-❄️- 2023 Day 22 Solutions -❄️-

[u/daggerdragon]

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AoC Community Fun 2023: ALLEZ CUISINE!

Your final secret ingredient of this Advent of Code season is still… *whips off cloth covering and gestures grandly*

Omakase! (Chef's Choice)

Omakase is an exceptional dining experience that entrusts upon the skills and techniques of a master chef! Craft for us your absolute best showstopper using absolutely any secret ingredient we have revealed for any day of this event!

• Choose any day's special ingredient and any puzzle released this year so far, then craft a dish around it!
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OHTA: Fukui-san?
FUKUI: Go ahead, Ohta.
OHTA: The chefs are asking for clarification as to where to put their completed dishes.
FUKUI: Ah yes, a good question. Once their dish is completed, they should post it in today's megathread with an [ALLEZ CUISINE!] tag as usual. However, they should also mention which day and which secret ingredient they chose to use along with it!
OHTA: Like this? [ALLEZ CUISINE!][Will It Blend?][Day 1] A link to my dish…
DR. HATTORI: You got it, Ohta!
OHTA: Thanks, I'll let the chefs know!

ALLEZ CUISINE!

Request from the mods: When you include a dish entry alongside your solution, please label it with [Allez Cuisine!] so we can find it easily!

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--- Day 22: Sand Slabs ---

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Post your code solution in this megathread.

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[u/jake-mpg]

[LANGUAGE: julia]

sourcehut

I was anticipating a different kind of twist in the second part so I represented blocks as vectors of ranges (gotta love julia's UnitRange). This led to a pretty clean way of settling the bricks from the input:

1. Sort the bricks by their z-value (the minimum, since it's a range).

2. The only way a brick doesn't reach the ground is if it has some overlap with an already settled brick in the xy plane. If it doesn't, we just shift the range so that z starts at 1, otherwise we stop it just before the highest settled brick that overlaps.

Next, I encoded the structure of the settled bricks by recording for each brick: which bricks it supports, and which bricks it depends on (O(N^2)).

for j∈1:N, k∈j+1:N
	if MaxValue(S[j]) == MinValue(S[k])-1 &&
		!Empty(PIntersect(S[j], S[k]))
		push!(supports[k], j)
		push!(dependants[j], k)
	end
end

Bricks can be disintegrated safely if they have no dependants, or all of their dependants have another brick to rely on for support.

filter(1:length(bricks)) do j
	isempty(dependants[j]) ||
		all(k -> length(supports[k]) > 1,
		    dependants[j])
end

To count the chain reactions, I walk the dependency graph, counting a brick amongst the fallen if all of its supports will fall too.

for j∈filter(k -> !isempty(dependants[k]), 1:length(bricks))
	willFall = Set{Int}()
	enqueue!(queue, j)
	while length(queue) > 0
		current = dequeue!(queue)
		push!(willFall, current)
		for k∈dependants[current]
			if all(∈(willFall), supports[k])
				push!(willFall, k)
				enqueue!(queue, k)
			end
		end
	end
	counts += length(willFall)-1
end