Advent of Code 2022, Day 5

Ramen Junkie December 5, 2022 0Comment

Day 05 – Moving Crates

The Problem Part 1:

The expedition can depart as soon as the final supplies have been unloaded from the ships. Supplies are stored in stacks of marked crates, but because the needed supplies are buried under many other crates, the crates need to be rearranged.

The ship has a giant cargo crane capable of moving crates between stacks. To ensure none of the crates get crushed or fall over, the crane operator will rearrange them in a series of carefully-planned steps. After the crates are rearranged, the desired crates will be at the top of each stack.

The Elves don’t want to interrupt the crane operator during this delicate procedure, but they forgot to ask her which crate will end up where, and they want to be ready to unload them as soon as possible so they can embark.

They do, however, have a drawing of the starting stacks of crates and the rearrangement procedure (your puzzle input).

After the rearrangement procedure completes, what crate ends up on top of each stack?

And I guessed what Part 2 would be, before it was even revealed:

Some mud was covering the writing on the side of the crane, and you quickly wipe it away. The crane isn’t a CrateMover 9000 – it’s a CrateMover 9001.

The CrateMover 9001 is notable for many new and exciting features: air conditioning, leather seats, an extra cup holder, and the ability to pick up and move multiple crates at once.

I’m going to throw the sample input in here as well, because getting through this was the hardest part.

    [D]    
[N] [C]    
[Z] [M] [P]
1   2   3

move 1 from 2 to 1
move 3 from 1 to 3
move 2 from 2 to 1
move 1 from 1 to 2

It’s essentially, two inputs one file.   A “drawing” of the stack, and a series of moves.  Splitting these up was simple enough, there is a double space, and split() works on “\n\n”.  What was hard, was manipulating the Crate Stack drawing into a list of lists so it was easy to work with on the moves function.

I’ll post the whole code, but I want to look at this function, because it was the hardest part and the most interesting thing here.

def pivot(crates):
    crates.pop(-1)
    processed = []
    for row in crates:
        row = row.replace('    ', ' [ ]')
        # print(len(row))
        while len(row) < len(crates[-1]):
            row += ' [ ]'
        row = row[1:-1].split('] [')
        # print(row)
        processed.append(row)
    # print(processed)
    return list(zip(*processed[::-1]))

What I am looking for, is a list of lists, for each “stack”.  The problem is, my input is “vertical” instead of “horizontal” and it has gaps.  I considered that there may be a way to do this with Pandas and Data Frames, but a lot of these puzzles really feel like the fun part is solving them without a bunch of complicated imports.  

The first step is to remove the bottom row, it’s just column labels, I don’t care about that part.

After a lot of consideration for how to handle the “blanks” I realized the best way was to convert them into “Blank Crates”.  Which is where the “row = row.replace(‘    ‘, ‘ [ ]’)” section comes in.  I also found that for the pivot later, I would need to fill in any blank spaces at the end of the rows as well, which si where the following while loop comes in.  

With the array filled in “fully” with blanks, I can straight split each row on “] [” and get a list, which is appended to “processed” as a fresh rwo, giving me a list of lists, but it’s still veetical.  I went out looking for a “pivot” function.  For anyone not familiar with this term, a super layman explanation is, a Pivot Table, is essentially a new table, based on values of an old table.  It’s been ‘Pivoted”.  A list of lists, is essentiallyy just “a table” of data,  A series of rows and columns.   All I want to do is literally pivot it so X becomes Y, because it makes things easier to “move crates” later with split() and pop() functions.  What I found was this:

list(zip(*processed[::-1]))

That handy Zip function again, which will basically just, do what I want.  

The only other real problem I had with this puzzle was on Part 2.  I tried several methods I found online, but I could not get a second copy of my original list.  It seems Python does some “Screwy Bullshit” with it’s lists where copies are not copies.  I kept getting errors trying to run my Part 2 code, index out of range.  Well, it was starting with the modified version of the original crates list.  I tried crates2 = crates[:], and crates2 = crates.copy() and crates2 = copy.copy() and nothing actually did what I needed.  

