Combine AoC Repos

This commit is contained in:
2016-12-16 16:21:15 -06:00
parent 5977b28d73
commit 105dbd1ff7
151 changed files with 9081 additions and 1 deletions

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R3, L5, R2, L1, L2, R5, L2, R2, L2, L2, L1, R2, L2, R4, R4, R1, L2, L3, R3, L1, R2, L2, L4, R4, R5, L3, R3, L3, L3, R4, R5, L3, R3, L5, L1, L2, R2, L1, R3, R1, L1, R187, L1, R2, R47, L5, L1, L2, R4, R3, L3, R3, R4, R1, R3, L1, L4, L1, R2, L1, R4, R5, L1, R77, L5, L4, R3, L2, R4, R5, R5, L2, L2, R2, R5, L2, R194, R5, L2, R4, L5, L4, L2, R5, L3, L2, L5, R5, R2, L3, R3, R1, L4, R2, L1, R5, L1, R5, L1, L1, R3, L1, R5, R2, R5, R5, L4, L5, L5, L5, R3, L2, L5, L4, R3, R1, R1, R4, L2, L4, R5, R5, R4, L2, L2, R5, R5, L5, L2, R4, R4, L4, R1, L3, R1, L1, L1, L1, L4, R5, R4, L4, L4, R5, R3, L2, L2, R3, R1, R4, L3, R1, L4, R3, L3, L2, R2, R2, R2, L1, L4, R3, R2, R2, L3, R2, L3, L2, R4, L2, R3, L4, R5, R4, R1, R5, R3

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package main
import (
"fmt"
"io"
"log"
"math"
"strconv"
"strings"
)
const (
north = iota
east
south
west
)
type pos struct {
dir int
x int
y int
}
func (s *pos) TurnLeft() {
s.dir++
if s.dir > west {
s.dir = north
}
}
func (s *pos) TurnRight() {
s.dir--
if s.dir < north {
s.dir = west
}
}
func (s *pos) CoordsToString() string {
return strconv.Itoa(s.x) + "-" + strconv.Itoa(s.y)
}
func (s *pos) DistanceTravelled() int {
return int(math.Abs(float64(s.y)) + math.Abs(float64(s.x)))
}
func (s *pos) Walk(dist int) {
switch s.dir {
case north:
s.y += dist
case east:
s.x += dist
case south:
s.y -= dist
case west:
s.x -= dist
}
}
func main() {
var input string
var readInp string
travelLog := make(map[string]int)
firstRepeat := false
currPos := pos{dir: north, x: 0, y: 0}
travelLog[currPos.CoordsToString()]++
for {
_, err := fmt.Scan(&readInp)
if err != nil {
if err != io.EOF {
log.Fatal(err)
}
break
}
input += readInp
}
for _, k := range strings.Split(input, ",") {
k = strings.Trim(k, " ")
if k[0] == 'R' {
currPos.TurnRight()
} else {
currPos.TurnLeft()
}
tDist := atoi(k[1:])
for mvI := 0; mvI < tDist; mvI++ {
currPos.Walk(1)
if !firstRepeat && travelLog[currPos.CoordsToString()] > 0 {
fmt.Println(" First Repeated Position: " + currPos.CoordsToString())
fmt.Println(" Distance: ", currPos.DistanceTravelled())
firstRepeat = true
}
travelLog[currPos.CoordsToString()]++
}
}
fmt.Println("End Position: ", currPos.x, ",", currPos.y)
fmt.Println(" Distance: ", currPos.DistanceTravelled())
}
func atoi(inp string) int {
var err error
var i int
if i, err = strconv.Atoi(inp); err != nil {
log.Fatal(err.Error())
}
return i
}

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Advent of Code
--- Day 1: No Time for a Taxicab ---
Santa's sleigh uses a very high-precision clock to guide its movements, and the clock's oscillator is regulated by stars. Unfortunately, the stars have been stolen... by the Easter Bunny. To save Christmas, Santa needs you to
retrieve 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!
You're airdropped near Easter Bunny Headquarters in a city somewhere. "Near", unfortunately, is as close as you can get - the instructions on the Easter Bunny Recruiting Document the Elves intercepted start here, and nobody had time
to work them out further.
The Document indicates that you should start at the given coordinates (where you just landed) and face North. Then, follow the provided sequence: either turn left (L) or right (R) 90 degrees, then walk forward the given number of
blocks, ending at a new intersection.
There's no time to follow such ridiculous instructions on foot, though, so you take a moment and work out the destination. Given that you can only walk on the street grid of the city, how far is the shortest path to the destination?
For example:
 Following R2, L3 leaves you 2 blocks East and 3 blocks North, or 5 blocks away.
 R2, R2, R2 leaves you 2 blocks due South of your starting position, which is 2 blocks away.
 R5, L5, R5, R3 leaves you 12 blocks away.
How many blocks away is Easter Bunny HQ?
Your puzzle answer was ___.
--- Part Two ---
Then, you notice the instructions continue on the back of the Recruiting Document. Easter Bunny HQ is actually at the first location you visit twice.
For example, if your instructions are R8, R4, R4, R8, the first location you visit twice is 4 blocks away, due East.
How many blocks away is the first location you visit twice?
Your puzzle answer was ___.
If you still want to see it, you can get your puzzle input.
References
Visible links
. http://adventofcode.com/
. http://adventofcode.com/2016/about
. http://adventofcode.com/2016/support
. http://adventofcode.com/2016/events
. http://adventofcode.com/2016/settings
. http://adventofcode.com/2016/auth/logout
. http://adventofcode.com/2016
. http://adventofcode.com/2016
. http://adventofcode.com/2016/leaderboard
. http://adventofcode.com/2016/stats
. http://adventofcode.com/2016/sponsors
. http://adventofcode.com/2016/sponsors
. https://en.wikipedia.org/wiki/Taxicab_geometry
. http://adventofcode.com/2016
. http://adventofcode.com/2016/day/1/input

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DULUDRDDDRLUDURUUULRRRURDRDULRUDDUDRULUDDUDRLDULRRLRDRUDUUULUUDLRURDUDDDDRDLLLLULRDLDRDLRLULRUURDDUULUDLRURRDDRDDRDDLDRDLLUURDRUULRRURURRDLRLLLUDULULULULUDRLLRUDUURLDRLRLRDRRDRLLLDURRDULDURDDRLURRDURLRRRLDLLLDRUUURLRDLDLLLLRDURRLDLULRLDDLDLURLRRDDRUDDUULRURRUDLRDLDUURDDDDRLRURUDULUDLRRLLLLLRDRURLLDLDULUUDLUDDDRLLDRRUDLLURRUUDDRRLLRRLDDDURLDRDRLURRRRDRRRDDUDULULDURRUUURRRDULUUUDDRULDRLLRDLDURLURRLLRUUUULRDURLLDDRLLDLRLRULUUDRURUDLLURUDDRDURLRDRRRDURLDDRDRLRLLURULUUULUDDDULDLRDDDRDLLRRLDRDULLUUUDLDDLDDDLLLLLLLDUDURURDURDRUURRRDDRDUDLULDURDUDURDDDRULDURURURRLURLURLUURLULDLLRUULURDDRLRDDLRDLRRR
LUURLRUDRRUDLLDLUDDURULURLUUDUUDDRLUULRDUDDUULDUUDRURDDRRDRLULLRDRDLRLLUURRUULRLDRULUDLDUUDDDRDDLRDLULDRLDUULDLRDLLLDLDLRDUULUDURRULLRLDUDRLLLULUUUULUUDUUURRRDULLUURUDRRLDURRUULDRDULDUDRDUUULUUDDRLUDRLDLDRUUURDLDUDRUDUURLLRRLRLLRRLDULDDULUDUUURULDDUDUDRURRDLULRUDDURDLDLLRRRLDRLULLLRUULDUDLUUDURRLLLRLUDURRDDLDRDDDLURDLDRRUDUDLUDULULRUUUDLUURLLRLDDLURULDURDLRRDDDDURLDDLLDDULLLRLDLDULDUUDDRLDUURDDLDLUUDULRRLRLUURURUURLRLURUURLDRUURLLRDDUUUDULUDDDRDRLDRDRRLRLDULLRRUDLURULULRDRURURLULDUDLRURLRDDRULDDLRD
LUDRULUULRRDDDDRRDUURUDDRLDDLDRDURRURULRDLDLDUUDRRDUUDUDLLLRRLDUDDRLDDLRRLRDRLUDLULUDDUUDULDUUULUDLDDURLDURUDLDRUUDRLRRLDLDDULDUUDDLDDLLURDRLRUURDDRUDDUDLDRRLRUDRUULRRRLRULULURDLRRURDRLRULDDDRDUULLURUUUURUDDLRRRRRDURLULDLUULUDRRUDUDRRDDRURDURLRLUDDLDLRRULUDLDDRLDDLDDDLLLLRDLLUULDDLULDLDRDDUDLURUDLDLDDRRUUDDDLRLLLDRRDDDUURDUDURUURRDRLLDUDLDUULLDLDLLUULLRRULDLDRURLDULDRUURDURRURDLRDLLLDRRUDRUUDRURLUDDRURLDURRDLUUDLUUDULLLDDDDRRDLLLDLURULDDRDLUUURRDRRUUDDUL
DUUULDUDDDURLLULDDLLUDURLLLURULULURUURDRURLRULLLLDRDDULRRDRRLLLRDDDUULLRRURRULLDDURRRLRDDLULDULLDUDLURRDLDDLURDLRLLDRURLLRLLRRRDRRRURURUUDDLLDDLDDDLRLURUUUULRDLUDDDURLLDDRLDRRLLUDUUULRLLDRRRLRUUDLDUULRLUDRULLLLDUDLLUUDDRUURLURUDRDDDLRURUDRLULLULUUDLDURDULRRDRLDURUULRDRRRDRDRRLRLRDDUULLRDLDURDDDULURRLULDDURDURDDUDURDLLUUULUDULRDDLDRDRUDLLUURDLRDURURULURULLDRLLRRULDLULULDLULRURLRRLUDLLLRLUDLURLULDULDRLLLDLDDDDRDRLRRLRDULUUDULDDLDURDLLLDDDDLLUURRDURLDLUDDLULRUUUDDRRLDLLLRDLLDRRRDDLULLURDDRRRRLDLRLLLRL
LULLRRDURRLDUUDRRURLURURRRLRDRUULUULURLLURRDRULRDURDDDDUULLLLDUULDLULURDRLDLULULDRLLDLLRLRULURUDRUUDULRULLLUDRULUDRLLUDLDRRDRUUURURLRDURDRLRDDDURLURRDLRUUUDUURULULDLUULRDLRRRDRDRLLLDLRRDRLLDDULDRUDRRLULLRDLDUDDULRDDLULRURULRLLLULDLLLLRDLDRURUDUURURLDRLUULLDUDULUDDDULUDLRUDDUDLULLUULUUURULURRULRDDURDDLURLRRDRDLDULRLRDRRRULRDDDRLLDDDDRRRRDRDLULUURDURULDLRDULDUDLDURUDLUDLUDDDUDURDURDDURLLRUDUURRRUDRRRRULLLLDDDLUULLUULRRRULDLURDLULRULDRLR

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package main
import (
"bufio"
"fmt"
"log"
"os"
"strconv"
"strings"
)
func main() {
input := stdinToStringSlice()
var press []rune
keypad := initKeypad("diamond")
pos := '5'
for _, i := range input {
for _, j := range i {
pos = moveFinger(j, pos, keypad)
}
press = append(press, pos)
}
fmt.Println("Code: ", string(press))
}
func moveFinger(dir rune, start rune, keypad map[string]rune) rune {
// Find the starting button
var startX, startY int
for k, v := range keypad {
if v == start {
// Found it
strCoords := strings.Split(k, ";")
startY = atoi(strCoords[0])
startX = atoi(strCoords[1])
}
}
switch dir {
case 'U':
newY := startY - 1
if keypad[itoa(newY)+";"+itoa(startX)] != 0 {
return keypad[itoa(newY)+";"+itoa(startX)]
}
case 'R':
newX := startX + 1
if keypad[itoa(startY)+";"+itoa(newX)] != 0 {
return keypad[itoa(startY)+";"+itoa(newX)]
}
case 'D':
newY := startY + 1
if keypad[itoa(newY)+";"+itoa(startX)] != 0 {
return keypad[itoa(newY)+";"+itoa(startX)]
}
case 'L':
newX := startX - 1
if keypad[itoa(startY)+";"+itoa(newX)] != 0 {
return keypad[itoa(startY)+";"+itoa(newX)]
}
}
return start
}
func initKeypad(tp string) map[string]rune {
ret := make(map[string]rune)
switch tp {
case "diamond":
ret["0;2"] = '1'
ret["1;1"] = '2'
ret["1;2"] = '3'
ret["1;3"] = '4'
ret["2;0"] = '5'
ret["2;1"] = '6'
ret["2;2"] = '7'
ret["2;3"] = '8'
ret["2;4"] = '9'
ret["3;1"] = 'A'
ret["3;2"] = 'B'
ret["3;3"] = 'C'
ret["4;2"] = 'D'
case "square":
currKey := 1
for i := 0; i < 3; i++ {
for j := 0; j < 3; j++ {
ret[itoa(i)+";"+itoa(j)] = rune(currKey)
currKey++
}
}
}
return ret
}
func itoa(i int) string {
return strconv.Itoa(i)
}
func atoi(i string) int {
var ret int
var err error
if ret, err = strconv.Atoi(i); err != nil {
log.Fatal("Invalid Atoi")
}
return ret
}
func stdinToStringSlice() []string {
var input []string
scanner := bufio.NewScanner(os.Stdin)
for scanner.Scan() {
input = append(input, scanner.Text())
}
return input
}

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Advent of Code
--- Day 2: Bathroom Security ---
You arrive at Easter Bunny Headquarters under cover of darkness. However, you left in such a rush that you forgot to
use the bathroom! Fancy office buildings like this one usually have keypad locks on their bathrooms, so you search the
front desk for the code.
"In order to improve security," the document you find says, "bathroom codes will no longer be written down. Instead,
please memorize and follow the procedure below to access the bathrooms."
The document goes on to explain that each button to be pressed can be found by starting on the previous button and
moving to adjacent buttons on the keypad: U moves up, D moves down, L moves left, and R moves right. Each line of
instructions corresponds to one button, starting at the previous button (or, for the first line, the "5" button); press
whatever button you're on at the end of each line. If a move doesn't lead to a button, ignore it.
You can't hold it much longer, so you decide to figure out the code as you walk to the bathroom. You picture a keypad
like this:
1 2 3
4 5 6
7 8 9
Suppose your instructions are:
ULL
RRDDD
LURDL
UUUUD
 You start at "5" and move up (to "2"), left (to "1"), and left (you can't, and stay on "1"), so the first button is
1.
 Starting from the previous button ("1"), you move right twice (to "3") and then down three times (stopping at "9"
after two moves and ignoring the third), ending up with 9.
 Continuing from "9", you move left, up, right, down, and left, ending with 8.
 Finally, you move up four times (stopping at "2"), then down once, ending with 5.
So, in this example, the bathroom code is 1985.
Your puzzle input is the instructions from the document you found at the front desk. What is the bathroom code?
Your puzzle answer was ______.
--- Part Two ---
You finally arrive at the bathroom (it's a several minute walk from the lobby so visitors can behold the many fancy
conference rooms and water coolers on this floor) and go to punch in the code. Much to your bladder's dismay, the
keypad is not at all like you imagined it. Instead, you are confronted with the result of hundreds of man-hours of
bathroom-keypad-design meetings:
1
2 3 4
5 6 7 8 9
A B C
D
You still start at "5" and stop when you're at an edge, but given the same instructions as above, the outcome is very
different:
 You start at "5" and don't move at all (up and left are both edges), ending at 5.
 Continuing from "5", you move right twice and down three times (through "6", "7", "B", "D", "D"), ending at D.
 Then, from "D", you move five more times (through "D", "B", "C", "C", "B"), ending at B.
 Finally, after five more moves, you end at 3.
So, given the actual keypad layout, the code would be 5DB3.
Using the same instructions in your puzzle input, what is the correct bathroom code?
Your puzzle answer was _____.
Both parts of this puzzle are complete! They provide two gold stars: **
If you still want to see it, you can get your puzzle input.
References
Visible links
. http://adventofcode.com/
. http://adventofcode.com/2016/about
. http://adventofcode.com/2016/support
. http://adventofcode.com/2016/events
. http://adventofcode.com/2016/settings
. http://adventofcode.com/2016/auth/logout
. http://adventofcode.com/2016
. http://adventofcode.com/2016
. http://adventofcode.com/2016/leaderboard
. http://adventofcode.com/2016/stats
. http://adventofcode.com/2016/sponsors
. http://adventofcode.com/2016/sponsors
. http://adventofcode.com/2016
. http://adventofcode.com/2016/day/2/input

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package main
import (
"bufio"
"fmt"
"log"
"os"
"sort"
"strconv"
"strings"
)
func main() {
input := stdinToStringSlice()
puzzleOne(input)
puzzleTwo(input)
}
func puzzleOne(input []string) {
fmt.Println("=== Puzzle One ===")
var numGood int
var sides []int
for i := range input {
pts := strings.Fields(input[i])
sides = []int{atoi(pts[0]), atoi(pts[1]), atoi(pts[2])}
if isValidTriangle(sides) {
numGood++
}
}
fmt.Println("Found Good: ", numGood)
}
func puzzleTwo(input []string) {
fmt.Println("=== Puzzle Two ===")
var numGood int
var sides1, sides2, sides3 []int
for i := range input {
pts := strings.Fields(input[i])
sides1 = append(sides1, atoi(pts[0]))
sides2 = append(sides2, atoi(pts[1]))
sides3 = append(sides3, atoi(pts[2]))
if len(sides1) == 3 {
if isValidTriangle(sides1) {
numGood++
}
if isValidTriangle(sides2) {
numGood++
}
if isValidTriangle(sides3) {
numGood++
}
sides1 = []int{}
sides2 = []int{}
sides3 = []int{}
}
}
fmt.Println("Found Good: ", numGood)
}
func isValidTriangle(sides []int) bool {
sort.Ints(sides)
if sides[2] < (sides[1] + sides[0]) {
return true
}
return false
}
func itoa(i int) string {
return strconv.Itoa(i)
}
func atoi(i string) int {
var ret int
var err error
if ret, err = strconv.Atoi(i); err != nil {
log.Fatal("Invalid Atoi")
}
return ret
}
func stdinToStringSlice() []string {
var input []string
scanner := bufio.NewScanner(os.Stdin)
for scanner.Scan() {
input = append(input, scanner.Text())
}
return input
}

