thor/advent-of-code-go
1
0
Fork 0
mirror of https://github.com/Threnklyn/advent-of-code-go.git synced 2026-05-19 03:23:27 +02:00
Code Issues Packages Projects Releases Wiki Activity

cleanup effort pt 1

Browse Source
This commit is contained in:
alexchao26
2020-12-28 22:47:58 -05:00
parent 05ca29c8f2
commit feb7be9152
45 changed files with 434 additions and 3862 deletions
Download Patch File Download Diff File Expand all files Collapse all files
2018/day01/main.go
+46
View File
@@ -0,0 +1,46 @@
package main
import (
"flag"
"fmt"
"strings"
"github.com/alexchao26/advent-of-code-go/cast"
"github.com/alexchao26/advent-of-code-go/util"
)
func main() {
var part int
flag.IntVar(&part, "part", 1, "part 1 or 2")
flag.Parse()
fmt.Println("Running part", part)
ans := part1(util.ReadFile("./input.txt"))
fmt.Println("Output:", ans)
}
func part1(input string) int {
var sum int
for _, line := range strings.Split(input, "\n") {
sum += cast.ToInt(line)
}
return sum
}
func part2(input string) int {
var sum int
seen := map[int]bool{}
for {
for _, line := range strings.Split(input, "\n") {
sum += cast.ToInt(line)
if seen[sum] {
return sum
}
seen[sum] = true
}
}
panic("expect return from loop")
}
2018/day01/main_test.go
+41
View File
@@ -0,0 +1,41 @@
package main
import (
"testing"
"github.com/alexchao26/advent-of-code-go/util"
)
func Test_part1(t *testing.T) {
tests := []struct {
name string
input string
want int
}{
{"actual", util.ReadFile("input.txt"), 497},
}
for _, tt := range tests {
t.Run(tt.name, func(t *testing.T) {
if got := part1(tt.input); got != tt.want {
t.Errorf("part1() = %v, want %v", got, tt.want)
}
})
}
}
func Test_part2(t *testing.T) {
tests := []struct {
name string
input string
want int
}{
{"actual", util.ReadFile("input.txt"), 558},
}
for _, tt := range tests {
t.Run(tt.name, func(t *testing.T) {
if got := part2(tt.input); got != tt.want {
t.Errorf("part2() = %v, want %v", got, tt.want)
}
})
}
}
2018/day01/part1/main.go
-28
View File
@@ -1,28 +0,0 @@
package main
import (
"fmt"
"strconv"
"strings"
"github.com/alexchao26/advent-of-code-go/util"
)
func main() {
input := util.ReadFile("../input.txt")
sli := strings.Split(input, "\n")
var sum int
for _, instruction := range sli {
sign := instruction[:1]
num, _ := strconv.Atoi(instruction[1:])
if sign == "+" {
sum += num
} else {
sum -= num
}
}
fmt.Println("Final sum", sum)
}
2018/day01/part2/main.go
-35
View File
@@ -1,35 +0,0 @@
package main
import (
"fmt"
"strconv"
"strings"
"github.com/alexchao26/advent-of-code-go/util"
)
func main() {
input := util.ReadFile("../input.txt")
sli := strings.Split(input, "\n")
var sum int
seen := make(map[int]bool)
for {
for _, instruction := range sli {
sign := instruction[:1]
num, _ := strconv.Atoi(instruction[1:])
if sign == "+" {
sum += num
} else {
sum -= num
}
if seen[sum] {
fmt.Println("Number seen twice is:", sum)
return
}
seen[sum] = true
}
}
}
2018/day01/prompt.md
-53
View File
@@ -1,53 +0,0 @@
--- Day 1: Chronal Calibration ---
"We've detected some temporal anomalies," one of Santa's Elves at the Temporal Anomaly Research and Detection Instrument Station tells you. She sounded pretty worried when she called you down here. "At 500-year intervals into the past, someone has been changing Santa's history!"
"The good news is that the changes won't propagate to our time stream for another 25 days, and we have a device" - she attaches something to your wrist - "that will let you fix the changes with no such propagation delay. It's configured to send you 500 years further into the past every few days; that was the best we could do on such short notice."
"The bad news is that we are detecting roughly fifty anomalies throughout time; the device will indicate fixed anomalies with stars. The other bad news is that we only have one device and you're the best person for the job! Good lu--" She taps a button on the device and you suddenly feel like you're falling. To save Christmas, 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!
After feeling like you've been falling for a few minutes, you look at the device's tiny screen. "Error: Device must be calibrated before first use. Frequency drift detected. Cannot maintain destination lock." Below the message, the device shows a sequence of changes in frequency (your puzzle input). A value like +6 means the current frequency increases by 6; a value like -3 means the current frequency decreases by 3.
For example, if the device displays frequency changes of +1, -2, +3, +1, then starting from a frequency of zero, the following changes would occur:
Current frequency 0, change of +1; resulting frequency 1.
Current frequency 1, change of -2; resulting frequency -1.
Current frequency -1, change of +3; resulting frequency 2.
Current frequency 2, change of +1; resulting frequency 3.
In this example, the resulting frequency is 3.
Here are other example situations:
+1, +1, +1 results in 3
+1, +1, -2 results in 0
-1, -2, -3 results in -6
Starting with a frequency of zero, what is the resulting frequency after all of the changes in frequency have been applied?
Your puzzle answer was 497.
--- Part Two ---
You notice that the device repeats the same frequency change list over and over. To calibrate the device, you need to find the first frequency it reaches twice.
For example, using the same list of changes above, the device would loop as follows:
Current frequency 0, change of +1; resulting frequency 1.
Current frequency 1, change of -2; resulting frequency -1.
Current frequency -1, change of +3; resulting frequency 2.
Current frequency 2, change of +1; resulting frequency 3.
(At this point, the device continues from the start of the list.)
Current frequency 3, change of +1; resulting frequency 4.
Current frequency 4, change of -2; resulting frequency 2, which has already been seen.
In this example, the first frequency reached twice is 2. Note that your device might need to repeat its list of frequency changes many times before a duplicate frequency is found, and that duplicates might be found while in the middle of processing the list.
Here are other examples:
+1, -1 first reaches 0 twice.
+3, +3, +4, -2, -4 first reaches 10 twice.
-6, +3, +8, +5, -6 first reaches 5 twice.
+7, +7, -2, -7, -4 first reaches 14 twice.
What is the first frequency your device reaches twice?
Your puzzle answer was 558.
Both parts of this puzzle are complete! They provide two gold stars: **
2018/day02/main.go
+80
View File
@@ -0,0 +1,80 @@
package main
import (
"flag"
"fmt"
"strings"
"github.com/alexchao26/advent-of-code-go/util"
)
func main() {
var part int
flag.IntVar(&part, "part", 1, "part 1 or 2")
flag.Parse()
fmt.Println("Running part", part)
if part == 1 {
ans := part1(util.ReadFile("./input.txt"))
fmt.Println("Output:", ans)
} else {
ans := part2(util.ReadFile("./input.txt"))
fmt.Println("Output:", ans)
}
}
func part1(input string) int {
var twos, threes int
for _, boxID := range strings.Split(input, "\n") {
charCounts := getCharCount(boxID)
for _, v := range charCounts {
if v == 2 {
twos++
break
}
}
for _, v := range charCounts {
if v == 3 {
threes++
}
}
}
return twos * threes
}
func getCharCount(box string) map[rune]int {
chars := make(map[rune]int)
for _, c := range box {
chars[c]++
}
return chars
}
func part2(input string) string {
lines := strings.Split(input, "\n")
for i := 0; i < len(lines); i++ {
for j := i + 1; j < len(lines); j++ {
if sameChars := getSameCharacters(lines[i], lines[j]); sameChars != "" {
return sameChars
}
}
}
return ""
}
func getSameCharacters(str1, str2 string) string {
var mismatchSeen bool
var sameChars string
for i := 0; i < len(str1); i++ {
if str1[i] == str2[i] {
sameChars += string(str1[i])
} else if mismatchSeen {
// if a mismatch has already been seen, then it's 2 characters off
// return an empty string
return ""
} else {
mismatchSeen = true
}
}
return sameChars
}
2018/day02/main_test.go
+41
View File
@@ -0,0 +1,41 @@
package main
import (
"testing"
"github.com/alexchao26/advent-of-code-go/util"
)
func Test_part1(t *testing.T) {
tests := []struct {
name string
input string
want int
}{
{"actual", util.ReadFile("input.txt"), 5434},
}
for _, tt := range tests {
t.Run(tt.name, func(t *testing.T) {
if got := part1(tt.input); got != tt.want {
t.Errorf("part1() = %v, want %v", got, tt.want)
}
})
}
}
func Test_part2(t *testing.T) {
tests := []struct {
name string
input string
want string
}{
{"actual", util.ReadFile("input.txt"), "agimdjvlhedpsyoqfzuknpjwt"},
}
for _, tt := range tests {
t.Run(tt.name, func(t *testing.T) {
if got := part2(tt.input); got != tt.want {
t.Errorf("part2() = %v, want %v", got, tt.want)
}
})
}
}
2018/day02/part1/main.go
-51
View File
@@ -1,51 +0,0 @@
package main
import (
"fmt"
"strings"
"github.com/alexchao26/advent-of-code-go/util"
)
func main() {
input := util.ReadFile("../input.txt")
lines := strings.Split(input, "\n")
var twos, threes int
for _, boxID := range lines {
if hasTwo(boxID) {
twos++
}
if hasThree(boxID) {
threes++
}
}
fmt.Println("checksum", twos*threes)
}
func hasTwo(box string) bool {
chars := make(map[rune]int)
for _, c := range box {
chars[c]++
}
for _, v := range chars {
if v == 2 {
return true
}
}
return false
}
func hasThree(box string) bool {
chars := make(map[rune]int)
for _, c := range box {
chars[c]++
}
for _, v := range chars {
if v == 3 {
return true
}
}
return false
}
2018/day02/part2/main.go
-40
View File
@@ -1,40 +0,0 @@
package main
import (
"fmt"
"strings"
"github.com/alexchao26/advent-of-code-go/util"
)
// brute force it...
// assume only one valid answer
func main() {
input := util.ReadFile("../input.txt")
lines := strings.Split(input, "\n")
for i := 0; i < len(lines); i++ {
for j := i + 1; j < len(lines); j++ {
if sameChars := getSameCharacters(lines[i], lines[j]); sameChars != "" {
fmt.Println("common letters are", sameChars)
return
}
}
}
}
func getSameCharacters(str1, str2 string) string {
var mismatchSeen bool
var sameChars string
for i := 0; i < len(str1); i++ {
if str1[i] == str2[i] {
sameChars += string(str1[i])
} else if mismatchSeen {
// if a mismatch has already been seen, then it's 2 characters off
// return an empty string
return ""
} else {
mismatchSeen = true
}
}
return sameChars
}
2018/day02/prompt.md
-45
View File
@@ -1,45 +0,0 @@
--- Day 2: Inventory Management System ---
You stop falling through time, catch your breath, and check the screen on the device. "Destination reached. Current Year: 1518. Current Location: North Pole Utility Closet 83N10." You made it! Now, to find those anomalies.
Outside the utility closet, you hear footsteps and a voice. "...I'm not sure either. But now that so many people have chimneys, maybe he could sneak in that way?" Another voice responds, "Actually, we've been working on a new kind of suit that would let him fit through tight spaces like that. But, I heard that a few days ago, they lost the prototype fabric, the design plans, everything! Nobody on the team can even seem to remember important details of the project!"
"Wouldn't they have had enough fabric to fill several boxes in the warehouse? They'd be stored together, so the box IDs should be similar. Too bad it would take forever to search the warehouse for two similar box IDs..." They walk too far away to hear any more.
Late at night, you sneak to the warehouse - who knows what kinds of paradoxes you could cause if you were discovered - and use your fancy wrist device to quickly scan every box and produce a list of the likely candidates (your puzzle input).
To make sure you didn't miss any, you scan the likely candidate boxes again, counting the number that have an ID containing exactly two of any letter and then separately counting those with exactly three of any letter. You can multiply those two counts together to get a rudimentary checksum and compare it to what your device predicts.
For example, if you see the following box IDs:
abcdef contains no letters that appear exactly two or three times.
bababc contains two a and three b, so it counts for both.
abbcde contains two b, but no letter appears exactly three times.
abcccd contains three c, but no letter appears exactly two times.
aabcdd contains two a and two d, but it only counts once.
abcdee contains two e.
ababab contains three a and three b, but it only counts once.
Of these box IDs, four of them contain a letter which appears exactly twice, and three of them contain a letter which appears exactly three times. Multiplying these together produces a checksum of 4 * 3 = 12.
What is the checksum for your list of box IDs?
Your puzzle answer was 5434.
The first half of this puzzle is complete! It provides one gold star: *
--- Part Two ---
Confident that your list of box IDs is complete, you're ready to find the boxes full of prototype fabric.
The boxes will have IDs which differ by exactly one character at the same position in both strings. For example, given the following box IDs:
abcde
fghij
klmno
pqrst
fguij
axcye
wvxyz
The IDs abcde and axcye are close, but they differ by two characters (the second and fourth). However, the IDs fghij and fguij differ by exactly one character, the third (h and u). Those must be the correct boxes.
What letters are common between the two correct box IDs? (In the example above, this is found by removing the differing character from either ID, producing fgij.)
