations but I think 18 is a little more advanced and probably wouldn't touch it in an early CS class - it was fun to work through though :-).