So I just, commented out the Part 1 code so it wouldn’t run.

with open("Day05Input.txt") as file:
    data = file.read()

split_data = data.split('\n\n')
crates = split_data[0].split('\n')
moves = split_data[1].split('\n')

def pivot(crates):
    crates.pop(-1)
    processed = []
    for row in crates:
        row = row.replace('    ', ' [ ]')
        # print(len(row))
        while len(row) < len(crates[-1]):
            row += ' [ ]'
        row = row[1:-1].split('] [')
        # print(row)
        processed.append(row)
    # print(processed)
    return list(zip(*processed[::-1]))

def drop_blanks(crates):
    de_blanked = []
    for crate in crates:
        row = [i for i in crate if i != ' ']
        de_blanked.append(row)
    return de_blanked

def move_crates(crate_move, moves):
    for move in moves:
        steps = move.split(" ")
        how_many = int(steps[1])
        from_stack = int(steps[3])-1
        to_stack = int(steps[5])-1
        for i in range(0,how_many):
            # print(crate_move[from_stack])
            # print(crate_move[to_stack])
            crate_move[to_stack].append(crate_move[from_stack].pop())
        #print(steps)
    return crate_move

def move_crates_in_stacks(crates_stack, moves):
    for move in moves:
        steps = move.split(" ")
        how_many = int(steps[1])
        from_stack = int(steps[3])-1
        to_stack = int(steps[5])-1
        #print(crates_stack[from_stack])
        move_list = crates_stack[from_stack][-how_many:]
        #print(crates_stack[to_stack])
        for i in range(how_many):
            crates_stack[from_stack].pop()
            crates_stack[to_stack].append(move_list[i])
        #print(steps)
    return crates_stack

crates = pivot(crates)
crates = drop_blanks(crates)
#Part 1
#moved = move_crates(crates, moves)
moved2 = move_crates_in_stacks(crates, moves)

print(crates)
#print(len(crates))
#print(moves)

#Part 1
# for each in moved:
#     print(each[-1])
#svfdLGLWV

for each in moved2:
    print(each[-1])
#DCVTCVPCL

Advent of Code 2022, Day 4

Ramen Junkie December 4, 2022 0Comment

Day 4 – Cleaning Up The Landing Zone

The Problem Part 1:

Space needs to be cleared before the last supplies can be unloaded from the ships, and so several Elves have been assigned the job of cleaning up sections of the camp. Every section has a unique ID number, and each Elf is assigned a range of section IDs.

However, as some of the Elves compare their section assignments with each other, they’ve noticed that many of the assignments overlap. To try to quickly find overlaps and reduce duplicated effort, the Elves pair up and make a big list of the section assignments for each pair (your puzzle input).

Some of the pairs have noticed that one of their assignments fully contains the other. For example, 2-8 fully contains 3-7, and 6-6 is fully contained by 4-6. In pairs where one assignment fully contains the other, one Elf in the pair would be exclusively cleaning sections their partner will already be cleaning, so these seem like the most in need of reconsideration. In this example, there are 2 such pairs.

In how many assignment pairs does one range fully contain the other?

The Problem Part 2:

It seems like there is still quite a bit of duplicate work planned. Instead, the Elves would like to know the number of pairs that overlap at all.

These two problems use two seperate functions, but the same loop to solve both at once.  I’m particularly proud of my solution for checking on Overlap, though I’m not sure if it’s the best solution.  For example, with the sample data:

2-4,6-8
2-3,4-5
5-7,7-9
2-8,3-7
6-6,4-6
2-6,4-8

the first one 2-4, and 6-8, does not over lap at all, but in the 4th one, 2-8 completely contains 3-7.  What I used was simple math based logic, substrct the start numbers and end numbers.  With the lines 1 and 4, this would give:

2-6 = -4, 4-8 = -4
2-3 = -1, 8-7 = 1

Then, multiply each of these results together for each elf.

