For part 1, it took way longer than necessary since I read the
instructions over and over again. Gotta love AOC WOT.
Part 2 was a PITA. I got stuck at the last 2 examples since I at the
time did not find any clever way to determine which of the 2 outputs for
'1' mapped to c and f.
Since I already spent far too much time and effort, a try/except with
ValueError (digits.index will fail if c and f are flipped wrong) will do
for now.
There is probably a more efficient and smart way to do this.
Part 2 requires more clever code. In the first part, I kept track on
every single fish on order, much like the output of the example in the
puzzle.
However, by increasing the day count to 256, my code became obese and
would not run to the end: the OS killed it before it was done. It halted
around 150 days.
Browsing the sub reddit, I quickly realised that the current state of
each fish individually did not matter. For example:
3,4,3,1,2
This says 2 fishes has a timer set to 3, and 1 fish each has a timer of
1, 2 and 4.
Add one day, and we have
2,3,2,0,1
This says 2 fishes has a timer set to 2, and 1 fish each has a timer of
0, 1 and 3.
Notice how **the relative count does not change** (2, 1, 1, 1).
By using the excellent collections.counter, you get
>>> from collections import Counter
>>> Counter([3,4,3,1,2])
Counter({3: 2, 4: 1, 1: 1, 2: 1})
>>> Counter([2,3,2,0,1])
Counter({2: 2, 3: 1, 0: 1, 1: 1})
>>>
If visualized in a tabulat view, you get this:
0 1 2 3 4 5 6
----------------------------------
Initial 1 1 2 1
Day 1 1 1 2 1
The sequence moves **one step to the left each day**. Add another day,
and something extra happens.
0 1 2 3 4 5 6 7 8
----------------------------------------
Initial 1 1 2 1
Day 1 1 1 2 1
Day 2 1 2 1 1 1
2 things:
* A value higher than zero pops out on the left. By rules, 1 fish
has created 1 new fish, and resets its timer to 6.
* The new fish created has a timer of 8, since it will need 2 days
before it is able to start the create timer.
There are now 6 fishes.
Now, what happens on day 3?
0 1 2 3 4 5 6 7 8
----------------------------------------
Initial 1 1 2 1
Day 1 1 1 2 1
Day 2 1 2 1 1 1
Day 3 2 1 1 1 1 1
Now this happened:
* The fish on 0 creates a new fish, resets to 6.
* The created fish, once again, has a timer of 8.
* The fish created on day 2 decreases timer to 7.
There are now 7 fishes.
And so it continues. In Python, the most memory efficient and performant
method of leftpadding a list of values is to use collections.deque, with
the functions popleft and append.
Tricky part, got the test data correct at part 1 long before the actual input gace the correct answers. Pure luck I guess.
Part 2 I got stuck on, and resolved to visualization by a loop of
prints.
Yes, I do "".join(). Python is not always beautiful.
I began reusing part 1 for part 2, until I realised you cannot reuse the
Counter.most_common, and you need the actual values to be able to se
equal occourences.
I probably lost 5-15 minutes just to dribble with 3 levels of nested
objects. In GMT+1 before coffee, that cost me.
Part 2 was way uglier before some well motivated refactoring. Since all
tests and expected output were in place, refactoring was easy.
Felt a bit slow and rusty. Tooling was not set up properly on the
computer which decreased the flow.
Anyhow, fun first day! Could have been done more pythonic with list sequences, and readability would have increased with more use of sum() and lambdas. But this is not what Advent of Code is about.
