72 lines
2.7 KiB
Markdown
72 lines
2.7 KiB
Markdown
|
---
|
||
|
|
title: "Advent of Code 2021 - Day 6"
|
||
|
|
date: 2021-12-06T19:40:20+01:00
|
||
|
|
day: 6
|
||
|
|
stars: 2
|
||
|
|
---
|
||
|
|
|
||
|
|
Today's challenge is [here](https://adventofcode.com/2021/day/6). We have to find the amount of fish 'split' after some days. A problem based on [exponential growth](https://en.wikipedia.org/wiki/Exponential_growth).
|
||
|
|
|
||
|
|
When I started doing this challenge I first thought about using Object-Oriented Programming, but that didn't work out. When calculating large amounts of fish I started to run out of memory and the algorithm was too slow.
|
||
|
|
|
||
|
|
Loading the data:
|
||
|
|
```python
|
||
|
|
def load() -> list[int]:
|
||
|
|
with open('../.input/day06') as f:
|
||
|
|
return [int(line) for line in f.read().split(",")]
|
||
|
|
```
|
||
|
|
|
||
|
|
## Task 1
|
||
|
|
|
||
|
|
Pretty simple, we just need to advance the fish and then count the amount of fish. I created simple methods in the `LanternFish` class that do just that.
|
||
|
|
|
||
|
|
Here's a class that represents a fish:
|
||
|
|
```python
|
||
|
|
class LanternFish:
|
||
|
|
def __init__(self, cycle: int, current: int = None):
|
||
|
|
self.cycle = cycle
|
||
|
|
self.current = current or cycle
|
||
|
|
self.children = []
|
||
|
|
|
||
|
|
def advance(self) -> None:
|
||
|
|
self.current -= 1
|
||
|
|
[child.advance() for child in self.children]
|
||
|
|
if self.current == -1:
|
||
|
|
self.current = self.cycle
|
||
|
|
self.children.append(LanternFish(cycle=6, current=8))
|
||
|
|
|
||
|
|
def count(self) -> int:
|
||
|
|
return 1 + sum([child.count() for child in self.children])
|
||
|
|
```
|
||
|
|
|
||
|
|
Here we iterate through number of days and advance the fish. The way fishes are related to each other is that they all form a tree. When I advance a fish, I also advance all of its children. The amount of fish is the sum of the amount of fish in the tree.
|
||
|
|
|
||
|
|
```python
|
||
|
|
def solve1() -> int:
|
||
|
|
fishes = [LanternFish(cycle=6, current=init) for init in load()]
|
||
|
|
[[fish.advance() for fish in fishes] for _ in range(80)]
|
||
|
|
return sum([fish.count() for fish in fishes])
|
||
|
|
```
|
||
|
|
|
||
|
|
## Task 2
|
||
|
|
|
||
|
|
Here I took a different approach, I stored the fishes in a simple dictionary. The key is the day and the value is the number of fishes. I then iterate through the days and advance the fishes.
|
||
|
|
|
||
|
|
The way this works is kind of hard to explain really, but the key to it all is the `%` [modulo operator](https://en.wikipedia.org/wiki/Modulo_operation). The modulo operator returns the remainder of a division. In this case, the modulo operator is used to determine whether some fish should give birth.
|
||
|
|
|
||
|
|
```python
|
||
|
|
def solve2() -> int:
|
||
|
|
fishes = { i: 0 for i in range(7) }
|
||
|
|
newborns = { i: 0 for i in range(9) }
|
||
|
|
|
||
|
|
for init in load():
|
||
|
|
fishes[init] += 1
|
||
|
|
|
||
|
|
for epoch in range(256):
|
||
|
|
born_from_old = fishes[epoch % 7]
|
||
|
|
fishes[epoch % 7] += newborns[epoch % 9]
|
||
|
|
newborns[epoch % 9] += born_from_old
|
||
|
|
|
||
|
|
return sum(fishes.values()) + sum(newborns.values())
|
||
|
|
```
|