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Advent of Code 2019: Day 3

I'm doing Advent of Code this year. Below is my attempt at day 3:

Problem

Part One

--- Day 3: Crossed Wires ---

The gravity assist was successful, and you're well on your way to the Venus refuelling station. During the rush back on Earth, the fuel management system wasn't completely installed, so that's next on the priority list.

Opening the front panel reveals a jumble of wires. Specifically, two wires are connected to a central port and extend outward on a grid. You trace the path each wire takes as it leaves the central port, one wire per line of text (your puzzle input).

The wires twist and turn, but the two wires occasionally cross paths. To fix the circuit, you need to find the intersection point closest to the central port. Because the wires are on a grid, use the Manhattan distance for this measurement. While the wires do technically cross right at the central port where they both start, this point does not count, nor does a wire count as crossing with itself.

For example, if the first wire's path is R8,U5,L5,D3, then starting from the central port (o), it goes right 8, up 5, left 5, and finally down 3:

...........
...........
...........
....+----+.
....|....|.
....|....|.
....|....|.
.........|.
.o-------+.
...........

Then, if the second wire's path is U7,R6,D4,L4, it goes up 7, right 6, down 4, and left 4:

...........
.+-----+...
.|.....|...
.|..+--X-+.
.|..|..|.|.
.|.-X--+.|.
.|..|....|.
.|.......|.
.o-------+.
...........

These wires cross at two locations (marked X), but the lower-left one is closer to the central port: its distance is 3 + 3 = 6.

Here are a few more examples:

  • R75,D30,R83,U83,L12,D49,R71,U7,L72
    U62,R66,U55,R34,D71,R55,D58,R83 = distance 159

  • R98,U47,R26,D63,R33,U87,L62,D20,R33,U53,R51
    U98,R91,D20,R16,D67,R40,U7,R15,U6,R7 = distance 135

What is the Manhattan distance from the central port to the closest intersection?

 

Part Two

--- Part Two ---

It turns out that this circuit is very timing-sensitive; you actually need to minimize the signal delay.

To do this, calculate the number of steps each wire takes to reach each intersection; choose the intersection where the sum of both wires' steps is lowest. If a wire visits a position on the grid multiple times, use the steps value from the first time it visits that position when calculating the total value of a specific intersection.

The number of steps a wire takes is the total number of grid squares the wire has entered to get to that location, including the intersection being considered. Again consider the example from above:

...........
.+-----+...
.|.....|...
.|..+--X-+.
.|..|..|.|.
.|.-X--+.|.
.|..|....|.
.|.......|.
.o-------+.
...........

In the above example, the intersection closest to the central port is reached after 8+5+5+2 = 20 steps by the first wire and 7+6+4+3 = 20 steps by the second wire for a total of 20+20 = 40 steps.

However, the top-right intersection is better: the first wire takes only 8+5+2 = 15 and the second wire takes only 7+6+2 = 15, a total of 15+15 = 30 steps.

Here are the best steps for the extra examples from above:

  • R75,D30,R83,U83,L12,D49,R71,U7,L72
    U62,R66,U55,R34,D71,R55,D58,R83 = 610 steps
  • R98,U47,R26,D63,R33,U87,L62,D20,R33,U53,R51
    U98,R91,D20,R16,D67,R40,U7,R15,U6,R7 = 410 steps

What is the fewest combined steps the wires must take to reach an intersection?


Solution

with open(r"..\Inputs\day_3.txt") as file:
    wires = [line.rstrip().split(",") for line in file.readlines()]


def dist(p1: tuple, p2=(0, 0): tuple) -> int:
    """Calculates the Manhattan distance between two points"""
    return abs(p1[0] - p2[0]) + abs(p1[1] - p2[1])

def trace(step: str, pos: tuple) -> list:
    """Draws the line formed by applying a step (`step`) to a given coordinate (`pos`).

        Args:
            `step`: The step to be taken.
            `pos`: The position to which the step should be applied.
        Returns:
            (list): A list of coordinates that define the line.
    """
    shift = int(step[1:])
    direction = step[0]
    x = pos[0]
    y = pos[1]
    if direction == "D":
        line = [(x, y-j) for j in range(1, shift+1)]
    elif direction == "U":
        line = [(x, y+j) for j in range(1, shift+1)]
    elif direction == "L":
        line = [(x-i, y) for i in range(1, shift+1)]
    elif direction == "R":
        line = [(x+i, y) for i in range(1, shift+1)]
    return line

def draw(pos: tuple, path: list) -> list:
    """Draws the graph formed by following a sequence of steps (`path`) from a given starting position (`pos`)."""
    graph = []
    for step in path:
        line = trace(step, pos)
        pos = line[-1]  # Update `pos`.
        graph += line
    return graph

def intersect(paths: list) -> tuple:
    """Draws the graphs formed by following the two provided paths and finds their intersection.

        Args:
            `paths`: A list of two elements, corresponding to the paths for the two wires.
        Returns:
            (tuple): A 3-tuple representing the graphs of the three wires and their intersections.
                `graph_1 (list)`: Graph of the first wire.
                `graph_2 (list)`: Graph of the second wire.
                `crosses`: Set of points where the two wires cross each other.
    """
    graph_1, graph_2 = draw((0, 0), paths[0]), draw((0, 0), paths[1])
    crosses = set(graph_1) & set(graph_2)
    return crosses, graph_1, graph_2

def reaches(paths: list) -> dict:
    """Finds the distance travelled by each wire to their various points of intersection.

        Args:
            `paths`: A list of two elements, corresponding to the paths for the two wires.
        Returns:
            (dict): A dictionary mapping each point of intersection to the distances the two wires take to reach it.
    """
    crosses, graph_1, graph_2 = intersect(paths)
    dct = {point: [None, None] for point in crosses}
    for idx, pair in enumerate(graph_1):
        if pair in crosses:
            dct[pair][0] = dct[pair][0] or idx+1
    for idx, pair in enumerate(graph_2):
        if pair in crosses:
            dct[pair][1] = dct[pair][1] or idx+1
    return dct


def part_one() -> int:
    """Finds minimum Manhattan distance from origin of the intersection points."""
    return min(dist(cross) for cross in intersect(wires)[0])

def part_two() -> int:
    """Finds minimum distance travelled by the two wires to reach an intersection point."""
    return min(sum(pair) for pair in reaches(wires).values())

Notes

I don't consider myself a beginner in Python, however I am not proficient in it either. I guess I would be at the lower ends of intermediate. I am familiar with PEP 8 and have read it in its entirety, thus, any stylistic deviations I make are probably deliberate. Nevertheless, feel free to point them out if you feel a particular choice of mine was sufficiently erroneous. I am concerned with best practices and readability, but also the performance of my code. I am not sure what tradeoff is appropriate, and would appreciate feedback on the tradeoffs I did make.

For example, reaches() could have been rewritten as:

def reaches(paths: list) -> dict:
    """Finds the distance travelled by each wire to their various points of intersection.

        Args:
            `paths`: A list of two elements, corresponding to the paths for the two wires.
        Returns:
            (dict): A dictionary mapping each point of intersection to the distances the two wires take to reach it.
    """
    crosses, graph_1, graph_2 = intersect(paths)
    dct = {point: [None, None] for point in crosses}
    ln_1, ln_2 = len(graph_1), len(graph_2)
    if ln_1 < ln_2:
        graph_1 += [None] * (ln_2 - ln_1)
    else:
        graph_2 += [None] * (ln_1 - ln_2)
    for idx, pair in enumerate(zip(graph_1, graph_2)):
        if pair[0] in crosses:
            dct[pair[0]][0] = dct[pair[0]][0] or idx+1
        if pair[1] in crosses:
            dct[pair[1]][1] = dct[pair[1]][1] or idx+1
    return dct

Which eliminates one iteration and could be more performant (actual testing in my Jupyter notebook with %timeit did show a slight performance improvement (mean of 616 ms vs 592 ms), but I'm not sure it was significant), but might make the code less readable? I wasn't sure which tradeoff to make here.

My style tends to over utilise functions. This is partly because I genuinely think separating functionality into functions is a good thing, but is also an artifiact of my development practices. I tend to write the program in a Jupyter notebook (the ability to execute arbitrary code excerpts in semi isolated cells is a very helpful development aid and lends itself naturally to one function per cell (with the added benefit of being able to easily test functions independently)). I would welcome thoughts on this, but unless it is particularly egregious, I am unlikely to change it.

I am aware that the approach I took to solving this problem is not necessarily the most efficient (especially space wise), as each line could be represented by its start and end coordinates (and not as a list of all its coordinates), but I haven't worked out a satisfactory solution using that approach, so this is all I have for now.