2018/day03/main.go
+108
View File
@@ -0,0 +1,108 @@
package main
import (
"flag"
"fmt"
"strings"
"github.com/alexchao26/advent-of-code-go/cast"
"github.com/alexchao26/advent-of-code-go/util"
)
func main() {
var part int
flag.IntVar(&part, "part", 1, "part 1 or 2")
flag.Parse()
fmt.Println("Running part", part)
if part == 1 {
ans := part1(util.ReadFile("./input.txt"))
util.CopyToClipboard(fmt.Sprintf("%v", ans))
fmt.Println("Output:", ans)
} else {
ans := part2(util.ReadFile("./input.txt"))
util.CopyToClipboard(fmt.Sprintf("%v", ans))
fmt.Println("Output:", ans)
}
}
func part1(input string) int {
lines := strings.Split(input, "\n")
seen := make(map[string]bool)
counted := make(map[string]bool)
var overlap int
for _, line := range lines {
row := cast.ToInt(line[strings.Index(line, "@")+2 : strings.Index(line, ",")])
col := cast.ToInt(line[strings.Index(line, ",")+1 : strings.Index(line, ":")])
width := cast.ToInt(line[strings.Index(line, ":")+2 : strings.Index(line, "x")])
height := cast.ToInt(line[strings.Index(line, "x")+1:])
for i := 0; i < width; i++ {
for j := 0; j < height; j++ {
coords := fmt.Sprintf("%vx%v", row+i, col+j)
if seen[coords] && !counted[coords] {
overlap++
counted[coords] = true
}
seen[coords] = true
}
}
}
return overlap
}
func part2(input string) int {
lines := strings.Split(input, "\n")
coords := makeMapOfCoordinates(lines)
for _, line := range lines {
uniqueID := hasNoOverlap(coords, line)
if uniqueID != -1 {
return uniqueID
}
}
panic("expect return from loop")
}
func makeMapOfCoordinates(lines []string) map[string]int {
seen := make(map[string]int)
for _, line := range lines {
// ID := line[:strings.Index(line, " @")]
row := cast.ToInt(line[strings.Index(line, "@")+2 : strings.Index(line, ",")])
col := cast.ToInt(line[strings.Index(line, ",")+1 : strings.Index(line, ":")])
width := cast.ToInt(line[strings.Index(line, ":")+2 : strings.Index(line, "x")])
height := cast.ToInt(line[strings.Index(line, "x")+1:])
for i := 0; i < width; i++ {
for j := 0; j < height; j++ {
coords := fmt.Sprintf("%vx%v", row+i, col+j)
seen[coords]++
}
}
}
return seen
}
// if cut is unique returns the ID, otherwise -1
func hasNoOverlap(seen map[string]int, line string) int {
ID := cast.ToInt(line[1:strings.Index(line, " @")])
row := cast.ToInt(line[strings.Index(line, "@")+2 : strings.Index(line, ",")])
col := cast.ToInt(line[strings.Index(line, ",")+1 : strings.Index(line, ":")])
width := cast.ToInt(line[strings.Index(line, ":")+2 : strings.Index(line, "x")])
height := cast.ToInt(line[strings.Index(line, "x")+1:])
for i := 0; i < width; i++ {
for j := 0; j < height; j++ {
coords := fmt.Sprintf("%vx%v", row+i, col+j)
if seen[coords] != 1 {
return -1
}
}
}
return ID
}
2018/day03/main_test.go
+41
View File
@@ -0,0 +1,41 @@
package main
import (
"testing"
"github.com/alexchao26/advent-of-code-go/util"
)
func Test_part1(t *testing.T) {
tests := []struct {
name string
input string
want int
}{
{"actual", util.ReadFile("input.txt"), 118223},
}
for _, tt := range tests {
t.Run(tt.name, func(t *testing.T) {
if got := part1(tt.input); got != tt.want {
t.Errorf("part1() = %v, want %v", got, tt.want)
}
})
}
}
func Test_part2(t *testing.T) {
tests := []struct {
name string
input string
want int
}{
{"actual", util.ReadFile("input.txt"), 412},
}
for _, tt := range tests {
t.Run(tt.name, func(t *testing.T) {
if got := part2(tt.input); got != tt.want {
t.Errorf("part2() = %v, want %v", got, tt.want)
}
})
}
}
2018/day03/part1/main.go
-39
View File
@@ -1,39 +0,0 @@
package main
import (
"fmt"
"strconv"
"strings"
"github.com/alexchao26/advent-of-code-go/util"
)
func main() {
input := util.ReadFile("../input.txt")
lines := strings.Split(input, "\n")
seen := make(map[string]bool)
counted := make(map[string]bool)
var overlap int
for _, line := range lines {
// ID := line[:strings.Index(line, " @")]
row, _ := strconv.Atoi(line[strings.Index(line, "@")+2 : strings.Index(line, ",")])
col, _ := strconv.Atoi(line[strings.Index(line, ",")+1 : strings.Index(line, ":")])
width, _ := strconv.Atoi(line[strings.Index(line, ":")+2 : strings.Index(line, "x")])
height, _ := strconv.Atoi(line[strings.Index(line, "x")+1:])
for i := 0; i < width; i++ {
for j := 0; j < height; j++ {
coords := fmt.Sprintf("%vx%v", row+i, col+j)
if seen[coords] && !counted[coords] {
overlap++
counted[coords] = true
}
seen[coords] = true
}
}
}
fmt.Println("Overlapping cells:", overlap)
}
2018/day03/part2/main.go
-62
View File
@@ -1,62 +0,0 @@
package main
import (
"fmt"
"strconv"
"strings"
"github.com/alexchao26/advent-of-code-go/util"
)
func main() {
input := util.ReadFile("../input.txt")
lines := strings.Split(input, "\n")
coords := makeMapOfCoordinates(lines)
for _, line := range lines {
uniqueID := hasNoOverlap(coords, line)
if uniqueID != -1 {
fmt.Println("Unique Plan ID is:", uniqueID)
}
}
}
func makeMapOfCoordinates(lines []string) map[string]int {
seen := make(map[string]int)
for _, line := range lines {
// ID := line[:strings.Index(line, " @")]
row, _ := strconv.Atoi(line[strings.Index(line, "@")+2 : strings.Index(line, ",")])
col, _ := strconv.Atoi(line[strings.Index(line, ",")+1 : strings.Index(line, ":")])
width, _ := strconv.Atoi(line[strings.Index(line, ":")+2 : strings.Index(line, "x")])
height, _ := strconv.Atoi(line[strings.Index(line, "x")+1:])
for i := 0; i < width; i++ {
for j := 0; j < height; j++ {
coords := fmt.Sprintf("%vx%v", row+i, col+j)
seen[coords]++
}
}
}
return seen
}
// if cut is unique returns the ID, otherwise -1
func hasNoOverlap(seen map[string]int, line string) int {
ID, _ := strconv.Atoi(line[1:strings.Index(line, " @")])
row, _ := strconv.Atoi(line[strings.Index(line, "@")+2 : strings.Index(line, ",")])
col, _ := strconv.Atoi(line[strings.Index(line, ",")+1 : strings.Index(line, ":")])
width, _ := strconv.Atoi(line[strings.Index(line, ":")+2 : strings.Index(line, "x")])
height, _ := strconv.Atoi(line[strings.Index(line, "x")+1:])
for i := 0; i < width; i++ {
for j := 0; j < height; j++ {
coords := fmt.Sprintf("%vx%v", row+i, col+j)
if seen[coords] != 1 {
return -1
}
}
}
return ID
}
2018/day03/prompt.md
-51
View File
@@ -1,51 +0,0 @@
--- Day 3: No Matter How You Slice It ---
The Elves managed to locate the chimney-squeeze prototype fabric for Santa's suit (thanks to someone who helpfully wrote its box IDs on the wall of the warehouse in the middle of the night). Unfortunately, anomalies are still affecting them - nobody can even agree on how to cut the fabric.
The whole piece of fabric they're working on is a very large square - at least 1000 inches on each side.
Each Elf has made a claim about which area of fabric would be ideal for Santa's suit. All claims have an ID and consist of a single rectangle with edges parallel to the edges of the fabric. Each claim's rectangle is defined as follows:
The number of inches between the left edge of the fabric and the left edge of the rectangle.
The number of inches between the top edge of the fabric and the top edge of the rectangle.
The width of the rectangle in inches.
The height of the rectangle in inches.
A claim like #123 @ 3,2: 5x4 means that claim ID 123 specifies a rectangle 3 inches from the left edge, 2 inches from the top edge, 5 inches wide, and 4 inches tall. Visually, it claims the square inches of fabric represented by # (and ignores the square inches of fabric represented by .) in the diagram below:
...........
...........
...#####...
...#####...
...#####...
...#####...
...........
...........
...........
The problem is that many of the claims overlap, causing two or more claims to cover part of the same areas. For example, consider the following claims:
#1 @ 1,3: 4x4
#2 @ 3,1: 4x4
#3 @ 5,5: 2x2
Visually, these claim the following areas:
........
...2222.
...2222.
.11XX22.
.11XX22.
.111133.
.111133.
........
The four square inches marked with X are claimed by both 1 and 2. (Claim 3, while adjacent to the others, does not overlap either of them.)
If the Elves all proceed with their own plans, none of them will have enough fabric. How many square inches of fabric are within two or more claims?
Your puzzle answer was 118223.
The first half of this puzzle is complete! It provides one gold star: *
--- Part Two ---
Amidst the chaos, you notice that exactly one claim doesn't overlap by even a single square inch of fabric with any other claim. If you can somehow draw attention to it, maybe the Elves will be able to make Santa's suit after all!
For example, in the claims above, only claim 3 is intact after all claims are made.
What is the ID of the only claim that doesn't overlap?
2018/day04/part1/main.go → 2018/day04/main.go
+38 -7
View File
@@ -1,6 +1,7 @@
package main
import (
"flag"
"fmt"
"sort"
"strconv"
@@ -10,7 +11,16 @@ import (
)
func main() {
input := util.ReadFile("../input.txt")
var part int
flag.IntVar(&part, "part", 1, "part 1 or 2")
flag.Parse()
fmt.Println("Running part", part)
ans := part1(util.ReadFile("./input.txt"), part)
fmt.Println("Output:", ans)
}
func part1(input string, part int) int {
lines := strings.Split(input, "\n")
// sort inputs by time stamp, string sorting is sufficient
@@ -24,6 +34,8 @@ func main() {
// find which guard has slept the most total time
// then find which minute he is asleep at most frequently
mapIDGuard := make(map[int]*guard)
// for part 2, track each guard for the actual minute asleep
mapIDToMinutesArray := make(map[int]*[60]int)
lastGuardID := timeEntries[0].ID
for i, timeEntry := range timeEntries {
if timeEntry.ID != 0 {
@@ -39,15 +51,22 @@ func main() {
if mapIDGuard[lastGuardID] == nil {
mapIDGuard[lastGuardID] = &guard{}
}
// part 2 parsing
if mapIDToMinutesArray[lastGuardID] == nil {
mapIDToMinutesArray[lastGuardID] = &[60]int{}
}
mapIDGuard[lastGuardID].totalTimeAsleep += endTime - startTime
for startTime < endTime {
mapIDGuard[lastGuardID].minutesAsleep[startTime]++
mapIDToMinutesArray[lastGuardID][startTime]++
startTime++
}
}
}
}
if part == 1 {
// who sleeps the most
var IDOfSleepiestGuard, bestMinute, highestFreq int
for i, g := range mapIDGuard {
@@ -68,11 +87,24 @@ func main() {
}
// print ID * time (minute)
fmt.Printf("Sleepiest guard is %v, at minute %v.\ngithub.com/alexchao26/advent-of-code-go answer: %v\n",
IDOfSleepiestGuard,
bestMinute,
IDOfSleepiestGuard*bestMinute,
)
return IDOfSleepiestGuard * bestMinute
}
// part 2 stuff
var highestFreq, ID, bestMinute int
// find the minute they are the asleep the most
for i, arr := range mapIDToMinutesArray {
for min, freq := range arr {
if freq > highestFreq {
bestMinute = min
highestFreq = freq
ID = i
}
}
}
// print ID * time (minute)
return ID * bestMinute
}
type entry struct {
@@ -81,7 +113,6 @@ type entry struct {
}
func makeEntry(line string) entry {
fmt.Println(line)
var e entry
e.year, _ = strconv.Atoi(line[1:5])
e.month, _ = strconv.Atoi(line[6:8])
2018/day04/main_test.go
+26
View File
@@ -0,0 +1,26 @@
package main
import (
"testing"
"github.com/alexchao26/advent-of-code-go/util"
)
func Test_part1(t *testing.T) {
tests := []struct {
name string
input string
part int
want int
}{
{"actual", util.ReadFile("input.txt"), 1, 38813},
{"actual", util.ReadFile("input.txt"), 2, 141071},
}
for _, tt := range tests {
t.Run(tt.name, func(t *testing.T) {
if got := part1(tt.input, tt.part); got != tt.want {
t.Errorf("part1() = %v, want %v", got, tt.want)
}
})
}
}
2018/day04/part2/main.go
-100
View File
@@ -1,100 +0,0 @@
package main
import (
"fmt"
"sort"
"strconv"
"strings"
"github.com/alexchao26/advent-of-code-go/util"
)
func main() {
input := util.ReadFile("../input.txt")
lines := strings.Split(input, "\n")
// sort inputs by time stamp, string sorting is sufficient
sort.Strings(lines)
timeEntries := make([]entry, len(lines))
for i, line := range lines {
timeEntries[i] = makeEntry(line)
}
// find which guard has slept the most total time
// then find which minute he is asleep at most frequently
mapIDGuard := make(map[int]*[60]int)
lastGuardID := timeEntries[0].ID
for i, timeEntry := range timeEntries {
if timeEntry.ID != 0 {
lastGuardID = timeEntry.ID
} else {
// if the time entry is awake, then check the next one entry
// if the next entry is the same day, then assume its the same guard
// update that guard's stats
if !timeEntry.awake && i+1 != len(timeEntries) &&
timeEntries[i+1].day == timeEntry.day {
endTime := timeEntries[i+1].minute
startTime := timeEntry.minute
if mapIDGuard[lastGuardID] == nil {
mapIDGuard[lastGuardID] = &[60]int{}
}
for startTime < endTime {
mapIDGuard[lastGuardID][startTime]++
startTime++
}
}
}
}
var highestFreq, ID, bestMinute int
// find the minute they are the asleep the most
for i, arr := range mapIDGuard {
for min, freq := range arr {
if freq > highestFreq {
bestMinute = min
highestFreq = freq
ID = i
}
}
}
// print ID * time (minute)
fmt.Printf("Guard %v is asleep the most at minute %v.\ngithub.com/alexchao26/advent-of-code-go answer: %v\n",
ID,
bestMinute,
ID*bestMinute,
)
}
type entry struct {
ID, year, month, day, minute int
awake bool
}
func makeEntry(line string) entry {
fmt.Println(line)
var e entry
e.year, _ = strconv.Atoi(line[1:5])
e.month, _ = strconv.Atoi(line[6:8])
e.day, _ = strconv.Atoi(line[9:11])
hour, _ := strconv.Atoi(line[12:14])
// if started before midnight, zero out minute value
if hour != 0 {
e.minute = 0
} else {
e.minute, _ = strconv.Atoi(line[15:17])
}
// if the instruction is that the guard has fallen asleep, leave awake=false
if strings.Contains(line, "falls asleep") {
return e
}
e.awake = true
if strings.Contains(line, "Guard") {
e.ID, _ = strconv.Atoi(line[strings.Index(line, "#")+1 : strings.Index(line, " begins")])
}
return e
}
2018/day04/prompt.md
-60
View File
@@ -1,60 +0,0 @@
--- Day 4: Repose Record ---
You've sneaked into another supply closet - this time, it's across from the prototype suit manufacturing lab. You need to sneak inside and fix the issues with the suit, but there's a guard stationed outside the lab, so this is as close as you can safely get.
As you search the closet for anything that might help, you discover that you're not the first person to want to sneak in. Covering the walls, someone has spent an hour starting every midnight for the past few months secretly observing this guard post! They've been writing down the ID of the one guard on duty that night - the Elves seem to have decided that one guard was enough for the overnight shift - as well as when they fall asleep or wake up while at their post (your puzzle input).
For example, consider the following records, which have already been organized into chronological order:
[1518-11-01 00:00] Guard #10 begins shift
[1518-11-01 00:05] falls asleep
[1518-11-01 00:25] wakes up
[1518-11-01 00:30] falls asleep
[1518-11-01 00:55] wakes up
[1518-11-01 23:58] Guard #99 begins shift
[1518-11-02 00:40] falls asleep
[1518-11-02 00:50] wakes up
[1518-11-03 00:05] Guard #10 begins shift
[1518-11-03 00:24] falls asleep
[1518-11-03 00:29] wakes up
[1518-11-04 00:02] Guard #99 begins shift
[1518-11-04 00:36] falls asleep
[1518-11-04 00:46] wakes up
[1518-11-05 00:03] Guard #99 begins shift
[1518-11-05 00:45] falls asleep
[1518-11-05 00:55] wakes up
Timestamps are written using year-month-day hour:minute format. The guard falling asleep or waking up is always the one whose shift most recently started. Because all asleep/awake times are during the midnight hour (00:00 - 00:59), only the minute portion (00 - 59) is relevant for those events.
Visually, these records show that the guards are asleep at these times:
Date ID Minute
000000000011111111112222222222333333333344444444445555555555
012345678901234567890123456789012345678901234567890123456789
11-01 #10 .....####################.....#########################.....
11-02 #99 ........................................##########..........
11-03 #10 ........................#####...............................
11-04 #99 ....................................##########..............
11-05 #99 .............................................##########.....
The columns are Date, which shows the month-day portion of the relevant day; ID, which shows the guard on duty that day; and Minute, which shows the minutes during which the guard was asleep within the midnight hour. (The Minute column's header shows the minute's ten's digit in the first row and the one's digit in the second row.) Awake is shown as ., and asleep is shown as #.
Note that guards count as asleep on the minute they fall asleep, and they count as awake on the minute they wake up. For example, because Guard #10 wakes up at 00:25 on 1518-11-01, minute 25 is marked as awake.
If you can figure out the guard most likely to be asleep at a specific time, you might be able to trick that guard into working tonight so you can have the best chance of sneaking in. You have two strategies for choosing the best guard/minute combination.
Strategy 1: Find the guard that has the most minutes asleep. What minute does that guard spend asleep the most?
In the example above, Guard #10 spent the most minutes asleep, a total of 50 minutes (20+25+5), while Guard #99 only slept for a total of 30 minutes (10+10+10). Guard #10 was asleep most during minute 24 (on two days, whereas any other minute the guard was asleep was only seen on one day).
While this example listed the entries in chronological order, your entries are in the order you found them. You'll need to organize them before they can be analyzed.
What is the ID of the guard you chose multiplied by the minute you chose? (In the above example, the answer would be 10 * 24 = 240.)
Your puzzle answer was 38813.
The first half of this puzzle is complete! It provides one gold star: *
--- Part Two ---
Strategy 2: Of all guards, which guard is most frequently asleep on the same minute?
In the example above, Guard #99 spent minute 45 asleep more than any other guard or minute - three times in total. (In all other cases, any guard spent any minute asleep at most twice.)
What is the ID of the guard you chose multiplied by the minute you chose? (In the above example, the answer would be 99 * 45 = 4455.)
2018/day05/prompt.md
-61
View File
@@ -1,61 +0,0 @@
--- Day 5: Alchemical Reduction ---
You've managed to sneak in to the prototype suit manufacturing lab. The Elves are making decent progress, but are still struggling with the suit's size reduction capabilities.
While the very latest in 1518 alchemical technology might have solved their problem eventually, you can do better. You scan the chemical composition of the suit's material and discover that it is formed by extremely long polymers (one of which is available as your puzzle input).
The polymer is formed by smaller units which, when triggered, react with each other such that two adjacent units of the same type and opposite polarity are destroyed. Units' types are represented by letters; units' polarity is represented by capitalization. For instance, r and R are units with the same type but opposite polarity, whereas r and s are entirely different types and do not react.
