Toboggan Trajectory

Published: 2020-12-03 Original Prompt

Part 1

This is simpler than it looked at first glance. We only look at each row once and we need to loop around when the reach the end of the list. The modulo operator (%) is perfect for this:

trees = 0
width = len(self.input[0]) # only need to calculate once
for index, line in enumerate(self.input):
    if line[(index * 3) % width] == "#":
        trees += 1
return trees

Part 2

At first I thought part 2 was as easy as making my function generic and adding a skip if down > 1:

def count_trees(self, right: int, down: int) -> int:
    width = len(self.input[0])
    trees = 0

    for index, line in enumerate(self.input):
        if down > 1 and index % down != 0:
            continue

        if line[(index * right) % width] == "#":
            trees += 1
    return trees

For the example, this yielded [2, 7, 3, 4, 1] which is nearly right. We got 1 for Right 1, Down 2 instead of 2. Previously, we could use index to see how many lines we’ve looked at, but now that we’re skipping some lines, we’ll have to account for that in our indexing:

def count_trees(self, right: int, down: int) -> int:
    width = len(self.input[0])
    trees = 0

    for index, line in enumerate(self.input):
        if down > 1 and index % down != 0:
            continue

        if line[(index // down * right) % width] == "#":
            trees += 1

    return trees

This gives us a nice generic function that we can fulfill both parts of the puzzle with!