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Advent of Code
--- Day 3: Squares With Three Sides ---
Now that you can think clearly, you move deeper into the labyrinth of hallways and office
furniture that makes up this part of Easter Bunny HQ. This must be a graphic design
department; the walls are covered in specifications for triangles.
Or are they?
The design document gives the side lengths of each triangle it describes, but... 5 10 25? Some
of these aren't triangles. You can't help but mark the impossible ones.
In a valid triangle, the sum of any two sides must be larger than the remaining side. For
example, the "triangle" given above is impossible, because 5 + 10 is not larger than 25.
In your puzzle input, how many of the listed triangles are possible?
Your puzzle answer was ___.
The first half of this puzzle is complete! It provides one gold star: *
--- Part Two ---
Now that you've helpfully marked up their design documents, it occurs to you that triangles
are specified in groups of three vertically. Each set of three numbers in a column specifies a
triangle. Rows are unrelated.
For example, given the following specification, numbers with the same hundreds digit would be
part of the same triangle:
101 301 501
102 302 502
103 303 503
201 401 601
202 402 602
203 403 603
In your puzzle input, and instead reading by columns, how many of the listed triangles are
possible?
Although it hasn't changed, you can still get your puzzle input.
References
Visible links
. http://adventofcode.com/
. http://adventofcode.com/2016/about
. http://adventofcode.com/2016/support
. http://adventofcode.com/2016/events
. http://adventofcode.com/2016/settings
. http://adventofcode.com/2016/auth/logout
. http://adventofcode.com/2016
. http://adventofcode.com/2016
. http://adventofcode.com/2016/leaderboard
. http://adventofcode.com/2016/stats
. http://adventofcode.com/2016/sponsors
. http://adventofcode.com/2016/sponsors
. http://adventofcode.com/2016/day/3/input

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2016/day04/input Normal file
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hqcfqwydw-fbqijys-whqii-huiuqhsx-660[qhiwf]
oxjmxdfkd-pzxsbkdbo-erkq-ixyloxqlov-913[xodkb]
bpvctixr-eaphixr-vgphh-gthtpgrw-947[smrkl]
iwcjapey-lhwopey-cnwoo-wymqeoepekj-992[eowpy]
mvhkvbdib-agjrzm-zibdizzmdib-317[bizdm]
excdklvo-lkcuod-dbksxsxq-146[ztwya]
ocipgvke-ejqeqncvg-octmgvkpi-908[prmku]
ktwbhtvmbox-vetllbybxw-vtgwr-vhtmbgz-tvjnblbmbhg-579[uvnyc]
dpmpsgvm-tdbwfohfs-ivou-tijqqjoh-389[emdac]
forwcoqhwjs-pibbm-igsf-hsghwbu-532[bhswf]
uzfqdzmfuazmx-nmewqf-ogefayqd-eqdhuoq-664[qfdem]
fnjyxwrinm-yujbcrl-pajbb-uxprbcrlb-277[brjcl]
aoubshwq-dzoghwq-ufogg-fsoqeiwgwhwcb-714[nkrmy]
pbeebfvir-rtt-fnyrf-975[frbet]
bnknqetk-qzaahs-trdq-sdrshmf-235[mtcqz]
odiih-ljwmh-lxjcrwp-orwjwlrwp-927[wjlrh]
sxdobxkdsyxkv-bkllsd-cobfsmoc-302[sbdko]
gzefmnxq-omzpk-ymzmsqyqzf-352[saomt]
tvsnigxmpi-gerhc-gsexmrk-qerekiqirx-854[eirgx]
ktfitzbgz-vtgwr-ftgtzxfxgm-267[tgfzx]
lxuxaodu-npp-orwjwlrwp-563[pwlor]
oazegyqd-sdmpq-pkq-xmnadmfadk-352[damqk]
wfruflnsl-gzssd-hzxytrjw-xjwanhj-177[bgxsp]
pbybeshy-qlr-qrfvta-455[tmios]
xmrrq-udskkaxawv-vqw-esfsywewfl-918[fdqsb]
vhehkyne-vahvhetmx-ltexl-917[uvhmy]
molgbzqfib-ciltbo-obzbfsfkd-393[htayl]
veqtekmrk-jpsaiv-wlmttmrk-256[ewyhq]
cvabijtm-lgm-apqxxqvo-512[dinjm]
oaxadrgx-nmewqf-qzsuzqqduzs-456[oevtg]
vehmsegxmzi-veffmx-wepiw-880[emfiv]
fruurvlyh-fubrjhqlf-fdqgb-frdwlqj-ghvljq-413[cgkzy]
otzkxtgzoutgr-inuiurgzk-sgxqkzotm-774[gtzko]
hwbba-eqpuwogt-itcfg-tcddkv-ujkrrkpi-154[ktbcd]
pynffvsvrq-cynfgvp-tenff-ynobengbel-377[fnevy]
aoubshwq-qcbgiasf-ufors-qvcqczohs-hsqvbczcum-558[hypcz]
kzeed-xhfajsljw-mzsy-knsfshnsl-281[nsmtd]
hwdtljsnh-hfsid-htfynsl-ijufwyrjsy-177[hsfjy]
excdklvo-zvkcdsm-qbkcc-psxkxmsxq-900[yznml]
diozmivodjivg-xviyt-pnzm-oznodib-239[iodvz]
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vxupkizork-kmm-sgtgmksktz-280[yublv]
cvabijtm-kivlg-kwibqvo-twoqabqka-408[pgush]
hqcfqwydw-fbqijys-whqii-mehaixef-218[vzaur]
bpvctixr-rpcsn-rdpixcv-ldgzhwde-271[cifnu]
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lujbbrornm-bljenwpna-qdwc-fxatbqxy-589[bnajl]
eqpuwogt-itcfg-tcddkv-vgejpqnqia-258[besga]
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lhkhszqx-fqzcd-atmmx-lzqjdshmf-859[hmqzd]
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fab-eqodqf-rxaiqd-etubbuzs-612[bqade]
cxy-bnlanc-ljwmh-nwprwnnarwp-251[nwacl]
hdgdovmt-bmvyz-pinovwgz-ytz-omvdidib-239[qfmcj]
wsvsdkbi-qbkno-mkxni-mykdsxq-bokmaescsdsyx-328[skbdm]
njmjubsz-hsbef-gmpxfs-tijqqjoh-727[ykelf]
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cvabijtm-kivlg-ewzsapwx-538[posuz]
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wihmogyl-aluxy-vumeyn-zchuhwcha-110[eisnw]
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hqfxxnknji-kqtbjw-wjhjnansl-177[ctzqd]
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wrs-vhfuhw-exqqb-dqdobvlv-751[qvbdh]
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guahyncw-vumeyn-xypyfijgyhn-370[ynghu]
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rdchjbtg-vgpst-qphzti-gtrtxkxcv-817[mayne]
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vkppo-sqdto-vydqdsydw-114[pzbiy]
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rdchjbtg-vgpst-rwdrdapit-stepgibtci-271[tdgip]
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vetllbybxw-unggr-tgterlbl-631[mfwxo]
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vcibutulxiom-jfumncw-alumm-ijyluncihm-214[muicl]
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rdchjbtg-vgpst-rpcsn-rdpixcv-hidgpvt-765[stnfw]
buzahisl-lnn-thuhnltlua-955[oschg]
enzcntvat-ohaal-bcrengvbaf-793[anbce]
eqpuwogt-itcfg-uecxgpigt-jwpv-hkpcpekpi-362[pgcei]
avw-zljyla-qlssfilhu-dvyrzovw-175[lvasw]
iuruxlar-xgsvgmotm-inuiurgzk-zxgototm-982[mlnut]
tyepcyletzylw-prr-opalcexpye-925[boymz]
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tcrjjzwzvu-vxx-kirzezex-841[zxejr]
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vcibutulxiom-wbiwifuny-guleyncha-682[uzxms]
ejpanjwpekjwh-bqvvu-ywjzu-nayaerejc-628[jeawn]
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eza-dpncpe-clmmte-lylwjdtd-509[delcm]
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pybgmyargtc-aylbw-qcptgacq-600[oscut]
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gcfcnuls-aluxy-wuhxs-jolwbumcha-578[uclah]
pyknyegle-pybgmyargtc-aylbw-qfgnngle-470[gyeln]
oazegyqd-sdmpq-gzefmnxq-qss-geqd-fqefuzs-508[qesdf]
xjmmjndqz-mvwwdo-yzkvmohzio-551[ypzog]
zekvierkzferc-treup-uvgcfpdvek-789[stzno]
ejpanjwpekjwh-xqjju-odellejc-576[enmtc]
ltpedcxots-tvv-sthxvc-115[skptq]
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gzefmnxq-otaoaxmfq-emxqe-326[emqxa]
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frqvxphu-judgh-gbh-uhfhlylqj-153[hufgj]
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froruixo-hjj-zrunvkrs-777[synml]
jvuzbtly-nyhkl-jhukf-jvhapun-jvuahputlua-929[ndjmy]
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ktiaaqnqml-zijjqb-apqxxqvo-798[qaijx]
hqfxxnknji-hfsid-wjhjnansl-931[nhjfi]
xjmmjndqz-wpiit-vxlpdndodji-941[dijmn]
ksodcbwnsr-rms-cdsfohwcbg-896[xvuol]
eza-dpncpe-tyepcyletzylw-nsznzwlep-nzyeltyxpye-847[xydvf]
emixwvqhml-jiasmb-ivitgaqa-928[iamqv]
etyyx-idkkxadzm-ehmzmbhmf-313[josnm]
lhkhszqx-fqzcd-bgnbnkzsd-qdzbpthrhshnm-911[bqzra]
dzczkrip-xiruv-upv-wzeretzex-945[icynm]
wihmogyl-aluxy-mwupyhayl-bohn-lymyulwb-266[nuraz]
kmjezxodgz-xcjxjgvoz-zibdizzmdib-239[yzkgs]
hqfxxnknji-wfggny-hzxytrjw-xjwanhj-593[jnxhw]
oknkvcta-itcfg-eqpuwogt-itcfg-ecpfa-eqcvkpi-ucngu-986[cgtef]
ykhknbqh-oywrajcan-dqjp-qoan-paopejc-810[ondma]
nwilwcejc-ywjzu-ykwpejc-naoawnyd-238[zjwsh]
dzczkrip-xiruv-sleep-rercpjzj-451[wykfr]
gpewwmjmih-nippcfier-qerekiqirx-178[ieprm]
bqvvu-oywrajcan-dqjp-wjwhuoeo-420[jowaq]
kzgwomvqk-xtiabqk-oziaa-bziqvqvo-148[qaiko]
fab-eqodqf-eomhqzsqd-tgzf-fdmuzuzs-820[fqzde]
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vkrhzxgbv-unggr-tgterlbl-319[tsrkm]
wihmogyl-aluxy-wuhxs-uhufsmcm-526[uhmls]
nzydfxpc-rclop-awldetn-rcldd-pyrtyppctyr-951[pcdry]
egdytrixat-eaphixr-vgphh-pcpanhxh-921[hpaxe]
nwzekwypera-lhwopey-cnwoo-hkceopeyo-654[eowpy]
zovldbkfz-zlkprjbo-doxab-zxkav-ixyloxqlov-367[olxzb]
lgh-kwujwl-xmrrq-kusnwfywj-zmfl-hmjuzskafy-372[gmait]
ipvohghykvbz-jhukf-ylzlhyjo-357[awkcb]
dmybmsuzs-otaoaxmfq-dqmocgueufuaz-976[muaod]
zbytomdsvo-bkllsd-cdybkqo-796[eufzt]
sbqiiyvyut-fbqijys-whqii-iqbui-998[ebfqa]
qyujihctyx-wbiwifuny-guleyncha-838[ejitg]
ikhcxvmbex-unggr-kxvxbobgz-683[ejuzo]
hafgnoyr-ohaal-jbexfubc-923[bjmzn]
shmml-wryylorna-genvavat-455[almnr]
yknnkoera-xqjju-klanwpekjo-420[empdo]
upq-tfdsfu-kfmmzcfbo-efwfmpqnfou-415[nmfed]
xcitgcpixdcpa-rdchjbtg-vgpst-hrpktcvtg-wjci-stepgibtci-557[ctgip]
fydelmwp-nsznzwlep-opgpwzaxpye-769[pewzl]
glrcplyrgmlyj-cee-pcqcypaf-548[ymzlj]
xmtjbzidx-wpiit-xjiovdihzio-265[ztyda]
rwcnawjcrxwju-ljwmh-mnbrpw-901[wjrcm]
wlqqp-tyftfcrkv-ivtvzmzex-841[tvfqz]
thnulapj-wshzapj-nyhzz-zopwwpun-669[pzhnw]
bpvctixr-rdggdhxkt-hrpktcvtg-wjci-pcpanhxh-401[chptg]
eza-dpncpe-awldetn-rcldd-dlwpd-743[delpa]
pbybeshy-sybjre-npdhvfvgvba-299[bvyeh]
qmpmxevc-kvehi-jpsaiv-viwievgl-802[viemp]
jrncbavmrq-pnaql-pbngvat-qrcyblzrag-715[arbnq]
ugjjgkanw-wyy-kzahhafy-736[clxvm]
mwupyhayl-bohn-nluchcha-682[hacln]
qjopwxha-xwogap-ykjpwejiajp-108[jpawo]
avw-zljyla-jhukf-huhsfzpz-175[hzafj]
lzfmdshb-okzrshb-fqzrr-cdoknxldms-573[olwsf]
cqwdujys-sbqiiyvyut-uww-iuhlysui-426[cwfuy]
yaxsnlcrun-ljwmh-bqryyrwp-901[rylnw]
cebwrpgvyr-pelbtravp-enoovg-znantrzrag-455[raegn]
nbhofujd-qmbtujd-hsbtt-efwfmpqnfou-389[fbtud]
pynffvsvrq-pnaql-pbngvat-ynobengbel-507[nmyvz]
ltpedcxots-gpqqxi-ldgzhwde-739[bkapm]
nglmtuex-vahvhetmx-wxiehrfxgm-527[zwksp]
kgjgrypw-epybc-aylbw-amyrgle-qyjcq-626[ygabc]
yflexwxoalrp-avb-abmilvjbkq-445[siqmz]
jshzzpmplk-kfl-klclsvwtlua-331[lkpsz]
ujoon-eaphixr-vgphh-prfjxhxixdc-193[hyzjx]
dfcxsqhwzs-qobrm-zcuwghwqg-168[qwcgh]
bqvvu-ydkykhwpa-klanwpekjo-966[kapvw]
aoubshwq-pibbm-kcfygvcd-740[wnucy]

108
2016/day04/main.go Normal file
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package main
import (
"bufio"
"fmt"
"log"
"os"
"strconv"
)
const (
locChecksum = iota
locSector
locName
)
func main() {
input := stdinToStringSlice()
var sectorTotal int
for _, h := range input {
var chksum, sector, name string
var loc int
freq := make(map[rune]int)
for i := len(h) - 1; i >= 0; i-- {
switch h[i] {
case ']': // start reading checksum
loc = locChecksum
case '[': // start reading sector
loc = locSector
case '-':
if loc == locSector { // done with sector
loc = locName
}
default:
switch loc {
case locChecksum:
chksum = string(h[i]) + chksum
case locSector:
sector = string(h[i]) + sector
case locName:
name = string(h[i]) + name
freq[rune(h[i])]++
}
}
}
fmt.Println(decryptName(name, atoi(sector)), " :: ", sector)
var top rune
var realChecksum string
for i := 0; i < 5; i++ {
top, freq = findTop(freq)
realChecksum = realChecksum + string(top)
}
if realChecksum == chksum {
sectorTotal += atoi(sector)
}
}
fmt.Println("Sector Total: " + itoa(sectorTotal))
}
func decryptName(name string, rot int) string {
var decName string
for _, v := range name {
for k := 0; k < rot; k++ {
v++
if v == ('z' + 1) {
v = 'a'
}
}
decName = decName + string(v)
}
return decName
}
func findTop(inp map[rune]int) (rune, map[rune]int) {
var topCnt int
var topVal rune
for k, v := range inp {
if v > topCnt || (v == topCnt && k < topVal) {
topCnt = v
topVal = k
}
}
delete(inp, topVal)
return topVal, inp
}
func stdinToStringSlice() []string {
var input []string
scanner := bufio.NewScanner(os.Stdin)
for scanner.Scan() {
input = append(input, scanner.Text())
}
return input
}
func itoa(i int) string {
return strconv.Itoa(i)
}
func atoi(i string) int {
var ret int
var err error
if ret, err = strconv.Atoi(i); err != nil {
log.Fatal("Invalid Atoi")
}
return ret
}

69
2016/day04/problem Normal file
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Advent of Code
--- Day 4: Security Through Obscurity ---
Finally, you come across an information kiosk with a list of rooms. Of course, the list is
encrypted and full of decoy data, but the instructions to decode the list are barely hidden
nearby. Better remove the decoy data first.
Each room consists of an encrypted name (lowercase letters separated by dashes) followed by a
dash, a sector ID, and a checksum in square brackets.
A room is real (not a decoy) if the checksum is the five most common letters in the encrypted
name, in order, with ties broken by alphabetization. For example:
 aaaaa-bbb-z-y-x-123[abxyz] is a real room because the most common letters are a (5), b
(3), and then a tie between x, y, and z, which are listed alphabetically.
 a-b-c-d-e-f-g-h-987[abcde] is a real room because although the letters are all tied (1 of
each), the first five are listed alphabetically.
 not-a-real-room-404[oarel] is a real room.
 totally-real-room-200[decoy] is not.
Of the real rooms from the list above, the sum of their sector IDs is 1514.
What is the sum of the sector IDs of the real rooms?
Your puzzle answer was ___________.
The first half of this puzzle is complete! It provides one gold star: *
--- Part Two ---
With all the decoy data out of the way, it's time to decrypt this list and get moving.
The room names are encrypted by a state-of-the-art shift cipher, which is nearly unbreakable
without the right software. However, the information kiosk designers at Easter Bunny HQ were
not expecting to deal with a master cryptographer like yourself.
To decrypt a room name, rotate each letter forward through the alphabet a number of times
equal to the room's sector ID. A becomes B, B becomes C, Z becomes A, and so on. Dashes become
spaces.
For example, the real name for qzmt-zixmtkozy-ivhz-343 is very encrypted name.
What is the sector ID of the room where North Pole objects are stored?
Although it hasn't changed, you can still get your puzzle input.
Answer: _____________________ [ [Submit] ]
You can also [Shareon Twitter Google+ Reddit] this puzzle.
References
Visible links
. http://adventofcode.com/
. http://adventofcode.com/2016/about
. http://adventofcode.com/2016/support
. http://adventofcode.com/2016/events
. http://adventofcode.com/2016/settings
. http://adventofcode.com/2016/auth/logout
. http://adventofcode.com/2016
. http://adventofcode.com/2016
. http://adventofcode.com/2016/leaderboard
. http://adventofcode.com/2016/stats
. http://adventofcode.com/2016/sponsors
. http://adventofcode.com/2016/sponsors
. http://www.novetta.com/careers/#opportunities
. https://en.wikipedia.org/wiki/Caesar_cipher
. http://adventofcode.com/2016/day/4/input

