Of course it did, bending the knee so it can run rampant on people’s private data

Python

Had to rely on an external polygon library for this one. Part 1 could have been easily done without it but part 2 would be diffucult (you can even use the simplify function to count the number of straight edges in internal and external boundaries modulo checking the collinearity of the start and end of the boundary)

import numpy as np
from pathlib import Path
from shapely import box, union, MultiPolygon, Polygon, MultiLineString
cwd = Path(__file__).parent

def parse_input(file_path):
  with file_path.open("r") as fp:
    garden = list(map(list, fp.read().splitlines()))

  return np.array(garden)

def get_polygon(plant, garden):
  coords = list(map(tuple, list(np.argwhere(garden==plant))))
  for indc,coord in enumerate(coords):

    box_next = box(xmin=coord[0], ymin=coord[1], xmax=coord[0]+1,
                   ymax=coord[1]+1)

    if indc==0:
      poly = box_next
    else:
      poly = union(poly, box_next)

  if isinstance(poly, Polygon):
    poly = MultiPolygon([poly])

  return poly

def are_collinear(coords, tol=None):
    coords = np.array(coords, dtype=float)
    coords -= coords[0]
    return np.linalg.matrix_rank(coords, tol=tol)==1

def simplify_boundary(boundary):

  # if the object has internal and external boundaries then split them
  # and recurse
  if isinstance(boundary, MultiLineString):
    coordinates = []
    for b in boundary.geoms:
      coordinates.append(simplify_boundary(b))
    return list(np.concat(coordinates, axis=0))

  simple_boundary = boundary.simplify(0)
  coords = [np.array(x) for x in list(simple_boundary.coords)[:-1]]
  resolved = False

  while not resolved:

    end_side=\
    np.concat([x[:,None] for x in [coords[-1], coords[0], coords[1]]], axis=1)

    if  are_collinear(end_side.T):
      coords = coords[1:]
    else:
      resolved = True

  return coords

def solve_problem(file_name):

  garden = parse_input(Path(cwd, file_name))
  unique_plants = set(garden.flatten())
  total_price = 0
  discounted_total_price = 0

  for plant in unique_plants:

    polygon = get_polygon(plant, garden)

    for geom in polygon.geoms:
      coordinates = simplify_boundary(geom.boundary)
      total_price += geom.area*geom.length
      discounted_total_price += geom.area*len(coordinates)

  return int(total_price), int(discounted_total_price)

Python

I initially cached the calculate_next function but honestly number of unique numbers don’t grow that much (couple thousands) so I did not feel a difference when I removed the cache. Using a dict just blazes through the problem.

from pathlib import Path
from collections import defaultdict
cwd = Path(__file__).parent

def parse_input(file_path):
  with file_path.open("r") as fp:
    numbers = list(map(int, fp.read().splitlines()[0].split(' ')))

  return numbers

def calculate_next(val):

  if val == 0:
    return [1]
  if (l:=len(str(val)))%2==0:
    return [int(str(val)[:int(l/2)]), int(str(val)[int(l/2):])]
  else:
    return [2024*val]

def solve_problem(file_name, nblinks):

  numbers = parse_input(Path(cwd, file_name))
  nvals = 0

  for indt, node in enumerate(numbers):

    last_nodes = {node:1}
    counter = 0

    while counter<nblinks:
      new_nodes = defaultdict(int)

      for val,count in last_nodes.items():
        val_next_nodes = calculate_next(val)

        for node in val_next_nodes:
          new_nodes[node] += count

      last_nodes = new_nodes
      counter += 1
    nvals += sum(last_nodes.values())

  return nvals

I always likened those creatures to raw chickens

LLMs in a robot that talks to a child? Surely nothing can go wrong with that.

nope, not true. some cats simply have a fetish for tap water. you can fill the bowl to a brim but they will still run to the tap the moment they hear someone is there.

Python

Not surprisingly, trees

import numpy as np
from pathlib import Path

cwd = Path(__file__).parent

cross = np.array([[-1,0],[1,0],[0,-1],[0,1]])

class Node():
  def __init__(self, coord, parent):
    self.coord = coord
    self.parent = parent

  def __repr__(self):
    return f"{self.coord}"

def parse_input(file_path):

  with file_path.open("r") as fp:
    data = list(map(list, fp.read().splitlines()))

  return np.array(data, dtype=int)

def find_neighbours(node_pos, grid):

  I = list(filter(lambda x: all([c>=0 and o-c>0 for c,o in zip(x,grid.shape)]),
                  list(cross + node_pos)))

  candidates = grid[tuple(np.array(I).T)]
  J = np.argwhere(candidates-grid[tuple(node_pos)]==1).flatten()

  return list(np.array(I).T[:, J].T)

def construct_tree_paths(grid):

  roots = list(np.argwhere(grid==0))
  trees = []

  for root in roots:

    levels = [[Node(root, None)]]
    while len(levels[-1])>0 or len(levels)==1:
      levels.append([Node(node, root) for root in levels[-1] for node in
                     find_neighbours(root.coord, grid)])
    trees.append(levels)

  return trees

def trace_back(trees, grid):

  paths = []

  for levels in trees:
    for node in levels[-2]:

      path = ""
      while node is not None:
        coord = ",".join(node.coord.astype(str))
        path += f"{coord} "
        node = node.parent
      paths.append(path)

  return paths

def solve_problem(file_name):

  grid = parse_input(Path(cwd, file_name))
  trees = construct_tree_paths(grid)
  trails = trace_back(trees, grid)
  ntrails = len(set(trails))
  nreached = sum([len(set([tuple(x.coord) for x in levels[-2]])) for levels in trees])

  return nreached, ntrails

I think this model exists. waitrose in UK might be one of them, not sure though.