For example:
In aA, a and A react, leaving nothing behind.
In abBA, bB destroys itself, leaving aA. As above, this then destroys itself, leaving nothing.
In abAB, no two adjacent units are of the same type, and so nothing happens.
In aabAAB, even though aa and AA are of the same type, their polarities match, and so nothing happens.
Now, consider a larger example, dabAcCaCBAcCcaDA:
dabAcCaCBAcCcaDA The first 'cC' is removed.
dabAaCBAcCcaDA This creates 'Aa', which is removed.
dabCBAcCcaDA Either 'cC' or 'Cc' are removed (the result is the same).
dabCBAcaDA No further actions can be taken.
After all possible reactions, the resulting polymer contains 10 units.
How many units remain after fully reacting the polymer you scanned? (Note: in this puzzle and others, the input is large; if you copy/paste your input, make sure you get the whole thing.)
--- Part Two ---
Time to improve the polymer.
One of the unit types is causing problems; it's preventing the polymer from collapsing as much as it should. Your goal is to figure out which unit type is causing the most problems, remove all instances of it (regardless of polarity), fully react the remaining polymer, and measure its length.
For example, again using the polymer dabAcCaCBAcCcaDA from above:
Removing all A/a units produces dbcCCBcCcD. Fully reacting this polymer produces dbCBcD, which has length 6.
Removing all B/b units produces daAcCaCAcCcaDA. Fully reacting this polymer produces daCAcaDA, which has length 8.
Removing all C/c units produces dabAaBAaDA. Fully reacting this polymer produces daDA, which has length 4.
Removing all D/d units produces abAcCaCBAcCcaA. Fully reacting this polymer produces abCBAc, which has length 6.
In this example, removing all C/c units was best, producing the answer 4.
What is the length of the shortest polymer you can produce by removing all units of exactly one type and fully reacting the result?
2018/day06/prompt.md
-115
View File
@@ -1,115 +0,0 @@
--- Day 6: Chronal Coordinates ---
The device on your wrist beeps several times, and once again you feel like you're falling.
"Situation critical," the device announces. "Destination indeterminate. Chronal interference detected. Please specify new target coordinates."
The device then produces a list of coordinates (your puzzle input). Are they places it thinks are safe or dangerous? It recommends you check manual page 729. The Elves did not give you a manual.
If they're dangerous, maybe you can minimize the danger by finding the coordinate that gives the largest distance from the other points.
Using only the Manhattan distance, determine the area around each coordinate by counting the number of integer X,Y locations that are closest to that coordinate (and aren't tied in distance to any other coordinate).
Your goal is to find the size of the largest area that isn't infinite. For example, consider the following list of coordinates:
1, 1
1, 6
8, 3
3, 4
5, 5
8, 9
If we name these coordinates A through F, we can draw them on a grid, putting 0,0 at the top left:
..........
.A........
..........
........C.
...D......
.....E....
.B........
..........
..........
........F.
This view is partial - the actual grid extends infinitely in all directions. Using the Manhattan distance, each location's closest coordinate can be determined, shown here in lowercase:
aaaaa.cccc
aAaaa.cccc
aaaddecccc
aadddeccCc
..dDdeeccc
bb.deEeecc
bBb.eeee..
bbb.eeefff
bbb.eeffff
bbb.ffffFf
Locations shown as . are equally far from two or more coordinates, and so they don't count as being closest to any.
In this example, the areas of coordinates A, B, C, and F are infinite - while not shown here, their areas extend forever outside the visible grid. However, the areas of coordinates D and E are finite: D is closest to 9 locations, and E is closest to 17 (both including the coordinate's location itself). Therefore, in this example, the size of the largest area is 17.
What is the size of the largest area that isn't infinite?
--- Part Two ---
On the other hand, if the coordinates are safe, maybe the best you can do is try to find a region near as many coordinates as possible.
For example, suppose you want the sum of the Manhattan distance to all of the coordinates to be less than 32. For each location, add up the distances to all of the given coordinates; if the total of those distances is less than 32, that location is within the desired region. Using the same coordinates as above, the resulting region looks like this:
..........
.A........
..........
...###..C.
..#D###...
..###E#...
.B.###....
..........
..........
........F.
In particular, consider the highlighted location 4,3 located at the top middle of the region. Its calculation is as follows, where abs() is the absolute value function:
Distance to coordinate A: abs(4-1) + abs(3-1) =  5
Distance to coordinate B: abs(4-1) + abs(3-6) =  6
Distance to coordinate C: abs(4-8) + abs(3-3) =  4
Distance to coordinate D: abs(4-3) + abs(3-4) =  2
Distance to coordinate E: abs(4-5) + abs(3-5) =  3
Distance to coordinate F: abs(4-8) + abs(3-9) = 10
Total distance: 5 + 6 + 4 + 2 + 3 + 10 = 30
Because the total distance to all coordinates (30) is less than 32, the location is within the region.
This region, which also includes coordinates D and E, has a total size of 16.
Your actual region will need to be much larger than this example, though, instead including all locations with a total distance of less than 10000.
What is the size of the region containing all locations which have a total distance to all given coordinates of less than 10000?
2018/day07/prompt.md
-98
View File
@@ -1,98 +0,0 @@
--- Day 7: The Sum of Its Parts ---
You find yourself standing on a snow-covered coastline; apparently, you landed a little off course. The region is too hilly to see the North Pole from here, but you do spot some Elves that seem to be trying to unpack something that washed ashore. It's quite cold out, so you decide to risk creating a paradox by asking them for directions.
"Oh, are you the search party?" Somehow, you can understand whatever Elves from the year 1018 speak; you assume it's Ancient Nordic Elvish. Could the device on your wrist also be a translator? "Those clothes don't look very warm; take this." They hand you a heavy coat.
"We do need to find our way back to the North Pole, but we have higher priorities at the moment. You see, believe it or not, this box contains something that will solve all of Santa's transportation problems - at least, that's what it looks like from the pictures in the instructions." It doesn't seem like they can read whatever language it's in, but you can: "Sleigh kit. Some assembly required."
"'Sleigh'? What a wonderful name! You must help us assemble this 'sleigh' at once!" They start excitedly pulling more parts out of the box.
The instructions specify a series of steps and requirements about which steps must be finished before others can begin (your puzzle input). Each step is designated by a single letter. For example, suppose you have the following instructions:
Step C must be finished before step A can begin.
Step C must be finished before step F can begin.
Step A must be finished before step B can begin.
Step A must be finished before step D can begin.
Step B must be finished before step E can begin.
Step D must be finished before step E can begin.
Step F must be finished before step E can begin.
Visually, these requirements look like this:
-->A--->B--
/ \ \
C -->D----->E
\ /
---->F-----
Your first goal is to determine the order in which the steps should be completed. If more than one step is ready, choose the step which is first alphabetically. In this example, the steps would be completed as follows:
Only C is available, and so it is done first.
Next, both A and F are available. A is first alphabetically, so it is done next.
Then, even though F was available earlier, steps B and D are now also available, and B is the first alphabetically of the three.
After that, only D and F are available. E is not available because only some of its prerequisites are complete. Therefore, D is completed next.
F is the only choice, so it is done next.
Finally, E is completed.
So, in this example, the correct order is CABDFE.
In what order should the steps in your instructions be completed?
--- Part Two ---
As you're about to begin construction, four of the Elves offer to help. "The sun will set soon; it'll go faster if we work together." Now, you need to account for multiple people working on steps simultaneously. If multiple steps are available, workers should still begin them in alphabetical order.
Each step takes 60 seconds plus an amount corresponding to its letter: A=1, B=2, C=3, and so on. So, step A takes 60+1=61 seconds, while step Z takes 60+26=86 seconds. No time is required between steps.
To simplify things for the example, however, suppose you only have help from one Elf (a total of two workers) and that each step takes 60 fewer seconds (so that step A takes 1 second and step Z takes 26 seconds). Then, using the same instructions as above, this is how each second would be spent:
Second Worker 1 Worker 2 Done
0 C .
1 C .
2 C .
3 A F C
4 B F CA
5 B F CA
6 D F CAB
7 D F CAB
8 D F CAB
9 D . CABF
10 E . CABFD
11 E . CABFD
12 E . CABFD
13 E . CABFD
14 E . CABFD
15 . . CABFDE
Each row represents one second of time. The Second column identifies how many seconds have passed as of the beginning of that second. Each worker column shows the step that worker is currently doing (or . if they are idle). The Done column shows completed steps.
Note that the order of the steps has changed; this is because steps now take time to finish and multiple workers can begin multiple steps simultaneously.
In this example, it would take 15 seconds for two workers to complete these steps.
With 5 workers and the 60+ second step durations described above, how long will it take to complete all of the steps?
2018/day08/prompt.md
-82
View File
@@ -1,82 +0,0 @@
--- Day 8: Memory Maneuver ---
The sleigh is much easier to pull than you'd expect for something its weight. Unfortunately, neither you nor the Elves know which way the North Pole is from here.
You check your wrist device for anything that might help. It seems to have some kind of navigation system! Activating the navigation system produces more bad news: "Failed to start navigation system. Could not read software license file."
The navigation system's license file consists of a list of numbers (your puzzle input). The numbers define a data structure which, when processed, produces some kind of tree that can be used to calculate the license number.
The tree is made up of nodes; a single, outermost node forms the tree's root, and it contains all other nodes in the tree (or contains nodes that contain nodes, and so on).
Specifically, a node consists of:
A header, which is always exactly two numbers:
The quantity of child nodes.
The quantity of metadata entries.
Zero or more child nodes (as specified in the header).
One or more metadata entries (as specified in the header).
Each child node is itself a node that has its own header, child nodes, and metadata. For example:
2 3 0 3 10 11 12 1 1 0 1 99 2 1 1 2
A----------------------------------
B----------- C-----------
D-----
In this example, each node of the tree is also marked with an underline starting with a letter for easier identification. In it, there are four nodes:
A, which has 2 child nodes (B, C) and 3 metadata entries (1, 1, 2).
B, which has 0 child nodes and 3 metadata entries (10, 11, 12).
C, which has 1 child node (D) and 1 metadata entry (2).
D, which has 0 child nodes and 1 metadata entry (99).
The first check done on the license file is to simply add up all of the metadata entries. In this example, that sum is 1+1+2+10+11+12+2+99=138.
What is the sum of all metadata entries?
--- Part Two ---
The second check is slightly more complicated: you need to find the value of the root node (A in the example above).
The value of a node depends on whether it has child nodes.
If a node has no child nodes, its value is the sum of its metadata entries. So, the value of node B is 10+11+12=33, and the value of node D is 99.
However, if a node does have child nodes, the metadata entries become indexes which refer to those child nodes. A metadata entry of 1 refers to the first child node, 2 to the second, 3 to the third, and so on. The value of this node is the sum of the values of the child nodes referenced by the metadata entries. If a referenced child node does not exist, that reference is skipped. A child node can be referenced multiple time and counts each time it is referenced. A metadata entry of 0 does not refer to any child node.
For example, again using the above nodes:
Node C has one metadata entry, 2. Because node C has only one child node, 2 references a child node which does not exist, and so the value of node C is 0.
Node A has three metadata entries: 1, 1, and 2. The 1 references node A's first child node, B, and the 2 references node A's second child node, C. Because node B has a value of 33 and node C has a value of 0, the value of node A is 33+33+0=66.
So, in this example, the value of the root node is 66.
What is the value of the root node?
2018/day09/prompt.md
-73
View File
@@ -1,73 +0,0 @@
--- Day 9: Marble Mania ---
You talk to the Elves while you wait for your navigation system to initialize. To pass the time, they introduce you to their favorite marble game.
The Elves play this game by taking turns arranging the marbles in a circle according to very particular rules. The marbles are numbered starting with 0 and increasing by 1 until every marble has a number.
First, the marble numbered 0 is placed in the circle. At this point, while it contains only a single marble, it is still a circle: the marble is both clockwise from itself and counter-clockwise from itself. This marble is designated the current marble.
Then, each Elf takes a turn placing the lowest-numbered remaining marble into the circle between the marbles that are 1 and 2 marbles clockwise of the current marble. (When the circle is large enough, this means that there is one marble between the marble that was just placed and the current marble.) The marble that was just placed then becomes the current marble.
However, if the marble that is about to be placed has a number which is a multiple of 23, something entirely different happens. First, the current player keeps the marble they would have placed, adding it to their score. In addition, the marble 7 marbles counter-clockwise from the current marble is removed from the circle and also added to the current player's score. The marble located immediately clockwise of the marble that was removed becomes the new current marble.
For example, suppose there are 9 players. After the marble with value 0 is placed in the middle, each player (shown in square brackets) takes a turn. The result of each of those turns would produce circles of marbles like this, where clockwise is to the right and the resulting current marble is in parentheses:
[-] (0)
[1] 0 (1)
[2] 0 (2) 1
[3] 0 2 1 (3)
[4] 0 (4) 2 1 3
[5] 0 4 2 (5) 1 3
[6] 0 4 2 5 1 (6) 3
[7] 0 4 2 5 1 6 3 (7)
[8] 0 (8) 4 2 5 1 6 3 7
[9] 0 8 4 (9) 2 5 1 6 3 7
[1] 0 8 4 9 2(10) 5 1 6 3 7
[2] 0 8 4 9 2 10 5(11) 1 6 3 7
[3] 0 8 4 9 2 10 5 11 1(12) 6 3 7
[4] 0 8 4 9 2 10 5 11 1 12 6(13) 3 7
[5] 0 8 4 9 2 10 5 11 1 12 6 13 3(14) 7
[6] 0 8 4 9 2 10 5 11 1 12 6 13 3 14 7(15)
[7] 0(16) 8 4 9 2 10 5 11 1 12 6 13 3 14 7 15
[8] 0 16 8(17) 4 9 2 10 5 11 1 12 6 13 3 14 7 15
[9] 0 16 8 17 4(18) 9 2 10 5 11 1 12 6 13 3 14 7 15
[1] 0 16 8 17 4 18 9(19) 2 10 5 11 1 12 6 13 3 14 7 15
[2] 0 16 8 17 4 18 9 19 2(20)10 5 11 1 12 6 13 3 14 7 15
[3] 0 16 8 17 4 18 9 19 2 20 10(21) 5 11 1 12 6 13 3 14 7 15
[4] 0 16 8 17 4 18 9 19 2 20 10 21 5(22)11 1 12 6 13 3 14 7 15
[5] 0 16 8 17 4 18(19) 2 20 10 21 5 22 11 1 12 6 13 3 14 7 15
[6] 0 16 8 17 4 18 19 2(24)20 10 21 5 22 11 1 12 6 13 3 14 7 15
[7] 0 16 8 17 4 18 19 2 24 20(25)10 21 5 22 11 1 12 6 13 3 14 7 15
The goal is to be the player with the highest score after the last marble is used up. Assuming the example above ends after the marble numbered 25, the winning score is 23+9=32 (because player 5 kept marble 23 and removed marble 9, while no other player got any points in this very short example game).
Here are a few more examples:
10 players; last marble is worth 1618 points: high score is 8317
13 players; last marble is worth 7999 points: high score is 146373
17 players; last marble is worth 1104 points: high score is 2764
21 players; last marble is worth 6111 points: high score is 54718
30 players; last marble is worth 5807 points: high score is 37305
What is the winning Elf's score?
--- Part Two ---
Amused by the speed of your answer, the Elves are curious:
What would the new winning Elf's score be if the number of the last marble were 100 times larger?
2018/day11/prompt.md
-102
View File
@@ -1,102 +0,0 @@
--- Day 11: Chronal Charge ---
You watch the Elves and their sleigh fade into the distance as they head toward the North Pole.
Actually, you're the one fading. The falling sensation returns.
The low fuel warning light is illuminated on your wrist-mounted device. Tapping it once causes it to project a hologram of the situation: a 300x300 grid of fuel cells and their current power levels, some negative. You're not sure what negative power means in the context of time travel, but it can't be good.
Each fuel cell has a coordinate ranging from 1 to 300 in both the X (horizontal) and Y (vertical) direction. In X,Y notation, the top-left cell is 1,1, and the top-right cell is 300,1.
The interface lets you select any 3x3 square of fuel cells. To increase your chances of getting to your destination, you decide to choose the 3x3 square with the largest total power.
The power level in a given fuel cell can be found through the following process:
Find the fuel cell's rack ID, which is its X coordinate plus 10.
Begin with a power level of the rack ID times the Y coordinate.
Increase the power level by the value of the grid serial number (your puzzle input).
Set the power level to itself multiplied by the rack ID.