1
2016/day05/input Normal file
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@@ -0,0 +1 @@
ojvtpuvg

118
2016/day05/main.go Normal file
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@@ -0,0 +1,118 @@
package main
import (
"bufio"
"crypto/md5"
"fmt"
"log"
"math/rand"
"os"
"strconv"
"strings"
"time"
)
func main() {
input := stdinToString()
partOne(input)
partTwo(input)
}
func partOne(input string) {
complete := false
idx := 1
password := ""
for !complete {
if idx%1000000 == 0 {
fmt.Print("O")
} else if idx%100000 == 0 {
fmt.Print(".")
}
sum := fmt.Sprintf("%x", md5.Sum([]byte(input+itoa(idx))))
if strings.HasPrefix(sum, "00000") {
fmt.Println("d")
fmt.Println(itoa(idx))
password += string(sum[5])
}
idx++
if len(password) == 8 {
complete = true
}
}
fmt.Println("")
fmt.Println(password)
}
func partTwo(input string) {
fmt.Println("Cracking the Password, hold on to your butts: ")
rand.Seed(time.Now().UTC().UnixNano())
complete := false
idx := 1469580
password := []byte{0, 0, 0, 0, 0, 0, 0, 0}
printPassword(password)
fnd := 0
for !complete {
sum := fmt.Sprintf("%x", md5.Sum([]byte(input+itoa(idx))))
if strings.HasPrefix(sum, "00000") {
pos, err := strconv.Atoi(string(sum[5]))
if err == nil && pos < 8 && password[pos] == 0 {
password[pos] = sum[6]
fnd++
if fnd == 8 {
complete = true
}
}
}
if idx%100000 == 0 {
clearPassword()
printPassword(password)
}
idx++
}
clearPassword()
printPassword(password)
fmt.Println("")
fmt.Println("!~CRACK'D~!")
}
func clearPassword() {
fmt.Print("")
}
func printPassword(pw []byte) {
for i := 0; i < len(pw); i++ {
if pw[i] == 0 {
// Print random char
rndBt := byte(97 + rand.Intn(122-97))
if rndBt%3 == 0 {
fmt.Print(itoa(int(rndBt % 10)))
} else {
fmt.Print(string(byte(97 + rand.Intn(122-97))))
}
} else {
fmt.Print(string(pw[i]))
}
}
}
func stdinToString() string {
var input string
scanner := bufio.NewScanner(os.Stdin)
for scanner.Scan() {
input = input + scanner.Text()
}
return input
}
func itoa(i int) string {
return strconv.Itoa(i)
}
func atoi(i string) int {
var ret int
var err error
if ret, err = strconv.Atoi(i); err != nil {
log.Fatal("Invalid Atoi")
}
return ret
}

71
2016/day05/problem Normal file
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@@ -0,0 +1,71 @@
Advent of Code
--- Day 5: How About a Nice Game of Chess? ---
You are faced with a security door designed by Easter Bunny engineers that seem to have acquired most of their security knowledge by watching hacking movies.
The eight-character password for the door is generated one character at a time by finding the MD5 hash of some Door ID (your puzzle input) and an increasing integer
index (starting with 0).
A hash indicates the next character in the password if its hexadecimal representation starts with five zeroes. If it does, the sixth character in the hash is the next
character of the password.
For example, if the Door ID is abc:
 The first index which produces a hash that starts with five zeroes is 3231929, which we find by hashing abc3231929; the sixth character of the hash, and thus the
first character of the password, is 1.
 5017308 produces the next interesting hash, which starts with 000008f82..., so the second character of the password is 8.
 The third time a hash starts with five zeroes is for abc5278568, discovering the character f.
In this example, after continuing this search a total of eight times, the password is 18f47a30.
Given the actual Door ID, what is the password?
Your puzzle answer was ________.
--- Part Two ---
As the door slides open, you are presented with a second door that uses a slightly more inspired security mechanism. Clearly unimpressed by the last version (in what
movie is the password decrypted in order?!), the Easter Bunny engineers have worked out a better solution.
Instead of simply filling in the password from left to right, the hash now also indicates the position within the password to fill. You still look for hashes that begin
with five zeroes; however, now, the sixth character represents the position (0-7), and the seventh character is the character to put in that position.
A hash result of 000001f means that f is the second character in the password. Use only the first result for each position, and ignore invalid positions.
For example, if the Door ID is abc:
 The first interesting hash is from abc3231929, which produces 0000015...; so, 5 goes in position 1: _5______.
 In the previous method, 5017308 produced an interesting hash; however, it is ignored, because it specifies an invalid position (8).
 The second interesting hash is at index 5357525, which produces 000004e...; so, e goes in position 4: _5__e___.
You almost choke on your popcorn as the final character falls into place, producing the password 05ace8e3.
Given the actual Door ID and this new method, what is the password? Be extra proud of your solution if it uses a cinematic "decrypting" animation.
Your puzzle answer was ________.
Your puzzle input was ojvtpuvg.
References
Visible links
. http://adventofcode.com/
. http://adventofcode.com/2016/about
. http://adventofcode.com/2016/support
. http://adventofcode.com/2016/events
. http://adventofcode.com/2016/settings
. http://adventofcode.com/2016/auth/logout
. http://adventofcode.com/2016
. http://adventofcode.com/2016
. http://adventofcode.com/2016/leaderboard
. http://adventofcode.com/2016/stats
. http://adventofcode.com/2016/sponsors
. http://adventofcode.com/2016/sponsors
. https://www.detroitlabs.com/careers
. https://en.wikipedia.org/wiki/Hackers_(film)
. https://en.wikipedia.org/wiki/WarGames
. https://en.wikipedia.org/wiki/MD5
. https://en.wikipedia.org/wiki/Hexadecimal
. https://www.youtube.com/watch?v=NHWjlCaIrQo&t=25
. http://adventofcode.com/2016

650
2016/day06/input Normal file
View File

@@ -0,0 +1,650 @@
vlmbsivg
wjskjzff
hmcodblh
nvohhghm
xainmzpz
frkckyrg
zqdmvmee
sbsqzlju
eqdnsvqi
xdmyqmxk
gchzzkci
unjukwld
fmnpwrnf
huvyqxkt
uxekmktg
pdzxhbfa
khlkjqhk
ravznzpy
btpqmfzt
tcnpghzn
yuplhxzr
ablaihve
gjdehjcu
dvmkpskx
lukbiacr
ttrzcksg
ecfehhzq
octwtlsc
colisfrx
svtipfzv
kaffxell
rsmdpuic
viaidaip
drbezwvb
tyhnjseu
weuopaqc
igmqdwiz
gerzveig
bnqltmlo
jgwcxybe
iaabqazu
bgflkfyt
boznsgas
agiyktkz
ivvnbvuv
tospvglb
vxfhcvkz
kshbpxws
uejxthhb
vyffdirv
ibfywjob
cpzlamql
pyzsyqvq
yrqdolxy
svyihgcn
vupxjagi
cncrrcuv
ydmvwcje
yrpqbwhn
jrgsqtpl
axzbhwaw
ggsjjmvb
qtdsykjo
cnjmerzo
lzqwtxms
yhhodrpp
ryiwahsd
kggpmwvl
lextjotq
tlxjbxef
fygwhizi
moomppou
ezavglhp
omylulyf
ahbasrtz
ccbblybl
ueyudhva
blrewxyn
ytdwdavt
rgpakvlz
bmlycpbm
ewbvxsrm
hcukzfwq
rkkixnbu
masthktz
soolsxwh
zwszloqc
wjizzawg
ojcfqylq
nnzmlcyh
pvpmbtxx
assogumc
quorbsbu
uywftned
jneaeojp
mdvwycom
cpoqljak
dgxinijd
oknnyivd
uclfujcu
tglfuqym
xixpvogp
ywpxxlzu
tzaooymi
srjmonav
pnfugqmd
lryrknay
ijyxsjyr
nscjqibj
xfbyavdq
qoqcxyek
txscvcir
rbbpybgj
tkhqbyup
gmqcfvgi
itftwfkt
uzjbjwgw
rmfoqkvx
qtnlthgs
szdpoddo
qunqnglj
vpyzxigu
yvgaspgn
ugzomjyi
xehwrqqa
gerohqzw
svumyqwg
mqyztgha
kxaoylrd
qlrivitr
mqyqiewf
jbmnazei
mxlxbhwt
blwoptvb
psdatjoo
ygetkaoq
wygpqzdo
hlclzfmf
jpbvivrw
pmukraot
swbpgume
dvkkfjph
vrodfocs
zufhfmqm
djwpfova
fkmbbzho
wecxogkb
nhkrcpkp
vjeqcaaf
xrqarugv
fwmcwuok
epbkndfk
zwpgujke
modqokkj
ikewchsd
jwrhzjlb
smlbnbef
mxbbtgqb
viwdgsdx
goqsuvzj
rbroiwxp
zdiptzal
ybfdenoy
oszdisbg
nvfjluan
lqsbzkzw
psuefwbt
yallonfl
ijiwdpgh
gglvyutq
vqoydnjd
qcryrkrs
aivepbza
ylhvocnp
itvgswxy
bztkautf
ptrfhgug
jxrrgihp
pemkoxua
gisyyymp
lojsmpxa
eqfowfiy
fjgiccrd
hbxkngmy
drqthobt
ikyiobsj
cfwzjnns
krhtlqjb
ypahmuqg
cckkvsls
cvtpertw
lokmfmqj
eicgeaew
cuenpdcs
dztpufap
qrwculkl
ynzjrfps
mucvocas
dhorhvqw
ldidkudb
xxpbrirx
ksudlgat
giljlkxr
jhyqnccn
pmcscebv
cwidmjza
rqqirtgy
zioenwys
xbjruyyl
asptiznh
jqpjaenx
lugmurst
kmerqmnn
dxyndveo
uuqchrez
upcfeanl
sercqwtv
gbyfvzic
zlozeixc
cqtrvust
qdjntgwm
xiqmxxpw
gksowkih
pltnircb
pvoulbsn
hjxqhgjg
tcewvoyn
apjfkhzx
mfnuzwxz
ovchtwej
efqrkewr
jwgxupik
gcwelfie
ammaagcn
nyhkwztj
zakrmiqp
jjrdohih
rnwxdbge
zeobprqj
zwwtpnsx
mdnygzxh
nkfluipm
efwgmsfk
jlnztudt
medtzrwv
wclfyxbb
jjetrasx
wmwuhaar
ijwzhaoh
xdlfrrnn
dvxecbjt
vwihvqzh
kgpagxiw
jvixdjuc
gfiweyxg
xejknscb
hxheizpd
dubgkfgr
wkxidfgc
tzxylcho
qvububkp
erlnhqov
weomsswt
rfpoxknw
itnvsxdi
cbnocofl
ldadkmsf
pwhhklgc
itudhryg
vlvtoljp
hjggmitm
mmzawuak
quigzvkq
rsmiptim
tsqytvda
pmprbqfm
eppsdefz
adcqcsvc
lwiyaddu
wwqpjubg
jilsipuz
wzrxuskh
laraxaqd
epuyflmt
mdclmqun
hkjrdlxu
tcilebni
rkrekccq
aazdcsnc
ngijyoqs
bymuwqsv
lsanabvd
ylnektqm
fthtunja
vvfmqqbe
shbajllw
xgdgwmdc
hrzmxpif
nskrdynl
ineiworp
nlgfboek
yrsztmql
dnrewurn
uwtdmpru
eygvelve
tfitkluc
iqpxtxpw
hztyqkrl
xastoukn
djbllswj
kcnpmzsu
kepgfdit
okvnsjlp
fhjqkxxt
cnoewgjm
uhyjofee
knpdrbij
rreiagox
hxkhodce
zmaxocze
wtpaqndq
jvcagcvr
uhkmigmp
xadnxjhe
uiaplibp
otkxrzix
yzedogvy
ekfnffhq
fakkmnxr
bztnaqpo
tmyttvbn
univukhd
caujacsj
flblfclw
czfogrtr
inxubopi
afwfgcpb
mycsrthv
kdwsxnay
koujdeqg
hirhfotk
wkxxvthj
xsosgzwi
bxilicsn
hniuyqyz
lwbgmzwc
sncerhsu
fuhbmtfk
kaftqhuy
pqvglqfj
bghwspjz
twijseky
dpfmjhis
narvsefe
eraymofq
qiopspmb
taehddxo
zttknqxx
thgkqpfz
dgufruwc
bunvatbf
xonactvk
axjukwjn
opbjnbxx
fwvipkwm
etxwpjaj
hqdclxoq
fsnugidt
zmbpypye
hoqsimwa
waghzrcr
njymehaz
zknxntkj
ntjtbato
rqhupkqa
jnjxdtdf
eubzelni
astkohuv
ptfvacsg
bzjhupen
ijeukuux
abiyvjke
psbjwvcf
sucqapsu
axbjujok
oxvdokyb
olsyowug
nitvqkcl
njswglrr
ovheurur
ayedpsqo
vvclemhs
wjrfsvyk
fdhcqdru
qrkvvfay
kuhlridl
yxnrykeh
ctkznbzx
pfaojeww
lobsvohy
wutlpeak
nhppjsax
yihmsdmd
iszriptf
qdnhytkn
vooshrhs
wzogjdin
tkseeclg
jxzgiyxq
osefrqid
ixgfrbhq
zkggjudf
dyxgeond
rpwplncj
ndpxtrgc
ubuheera
migvasdi
acgygdxb
ccxcimgx
lvdpvhdr
ghvhyfmz
dqtheasu
ebjozdgp
tluznvkr
qhvzqrzm
qbjpiipl
bplqnbti
kizmxvno
uobfxakf
syllmscz
gghcqygd
snkwjojf
vdcvdpno
qussueej
zmlxuulr
htqbwjve
ixynqepo
ovaelpay
eywkbzsc
qwoqmtig
prkxadem
jfsscvhm
hqslphpy
bfiiyruu
fmkizdns
vpsisoaf
couzrtsv
wycwxxlh
upvcymri
hqlqfvbm
wzaeftmu
boxdzesc
lmcuplex
xyuifcev
fqgvggmq
seykzeql
fewgnzoq
qjmsxyto
ebckyjih
jzrbwgjj
vwojkmpc
sqdqaffc
yukfwymc
yjptcsmz
tencuncy
rbtxxpuv
yiffiway
eefyzfqb
moxjnvoq
wijspyho
fdtsjins
rxqcbsgd
oamredol
baeulkgv
rggwhexz
qfkkitpv
uexaisra
cqwjlwxi
tjavidpj
efbavnsa
uxnnkruo
wriebzoz
wexcdrht
hzmyocox
qseogxfr
fflfanfi
vpaiudnq
olvuxneh
fcbsytua
sfudzkhx
xghrupyg
aeclhwom
byobwacz
rsditwjr
lvvqfcup
lflatigg
kuojuttb
aouuzdnw
klbwzjlw
zkambbfg
xdevxqbr
zqdlejop
vfhbkyik
zeyujeql
skvhtden
ddwgfxck
lnrqrbkk
rtnzbizt
uaxpghyx
asybsmyx
djdcnnpx
ykwhiyjk
zhvodxrh
bmzhumtw
aoutwtdv
mjadhrvo
tkugwiep
ystjfami
bfqqcmtb
bbvmzwdo
hgufdhpb
hhmamhwy
gpasfwju
zhpxvhjf
vzqggwtj
kbjrhlrm
inhayugt
fvspcluq
grxbmozt
muohwoyd
nigtpaty
ozzvzejw
nyyngjme
jmtoeyly
qyxhdpaq
iowatjzd
iszcablv
ocgavedu
xbnzrfqv
swzzzzwu
stqfxgoo
xpmubjxr
katkezfz
pudaitbg
ybarabzd
mpvqbzsr
rhinjfpm
snevfkqk
pwwccyyu
ocljrwko
dxdryyvq
rpqgkeol
wvqbnkwc
hgnwlgua
mcqmblme
fhzgsdch
kdxugdug
oltmtill
sbduaius
cqauyawa
xfznpmvh
dmivepyp
dgsngwbw
mbgsryyh
aojpsuvb
igvrbclq
lduxmyiy
dhomqxlb
myszbzrf
kbadlpoy
ftnumgci
xwzbhyfz
lzutwqtv
gimhovca
zfudpuny
nqkrsatc
uldebzzq
olzoesdw
nchgfvll
oopwvqlh
jhsbnnjw
siyclmze
frhzjbnv
nanvvjht
wtucekgs
xiyyglkr
zkakjomw
fdeeoopf
tlctovbh
cnqwlmkh
zreoslve
lcexflbp
erkaxhyf
uflifmci
byfovhfn
pmlsjebs
rmdncvno
jzxsteml
ehromnts
wwrgqdmv
gydqxmbb
kjjnrsyf
okzknoil
pqmjbnhz
rbgydndw
jfmclxjm
bfotvtud
mftuqeei
lhchcrja
czperrwe
hyymnbfh
gtmctpcn
ooteayaj
ipxmxdrg
lxljcbmk
kexfnhaz
zwvwqxwd
ccriznpe
gadlykes
nuagvrrv
xgvdaxfk
gnkzxirj
vdmrccgf
vhgjcski
qpdhafox
oijrjuwm
qxtyktrs
mkyalzfa
tnmunnod
ukjvbgjs
cywsgmfc
uqkzjgbn
ppjifbhj
gpwlnflu
alabzvvi
vtygcqph
ezxtvdqo
hzeejxxe
aavwflnw
nrfqefxa
dsgysuua
pikwyfym
nesdcxdi
ghqlicny
jcvekqdr
cyeiicma
slfxqsdt
dcrcwoek
qdmtpdxr
vlzjoqfm

55
2016/day06/main.go Normal file
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package main
import (
"bufio"
"fmt"
"os"
)
func main() {
input := stdinToStringSlice()
var partOne, partTwo string
for j := range input[0] {
freq := make(map[byte]int)
for i := range input {
freq[input[i][j]]++
}
partOne += getTopByte(freq)
partTwo += getBottomByte(freq)
}
fmt.Println("Answer to Part 1: " + partOne)
fmt.Println("Answer to Part 2: " + partTwo)
}
func getTopByte(f map[byte]int) string {
var topByte byte
var topByteCnt int
for k, v := range f {
if v > topByteCnt {
topByteCnt = v
topByte = k
}
}
return string(topByte)
}
func getBottomByte(f map[byte]int) string {
var botByte byte
var botByteCnt int
for k, v := range f {
if v < botByteCnt || botByte == 0 {
botByteCnt = v
botByte = k
}
}
return string(botByte)
}
func stdinToStringSlice() []string {
var input []string
scanner := bufio.NewScanner(os.Stdin)
for scanner.Scan() {
input = append(input, scanner.Text())
}
return input
}

