Advent of Code

Advent of Code

Advent of Code 2020 – Day 3

Ramen Junkie 0Comment

Day 3 was the first that really presented some problems, and I admit, I did a “cheaty trick” to solve it. Part 1 and 2 use the same code as well, because Part 2 was a simple variation of Part 1.

For Day 3, you get a “map” of a forest with Trees. You have to count how many trees will need to be avoided for a particular path through the forest. The path is a straight line along a particular slope, (over right, down some). here is the sample forest:

..##.......
#...#...#..
.#....#..#.
..#.#...#.#
.#...##..#.
..#.##.....
.#.#.#....#
.#........#
#.##...#...
#...##....#
.#..#...#.#

So, the problem I ran into, is that the hash (#) in Python, creates comments. I am sure there is a way, but I could not figure out how to escape out the hash to compare and count it in an ‘if’ statement. So what I did instead of was replace all the #s in my map code, with ‘T’s. I can look for ‘T’s all day.

With that, here is a my solution: 

def split_str(s):
  return [ch for ch in s]

with open('day3datab.txt') as f:
    lines = [line.rstrip() for line in f]

trees = 0
slopex = 1
slopey = 2
posx=0
posy=0
line_loop=len(lines[1])
distance=len(lines)-2
#print distance

while (posy<distance):
  #print trees
  posx=posx+slopex
  posy=posy+slopey
  #print str(posy)+"/"+str(distance)
  if(posx>=line_loop):
    posx=posx-line_loop
  #print str(posx)+","+str(posy)
  print str(lines[posy][posx])
  whatis=str(lines[posy][posx])
  if (whatis == "T"):
    trees+=1
  print trees

The way the problem works is, the forest repeats to the right infinitely. So step one was getting the width of the forest repeat sequence, so I could loop my position back around as needed. Otherwise it’s a simple matter of updating the position of ‘row’ and ‘col’, checking if it needs to loop (is row position greater than the loop, if so subtract the loop value), then check for a T(ree).

Part 2 was the same problem, except you check several slopes and multiply the results together. To solve this I just adjusted my slope variables and then used a calculator to multiply the results.

I could have changed the code to ask for the slope each run, or even have it loop through several slopes until you tell it to end, when it would multiply the results itself, but frankly, I don’t have that kind of time. This is a fun side project to do while watching TV at night.

Advent of Code

Advent of Code 2020 – Day 2

Ramen Junkie 0Comment

Day two is a little more complicated than Day 1 was. Today’s challenge is to take a blob of passwords, and verify if they are acceptable or not, per the “company standard” at the time of the creation of each password.

For example:

1-3 a: abcde
1-3 b: cdefg
2-9 c: ccccccccc

For part 1, on line 1, the letter a should appear at least 1 time and at most 3 times. For line 2, the letter b should appear at least 1 time and at most 3 times. For line 3, the letter c should appear at least 2 times and at most 9 times.

My solution for Day 2, Part 1 is below:

number_valid = 0
toolow = 0
toohigh = 0
number_of_entries = 0

with open('day2data.txt') as f:
    lines = [line.rstrip() for line in f]

for x in lines:
  dashloc = x.find("-")
  spaceloc = x.find(" ")
  colloc = x.find(":")
  mincount=int(x[0:dashloc])
  maxcount=int(x[int(dashloc)+1:spaceloc])
  limitchar=x[int(spaceloc+1):int(colloc)]
  password = x[int(colloc+2):]
  checkvalue = int(password.count(limitchar))
  if(mincount <= checkvalue <= maxcount):
    number_valid=number_valid+1
  if(checkvalue<mincount):
    toolow+=1
  if(checkvalue>maxcount):
    toohigh+=1
  if(mincount>=maxcount):
    print "Problem?"

print number_valid
#print toolow
#print toohigh
#print number_valid+toolow+toohigh

On a note, I’ve got some debugging bits still in there, that are commented out. After reading the file in, I’ve initialized some variables I’ll be using in the code. Then I start looping through the data.