Keep only the hundreds digit of the power level (so 12345 becomes 3; numbers with no hundreds digit become 0).
Subtract 5 from the power level.
For example, to find the power level of the fuel cell at 3,5 in a grid with serial number 8:
The rack ID is 3 + 10 = 13.
The power level starts at 13 * 5 = 65.
Adding the serial number produces 65 + 8 = 73.
Multiplying by the rack ID produces 73 * 13 = 949.
The hundreds digit of 949 is 9.
Subtracting 5 produces 9 - 5 = 4.
So, the power level of this fuel cell is 4.
Here are some more example power levels:
Fuel cell at  122,79, grid serial number 57: power level -5.
Fuel cell at 217,196, grid serial number 39: power level  0.
Fuel cell at 101,153, grid serial number 71: power level  4.
Your goal is to find the 3x3 square which has the largest total power. The square must be entirely within the 300x300 grid. Identify this square using the X,Y coordinate of its top-left fuel cell. For example:
For grid serial number 18, the largest total 3x3 square has a top-left corner of 33,45 (with a total power of 29); these fuel cells appear in the middle of this 5x5 region:
-2 -4 4 4 4
-4 4 4 4 -5
4 3 3 4 -4
1 1 2 4 -3
-1 0 2 -5 -2
For grid serial number 42, the largest 3x3 square's top-left is 21,61 (with a total power of 30); they are in the middle of this region:
-3 4 2 2 2
-4 4 3 3 4
-5 3 3 4 -4
4 3 3 4 -3
3 3 3 -5 -1
What is the X,Y coordinate of the top-left fuel cell of the 3x3 square with the largest total power?
--- Part Two ---
You discover a dial on the side of the device; it seems to let you select a square of any size, not just 3x3. Sizes from 1x1 to 300x300 are supported.
Realizing this, you now must find the square of any size with the largest total power. Identify this square by including its size as a third parameter after the top-left coordinate: a 9x9 square with a top-left corner of 3,5 is identified as 3,5,9.
For example:
For grid serial number 18, the largest total square (with a total power of 113) is 16x16 and has a top-left corner of 90,269, so its identifier is 90,269,16.
For grid serial number 42, the largest total square (with a total power of 119) is 12x12 and has a top-left corner of 232,251, so its identifier is 232,251,12.
What is the X,Y,size identifier of the square with the largest total power?
2018/day12/prompt.md
-93
View File
@@ -1,93 +0,0 @@
--- Day 12: Subterranean Sustainability ---
The year 518 is significantly more underground than your history books implied. Either that, or you've arrived in a vast cavern network under the North Pole.
After exploring a little, you discover a long tunnel that contains a row of small pots as far as you can see to your left and right. A few of them contain plants - someone is trying to grow things in these geothermally-heated caves.
The pots are numbered, with 0 in front of you. To the left, the pots are numbered -1, -2, -3, and so on; to the right, 1, 2, 3.... Your puzzle input contains a list of pots from 0 to the right and whether they do (#) or do not (.) currently contain a plant, the initial state. (No other pots currently contain plants.) For example, an initial state of #..##.... indicates that pots 0, 3, and 4 currently contain plants.
Your puzzle input also contains some notes you find on a nearby table: someone has been trying to figure out how these plants spread to nearby pots. Based on the notes, for each generation of plants, a given pot has or does not have a plant based on whether that pot (and the two pots on either side of it) had a plant in the last generation. These are written as LLCRR => N, where L are pots to the left, C is the current pot being considered, R are the pots to the right, and N is whether the current pot will have a plant in the next generation. For example:
A note like ..#.. => . means that a pot that contains a plant but with no plants within two pots of it will not have a plant in it during the next generation.
A note like ##.## => . means that an empty pot with two plants on each side of it will remain empty in the next generation.
A note like .##.# => # means that a pot has a plant in a given generation if, in the previous generation, there were plants in that pot, the one immediately to the left, and the one two pots to the right, but not in the ones immediately to the right and two to the left.
It's not clear what these plants are for, but you're sure it's important, so you'd like to make sure the current configuration of plants is sustainable by determining what will happen after 20 generations.
For example, given the following input:
initial state: #..#.#..##......###...###
...## => #
..#.. => #
.#... => #
.#.#. => #
.#.## => #
.##.. => #
.#### => #
#.#.# => #
#.### => #
##.#. => #
##.## => #
###.. => #
###.# => #
####. => #
For brevity, in this example, only the combinations which do produce a plant are listed. (Your input includes all possible combinations.) Then, the next 20 generations will look like this:
1 2 3
0 0 0 0
0: ...#..#.#..##......###...###...........
1: ...#...#....#.....#..#..#..#...........
2: ...##..##...##....#..#..#..##..........
3: ..#.#...#..#.#....#..#..#...#..........
4: ...#.#..#...#.#...#..#..##..##.........
5: ....#...##...#.#..#..#...#...#.........
6: ....##.#.#....#...#..##..##..##........
7: ...#..###.#...##..#...#...#...#........
8: ...#....##.#.#.#..##..##..##..##.......
9: ...##..#..#####....#...#...#...#.......
10: ..#.#..#...#.##....##..##..##..##......
11: ...#...##...#.#...#.#...#...#...#......
12: ...##.#.#....#.#...#.#..##..##..##.....
13: ..#..###.#....#.#...#....#...#...#.....
14: ..#....##.#....#.#..##...##..##..##....
15: ..##..#..#.#....#....#..#.#...#...#....
16: .#.#..#...#.#...##...#...#.#..##..##...
17: ..#...##...#.#.#.#...##...#....#...#...
18: ..##.#.#....#####.#.#.#...##...##..##..
19: .#..###.#..#.#.#######.#.#.#..#.#...#..
20: .#....##....#####...#######....#.#..##.
The generation is shown along the left, where 0 is the initial state. The pot numbers are shown along the top, where 0 labels the center pot, negative-numbered pots extend to the left, and positive pots extend toward the right. Remember, the initial state begins at pot 0, which is not the leftmost pot used in this example.
After one generation, only seven plants remain. The one in pot 0 matched the rule looking for ..#.., the one in pot 4 matched the rule looking for .#.#., pot 9 matched .##.., and so on.
In this example, after 20 generations, the pots shown as # contain plants, the furthest left of which is pot -2, and the furthest right of which is pot 34. Adding up all the numbers of plant-containing pots after the 20th generation produces 325.
After 20 generations, what is the sum of the numbers of all pots which contain a plant?
--- Part Two ---
You realize that 20 generations aren't enough. After all, these plants will need to last another 1500 years to even reach your timeline, not to mention your future.
After fifty billion (50000000000) generations, what is the sum of the numbers of all pots which contain a plant?
2018/day13/prompt.md
-235
View File
@@ -1,235 +0,0 @@
--- Day 13: Mine Cart Madness ---
A crop of this size requires significant logistics to transport produce, soil, fertilizer, and so on. The Elves are very busy pushing things around in carts on some kind of rudimentary system of tracks they've come up with.
Seeing as how cart-and-track systems don't appear in recorded history for another 1000 years, the Elves seem to be making this up as they go along. They haven't even figured out how to avoid collisions yet.
You map out the tracks (your puzzle input) and see where you can help.
Tracks consist of straight paths (| and -), curves (/ and \), and intersections (+). Curves connect exactly two perpendicular pieces of track; for example, this is a closed loop:
/----\
| |
| |
\----/
Intersections occur when two perpendicular paths cross. At an intersection, a cart is capable of turning left, turning right, or continuing straight. Here are two loops connected by two intersections:
/-----\
| |
| /--+--\
| | | |
\--+--/ |
| |
\-----/
Several carts are also on the tracks. Carts always face either up (^), down (v), left (<), or right (>). (On your initial map, the track under each cart is a straight path matching the direction the cart is facing.)
Each time a cart has the option to turn (by arriving at any intersection), it turns left the first time, goes straight the second time, turns right the third time, and then repeats those directions starting again with left the fourth time, straight the fifth time, and so on. This process is independent of the particular intersection at which the cart has arrived - that is, the cart has no per-intersection memory.
Carts all move at the same speed; they take turns moving a single step at a time. They do this based on their current location: carts on the top row move first (acting from left to right), then carts on the second row move (again from left to right), then carts on the third row, and so on. Once each cart has moved one step, the process repeats; each of these loops is called a tick.
For example, suppose there are two carts on a straight track:
| | | | |
v | | | |
| v v | |
| | | v X
| | ^ ^ |
^ ^ | | |
| | | | |
First, the top cart moves. It is facing down (v), so it moves down one square. Second, the bottom cart moves. It is facing up (^), so it moves up one square. Because all carts have moved, the first tick ends. Then, the process repeats, starting with the first cart. The first cart moves down, then the second cart moves up - right into the first cart, colliding with it! (The location of the crash is marked with an X.) This ends the second and last tick.
Here is a longer example:
/->-\
| | /----\
| /-+--+-\ |
| | | | v |
\-+-/ \-+--/
\------/
/-->\
| | /----\
| /-+--+-\ |
| | | | | |
\-+-/ \->--/
\------/
/---v
| | /----\
| /-+--+-\ |
| | | | | |
\-+-/ \-+>-/
\------/
/---\
| v /----\
| /-+--+-\ |
| | | | | |
\-+-/ \-+->/
\------/
/---\
| | /----\
| /->--+-\ |
| | | | | |
\-+-/ \-+--^
\------/
/---\
| | /----\
| /-+>-+-\ |
| | | | | ^
\-+-/ \-+--/
\------/
/---\
| | /----\
| /-+->+-\ ^
| | | | | |
\-+-/ \-+--/
\------/
/---\
| | /----<
| /-+-->-\ |
| | | | | |
\-+-/ \-+--/
\------/
/---\
| | /---<\
| /-+--+>\ |
| | | | | |
\-+-/ \-+--/
\------/
/---\
| | /--<-\
| /-+--+-v |
| | | | | |
\-+-/ \-+--/
\------/
/---\
| | /-<--\
| /-+--+-\ |
| | | | v |
\-+-/ \-+--/
\------/
/---\
| | /<---\
| /-+--+-\ |
| | | | | |
\-+-/ \-<--/
\------/
/---\
| | v----\
| /-+--+-\ |
| | | | | |
\-+-/ \<+--/
\------/
/---\
| | /----\
| /-+--v-\ |
| | | | | |
\-+-/ ^-+--/
\------/
/---\
| | /----\
| /-+--+-\ |
| | | X | |
\-+-/ \-+--/
\------/
After following their respective paths for a while, the carts eventually crash. To help prevent crashes, you'd like to know the location of the first crash. Locations are given in X,Y coordinates, where the furthest left column is X=0 and the furthest top row is Y=0:
111
0123456789012
0/---\
1| | /----\
2| /-+--+-\ |
3| | | X | |
4\-+-/ \-+--/
5 \------/
In this example, the location of the first crash is 7,3.
--- Part Two ---
There isn't much you can do to prevent crashes in this ridiculous system. However, by predicting the crashes, the Elves know where to be in advance and instantly remove the two crashing carts the moment any crash occurs.
They can proceed like this for a while, but eventually, they're going to run out of carts. It could be useful to figure out where the last cart that hasn't crashed will end up.
For example:
/>-<\
| |
| /<+-\
| | | v
\>+</ |
| ^
\<->/
/---\
| |
| v-+-\
| | | |
\-+-/ |
| |
^---^
/---\
| |
| /-+-\
| v | |
\-+-/ |
^ ^
\---/
/---\
| |
| /-+-\
| | | |
\-+-/ ^
| |
\---/
After four very expensive crashes, a tick ends with only one cart remaining; its final location is 6,4.
What is the location of the last cart at the end of the first tick where it is the only cart left?
2018/day14/prompt.md
-70
View File
@@ -1,70 +0,0 @@
--- Day 14: Chocolate Charts ---
You finally have a chance to look at all of the produce moving around. Chocolate, cinnamon, mint, chili peppers, nutmeg, vanilla... the Elves must be growing these plants to make hot chocolate! As you realize this, you hear a conversation in the distance. When you go to investigate, you discover two Elves in what appears to be a makeshift underground kitchen/laboratory.
The Elves are trying to come up with the ultimate hot chocolate recipe; they're even maintaining a scoreboard which tracks the quality score (0-9) of each recipe.
Only two recipes are on the board: the first recipe got a score of 3, the second, 7. Each of the two Elves has a current recipe: the first Elf starts with the first recipe, and the second Elf starts with the second recipe.
To create new recipes, the two Elves combine their current recipes. This creates new recipes from the digits of the sum of the current recipes' scores. With the current recipes' scores of 3 and 7, their sum is 10, and so two new recipes would be created: the first with score 1 and the second with score 0. If the current recipes' scores were 2 and 3, the sum, 5, would only create one recipe (with a score of 5) with its single digit.
The new recipes are added to the end of the scoreboard in the order they are created. So, after the first round, the scoreboard is 3, 7, 1, 0.
After all new recipes are added to the scoreboard, each Elf picks a new current recipe. To do this, the Elf steps forward through the scoreboard a number of recipes equal to 1 plus the score of their current recipe. So, after the first round, the first Elf moves forward 1 + 3 = 4 times, while the second Elf moves forward 1 + 7 = 8 times. If they run out of recipes, they loop back around to the beginning. After the first round, both Elves happen to loop around until they land on the same recipe that they had in the beginning; in general, they will move to different recipes.
Drawing the first Elf as parentheses and the second Elf as square brackets, they continue this process:
(3)[7]
(3)[7] 1 0
3 7 1 [0](1) 0
3 7 1 0 [1] 0 (1)
(3) 7 1 0 1 0 [1] 2
3 7 1 0 (1) 0 1 2 [4]
3 7 1 [0] 1 0 (1) 2 4 5
3 7 1 0 [1] 0 1 2 (4) 5 1
3 (7) 1 0 1 0 [1] 2 4 5 1 5
3 7 1 0 1 0 1 2 [4](5) 1 5 8
3 (7) 1 0 1 0 1 2 4 5 1 5 8 [9]
3 7 1 0 1 0 1 [2] 4 (5) 1 5 8 9 1 6
3 7 1 0 1 0 1 2 4 5 [1] 5 8 9 1 (6) 7
3 7 1 0 (1) 0 1 2 4 5 1 5 [8] 9 1 6 7 7
3 7 [1] 0 1 0 (1) 2 4 5 1 5 8 9 1 6 7 7 9
3 7 1 0 [1] 0 1 2 (4) 5 1 5 8 9 1 6 7 7 9 2
The Elves think their skill will improve after making a few recipes (your puzzle input). However, that could take ages; you can speed this up considerably by identifying the scores of the ten recipes after that. For example:
If the Elves think their skill will improve after making 9 recipes, the scores of the ten recipes after the first nine on the scoreboard would be 5158916779 (highlighted in the last line of the diagram).
After 5 recipes, the scores of the next ten would be 0124515891.
After 18 recipes, the scores of the next ten would be 9251071085.
After 2018 recipes, the scores of the next ten would be 5941429882.
What are the scores of the ten recipes immediately after the number of recipes in your puzzle input?
--- Part Two ---
As it turns out, you got the Elves' plan backwards. They actually want to know how many recipes appear on the scoreboard to the left of the first recipes whose scores are the digits from your puzzle input.
51589 first appears after 9 recipes.
01245 first appears after 5 recipes.
92510 first appears after 18 recipes.
59414 first appears after 2018 recipes.
How many recipes appear on the scoreboard to the left of the score sequence in your puzzle input?
2018/day16/prompt.md
-122
View File
@@ -1,122 +0,0 @@
--- Day 16: Chronal Classification ---
As you see the Elves defend their hot chocolate successfully, you go back to falling through time. This is going to become a problem.
If you're ever going to return to your own time, you need to understand how this device on your wrist works. You have a little while before you reach your next destination, and with a bit of trial and error, you manage to pull up a programming manual on the device's tiny screen.
According to the manual, the device has four registers (numbered 0 through 3) that can be manipulated by instructions containing one of 16 opcodes. The registers start with the value 0.
Every instruction consists of four values: an opcode, two inputs (named A and B), and an output (named C), in that order. The opcode specifies the behavior of the instruction and how the inputs are interpreted. The output, C, is always treated as a register.
In the opcode descriptions below, if something says "value A", it means to take the number given as A literally. (This is also called an "immediate" value.) If something says "register A", it means to use the number given as A to read from (or write to) the register with that number. So, if the opcode addi adds register A and value B, storing the result in register C, and the instruction addi 0 7 3 is encountered, it would add 7 to the value contained by register 0 and store the sum in register 3, never modifying registers 0, 1, or 2 in the process.
Many opcodes are similar except for how they interpret their arguments. The opcodes fall into seven general categories:
Addition:
addr (add register) stores into register C the result of adding register A and register B.
addi (add immediate) stores into register C the result of adding register A and value B.
