## Load the packages required
library(tidyverse)
library(zoo)After saving Christmas five years in a row, you’ve decided to take a vacation at a nice resort on a tropical island. Surely, Christmas will go on without you.
The tropical island has its own currency and is entirely cash-only. The gold coins used there have a little picture of a starfish; the locals just call them stars. None of the currency exchanges seem to have heard of them, but somehow, you’ll need to find fifty of these coins by the time you arrive so you can pay the deposit on your room.
To save your vacation, you need to get all fifty stars by December 25th.
Collect stars by solving puzzles. Two puzzles will be made available on each day in the Advent calendar; the second puzzle is unlocked when you complete the first. Each puzzle grants one star. Good luck!
Before you leave, the Elves in accounting just need you to fix your expense report (your puzzle input); apparently, something isn’t quite adding up.
Specifically, they need you to find the two entries that sum to 2020 and then multiply those two numbers together.
For example, suppose your expense report contained the following:
1721 979 366 299 675 1456 In this list, the two entries that sum to 2020 are 1721 and 299. Multiplying them together produces 1721 * 299 = 514579, so the correct answer is 514579.
Of course, your expense report is much larger. Find the two entries that sum to 2020; what do you get if you multiply them together?
## Read in the data
df <- read.table("Day1/input.txt")
input_1 <- df$V1
input_2 <- df$V1
tab1 <- expand_grid(input_1, input_2)
result1 <- tab1 %>%
mutate(jumla = input_1 + input_2) %>%
filter(jumla == 2020) %>%
select(input_1, input_2) %>%
sample_n(1) %>%
mutate(product = input_1 * input_2) %>%
pull(product)The Elves in accounting are thankful for your help; one of them even offers you a starfish coin they had left over from a past vacation. They offer you a second one if you can find three numbers in your expense report that meet the same criteria.
Using the above example again, the three entries that sum to 2020 are 979, 366, and 675. Multiplying them together produces the answer, 241861950.
In your expense report, what is the product of the three entries that sum to 2020?
input_3 <- df$V1
tab2 <- expand_grid(input_1, input_2, input_3)
result2 <- tab2 %>%
mutate(jumla = input_1 + input_2 + input_3) %>%
filter(jumla == 2020) %>%
select(input_1, input_2, input_3) %>%
sample_n(1) %>%
mutate(product = input_1 * input_2 * input_3) %>%
pull(product)Your flight departs in a few days from the coastal airport; the easiest way down to the coast from here is via toboggan.
The shopkeeper at the North Pole Toboggan Rental Shop is having a bad day. “Something’s wrong with our computers; we can’t log in!” You ask if you can take a look.
Their password database seems to be a little corrupted: some of the passwords wouldn’t have been allowed by the Official Toboggan Corporate Policy that was in effect when they were chosen.
To try to debug the problem, they have created a list (your puzzle input) of passwords (according to the corrupted database) and the corporate policy when that password was set.
For example, suppose you have the following list:
1-3 a: abcde 1-3 b: cdefg 2-9 c: ccccccccc Each line gives the password policy and then the password. The password policy indicates the lowest and highest number of times a given letter must appear for the password to be valid. For example, 1-3 a means that the password must contain a at least 1 time and at most 3 times.
In the above example, 2 passwords are valid. The middle password, cdefg, is not; it contains no instances of b, but needs at least 1. The first and third passwords are valid: they contain one a or nine c, both within the limits of their respective policies.
How many passwords are valid according to their policies?
## Read in the datasets
df <- read.table("Day2/input_day2.txt")
## Rename the columns into `condition`, `letter`, `password`
colnames(df) <- c("condition", "letter", "password")
## Split the condition into `left_condition` and `right_condition`.
df <- df %>%
separate(condition,
into = c("left_condition", "right_condition"),
sep = "-", remove = FALSE)
## Convert the `left_condition` and `right_condition` into numerics
df <- df %>%
mutate(across(c("left_condition", "right_condition"), ~as.numeric(.)))
