Should've been this meme format

I'm surprised that it isn't "Canadian". The nationality suffix is often used to denote an animal's native area (eg: Indian Elephant, etc)

It's coolyori because it is the cool version of Sayori (from DDLC)

Python

The DP solution to go column by column and compute the min flaps for each cell is correct but too slow. Fortunately this problem can be solved using MATH

# Uses math to calculate the minimum number of flaps needed to reach the last passage at any height.
# This solution relies on a couple of facts:
#   1. The number of flaps needed to reach a passage at (x, y) is ceil((x + y) / 2).
#       Proof:
#       - Each flap increases your height by 1 and each glide decreases it by 1
#       - Suppose you flapped f times and covered a horizontal distance of x
#       - So your final height would be y = f - (x - f) = 2f - x
#       - Now suppose a passage at position p_x begins at height p_y. To enter it, your height must be, y >= p_y
#       - Substituting y, 2f - x >= p_y => 2f - p_x >= p_y
#       - Solving for f, f >= (p_x + p_y) / 2
#       - Since f is an integer, f = ceil((p_x + p_y) / 2)
#   2. For each wall, the lowest opening gives the smallest lower bound on the cumulative number of flaps.
#      The maximum of these bounds is necessary because an earlier high wall may require more flaps than the last wall.
#   3. For this input, that maximum lower bound is attainable through all the walls, so it is the optimal answer.
#      This is not true in general: an opening's upper edge or the distance between walls can make the lowest opening unreachable.
def flapsMath(data: str):
    flaps = 0

    passages = [[int(p) for p in passage.split(',')] for passage in data.splitlines()]
    passages.sort(key=lambda p: (p[0], p[1]))  # Sort passages so that we always have the lowest opening first
    last_seen_x = 0

    for x, y, _ in passages:
        if x != last_seen_x:
            last_seen_x = x
            # calculate the minimum number of flaps needed to cross this passage at its lowest opening
            # (x + y + 1) // 2 is equivalent to ceil((x + y) / 2) for integers
            flaps = max(flaps, (x + y + 1) // 2)

    return flaps

Python

Couldn't finish the series when it was released but I'm returning to finish it now. The set of free branches of part3 is too large for brute-force but you can exploit the quirk in the input where each free branch only contributes positively or negatively.

from collections import defaultdict
from dataclasses import dataclass
import re

# regex to match numbers in the input data
MATCH_NUMS_PATTERN = re.compile(r"(-?\d+)")

# Plant state class
@dataclass
class Plant:
    id: int
    thickness: int
    # is_free indicates whether the plant has a free branch.
    # it is also used to turn the effect of free branches on or off.
    is_free: bool = False

# Parses the plant input data into a list of Plant objects and a graph representing the connections between plants
# The graph root is the the final plant and the leaves are the free branches. 
def parse_plants(data: str):
    plants: list[Plant] = []
    graph = defaultdict(list)

    # Divide the input into blocks for each plant
    for block in data.split("\n\n"):
        # line iterator to control consumption of lines in the block 
        lines_iter = iter(block.splitlines())

        # get the plant's id and thickness from the first line of the block
        id, thickness = map(int, re.findall(MATCH_NUMS_PATTERN, next(lines_iter)))
        curr_plant = Plant(id, thickness)
        plants.append(curr_plant)

        # parse the remaining lines in the block to get the plant's branches
        for line in lines_iter:
            if line.startswith("- free"):
                curr_plant.is_free = True
            else:
                from_plant, thickness = map(int, re.findall(MATCH_NUMS_PATTERN, line))
                graph[curr_plant.id].append((from_plant, thickness))
    
    return plants, graph

# Recursively calculates the energy for a given plant.
# Naive implementation with no memoization, but enough for the input size.
def get_energy_at(plants: list[Plant], graph: dict[int, list[tuple[int, int]]], plant_id: int):
    plant: Plant = plants[plant_id-1]
    energy = 0

    # if the plant has a free branch, its energy is equal to its thickness.
    # otherwise, its energy is the sum of the incoming energy from its branches, multiplied by the thickness of each branch.
    if plant.is_free:
        energy = plant.thickness
    else:
        for from_plant, thickness in graph[plant_id]:
            energy += thickness * get_energy_at(plants, graph, from_plant)

