# AoC 2021 Day 14: Polymerizationinator

### Source: Extended Polymerization

#### Part 1: Given a complete function f(a, b) -> c where any string ab becomes acb and an input string, apply the function at the same time to every (overlapping) pair of letters. Repeat this time times. Report the difference in counts between the most and least common letters in the final situation.

Okay. First try, let’s just solve this directly. Although, since we know that we’re going to be inserting into the middle of the list constantly, we can be slightly more efficient by using a linked list:


PolyMap = Mapping[Tuple[str, str], str]

@dataclass
class Node(Iterable):
value: str
next: Optional['Node']

def expand_via(self, map: PolyMap) -> 'Node':
if self.next is None:
return self

for a, b in zip(self, self.next):
if (a.value, b.value) in map:
a.next = Node(map[a.value, b.value], b)

return self

@staticmethod
def from_iter(iter: Iterable) -> 'Node':
previous = None

for el in iter:
n = Node(el, None)

if previous:
previous.next = n
else:

previous = n

def __iter__(self):
self.iter_node = self
return self

def __next__(self):
if not self.iter_node:
raise StopIteration

result = self.iter_node
self.iter_node = self.iter_node.next
return result

def __str__(self):
return ''.join(self)

def load(file: TextIO) -> Tuple[Node, PolyMap]:

map = {
(line[0], line[1]): line[6]
for line in file
}

return ls, map


Okay, there are basically three interesting parts here.

• First, the from_iter function. That will take any iterable and make a linked list out of it, returning the head of that list. It does this by creating each Node in turn and attaching it to the previously created node.

• Next, the __iter__ and __next__ functions. Together, these make Node iterable, so that we can do things like for n in ls: ... where ls is the head of a linked list.

• Finally, expand_via. That will take in a mapping of pairs to a single character and expand the current linked list with it. Specifically, for any characters a and b in the linked list, insert a new node between them with value map[a, b]. Pretty cool right?

• While we’re at it, add a function to load the puzzle definitions from a file: the initial linked list (a string is iterable) and the series of mappings. They’re always single characters in a set format, so we can hardcode where they are in the string. Perfect? No. Workable for now, sure.

With all that, we can write a direct solution to the puzzle:


@app.command()
def direct(file: typer.FileText, steps: int):

import time
start = time.time()

for i in range(steps):
logging.info(f'Generation {i} calculate after {time.time() - start:02f} sec, has {len(list(ls))} elements')
ls.expand_via(map)

counts = Counter(n.value for n in ls)
logging.info(f'{counts=}')

most, _ = max((qty, c) for (c, qty) in counts.items())
least, _ = min((qty, c) for (c, qty) in counts.items())

print(most - least)


(There’s a reason I named this direct rather than part1)

It actually works great:

$python3 polymerizationinator.py direct input.txt 10 2797  So that’s all she wrote, right? Well… #### Part 2: Do the same for 40 iterations (instead of 10). Okay. We have a problem. $ python3 polymerizationinator.py direct input.txt 10
2797
# time 76895250ns / 0.08s

$python3 polymerizationinator.py direct input.txt 11 5437 # time 77964125ns / 0.08s$ python3 polymerizationinator.py direct input.txt 12
10672
# time 123601208ns / 0.12s

$python3 polymerizationinator.py direct input.txt 13 21340 # time 212642208ns / 0.21s$ python3 polymerizationinator.py direct input.txt 14
42836
# time 409988583ns / 0.41s

$python3 polymerizationinator.py direct input.txt 15 85972 # time 795902958ns / 0.80s$ python3 polymerizationinator.py direct input.txt 16
172844
# time 1567806500ns / 1.57s

$python3 polymerizationinator.py direct input.txt 17 347059 # time 3146193833ns / 3.15s$ python3 polymerizationinator.py direct input.txt 18
696285
# time 6848070667ns / 6.85s

$python3 polymerizationinator.py direct input.txt 19 1395343 # time 12827071459ns / 12.83s$ python3 polymerizationinator.py direct input.txt 20
2793845
# time 35055311792ns / 35.06s


It’s quick enough at small problems, but as they mentioned in the problem, it grows exponentially. Each iteration we add roughly doubles the time. And we’re already up to 35 seconds after just 20. To double that another 20 times… we’re talking a runtime of roughly a year. I’m… not going to wait that long. Instead, let’s try something totally different. Rather than actually calculate the strings, we’re going to recursively calculate how many of each value there are for any given pair of letters and how far we still need to split them.