Wow I got thrashed by chatgpt. Strictly speaking that is correct, it is more akin to Tree Search. But even then not strictly because in tree search you are searching through a list of objects that is known, you build a tree out of it and based on some conditions eliminate half of the remaining tree each time. Here I have some state space (as chatgpt claims!) and again based on applying certain conditions, I eliminate some portion of the search space successively (so I dont have to evaluate that part of the tree anymore). To me both are very similar in spirit as both methods avoid evaluating some function on all the possible inputs and successively chops off a fraction of the search space. To be more correct I will atleast replace it with tree search though, thanks. And thanks for taking a close look at my solution and improving it.

there is no monthly interest in regular accounts here unless you put it in a savers account. but yes I do and I know that my pension usually invests the money too without much flexibility on where to invest it in. so unless you are Ron Swanson there is no complete disconnection from this web. but the fix is easy: all I have to say is “individually major share holder” since those will be the ones deciding about company policies not me.

I think any company whose only shareholders are made up of people holding 0.000001% in shares wont suffer from the same consequences a company does when there are shareholders like %10, 20, 30 etc. Same difference between having billions or hundred thousands.

well I sure am somehow buying products that probably goes to one of the shareholders’ pockets that is for sure.

they are all small compared to the shareholders who likely decide the company policies more than the CEOs

thanks that indeed is a Rosetta stone

If you havent already done so, doing it in the form of “tree search”, the code completes in the blink of an eye (though on a high end cpu 11th Gen Intel® Core™ i7-11800H @ 2.30GHz). posted the code below

I feel like I need a Rosetta stone to read this code

Python

It is a tree search

def parse_input(path):

  with path.open("r") as fp:
    lines = fp.read().splitlines()

  roots = [int(line.split(':')[0]) for line in lines]
  node_lists = [[int(x)  for x in line.split(':')[1][1:].split(' ')] for line in lines]

  return roots, node_lists

def construct_tree(root, nodes, include_concat):

  levels = [[] for _ in range(len(nodes)+1)]
  levels[0] = [(str(root), "")]
  # level nodes are tuples of the form (val, operation) where both are str
  # val can be numerical or empty string
  # operation can be *, +, || or empty string

  for indl, level in enumerate(levels[1:], start=1):

    node = nodes[indl-1]

    for elem in levels[indl-1]:

      if elem[0]=='':
        continue

      if elem[0][-len(str(node)):] == str(node) and include_concat:
        levels[indl].append((elem[0][:-len(str(node))], "||"))
      if (a:=int(elem[0]))%(b:=int(node))==0:
        levels[indl].append((str(int(a/b)), '*'))
      if (a:=int(elem[0])) - (b:=int(node))>0:
        levels[indl].append((str(a - b), "+"))

  return levels[-1]

def solve_problem(file_name, include_concat):

  roots, node_lists = parse_input(Path(cwd, file_name))
  valid_roots = []

  for root, nodes in zip(roots, node_lists):

    top = construct_tree(root, nodes[::-1], include_concat)

    if any((x[0]=='1' and x[1]=='*') or (x[0]=='0' and x[1]=='+') or
           (x[0]=='' and x[1]=='||') for x in top):

      valid_roots.append(root)

  return sum(valid_roots)

I did a similar approach (place obstacles on guards path). Takes about 80s 10-15s in 11th Gen Intel® Core™ i7-11800H. Motivated by the code above, I also restricted the search to start right before the obstacle rather than the whole path which took it down from 80s to 10-15s

Well also because he was likely just a pawn. at the root of all this are the shareholders who are billionaires and who likely make the calls regarding company policies. this guy was likely just their lapdog. so even though a rule of no more than 500mil would not deal with this guy, it would definitely have prevented the existence of a parasite company of this scale. I would still say though 50 mil should be sufficient. It will allow you almost all the reasonable luxuries you can imagine if that is your thing.

nice one thanks

Python

sort using a compare function

from math import floor
from pathlib import Path
from functools import cmp_to_key
cwd = Path(__file__).parent

def parse_protocol(path):

  with path.open("r") as fp:
    data = fp.read().splitlines()

  rules = data[:data.index('')]
  page_to_rule = {r.split('|')[0]:[] for r in rules}
  [page_to_rule[r.split('|')[0]].append(r.split('|')[1]) for r in rules]

  updates = list(map(lambda x: x.split(','), data[data.index('')+1:]))

  return page_to_rule, updates

def sort_pages(pages, page_to_rule):

  compare_pages = lambda page1, page2:\
    0 if page1 not in page_to_rule or page2 not in page_to_rule[page1] else -1

  return sorted(pages, key = cmp_to_key(compare_pages))

def solve_problem(file_name, fix):

  page_to_rule, updates = parse_protocol(Path(cwd, file_name))

  to_print = [temp_p[int(floor(len(pages)/2))] for pages in updates
              if (not fix and (temp_p:=pages) == sort_pages(pages, page_to_rule))
              or (fix and (temp_p:=sort_pages(pages, page_to_rule)) != pages)]

  return sum(map(int,to_print))