-4 * -4 = 16
-1 * 1 = -1

For all lines, if the result is greater than 0, it’s not 100% inclusive.  From the above, 16, is not, -1 is.  This is also true of line 5, which will result in 0.  A zero results if either the start or end are equal, which will always be completely inclusive.

with open("Day04Input.txt") as file:
    data = file.read()

worklist = data.split("\n")

def check_range(elves):
    work = elves.split(",")
    elf1 = work[0].split("-")
    elf2 = work[1].split("-")
    for i in range(int(elf1[0]),int(elf1[1])+1):
        if i in range(int(elf2[0]),int(elf2[1])+1):
            return 1
    return 0

def check_overlap(elves):
    work = elves.split(",")
    elf1 = work[0].split("-")
    elf2 = work[1].split("-")
    if (int(elf1[0])-int(elf2[0]))*(int(elf1[1])-int(elf2[1])) > 0:
        return 0
    return 1

total = 0
range_total = 0

for each in worklist:
    total += check_overlap(each)
    range_total += check_range(each)

print(total)
#485

print(range_total)
# 658 LOW
#857

Advent of Code 2022, Day 3

Ramen Junkie December 3, 2022 0Comment

Day 3 – Sorting the Food Into Sacks

The Problem Part 1:

One Elf has the important job of loading all of the rucksacks with supplies for the jungle journey. Unfortunately, that Elf didn’t quite follow the packing instructions, and so a few items now need to be rearranged.

Each rucksack has two large compartments. All items of a given type are meant to go into exactly one of the two compartments. The Elf that did the packing failed to follow this rule for exactly one item type per rucksack.

The Elves have made a list of all of the items currently in each rucksack (your puzzle input), but they need your help finding the errors. Every item type is identified by a single lowercase or uppercase letter (that is, a and A refer to different types of items).

The list of items for each rucksack is given as characters all on a single line. A given rucksack always has the same number of items in each of its two compartments, so the first half of the characters represent items in the first compartment, while the second half of the characters represent items in the second compartment.

Part 2 Problem:

For safety, the Elves are divided into groups of three. Every Elf carries a badge that identifies their group. For efficiency, within each group of three Elves, the badge is the only item type carried by all three Elves. That is, if a group’s badge is item type B, then all three Elves will have item type B somewhere in their rucksack, and at most two of the Elves will be carrying any other item type.

The problem is that someone forgot to put this year’s updated authenticity sticker on the badges. All of the badges need to be pulled out of the rucksacks so the new authenticity stickers can be attached.

Additionally, nobody wrote down which item type corresponds to each group’s badges. The only way to tell which item type is the right one is by finding the one item type that is common between all three Elves in each group.

So, the actual input on a lot of these puzzles is a numerical value, for this problem, it wanted the totals for the Food ID and Badge IDs, based on 1-26 and 27-52, depending on the value.  The simple way of summing these is the list at the top.  Each value is just the index of that list.

I also have this really messy line for ” contents = [[each[:int(len(each)/2)],each[int(len(each)/2):]] for each in data.split(‘\n’)]” in there, to break up each line into a list of lists.  Split can’t be used because each line is just a long list of jumbled letters.  The next line uses a function that I came across looking for an easy way to group everyone up into blocks of 3 for the second part.  I found the “zip” function, which is pretty cool and I’ve used it later in another puzzle as well.

The original Part 1 version of this code just ran through the individual contents variable.  But when I added in Part 2, with the 3 Elf Groups, I modified it to run off of groups of 3, while still solving both problems at once.  The actual finding of the items and badges was not hard, it’s just checking to see if values of one string are in another string using a loop.  A List Comprehension would work too, but for simple loops like this I find them more readable.

with open("Day03Input.txt") as file:
    data = file.read()

letters = [
    'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o',
    'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z', 'A', 'B', 'C', 'D',
    'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S',
    'T', 'U', 'V', 'W', 'X', 'Y', 'Z']

contents = [[each[:int(len(each)/2)],each[int(len(each)/2):]] for each in data.split('\n')]

groups = zip(*[iter(contents)]*3)
def find_item(elf):
    for i in elf[0]:
        if i in elf[1]:
            return letters.index(i) + 1

def find_badge(group):
    elves = [group[0][0]+group[0][1],group[1][0]+group[1][1],group[2][0]+group[2][1]]
    for i in elves[0]:
        if i in elves[1] and i in elves[2]:
            return letters.index(i)+1

priority_sum = 0
badge_sum = 0

for each_group in groups:
    for elf in each_group:
        priority_sum += find_item(elf)
    badge_sum += find_badge(each_group)

print(priority_sum)
# 7526 Too Low
# 7826
print(badge_sum)
# 2577