## Remove the colon from the `letter`
df <- df %>%
mutate(letter = trimws(gsub(":", "", letter)))## Generate a variable `letter_counter` that contains the number of times the
## letter appears in the password
df1 <- df %>%
mutate(letter_counter = str_count(password, letter))
## Generate a variable `validity`, that checks whether the password condition has
## been met or not.
df1 <- df1 %>%
mutate(validity = ifelse(letter_counter >= left_condition &
letter_counter <= right_condition, "valid", "not valid" ))
## How many passwords are valid?
valid_passwords <- df1 %>%
filter(validity == "valid") %>%
select(validity) %>%
count() %>%
pull(n)While it appears you validated the passwords correctly, they don’t seem to be what the Official Toboggan Corporate Authentication System is expecting.
The shopkeeper suddenly realizes that he just accidentally explained the password policy rules from his old job at the sled rental place down the street! The Official Toboggan Corporate Policy actually works a little differently.
Each policy actually describes two positions in the password, where 1 means the first character, 2 means the second character, and so on. (Be careful; Toboggan Corporate Policies have no concept of “index zero”!) Exactly one of these positions must contain the given letter. Other occurrences of the letter are irrelevant for the purposes of policy enforcement.
Given the same example list from above:
1-3 a: abcde is valid: position 1 contains a and position 3 does not. 1-3 b: cdefg is invalid: neither position 1 nor position 3 contains b. 2-9 c: ccccccccc is invalid: both position 2 and position 9 contain c.
How many passwords are valid according to the new interpretation of the policies?
## Generate variables that indicate the letters that are in the positions given
df2 <- df %>%
mutate(left_position_letter = substr(password, left_condition, left_condition),
right_position_letter = substr(password, right_condition, right_condition))
df2 <- df2 %>%
## only one of the letters should match the positions, not both
mutate(diclaimer = ifelse(letter == left_position_letter &
letter == right_position_letter, "not allowed", "okay")) %>%
mutate(validity = ifelse((letter == left_position_letter|
letter == right_position_letter) &
diclaimer != "not allowed",
"valid", "not valid"))
## Generate the number of passwords that are valid
valid_passwords2 <- df2 %>%
filter(validity == "valid") %>%
select(validity) %>%
count() %>%
pull(n)You arrive at the airport only to realize that you grabbed your North Pole Credentials instead of your passport. While these documents are extremely similar, North Pole Credentials aren’t issued by a country and therefore aren’t actually valid documentation for travel in most of the world.
It seems like you’re not the only one having problems, though; a very long line has formed for the automatic passport scanners, and the delay could upset your travel itinerary.
Due to some questionable network security, you realize you might be able to solve both of these problems at the same time.
The automatic passport scanners are slow because they’re having trouble detecting which passports have all required fields. The expected fields are as follows:
byr (Birth Year)
iyr (Issue Year)
eyr (Expiration Year)
hgt (Height)
hcl (Hair Color)
ecl (Eye Color)
pid (Passport ID)
cid (Country ID)
Passport data is validated in batch files (your puzzle input). Each passport is represented as a sequence of key:value pairs separated by spaces or newlines. Passports are separated by blank lines.
Here is an example batch file containing four passports:
ecl:gry pid:860033327 eyr:2020 hcl:#fffffd byr:1937 iyr:2017 cid:147 hgt:183cm
iyr:2013 ecl:amb cid:350 eyr:2023 pid:028048884 hcl:#cfa07d byr:1929
hcl:#ae17e1 iyr:2013 eyr:2024 ecl:brn pid:760753108 byr:1931 hgt:179cm
hcl:#cfa07d eyr:2025 pid:166559648 iyr:2011 ecl:brn hgt:59in The first passport is valid - all eight fields are present. The second passport is invalid - it is missing hgt (the Height field).
The third passport is interesting; the only missing field is cid, so it looks like data from North Pole Credentials, not a passport at all! Surely, nobody would mind if you made the system temporarily ignore missing cid fields. Treat this “passport” as valid.
