 My attempts to work through the 2021 Advent of Code problems.
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```commit 1382a599ad70cc4124cc5c7946c6ff39d5bd0556
Date:   Tue,  7 Dec 2021 17:32:45 -0500

Clarifying the comments a bit

Diffstat:
Mday07_part1.py | 22++++++++++++----------
```
```1 file changed, 12 insertions(+), 10 deletions(-)
diff --git a/day07_part1.py b/day07_part1.py
@@ -6,16 +6,18 @@ minimize fuel"""
# 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.
+# 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, the subs with positions less than or equal to x_m would each
+# spend the same amount of additional fuel to reach x_o, while the subs with
+# positions greater than x_o would each save that same amount of fuel (no need
+# to worry about any 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 (i.e., there are
+# more subs spending extra fuel than there are saving fuel). Since the same
+# logic applies to an x_o less than x_m, we have shown by contradition that x_m
+# must be the best position.

from utils import get_puzzle_input, convert_int_line_to_series

```