Multiplication:
mulr (multiply register) stores into register C the result of multiplying register A and register B.
muli (multiply immediate) stores into register C the result of multiplying register A and value B.
Bitwise AND:
banr (bitwise AND register) stores into register C the result of the bitwise AND of register A and register B.
bani (bitwise AND immediate) stores into register C the result of the bitwise AND of register A and value B.
Bitwise OR:
borr (bitwise OR register) stores into register C the result of the bitwise OR of register A and register B.
bori (bitwise OR immediate) stores into register C the result of the bitwise OR of register A and value B.
Assignment:
setr (set register) copies the contents of register A into register C. (Input B is ignored.)
seti (set immediate) stores value A into register C. (Input B is ignored.)
Greater-than testing:
gtir (greater-than immediate/register) sets register C to 1 if value A is greater than register B. Otherwise, register C is set to 0.
gtri (greater-than register/immediate) sets register C to 1 if register A is greater than value B. Otherwise, register C is set to 0.
gtrr (greater-than register/register) sets register C to 1 if register A is greater than register B. Otherwise, register C is set to 0.
Equality testing:
eqir (equal immediate/register) sets register C to 1 if value A is equal to register B. Otherwise, register C is set to 0.
eqri (equal register/immediate) sets register C to 1 if register A is equal to value B. Otherwise, register C is set to 0.
eqrr (equal register/register) sets register C to 1 if register A is equal to register B. Otherwise, register C is set to 0.
Unfortunately, while the manual gives the name of each opcode, it doesn't seem to indicate the number. However, you can monitor the CPU to see the contents of the registers before and after instructions are executed to try to work them out. Each opcode has a number from 0 through 15, but the manual doesn't say which is which. For example, suppose you capture the following sample:
Before: [3, 2, 1, 1]
9 2 1 2
After: [3, 2, 2, 1]
This sample shows the effect of the instruction 9 2 1 2 on the registers. Before the instruction is executed, register 0 has value 3, register 1 has value 2, and registers 2 and 3 have value 1. After the instruction is executed, register 2's value becomes 2.
The instruction itself, 9 2 1 2, means that opcode 9 was executed with A=2, B=1, and C=2. Opcode 9 could be any of the 16 opcodes listed above, but only three of them behave in a way that would cause the result shown in the sample:
Opcode 9 could be mulr: register 2 (which has a value of 1) times register 1 (which has a value of 2) produces 2, which matches the value stored in the output register, register 2.
Opcode 9 could be addi: register 2 (which has a value of 1) plus value 1 produces 2, which matches the value stored in the output register, register 2.
Opcode 9 could be seti: value 2 matches the value stored in the output register, register 2; the number given for B is irrelevant.
None of the other opcodes produce the result captured in the sample. Because of this, the sample above behaves like three opcodes.
You collect many of these samples (the first section of your puzzle input). The manual also includes a small test program (the second section of your puzzle input) - you can ignore it for now.
Ignoring the opcode numbers, how many samples in your puzzle input behave like three or more opcodes?
--- Part Two ---
Using the samples you collected, work out the number of each opcode and execute the test program (the second section of your puzzle input).
What value is contained in register 0 after executing the test program?
2018/day17/prompt.md
-195
View File
@@ -1,195 +0,0 @@
--- Day 17: Reservoir Research ---
You arrive in the year 18. If it weren't for the coat you got in 1018, you would be very cold: the North Pole base hasn't even been constructed.
Rather, it hasn't been constructed yet. The Elves are making a little progress, but there's not a lot of liquid water in this climate, so they're getting very dehydrated. Maybe there's more underground?
You scan a two-dimensional vertical slice of the ground nearby and discover that it is mostly sand with veins of clay. The scan only provides data with a granularity of square meters, but it should be good enough to determine how much water is trapped there. In the scan, x represents the distance to the right, and y represents the distance down. There is also a spring of water near the surface at x=500, y=0. The scan identifies which square meters are clay (your puzzle input).
For example, suppose your scan shows the following veins of clay:
x=495, y=2..7
y=7, x=495..501
x=501, y=3..7
x=498, y=2..4
x=506, y=1..2
x=498, y=10..13
x=504, y=10..13
y=13, x=498..504
Rendering clay as #, sand as ., and the water spring as +, and with x increasing to the right and y increasing downward, this becomes:
44444455555555
99999900000000
45678901234567
0 ......+.......
1 ............#.
2 .#..#.......#.
3 .#..#..#......
4 .#..#..#......
5 .#.....#......
6 .#.....#......
7 .#######......
8 ..............
9 ..............
10 ....#.....#...
11 ....#.....#...
12 ....#.....#...
13 ....#######...
The spring of water will produce water forever. Water can move through sand, but is blocked by clay. Water always moves down when possible, and spreads to the left and right otherwise, filling space that has clay on both sides and falling out otherwise.
For example, if five squares of water are created, they will flow downward until they reach the clay and settle there. Water that has come to rest is shown here as ~, while sand through which water has passed (but which is now dry again) is shown as |:
......+.......
......|.....#.
.#..#.|.....#.
.#..#.|#......
.#..#.|#......
.#....|#......
.#~~~~~#......
.#######......
..............
..............
....#.....#...
....#.....#...
....#.....#...
....#######...
Two squares of water can't occupy the same location. If another five squares of water are created, they will settle on the first five, filling the clay reservoir a little more:
......+.......
......|.....#.
.#..#.|.....#.
.#..#.|#......
.#..#.|#......
.#~~~~~#......
.#~~~~~#......
.#######......
..............
..............
....#.....#...
....#.....#...
....#.....#...
....#######...
Water pressure does not apply in this scenario. If another four squares of water are created, they will stay on the right side of the barrier, and no water will reach the left side:
......+.......
......|.....#.
.#..#.|.....#.
.#..#~~#......
.#..#~~#......
.#~~~~~#......
.#~~~~~#......
.#######......
..............
..............
....#.....#...
....#.....#...
....#.....#...
....#######...
At this point, the top reservoir overflows. While water can reach the tiles above the surface of the water, it cannot settle there, and so the next five squares of water settle like this:
......+.......
......|.....#.
.#..#||||...#.
.#..#~~#|.....
.#..#~~#|.....
.#~~~~~#|.....
.#~~~~~#|.....
.#######|.....
........|.....
........|.....
....#...|.#...
....#...|.#...
....#~~~~~#...
....#######...
Note especially the leftmost |: the new squares of water can reach this tile, but cannot stop there. Instead, eventually, they all fall to the right and settle in the reservoir below.
After 10 more squares of water, the bottom reservoir is also full:
......+.......
......|.....#.
.#..#||||...#.
.#..#~~#|.....
.#..#~~#|.....
.#~~~~~#|.....
.#~~~~~#|.....
.#######|.....
........|.....
........|.....
....#~~~~~#...
....#~~~~~#...
....#~~~~~#...
....#######...
Finally, while there is nowhere left for the water to settle, it can reach a few more tiles before overflowing beyond the bottom of the scanned data:
......+....... (line not counted: above minimum y value)
......|.....#.
.#..#||||...#.
.#..#~~#|.....
.#..#~~#|.....
.#~~~~~#|.....
.#~~~~~#|.....
.#######|.....
........|.....
...|||||||||..
...|#~~~~~#|..
...|#~~~~~#|..
...|#~~~~~#|..
...|#######|..
...|.......|.. (line not counted: below maximum y value)
...|.......|.. (line not counted: below maximum y value)
...|.......|.. (line not counted: below maximum y value)
How many tiles can be reached by the water? To prevent counting forever, ignore tiles with a y coordinate smaller than the smallest y coordinate in your scan data or larger than the largest one. Any x coordinate is valid. In this example, the lowest y coordinate given is 1, and the highest is 13, causing the water spring (in row 0) and the water falling off the bottom of the render (in rows 14 through infinity) to be ignored.
So, in the example above, counting both water at rest (~) and other sand tiles the water can hypothetically reach (|), the total number of tiles the water can reach is 57.
How many tiles can the water reach within the range of y values in your scan?
--- Part Two ---
After a very long time, the water spring will run dry. How much water will be retained?
In the example above, water that won't eventually drain out is shown as ~, a total of 29 tiles.
How many water tiles are left after the water spring stops producing water and all remaining water not at rest has drained?
2018/day18/prompt.md
-176
View File
@@ -1,176 +0,0 @@
--- Day 18: Settlers of The North Pole ---
On the outskirts of the North Pole base construction project, many Elves are collecting lumber.
The lumber collection area is 50 acres by 50 acres; each acre can be either open ground (.), trees (|), or a lumberyard (#). You take a scan of the area (your puzzle input).
Strange magic is at work here: each minute, the landscape looks entirely different. In exactly one minute, an open acre can fill with trees, a wooded acre can be converted to a lumberyard, or a lumberyard can be cleared to open ground (the lumber having been sent to other projects).
The change to each acre is based entirely on the contents of that acre as well as the number of open, wooded, or lumberyard acres adjacent to it at the start of each minute. Here, "adjacent" means any of the eight acres surrounding that acre. (Acres on the edges of the lumber collection area might have fewer than eight adjacent acres; the missing acres aren't counted.)
In particular:
An open acre will become filled with trees if three or more adjacent acres contained trees. Otherwise, nothing happens.
An acre filled with trees will become a lumberyard if three or more adjacent acres were lumberyards. Otherwise, nothing happens.
An acre containing a lumberyard will remain a lumberyard if it was adjacent to at least one other lumberyard and at least one acre containing trees. Otherwise, it becomes open.
These changes happen across all acres simultaneously, each of them using the state of all acres at the beginning of the minute and changing to their new form by the end of that same minute. Changes that happen during the minute don't affect each other.
For example, suppose the lumber collection area is instead only 10 by 10 acres with this initial configuration:
Initial state:
.#.#...|#.
.....#|##|
.|..|...#.
..|#.....#
#.#|||#|#|
...#.||...
.|....|...
||...#|.#|
|.||||..|.
...#.|..|.
After 1 minute:
.......##.
......|###
.|..|...#.
..|#||...#
..##||.|#|
...#||||..
||...|||..
|||||.||.|
||||||||||
....||..|.
After 2 minutes:
.......#..
......|#..
.|.|||....
..##|||..#
..###|||#|
...#|||||.
|||||||||.
||||||||||
||||||||||
.|||||||||
After 3 minutes:
.......#..
....|||#..
.|.||||...
..###|||.#
...##|||#|
.||##|||||
||||||||||
||||||||||
||||||||||
||||||||||
After 4 minutes:
.....|.#..
...||||#..
.|.#||||..
..###||||#
...###||#|
|||##|||||
||||||||||
||||||||||
||||||||||
||||||||||
After 5 minutes:
....|||#..
...||||#..
.|.##||||.
..####|||#
.|.###||#|
|||###||||
||||||||||
||||||||||
||||||||||
||||||||||
After 6 minutes:
...||||#..
...||||#..
.|.###|||.
..#.##|||#
|||#.##|#|
|||###||||
||||#|||||
||||||||||
||||||||||
||||||||||
After 7 minutes:
...||||#..
..||#|##..
.|.####||.
||#..##||#
||##.##|#|
|||####|||
|||###||||
||||||||||
||||||||||
||||||||||
After 8 minutes:
..||||##..
..|#####..
|||#####|.
||#...##|#
||##..###|
||##.###||
|||####|||
||||#|||||
||||||||||
||||||||||
After 9 minutes:
..||###...
.||#####..
||##...##.
||#....###
|##....##|
||##..###|
||######||
|||###||||
||||||||||
||||||||||
After 10 minutes:
.||##.....
||###.....
||##......
|##.....##
|##.....##
|##....##|
||##.####|
||#####|||
||||#|||||
||||||||||
After 10 minutes, there are 37 wooded acres and 31 lumberyards. Multiplying the number of wooded acres by the number of lumberyards gives the total resource value after ten minutes: 37 * 31 = 1147.
What will the total resource value of the lumber collection area be after 10 minutes?
--- Part Two ---
This important natural resource will need to last for at least thousands of years. Are the Elves collecting this lumber sustainably?
What will the total resource value of the lumber collection area be after 1000000000 minutes?
2018/day19/prompt.md
-70
View File
@@ -1,70 +0,0 @@
--- Day 19: Go With The Flow ---
With the Elves well on their way constructing the North Pole base, you turn your attention back to understanding the inner workings of programming the device.
You can't help but notice that the device's opcodes don't contain any flow control like jump instructions. The device's manual goes on to explain:
"In programs where flow control is required, the instruction pointer can be bound to a register so that it can be manipulated directly. This way, setr/seti can function as absolute jumps, addr/addi can function as relative jumps, and other opcodes can cause truly fascinating effects."
This mechanism is achieved through a declaration like #ip 1, which would modify register 1 so that accesses to it let the program indirectly access the instruction pointer itself. To compensate for this kind of binding, there are now six registers (numbered 0 through 5); the five not bound to the instruction pointer behave as normal. Otherwise, the same rules apply as the last time you worked with this device.
When the instruction pointer is bound to a register, its value is written to that register just before each instruction is executed, and the value of that register is written back to the instruction pointer immediately after each instruction finishes execution. Afterward, move to the next instruction by adding one to the instruction pointer, even if the value in the instruction pointer was just updated by an instruction. (Because of this, instructions must effectively set the instruction pointer to the instruction before the one they want executed next.)
The instruction pointer is 0 during the first instruction, 1 during the second, and so on. If the instruction pointer ever causes the device to attempt to load an instruction outside the instructions defined in the program, the program instead immediately halts. The instruction pointer starts at 0.
It turns out that this new information is already proving useful: the CPU in the device is not very powerful, and a background process is occupying most of its time. You dump the background process' declarations and instructions to a file (your puzzle input), making sure to use the names of the opcodes rather than the numbers.
For example, suppose you have the following program:
#ip 0
seti 5 0 1
seti 6 0 2
addi 0 1 0
addr 1 2 3
setr 1 0 0
seti 8 0 4
seti 9 0 5
When executed, the following instructions are executed. Each line contains the value of the instruction pointer at the time the instruction started, the values of the six registers before executing the instructions (in square brackets), the instruction itself, and the values of the six registers after executing the instruction (also in square brackets).
ip=0 [0, 0, 0, 0, 0, 0] seti 5 0 1 [0, 5, 0, 0, 0, 0]
ip=1 [1, 5, 0, 0, 0, 0] seti 6 0 2 [1, 5, 6, 0, 0, 0]
ip=2 [2, 5, 6, 0, 0, 0] addi 0 1 0 [3, 5, 6, 0, 0, 0]
ip=4 [4, 5, 6, 0, 0, 0] setr 1 0 0 [5, 5, 6, 0, 0, 0]
ip=6 [6, 5, 6, 0, 0, 0] seti 9 0 5 [6, 5, 6, 0, 0, 9]
In detail, when running this program, the following events occur:
The first line (#ip 0) indicates that the instruction pointer should be bound to register 0 in this program. This is not an instruction, and so the value of the instruction pointer does not change during the processing of this line.
The instruction pointer contains 0, and so the first instruction is executed (seti 5 0 1). It updates register 0 to the current instruction pointer value (0), sets register 1 to 5, sets the instruction pointer to the value of register 0 (which has no effect, as the instruction did not modify register 0), and then adds one to the instruction pointer.
The instruction pointer contains 1, and so the second instruction, seti 6 0 2, is executed. This is very similar to the instruction before it: 6 is stored in register 2, and the instruction pointer is left with the value 2.
The instruction pointer is 2, which points at the instruction addi 0 1 0. This is like a relative jump: the value of the instruction pointer, 2, is loaded into register 0. Then, addi finds the result of adding the value in register 0 and the value 1, storing the result, 3, back in register 0. Register 0 is then copied back to the instruction pointer, which will cause it to end up 1 larger than it would have otherwise and skip the next instruction (addr 1 2 3) entirely. Finally, 1 is added to the instruction pointer.
The instruction pointer is 4, so the instruction setr 1 0 0 is run. This is like an absolute jump: it copies the value contained in register 1, 5, into register 0, which causes it to end up in the instruction pointer. The instruction pointer is then incremented, leaving it at 6.
The instruction pointer is 6, so the instruction seti 9 0 5 stores 9 into register 5. The instruction pointer is incremented, causing it to point outside the program, and so the program ends.
What value is left in register 0 when the background process halts?
--- Part Two ---
A new background process immediately spins up in its place. It appears identical, but on closer inspection, you notice that this time, register 0 started with the value 1.
What value is left in register 0 when this new background process halts?
2018/day22/prompt.md
-373
View File
@@ -1,373 +0,0 @@
--- Day 22: Mode Maze ---
This is it, your final stop: the year -483. It's snowing and dark outside; the only light you can see is coming from a small cottage in the distance. You make your way there and knock on the door.