The fourth passport is missing two fields, cid and byr. Missing cid is fine, but missing any other field is not, so this passport is invalid.
According to the above rules, your improved system would report 2 valid passports.
Count the number of valid passports - those that have all required fields. Treat cid as optional. In your batch file, how many passports are valid?
## Read in the data
df <- data.frame(V1 = read_lines("Day4/input_day4.txt"))
## Generate a variable that shows the passport id
df2 <- df %>%
mutate(blank = ifelse(V1 == "", "blank", NA)) %>%
mutate(counter = seq_along(blank),
passportid = ifelse(blank == "blank", paste0(blank, counter),NA),
passportid = na.locf0(passportid, fromLast = TRUE)) %>%
filter(blank != "blank" |is.na(blank)) %>%
select(-blank, -counter)
## Collapse the records into one record
df2 <- df2 %>%
group_by(passportid) %>%
mutate(single_record = paste(V1, collapse = " ")) %>%
distinct(passportid, single_record)
## Separate the record into different variables
df2 <- df2 %>%
separate(single_record, into = paste0("C_", 1:8), sep = " ")
## Reshape the data
df2 <- df2 %>%
pivot_longer(names_to = "field_no",
values_to = "field",
cols = starts_with("C_"),
values_drop_na = TRUE)
## Separate the field into field_name and field_value
df2 <- df2 %>%
separate(field, into = c("field_name", "field_value"), sep = ":") %>%
mutate(field_name = trimws(field_name)) %>%
select(-field_no)
## Generate a variable that counts the number of fields per passport id
df2 <- df2 %>%
group_by(passportid) %>%
mutate(field_count = length(unique(field_name)))
## Check validity
df2 <- df2 %>%
group_by(passportid) %>%
mutate(validity = ifelse(field_count == 8, "valid",
ifelse(field_count == 7 & any(field_name == "cid") == FALSE,
"valid", "invalid")))
## Calculate the number of valid passports
valid_passports1 <- df2 %>%
distinct(passportid, validity) %>%
group_by(validity) %>%
count() %>%
filter(validity == "valid") %>%
pull(n)The line is moving more quickly now, but you overhear airport security talking about how passports with invalid data are getting through. Better add some data validation, quick!
You can continue to ignore the cid field, but each other field has strict rules about what values are valid for automatic validation:
byr (Birth Year) - four digits; at least 1920 and at most 2002.
iyr (Issue Year) - four digits; at least 2010 and at most 2020.
eyr (Expiration Year) - four digits; at least 2020 and at most 2030.
hgt (Height) - a number followed by either cm or in:
If cm, the number must be at least 150 and at most 193.
If in, the number must be at least 59 and at most 76.
hcl (Hair Color) - a # followed by exactly six characters 0-9 or a-f.
ecl (Eye Color) - exactly one of: amb blu brn gry grn hzl oth.
pid (Passport ID) - a nine-digit number, including leading zeroes.
cid (Country ID) - ignored, missing or not.
Your job is to count the passports where all required fields are both present and valid according to the above rules. Here are some example values:
byr valid: 2002
byr invalid: 2003
hgt valid: 60in
hgt valid: 190cm
hgt invalid: 190in
hgt invalid: 190
hcl valid: #123abc
hcl invalid: #123abz
hcl invalid: 123abc
ecl valid: brn
ecl invalid: wat
pid valid: 000000001
pid invalid: 0123456789
Here are some invalid passports:
eyr:1972 cid:100
hcl:#18171d ecl:amb hgt:170 pid:186cm iyr:2018 byr:1926
iyr:2019
hcl:#602927 eyr:1967 hgt:170cm
ecl:grn pid:012533040 byr:1946
hcl:dab227 iyr:2012
ecl:brn hgt:182cm pid:021572410 eyr:2020 byr:1992 cid:277
hgt:59cm ecl:zzz
eyr:2038 hcl:74454a iyr:2023
pid:3556412378 byr:2007
Here are some valid passports:
pid:087499704 hgt:74in ecl:grn iyr:2012 eyr:2030 byr:1980 hcl:#623a2f
eyr:2029 ecl:blu cid:129 byr:1989 iyr:2014 pid:896056539 hcl:#a97842 hgt:165cm
hcl:#888785 hgt:164cm byr:2001 iyr:2015 cid:88 pid:545766238 ecl:hzl eyr:2022
iyr:2010 hgt:158cm hcl:#b6652a ecl:blu byr:1944 eyr:2021 pid:093154719
Count the number of valid passports - those that have all required fields and valid values. Continue to treat cid as optional. In your batch file, how many passports are valid?