This is where my janky code really gets to shine, as I am sure there is a better way to handle this. First step is to locate the special characters separating the data, from the exampe, “1-3 a:abcde”, these are a dash, a space and a colon. With these located, I’ve extracted the min and max values, the letter I’m tracking, and the password itself.

After getting these core values, it’s simple enough to count the occurrences of the letter, then verify it against the min/max values and count how many are valid.

I had a bit of trouble at first because I had used “password = x[int(colloc+2):-1]” instead of “password = x[int(colloc+2):]”. I’m not a master of a lot of programming languages, but I am familiar with enough of them that I get the syntax confused a lot (I keep forgetting the colon on ifs and loops in Python). I forget which language uses -1 for “go to tot he end of a range”, but I am pretty sure there is one, because I did this a lot at first.

Part 2 changes things up a bit. Instead of the first numbers representing a range of how many, it demands that the special letter appears at one of those two positions, but not both.

My solution for part 2 is below:

number_valid = 0
number_of_entries = 0

with open('day2data.txt') as f:
    lines = [line.rstrip() for line in f]

for x in lines:
  first=0
  second=0
  dashloc = x.find("-")
  spaceloc = x.find(" ")
  colloc = x.find(":")
  mincount=int(x[0:dashloc])-1
  maxcount=int(x[int(dashloc)+1:spaceloc])-1
  limitchar=x[int(spaceloc+1):int(colloc)]
  password = x[int(colloc+2):]
  if(password[mincount]==limitchar):
    first=1
  if(password[maxcount]==limitchar):
    second=1
  if ((first==1) or (second==1)):
    if(first !=second):
      number_valid+=1

print number_valid
#print toolow
#print toohigh
#print number_valid+toolow+toohigh

More sloppy code, i didn’t even change the variable names. Now, instead of counting and comparing, I’m checking for the special character in each position, and toggling a variable if it’s there or not, then making sure it’s nor present in both. This is where the new variables “first” and “second” come in.

This biggest challenge on Day 2 was that it required more manipulation of the input data.

Github Repository of my Solutions

Advent of Code

Advent of Code 2020 – Day 1

Ramen Junkie 0Comment

So, I want to say up front, I don’t know if I will finish this, but I plan to try. Also, while I intend to publish these posts on the respective day of each challenge, I may not actually DO the challenge day of. That is to say, some of these, probably most of these, will be back dated.

Advent of Code is a little 25 day advent calendar of code based challenges. I heard it mentioned on TWIT by Leo Laporte. It can be done in any language or system. There are people who solve these using game engines and such. As this is my first go, and I am not a “professional programmer”, I am doing in in sloppy Python. I’ll be posting my solutions in a Github Repo, which means this challenge serves a second purpose of helping me have an excuse to learn how to better use Github.

The first day’s challenge is pretty simple. Given a list of numbers, figure out which ones add up to 2020 and then multiply them together. Each day has two challenges, based on the same base data set. The data set for each person seems to be different. Here is my solution for Day 1 Part 1:

with open('day1data.txt') as f:
    lines = [line.rstrip() for line in f]

for x in lines:
  for y in lines:
    if((int(x)+int(y))==2020):
      print int(x)*int(y)

Each day of this puzzle (so far) involves reading in a data file, and working with it. My solution involves looping each number, and multiplying it by each other number, and checking for if they sum 2020, then posting the result of multiplying the two numbers.

Part two is essentially the same except it involves three numbers instead of two. This pretty much just meant adding another layer of loop to my loops.

with open('day1data.txt') as f:
    lines = [line.rstrip() for line in f]

for x in lines:
  for y in lines:
    for z in lines:
      if((int(x)+int(y)+int(z))==2020):
        print int(x)*int(y)*int(z)

This first days’ challenge was pretty simple. The rest of the days, not so much.

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