@xavdid does Advent of Code

Transparent Origami

Published: 2021-12-13 Original Prompt

Part 1

Wow, grid heavy puzzles this year. Because we don’t need to calculate neighbors (or the empty space), we can store the grid as set of Points (which is in turn a Tuple[int, int]). This makes it fast to iterate and, when we re-add a point to the grid (when it folds onto itself), it doesn’t create a duplicate entry. Let’s do some parsing!

from typing import Tuple, Set, cast, List
Point = Tuple[int, int]
Grid = Set[Point]
# there's only one empty line, so that's the divider
break_index = self.input.index("")
dots: Grid = {
cast(Point, tuple(map(int, s.split(",")))) for s in self.input[:break_index]

That dots is a little dense, but the gist is:

  1. split the line by ,
  2. for the 2-item array (["6", "10"]), call int on each using the map function
  3. Tell Python that this tuple (which, as far as it knows, could be any length) is a 2-tuple full of its with the cast function
  4. Wrap that in a set comprehension- note that it’s wrapped in curly braces, just like a dict comprehension. But, there’s no key: value pattern; it’s more like declaring a set with {1, 2, 3}

Before we get to parsing the fold instructions, I found it helpful to figure out how we were going to use those instructions with our grid. To know we’re doing that right, then we should be able to print the grid. It’s a pretty standard 2d iteration:

def print_grid(dots: Grid):
max_x = max(x[0] for x in dots)
max_y = max(y[1] for y in dots)
for y in range(max_y + 1):
for x in range(max_x + 1):
print("#" if (x, y) in dots else ".", end="")
print() # newline!
print() # space after the print

Passing the end kwarg to print lets us skip the regular newline. We also add 1 to the range since it stops before the end value otherwise.

Now we’re ready to do some folding. Folds happen in two directions, either across a vertical or horizontal line. In each mode, only one of the values (x or y) will change; the other stays the same. So we’ll be creating a new set, adding half of the points as is (the left or top halves depending on direction) and a modified version of the other half. For the points that we’re modifying, I made variables to note which index in the Point tuple is changing. That way, it works no matter which direction the fold is. Here’s how that looks:

def fold_grid(dots: Grid, horiz: bool, val: int) -> Grid:
result: Grid = set()
modified_index, same_index = (1, 0) if horiz else (0, 1)
for p in dots:
# if being folded onto, no change
if p[modified_index] < val:
updated_point = [-1, -1]
# one half of the points is unmodified
updated_point[same_index] = p[same_index]
# the other half changes based on its distance to the line
updated_point[modified_index] = 2 * val - p[modified_index]
result.add(cast(Point, tuple(updated_point)))
return result

In the half that’s being folded onto, nothing changes. We add those to the result and continue. Then, the index that’s not changing (eg the horizontal value in a vertical fold) is copied over. Last is the index that does change, which loses its distance to the fold line twice (eg. 14 -> 7 and then 7 -> 0 in the example).

We can fold our grid and see the result. All that’s left is to be able to parse the fold instructions. The odd shape lends itself well to Regular Expressions (aka regex). This is a specially structured string that looks for patterns in other strings. It’s a super powerful feature found in most programming languages.

In our case, we’re looking for:

  1. the letter x or the letter y
  2. followed by an =
  3. followed by a number of any length

In regex, that’s: (x|y)=(\d+). The parens denote capture groups, or the specific parts of a match we’re interested in. x|y means x or y. = means give us literally that character. \d means a digit and + is “of 1+ length”, which is what we want! You can see an interactive demo of this on regex101. Here’s the Python function:

import re
def parse_folds(folds: List[str]) -> List[Tuple[bool, int]]:
result: List[Tuple[bool, int]] = []
for fold in folds:
fold_desc = re.search(r"(x|y)=(\d+)", fold)
assert fold_desc
result.append((fold_desc.group(1) == "y", int(fold_desc.group(2))))
return result

The result of a regex search in Python is either a match object (if it finds) or None if it doesn’t. Since we know our input is well-formatted and this will always work, we can use assert to tell any type-checking tools (and Python itself) that “We promise we’ll find something - error if we don’t!“.

Each capture group is accessed with .group(), 1-indexed. So our fold result is the tuple (True, 7) for fold along y=7. A list of those describes all our folds.

Finally, we’ve got all the pieces we need to do part 1! Here is the whole thing:

break_index = self.input.index("")
dots: Grid = {
cast(Point, tuple(map(int, s.split(",")))) for s in self.input[:break_index]
folds = parse_folds(self.input[break_index + 1 :])
dots = fold_grid(dots, *folds[0])
return len(dots)

Part 2

And, now we have to do the rest. Luckily, we’ve got all we need already!

dots = fold_grid(dots, *folds[0])
size_after_first_fold = len(dots)
for fold_ins in folds[1:]:
dots = fold_grid(dots, *fold_ins)
return size_after_first_fold, "look at the ascii art"

This is the rare example where we can’t (conveniently) parse out the answer in the code. So we’ll just tell the human reader to interpret it for themselves. :boom: