# Advent of Code - Day 3: Toboggan Trajectory¶

## Part One¶

With the toboggan login problems resolved, you set off toward the airport. While travel by toboggan might be easy, it’s certainly not safe: there’s very minimal steering and the area is covered in trees. You’ll need to see which angles will take you near the fewest trees.

Due to the local geology, trees in this area only grow on exact integer coordinates in a grid. You make a map (your puzzle input) of the open squares (.) and trees (#) you can see. For example:

```
..##.......
#...#...#..
.#....#..#.
..#.#...#.#
.#...##..#.
..#.##.....
.#.#.#....#
.#........#
#.##...#...
#...##....#
.#..#...#.#
```

These aren’t the only trees, though; due to something you read about once involving arboreal genetics and biome stability, the same pattern repeats to the right many times:

```
..##.........##.........##.........##.........##.........##....... --->
#...#...#..#...#...#..#...#...#..#...#...#..#...#...#..#...#...#..
.#....#..#..#....#..#..#....#..#..#....#..#..#....#..#..#....#..#.
..#.#...#.#..#.#...#.#..#.#...#.#..#.#...#.#..#.#...#.#..#.#...#.#
.#...##..#..#...##..#..#...##..#..#...##..#..#...##..#..#...##..#.
..#.##.......#.##.......#.##.......#.##.......#.##.......#.##..... --->
.#.#.#....#.#.#.#....#.#.#.#....#.#.#.#....#.#.#.#....#.#.#.#....#
.#........#.#........#.#........#.#........#.#........#.#........#
#.##...#...#.##...#...#.##...#...#.##...#...#.##...#...#.##...#...
#...##....##...##....##...##....##...##....##...##....##...##....#
.#..#...#.#.#..#...#.#.#..#...#.#.#..#...#.#.#..#...#.#.#..#...#.# --->
```

You start on the open square (.) in the top-left corner and need to reach the bottom (below the bottom-most row on your map).

The toboggan can only follow a few specific slopes (you opted for a cheaper model that prefers rational numbers); start by counting all the trees you would encounter for the slope right 3, down 1:

From your starting position at the top-left, check the position that is right 3 and down 1. Then, check the position that is right 3 and down 1 from there, and so on until you go past the bottom of the map.

The locations you’d check in the above example are marked here with O where there was an open square and X where there was a tree:

```
..##.........##.........##.........##.........##.........##....... --->
#..O#...#..#...#...#..#...#...#..#...#...#..#...#...#..#...#...#..
.#....X..#..#....#..#..#....#..#..#....#..#..#....#..#..#....#..#.
..#.#...#O#..#.#...#.#..#.#...#.#..#.#...#.#..#.#...#.#..#.#...#.#
.#...##..#..X...##..#..#...##..#..#...##..#..#...##..#..#...##..#.
..#.##.......#.X#.......#.##.......#.##.......#.##.......#.##..... --->
.#.#.#....#.#.#.#.O..#.#.#.#....#.#.#.#....#.#.#.#....#.#.#.#....#
.#........#.#........X.#........#.#........#.#........#.#........#
#.##...#...#.##...#...#.X#...#...#.##...#...#.##...#...#.##...#...
#...##....##...##....##...#X....##...##....##...##....##...##....#
.#..#...#.#.#..#...#.#.#..#...X.#.#..#...#.#.#..#...#.#.#..#...#.# --->
```

In this example, traversing the map using this slope would cause you to encounter 7 trees.

Starting at the top-left corner of your map and following a slope of right 3 and
down 1, **how many trees would you encounter?**

### Load Input data¶

Note

The input data can be found here. :::

```
with open("../../data/advent-of-code/2020/day-3-input") as fid:
data = fid.readlines()
data = [list(x.strip()) for x in data if x != "\n"]
print(len(data))
print(data[0])
```

```
323
['.', '.', '.', '.', '.', '.', '.', '.', '#', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '#', '.', '.', '.', '.', '.', '.', '.', '.']
```

### Solution¶

```
def tree_counter(terrain, slope, tree="#"):
"""Count how many trees we encounter"""
right_step, down_step = slope
tree_count, idx = 0, 0
for row in terrain[down_step::down_step]:
idx += right_step
indexer = idx % len(row)
if row[indexer] == tree:
tree_count += 1
return tree_count
```

Let’s create a small test sample to use for validating the solution above:

```
test_data = """..##.........##.........##.........##.........##.........##.......
#...#...#..#...#...#..#...#...#..#...#...#..#...#...#..#...#...#..
.#....#..#..#....#..#..#....#..#..#....#..#..#....#..#..#....#..#.
..#.#...#.#..#.#...#.#..#.#...#.#..#.#...#.#..#.#...#.#..#.#...#.#
.#...##..#..#...##..#..#...##..#..#...##..#..#...##..#..#...##..#.
..#.##.......#.##.......#.##.......#.##.......#.##.......#.##.....
.#.#.#....#.#.#.#....#.#.#.#....#.#.#.#....#.#.#.#....#.#.#.#....#
.#........#.#........#.#........#.#........#.#........#.#........#
#.##...#...#.##...#...#.##...#...#.##...#...#.##...#...#.##...#...
#...##....##...##....##...##....##...##....##...##....##...##....#
.#..#...#.#.#..#...#.#.#..#...#.#.#..#...#.#.#..#...#.#.#..#...#.#""".splitlines()
test_data = [list(line) for line in test_data]
```

Running the `tree_counter`

function with a slope of `(3, 1)`

produces the right
output:

```
tree_counter(test_data, slope=(3, 1))
```

```
7
```

Now that we are somewhat confident about the validity of our solution, let’s use our full input data to count how many trees we will encounter:

```
print(f"Found {tree_counter(data, slope=(3, 1))} Trees")
```

```
Found 148 Trees
```

## Part Two¶

Time to check the rest of the slopes - you need to minimize the probability of a sudden arboreal stop, after all.

Determine the number of trees you would encounter if, for each of the following slopes, you start at the top-left corner and traverse the map all the way to the bottom:

Right 1, down 1.

Right 3, down 1. (This is the slope you already checked.)

Right 5, down 1.

Right 7, down 1.

Right 1, down 2.

In the above example, these slopes would find 2, 7, 3, 4, and 2 tree(s) respectively; multiplied together, these produce the answer 336.

**What do you get if you multiply together the number of trees encountered on
each of the listed slopes?**

### Solution¶

For part two, all we have to do is compute the number of trees for each slope,
and use the `math.prod`

function to multiply together the number of trees found.
Here’s a code cell that demonstrates that this logic works as expected when
using the test sample data:

```
import math
slopes = [(1, 1), (3, 1), (5, 1), (7, 1), (1, 2)]
counts = [tree_counter(test_data, slope) for slope in slopes]
counts, math.prod(counts)
```

```
([2, 7, 3, 4, 2], 336)
```

It’s now time to get the value for the full input data:

```
counts = [tree_counter(data, slope) for slope in slopes]
counts, math.prod(counts)
print(
f"Counts for the given slopes {slopes}: \n\t\t\t\t{counts}\nThe total number of trees encountered: {math.prod(counts)}"
)
```

```
Counts for the given slopes [(1, 1), (3, 1), (5, 1), (7, 1), (1, 2)]:
[50, 148, 53, 64, 29]
The total number of trees encountered: 727923200
```