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Advent of Code 2019 - Day 6

#2019#advent-of-code
bjorng

bjorng

Erlang Core Team

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

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. :slight_smile:

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Post #18
bjorng

bjorng

Erlang Core Team

Here is my solution.

5
Post #2
sb8244

sb8244

Author of Real-Time Phoenix

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.

4
Post #3
sasajuric

sasajuric

Author of Elixir In Action

Here’s my solution, also powered by digraph.

For the fun of it, I also implemented a pure functional version.

4
Post #7
JEG2

JEG2

Author of Designing Elixir Systems with OTP

I love an excuse to use digraph!

3
Post #17

Where Next?

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