def recursive(file: typer.FileText, steps: int):

# Recursively count all elements that will be returned from the character pair a,b at depth
# This will use the mapping specified in map above
# This will recur depth += 1 each time until depth = steps (so will always terminate)
def count(a, b, depth):
logging.info(f'{" " * depth} > count({a}, {b}, {depth})')

if depth == steps:
result = {a: 1}
else:
result = {}

for left, right in [(a, map[a, b]), (map[a, b], b)]:
for k, v in count(left, right, depth + 1).items():
result[k] = result.get(k, 0) + v

logging.info(f'{" " * depth} < count({a}, {b}, {depth}) = {result}')
return result

# Recursively figure out the counts for each pair of elements
# The rightmost element is never counted, so add it at the end
counts: MutableMapping[str, int] = {}
if ls.next is not None:
for a, b in zip(ls, ls.next):
for k, v in count(a.value, b.value, 0).items():
counts[k] = counts.get(k, 0) + v
counts[b.value] = counts.get(b.value, 0) + 1

logging.info(f'{counts=}')

most, _ = max((qty, c) for (c, qty) in counts.items())
least, _ = min((qty, c) for (c, qty) in counts.items())

print(most - least)


The entire strength of this function is the count helper function. Like I said (both above and in the comment), this will take a pair of letters (the a and b) and a current depth and it will guarantee that it will return the total count of each letter in the final output between that a and b (including the a, but not the b, I’ll come back to that).

With that guarantee, that means that we can use the function recursively. We know that if we have count(a, b, 1) and the mapping map[a, b] = c, that means that we can guarantee:

count(a, b, n) = count(a, map(a, b), n+1) + count(map(a, b), b, n+1)

Since n is always increasing and we have a base case of n = the final depth, our recursion is guaranteed to terminate.

$python3 polymerizationinator.py direct input.txt 15 85972 # time 800473667ns / 0.80s$ python3 polymerizationinator.py recursive input.txt 15
85972
# time 2147194583ns / 2.15s


Yay! It gives me the same answer. But… it’s slower. Turns out, the linked list solution is actually pretty fast and we’re paying a lot for each recursive function call (relatively speaking).

BUT (and you knew there would be a but, otherwise, why would I be writing this in the first place?), we can do better:

from functools import cache

def recursive(file: typer.FileText, steps: int):

# Recursively count all elements that will be returned from the character pair a,b at depth
# This will use the mapping specified in map above
# This will recur depth += 1 each time until depth = steps (so will always terminate)
# The @cache will make sure that we only recalculate a given a/b/depth triple once
@cache
def count(a, b, depth):
logging.info(f'{" " * depth} > count({a}, {b}, {depth})')

...

...

$python3 polymerizationinator.py recursive input.txt 15 85972 # time 2147194583ns / 2.15s$ python3 polymerizationinator.py --cache recursive input.txt 15
85972
# time 43858541ns / 0.04s


Now that is what I’m talking about.

So what in the world just happened? Well, when we’re recursively building up all of these counts, we’re going to run into count(a, b, n) more than once in the same layer. Especially as the sequence gets longer and longer, there are going to be many duplicates. And what’s more, once you hit one of those cases, all of the recursive calls from that point on will be the same. What cache does is tell the computer to calculate a given answer (for input a, b, depth) once and store (memoize) the answer. So we only have to do each branch once. This cuts out a huge amount of work.

And… it lets us run the full 40 really quickly:

\$ python3 polymerizationinator.py --cache recursive input.txt 40
2926813379532
# time 50570625ns / 0.05s


A bit less than a year, no?

I love problems like this. Most of the time, optimization doesn’t really matter. For example, if you know you’re never going to run this simulation more than 10-15 generations. But every once in a while, you bump into a truly exponential function and have to run it long enough for it to explode. Then things get interesting. And that’s when knowing about things like recursion and memoization come in very handy.

Cool, right?

Or right. Coming back to that thing. Why do we only add b at the end outside of the recursion? Because we’re always overlapping. So when you go from ababab (where ab -> c and ba -> d), you’re going to have recursive calls generating acb, bda, acb, bda, acb. Each time, the last character of one is the first of the next, so we can’t count that or we’ll end up overcounting. But that means we never do count the last b of the last acb (the only one that doesn’t overlap).

As they say, there are two hard things in computer science: caching, naming things, and off by one errors.

Pretty sure I managed to hit all of them in this post!