Advent of Code

Advent of Code

Advent of Code 2024 – Day 01 – Historian Hysteria

Ramen Junkie 1Comment

Apparently I only attempt do Advent of Code every other year.  I had this vague thought about doing this thing in JavaScript this year, but considering I am way better at Python than JS, and I have yet to actually complete one of these 100%, i should stick with what am good-ish at.  

The early day are always stupid easy.  It’s usually around halfway they have some super goofy and hard problem that I get stuck on.  The main change this year is doing it through the Bash terminal.  Though I think I did it this way in 2020.  I may swap later, who knows.  The only real annoyance is opening and closing files to test run.

I also have not done a lot of coding in a while, so I had to search up on how to do a few things.  It was a good starter scenario though, it used a lot of things I have forgotten.

I don’t do anything super fancy with my code on these, I have seen some annoyingly unreadable one liners a few times.  I always prefer clean code.  The thing runs in essentially zero time anyway, I don’t need to be “milliseconds more efficient”.

Anyway, here is my solution for part 1 and 2 of Day 1.

with open("Day01Input.txt") as file:
    data = file.read()

sets = [each.split(" ") for each in data.split("\n")]
sets.pop()

list1 = []
list2 = []

for each in sets:
   list1.append(int(each[0]))
   list2.append(int(each[1]))

list1.sort()
list2.sort()

total = 0
total2 = 0
for i in range(len(list1)):
    total += abs(list1[i]-list2[i])
    total2 += list1[i]*list2.count(list1[i])

print(total)
print(total2)
Advent of Code

Advent of Code 2022, I’m Done

Ramen Junkie 0Comment

Well, I made it farther than my last “in real time attempt” in 2020 by 3 starts. I may check in one the puzzles each day, but my experience is, they only get more complex as time goes on, so I doubt I’ll be completing any more of them. Each day is starting to take a lot more time to solve out, the solutions are getting a lot more finicky to produce. We’ve also reached the point where the puzzle inputs also feel ridiculously obtuse. Like the Day 15 puzzle, where every number was in the millions, basically, for the only purpose of making everything slow without some sort of magic reduction math. Though skimming through other’s solutions, there didn’t seem to really BE any “magic reduction” option there. \

Which is fine. It’s not supposed to be easy. I don’t expect it to be easy.

But I have long ago accepted that things I’m doing for relaxation or enjoyment, should at least be relaxing and enjoyable. And These puzzles have reached a point where the amount of enjoyment and relaxation I get from them is no longer worthwhile.

So I’m choosing to end this year’s journey here.

Maybe I’ll go back and finish them some day, but more at my own leisure. I mean, I had started doing the old 2015 puzzles in the week leading up to this year’s event. I was never doing this in any attempt to get on the leader boards or anything anyway, hell I didn’t even start most day’s puzzles until the day was half over or later.

For what it’s worth, i did make a strong attempt on Day 15 but I just could not get it to output the correct answer, and I’m not real sure why. I couldn’t even get the sample input to work out, I was always one off. It’s possible, and likely, I was counting the space where the beacon existed, but my actual input data was off by a little over 1 million, and there are not 1 million beacons on the board. Plus it was 1 million under, where my sample input solution was 1 over.

I’m not even attempting today’s, for Day 16. I can see the logic needed, but the nuance to accomplish it will just take me too long to code out and like I said above, enjoyment and relaxation is the point. I don’t need to add hours of stress to my day.

Advent of Code

Advent of Code 2020 – Day 13

Ramen Junkie 0Comment

Today’s lesson in extremely, incredibly, stupidly, inefficient methodology, is brought to you by, the letter “i” and the number “640856202464541”. Day 13 consisted of calculating bus departure times. Part 1 was stupidly simple. You have a list of buses, each bus makes a regular round trip to wherever, and the round trip always takes the same amount of minutes. At some point in the past, the buses all left at once.

Say the buses were 4, 5, and 6, each number also being the minutes for a round trip. Every 4 minutes, bus 4 returns, every 5 minutes bus 5 returns, every 6 minutes, bus 6 returns. If you are available to leave at 9 minutes, which bus leaves first. In this example, the first bus to return would be 5 (5*2=10)

Part 1 was easy, start at your departure time, check to see if any of the numbers divide into it evenly, if not, increment by one and repeat.

with open('day13data.txt') as f:
    lines = [line.rstrip() for line in f]

time = int(lines[0])
raw = lines[1].split(',')
busses= [] 

for i in raw:
  if i != 'x':
    busses.append(int(i))

print time
print busses

for x in busses:
  bus_time=x*(round((float(time)/float(x))+.5))
  wait=bus_time-time
  print "For bus "+str(x)+": "+str(bus_time)+" and you will wait: "+str(wait)+" Ans: "+str(x*wait)

The trickiest part is that the data contained non existent buses, labeled with an “x”. These are not a factor for part 1, so they just get stripped out.

Part 2 is where the pain in the ass was. For Part 2, you have to find a starting time where each bus, will leave, one minute apart, according to their offset. This includes the “missing” buses. So for the quick example I had above of 4,5,6, this time would occur at well, 4 minutes, because I made them sequential, but the next one would occur at 64,65,66 (16*4, 13*5, 11*6). This gets even more complicated when you add in the blanks, for example, 4,x,5,x,6 occurs at 68,70,72. As you spread the numbers apart and rearrange them to be non sequential, it complicates everything more.

I actually figured out the methodology pretty quickly, and wrote a working, good bit of code pretty quickly, my problem, was picking a starting point. I am sure there is some numerology methods (I am pretty sure you can do something using remainders), to calculate even an approximate starting point. In my standard of ugly code, I was just brute forcing it.

with open('day13data.txt') as f:
    lines = [line.rstrip() for line in f]

lines_array = lines[1].split(',')
times = []
counter=0
#First for sample Data Set day13datab.txt, second for real data
#multiplier=152600
multiplier=15630639084400
offsets=[]
busses=[]

for i in lines_array:
  if i != 'x':
    busses.append(i)
    offsets.append(lines_array.index(i))

#print busses
#print offsets

while (1):
  for i in offsets:
    times.append(((int(busses[0]))*multiplier)+i)

  #print times
  counter=1
  for j in busses[1:]:
    #print j
    if (times[counter]/float(j)).is_integer():
      counter+=1
    else:
      break
        
  #print times
  #print counter
  if counter == len(busses):
    break
  counter = 0
  times =[]
  multiplier+=1
  #print multiplier

print times

The problem is, this takes a very, very, very, very, very, long time. I left this code running for hours on my server and it didn’t finish, I honestly feel like it could very likely run for days, weeks, years, and never finish. The iteration on my multiplier that gives the correct set, for my data (every person has a different data set) was 15,630,639,084,501. 15 TRILLION.

So I fudged it a bit, because I was pretty sure my code was good, but I didn’t have an eternity to wait. So I found someone else’s code, found the correct answer, then worked out my starting offset from there. Sure enough, when I start at 15,630,6390,084,400. It quickly finds the correct answer.

If I needed this code for something important, I would certainly try to clean it up. I even started working on an iteration that would sort out my bus list to start with the largest number, instead of iterating the list every (low value) it would iterate every (high value), which in my case I believe was something like 15 versus 900. This was trickier than a straight sort though because I had a second list of offsets that I needed to also re order. I got this code working for the sample data set, but it failed in my proper data set because, while the sample data set started with the lowest bus number, the actual data set did not, and I had not calculated for that.

PS, yes, I misspelled “busses” in my code

« Previous Page