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.

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)