69
2016/day06/problem Normal file
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Advent of Code
--- Day 6: Signals and Noise ---
Something is jamming your communications with Santa. Fortunately, your signal is only partially jammed, and protocol in situations like this is to switch to
a simple repetition code to get the message through.
In this model, the same message is sent repeatedly. You've recorded the repeating message signal (your puzzle input), but the data seems quite corrupted -
almost too badly to recover. Almost.
All you need to do is figure out which character is most frequent for each position. For example, suppose you had recorded the following messages:
eedadn
drvtee
eandsr
raavrd
atevrs
tsrnev
sdttsa
rasrtv
nssdts
ntnada
svetve
tesnvt
vntsnd
vrdear
dvrsen
enarar
The most common character in the first column is e; in the second, a; in the third, s, and so on. Combining these characters returns the error-corrected
message, easter.
Given the recording in your puzzle input, what is the error-corrected version of the message being sent?
Your puzzle answer was ________.
--- Part Two ---
Of course, that would be the message - if you hadn't agreed to use a modified repetition code instead.
In this modified code, the sender instead transmits what looks like random data, but for each character, the character they actually want to send is
slightly less likely than the others. Even after signal-jamming noise, you can look at the letter distributions in each column and choose the least common
letter to reconstruct the original message.
In the above example, the least common character in the first column is a; in the second, d, and so on. Repeating this process for the remaining characters
produces the original message, advent.
Given the recording in your puzzle input and this new decoding methodology, what is the original message that Santa is trying to send?
Your puzzle answer was ________.
References
Visible links
. http://adventofcode.com/
. http://adventofcode.com/2016/about
. http://adventofcode.com/2016/support
. http://adventofcode.com/2016/events
. http://adventofcode.com/2016/settings
. http://adventofcode.com/2016/auth/logout
. http://adventofcode.com/2016
. http://adventofcode.com/2016
. http://adventofcode.com/2016/leaderboard
. http://adventofcode.com/2016/stats
. http://adventofcode.com/2016/sponsors
. http://adventofcode.com/2016/sponsors
. https://en.wikipedia.org/wiki/Repetition_code
. http://adventofcode.com/2016
. http://adventofcode.com/2016/day/6/input

2000
2016/day07/input Normal file

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95
2016/day07/main.go Normal file
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package main
import (
"bufio"
"fmt"
"os"
"regexp"
"strings"
)
func main() {
ips := stdinToStringSlice()
var allowTLS []string
var allowSSL []string
for _, ip := range ips {
if hasTLS(ip) {
allowTLS = append(allowTLS, ip)
}
if hasSSL(ip) {
allowSSL = append(allowSSL, ip)
}
}
fmt.Println("IPs supporting TLS: ", len(allowTLS))
fmt.Println("IPs supporting SSL: ", len(allowSSL))
}
func hasTLS(ip string) bool {
bad := false
hprSq := getHypernetSqns(ip)
for _, hpr := range hprSq {
if hasABBA(hpr) {
return false
}
}
if !bad && hasABBA(ip) {
return true
}
return false
}
func hasSSL(ip string) bool {
hprSq := getHypernetSqns(ip)
rest := cutHypernetSqns(ip, hprSq)
for i := range rest {
if len(rest) > i+2 {
if rest[i] == rest[i+2] && rest[i] != rest[i+1] {
bab := string(rest[i+1]) + string(rest[i]) + string(rest[i+1])
for j := range hprSq {
if strings.Contains(hprSq[j], bab) {
return true
}
}
}
}
}
return false
}
func getHypernetSqns(ip string) []string {
var ret []string
re := regexp.MustCompile(`\[[^\[]*\]`)
s := re.FindAll([]byte(ip), -1)
for i := range s {
ret = append(ret, string(s[i][1:len(s[i])-1]))
}
return ret
}
func cutHypernetSqns(ip string, hpn []string) string {
for i := range hpn {
ip = strings.Replace(ip, hpn[i], "", -1)
}
return ip
}
func hasABBA(ip string) bool {
for j := range ip {
if len(ip) > j+3 {
if ip[j] == ip[j+3] && ip[j+1] == ip[j+2] && ip[j] != ip[j+1] {
return true
}
}
}
return false
}
func stdinToStringSlice() []string {
var input []string
scanner := bufio.NewScanner(os.Stdin)
for scanner.Scan() {
input = append(input, scanner.Text())
}
return input
}

62
2016/day07/problem Normal file
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Advent of Code
--- Day 7: Internet Protocol Version 7 ---
While snooping around the local network of EBHQ, you compile a list of IP addresses (they're IPv7, of course; IPv6 is much too limited).
You'd like to figure out which IPs support TLS (transport-layer snooping).
An IP supports TLS if it has an Autonomous Bridge Bypass Annotation, or ABBA. An ABBA is any four-character sequence which consists of a
pair of two different characters followed by the reverse of that pair, such as xyyx or abba. However, the IP also must not have an ABBA
within any hypernet sequences, which are contained by square brackets.
For example:
 abba[mnop]qrst supports TLS (abba outside square brackets).
 abcd[bddb]xyyx does not support TLS (bddb is within square brackets, even though xyyx is outside square brackets).
 aaaa[qwer]tyui does not support TLS (aaaa is invalid; the interior characters must be different).
 ioxxoj[asdfgh]zxcvbn supports TLS (oxxo is outside square brackets, even though it's within a larger string).
How many IPs in your puzzle input support TLS?
Your puzzle answer was _____.
--- Part Two ---
You would also like to know which IPs support SSL (super-secret listening).
An IP supports SSL if it has an Area-Broadcast Accessor, or ABA, anywhere in the supernet sequences (outside any square bracketed sections),
and a corresponding Byte Allocation Block, or BAB, anywhere in the hypernet sequences. An ABA is any three-character sequence which consists
of the same character twice with a different character between them, such as xyx or aba. A corresponding BAB is the same characters but in
reversed positions: yxy and bab, respectively.
For example:
 aba[bab]xyz supports SSL (aba outside square brackets with corresponding bab within square brackets).
 xyx[xyx]xyx does not support SSL (xyx, but no corresponding yxy).
 aaa[kek]eke supports SSL (eke in supernet with corresponding kek in hypernet; the aaa sequence is not related, because the interior
character must be different).
 zazbz[bzb]cdb supports SSL (zaz has no corresponding aza, but zbz has a corresponding bzb, even though zaz and zbz overlap).
How many IPs in your puzzle input support SSL?
Your puzzle answer was _______.
References
Visible links
. http://adventofcode.com/
. http://adventofcode.com/2016/about
. http://adventofcode.com/2016/support
. http://adventofcode.com/2016/events
. http://adventofcode.com/2016/settings
. http://adventofcode.com/2016/auth/logout
. http://adventofcode.com/2016
. http://adventofcode.com/2016
. http://adventofcode.com/2016/leaderboard
. http://adventofcode.com/2016/stats
. http://adventofcode.com/2016/sponsors
. http://adventofcode.com/2016/sponsors
. https://en.wikipedia.org/wiki/IP_address
. https://en.wikipedia.org/wiki/IPv6
. http://adventofcode.com/2016
. http://adventofcode.com/2016/day/7/input

658
2016/day08-add/main.go Normal file
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package main
import (
"fmt"
"os"
"strconv"
"strings"
)
/* This program generates instructions to display words
* on the screen from day 8
* It assumes the screen size is 50x6
*/
func main() {
if len(os.Args) < 2 {
fmt.Println("Expected argument with word to convert.")
os.Exit(1)
}
wrd := strings.ToUpper(os.Args[1])
for i := len(wrd) - 1; i >= 0; i-- {
printInst(runeToInst(rune(wrd[i])))
if i > 0 {
printInst(shiftCols(2))
}
}
}
func printInst(i []string) {
for j := range i {
fmt.Println(i[j])
}
}
func runeToInst(r rune) []string {
var ret []string
switch r {
case 'A':
/*
##
# #
# #
####
# #
# #
*/
ret = append(ret, "rect 1x5")
ret = append(ret, "rotate column x=0 by 1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 3")
ret = append(ret, "rect 1x1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 3")
ret = append(ret, "rect 1x1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x5")
ret = append(ret, "rotate column x=0 by 1")
case 'B':
/*
###
# #
###
# #
# #
###
*/
ret = append(ret, "rect 1x2")
ret = append(ret, "rotate column x=0 by 2")
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 3")
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 2")
ret = append(ret, "rect 1x1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 3")
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 2")
ret = append(ret, "rect 1x1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x6")
case 'C':
/*
##
# #
#
#
# #
##
*/
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 3")
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 5")
ret = append(ret, "rect 1x1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 5")
ret = append(ret, "rect 1x1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x4")
ret = append(ret, "rotate column x=0 by 1")
case 'D':
/*
###
# #
# #
# #
# #
###
*/
ret = append(ret, "rect 1x4")
ret = append(ret, "rotate column x=0 by 1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 5")
ret = append(ret, "rect 1x1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 5")
ret = append(ret, "rect 1x1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x6")
case 'E':
/*
####
#
###
#
#
####
*/
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 5")
ret = append(ret, "rect 1x1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 3")
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 2")
ret = append(ret, "rect 1x1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 3")
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 2")
ret = append(ret, "rect 1x1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x6")
case 'F':
/*
####
#
###
#
#
#
*/
ret = append(ret, shiftCols(5)...)
ret = append(ret, "rect 3x1")
ret = append(ret, "rotate column x=2 by 2")
ret = append(ret, "rotate column x=1 by 2")
ret = append(ret, "rotate column x=0 by 2")
ret = append(ret, "rect 4x1")
ret = append(ret, "rect 1x6")
case 'G':
/*
###
#
# ##
# #
# #
##
*/
ret = append(ret, "rect 1x3")
ret = append(ret, "rotate column x=0 by 2")
ret = append(ret, "rect 1x1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 3")
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 2")
ret = append(ret, "rect 1x1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 5")
ret = append(ret, "rect 1x1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x4")
ret = append(ret, "rotate column x=0 by 1")
case 'H':
/*
# #
# #
####
# #
# #
# #
*/
ret = append(ret, "rect 1x6")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 2")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 2")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x6")
case 'I':
/*
###
#
#
#
#
###
*/
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 5")
ret = append(ret, "rect 1x1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x6")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 5")
ret = append(ret, "rect 1x1")
case 'J':
/*
#
#
#
# #
# #
##
*/
ret = append(ret, "rect 1x5")
ret = append(ret, shiftCols(2)...)
ret = append(ret, "rect 2x1")
ret = append(ret, "rotate column x=1 by 5")
ret = append(ret, "rotate column x=0 by 5")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x2")
ret = append(ret, "rotate column x=0 by 3")
case 'K':
/*
# #
# #
##
# #
# #
# #
*/
ret = append(ret, "rect 1x2")
ret = append(ret, "rotate column x=0 by 4")
ret = append(ret, "rect 1x1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 2")
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 2")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x6")
case 'L':
/*
#
#
#
#
#
####
*/
ret = append(ret, shiftCols(3)...)
ret = append(ret, "rect 4x1")
ret = append(ret, "rotate column x=0 by 5")
ret = append(ret, "rotate column x=1 by 5")
ret = append(ret, "rotate column x=2 by 5")
ret = append(ret, "rotate column x=3 by 5")
ret = append(ret, "rect 1x6")
case 'M':
/*
# #
####
####
# #
# #
# #
*/
ret = append(ret, "rect 1x6")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x2")
ret = append(ret, "rotate column x=0 by 1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x2")
ret = append(ret, "rotate column x=0 by 1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x6")
case 'N':
/*
# #
## #
# ##
# ##
# #
# #
*/
ret = append(ret, "rect 1x6")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x2")
ret = append(ret, "rotate column x=0 by 2")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x6")
case 'O':
/*
##
# #
# #
# #
# #
##
*/
ret = append(ret, "rect 1x4")
ret = append(ret, "rotate column x=0 by 1")
ret = append(ret, shiftCols(2)...)
ret = append(ret, "rect 2x1")
ret = append(ret, "rotate column x=1 by 5")
ret = append(ret, "rotate column x=0 by 5")
ret = append(ret, "rect 2x1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x4")
ret = append(ret, "rotate column x=0 by 1")
case 'P':
/*
###
# #
###
#
#
#
*/
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 1")
ret = append(ret, shiftCols(2)...)
ret = append(ret, "rect 2x1")
ret = append(ret, "rotate column x=1 by 2")
ret = append(ret, "rotate column x=0 by 2")
ret = append(ret, "rect 2x1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x6")
case 'Q':
/*
##
# #
# #
# #
# ##
###
*/
ret = append(ret, "rect 1x5")
ret = append(ret, "rotate column x=0 by 1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x2")
ret = append(ret, "rotate column x=0 by 4")
ret = append(ret, "rect 1x1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 5")
ret = append(ret, "rect 1x1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x4")
ret = append(ret, "rotate column x=0 by 1")
case 'R':
/*
###
# #
# #
###
# #
# #
*/
ret = append(ret, "rect 1x2")
ret = append(ret, "rotate column x=0 by 3")
ret = append(ret, "rect 1x2")
ret = append(ret, "rotate column x=0 by 1")
ret = append(ret, shiftCols(2)...)
ret = append(ret, "rect 2x1")
ret = append(ret, "rotate column x=1 by 3")
ret = append(ret, "rotate column x=0 by 3")
ret = append(ret, "rect 2x1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x6")
case 'S':
/*
###
#
#
##
#
###
*/
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 4")
ret = append(ret, "rect 1x1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 2")
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 3")
ret = append(ret, "rect 1x1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 2")
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 3")
ret = append(ret, "rect 1x1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x2")
ret = append(ret, "rotate column x=0 by 1")
case 'T':
/*
####
#
#
#
#
#
*/
ret = append(ret, shiftCols(2)...)
ret = append(ret, "rect 3x1")
ret = append(ret, "rect 1x6")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x1")
case 'U':
/*
# #
# #
# #
# #
# #
##
*/
ret = append(ret, "rect 1x5")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 5")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 5")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x5")
case 'V':
/*
# #
# #
# #
##
##
##
*/
ret = append(ret, "rect 1x3")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x3")
ret = append(ret, "rotate column x=0 by 3")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x3")
ret = append(ret, "rotate column x=0 by 3")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x3")
case 'W':
/*
# #
# #
# #
####
####
# #
*/
ret = append(ret, "rect 1x6")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x2")
ret = append(ret, "rotate column x=0 by 3")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x2")
ret = append(ret, "rotate column x=0 by 3")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x6")
case 'X':
/*
# #
# #
##
##
# #
# #
*/
ret = append(ret, "rect 1x2")
ret = append(ret, "rotate column x=0 by 4")
ret = append(ret, "rect 1x2")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x2")
ret = append(ret, "rotate column x=0 by 2")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x2")
ret = append(ret, "rotate column x=0 by 2")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x2")
ret = append(ret, "rotate column x=0 by 4")
ret = append(ret, "rect 1x2")
case 'Y':
/*
# #
# #
###
#
#
##
*/
ret = append(ret, "rect 1x5")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 3")
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 2")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 3")
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 2")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x2")
case 'Z':
/*
####
# #
#
#
# #
####
*/
ret = append(ret, "rect 1x2")
ret = append(ret, "rotate column x=0 by 4")
ret = append(ret, "rect 1x2")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 3")
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 2")
ret = append(ret, "rect 1x1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 2")
ret = append(ret, "rect 1x1")
ret = append(ret, "rotate column x=0 by 3")
ret = append(ret, "rect 1x1")
ret = append(ret, shiftCols(1)...)
ret = append(ret, "rect 1x2")
ret = append(ret, "rotate column x=0 by 4")
ret = append(ret, "rect 1x2")
case ' ':
ret = append(ret, shiftCols(4)...)
default:
ret = append(ret, shiftCols(4)...)
ret = append(ret, "rect 4x1")
ret = append(ret, "rotate column x=3 by 2")
ret = append(ret, "rotate column x=2 by 2")
ret = append(ret, "rotate column x=1 by 2")
ret = append(ret, "rotate column x=0 by 2")
}
return ret
}
func shiftCols(dist int) []string {
var ret []string
for i := 0; i < 6; i++ {
ret = append(ret, "rotate row y="+itoa(i)+" by "+itoa(dist))
}
return ret
}
func shiftRows(dist int) []string {
var ret []string
for i := 0; i < 50; i++ {
ret = append(ret, "rotate col x="+itoa(i)+" by "+itoa(dist))
}
return ret
}
func itoa(i int) string {
return strconv.Itoa(i)
}