A portly man with a large, white beard answers the door and invites you inside. For someone living near the North Pole in -483, he must not get many visitors, but he doesn't act surprised to see you. Instead, he offers you some milk and cookies.
After talking for a while, he asks a favor of you. His friend hasn't come back in a few hours, and he's not sure where he is. Scanning the region briefly, you discover one life signal in a cave system nearby; his friend must have taken shelter there. The man asks if you can go there to retrieve his friend.
The cave is divided into square regions which are either dominantly rocky, narrow, or wet (called its type). Each region occupies exactly one coordinate in X,Y format where X and Y are integers and zero or greater. (Adjacent regions can be the same type.)
The scan (your puzzle input) is not very detailed: it only reveals the depth of the cave system and the coordinates of the target. However, it does not reveal the type of each region. The mouth of the cave is at 0,0.
The man explains that due to the unusual geology in the area, there is a method to determine any region's type based on its erosion level. The erosion level of a region can be determined from its geologic index. The geologic index can be determined using the first rule that applies from the list below:
The region at 0,0 (the mouth of the cave) has a geologic index of 0.
The region at the coordinates of the target has a geologic index of 0.
If the region's Y coordinate is 0, the geologic index is its X coordinate times 16807.
If the region's X coordinate is 0, the geologic index is its Y coordinate times 48271.
Otherwise, the region's geologic index is the result of multiplying the erosion levels of the regions at X-1,Y and X,Y-1.
A region's erosion level is its geologic index plus the cave system's depth, all modulo 20183. Then:
If the erosion level modulo 3 is 0, the region's type is rocky.
If the erosion level modulo 3 is 1, the region's type is wet.
If the erosion level modulo 3 is 2, the region's type is narrow.
For example, suppose the cave system's depth is 510 and the target's coordinates are 10,10. Using % to represent the modulo operator, the cavern would look as follows:
At 0,0, the geologic index is 0. The erosion level is (0 + 510) % 20183 = 510. The type is 510 % 3 = 0, rocky.
At 1,0, because the Y coordinate is 0, the geologic index is 1 * 16807 = 16807. The erosion level is (16807 + 510) % 20183 = 17317. The type is 17317 % 3 = 1, wet.
At 0,1, because the X coordinate is 0, the geologic index is 1 * 48271 = 48271. The erosion level is (48271 + 510) % 20183 = 8415. The type is 8415 % 3 = 0, rocky.
At 1,1, neither coordinate is 0 and it is not the coordinate of the target, so the geologic index is the erosion level of 0,1 (8415) times the erosion level of 1,0 (17317), 8415 * 17317 = 145722555. The erosion level is (145722555 + 510) % 20183 = 1805. The type is 1805 % 3 = 2, narrow.
At 10,10, because they are the target's coordinates, the geologic index is 0. The erosion level is (0 + 510) % 20183 = 510. The type is 510 % 3 = 0, rocky.
Drawing this same cave system with rocky as ., wet as =, narrow as |, the mouth as M, the target as T, with 0,0 in the top-left corner, X increasing to the right, and Y increasing downward, the top-left corner of the map looks like this:
M=.|=.|.|=.|=|=.
.|=|=|||..|.=...
.==|....||=..|==
=.|....|.==.|==.
=|..==...=.|==..
=||.=.=||=|=..|=
|.=.===|||..=..|
|..==||=.|==|===
.=..===..=|.|||.
.======|||=|=.|=
.===|=|===T===||
=|||...|==..|=.|
=.=|=.=..=.||==|
||=|=...|==.=|==
|=.=||===.|||===
||.|==.|.|.||=||
Before you go in, you should determine the risk level of the area. For the rectangle that has a top-left corner of region 0,0 and a bottom-right corner of the region containing the target, add up the risk level of each individual region: 0 for rocky regions, 1 for wet regions, and 2 for narrow regions.
In the cave system above, because the mouth is at 0,0 and the target is at 10,10, adding up the risk level of all regions with an X coordinate from 0 to 10 and a Y coordinate from 0 to 10, this total is 114.
What is the total risk level for the smallest rectangle that includes 0,0 and the target's coordinates?
--- Part Two ---
Okay, it's time to go rescue the man's friend.
As you leave, he hands you some tools: a torch and some climbing gear. You can't equip both tools at once, but you can choose to use neither.
Tools can only be used in certain regions:
In rocky regions, you can use the climbing gear or the torch. You cannot use neither (you'll likely slip and fall).
In wet regions, you can use the climbing gear or neither tool. You cannot use the torch (if it gets wet, you won't have a light source).
In narrow regions, you can use the torch or neither tool. You cannot use the climbing gear (it's too bulky to fit).
You start at 0,0 (the mouth of the cave) with the torch equipped and must reach the target coordinates as quickly as possible. The regions with negative X or Y are solid rock and cannot be traversed. The fastest route might involve entering regions beyond the X or Y coordinate of the target.
You can move to an adjacent region (up, down, left, or right; never diagonally) if your currently equipped tool allows you to enter that region. Moving to an adjacent region takes one minute. (For example, if you have the torch equipped, you can move between rocky and narrow regions, but cannot enter wet regions.)
You can change your currently equipped tool or put both away if your new equipment would be valid for your current region. Switching to using the climbing gear, torch, or neither always takes seven minutes, regardless of which tools you start with. (For example, if you are in a rocky region, you can switch from the torch to the climbing gear, but you cannot switch to neither.)
Finally, once you reach the target, you need the torch equipped before you can find him in the dark. The target is always in a rocky region, so if you arrive there with climbing gear equipped, you will need to spend seven minutes switching to your torch.
For example, using the same cave system as above, starting in the top left corner (0,0) and moving to the bottom right corner (the target, 10,10) as quickly as possible, one possible route is as follows, with your current position marked X:
Initially:
X=.|=.|.|=.|=|=.
.|=|=|||..|.=...
.==|....||=..|==
=.|....|.==.|==.
=|..==...=.|==..
=||.=.=||=|=..|=
|.=.===|||..=..|
|..==||=.|==|===
.=..===..=|.|||.
.======|||=|=.|=
.===|=|===T===||
=|||...|==..|=.|
=.=|=.=..=.||==|
||=|=...|==.=|==
|=.=||===.|||===
||.|==.|.|.||=||
Down:
M=.|=.|.|=.|=|=.
X|=|=|||..|.=...
.==|....||=..|==
=.|....|.==.|==.
=|..==...=.|==..
=||.=.=||=|=..|=
|.=.===|||..=..|
|..==||=.|==|===
.=..===..=|.|||.
.======|||=|=.|=
.===|=|===T===||
=|||...|==..|=.|
=.=|=.=..=.||==|
||=|=...|==.=|==
|=.=||===.|||===
||.|==.|.|.||=||
Right:
M=.|=.|.|=.|=|=.
.X=|=|||..|.=...
.==|....||=..|==
=.|....|.==.|==.
=|..==...=.|==..
=||.=.=||=|=..|=
|.=.===|||..=..|
|..==||=.|==|===
.=..===..=|.|||.
.======|||=|=.|=
.===|=|===T===||
=|||...|==..|=.|
=.=|=.=..=.||==|
||=|=...|==.=|==
|=.=||===.|||===
||.|==.|.|.||=||
Switch from using the torch to neither tool:
M=.|=.|.|=.|=|=.
.X=|=|||..|.=...
.==|....||=..|==
=.|....|.==.|==.
=|..==...=.|==..
=||.=.=||=|=..|=
|.=.===|||..=..|
|..==||=.|==|===
.=..===..=|.|||.
.======|||=|=.|=
.===|=|===T===||
=|||...|==..|=.|
=.=|=.=..=.||==|
||=|=...|==.=|==
|=.=||===.|||===
||.|==.|.|.||=||
Right 3:
M=.|=.|.|=.|=|=.
.|=|X|||..|.=...
.==|....||=..|==
=.|....|.==.|==.
=|..==...=.|==..
=||.=.=||=|=..|=
|.=.===|||..=..|
|..==||=.|==|===
.=..===..=|.|||.
.======|||=|=.|=
.===|=|===T===||
=|||...|==..|=.|
=.=|=.=..=.||==|
||=|=...|==.=|==
|=.=||===.|||===
||.|==.|.|.||=||
Switch from using neither tool to the climbing gear:
M=.|=.|.|=.|=|=.
.|=|X|||..|.=...
.==|....||=..|==
=.|....|.==.|==.
=|..==...=.|==..
=||.=.=||=|=..|=
|.=.===|||..=..|
|..==||=.|==|===
.=..===..=|.|||.
.======|||=|=.|=
.===|=|===T===||
=|||...|==..|=.|
=.=|=.=..=.||==|
||=|=...|==.=|==
|=.=||===.|||===
||.|==.|.|.||=||
Down 7:
M=.|=.|.|=.|=|=.
.|=|=|||..|.=...
.==|....||=..|==
=.|....|.==.|==.
=|..==...=.|==..
=||.=.=||=|=..|=
|.=.===|||..=..|
|..==||=.|==|===
.=..X==..=|.|||.
.======|||=|=.|=
.===|=|===T===||
=|||...|==..|=.|
=.=|=.=..=.||==|
||=|=...|==.=|==
|=.=||===.|||===
||.|==.|.|.||=||
Right:
M=.|=.|.|=.|=|=.
.|=|=|||..|.=...
.==|....||=..|==
=.|....|.==.|==.
=|..==...=.|==..
=||.=.=||=|=..|=
|.=.===|||..=..|
|..==||=.|==|===
.=..=X=..=|.|||.
.======|||=|=.|=
.===|=|===T===||
=|||...|==..|=.|
=.=|=.=..=.||==|
||=|=...|==.=|==
|=.=||===.|||===
||.|==.|.|.||=||
Down 3:
M=.|=.|.|=.|=|=.
.|=|=|||..|.=...
.==|....||=..|==
=.|....|.==.|==.
=|..==...=.|==..
=||.=.=||=|=..|=
|.=.===|||..=..|
|..==||=.|==|===
.=..===..=|.|||.
.======|||=|=.|=
.===|=|===T===||
=|||.X.|==..|=.|
=.=|=.=..=.||==|
||=|=...|==.=|==
|=.=||===.|||===
||.|==.|.|.||=||
Right:
M=.|=.|.|=.|=|=.
.|=|=|||..|.=...
.==|....||=..|==
=.|....|.==.|==.
=|..==...=.|==..
=||.=.=||=|=..|=
|.=.===|||..=..|
|..==||=.|==|===
.=..===..=|.|||.
.======|||=|=.|=
.===|=|===T===||
=|||..X|==..|=.|
=.=|=.=..=.||==|
||=|=...|==.=|==
|=.=||===.|||===
||.|==.|.|.||=||
Down:
M=.|=.|.|=.|=|=.
.|=|=|||..|.=...
.==|....||=..|==
=.|....|.==.|==.
=|..==...=.|==..
=||.=.=||=|=..|=
|.=.===|||..=..|
|..==||=.|==|===
.=..===..=|.|||.
.======|||=|=.|=
.===|=|===T===||
=|||...|==..|=.|
=.=|=.X..=.||==|
||=|=...|==.=|==
|=.=||===.|||===
||.|==.|.|.||=||
Right 4:
M=.|=.|.|=.|=|=.
.|=|=|||..|.=...
.==|....||=..|==
=.|....|.==.|==.
=|..==...=.|==..
=||.=.=||=|=..|=
|.=.===|||..=..|
|..==||=.|==|===
.=..===..=|.|||.
.======|||=|=.|=
.===|=|===T===||
=|||...|==..|=.|
=.=|=.=..=X||==|
||=|=...|==.=|==
|=.=||===.|||===
||.|==.|.|.||=||
Up 2:
M=.|=.|.|=.|=|=.
.|=|=|||..|.=...
.==|....||=..|==
=.|....|.==.|==.
=|..==...=.|==..
=||.=.=||=|=..|=
|.=.===|||..=..|
|..==||=.|==|===
.=..===..=|.|||.
.======|||=|=.|=
.===|=|===X===||
=|||...|==..|=.|
=.=|=.=..=.||==|
||=|=...|==.=|==
|=.=||===.|||===
||.|==.|.|.||=||
Switch from using the climbing gear to the torch:
M=.|=.|.|=.|=|=.
.|=|=|||..|.=...
.==|....||=..|==
=.|....|.==.|==.
=|..==...=.|==..
=||.=.=||=|=..|=
|.=.===|||..=..|
|..==||=.|==|===
.=..===..=|.|||.
.======|||=|=.|=
.===|=|===X===||
=|||...|==..|=.|
=.=|=.=..=.||==|
||=|=...|==.=|==
|=.=||===.|||===
||.|==.|.|.||=||
This is tied with other routes as the fastest way to reach the target: 45 minutes. In it, 21 minutes are spent switching tools (three times, seven minutes each) and the remaining 24 minutes are spent moving.
What is the fewest number of minutes you can take to reach the target?
2018/day24/prompt.md
-378
View File
@@ -1,378 +0,0 @@
--- Day 24: Immune System Simulator 20XX ---
After a weird buzzing noise, you appear back at the man's cottage. He seems relieved to see his friend, but quickly notices that the little reindeer caught some kind of cold while out exploring.
The portly man explains that this reindeer's immune system isn't similar to regular reindeer immune systems:
The immune system and the infection each have an army made up of several groups; each group consists of one or more identical units. The armies repeatedly fight until only one army has units remaining.
Units within a group all have the same hit points (amount of damage a unit can take before it is destroyed), attack damage (the amount of damage each unit deals), an attack type, an initiative (higher initiative units attack first and win ties), and sometimes weaknesses or immunities. Here is an example group:
18 units each with 729 hit points (weak to fire; immune to cold, slashing)
with an attack that does 8 radiation damage at initiative 10
Each group also has an effective power: the number of units in that group multiplied by their attack damage. The above group has an effective power of 18 * 8 = 144. Groups never have zero or negative units; instead, the group is removed from combat.
Each fight consists of two phases: target selection and attacking.
During the target selection phase, each group attempts to choose one target. In decreasing order of effective power, groups choose their targets; in a tie, the group with the higher initiative chooses first. The attacking group chooses to target the group in the enemy army to which it would deal the most damage (after accounting for weaknesses and immunities, but not accounting for whether the defending group has enough units to actually receive all of that damage).
If an attacking group is considering two defending groups to which it would deal equal damage, it chooses to target the defending group with the largest effective power; if there is still a tie, it chooses the defending group with the highest initiative. If it cannot deal any defending groups damage, it does not choose a target. Defending groups can only be chosen as a target by one attacking group.
At the end of the target selection phase, each group has selected zero or one groups to attack, and each group is being attacked by zero or one groups.
During the attacking phase, each group deals damage to the target it selected, if any. Groups attack in decreasing order of initiative, regardless of whether they are part of the infection or the immune system. (If a group contains no units, it cannot attack.)
The damage an attacking group deals to a defending group depends on the attacking group's attack type and the defending group's immunities and weaknesses. By default, an attacking group would deal damage equal to its effective power to the defending group. However, if the defending group is immune to the attacking group's attack type, the defending group instead takes no damage; if the defending group is weak to the attacking group's attack type, the defending group instead takes double damage.
The defending group only loses whole units from damage; damage is always dealt in such a way that it kills the most units possible, and any remaining damage to a unit that does not immediately kill it is ignored. For example, if a defending group contains 10 units with 10 hit points each and receives 75 damage, it loses exactly 7 units and is left with 3 units at full health.
After the fight is over, if both armies still contain units, a new fight begins; combat only ends once one army has lost all of its units.
