Challenge accepted. Let’s do it!

First things first, let’s generate a list of all official Arch Linux packages. Fortunately, pacman, the best pragmatic package manager in existence, makes this a breeze.

Great, now let’s install it all!

sudo pacman -S $(pacman -Sql)

10 seconds later, you’ll find yourself with… unresolvable package conflicts detected?

OK, fine, let’s disable dependency checking then:

sudo pacman -Sdd $(pacman -Sql)

Nope, didn’t work. We have to do something about the conflicting packages!

We could resolve all the conflicts manually with an hour of work… or we could write a program!

Time for some algorithms!

It’s time to put our algorithms knowledge to good use. This is just a graph We can think of each package as a node in a graph and each conflict is an edge. Since we don’t care about dependency checks (which would make for a likely broken system but who cares), we don’t need to add any other edges to the graph.

For each edge, we need to pick at most one package, but not both. That sounds a lot like a maximum independent set!

Wait… it’s NP hard though? And we have up to 12000 nodes, so we’ll never be able to find the answer before the heat death of the universe, right?

Well, do we have 12000 connected nodes? No, since the largest connected component is probably only a few nodes. We aren’t going to have hundreds or thousands of packages all conflicting with each other. Therefore, we don’t need to prove P = NP to be able to find the maximum independent set of this particular graph. Phew!

Implementing this in Julia

We’re going to use Julia for implementing this algorithm, because Julia is Python but better. We first need to get a list of all packages:

pkgname = split(read(`pacman -Sql`, String))

N = length(pkgname)

Now, we’ll get info about each package, using multithreading to speed things up: (this is the slowest part of the program, actually)

struct Package
    provides::Vector{String}
    conflicts::Vector{String}
    size::Float64
end

pkginfo = Vector{Package}(undef, N)

Threads.@threads for i = 1:N
    pkg = pkgname[i]
    info = map(x -> split(replace(split(x, "n")[1], "None" => "")), split(read(`pacman -Si $pkg`, String), " : "))
    push!(info[10], pkg)
    pkginfo[i] = Package(info[10], info[13], parse(Float64, info[16][1]))
end

We need special handling for virtual packages:

providedby = Dict{String, Vector{Int}}()

for i = 1:N
	for p in pkginfo[i].provides
		p = split(p, "=")[1]
		if !(p in keys(providedby))
			providedby[p] = Vector{Int}()
		end
		push!(providedby[p], i)
	end
end

We can use this to construct the graph:

G = [Set{Int}() for i = 1:N]

for i = 1:N
	for p in pkginfo[i].conflicts
		if p in keys(providedby)
			for j in providedby[p]
				if j != i
					push!(G[i], j)
					push!(G[j], i)
				end
			end
		end
	end
end

Now we can find each connected component using BFS, and brute-force the maximum independent set by trying every subset of the nodes in that component. It’s implemented here using some bit manipulation trickery. Since the components are all very small, the brute-force finishes really quickly!

ans = BitSet(1:N)

used = BitSet()

for i = 1:N
	if !(i in used)
		push!(used, i)
		component = Vector{Int}()
		queue = Vector{Int}([i])
		while !isempty(queue)
			u = popfirst!(queue)
			push!(component, u)
			for v in G[u]
				if !(v in used)
					push!(used, v)
					push!(queue, v)
				end
			end
		end

		M = length(component)
		best = (0, 0.0, 0)
		for m = 1:(1<<M)-1
			good = true
			for j = 1:M
				if (m>>(j-1))&1 == 1
					for k = j+1:M
						if (m>>(k-1))&1 == 1 && component[j] in G[component[k]]
							good = false
						end
					end
				end
			end
			if !good
				continue
			end

			cnt = lengt