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rect 1x1
rotate row y=0 by 5
rect 1x1
rotate row y=0 by 5
rect 1x1
rotate row y=0 by 5
rect 1x1
rotate row y=0 by 5
rect 1x1
rotate row y=0 by 2
rect 1x1
rotate row y=0 by 2
rect 1x1
rotate row y=0 by 3
rect 1x1
rotate row y=0 by 3
rect 2x1
rotate row y=0 by 2
rect 1x1
rotate row y=0 by 3
rect 2x1
rotate row y=0 by 2
rect 1x1
rotate row y=0 by 3
rect 2x1
rotate row y=0 by 5
rect 4x1
rotate row y=0 by 5
rotate column x=0 by 1
rect 4x1
rotate row y=0 by 10
rotate column x=5 by 2
rotate column x=0 by 1
rect 9x1
rotate row y=2 by 5
rotate row y=0 by 5
rotate column x=0 by 1
rect 4x1
rotate row y=2 by 5
rotate row y=0 by 5
rotate column x=0 by 1
rect 4x1
rotate column x=40 by 1
rotate column x=27 by 1
rotate column x=22 by 1
rotate column x=17 by 1
rotate column x=12 by 1
rotate column x=7 by 1
rotate column x=2 by 1
rotate row y=2 by 5
rotate row y=1 by 3
rotate row y=0 by 5
rect 1x3
rotate row y=2 by 10
rotate row y=1 by 7
rotate row y=0 by 2
rotate column x=3 by 2
rotate column x=2 by 1
rotate column x=0 by 1
rect 4x1
rotate row y=2 by 5
rotate row y=1 by 3
rotate row y=0 by 3
rect 1x3
rotate column x=45 by 1
rotate row y=2 by 7
rotate row y=1 by 10
rotate row y=0 by 2
rotate column x=3 by 1
rotate column x=2 by 2
rotate column x=0 by 1
rect 4x1
rotate row y=2 by 13
rotate row y=0 by 5
rotate column x=3 by 1
rotate column x=0 by 1
rect 4x1
rotate row y=3 by 10
rotate row y=2 by 10
rotate row y=0 by 5
rotate column x=3 by 1
rotate column x=2 by 1
rotate column x=0 by 1
rect 4x1
rotate row y=3 by 8
rotate row y=0 by 5
rotate column x=3 by 1
rotate column x=2 by 1
rotate column x=0 by 1
rect 4x1
rotate row y=3 by 17
rotate row y=2 by 20
rotate row y=0 by 15
rotate column x=13 by 1
rotate column x=12 by 3
rotate column x=10 by 1
rotate column x=8 by 1
rotate column x=7 by 2
rotate column x=6 by 1
rotate column x=5 by 1
rotate column x=3 by 1
rotate column x=2 by 2
rotate column x=0 by 1
rect 14x1
rotate row y=1 by 47
rotate column x=9 by 1
rotate column x=4 by 1
rotate row y=3 by 3
rotate row y=2 by 10
rotate row y=1 by 8
rotate row y=0 by 5
rotate column x=2 by 2
rotate column x=0 by 2
rect 3x2
rotate row y=3 by 12
rotate row y=2 by 10
rotate row y=0 by 10
rotate column x=8 by 1
rotate column x=7 by 3
rotate column x=5 by 1
rotate column x=3 by 1
rotate column x=2 by 1
rotate column x=1 by 1
rotate column x=0 by 1
rect 9x1
rotate row y=0 by 20
rotate column x=46 by 1
rotate row y=4 by 17
rotate row y=3 by 10
rotate row y=2 by 10
rotate row y=1 by 5
rotate column x=8 by 1
rotate column x=7 by 1
rotate column x=6 by 1
rotate column x=5 by 1
rotate column x=3 by 1
rotate column x=2 by 2
rotate column x=1 by 1
rotate column x=0 by 1
rect 9x1
rotate column x=32 by 4
rotate row y=4 by 33
rotate row y=3 by 5
rotate row y=2 by 15
rotate row y=0 by 15
rotate column x=13 by 1
rotate column x=12 by 3
rotate column x=10 by 1
rotate column x=8 by 1
rotate column x=7 by 2
rotate column x=6 by 1
rotate column x=5 by 1
rotate column x=3 by 1
rotate column x=2 by 1
rotate column x=1 by 1
rotate column x=0 by 1
rect 14x1
rotate column x=39 by 3
rotate column x=35 by 4
rotate column x=20 by 4
rotate column x=19 by 3
rotate column x=10 by 4
rotate column x=9 by 3
rotate column x=8 by 3
rotate column x=5 by 4
rotate column x=4 by 3
rotate row y=5 by 5
rotate row y=4 by 5
rotate row y=3 by 33
rotate row y=1 by 30
rotate column x=48 by 1
rotate column x=47 by 5
rotate column x=46 by 5
rotate column x=45 by 1
rotate column x=43 by 1
rotate column x=38 by 3
rotate column x=37 by 3
rotate column x=36 by 5
rotate column x=35 by 1
rotate column x=33 by 1
rotate column x=32 by 5
rotate column x=31 by 5
rotate column x=30 by 1
rotate column x=23 by 4
rotate column x=22 by 3
rotate column x=21 by 3
rotate column x=20 by 1
rotate column x=12 by 2
rotate column x=11 by 2
rotate column x=3 by 5
rotate column x=2 by 5
rotate column x=1 by 3
rotate column x=0 by 4

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package main
import (
"bufio"
"errors"
"fmt"
"log"
"os"
"strconv"
"strings"
"time"
)
func main() {
width := 50
height := 6
if len(os.Args) >= 2 {
width = atoi(os.Args[1])
}
if len(os.Args) >= 3 {
height = atoi(os.Args[2])
}
input := stdinToStringSlice()
d := CreateDisplay(width, height)
for idx, ins := range input {
fmt.Println(idx, ins)
d.ClearScreen()
d.ProcInstruction(ins)
d.PrintScreen()
time.Sleep(time.Millisecond * 25)
}
fmt.Println("Voltage Used: ", d.GetVoltage())
}
type Display struct {
Width int
Height int
Screen [][]bool
}
func CreateDisplay(w, h int) *Display {
d := Display{Width: w, Height: h}
for i := 0; i < h; i++ {
var row []bool
for j := 0; j < w; j++ {
row = append(row, false)
}
d.Screen = append(d.Screen, row)
}
return &d
}
func (d *Display) ClearScreen() {
fmt.Print("\033[H\033[2J")
}
func (d *Display) PrintScreen() {
fmt.Print("\u250C")
for i := 0; i < d.Width; i++ {
fmt.Print("\u2500")
}
fmt.Println("\u2510")
for y := range d.Screen {
fmt.Print("\u2502")
for x := range d.Screen[y] {
if d.Screen[y][x] {
fmt.Print("\u2588")
} else {
fmt.Print(" ")
}
}
fmt.Println("\u2502")
}
fmt.Print("\u2514")
for i := 0; i < d.Width; i++ {
fmt.Print("\u2500")
}
fmt.Println("\u2518")
}
func (d *Display) SetWidth(w int) {
d.Width = w
}
func (d *Display) SetHeight(h int) {
d.Height = h
}
func (d *Display) ProcInstruction(ins string) {
pts := strings.Fields(ins)
switch pts[0] {
case "rect":
if len(pts) < 2 {
return
}
dim := strings.Split(pts[1], "x")
bldX := atoi(dim[0])
bldY := atoi(dim[1])
d.CreateRect(bldX, bldY)
case "rotate":
if len(pts) < 5 {
return
}
args := strings.Split(pts[2], "=")
switch pts[1] {
case "row":
if args[0] != "y" {
return
}
d.RotateRow(atoi(args[1]), atoi(pts[4]))
case "column":
if args[0] != "x" {
return
}
d.RotateCol(atoi(args[1]), atoi(pts[4]))
}
}
}
func (d *Display) CreateRect(x, y int) {
if y >= d.Height && x >= d.Width {
fmt.Println(" Error creating rect", x, y)
return
}
for i := 0; i < y; i++ {
for j := 0; j < x; j++ {
d.Screen[i][j] = true
}
}
}
func (d *Display) RotateRow(y, dist int) {
orig, err := d.GetRow(y)
if err != nil {
fmt.Println(err.Error())
os.Exit(1)
return
}
for i := range orig {
d.Screen[y][(i+dist)%d.Width] = orig[i]
}
}
func (d *Display) RotateCol(x, dist int) {
orig, err := d.GetCol(x)
if err != nil {
fmt.Println(err.Error())
os.Exit(1)
return
}
for i := range orig {
d.Screen[(i+dist)%d.Height][x] = orig[i]
}
}
func (d *Display) GetRow(y int) ([]bool, error) {
var ret []bool
if d.Height <= y {
return ret, errors.New("Invalid Row Requested")
}
for i := 0; i < d.Width; i++ {
ret = append(ret, d.Screen[y][i])
}
return ret, nil
}
func (d *Display) GetCol(x int) ([]bool, error) {
var ret []bool
if d.Width <= x {
return ret, errors.New("Invalid Column Requested")
}
for i := 0; i < d.Height; i++ {
ret = append(ret, d.Screen[i][x])
}
return ret, nil
}
func (d *Display) GetVoltage() int {
var ret int
for i := range d.Screen {
for j := range d.Screen[i] {
if d.Screen[i][j] {
ret++
}
}
}
return ret
}
func stdinToStringSlice() []string {
var input []string
scanner := bufio.NewScanner(os.Stdin)
for scanner.Scan() {
input = append(input, scanner.Text())
}
return input
}
func atoi(i string) int {
var ret int
var err error
if ret, err = strconv.Atoi(i); err != nil {
log.Fatal("Invalid Atoi")
}
return ret
}

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Advent of Code
--- Day 8: Two-Factor Authentication ---
You come across a door implementing what you can only assume is an implementation of two-factor authentication
after a long game of requirements telephone.
To get past the door, you first swipe a keycard (no problem; there was one on a nearby desk). Then, it displays a
code on a little screen, and you type that code on a keypad. Then, presumably, the door unlocks.
Unfortunately, the screen has been smashed. After a few minutes, you've taken everything apart and figured out how
it works. Now you just have to work out what the screen would have displayed.
The magnetic strip on the card you swiped encodes a series of instructions for the screen; these instructions are
your puzzle input. The screen is 50 pixels wide and 6 pixels tall, all of which start off, and is capable of three
somewhat peculiar operations:
 rect AxB turns on all of the pixels in a rectangle at the top-left of the screen which is A wide and B tall.
 rotate row y=A by B shifts all of the pixels in row A (0 is the top row) right by B pixels. Pixels that would
fall off the right end appear at the left end of the row.
 rotate column x=A by B shifts all of the pixels in column A (0 is the left column) down by B pixels. Pixels
that would fall off the bottom appear at the top of the column.
For example, here is a simple sequence on a smaller screen:
 rect 3x2 creates a small rectangle in the top-left corner:
###....
###....
.......
 rotate column x=1 by 1 rotates the second column down by one pixel:
#.#....
###....
.#.....
 rotate row y=0 by 4 rotates the top row right by four pixels:
....#.#
###....
.#.....
 rotate column x=1 by 1 again rotates the second column down by one pixel, causing the bottom pixel to wrap back
to the top:
.#..#.#
#.#....
.#.....
As you can see, this display technology is extremely powerful, and will soon dominate the
tiny-code-displaying-screen market. That's what the advertisement on the back of the display tries to convince you,
anyway.
There seems to be an intermediate check of the voltage used by the display: after you swipe your card, if the
screen did work, how many pixels should be lit?
Your puzzle answer was _____.
--- Part Two ---
You notice that the screen is only capable of displaying capital letters; in the font it uses, each letter is 5
pixels wide and 6 tall.
After you swipe your card, what code is the screen trying to display?
Your puzzle answer was _____________.
References
Visible links
. http://adventofcode.com/
. http://adventofcode.com/2016/about
. http://adventofcode.com/2016/support
. http://adventofcode.com/2016/events
. http://adventofcode.com/2016/settings
. http://adventofcode.com/2016/auth/logout
. http://adventofcode.com/2016
. http://adventofcode.com/2016
. http://adventofcode.com/2016/leaderboard
. http://adventofcode.com/2016/stats
. http://adventofcode.com/2016/sponsors
. http://adventofcode.com/2016/sponsors
. https://en.wikipedia.org/wiki/Multi-factor_authentication
. https://en.wikipedia.org/wiki/Requirement
. https://en.wikipedia.org/wiki/Chinese_whispers
. https://www.google.com/search?q=tiny+lcd&tbm=isch
. http://adventofcode.com/2016
. http://adventofcode.com/2016/day/8/input

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package main
import (
"bufio"
"bytes"
"fmt"
"io/ioutil"
"log"
"os"
"regexp"
"strconv"
)
var re *regexp.Regexp
func main() {
var input []byte
var noDecompress bool
var printSize bool
if len(os.Args) > 1 {
for i := 1; i < len(os.Args); i++ {
if os.Args[i][0] != '-' {
// Assume it's a filename
// Read from the given filename
input = fileToByteSlice(os.Args[1])
} else {
switch os.Args[i] {
case "-size":
printSize = true
case "-dry":
// -dry implies printSize
printSize = true
noDecompress = true
}
}
}
}
if len(input) == 0 {
// Must be stdin
input = stdinToByteSlice()
}
input = bytes.TrimSpace(input)
// re matches compression markers and submatches the values we care about
re = regexp.MustCompile(`\((\d*)x(\d*)\)`)
if noDecompress {
cnt := depthDecompressLength(input)
fmt.Println("Total Decompressed Bytes:", cnt)
os.Exit(0)
}
output := depthDecompress(input)
fmt.Println(string(output))
if printSize {
fmt.Println("Total Decompressed Bytes:", len(output))
}
}
// depthDecompressLength counts how many uncompressed bytes are
// in the byte slice by unravelling the compression levels
// recursively
func depthDecompressLength(cmp []byte) uint64 {
mrkParts := re.FindStringSubmatch(string(cmp))
if len(mrkParts) < 3 {
// No compressionmarker, just return cmp length
return uint64(len(cmp))
}
marker, mrkBytes, mrkDupe := []byte(mrkParts[0]), atoi(mrkParts[1]), atoi(mrkParts[2])
mrkPos := bytes.Index(cmp, marker)
if mrkPos > 0 {
cmp = cmp[mrkPos:]
}
recurBytes := bytes.TrimPrefix(cmp, marker)[:mrkBytes]
remainder := bytes.TrimPrefix(cmp, append(marker, recurBytes...))
return uint64(mrkPos) + (depthDecompressLength(recurBytes) * uint64(mrkDupe)) + depthDecompressLength(remainder)
}
func depthDecompress(cmp []byte) []byte {
mrkParts := re.FindStringSubmatch(string(cmp))
if len(mrkParts) < 3 {
// No compression marker, just return cmp length
return cmp
}
marker, mrkBytes, mrkDupe := []byte(mrkParts[0]), atoi(mrkParts[1]), atoi(mrkParts[2])
mrkPos := bytes.Index(cmp, marker)
var ret []byte
if mrkPos > 0 {
ret = cmp[:mrkPos]
cmp = cmp[mrkPos:]
}
recurBytes := bytes.TrimPrefix(cmp, marker)[:mrkBytes]
remainder := bytes.TrimPrefix(cmp, append(marker, recurBytes...))
ret = append(ret, bytes.Repeat(depthDecompress(recurBytes), mrkDupe)...)
return append(ret, depthDecompress(remainder)...)
}
func fileToByteSlice(fn string) []byte {
var c []byte
var err error
c, err = ioutil.ReadFile(fn)
if err != nil {
fmt.Println("Unable to read file: " + fn)
os.Exit(1)
}
return c
}
func stdinToByteSlice() []byte {
var input string
scanner := bufio.NewScanner(os.Stdin)
for scanner.Scan() {
input += scanner.Text()
}
return []byte(input)
}
func atoi(i string) int {
var ret int
var err error
if ret, err = strconv.Atoi(i); err != nil {
log.Fatal("Invalid Atoi")
}
return ret
}

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Advent of Code
--- Day 9: Explosives in Cyberspace ---
Wandering around a secure area, you come across a datalink port to a new part of the network. After briefly scanning it for
interesting files, you find one file in particular that catches your attention. It's compressed with an experimental format, but
fortunately, the documentation for the format is nearby.
The format compresses a sequence of characters. Whitespace is ignored. To indicate that some sequence should be repeated, a marker is
added to the file, like (10x2). To decompress this marker, take the subsequent 10 characters and repeat them 2 times. Then, continue
reading the file after the repeated data. The marker itself is not included in the decompressed output.
If parentheses or other characters appear within the data referenced by a marker, that's okay - treat it like normal data, not a
marker, and then resume looking for markers after the decompressed section.
For example:
 ADVENT contains no markers and decompresses to itself with no changes, resulting in a decompressed length of 6.
 A(1x5)BC repeats only the B a total of 5 times, becoming ABBBBBC for a decompressed length of 7.
• (3x3)XYZ becomes XYZXYZXYZ for a decompressed length of 9.
 A(2x2)BCD(2x2)EFG doubles the BC and EF, becoming ABCBCDEFEFG for a decompressed length of 11.
• (6x1)(1x3)A simply becomes (1x3)A - the (1x3) looks like a marker, but because it's within a data section of another marker, it
is not treated any differently from the A that comes after it. It has a decompressed length of 6.
 X(8x2)(3x3)ABCY becomes X(3x3)ABC(3x3)ABCY (for a decompressed length of 18), because the decompressed data from the (8x2) marker
(the (3x3)ABC) is skipped and not processed further.
What is the decompressed length of the file (your puzzle input)? Don't count whitespace.
Your puzzle answer was __________.
--- Part Two ---
Apparently, the file actually uses version two of the format.
In version two, the only difference is that markers within decompressed data are decompressed. This, the documentation explains,
provides much more substantial compression capabilities, allowing many-gigabyte files to be stored in only a few kilobytes.
For example:
• (3x3)XYZ still becomes XYZXYZXYZ, as the decompressed section contains no markers.
 X(8x2)(3x3)ABCY becomes XABCABCABCABCABCABCY, because the decompressed data from the (8x2) marker is then further decompressed,
thus triggering the (3x3) marker twice for a total of six ABC sequences.
• (27x12)(20x12)(13x14)(7x10)(1x12)A decompresses into a string of A repeated 241920 times.
• (25x3)(3x3)ABC(2x3)XY(5x2)PQRSTX(18x9)(3x2)TWO(5x7)SEVEN becomes 445 characters long.
Unfortunately, the computer you brought probably doesn't have enough memory to actually decompress the file; you'll have to come up
with another way to get its decompressed length.
What is the decompressed length of the file using this improved format?
Although it hasn't changed, you can still get your puzzle input.
Answer: _____________________ [ [Submit] ]
References
Visible links
. http://adventofcode.com/
. http://adventofcode.com/2016/about
. http://adventofcode.com/2016/support
. http://adventofcode.com/2016/events
. http://adventofcode.com/2016/settings
. http://adventofcode.com/2016/auth/logout
. http://adventofcode.com/2016
. http://adventofcode.com/2016
. http://adventofcode.com/2016/leaderboard
. http://adventofcode.com/2016/stats
. http://adventofcode.com/2016/sponsors
. http://adventofcode.com/2016/sponsors
. http://adventofcode.com/2016/day/9/input

