My attempts to work through the 2021 Advent of Code problems.
git clone
Log | Files | Refs | README | LICENSE

commit e13970376c770cdd6c5fe6a2a50f3a99a4e01387
parent 726e9f7efd14ce878e96b649fe4ffc41fa8cb4d4
Author: Eamon Caddigan <>
Date:   Tue,  7 Dec 2021 13:25:09 -0500

I feel like I owe folks an explanation of why this solution is correct

Diffstat: | 17++++++++++++++++-
1 file changed, 16 insertions(+), 1 deletion(-)

diff --git a/ b/ @@ -1,7 +1,22 @@ #!/usr/bin/env python -"""Advent of Code 2021, day 7 (part 2): optimize 'crab submarine' alignment to +"""Advent of Code 2021, day 7 (part 1): optimize 'crab submarine' alignment to minimize fuel""" +# How do we know that the median minimizes sum(abs(start_positions - +# end_position))? Let's do a quick and dirty proof in the Advent of Code +# universe. +# Suppose that we've already moved all the crab submarines to the median +# position, x_m, but there's actually a better position that's greater than +# x_m, which we'll call x_o. If we had moved the submarines to position x_o +# instead of x_m, all of the subs with positions less than or equal to x_m +# would spend the same amount of additional fuel, and all of the subs with +# positions greater than or equal to x_o would save that same amount of fuel +# (we can forget about the subs between x_m and x_o for now). However, by the +# definition of the median, the majority of subs have positions less than or +# equal to x_m, which means moving to x_o will necessarily cost the crab sub +# fleet more fuel than it saves. The same logic applies to an x_o less than +# x_m, so we have proven by contradition that x_m must be the best position. + from utils import get_puzzle_input, convert_int_line_to_series def find_minimum_fuel_use(positions):