For example, consider the following armies:
Immune System:
17 units each with 5390 hit points (weak to radiation, bludgeoning) with
an attack that does 4507 fire damage at initiative 2
989 units each with 1274 hit points (immune to fire; weak to bludgeoning,
slashing) with an attack that does 25 slashing damage at initiative 3
Infection:
801 units each with 4706 hit points (weak to radiation) with an attack
that does 116 bludgeoning damage at initiative 1
4485 units each with 2961 hit points (immune to radiation; weak to fire,
cold) with an attack that does 12 slashing damage at initiative 4
If these armies were to enter combat, the following fights, including details during the target selection and attacking phases, would take place:
Immune System:
Group 1 contains 17 units
Group 2 contains 989 units
Infection:
Group 1 contains 801 units
Group 2 contains 4485 units
Infection group 1 would deal defending group 1 185832 damage
Infection group 1 would deal defending group 2 185832 damage
Infection group 2 would deal defending group 2 107640 damage
Immune System group 1 would deal defending group 1 76619 damage
Immune System group 1 would deal defending group 2 153238 damage
Immune System group 2 would deal defending group 1 24725 damage
Infection group 2 attacks defending group 2, killing 84 units
Immune System group 2 attacks defending group 1, killing 4 units
Immune System group 1 attacks defending group 2, killing 51 units
Infection group 1 attacks defending group 1, killing 17 units
Immune System:
Group 2 contains 905 units
Infection:
Group 1 contains 797 units
Group 2 contains 4434 units
Infection group 1 would deal defending group 2 184904 damage
Immune System group 2 would deal defending group 1 22625 damage
Immune System group 2 would deal defending group 2 22625 damage
Immune System group 2 attacks defending group 1, killing 4 units
Infection group 1 attacks defending group 2, killing 144 units
Immune System:
Group 2 contains 761 units
Infection:
Group 1 contains 793 units
Group 2 contains 4434 units
Infection group 1 would deal defending group 2 183976 damage
Immune System group 2 would deal defending group 1 19025 damage
Immune System group 2 would deal defending group 2 19025 damage
Immune System group 2 attacks defending group 1, killing 4 units
Infection group 1 attacks defending group 2, killing 143 units
Immune System:
Group 2 contains 618 units
Infection:
Group 1 contains 789 units
Group 2 contains 4434 units
Infection group 1 would deal defending group 2 183048 damage
Immune System group 2 would deal defending group 1 15450 damage
Immune System group 2 would deal defending group 2 15450 damage
Immune System group 2 attacks defending group 1, killing 3 units
Infection group 1 attacks defending group 2, killing 143 units
Immune System:
Group 2 contains 475 units
Infection:
Group 1 contains 786 units
Group 2 contains 4434 units
Infection group 1 would deal defending group 2 182352 damage
Immune System group 2 would deal defending group 1 11875 damage
Immune System group 2 would deal defending group 2 11875 damage
Immune System group 2 attacks defending group 1, killing 2 units
Infection group 1 attacks defending group 2, killing 142 units
Immune System:
Group 2 contains 333 units
Infection:
Group 1 contains 784 units
Group 2 contains 4434 units
Infection group 1 would deal defending group 2 181888 damage
Immune System group 2 would deal defending group 1 8325 damage
Immune System group 2 would deal defending group 2 8325 damage
Immune System group 2 attacks defending group 1, killing 1 unit
Infection group 1 attacks defending group 2, killing 142 units
Immune System:
Group 2 contains 191 units
Infection:
Group 1 contains 783 units
Group 2 contains 4434 units
Infection group 1 would deal defending group 2 181656 damage
Immune System group 2 would deal defending group 1 4775 damage
Immune System group 2 would deal defending group 2 4775 damage
Immune System group 2 attacks defending group 1, killing 1 unit
Infection group 1 attacks defending group 2, killing 142 units
Immune System:
Group 2 contains 49 units
Infection:
Group 1 contains 782 units
Group 2 contains 4434 units
Infection group 1 would deal defending group 2 181424 damage
Immune System group 2 would deal defending group 1 1225 damage
Immune System group 2 would deal defending group 2 1225 damage
Immune System group 2 attacks defending group 1, killing 0 units
Infection group 1 attacks defending group 2, killing 49 units
Immune System:
No groups remain.
Infection:
Group 1 contains 782 units
Group 2 contains 4434 units
In the example above, the winning army ends up with 782 + 4434 = 5216 units.
You scan the reindeer's condition (your puzzle input); the white-bearded man looks nervous. As it stands now, how many units would the winning army have?
--- Part Two ---
Things aren't looking good for the reindeer. The man asks whether more milk and cookies would help you think.
If only you could give the reindeer's immune system a boost, you might be able to change the outcome of the combat.
A boost is an integer increase in immune system units' attack damage. For example, if you were to boost the above example's immune system's units by 1570, the armies would instead look like this:
Immune System:
17 units each with 5390 hit points (weak to radiation, bludgeoning) with
an attack that does 6077 fire damage at initiative 2
989 units each with 1274 hit points (immune to fire; weak to bludgeoning,
slashing) with an attack that does 1595 slashing damage at initiative 3
Infection:
801 units each with 4706 hit points (weak to radiation) with an attack
that does 116 bludgeoning damage at initiative 1
4485 units each with 2961 hit points (immune to radiation; weak to fire,
cold) with an attack that does 12 slashing damage at initiative 4
With this boost, the combat proceeds differently:
Immune System:
Group 2 contains 989 units
Group 1 contains 17 units
Infection:
Group 1 contains 801 units
Group 2 contains 4485 units
Infection group 1 would deal defending group 2 185832 damage
Infection group 1 would deal defending group 1 185832 damage
Infection group 2 would deal defending group 1 53820 damage
Immune System group 2 would deal defending group 1 1577455 damage
Immune System group 2 would deal defending group 2 1577455 damage
Immune System group 1 would deal defending group 2 206618 damage
Infection group 2 attacks defending group 1, killing 9 units
Immune System group 2 attacks defending group 1, killing 335 units
Immune System group 1 attacks defending group 2, killing 32 units
Infection group 1 attacks defending group 2, killing 84 units
Immune System:
Group 2 contains 905 units
Group 1 contains 8 units
Infection:
Group 1 contains 466 units
Group 2 contains 4453 units
Infection group 1 would deal defending group 2 108112 damage
Infection group 1 would deal defending group 1 108112 damage
Infection group 2 would deal defending group 1 53436 damage
Immune System group 2 would deal defending group 1 1443475 damage
Immune System group 2 would deal defending group 2 1443475 damage
Immune System group 1 would deal defending group 2 97232 damage
Infection group 2 attacks defending group 1, killing 8 units
Immune System group 2 attacks defending group 1, killing 306 units
Infection group 1 attacks defending group 2, killing 29 units
Immune System:
Group 2 contains 876 units
Infection:
Group 2 contains 4453 units
Group 1 contains 160 units
Infection group 2 would deal defending group 2 106872 damage
Immune System group 2 would deal defending group 2 1397220 damage
Immune System group 2 would deal defending group 1 1397220 damage
Infection group 2 attacks defending group 2, killing 83 units
Immune System group 2 attacks defending group 2, killing 427 units
After a few fights...
Immune System:
Group 2 contains 64 units
Infection:
Group 2 contains 214 units
Group 1 contains 19 units
Infection group 2 would deal defending group 2 5136 damage
Immune System group 2 would deal defending group 2 102080 damage
Immune System group 2 would deal defending group 1 102080 damage
Infection group 2 attacks defending group 2, killing 4 units
Immune System group 2 attacks defending group 2, killing 32 units
Immune System:
Group 2 contains 60 units
Infection:
Group 1 contains 19 units
Group 2 contains 182 units
Infection group 1 would deal defending group 2 4408 damage
Immune System group 2 would deal defending group 1 95700 damage
Immune System group 2 would deal defending group 2 95700 damage
Immune System group 2 attacks defending group 1, killing 19 units
Immune System:
Group 2 contains 60 units
Infection:
Group 2 contains 182 units
Infection group 2 would deal defending group 2 4368 damage
Immune System group 2 would deal defending group 2 95700 damage
Infection group 2 attacks defending group 2, killing 3 units
Immune System group 2 attacks defending group 2, killing 30 units
After a few more fights...
Immune System:
Group 2 contains 51 units
Infection:
Group 2 contains 40 units
Infection group 2 would deal defending group 2 960 damage
Immune System group 2 would deal defending group 2 81345 damage
Infection group 2 attacks defending group 2, killing 0 units
Immune System group 2 attacks defending group 2, killing 27 units
Immune System:
Group 2 contains 51 units
Infection:
Group 2 contains 13 units
Infection group 2 would deal defending group 2 312 damage
Immune System group 2 would deal defending group 2 81345 damage
Infection group 2 attacks defending group 2, killing 0 units
Immune System group 2 attacks defending group 2, killing 13 units
Immune System:
Group 2 contains 51 units
Infection:
No groups remain.
This boost would allow the immune system's armies to win! It would be left with 51 units.
You don't even know how you could boost the reindeer's immune system or what effect it might have, so you need to be cautious and find the smallest boost that would allow the immune system to win.
How many units does the immune system have left after getting the smallest boost it needs to win?
2020/day01/prompt.md
View File
2020/day02/prompt.md
-54
View File
@@ -1,54 +0,0 @@
--- Day 2: Password Philosophy ---
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?
--- Part Two ---
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?
2020/day03/prompt.md
-92
View File
@@ -1,92 +0,0 @@
--- Day 3: Toboggan Trajectory ---
With the toboggan login problems resolved, you set off toward the airport. While travel by toboggan might be easy, it's certainly not safe: there's very minimal steering and the area is covered in trees. You'll need to see which angles will take you near the fewest trees.
Due to the local geology, trees in this area only grow on exact integer coordinates in a grid. You make a map (your puzzle input) of the open squares (.) and trees (#) you can see. For example:
..##.......
#...#...#..
.#....#..#.
..#.#...#.#
.#...##..#.
..#.##.....
.#.#.#....#
.#........#
#.##...#...
#...##....#
.#..#...#.#
These aren't the only trees, though; due to something you read about once involving arboreal genetics and biome stability, the same pattern repeats to the right many times:
..##.........##.........##.........##.........##.........##....... --->
#...#...#..#...#...#..#...#...#..#...#...#..#...#...#..#...#...#..
.#....#..#..#....#..#..#....#..#..#....#..#..#....#..#..#....#..#.
..#.#...#.#..#.#...#.#..#.#...#.#..#.#...#.#..#.#...#.#..#.#...#.#
.#...##..#..#...##..#..#...##..#..#...##..#..#...##..#..#...##..#.
..#.##.......#.##.......#.##.......#.##.......#.##.......#.##..... --->
.#.#.#....#.#.#.#....#.#.#.#....#.#.#.#....#.#.#.#....#.#.#.#....#
.#........#.#........#.#........#.#........#.#........#.#........#
#.##...#...#.##...#...#.##...#...#.##...#...#.##...#...#.##...#...
#...##....##...##....##...##....##...##....##...##....##...##....#
.#..#...#.#.#..#...#.#.#..#...#.#.#..#...#.#.#..#...#.#.#..#...#.# --->
You start on the open square (.) in the top-left corner and need to reach the bottom (below the bottom-most row on your map).
The toboggan can only follow a few specific slopes (you opted for a cheaper model that prefers rational numbers); start by counting all the trees you would encounter for the slope right 3, down 1:
From your starting position at the top-left, check the position that is right 3 and down 1. Then, check the position that is right 3 and down 1 from there, and so on until you go past the bottom of the map.
The locations you'd check in the above example are marked here with O where there was an open square and X where there was a tree:
..##.........##.........##.........##.........##.........##....... --->
#..O#...#..#...#...#..#...#...#..#...#...#..#...#...#..#...#...#..
.#....X..#..#....#..#..#....#..#..#....#..#..#....#..#..#....#..#.
..#.#...#O#..#.#...#.#..#.#...#.#..#.#...#.#..#.#...#.#..#.#...#.#
.#...##..#..X...##..#..#...##..#..#...##..#..#...##..#..#...##..#.
..#.##.......#.X#.......#.##.......#.##.......#.##.......#.##..... --->
.#.#.#....#.#.#.#.O..#.#.#.#....#.#.#.#....#.#.#.#....#.#.#.#....#
.#........#.#........X.#........#.#........#.#........#.#........#
#.##...#...#.##...#...#.X#...#...#.##...#...#.##...#...#.##...#...
#...##....##...##....##...#X....##...##....##...##....##...##....#
.#..#...#.#.#..#...#.#.#..#...X.#.#..#...#.#.#..#...#.#.#..#...#.# --->
In this example, traversing the map using this slope would cause you to encounter 7 trees.
Starting at the top-left corner of your map and following a slope of right 3 and down 1, how many trees would you encounter?
--- Part Two ---
Time to check the rest of the slopes - you need to minimize the probability of a sudden arboreal stop, after all.
Determine the number of trees you would encounter if, for each of the following slopes, you start at the top-left corner and traverse the map all the way to the bottom:
Right 1, down 1.
Right 3, down 1. (This is the slope you already checked.)
Right 5, down 1.
Right 7, down 1.
Right 1, down 2.
In the above example, these slopes would find 2, 7, 3, 4, and 2 tree(s) respectively; multiplied together, these produce the answer 336.
What do you get if you multiply together the number of trees encountered on each of the listed slopes?
2020/day04/prompt.md
-149
View File
@@ -1,149 +0,0 @@
--- Day 4: Passport Processing ---
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?
--- Part Two ---
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?
2020/day05/prompt.md
-74
View File
@@ -1,74 +0,0 @@
--- Day 5: Binary Boarding ---
You board your plane only to discover a new problem: you dropped your boarding pass! You aren't sure which seat is yours, and all of the flight attendants are busy with the flood of people that suddenly made it through passport control.
You write a quick program to use your phone's camera to scan all of the nearby boarding passes (your puzzle input); perhaps you can find your seat through process of elimination.
Instead of zones or groups, this airline uses binary space partitioning to seat people. A seat might be specified like FBFBBFFRLR, where F means "front", B means "back", L means "left", and R means "right".
The first 7 characters will either be F or B; these specify exactly one of the 128 rows on the plane (numbered 0 through 127). Each letter tells you which half of a region the given seat is in. Start with the whole list of rows; the first letter indicates whether the seat is in the front (0 through 63) or the back (64 through 127). The next letter indicates which half of that region the seat is in, and so on until you're left with exactly one row.
For example, consider just the first seven characters of FBFBBFFRLR:
Start by considering the whole range, rows 0 through 127.
F means to take the lower half, keeping rows 0 through 63.
B means to take the upper half, keeping rows 32 through 63.
F means to take the lower half, keeping rows 32 through 47.
B means to take the upper half, keeping rows 40 through 47.
B keeps rows 44 through 47.
F keeps rows 44 through 45.
The final F keeps the lower of the two, row 44.
The last three characters will be either L or R; these specify exactly one of the 8 columns of seats on the plane (numbered 0 through 7). The same process as above proceeds again, this time with only three steps. L means to keep the lower half, while R means to keep the upper half.
For example, consider just the last 3 characters of FBFBBFFRLR:
Start by considering the whole range, columns 0 through 7.
R means to take the upper half, keeping columns 4 through 7.
L means to take the lower half, keeping columns 4 through 5.
The final R keeps the upper of the two, column 5.
So, decoding FBFBBFFRLR reveals that it is the seat at row 44, column 5.
Every seat also has a unique seat ID: multiply the row by 8, then add the column. In this example, the seat has ID 44 * 8 + 5 = 357.
Here are some other boarding passes:
BFFFBBFRRR: row 70, column 7, seat ID 567.
FFFBBBFRRR: row 14, column 7, seat ID 119.
BBFFBBFRLL: row 102, column 4, seat ID 820.
As a sanity check, look through your list of boarding passes. What is the highest seat ID on a boarding pass?
--- Part Two ---
Ding! The "fasten seat belt" signs have turned on. Time to find your seat.
It's a completely full flight, so your seat should be the only missing boarding pass in your list. However, there's a catch: some of the seats at the very front and back of the plane don't exist on this aircraft, so they'll be missing from your list as well.
Your seat wasn't at the very front or back, though; the seats with IDs +1 and -1 from yours will be in your list.
What is the ID of your seat?
2020/day06/prompt.md
-104
View File
@@ -1,104 +0,0 @@
--- Day 6: Custom Customs ---
As your flight approaches the regional airport where you'll switch to a much larger plane, customs declaration forms are distributed to the passengers.
The form asks a series of 26 yes-or-no questions marked a through z. All you need to do is identify the questions for which anyone in your group answers "yes". Since your group is just you, this doesn't take very long.
However, the person sitting next to you seems to be experiencing a language barrier and asks if you can help. For each of the people in their group, you write down the questions for which they answer "yes", one per line. For example:
abcx
abcy
abcz
In this group, there are 6 questions to which anyone answered "yes": a, b, c, x, y, and z. (Duplicate answers to the same question don't count extra; each question counts at most once.)
Another group asks for your help, then another, and eventually you've collected answers from every group on the plane (your puzzle input). Each group's answers are separated by a blank line, and within each group, each person's answers are on a single line. For example:
abc
a
b
c
ab
ac
a
a
a
a
b
This list represents answers from five groups:
The first group contains one person who answered "yes" to 3 questions: a, b, and c.
The second group contains three people; combined, they answered "yes" to 3 questions: a, b, and c.
The third group contains two people; combined, they answered "yes" to 3 questions: a, b, and c.
The fourth group contains four people; combined, they answered "yes" to only 1 question, a.
The last group contains one person who answered "yes" to only 1 question, b.
In this example, the sum of these counts is 3 + 3 + 3 + 1 + 1 = 11.
For each group, count the number of questions to which anyone answered "yes". What is the sum of those counts?