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bot 59 gives low to bot 176 and high to bot 120
bot 92 gives low to bot 42 and high to bot 187
value 31 goes to bot 114
bot 182 gives low to bot 49 and high to bot 176
bot 17 gives low to bot 181 and high to bot 162
bot 36 gives low to bot 118 and high to bot 121
bot 118 gives low to bot 164 and high to bot 55
bot 172 gives low to bot 79 and high to bot 123
bot 51 gives low to bot 60 and high to bot 31
bot 48 gives low to bot 107 and high to bot 58
bot 142 gives low to output 6 and high to bot 35
bot 133 gives low to output 4 and high to bot 47
bot 134 gives low to bot 122 and high to bot 66
bot 106 gives low to bot 155 and high to bot 99
bot 77 gives low to bot 93 and high to bot 84
bot 9 gives low to bot 173 and high to bot 197
bot 64 gives low to bot 123 and high to bot 48
bot 177 gives low to bot 21 and high to bot 132
bot 94 gives low to bot 6 and high to bot 25
bot 126 gives low to bot 193 and high to bot 56
bot 74 gives low to bot 187 and high to bot 125
bot 80 gives low to bot 41 and high to bot 191
bot 62 gives low to bot 157 and high to bot 138
bot 66 gives low to bot 1 and high to bot 209
bot 90 gives low to bot 104 and high to bot 34
bot 68 gives low to bot 23 and high to bot 87
bot 121 gives low to bot 55 and high to bot 126
bot 122 gives low to bot 137 and high to bot 1
bot 209 gives low to bot 168 and high to bot 26
bot 141 gives low to bot 170 and high to bot 6
bot 149 gives low to bot 62 and high to bot 13
bot 120 gives low to bot 179 and high to bot 71
bot 160 gives low to bot 194 and high to bot 151
bot 86 gives low to bot 96 and high to bot 106
value 13 goes to bot 9
bot 180 gives low to bot 189 and high to bot 27
value 67 goes to bot 88
bot 169 gives low to bot 99 and high to bot 159
bot 56 gives low to bot 98 and high to bot 147
bot 197 gives low to bot 174 and high to bot 81
bot 57 gives low to bot 113 and high to bot 179
bot 39 gives low to bot 115 and high to bot 3
bot 79 gives low to bot 22 and high to bot 40
bot 161 gives low to output 14 and high to bot 185
bot 21 gives low to bot 114 and high to bot 119
bot 136 gives low to bot 28 and high to bot 158
bot 105 gives low to bot 89 and high to bot 19
bot 168 gives low to bot 126 and high to bot 26
bot 193 gives low to bot 64 and high to bot 98
bot 186 gives low to bot 86 and high to bot 178
value 11 goes to bot 165
bot 33 gives low to bot 116 and high to bot 150
bot 32 gives low to bot 154 and high to bot 206
bot 166 gives low to bot 33 and high to bot 139
value 7 goes to bot 63
bot 203 gives low to bot 172 and high to bot 64
bot 200 gives low to bot 94 and high to bot 25
value 43 goes to bot 76
bot 145 gives low to bot 103 and high to bot 128
bot 119 gives low to bot 186 and high to bot 97
bot 12 gives low to bot 31 and high to bot 4
bot 23 gives low to bot 198 and high to bot 171
bot 34 gives low to bot 10 and high to bot 20
bot 198 gives low to bot 43 and high to bot 17
bot 50 gives low to output 1 and high to bot 127
bot 155 gives low to bot 191 and high to bot 32
bot 206 gives low to bot 12 and high to bot 43
bot 96 gives low to bot 80 and high to bot 155
bot 93 gives low to bot 44 and high to bot 70
bot 24 gives low to bot 85 and high to bot 83
bot 30 gives low to bot 159 and high to bot 68
bot 55 gives low to bot 203 and high to bot 193
bot 199 gives low to bot 68 and high to bot 135
bot 170 gives low to bot 97 and high to bot 5
bot 65 gives low to bot 152 and high to bot 194
bot 43 gives low to bot 4 and high to bot 181
bot 113 gives low to output 9 and high to bot 161
bot 81 gives low to bot 141 and high to bot 94
value 29 goes to bot 7
bot 46 gives low to bot 175 and high to bot 195
value 47 goes to bot 21
value 23 goes to bot 42
bot 13 gives low to bot 138 and high to bot 61
bot 135 gives low to bot 87 and high to bot 111
bot 194 gives low to bot 190 and high to bot 82
value 73 goes to bot 109
bot 154 gives low to bot 51 and high to bot 12
bot 1 gives low to bot 18 and high to bot 209
bot 98 gives low to bot 48 and high to bot 45
bot 147 gives low to bot 45 and high to bot 95
bot 47 gives low to output 19 and high to bot 152
bot 26 gives low to bot 56 and high to bot 147
bot 179 gives low to bot 161 and high to bot 71
bot 148 gives low to bot 204 and high to bot 137
bot 5 gives low to bot 67 and high to bot 85
bot 174 gives low to bot 132 and high to bot 141
bot 8 gives low to bot 13 and high to bot 75
bot 82 gives low to bot 146 and high to bot 22
bot 123 gives low to bot 40 and high to bot 107
bot 99 gives low to bot 32 and high to bot 201
bot 41 gives low to bot 196 and high to bot 192
bot 139 gives low to bot 150 and high to bot 153
bot 11 gives low to output 16 and high to bot 113
bot 72 gives low to bot 65 and high to bot 160
bot 195 gives low to bot 133 and high to bot 183
bot 54 gives low to output 12 and high to output 10
bot 158 gives low to bot 102 and high to bot 110
bot 112 gives low to bot 19 and high to bot 118
bot 31 gives low to bot 208 and high to bot 143
bot 167 gives low to bot 7 and high to bot 96
bot 63 gives low to bot 92 and high to bot 74
bot 116 gives low to bot 20 and high to bot 131
bot 184 gives low to bot 39 and high to bot 3
bot 162 gives low to bot 205 and high to bot 39
bot 108 gives low to output 11 and high to bot 175
value 53 goes to bot 207
bot 111 gives low to bot 202 and high to bot 184
bot 25 gives low to bot 24 and high to bot 83
value 71 goes to bot 77
bot 69 gives low to bot 142 and high to bot 0
bot 146 gives low to output 13 and high to bot 53
bot 7 gives low to bot 76 and high to bot 80
bot 131 gives low to bot 73 and high to bot 204
bot 102 gives low to bot 195 and high to bot 117
bot 76 gives low to bot 165 and high to bot 41
bot 153 gives low to bot 148 and high to bot 122
bot 208 gives low to bot 90 and high to bot 163
bot 70 gives low to bot 144 and high to bot 78
bot 125 gives low to bot 8 and high to bot 156
bot 83 gives low to bot 199 and high to bot 135
bot 75 gives low to bot 61 and high to bot 104
bot 67 gives low to bot 169 and high to bot 30
bot 14 gives low to bot 81 and high to bot 200
bot 159 gives low to bot 201 and high to bot 23
value 3 goes to bot 93
bot 110 gives low to bot 117 and high to bot 89
bot 128 gives low to bot 129 and high to bot 182
bot 87 gives low to bot 171 and high to bot 111
bot 45 gives low to bot 58 and high to bot 95
bot 4 gives low to bot 143 and high to bot 166
bot 60 gives low to bot 156 and high to bot 208
bot 27 gives low to bot 108 and high to bot 46
bot 42 gives low to bot 207 and high to bot 149
bot 117 gives low to bot 183 and high to bot 72
bot 115 gives low to bot 153 and high to bot 134
bot 140 gives low to bot 125 and high to bot 60
bot 173 gives low to bot 177 and high to bot 174
bot 138 gives low to bot 180 and high to bot 52
bot 100 gives low to bot 38 and high to bot 59
value 41 goes to bot 173
value 59 goes to bot 177
bot 165 gives low to bot 63 and high to bot 196
bot 84 gives low to bot 70 and high to bot 78
bot 2 gives low to bot 160 and high to bot 91
value 61 goes to bot 29
bot 114 gives low to bot 109 and high to bot 186
bot 205 gives low to bot 139 and high to bot 115
bot 175 gives low to output 17 and high to bot 133
bot 176 gives low to bot 57 and high to bot 120
bot 107 gives low to bot 124 and high to bot 15
bot 52 gives low to bot 27 and high to bot 28
bot 103 gives low to bot 50 and high to bot 129
bot 150 gives low to bot 131 and high to bot 148
bot 16 gives low to output 20 and high to bot 189
bot 190 gives low to output 18 and high to bot 146
bot 157 gives low to bot 16 and high to bot 180
bot 10 gives low to bot 158 and high to bot 130
bot 202 gives low to bot 162 and high to bot 184
bot 88 gives low to bot 77 and high to bot 84
bot 188 gives low to bot 128 and high to bot 38
bot 58 gives low to bot 15 and high to bot 101
bot 171 gives low to bot 17 and high to bot 202
bot 97 gives low to bot 178 and high to bot 67
bot 163 gives low to bot 34 and high to bot 116
bot 124 gives low to bot 0 and high to bot 145
bot 71 gives low to bot 185 and high to bot 54
bot 78 gives low to bot 14 and high to bot 200
bot 101 gives low to bot 188 and high to bot 100
bot 189 gives low to output 7 and high to bot 108
bot 95 gives low to bot 101 and high to bot 100
bot 0 gives low to bot 35 and high to bot 103
bot 207 gives low to bot 37 and high to bot 62
bot 49 gives low to bot 11 and high to bot 57
bot 85 gives low to bot 30 and high to bot 199
bot 89 gives low to bot 72 and high to bot 2
bot 3 gives low to bot 134 and high to bot 66
bot 181 gives low to bot 166 and high to bot 205
bot 91 gives low to bot 151 and high to bot 172
value 17 goes to bot 167
bot 20 gives low to bot 130 and high to bot 73
bot 196 gives low to bot 74 and high to bot 140
bot 18 gives low to bot 121 and high to bot 168
bot 185 gives low to output 15 and high to bot 54
bot 178 gives low to bot 106 and high to bot 169
bot 129 gives low to bot 127 and high to bot 49
bot 19 gives low to bot 2 and high to bot 164
bot 15 gives low to bot 145 and high to bot 188
bot 144 gives low to bot 197 and high to bot 14
bot 201 gives low to bot 206 and high to bot 198
bot 164 gives low to bot 91 and high to bot 203
bot 73 gives low to bot 105 and high to bot 112
bot 191 gives low to bot 192 and high to bot 154
bot 109 gives low to bot 167 and high to bot 86
bot 151 gives low to bot 82 and high to bot 79
bot 53 gives low to output 2 and high to bot 142
bot 37 gives low to bot 29 and high to bot 157
value 2 goes to bot 44
bot 204 gives low to bot 112 and high to bot 36
bot 40 gives low to bot 69 and high to bot 124
bot 22 gives low to bot 53 and high to bot 69
bot 104 gives low to bot 136 and high to bot 10
value 19 goes to bot 88
bot 127 gives low to output 5 and high to bot 11
bot 183 gives low to bot 47 and high to bot 65
bot 192 gives low to bot 140 and high to bot 51
bot 38 gives low to bot 182 and high to bot 59
bot 61 gives low to bot 52 and high to bot 136
bot 156 gives low to bot 75 and high to bot 90
value 37 goes to bot 37
bot 28 gives low to bot 46 and high to bot 102
bot 187 gives low to bot 149 and high to bot 8
bot 132 gives low to bot 119 and high to bot 170
bot 44 gives low to bot 9 and high to bot 144
bot 29 gives low to output 0 and high to bot 16
bot 6 gives low to bot 5 and high to bot 24
bot 137 gives low to bot 36 and high to bot 18
bot 130 gives low to bot 110 and high to bot 105
value 5 goes to bot 92
bot 35 gives low to output 3 and high to bot 50
bot 152 gives low to output 8 and high to bot 190
bot 143 gives low to bot 163 and high to bot 33

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package main
import (
"bufio"
"fmt"
"log"
"os"
"strconv"
"strings"
)
var allBots []Bot
var watch1, watch2 int
//var outputBins map[int][]int
func main() {
input := stdinToStringSlice()
watch1 = 61
watch2 = 17
//outputBins := make(map[int][]int)
for i := range input {
inst := strings.Fields(input[i])
if inst[0] == "bot" {
// And instruction
// make sure we have this bot
botNum := inst[0] + inst[1]
botIdx := findBotIdx(botNum)
allBots[botIdx].giveLowTo = inst[5] + inst[6]
allBots[botIdx].giveHighTo = inst[10] + inst[11]
}
}
// Run sim
for i := range input {
inst := strings.Fields(input[i])
if inst[0] == "value" {
val := atoi(inst[1])
botNum := inst[4] + inst[5]
botIdx := findBotIdx(botNum)
allBots[botIdx].chips = append(allBots[botIdx].chips, val)
allBots[botIdx].Trigger()
}
}
for i := range allBots {
fmt.Println(allBots[i])
}
}
func findBotIdx(id string) int {
fmt.Println("Find bot", id)
for i := range allBots {
if allBots[i].id == id {
return i
}
}
// We didn't find the bot
b := new(Bot)
b.id = id
allBots = append(allBots, *b)
return len(allBots) - 1
}
type Bot struct {
id string
chips []int
giveLowTo string
giveHighTo string
}
func (b *Bot) Trigger() {
if strings.HasPrefix(b.id, "output") {
return
}
if len(b.chips) == 2 {
low := b.chips[0]
high := b.chips[1]
if b.chips[0] > b.chips[1] {
high = b.chips[0]
low = b.chips[1]
}
giveLowIdx := findBotIdx(b.giveLowTo)
allBots[giveLowIdx].chips = append(allBots[giveLowIdx].chips, low)
giveHighIdx := findBotIdx(b.giveHighTo)
allBots[giveHighIdx].chips = append(allBots[giveHighIdx].chips, high)
allBots[giveLowIdx].Trigger()
allBots[giveHighIdx].Trigger()
}
}
func checkWatches(b *Bot) {
if b.chips[0] == watch1 && b.chips[1] == watch2 || b.chips[0] == watch1 && b.chips[1] == watch2 {
fmt.Println("Watch triggered:", b.id)
}
}
func stdinToStringSlice() []string {
var input []string
scanner := bufio.NewScanner(os.Stdin)
for scanner.Scan() {
input = append(input, scanner.Text())
}
return input
}
func itoa(i int) string {
return strconv.Itoa(i)
}
func atoi(i string) int {
var ret int
var err error
if ret, err = strconv.Atoi(i); err != nil {
log.Fatal("Invalid Atoi")
}
return ret
}

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Advent of Code
--- Day 10: Balance Bots ---
You come upon a factory in which many robots are zooming around handing small microchips to
each other.
Upon closer examination, you notice that each bot only proceeds when it has two microchips,
and once it does, it gives each one to a different bot or puts it in a marked "output" bin.
Sometimes, bots take microchips from "input" bins, too.
Inspecting one of the microchips, it seems like they each contain a single number; the bots
must use some logic to decide what to do with each chip. You access the local control computer
and download the bots' instructions (your puzzle input).
Some of the instructions specify that a specific-valued microchip should be given to a
specific bot; the rest of the instructions indicate what a given bot should do with its
lower-value or higher-value chip.
For example, consider the following instructions:
value 5 goes to bot 2
bot 2 gives low to bot 1 and high to bot 0
value 3 goes to bot 1
bot 1 gives low to output 1 and high to bot 0
bot 0 gives low to output 2 and high to output 0
value 2 goes to bot 2
 Initially, bot 1 starts with a value-3 chip, and bot 2 starts with a value-2 chip and a
value-5 chip.
 Because bot 2 has two microchips, it gives its lower one (2) to bot 1 and its higher one
(5) to bot 0.
 Then, bot 1 has two microchips; it puts the value-2 chip in output 1 and gives the value-3
chip to bot 0.
 Finally, bot 0 has two microchips; it puts the 3 in output 2 and the 5 in output 0.
In the end, output bin 0 contains a value-5 microchip, output bin 1 contains a value-2
microchip, and output bin 2 contains a value-3 microchip. In this configuration, bot number 2
is responsible for comparing value-5 microchips with value-2 microchips.
Based on your instructions, what is the number of the bot that is responsible for comparing
value-61 microchips with value-17 microchips?
Your puzzle answer was ______.
--- Part Two ---
What do you get if you multiply together the values of one chip in each of outputs 0, 1, and
2?
Your puzzle answer was _______.
References
Visible links
. http://adventofcode.com/
. http://adventofcode.com/2016/about
. http://adventofcode.com/2016/support
. http://adventofcode.com/2016/events
. http://adventofcode.com/2016/settings
. http://adventofcode.com/2016/auth/logout
. http://adventofcode.com/2016
. http://adventofcode.com/2016
. http://adventofcode.com/2016/leaderboard
. http://adventofcode.com/2016/stats
. http://adventofcode.com/2016/sponsors
. http://adventofcode.com/2016/sponsors
. https://www.youtube.com/watch?v=JnkMyfQ5YfY&t=40
. http://adventofcode.com/2016
. http://adventofcode.com/2016/day/10/input

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The first floor contains a thulium generator, a thulium-compatible microchip, a plutonium generator, and a strontium generator.
The second floor contains a plutonium-compatible microchip, and a strontium-compatible microchip.
The third floor contains a promethium generator, a promethium-compatible microchip, a ruthenium generator, and a ruthenium-compatible microchip.
The fourth floor contains nothing relevant.

176
2016/day11-add/main.go Normal file
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package main
import (
"bufio"
"errors"
"fmt"
"os"
"strconv"
"strings"
"github.com/fatih/color"
termbox "github.com/nsf/termbox-go"
)
var elevatorFloor int
var floorInv [][]string
var componentRegistry []string
var cursorPosX, cursorPosY int
var selectItems []string
// This code requires oxford commas in the input
func main() {
input := stdinToStringSlice()
for i := range input {
listIdx := strings.Index(input[i], "a ")
if listIdx == -1 {
floorInv = append(floorInv, []string{})
continue
}
input[i] = input[i][listIdx:]
floorInv = append(floorInv, strings.Split(input[i], ", "))
}
for i := range floorInv {
for j := range floorInv[i] {
tp := "G"
if strings.Contains(floorInv[i][j], "microchip") {
tp = "M"
}
floorInv[i][j] = strings.TrimPrefix(floorInv[i][j], "a ")
floorInv[i][j] = strings.TrimPrefix(floorInv[i][j], "and a ")
ele := strings.ToUpper(floorInv[i][j][:2])
floorInv[i][j] = ele + ":" + tp
componentRegistry = append(componentRegistry, floorInv[i][j])
}
}
err := termbox.Init()
if err != nil {
fmt.Println("Error initializing termbox")
os.Exit(1)
}
defer termbox.Close()
eventQueue := make(chan termbox.Event)
go func() {
for {
eventQueue <- termbox.PollEvent()
}
}()
PrintScreen()
var doQuit bool
for {
select {
case ev := <-eventQueue:
if ev.Type == termbox.EventKey {
switch {
case ev.Key == termbox.KeyArrowLeft:
if cursorPosX > 0 {
cursorPosX--
}
case ev.Key == termbox.KeyArrowRight:
if cursorPosX < len(componentRegistry)-1 {
cursorPosX++
}
case ev.Ch == 'q':
doQuit = true
}
}
ClearScreen()
PrintScreen()
}
if doQuit {
break
}
}
//fmt.Println("Registered Components: ", componentRegistry)
// a microchip without it's generator cannot be on the same floor
// as a different microchip that _does_ have it's generator
// Look at what we have on the lowest floor with items
// For each microchip, find it's generator
// if it's on this floor, take them both up a floor
// If we have microchips, see where their generators are
// If they are with us, take both and move up
// If they are one floor above,
}
func SelectItem() error {
return nil
}
func FloorHasComponent(flr int, cmp string) bool {
for i := range floorInv[flr] {
if floorInv[flr][i] == cmp {
return true
}
}
return false
}
func MoveElevator(dir int, take []string) error {
if len(take) < 1 || len(take) > 2 {
return errors.New("Error tried to move with invalid number of items: " + itoa(len(take)))
}
// Are the requested items on the current floor?
// If so, move them to the floor we're going to
for j := range take {
if FloorHasComponent(elevatorFloor, take[j]) {
// TODO: Move the component
floorInv[elevatorFloor+dir] = append(floorInv[elevatorFloor+dir], take[j])
} else {
return errors.New("Elevator tried to take " + take[j] + " but it's not on floor " + itoa(elevatorFloor+1))
}
}
return nil
}
func ClearScreen() {
fmt.Print("\033[H\033[2J")
}
func PrintScreen() {
c := color.New(color.FgCyan)
w := color.New(color.FgWhite)
g := color.New(color.FgGreen)
g.Add(color.Bold)
for i := len(floorInv); i > 0; i-- {
c.Printf("F%d ", i)
if elevatorFloor == i-1 {
c.Print("E ")
} else {
w.Print(". ")
}
for cr := range componentRegistry {
out := ". "
for j := range floorInv[i-1] {
if floorInv[i-1][j] == componentRegistry[cr] {
out = componentRegistry[cr] + " "
break
}
}
if elevatorFloor == i-1 && cursorPosX == cr {
g.Print(out)
} else {
w.Print(out)
}
}
w.Println("")
}
w.Println("")
}
func stdinToStringSlice() []string {
var input []string
scanner := bufio.NewScanner(os.Stdin)
for scanner.Scan() {
input = append(input, scanner.Text())
}
return input
}
func itoa(i int) string {
return strconv.Itoa(i)
}

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The first floor contains a hydrogen-compatible microchip, and a lithium-compatible microchip.
The second floor contains a hydrogen generator.
The third floor contains a lithium generator.
The fourth floor contains nothing relevant.

