bjorng
Advent of Code 2019 - Day 6
Note: This topic is to talk about Day 6 of the Advent of Code 2019.
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aaronnamba
A lot of people seem to have interpreted this as a graph problem. While not technically incorrect, it’s much simpler if you think of it as a tree. To solve part 2, forget Dijkstra, just find the common ancestor of YOU and SAN. 
bjorng
Here is my solution.
sb8244
Wait, digraph has the full answer in it??? Good to know for the future. I like your pt1 solution too.
My Solution ended up doing DFS/BFS manually. I got tripped up a bit by trying to do BFS without an Enumeration, and then I remembered that is a real hassle and that Enum.map would work great.
sasajuric
Here’s my solution, also powered by digraph.
For the fun of it, I also implemented a pure functional version.
JEG2
I love an excuse to use digraph!
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