    # energy only moves through the plant if it is less than or equal to the plant's thickness.
    return energy if plant.thickness <= energy else 0

# Part 1 is simple: just calculate the energy at the final plant with all free branches on.
def part1(data: str) -> int:
    plants, graph = parse_plants(data)
    return get_energy_at(plants, graph, len(plants))

# Part 2: use the boolean data to turn free branches on or off and calculate the energy at the final plant for each configuration.
def part2(data: str) -> int:
    # split the input data into plant data and boolean data
    plant_data, bool_data = data.split("\n\n\n")
    plants, graph = parse_plants(plant_data)

    all_energy = 0
    for line in bool_data.splitlines():
        # transform the boolean string into a list of integers and set the is_free attribute of each plant accordingly
        bools = map(int, line.split(' '))
        for i, b in enumerate(bools):
            plants[i].is_free = b == 1

        all_energy += get_energy_at(plants, graph, len(plants))
    return all_energy

# Part 3: calculate the maximum possible energy at the final plant, 
#   then calculate the cumulative difference in energy between the maximum and each provided configuration of free branches.
# To calculate the maximum possible energy:
#   First, I tried to progressively turn free plants off or on but that doesn't work and the energy stays at 0
#   Since this is a set of constraints, this can be solved by SMT solvers like z3
#   However, there is a quirk in the input data that allows for a simpler solution:
#       Each free branch contributes either positively or negatively ONLY
#       So we can simply turn off all free branches that contribute negatively and get the max energy.
# I don't like this solution because it relies on a quirk in the input data and doesn't work for all inputs,
#   even the sample data
def part3(data: str) -> int:
    # split the input data into plant data and boolean data
    plant_data, bool_data = data.split("\n\n\n")
    plants, graph = parse_plants(plant_data)

    # calculate the contribution of each free branch
    plant_contrib = defaultdict(int)
    for plant in plants:
        # free branches won't have any outgoing edges
        if plant.is_free:
            continue

        # for a non-leaf plant, we cumulate the contribution of each of its free branches
        for from_plant, thickness in graph[plant.id]:
            if not plants[from_plant-1].is_free:
                continue

            # assert our assumption about the input data that 
            #   each free branch contributes either positively or negatively ONLY
            if plant_contrib[from_plant]:
                assert (plant_contrib[from_plant] < 0) == (thickness < 0), (
                    "this approach only works if all free branches contribute "
                    "either positively or negatively ONLY"
                )
            
            plant_contrib[from_plant] += thickness

    # turn off all free branches that contribute negatively
    for id, contrib in plant_contrib.items():
        if contrib >= 0:
            continue
        plants[id-1].is_free = False

    # get max energy for this configuration
    max_energy = get_energy_at(plants, graph, len(plants))

    # calculate the cumulative difference in energy between the maximum and 
    #   each provided configuration of free branches.
    energy_diff = 0
    for line in bool_data.splitlines():
        bools = map(int, line.split(' '))
        for i, b in enumerate(bools):
            plants[i].is_free = b == 1

        dd_energy = get_energy_at(plants, graph, len(plants))
        # we skip configurations that do not activate the final plant
        if dd_energy == 0:
            continue
        energy_diff += max_energy - dd_energy
    
    return energy_diff

I hate that I can visualize this greentext so easily.

GPT doesn't really learn from people, it's the over-correction by OpenAI in the name of "safety" which is likely to have caused this.

I have no respect for people shoplifting non-essential items (like makeup, etc) but essential items like food is a different story.

Yes, but I'm tired of babysitting.

I pirate AAA, I buy indie.

The caterpillar stage. Wikipedia says that's 6-7 weeks long.

Even if it was true, would it be worth it? Guy must have had to miss out on so many milestones in his and his loved ones lives.

Life is just one big fight against entropy

The sequel to PLEASE DRINK VERIFICATION CAN

$8.3 trlllion in wealth

That is like 10 times the US yearly military budget. Where are you getting these numbers from?

feed her a stew that blinds her for 1 day

It's edited

What, your printer doesn't have a full keyboard under its battery? You've gotta get with the times my man.

All I can think about is that atrocious braces indentation

Never been more glad to have left that sinking ship early