4
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The first floor contains a thulium generator, a thulium-compatible microchip, a plutonium generator, and a strontium generator.
The second floor contains a plutonium-compatible microchip, and a strontium-compatible microchip.
The third floor contains a promethium generator, a promethium-compatible microchip, a ruthenium generator, and a ruthenium-compatible microchip.
The fourth floor contains nothing relevant.

4
2016/day11/input-part2 Normal file
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The first floor contains a elerium generator, a elerium-compatible microchip, a dilithium generator, a dilitium-compatible microchip, a thulium generator, a thulium-compatible microchip, a plutonium generator, and a strontium generator.
The second floor contains a plutonium-compatible microchip, and a strontium-compatible microchip.
The third floor contains a promethium generator, a promethium-compatible microchip, a ruthenium generator, and a ruthenium-compatible microchip.
The fourth floor contains nothing relevant.

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package main
import (
"bufio"
"fmt"
"os"
"strconv"
"strings"
"time"
"github.com/fatih/color"
)
// This is a purely mathematical solution to this problem
// We don't care about what the elements are.
// Loop on each floor and consider:
// 1: We have less than 3 items on the floor
// So move them all up; 1 total move
// 2: We have more than 3 items on the floor
// So move two up and one back down; 2 total moves
// Setting visual to true will output what's happening
var visual bool
func main() {
input := stdinToStringSlice()
var numMoves int
if !visual {
var lvlItmCnts []int
for i := range input {
lvlItmCnts = append(lvlItmCnts, strings.Count(input[i], "generator")+strings.Count(input[i], "microchip"))
}
for i := 0; i < 3; i++ {
for lvlItmCnts[i] > 0 {
if lvlItmCnts[i] < 3 {
lvlItmCnts[i+1] = lvlItmCnts[i+1] + lvlItmCnts[i]
lvlItmCnts[i] = 0
numMoves++
} else {
lvlItmCnts[i] = lvlItmCnts[i] - 1
lvlItmCnts[i+1] = lvlItmCnts[i+1] + 1
numMoves = numMoves + 2
}
}
}
} else {
// Visual mode spits a lot more stuff out, so it has
// to track a lot more things...
numMoves = VisualMode(input)
}
fmt.Println("Number of moves:", numMoves)
}
func VisualMode(input []string) int {
// This code requires oxford commas in the input :)
var elevatorFloor int
var floorInv [][]string
var componentRegistry []string
for i := range input {
listIdx := strings.Index(input[i], "a ")
if listIdx == -1 {
floorInv = append(floorInv, []string{})
continue
}
input[i] = input[i][listIdx:]
floorInv = append(floorInv, strings.Split(input[i], ", "))
}
for i := range floorInv {
elevatorFloor = i
for j := range floorInv[i] {
tp := "G"
if strings.Contains(floorInv[i][j], "microchip") {
tp = "M"
}
floorInv[i][j] = strings.TrimPrefix(floorInv[i][j], "a ")
floorInv[i][j] = strings.TrimPrefix(floorInv[i][j], "and a ")
ele := strings.ToUpper(floorInv[i][j][:2])
floorInv[i][j] = ele + ":" + tp
componentRegistry = append(componentRegistry, floorInv[i][j])
}
}
var numMoves int
var pt1, pt2 string
for i := 0; i < len(floorInv)-1; i++ {
for len(floorInv[i]) > 1 {
if len(floorInv[i]) > 1 {
// If there are exactly two items on the floor, move them up
if len(floorInv[i]) == 2 {
pt1, pt2, floorInv[i] = floorInv[i][0], floorInv[i][1], floorInv[i][2:]
floorInv[i+1] = append(floorInv[i+1], pt1, pt2)
numMoves++
} else if len(floorInv[i]) > 2 {
// If more than two, move one up, but add 2 to the move
// To simulate, moving two up then one back down
pt1, floorInv[i] = floorInv[i][0], floorInv[i][1:]
floorInv[i+1] = append(floorInv[i+1], pt1)
numMoves = numMoves + 2
}
} else {
// Only one left on floor, move it up
pt1, floorInv[i] = floorInv[i][0], []string{}
floorInv[i+1] = append(floorInv[i+1], pt1, pt2)
numMoves++
}
if visual {
ClearScreen()
PrintScreen(floorInv, elevatorFloor, componentRegistry)
time.Sleep(time.Millisecond * 250)
}
}
}
return numMoves
}
func ClearScreen() {
fmt.Print("\033[H\033[2J")
}
func PrintScreen(floorInv [][]string, elevatorFloor int, componentRegistry []string) {
c := color.New(color.FgCyan)
w := color.New(color.FgWhite)
for i := len(floorInv); i > 0; i-- {
c.Printf("F%d ", i)
if elevatorFloor == i-1 {
c.Print("E ")
} else {
w.Print(". ")
}
for cr := range componentRegistry {
var fnd bool
for j := range floorInv[i-1] {
if floorInv[i-1][j] == componentRegistry[cr] {
fnd = true
fmt.Print(componentRegistry[cr] + " ")
break
}
}
if !fnd {
fmt.Print(". ")
}
}
fmt.Println()
}
}
func stdinToStringSlice() []string {
var input []string
scanner := bufio.NewScanner(os.Stdin)
for scanner.Scan() {
input = append(input, scanner.Text())
}
return input
}
func itoa(i int) string {
return strconv.Itoa(i)
}

23
2016/day12/input Normal file
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cpy 1 a
cpy 1 b
cpy 26 d
jnz c 2
jnz 1 5
cpy 7 c
inc d
dec c
jnz c -2
cpy a c
inc a
dec b
jnz b -2
cpy c b
dec d
jnz d -6
cpy 19 c
cpy 14 d
inc a
dec d
jnz d -2
dec c
jnz c -5

78
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package main
import (
"bufio"
"fmt"
"log"
"os"
"strconv"
"strings"
)
var regs = map[string]int{
"a": 0,
"b": 0,
"c": 1,
"d": 0,
}
func main() {
input := stdinToStringSlice()
curr := 0
for curr < len(input) {
ins := strings.Fields(input[curr])
curr++
switch ins[0] {
case "cpy":
src := ins[1]
dst := ins[2]
// check if the src is a register
if v, ok := regs[src]; ok {
regs[dst] = v
} else {
// It's not, must be an int
regs[dst] = atoi(src)
}
case "inc":
regs[ins[1]] = regs[ins[1]] + 1
case "dec":
regs[ins[1]] = regs[ins[1]] - 1
case "jnz":
v, ok := regs[ins[1]]
if !ok {
v = atoi(ins[1])
}
if v != 0 {
// Subtract 1 from the jump because we incremented already
curr += atoi(ins[2]) - 1
}
}
}
PrintState()
}
// Fancy State Printing
func PrintState() {
datLine := fmt.Sprint("\u2502 a:", regs["a"], " b:", regs["b"], " c:", regs["c"], " d:", regs["d"], " \u2502")
fmt.Println("\u250C" + strings.Repeat("\u2500", len(datLine)-6) + "\u2510")
fmt.Println(datLine)
fmt.Println("\u2514" + strings.Repeat("\u2500", len(datLine)-6) + "\u2518")
}
func stdinToStringSlice() []string {
var input []string
scanner := bufio.NewScanner(os.Stdin)
for scanner.Scan() {
input = append(input, scanner.Text())
}
return input
}
func atoi(i string) int {
var ret int
var err error
if ret, err = strconv.Atoi(i); err != nil {
log.Fatal("Invalid Atoi:" + i)
}
return ret
}

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Advent of Code
--- Day 12: Leonardo's Monorail ---
You finally reach the top floor of this building: a garden with a slanted glass ceiling. Looks like there are no more
stars to be had.
While sitting on a nearby bench amidst some tiger lilies, you manage to decrypt some of the files you extracted from
the servers downstairs.
According to these documents, Easter Bunny HQ isn't just this building - it's a collection of buildings in the nearby
area. They're all connected by a local monorail, and there's another building not far from here! Unfortunately, being
night, the monorail is currently not operating.
You remotely connect to the monorail control systems and discover that the boot sequence expects a password. The
password-checking logic (your puzzle input) is easy to extract, but the code it uses is strange: it's assembunny code
designed for the new computer you just assembled. You'll have to execute the code and get the password.
The assembunny code you've extracted operates on four registers (a, b, c, and d) that start at 0 and can hold any
integer. However, it seems to make use of only a few instructions:
 cpy x y copies x (either an integer or the value of a register) into register y.
 inc x increases the value of register x by one.
 dec x decreases the value of register x by one.
 jnz x y jumps to an instruction y away (positive means forward; negative means backward), but only if x is not
zero.
The jnz instruction moves relative to itself: an offset of -1 would continue at the previous instruction, while an
offset of 2 would skip over the next instruction.
For example:
cpy 41 a
inc a
inc a
dec a
jnz a 2
dec a
The above code would set register a to 41, increase its value by 2, decrease its value by 1, and then skip the last
dec a (because a is not zero, so the jnz a 2 skips it), leaving register a at 42. When you move past the last
instruction, the program halts.
After executing the assembunny code in your puzzle input, what value is left in register a?
Your puzzle answer was __________.
--- Part Two ---
As you head down the fire escape to the monorail, you notice it didn't start; register c needs to be initialized to
the position of the ignition key.
If you instead initialize register c to be 1, what value is now left in register a?
Your puzzle answer was ___________.
References
Visible links
. http://adventofcode.com/
. http://adventofcode.com/2016/about
. http://adventofcode.com/2016/support
. http://adventofcode.com/2016/events
. http://adventofcode.com/2016/settings
. http://adventofcode.com/2016/auth/logout
. http://adventofcode.com/2016
. http://adventofcode.com/2016
. http://adventofcode.com/2016/leaderboard
. http://adventofcode.com/2016/stats
. http://adventofcode.com/2016/sponsors
. http://adventofcode.com/2016/sponsors
. https://www.google.com/search?q=tiger+lilies&tbm=isch
. http://adventofcode.com/2016/day/11
. https://en.wikipedia.org/wiki/Processor_register
. https://en.wikipedia.org/wiki/Integer
. https://en.wikipedia.org/wiki/Instruction_set
. http://adventofcode.com/2016
. http://adventofcode.com/2016/day/12/input

341
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package main
import (
"fmt"
"log"
"os"
"strconv"
"strings"
"time"
"github.com/fatih/color"
termbox "github.com/nsf/termbox-go"
)
var tWidth, tHeight int
// Puzzle 1 Input: 1364 31 39
// Puzzle 1 Test Input: 10 7 4
func main() {
mode := "solve"
if len(os.Args) < 4 {
fmt.Println("Usage: day13 <seed> <dest-x> <dest-y>")
os.Exit(1)
}
seed := atoi(os.Args[1])
destX, destY := atoi(os.Args[2]), atoi(os.Args[3])
if len(os.Args) >= 5 {
mode = os.Args[4]
}
f := CreateFloor(1, 1, destX, destY, seed)
if err := termbox.Init(); err != nil {
panic(err)
}
tWidth, tHeight = termbox.Size()
termbox.Close()
switch mode {
case "solve":
if f.Solve(f.start.x, f.start.y, 0, true) {
f.dispCoord = f.end
}
ClearScreen()
f.Print()
fmt.Println("Shortest Path:", len(f.solvePath.coords))
case "walk":
dist := 50
f.Walk(f.start.x, f.start.y, 0, dist, true)
fmt.Println("Within", dist, "steps: ", len(f.testedPath.coords))
}
}
type Coord struct {
x, y, dist int
}
func (c *Coord) Is(x, y int) bool {
return c.x == x && c.y == y
}
func (c *Coord) Equals(t *Coord) bool {
return c.x == t.x && c.y == t.y
}
func NewCoord(x, y int) *Coord {
return &Coord{x, y, -1}
}
type Path struct {
coords []Coord
}
func (p *Path) Append(c Coord) {
p.coords = append(p.coords, c)
}
func (p *Path) ContainsCoord(x, y int) bool {
for i := range p.coords {
if p.coords[i].Is(x, y) {
return true
}
}
return false
}
func (p *Path) GetCoordAt(x, y int) *Coord {
for i := range p.coords {
if p.coords[i].Is(x, y) {
return &p.coords[i]
}
}
return nil
}
type Floor struct {
start *Coord
end *Coord
seed int
testedPath Path
solvePath Path
dispCoord *Coord
}
func CreateFloor(stX, stY, endX, endY, seed int) *Floor {
f := Floor{
start: NewCoord(stX, stY),
end: NewCoord(endX, endY),
seed: seed,
}
return &f
}
func (f *Floor) Walk(x, y, dist, maxDist int, print bool) {
wrkCoord := Coord{x, y, dist}
if f.IsWall(x, y) || f.testedPath.ContainsCoord(x, y) {
return
}
if dist == maxDist {
f.testedPath.Append(wrkCoord)
return
}
if print {
f.dispCoord = &wrkCoord
ClearScreen()
f.Print()
fmt.Println("Tested Spots:", len(f.testedPath.coords))
time.Sleep(time.Millisecond * 70)
}
if !f.IsWall(x-1, y) {
if t := f.testedPath.GetCoordAt(x-1, y); t != nil {
if t.dist+1 < wrkCoord.dist {
return
}
}
}
if !f.IsWall(x+1, y) {
if t := f.testedPath.GetCoordAt(x+1, y); t != nil {
if t.dist+1 < wrkCoord.dist {
return
}
}
}
if !f.IsWall(x, y-1) {
if t := f.testedPath.GetCoordAt(x, y-1); t != nil {
if t.dist+1 < wrkCoord.dist {
return
}
}
}
if !f.IsWall(x, y+1) {
if t := f.testedPath.GetCoordAt(x, y+1); t != nil {
if t.dist+1 < wrkCoord.dist {
return
}
}
}
f.testedPath.Append(wrkCoord)
// Try intelligently first
// (Attempt to move towards the exit)
if x > 0 {
f.Walk(x-1, y, wrkCoord.dist+1, maxDist, print)
}
if y > 0 {
f.Walk(x, y-1, wrkCoord.dist+1, maxDist, print)
}
f.Walk(x+1, y, wrkCoord.dist+1, maxDist, print)
f.Walk(x, y+1, wrkCoord.dist+1, maxDist, print)
}
func (f *Floor) Solve(x, y, dist int, print bool) bool {
wrkCoord := Coord{x, y, dist}
if f.end.Is(x, y) {
return true
}
if f.IsWall(x, y) || f.testedPath.ContainsCoord(x, y) {
return false
}
// Test if there is a shorter path to this coordinate
if !f.IsWall(x-1, y) {
if t := f.testedPath.GetCoordAt(x-1, y); t != nil {
if t.dist+1 < wrkCoord.dist {
return false
}
}
}
if !f.IsWall(x+1, y) {
if t := f.testedPath.GetCoordAt(x+1, y); t != nil {
if t.dist+1 < wrkCoord.dist {
return false
}
}
}
if !f.IsWall(x, y-1) {
if t := f.testedPath.GetCoordAt(x, y-1); t != nil {
if t.dist+1 < wrkCoord.dist {
return false
}
}
}
if !f.IsWall(x, y+1) {
if t := f.testedPath.GetCoordAt(x, y+1); t != nil {
if t.dist+1 < wrkCoord.dist {
return false
}
}
}
if print {
f.dispCoord = &wrkCoord
ClearScreen()
f.Print()
fmt.Println("Tested Spots:", len(f.testedPath.coords))
time.Sleep(time.Millisecond * 70)
}
f.testedPath.Append(wrkCoord)
// Try intelligently first
// (Attempt to move towards the exit)
if x > f.end.x && x > 0 {
if f.Solve(x-1, y, wrkCoord.dist+1, print) {
f.solvePath.Append(wrkCoord)
return true
}
}
if y > f.end.y && y > 0 {
if f.Solve(x, y-1, wrkCoord.dist+1, print) {
f.solvePath.Append(wrkCoord)
return true
}
}
if x < f.end.x {
if f.Solve(x+1, y, wrkCoord.dist+1, print) {
f.solvePath.Append(wrkCoord)
return true
}
}
if y < f.end.y {
if f.Solve(x, y+1, wrkCoord.dist+1, print) {
f.solvePath.Append(wrkCoord)
return true
}
}
// Intelligence failed us... Just find a move
if x > 0 {
if f.Solve(x-1, y, wrkCoord.dist+1, print) {
f.solvePath.Append(wrkCoord)
return true
}
}
if y > 0 {
if f.Solve(x, y-1, wrkCoord.dist+1, print) {
f.solvePath.Append(wrkCoord)
return true
}
}
// This is where it gets shaky...
// Since we have an infinite maze, this could run forever
// So we have a hard cutoff at:
var MaxInt = int(^uint(0) >> 1)
if len(f.testedPath.coords) >= MaxInt {
fmt.Println("ERROR: Couldn't find a path.")
os.Exit(1)
}
if f.Solve(x+1, y, wrkCoord.dist+1, print) {
f.solvePath.Append(wrkCoord)
return true
}
if f.Solve(x, y+1, wrkCoord.dist+1, print) {
f.solvePath.Append(wrkCoord)
return true
}
return false
}
func (f *Floor) IsWall(x, y int) bool {
sum := (x*x + 3*x + 2*x*y + y + y*y + f.seed)
s := fmt.Sprintf("%b", sum)
if strings.Count(s, "1")%2 == 0 {
return false
}
return true
}
func (f *Floor) Print() {
wall := color.New(color.BgWhite)
space := color.New(color.FgWhite)
g := color.New(color.BgGreen).Add(color.FgBlack)
g.Add(color.Bold)
r := color.New(color.BgRed)
b := color.New(color.BgBlue).Add(color.BgBlack)
b.Add(color.Bold)
topY, topX := tHeight, tWidth
botY, botX := 0, 0
// We want to center approx 20x20 on our current location
// f.testedPath[len(f.testedPath)-1]?
if len(f.testedPath.coords) > 0 {
cntrCoord := f.testedPath.coords[len(f.testedPath.coords)-1]
if topY < cntrCoord.y+(tHeight/2) {
topY = cntrCoord.y + (tHeight / 2)
}
if topY > tHeight {
botY = topY - tHeight
}
if topX < cntrCoord.x+(tWidth/2) {
topX = cntrCoord.x + (tWidth / 2)
}
if topX > tWidth {
botX = topX - tWidth
}
}
for y := botY; y < topY; y++ {
for x := botX; x < topX; x++ {
if f.dispCoord != nil && f.dispCoord.Is(x, y) {
g.Print("O")
} else if f.solvePath.ContainsCoord(x, y) {
g.Print(".")
} else if f.testedPath.ContainsCoord(x, y) {
r.Print(" ")
} else {
if f.end.Is(x, y) {
b.Print("X")
} else if f.IsWall(x, y) {
wall.Print(" ")
} else {
space.Print(".")
}
}
}
fmt.Println()
}
}
func ClearScreen() {
fmt.Print("\033[H\033[2J")
}
func atoi(i string) int {
var ret int
var err error
if ret, err = strconv.Atoi(i); err != nil {
log.Fatal("Invalid Atoi")
}
return ret
}