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  • \$\begingroup\$ @Peilonrayz I removed the performance tag and stated that I didn't think I was a beginner? \$\endgroup\$ – Tobi Alafin Dec 5 '19 at 7:32
  • \$\begingroup\$ Oops, I misread that :( \$\endgroup\$ – Peilonrayz Dec 5 '19 at 7:39
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I couldn't find any drawbacks in your code. Maybe one nitpicking:

instead of

for idx, pair in enumerate(graph_1):
    if pair in crosses:
        dct[pair][0] = dct[pair][0] or idx+1

you can do

for idx, coords in enumerate(graph_1, 1): 
    if coords in crosses:
        dct[coords][0] = dct[coords][0] or idx

But I can suggest another approach - storing the length of wire while tracing, instead of calculating it afterward. It allows to eliminate the reaches function at all. I tried to add this functionality into your solution, but too much changes are required - it is easier to write from the beginning.

I have my own solution, which uses this approach. It is different from your in many aspects:

  • finding intersections happens at once for both part of the task. They are stored inside the Grid object.
  • every point of wire has the length and x, y coordinates, so why not to store this information together? I store these values into wire's dictionary in form of (y, x) : length.
  • using a dictionary instead of switch like if-else construction for R, L, U, D commands execution.

crossed_wires.py

class Grid:
    def __init__(self):
        self.commands = {
                            'R' : {"axis" : 'x', "step" : +1},
                            'L' : {"axis" : 'x', "step" : -1},
                            'U' : {"axis" : 'y', "step" : +1},
                            'D' : {"axis" : 'y', "step" : -1}
                        }
    def trace_wires(self, wire_1, wire_2):
        self.wire_1_coords = self.wire_coords_with_length(wire_1)
        self.wire_2_coords = self.wire_coords_with_length(wire_2)
        self.cross_coords = self.wires_intersections()

    def wire_coords_with_length(self, path):
        coords = {'x' : 0, 'y' : 0}
        wire_coords = {}
        wire_length = 0
        for move in path.split(','):
            direction = move[0] 
            magnitude = int(move[1:])

            axis = self.commands[direction]["axis"]
            step = self.commands[direction]["step"]

            start_axis_value = coords[axis]
            end_axis_value = start_axis_value + magnitude * step
            for new_axis_value in range(start_axis_value, end_axis_value + step, step):
                coords[axis] = new_axis_value
                new_coords = (coords['y'], coords['x'])

                if new_coords not in wire_coords:
                    wire_coords[new_coords] = wire_length
                    wire_length += 1

        return wire_coords

    def wires_intersections(self):
        intersections = self.wire_1_coords.keys() & self.wire_2_coords.keys()
        intersections.remove((0, 0))
        return intersections

    def min_manhattan_dst(self):
        return min(abs(coords[0]) + abs(coords[1]) for coords in self.cross_coords)

    def min_wire_lens(self):
        return min(self.wire_1_coords[coords] + self.wire_2_coords[coords] for coords in self.cross_coords)

wire_1 = input()
wire_2 = input()

grid = Grid()
grid.trace_wires(wire_1, wire_2)

print(grid.min_wire_lens())
print(grid.min_manhattan_dst())

It gets data from the standard input, so in the Linux the usage will be:

crossed_wires.py < input.txt

My solution is a little slower than your (for each part separately), the difference is about 50 ms. Probable it is caused by building a class.

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  • 1
    \$\begingroup\$ In wire_coords_with_length(), shouldn't wire_length be updated even if new_coords are already in wire_coords? \$\endgroup\$ – RootTwo Dec 8 '19 at 7:04
  • \$\begingroup\$ @RootTwo From the point of wire - should, because wire length have increased. It turns out this is not wire length, but the shortest length to the specific coordinates. Thus, wire_length should be renamed to something like min_wire_len_to_coords. \$\endgroup\$ – MiniMax Dec 8 '19 at 10:24
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I would suggest coding trace as a generator that takes a path and yields the sequence of coordinates on the path.

def trace(path):
    step = {'U':(0,1), 'D':(0,-1),
            'R':(1,0), 'L':(-1,0)}

    x = y = 0

    for segment in path:
        dx,dy = step[segment[0]]
        length = int(segment[1:])

        for n in range(length):
            x += dx
            y += dy
            yield x, y

Then code draw to build a dict mapping coords to the wire length to reach that coord.

def draw(path):
    coords = {}

    for path_length, coord in enumerate(trace(path), 1):
        if coord not in coords:
            coords[coord] = path_length

    return coords

Using dicts, let's you use the keys like sets, making it easy to find the intersecting points.

coords1 = draw(path1.split(','))
coords2 = draw(path2.split(','))

intersections = coords1.keys() & coords2.keys()

And the answers:

min_manhattan = min(abs(x)+abs(y) for x,y in intersections)

min_wire = min(coords1[coord] + coords2[coord] for coord in intersections)
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