Python

# pairwise helps picking consecutive values easier
#   [A,B,C,D] -> [AB, BC, CD]
# combinations will give me combinations of elements in a sequence
#   [A,B,C,D] -> [AB, AC, AD, BC, BD, CD]
from itertools import pairwise, combinations


# line will pass thorough the center if the points are on opposite ends,
#   i.e., total points / 2 distance away
def part1(data: str, nail_count: int = 32):
    opposite_dist = nail_count // 2
    nodes = [int(n) for n in data.split(",")]

    center_passes = 0
    for a, b in pairwise(nodes):
        if abs(a - b) == opposite_dist:
            center_passes += 1
    return center_passes


assert part1("1,5,2,6,8,4,1,7,3", 8) == 4


# BRUTEFORCE: take every two points and check if they intersect
def part2(data: str, nail_count: int = 32):
    def intersects(s1, s2):
        # expand the points
        (x1, x2), (y1, y2) = s1, s2
        # make sure x1 is the smaller nail and x2 is the bigger one
        if x1 > x2:
            x1, x2 = x2, x1
        # there is an intersection if one point lies bwtween x1 and x2
        #  and the other lies outside it
        if x1 < y1 < x2 and (y2 > x2 or y2 < x1):
            return True
        if x1 < y2 < x2 and (y1 > x2 or y1 < x1):
            return True
        return False

    nodes = [int(n) for n in data.split(",")]
    knots = 0
    for s1, s2 in combinations(pairwise(nodes), 2):
        if intersects(s1, s2):
            knots += 1

    return knots


assert part2("1,5,2,6,8,4,1,7,3,5,7,8,2", 8) == 21

from collections import defaultdict


# This is better than bruteforce
def part3(data: str, nail_count: int = 256):
    connections = defaultdict(list)
    nodes = [int(n) for n in data.split(",")]

    # record all connections, both ways
    for a, b in pairwise(nodes):
        connections[a].append(b)
        connections[b].append(a)

    max_cuts = 0
    for node_a in range(1, nail_count + 1):
        cuts = 0
        for node_b in range(node_a + 2, nail_count + 1):
            # add new cuts for the node we just added between a and b
            for other_node in connections[node_b - 1]:
                if other_node > node_b or other_node < node_a:
                    cuts += 1
            # also add any cuts for threads that go between node a and b
            cuts += connections[node_a].count(node_b)

            # remove cuts that were going from BETWEEN node_a and node_b-1 (prev node_b) to node_b
            for other_node in connections[node_b]:
                if node_a < other_node < node_b:
                    cuts -= 1
            # also remove any cuts for threads that go between node a and b-1
            cuts -= connections[node_a].count(node_b - 1)

            max_cuts = max(max_cuts, cuts)

    return max_cuts


assert part3("1,5,2,6,8,4,1,7,3,6", 8) == 7

Python

Used sliding window for part 3. The off-by-one difference between i and j tripped me up for a while.

def part1(data: str, mentor: str = 'A'):
    mentors = 0
    pairs = 0
    for p in data:
        if p == mentor:
            mentors += 1
        elif p == mentor.lower():
            pairs += mentors
    return pairs

assert part1("ABabACacBCbca") == 5

def part2(data: str):
    all_pairs = 0
    for mentor in 'ABC':
        all_pairs += part1(data, mentor)
    return all_pairs    

assert part2("ABabACacBCbca") == 11

from collections import defaultdict

def part3(data: str, distance_limit: int = 1000, repeat: int = 1000):
    n = len(data)
    N = len(data) * repeat

    pairs = 0
    person_count = defaultdict(int)
    curr = 0

    # initialize the first window excluding the right boundary
    for j in range(distance_limit):
        person_count[data[j % n]] += 1

    for curr in range(N):
        # move left boundary (if applicable)
        if (i := curr - distance_limit - 1) >= 0:
            person_count[data[i % n]] -= 1
        # move right boundary (if applicable)
        if (j := curr + distance_limit) < N:
            person_count[data[j % n]] += 1
        # if mentee, record pairs
        if data[curr % n].islower():
            pairs += person_count[data[curr % n].upper()]

    return pairs
    

assert (t := part3("AABCBABCABCabcabcABCCBAACBCa", 10, 1)) == 34, f"Expected 34 but got {t}"
assert (t := part3("AABCBABCABCabcabcABCCBAACBCa", 10, 2)) == 72, f"Expected 72 but got {t}"
assert (t := part3("AABCBABCABCabcabcABCCBAACBCa", 1000, 1000)) == 3442321, f"Expected 3442321 but got {t}"-

Never been more glad to have left that sinking ship early

Python

# snipped read_input implementation
data = read_input("input_p3.txt")

# part 1 and 2 can be solved using a calculator 
# with only the first and last gears

def part3(data):
    shafts = data.splitlines()

    rotations = 100 * int(shafts[0])
    for shaft in shafts[1:-1]:
        gear1, gear2 = [int(g) for g in shaft.split('|')]
        rotations *= gear2 / gear1
    rotations /= int(shafts[-1])
    
    return int(rotations)

assert part3("""5
7|21
18|36
27|27
10|50
10|50
11""") == 6818

print(part3(data))

"My chicks meow but that's because they're bilingual"

You don't know ???

Eh, I agree with the idea for Superman, but it's a little different for Batman. Someone who watches Batman use his advanced gadgets and hi-tech vehicles can at least surmise that he has access to a lot of money.

He might've chilled out a bit now, but he basically started as a saiyan supremacist.

HESOYAM is burned into mine

It's edited

Thank you Dave! Now we can finally put one of the world's greatest questions to rest.

But what happens when you put Ross's face on Nicolas Cage?