AoC 2021 Day 5: Linear Avoidinator

Source: Hydrothermal Venture

Part 1: Given a list of lines, find the number of integer points which are covered by more than one line (ignore non-vertical and non-horizontal lines).

Okay. Start with the data structures:

@dataclass(frozen=True)
class Point:
    x: int
    y: int


@dataclass(frozen=True)
class Line:
    p1: Point
    p2: Point

    def is_vertical(self):
        return self.p1.x == self.p2.x

    def is_horizontal(self):
        return self.p1.y == self.p2.y

    def is_orthagonal(self):
        return self.is_vertical() or self.is_horizontal()

    def points(self):
        # TODO: handle lines that aren't vertical, horizontal, or diagonal

        xd = 0 if self.p1.x == self.p2.x else (1 if self.p1.x < self.p2.x else -1)
        yd = 0 if self.p1.y == self.p2.y else (1 if self.p1.y < self.p2.y else -1)

        p = self.p1
        while p != self.p2:
            yield p
            p = Point(p.x + xd, p.y + yd)

        yield p

Dataclasses are great. They give you constructors and a bunch of other things for free. On top of that, if you specify frozen=True, making them immutable, you also get hashable types for free (which I’ll use in the problem).

Perhaps the most interesting bit here is the function that will iterate through the points in a List. Specifically, it will figure out the x and y delta (xd and yd) and repeatedly add that until you hit the end point.

Note: this only works for lines that are vertical, horizontal, or diagonal (at 45 degrees). Anything else needs a better line drawing algorithm (of which there are a few). If we need it, I’ll implement it.

Next, use that to parse:

def parse(file: TextIO) -> List[Line]:
    result = []

    for line in file:
        x1, y1, x2, y2 = [int(v) for v in line.replace(' -> ', ',').split(',')]
        result.append(Line(Point(x1, y1), Point(x2, y2)))

    return result

The input format is x1,y1 -> x2,y2, but it’s easier to split and convert if we do it all directly. There are a few other ways we could have done this: splitting on anything non-numeric or using a regular expression / something else for parsing directly. But I think this is clear enough.

And with all that, the problem is actually pretty short:

def part1(file: typer.FileText):

    lines = parse(file)
    counter: MutableMapping[Point, int] = collections.Counter()

    for line in lines:
        if not line.is_orthagonal():
            continue

        for point in line.points():
            counter[point] += 1

    print(sum(1 if count > 1 else 0 for point, count in counter.items()))

We’ll use the built in collections.Counter datatype, since that’s exactly what we’re doing: counting things. Then just iterate over every line, skip the non-orthagonal ones, iterate over every point, and count them up. At the end, print the number that we saw more than once. Et voila.

$ python3 linear-avoidinator.py part1 input.txt
5632

Part 2: Repeat, also including lines at 45 degrees.

That’s actually easier, since I already did the work to calculate diagonal lines when I wrote the Line class. Just drop the is_orthagonal clause:

def part2(file: typer.FileText):

    lines = parse(file)
    counter: MutableMapping[Point, int] = collections.Counter()

    for line in lines:
        for point in line.points():
            counter[point] += 1

    print(sum(1 if count > 1 else 0 for point, count in counter.items()))

And run it:

$ python3 linear-avoidinator.py part2 input.txt
22213

Not bad.

Timing

--- Day 5: Hydrothermal Venture ---

$ python3 linear-avoidinator.py part1 input.txt
5632
# time 170506875ns / 0.17s

$ python3 linear-avoidinator.py part2 input.txt
22213
# time 298739417ns / 0.30s

Okay, we’re starting to slow down. Iterating through a whole bunch of points, creating new objects for each, and storing all that in a Counter is not super efficient. But if it runs in a second on the largest data set we’re practically going to throw at it…

🤷

I’m perfectly all right with anything under a second and once problems get harder, I probably won’t fight too hard if they stay under a minute unless optimizing is unusually interesting. After that, I’ll probably spend some time on it.