--- Part Two ---
As you finish the last group's customs declaration, you notice that you misread one word in the instructions:
You don't need to identify the questions to which anyone answered "yes"; you need to identify the questions to which everyone answered "yes"!
Using the same example as above:
abc
a
b
c
ab
ac
a
a
a
a
b
This list represents answers from five groups:
In the first group, everyone (all 1 person) answered "yes" to 3 questions: a, b, and c.
In the second group, there is no question to which everyone answered "yes".
In the third group, everyone answered yes to only 1 question, a. Since some people did not answer "yes" to b or c, they don't count.
In the fourth group, everyone answered yes to only 1 question, a.
In the fifth group, everyone (all 1 person) answered "yes" to 1 question, b.
In this example, the sum of these counts is 3 + 0 + 1 + 1 + 1 = 6.
For each group, count the number of questions to which everyone answered "yes". What is the sum of those counts?
2020/day07/prompt.md
-85
View File
@@ -1,85 +0,0 @@
--- Day 7: Handy Haversacks ---
You land at the regional airport in time for your next flight. In fact, it looks like you'll even have time to grab some food: all flights are currently delayed due to issues in luggage processing.
Due to recent aviation regulations, many rules (your puzzle input) are being enforced about bags and their contents; bags must be color-coded and must contain specific quantities of other color-coded bags. Apparently, nobody responsible for these regulations considered how long they would take to enforce!
For example, consider the following rules:
light red bags contain 1 bright white bag, 2 muted yellow bags.
dark orange bags contain 3 bright white bags, 4 muted yellow bags.
bright white bags contain 1 shiny gold bag.
muted yellow bags contain 2 shiny gold bags, 9 faded blue bags.
shiny gold bags contain 1 dark olive bag, 2 vibrant plum bags.
dark olive bags contain 3 faded blue bags, 4 dotted black bags.
vibrant plum bags contain 5 faded blue bags, 6 dotted black bags.
faded blue bags contain no other bags.
dotted black bags contain no other bags.
These rules specify the required contents for 9 bag types. In this example, every faded blue bag is empty, every vibrant plum bag contains 11 bags (5 faded blue and 6 dotted black), and so on.
You have a shiny gold bag. If you wanted to carry it in at least one other bag, how many different bag colors would be valid for the outermost bag? (In other words: how many colors can, eventually, contain at least one shiny gold bag?)
In the above rules, the following options would be available to you:
A bright white bag, which can hold your shiny gold bag directly.
A muted yellow bag, which can hold your shiny gold bag directly, plus some other bags.
A dark orange bag, which can hold bright white and muted yellow bags, either of which could then hold your shiny gold bag.
A light red bag, which can hold bright white and muted yellow bags, either of which could then hold your shiny gold bag.
So, in this example, the number of bag colors that can eventually contain at least one shiny gold bag is 4.
How many bag colors can eventually contain at least one shiny gold bag? (The list of rules is quite long; make sure you get all of it.)
--- Part Two ---
It's getting pretty expensive to fly these days - not because of ticket prices, but because of the ridiculous number of bags you need to buy!
Consider again your shiny gold bag and the rules from the above example:
faded blue bags contain 0 other bags.
dotted black bags contain 0 other bags.
vibrant plum bags contain 11 other bags: 5 faded blue bags and 6 dotted black bags.
dark olive bags contain 7 other bags: 3 faded blue bags and 4 dotted black bags.
So, a single shiny gold bag must contain 1 dark olive bag (and the 7 bags within it) plus 2 vibrant plum bags (and the 11 bags within each of those): 1 + 1*7 + 2 + 2*11 = 32 bags!
Of course, the actual rules have a small chance of going several levels deeper than this example; be sure to count all of the bags, even if the nesting becomes topologically impractical!
Here's another example:
shiny gold bags contain 2 dark red bags.
dark red bags contain 2 dark orange bags.
dark orange bags contain 2 dark yellow bags.
dark yellow bags contain 2 dark green bags.
dark green bags contain 2 dark blue bags.
dark blue bags contain 2 dark violet bags.
dark violet bags contain no other bags.
In this example, a single shiny gold bag must contain 126 other bags.
How many individual bags are required inside your single shiny gold bag?
2020/day08/prompt.md
-111
View File
@@ -1,111 +0,0 @@
--- Day 8: Handheld Halting ---
Your flight to the major airline hub reaches cruising altitude without incident. While you consider checking the in-flight menu for one of those drinks that come with a little umbrella, you are interrupted by the kid sitting next to you.
Their handheld game console won't turn on! They ask if you can take a look.
You narrow the problem down to a strange infinite loop in the boot code (your puzzle input) of the device. You should be able to fix it, but first you need to be able to run the code in isolation.
The boot code is represented as a text file with one instruction per line of text. Each instruction consists of an operation (acc, jmp, or nop) and an argument (a signed number like +4 or -20).
acc increases or decreases a single global value called the accumulator by the value given in the argument. For example, acc +7 would increase the accumulator by 7. The accumulator starts at 0. After an acc instruction, the instruction immediately below it is executed next.
jmp jumps to a new instruction relative to itself. The next instruction to execute is found using the argument as an offset from the jmp instruction; for example, jmp +2 would skip the next instruction, jmp +1 would continue to the instruction immediately below it, and jmp -20 would cause the instruction 20 lines above to be executed next.
nop stands for No OPeration - it does nothing. The instruction immediately below it is executed next.
For example, consider the following program:
nop +0
acc +1
jmp +4
acc +3
jmp -3
acc -99
acc +1
jmp -4
acc +6
These instructions are visited in this order:
nop +0 | 1
acc +1 | 2, 8(!)
jmp +4 | 3
acc +3 | 6
jmp -3 | 7
acc -99 |
acc +1 | 4
jmp -4 | 5
acc +6 |
First, the nop +0 does nothing. Then, the accumulator is increased from 0 to 1 (acc +1) and jmp +4 sets the next instruction to the other acc +1 near the bottom. After it increases the accumulator from 1 to 2, jmp -4 executes, setting the next instruction to the only acc +3. It sets the accumulator to 5, and jmp -3 causes the program to continue back at the first acc +1.
This is an infinite loop: with this sequence of jumps, the program will run forever. The moment the program tries to run any instruction a second time, you know it will never terminate.
Immediately before the program would run an instruction a second time, the value in the accumulator is 5.
Run your copy of the boot code. Immediately before any instruction is executed a second time, what value is in the accumulator?
--- Part Two ---
After some careful analysis, you believe that exactly one instruction is corrupted.
Somewhere in the program, either a jmp is supposed to be a nop, or a nop is supposed to be a jmp. (No acc instructions were harmed in the corruption of this boot code.)
The program is supposed to terminate by attempting to execute an instruction immediately after the last instruction in the file. By changing exactly one jmp or nop, you can repair the boot code and make it terminate correctly.
For example, consider the same program from above:
nop +0
acc +1
jmp +4
acc +3
jmp -3
acc -99
acc +1
jmp -4
acc +6
If you change the first instruction from nop +0 to jmp +0, it would create a single-instruction infinite loop, never leaving that instruction. If you change almost any of the jmp instructions, the program will still eventually find another jmp instruction and loop forever.
However, if you change the second-to-last instruction (from jmp -4 to nop -4), the program terminates! The instructions are visited in this order:
nop +0 | 1
acc +1 | 2
jmp +4 | 3
acc +3 |
jmp -3 |
acc -99 |
acc +1 | 4
nop -4 | 5
acc +6 | 6
After the last instruction (acc +6), the program terminates by attempting to run the instruction below the last instruction in the file. With this change, after the program terminates, the accumulator contains the value 8 (acc +1, acc +1, acc +6).
Fix the program so that it terminates normally by changing exactly one jmp (to nop) or nop (to jmp). What is the value of the accumulator after the program terminates?
2020/day09/prompt.md
-105
View File
@@ -1,105 +0,0 @@
--- Day 9: Encoding Error ---
With your neighbor happily enjoying their video game, you turn your attention to an open data port on the little screen in the seat in front of you.
Though the port is non-standard, you manage to connect it to your computer through the clever use of several paperclips. Upon connection, the port outputs a series of numbers (your puzzle input).
The data appears to be encrypted with the eXchange-Masking Addition System (XMAS) which, conveniently for you, is an old cypher with an important weakness.
XMAS starts by transmitting a preamble of 25 numbers. After that, each number you receive should be the sum of any two of the 25 immediately previous numbers. The two numbers will have different values, and there might be more than one such pair.
For example, suppose your preamble consists of the numbers 1 through 25 in a random order. To be valid, the next number must be the sum of two of those numbers:
26 would be a valid next number, as it could be 1 plus 25 (or many other pairs, like 2 and 24).
49 would be a valid next number, as it is the sum of 24 and 25.
100 would not be valid; no two of the previous 25 numbers sum to 100.
50 would also not be valid; although 25 appears in the previous 25 numbers, the two numbers in the pair must be different.
Suppose the 26th number is 45, and the first number (no longer an option, as it is more than 25 numbers ago) was 20. Now, for the next number to be valid, there needs to be some pair of numbers among 1-19, 21-25, or 45 that add up to it:
26 would still be a valid next number, as 1 and 25 are still within the previous 25 numbers.
65 would not be valid, as no two of the available numbers sum to it.
64 and 66 would both be valid, as they are the result of 19+45 and 21+45 respectively.
Here is a larger example which only considers the previous 5 numbers (and has a preamble of length 5):
35
20
15
25
47
40
62
55
65
95
102
117
150
182
127
219
299
277
309
576
In this example, after the 5-number preamble, almost every number is the sum of two of the previous 5 numbers; the only number that does not follow this rule is 127.
The first step of attacking the weakness in the XMAS data is to find the first number in the list (after the preamble) which is not the sum of two of the 25 numbers before it. What is the first number that does not have this property?
--- Part Two ---
The final step in breaking the XMAS encryption relies on the invalid number you just found: you must find a contiguous set of at least two numbers in your list which sum to the invalid number from step 1.
Again consider the above example:
35
20
15
25
47
40
62
55
65
95
102
117
150
182
127
219
299
277
309
576
In this list, adding up all of the numbers from 15 through 40 produces the invalid number from step 1, 127. (Of course, the contiguous set of numbers in your actual list might be much longer.)
To find the encryption weakness, add together the smallest and largest number in this contiguous range; in this example, these are 15 and 47, producing 62.
What is the encryption weakness in your XMAS-encrypted list of numbers?
2020/day11/prompt.md
-260
View File
@@ -1,260 +0,0 @@
--- Day 11: Seating System ---
Your plane lands with plenty of time to spare. The final leg of your journey is a ferry that goes directly to the tropical island where you can finally start your vacation. As you reach the waiting area to board the ferry, you realize you're so early, nobody else has even arrived yet!
By modeling the process people use to choose (or abandon) their seat in the waiting area, you're pretty sure you can predict the best place to sit. You make a quick map of the seat layout (your puzzle input).
The seat layout fits neatly on a grid. Each position is either floor (.), an empty seat (L), or an occupied seat (#). For example, the initial seat layout might look like this:
L.LL.LL.LL
LLLLLLL.LL
L.L.L..L..
LLLL.LL.LL
L.LL.LL.LL
L.LLLLL.LL
..L.L.....
LLLLLLLLLL
L.LLLLLL.L
L.LLLLL.LL
Now, you just need to model the people who will be arriving shortly. Fortunately, people are entirely predictable and always follow a simple set of rules. All decisions are based on the number of occupied seats adjacent to a given seat (one of the eight positions immediately up, down, left, right, or diagonal from the seat). The following rules are applied to every seat simultaneously:
If a seat is empty (L) and there are no occupied seats adjacent to it, the seat becomes occupied.
If a seat is occupied (#) and four or more seats adjacent to it are also occupied, the seat becomes empty.
Otherwise, the seat's state does not change.
Floor (.) never changes; seats don't move, and nobody sits on the floor.
After one round of these rules, every seat in the example layout becomes occupied:
#.##.##.##
#######.##
#.#.#..#..
####.##.##
#.##.##.##
#.#####.##
..#.#.....
##########
#.######.#
#.#####.##
After a second round, the seats with four or more occupied adjacent seats become empty again:
#.LL.L#.##
#LLLLLL.L#
L.L.L..L..
#LLL.LL.L#
#.LL.LL.LL
#.LLLL#.##
..L.L.....
#LLLLLLLL#
#.LLLLLL.L
#.#LLLL.##
This process continues for three more rounds:
#.##.L#.##
#L###LL.L#
L.#.#..#..
#L##.##.L#
#.##.LL.LL
#.###L#.##
..#.#.....
#L######L#
#.LL###L.L
#.#L###.##
#.#L.L#.##
#LLL#LL.L#
L.L.L..#..
#LLL.##.L#
#.LL.LL.LL
#.LL#L#.##
..L.L.....
#L#LLLL#L#
#.LLLLLL.L
#.#L#L#.##
#.#L.L#.##
#LLL#LL.L#
L.#.L..#..
#L##.##.L#
#.#L.LL.LL
#.#L#L#.##
..L.L.....
#L#L##L#L#
#.LLLLLL.L
#.#L#L#.##
At this point, something interesting happens: the chaos stabilizes and further applications of these rules cause no seats to change state! Once people stop moving around, you count 37 occupied seats.
Simulate your seating area by applying the seating rules repeatedly until no seats change state. How many seats end up occupied?
--- Part Two ---
As soon as people start to arrive, you realize your mistake. People don't just care about adjacent seats - they care about the first seat they can see in each of those eight directions!
Now, instead of considering just the eight immediately adjacent seats, consider the first seat in each of those eight directions. For example, the empty seat below would see eight occupied seats:
.......#.
...#.....
.#.......
.........
..#L....#
....#....
.........
#........
...#.....
The leftmost empty seat below would only see one empty seat, but cannot see any of the occupied ones:
.............
.L.L.#.#.#.#.
.............
The empty seat below would see no occupied seats:
.##.##.
#.#.#.#
##...##
...L...
##...##
#.#.#.#
.##.##.
Also, people seem to be more tolerant than you expected: it now takes five or more visible occupied seats for an occupied seat to become empty (rather than four or more from the previous rules). The other rules still apply: empty seats that see no occupied seats become occupied, seats matching no rule don't change, and floor never changes.
Given the same starting layout as above, these new rules cause the seating area to shift around as follows:
L.LL.LL.LL
LLLLLLL.LL
L.L.L..L..
LLLL.LL.LL
L.LL.LL.LL
L.LLLLL.LL
..L.L.....
LLLLLLLLLL
L.LLLLLL.L
L.LLLLL.LL
#.##.##.##
#######.##
#.#.#..#..
####.##.##
#.##.##.##
#.#####.##
..#.#.....
##########
#.######.#
#.#####.##
#.LL.LL.L#
#LLLLLL.LL
L.L.L..L..
LLLL.LL.LL
L.LL.LL.LL
L.LLLLL.LL
..L.L.....
LLLLLLLLL#
#.LLLLLL.L
#.LLLLL.L#
#.L#.##.L#
#L#####.LL
L.#.#..#..
##L#.##.##
#.##.#L.##
#.#####.#L
..#.#.....
LLL####LL#
#.L#####.L
#.L####.L#
#.L#.L#.L#
#LLLLLL.LL
L.L.L..#..
##LL.LL.L#
L.LL.LL.L#
#.LLLLL.LL
..L.L.....
LLLLLLLLL#
#.LLLLL#.L
#.L#LL#.L#
#.L#.L#.L#
#LLLLLL.LL
L.L.L..#..
##L#.#L.L#
L.L#.#L.L#
#.L####.LL
..#.#.....
LLL###LLL#
#.LLLLL#.L
#.L#LL#.L#
#.L#.L#.L#
#LLLLLL.LL
L.L.L..#..
##L#.#L.L#
L.L#.LL.L#
#.LLLL#.LL
..#.L.....
LLL###LLL#
#.LLLLL#.L
#.L#LL#.L#
Again, at this point, people stop shifting around and the seating area reaches equilibrium. Once this occurs, you count 26 occupied seats.
Given the new visibility method and the rule change for occupied seats becoming empty, once equilibrium is reached, how many seats end up occupied?
scripts/template/template.go
+2 -3
View File
@@ -63,7 +63,7 @@ import (
"fmt"
"strings"
"github.com/alexchao26/advent-of-code-go/mathy"
"github.com/alexchao26/advent-of-code-go/cast"
"github.com/alexchao26/advent-of-code-go/util"
)
@@ -96,8 +96,7 @@ func part2(input string) int {
}
func parseInput(input string) (ans []int) {
lines := strings.Split(input, "\n")
for _, l := range lines {
for _, line := range strings.Split(input, "\n") {
ans = append(ans, cast.ToInt(l))
}
return ans