# Conditions:
## byr (Birth Year) - four digits; at least 1920 and at most 2002.
## iyr (Issue Year) - four digits; at least 2010 and at most 2020.
## eyr (Expiration Year) - four digits; at least 2020 and at most 2030.
## hgt (Height) - a number followed by either cm or in:
## If cm, the number must be at least 150 and at most 193.
## If in, the number must be at least 59 and at most 76.
## hcl (Hair Color) - a ## followed by exactly six characters 0-9 or a-f.
## ecl (Eye Color) - exactly one of: amb blu brn gry grn hzl oth.
## pid (Passport ID) - a nine-digit number, including leading zeroes.
## cid (Country ID) - ignored, missing or not.
df3 <- df2 %>%
mutate(field_value = trimws(field_value)) %>%
mutate(validity2 =
# byr (Birth Year) - four digits; at least 1920 and at most 2002.
ifelse(field_name == "byr" & nchar(field_value) == 4 &
between(as.numeric(field_value), 1920, 2002), "okay",
# iyr (Issue Year) - four digits; at least 2010 and at most 2020.
ifelse(field_name == "iyr" & nchar(field_value) == 4 &
between(as.numeric(field_value), 2010, 2020), "okay",
# eyr (Expiration Year) - four digits; at least 2020 and at most 2030.
ifelse(field_name == "eyr" & nchar(field_value) == 4 &
between(as.numeric(field_value), 2020, 2030), "okay",
# hgt (Height) - a number followed by either cm or in:
# If "cm", the number must be at least 150 and at most 193.
ifelse(field_name == "hgt" &
field_value %in% grep("cm", field_value, ignore.case = TRUE, value = TRUE) &
between(as.numeric(trimws(gsub("cm","", field_value))), 150, 193),"okay",
# If "in", the number must be at least 59 and at most 76.
ifelse(field_name == "hgt" &
field_value %in% grep("in", field_value,ignore.case = TRUE, value = TRUE) &
between(as.numeric(trimws(gsub("in","", field_value))), 59, 76),"okay",
# hcl (Hair Color) - a # followed by exactly six characters 0-9 or a-f.
ifelse(field_name == "hcl" &
field_value %in% grep("#", field_value,ignore.case = TRUE, value = TRUE) &
nchar(field_value) == 7, "okay",
# ecl (Eye Color) - exactly one of: amb blu brn gry grn hzl oth.
ifelse(field_name == "ecl" &
field_value %in% c("amb", "blu", "brn", "gry", "grn", "hzl", "oth") &
nchar(field_value) == 3, "okay",
# pid (Passport ID) - a nine-digit number, including leading zeroes.
ifelse(field_name == "pid" & nchar(gsub("[^0-9]+", "", field_value)) == 9, "okay",
# cid (Country ID) - ignored, missing or not.
ifelse(field_name == "cid", "okay", "invalid")))))))))) %>%
mutate(final_validity = ifelse(all(validity == "valid") & all(validity2 == "okay"),
"valid", "invalid"))
## Calculate the number of valid passports
valid_passports2 <- df3 %>%
distinct(passportid, final_validity) %>%
group_by(final_validity) %>%
count() %>%
filter(final_validity == "valid") %>%
pull(n)