Advent of Code 2019

[ThomasL]

Hi everybody! In case you did not know about Advent of Code yet:

Advent of Code is an Advent calendar of small programming puzzles for a variety of skill sets and skill levels that can be solved in any programming language you like. People use them as a speed contest, interview prep, company training, university coursework, practice problems, or to challenge each other.

You don't need a computer science background to participate - just a little programming knowledge and some problem solving skills will get you pretty far. Nor do you need a fancy computer; every problem has a solution that completes in at most 15 seconds on ten-year-old hardware.

The first puzzles will unlock on December 1st at midnight Eastern Time, which is very soon.

Looking forward to exchange thoughts and small hints ( NO FULL SOLUTIONS ).
If YOU think you MUST show us your code, otherwise you would die, PLEASE use the [spoiler ][/spoiler ] tags!

[Gribouillis]

What's the point of solving the puzzles if you don't share your solutions?

[perfringo]

I would say that first day tasks were very basic.

It's good to remember that no need for .strip() if converting single integer on row:

>>> int('5\n')
5

[ThomasL]

What's the point of solving the puzzles if you don't share your solutions?

Sharing is fine but i don´t want to be immediately spoilered opening this thread.
So everybody is invited to share their code but please use spoiler tags. :-)

[perfringo]

My solutions for day # 1

I tried to scrape data directly from input page (https://adventofcode.com/2019/day/1/input) however I was getting 400 so I just copied data into text file.

Puzzle # 1:
'sum calculation result for every row in file'

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with open('day_1_1_data.txt', 'r') as f:
    print(sum(int(line) // 3 - 2 for line in f))

Puzzle # 2 (used Python 3.8 and walrus operator)
'sum calculation results which are greater than zero for every row in file'

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with open('day_1_1_data.txt', 'r') as f:
    total = 0
    for num in f:
        while 0 < (num := int(num) // 3 - 2):
            total += num
    print(total)

[ThomasL]

I did puzzle one same way as perfringo.

For puzzle 2 i used a recursive approach as i´m still on python 3.7

My solutions for day 1:

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# part 1
with open("input.txt") as file:
    masses = [int(mass) for mass in file]

print(sum(mass // 3 - 2 for mass in masses))

# part 2
def calc_fuel(mass):
    if mass < 9: return 0
    fuel = mass // 3 - 2 
    return fuel + calc_fuel(fuel)

print(sum(calc_fuel(mass) for mass in masses))

[snippsat]

I tried to scrape data directly from input page (https://adventofcode.com/2019/day/1/input) however I was getting 400 so I just copied data into text file.

There is a way,could be a puzzle this to
When logged in the session cookie will be same for all days input.
Find session cookie in browser(inspect header).
Example:

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import requests

cookies = {'session': '536xxxxxxxxxx'}
day_1 = requests.post('https://adventofcode.com/2019/day/1/input', cookies=cookies)
result = sum(int(int(fuel) / 3 - 2) for fuel in day_1.text.strip().split('\n'))
print(result)

[perfringo]

There is a way,could be a puzzle this to

Thank you!

Day #2 puzzles I did in the copy-paste style but next days will access data directly.

Puzzle #1:
'for every slice of four check for opcode and break or apply calculation'

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import operator


def find_answer(data):
    calculate = {1: operator.add, 2: operator.mul}
    lst = data[:]
    for start in range(0, len(lst) + 1, 4):
        opcode, inp_1, inp_2, output = lst[start:start + 4]
        if opcode == 99:
            break
        else:
            lst[output] = calculate[opcode](lst[inp_1], lst[inp_2])

    return lst[0]

Puzzle #2:
'find permutation pair producing target output and return calculated value'

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import operator
from itertools import permutations


def find_answer(data, target):
    calculate = {1: operator.add, 2: operator.mul}
    for (noun, verb) in permutations(range(100), 2):
        lst = data[:]
        lst[1], lst[2] = noun, verb
        for pointer in range(0, len(lst) + 1, 4):
            opcode, par_1, par_2, par_3 = lst[pointer:pointer+4]
            if opcode == 99:
                break
            else:
                lst[par_3] = calculate[opcode](lst[par_1], lst[par_2])
        if lst[0] == target:
            return noun * 100 + verb

[perfringo]

Day #3

I was in real hurry with this one. So I decided to approach this 'systematically' in top-down style. I wanted to do some generator chaining but in order to get ready before my time was up I didn't follow this initial thought. Also - I did vectors decades ago, so I let shapely and numpy to do heavy lifting. This is brute-force and probably there is some math which allows to reach same result with less work.

Puzzel # 1:
'find intersection point with smallest Manhattan distance from start (0,0)'
For that:

• read data from webpage
  
• calculate relative moves
  
• calculate path using relative moves
  
• convert into shapely LineString
  
• find all intersections
  
• find smallest Manhattan distance from intersections
  
• output smallest Manhattan distance
  

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import requests
from shapely.geometry import LineString
import numpy as np

def main():
    moves_1, moves_2 = read_page(day=3)                                                # read moves from page
    wire_1, wire_2 = LineString(wire_path(moves_1)), LineString(wire_path(moves_2))    # calculate paths, convert to LineString
    wires_cross = wire_1.intersection(wire_2)                                          # find intersections
    closest_cross = get_closest(wires_cross)                                           # get smallest Manhattan distance
    print(f'Smallest Manhattan distance is {closest_cross} ')                          # output distance
    fewest_steps = shortest_distance(wire_1, wire_2, wires_cross)                      # find shortest distance to intersection
    print(f'Fewest steps to intersection is {fewest_steps}')                           # output shortest distance
    

def read_page(*,day):
    """
    Read data from webpage, convert to relative moves
    and yield two streams of moves.
    """
    cookie = {'session': '536****'}                                                    # replace with your own cookie
    content = requests.post(f'https://adventofcode.com/2019/day/{day}/input', cookies=cookie)
    series = (serie.split(',') for serie in content.text.split())
    yield from ((rel_move(item) for item in serie) for serie in series)


def rel_move(move):
    """
    Calculate relative moves.
    """
    replacements = str.maketrans('RULD', '  --')
    if move.startswith(('R', 'L')):
        return np.array([0, int(move.translate(replacements))])
    else:
        return np.array([int(move.translate(replacements)), 0])


def wire_path(moves):
    """
    Based on moves calculate wire path.
    Start with central port and add moves to last coordinate.
    """
    central_port = np.array([0, 0])
    path = [central_port]
    for move in moves:
        path.append(path[-1] + move)
    return path


def get_closest(coords):
    """
    Return Manhattan distance closest to central port (0, 0) which is not
    central port itself.
    """
    closest = min(abs(coord.x) + abs(coord.y) for coord in coords if all([coord.x != 0, coord.y !=0]))
    return int(closest)


def shortest_distance(first, second, intersections):
    """
    Calculate shortest distance excluding starting point.
    """
    distances = [first.project(intersection) + second.project(intersection) for intersection in intersections]
    return int(np.partition(distances, 1)[1])

if __name__ == '__main__':
    main()

Puzzle # 2:
Intersection points were already found. So:
'Find smallest distance to intersection point' (function 'shortest_distance' in code above)

[ndc85430]

I've been doing the problems in Clojure, but broadly my solution for day 3 is in the same vein as described above. The solution for day 4 was pretty short, too. Clojure is a functional language, so my solutions have lots of recursion and higher order functions (map and reduce have featured, for example).