Advent Of Code 2020 Days 1-5
Advent of Code 2020 Days 1-5
Advent is long over, but I’m working through Advent of Code 2020 at my own pace. Here are some of my thoughts and solutions.
These posts will be quite brief, just a few thoughts on each puzzle and the Python 3 code I used to solve it. All code on Github here. The code below is for Part 2 of each day, which often incorporates Part 1 in some way.
Day 1 - Report Repair
Thoughts
With the simple goal of finding the product of the three numbers in a list that add up to 2020. My first solution was horribly over-engineered. The solution below is a bit neater, but I still set a few unnecessary personal goals:
- The puzzle input I got from AoC had only one trio with a sum of 2020, and thus only one answer for the product. I wanted to check I could deal with a list of numbers where multiple trios added to 2020. So I used my own input for the below.
- Given that I would find multiple trios, I had to decide how to display the output. I decided to build a dictionary where the key is the trio, and the value is the product of the trio’s three elements.
- The advantage of a dictionary, I thought, is that it won’t accept multiple values for the same key.
- If I store the key as a
setof 3 integers, I thought, then when the code inevitably finds a duplicate set (compare{2001,18,1}and{1,2001,18}) , it will just harmlessly overwrite the product associated with that set with an equal value. - Actually….Python sets are mutable, so they aren’t hashable, so they can’t be used as dictionary keys. That’s why you’ll see
frozensetbelow, which is an immutable Python data type which otherwise behaves like aset.
Python Code
import sys
sum_ = 2020
with open(sys.argv[1]) as f:
lines = [int(l.rstrip('\n')) for l in f]
trios = {}
for line in lines:
for line2 in lines:
x = sum_ - line - line2
if x in lines and len({x, line, line2}) == 3:
trios[frozenset({x, line, line2})] = x*line*line2
for key, val in trios.items():
print(key, ":", val)
Example Output
frozenset({1856, 150, 14}) : 3897600
frozenset({1312, 694, 14}) : 12747392
frozenset({1674, 196, 150}) : 49215600
Day 2 - Password Philosophy
Thoughts
Not much excitement here, just counting up the number of valid passwords in a list, based on some rather eccentric password policies.
Python Code
import sys
with open(sys.argv[1]) as file:
lines = list()
for line in file:
line = line.replace(":", "")
line = line.replace("-", " ")
lines = lines + [line.strip().split()]
validcount = 0
for line in lines:
pos1 = int(line[0]) - 1
pos2 = int(line[1]) - 1
char = line[2]
password = line[3]
if password[pos1] == char:
if password[pos2] != char:
validcount += 1
elif password[pos2] == char:
validcount += 1
print(validcount)
Day 3 - Toboggan Trajectory
Thoughts
A puzzle about heading down a slope in a toboggan that can apparently survive colliding with a tree. Some posters on the subreddit decided to go the extra mile with graphics, even 3D representations in Unity! My solution is rather less ambitious.
Indexing from zero (e.g. list[0] is the first item in list) is handy when doing modular arithmetic.
e.g. in Python x % len(list) will convert any integer x into an index that’s within the bounds of list.
Hypothetically, if list[1] were the first item in list, you’d get an off-by-one error. For example, in a list with 30 items, 30 % len(list) is 0 and list[0] doesn’t exist! You’d need (x % len(list))+1 and nobody wants to deal with that.
Python Code
import sys
with open(sys.argv[1]) as file:
map_ = file.read().splitlines()
def trees(dx, dy):
x = 0
y = 0
xmax = len(map_[0])
ymax = len(map_)
count = 0
if map_[x][y] == "#":
count = 1
while y < ymax:
x = (x+dx)%(xmax)
y = (y+dy)
if y >= ymax:
break
if map_[y][x] != "." and map_[y][x] !="#":
print("error at x=",x, "y=",y)
break
elif map_[y][x] == "#":
count += 1
return count
print("product", trees(1,1)*trees(3,1)*trees(5,1)*trees(7,1)*trees(1,2) )
Day 4 - Passport Processing
Thoughts
Who doesn’t love a bit of Regex matching? Thank goodness for https://regex101.com/.
There was a salutary lesson here - while working on Part 2, I forgot to keep enforcing the requirements from Part 1 which were still relevant. Spent a fair while looking for a coding error which was actually an error understanding the requirements.
Python Code
import sys
import re
with open(sys.argv[1]) as file:
ports = file.read().split('\n\n')
ports = [re.findall(r'\S*:\S*', port) for port in ports]
countvalid = 0
req = {'ecl', 'pid', 'eyr', 'hcl', 'byr', 'iyr', 'hgt'}
for port in ports:
portvalid = True
present = set()
for field in port:
valid = True
present.add(field[:3])
x = field.partition(':')[2]
if field.startswith('byr'):
valid = 1920 <= int(x) <= 2002
elif field.startswith('iyr'):
x = int(x)
valid = 2010 <= int(x) <= 2020
elif field.startswith('eyr'):
valid = 2020 <= int(x) <= 2030
elif field.startswith('hgt'):
if field.endswith('cm'):
x = x.replace('cm', '')
valid = 150 <= int(x) <= 193
elif field.endswith('in'):
x = x.replace('in', '')
valid = 59 <= int(x) <= 76
else:
valid = False
elif field.startswith('hcl'):
valid = bool(re.fullmatch(r'#[0-9a-f]{6}', x))
elif field.startswith('ecl'):
valid = bool(re.fullmatch(r'amb|blu|brn|gry|grn|hzl|oth', x))
elif field.startswith('pid'):
valid = bool(re.fullmatch(r'[0-9]{9}', x))
portvalid = portvalid and valid
#print(field, valid)
present.discard('cid')
if present != req:
portvalid = False
if portvalid:
countvalid += 1
print(countvalid)
Day 5 - Binary Boarding
Thoughts
Nice quick one involving a very convenient plane with 128 rows and 8 columns of seats!
Python Code
import sys
with open(sys.argv[1]) as file:
lines = [line.rstrip('\n') for line in file]
rows = []
cols = []
IDs = []
for line in lines:
rows.append(line[:7])
cols.append(line[-3:])
rows = [row.replace("F","0").replace("B","1") for row in rows]
rows = [int(row,2) for row in rows] #convert from binary to decimal
cols = [col.replace("L","0").replace("R","1") for col in cols]
cols = [int(col,2) for col in cols] #convert from binary to decimal
for i in range(len(rows)):
IDs.append(rows[i]*8 + cols[i])
print(max(IDs))
IDs.sort()
for i in range(len(rows)):
if IDs[i] + 1 != IDs[i+1]:
print(IDs[i] + 1)
break