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Advent of Code
--- Day 13: A Maze of Twisty Little Cubicles ---
You arrive at the first floor of this new building to discover a much less welcoming environment than the shiny atrium of the
last one. Instead, you are in a maze of twisty little cubicles, all alike.
Every location in this area is addressed by a pair of non-negative integers (x,y). Each such coordinate is either a wall or an
open space. You can't move diagonally. The cube maze starts at 0,0 and seems to extend infinitely toward positive x and y;
negative values are invalid, as they represent a location outside the building. You are in a small waiting area at 1,1.
While it seems chaotic, a nearby morale-boosting poster explains, the layout is actually quite logical. You can determine
whether a given x,y coordinate will be a wall or an open space using a simple system:
 Find x*x + 3*x + 2*x*y + y + y*y.
 Add the office designer's favorite number (your puzzle input).
 Find the binary representation of that sum; count the number of bits that are 1.
 If the number of bits that are 1 is even, it's an open space.
 If the number of bits that are 1 is odd, it's a wall.
For example, if the office designer's favorite number were 10, drawing walls as # and open spaces as ., the corner of the
building containing 0,0 would look like this:
0123456789
0 .#.####.##
1 ..#..#...#
2 #....##...
3 ###.#.###.
4 .##..#..#.
5 ..##....#.
6 #...##.###
Now, suppose you wanted to reach 7,4. The shortest route you could take is marked as O:
0123456789
0 .#.####.##
1 .O#..#...#
2 #OOO.##...
3 ###O#.###.
4 .##OO#OO#.
5 ..##OOO.#.
6 #...##.###
Thus, reaching 7,4 would take a minimum of 11 steps (starting from your current location, 1,1).
What is the fewest number of steps required for you to reach 31,39?
Your puzzle answer was _____.
The first half of this puzzle is complete! It provides one gold star: *
--- Part Two ---
How many locations (distinct x,y coordinates, including your starting location) can you reach in at most 50 steps?
Your puzzle input is still 1364.
Answer: _____________________
References
Visible links
. http://adventofcode.com/
. http://adventofcode.com/2016/about
. http://adventofcode.com/2016/support
. http://adventofcode.com/2016/events
. http://adventofcode.com/2016/settings
. http://adventofcode.com/2016/auth/logout
. http://adventofcode.com/2016
. http://adventofcode.com/2016
. http://adventofcode.com/2016/leaderboard
. http://adventofcode.com/2016/stats
. http://adventofcode.com/2016/sponsors
. http://adventofcode.com/2016/sponsors
. https://en.wikipedia.org/wiki/Binary_number
. https://en.wikipedia.org/wiki/Bit

82
2016/day14/main.go Normal file
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package main
import (
"crypto/md5"
"fmt"
"log"
"os"
"strconv"
"strings"
)
var hashStore []string
var salt string
func main() {
if len(os.Args) < 3 {
fmt.Println("Usage: ./day14 <salt> <key-num>")
fmt.Println(" My input: ./day14 cuanljph 64")
os.Exit(1)
}
salt = os.Args[1]
keyNum := atoi(os.Args[2])
var foundKeys []string
tryIdx := 0
for len(foundKeys) < keyNum {
tst := getHash(tryIdx)
for i := range tst {
if len(tst) > i+3 {
if tst[i:i+3] == strings.Repeat(string(tst[i]), 3) {
// Found a triple, search the next 1000 for a quintuple
if searchForQuint(salt, tryIdx+1, string(tst[i])) {
foundKeys = append(foundKeys, tst)
}
// Once we test the first triplet, we're done with this one
break
}
}
}
tryIdx++
}
fmt.Println("Key:", getHash(tryIdx-1), "Idx:", tryIdx)
}
func getHash(idx int) string {
for len(hashStore)-1 < idx {
hashStore = append(
hashStore,
getStretchedHash(fmt.Sprintf("%x", md5.Sum([]byte(salt+itoa(len(hashStore)+1))))),
)
}
return hashStore[idx]
}
func getStretchedHash(s string) string {
for i := 0; i < 2016; i++ {
s = fmt.Sprintf("%x", md5.Sum([]byte(s)))
}
return s
}
func searchForQuint(salt string, startIdx int, srch string) bool {
for i := startIdx; i <= startIdx+1000; i++ {
tst := getHash(i)
if strings.Contains(tst, strings.Repeat(srch, 5)) {
return true
}
}
return false
}
func atoi(i string) int {
var ret int
var err error
if ret, err = strconv.Atoi(i); err != nil {
log.Fatal("Invalid Atoi")
}
return ret
}
func itoa(i int) string {
return strconv.Itoa(i)
}

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Advent of Code
--- Day 14: One-Time Pad ---
In order to communicate securely with Santa while you're on this mission, you've been using a one-time pad that you generate using a pre-agreed algorithm.
Unfortunately, you've run out of keys in your one-time pad, and so you need to generate some more.
To generate keys, you first get a stream of random data by taking the MD5 of a pre-arranged salt (your puzzle input) and an increasing integer index (starting with 0,
and represented in decimal); the resulting MD5 hash should be represented as a string of lowercase hexadecimal digits.
However, not all of these MD5 hashes are keys, and you need 64 new keys for your one-time pad. A hash is a key only if:
 It contains three of the same character in a row, like 777. Only consider the first such triplet in a hash.
 One of the next 1000 hashes in the stream contains that same character five times in a row, like 77777.
Considering future hashes for five-of-a-kind sequences does not cause those hashes to be skipped; instead, regardless of whether the current hash is a key, always
resume testing for keys starting with the very next hash.
For example, if the pre-arranged salt is abc:
 The first index which produces a triple is 18, because the MD5 hash of abc18 contains ...cc38887a5.... However, index 18 does not count as a key for your one-time
pad, because none of the next thousand hashes (index 19 through index 1018) contain 88888.
 The next index which produces a triple is 39; the hash of abc39 contains eee. It is also the first key: one of the next thousand hashes (the one at index 816)
contains eeeee.
 None of the next six triples are keys, but the one after that, at index 92, is: it contains 999 and index 200 contains 99999.
 Eventually, index 22728 meets all of the criteria to generate the 64th key.
So, using our example salt of abc, index 22728 produces the 64th key.
Given the actual salt in your puzzle input, what index produces your 64th one-time pad key?
Your puzzle input was cuanljph.
Your puzzle answer was ________.
--- Part Two ---
Of course, in order to make this process even more secure, you've also implemented key stretching.
Key stretching forces attackers to spend more time generating hashes. Unfortunately, it forces everyone else to spend more time, too.
To implement key stretching, whenever you generate a hash, before you use it, you first find the MD5 hash of that hash, then the MD5 hash of that hash, and so on, a
total of 2016 additional hashings. Always use lowercase hexadecimal representations of hashes.
For example, to find the stretched hash for index 0 and salt abc:
 Find the MD5 hash of abc0: 577571be4de9dcce85a041ba0410f29f.
 Then, find the MD5 hash of that hash: eec80a0c92dc8a0777c619d9bb51e910.
 Then, find the MD5 hash of that hash: 16062ce768787384c81fe17a7a60c7e3.
• ...repeat many times...
 Then, find the MD5 hash of that hash: a107ff634856bb300138cac6568c0f24.
So, the stretched hash for index 0 in this situation is a107ff.... In the end, you find the original hash (one use of MD5), then find the hash-of-the-previous-hash 2016
times, for a total of 2017 uses of MD5.
The rest of the process remains the same, but now the keys are entirely different. Again for salt abc:
 The first triple (222, at index 5) has no matching 22222 in the next thousand hashes.
 The second triple (eee, at index 10) hash a matching eeeee at index 89, and so it is the first key.
 Eventually, index 22551 produces the 64th key (triple fff with matching fffff at index 22859.
Given the actual salt in your puzzle input and using 2016 extra MD5 calls of key stretching, what index now produces your 64th one-time pad key?
Your puzzle answer was ________.
Your puzzle input was cuanljph.
References
Visible links
. http://adventofcode.com/
. http://adventofcode.com/2016/about
. http://adventofcode.com/2016/support
. http://adventofcode.com/2016/events
. http://adventofcode.com/2016/settings
. http://adventofcode.com/2016/auth/logout
. http://adventofcode.com/2016
. http://adventofcode.com/2016
. http://adventofcode.com/2016/leaderboard
. http://adventofcode.com/2016/stats
. http://adventofcode.com/2016/sponsors
. http://adventofcode.com/2016/sponsors
. https://en.wikipedia.org/wiki/One-time_pad
. https://en.wikipedia.org/wiki/Security_through_obscurity
. https://en.wikipedia.org/wiki/MD5
. https://en.wikipedia.org/wiki/Salt_(cryptography)
. https://en.wikipedia.org/wiki/MD5#Security
. https://en.wikipedia.org/wiki/Key_stretching
. http://adventofcode.com/2016

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Disc #1 has 17 positions; at time=0, it is at position 15.
Disc #2 has 3 positions; at time=0, it is at position 2.
Disc #3 has 19 positions; at time=0, it is at position 4.
Disc #4 has 13 positions; at time=0, it is at position 2.
Disc #5 has 7 positions; at time=0, it is at position 2.
Disc #6 has 5 positions; at time=0, it is at position 0.

118
2016/day15/main.go Normal file
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package main
import (
"bufio"
"fmt"
"log"
"os"
"strconv"
"strings"
)
func main() {
input := stdinToStringSlice()
var verbose, pt2 bool
var allDiscs []Disc
if len(os.Args) > 1 {
for i := range os.Args {
switch os.Args[i] {
case "-2": // Do part 2
pt2 = true
case "-v": // Verbose mode
verbose = true
}
}
}
for i := range input {
pts := strings.Fields(input[i])
id := atoi(pts[1][1:])
ttl := atoi(pts[3])
curr := atoi(pts[11][0 : len(pts[11])-1])
allDiscs = append(allDiscs, *CreateDisc(id, ttl, curr))
}
if pt2 {
d := CreateDisc(7, 11, 0)
allDiscs = append(allDiscs, *d)
}
var numTicks int
for !AllDiscsAlign(allDiscs) {
numTicks++
TickAllDiscs(allDiscs)
if verbose {
PrintStatus(allDiscs)
}
}
fmt.Println("Total Ticks: ", numTicks)
}
func PrintStatus(d []Disc) {
fmt.Print("[ ")
for i := range d {
fmt.Printf("(%2d/%2d) ", d[i].currPos, d[i].ttlPos)
}
fmt.Println("]")
}
func TickAllDiscs(d []Disc) {
for i := range d {
d[i].Tick()
}
}
func AllDiscsAlign(d []Disc) bool {
for i := range d {
if (d[i].currPos+d[i].id)%d[i].ttlPos != 0 {
return false
}
}
return true
}
type Disc struct {
id int
ttlPos int
currPos int
startPos int
numTicks int
bufTicks int
}
func CreateDisc(id, ttl, curr int) *Disc {
d := new(Disc)
d.id = id
d.ttlPos = ttl
d.currPos = curr
d.startPos = curr
return d
}
func (d *Disc) Tick() {
d.currPos = (d.currPos + 1) % d.ttlPos
d.numTicks++
}
func stdinToStringSlice() []string {
var input []string
scanner := bufio.NewScanner(os.Stdin)
for scanner.Scan() {
input = append(input, scanner.Text())
}
return input
}
func atoi(i string) int {
var ret int
var err error
if ret, err = strconv.Atoi(i); err != nil {
log.Fatal("Invalid Atoi")
}
return ret
}
func ClearScreen() {
fmt.Print("\033[H\033[2J")
}

65
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Advent of Code
--- Day 15: Timing is Everything ---
The halls open into an interior plaza containing a large kinetic sculpture. The sculpture is in a sealed enclosure and seems to involve a set of identical spherical capsules that are carried to the top and allowed to bounce through
the maze of spinning pieces.
Part of the sculpture is even interactive! When a button is pressed, a capsule is dropped and tries to fall through slots in a set of rotating discs to finally go through a little hole at the bottom and come out of the sculpture. If
any of the slots aren't aligned with the capsule as it passes, the capsule bounces off the disc and soars away. You feel compelled to get one of those capsules.
The discs pause their motion each second and come in different sizes; they seem to each have a fixed number of positions at which they stop. You decide to call the position with the slot 0, and count up for each position it reaches
next.
Furthermore, the discs are spaced out so that after you push the button, one second elapses before the first disc is reached, and one second elapses as the capsule passes from one disk to the one below it. So, if you push the button
at time=100, then the capsule reaches the top disc at time=101, the second disc at time=102, the third disc at time=103, and so on.
The button will only drop a capsule at an integer time - no fractional seconds allowed.
For example, at time=0, suppose you see the following arrangement:
Disc #1 has 5 positions; at time=0, it is at position 4.
Disc #2 has 2 positions; at time=0, it is at position 1.
If you press the button exactly at time=0, the capsule would start to fall; it would reach the first disc at time=1. Since the first disc was at position 4 at time=0, by time=1 it has ticked one position forward. As a five-position
disc, the next position is 0, and the capsule falls through the slot.
Then, at time=2, the capsule reaches the second disc. The second disc has ticked forward two positions at this point: it started at position 1, then continued to position 0, and finally ended up at position 1 again. Because there's
only a slot at position 0, the capsule bounces away.
If, however, you wait until time=5 to push the button, then when the capsule reaches each disc, the first disc will have ticked forward 5+1 = 6 times (to position 0), and the second disc will have ticked forward 5+2 = 7 times (also
to position 0). In this case, the capsule would fall through the discs and come out of the machine.
However, your situation has more than two discs; you've noted their positions in your puzzle input. What is the first time you can press the button to get a capsule?
Your puzzle answer was _________.
--- Part Two ---
After getting the first capsule (it contained a star! what great fortune!), the machine detects your success and begins to rearrange itself.
When it's done, the discs are back in their original configuration as if it were time=0 again, but a new disc with 11 positions and starting at position 0 has appeared exactly one second below the previously-bottom disc.
With this new disc, and counting again starting from time=0 with the configuration in your puzzle input, what is the first time you can press the button to get another capsule?
Although it hasn't changed, you can still get your puzzle input.
Answer: _____________________
References
Visible links
. http://adventofcode.com/
. http://adventofcode.com/2016/about
. http://adventofcode.com/2016/support
. http://adventofcode.com/2016/events
. http://adventofcode.com/2016/settings
. http://adventofcode.com/2016/auth/logout
. http://adventofcode.com/2016
. http://adventofcode.com/2016
. http://adventofcode.com/2016/leaderboard
. http://adventofcode.com/2016/stats
. http://adventofcode.com/2016/sponsors
. http://adventofcode.com/2016/sponsors
. https://youtu.be/IxDoO9oODOk?t=177
. http://adventofcode.com/2016/day/15/input

1
2016/day16/input Normal file
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00111101111101000

89
2016/day16/main.go Normal file
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package main
import (
"bufio"
"fmt"
"log"
"os"
"strconv"
)
// Part 1 Disk Size: 272
// Part 2 Disk Size: 35651584
func main() {
diskSize := 272
if len(os.Args) > 1 {
switch os.Args[1] {
case "-2":
// Do part two
diskSize = 35651584
default:
// manually set disk size
diskSize = atoi(os.Args[1])
}
}
input := stdinToString()
for len(input) < diskSize {
input = dragonCurve(input)
}
input = input[:diskSize]
fmt.Println(checksum(input))
}
func dragonCurve(inp string) string {
return inp + "0" + invert(reverse(inp))
}
func invert(inp string) string {
var ret []rune
for i := range inp {
if inp[i] == '1' {
ret = append(ret, '0')
} else {
ret = append(ret, '1')
}
}
return string(ret)
}
func reverse(inp string) string {
var ret []byte
for i := len(inp) - 1; i >= 0; i-- {
ret = append(ret, inp[i])
}
return string(ret)
}
func checksum(inp string) string {
var chkSum []rune
for i := 0; i < len(inp); i += 2 {
if inp[i] == inp[i+1] {
chkSum = append(chkSum, '1')
} else {
chkSum = append(chkSum, '0')
}
}
if len(chkSum)%2 == 0 {
// Length is even, repeat
return checksum(string(chkSum))
}
return string(chkSum)
}
func stdinToString() string {
var input string
scanner := bufio.NewScanner(os.Stdin)
for scanner.Scan() {
input += scanner.Text()
}
return input
}
func atoi(i string) int {
var ret int
var err error
if ret, err = strconv.Atoi(i); err != nil {
log.Fatal("Invalid Atoi")
}
return ret
}

95
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Advent of Code
--- Day 16: Dragon Checksum ---
You're done scanning this part of the network, but you've left traces of your presence. You need to
overwrite some disks with random-looking data to cover your tracks and update the local security system
with a new checksum for those disks.
For the data to not be suspiscious, it needs to have certain properties; purely random data will be
detected as tampering. To generate appropriate random data, you'll need to use a modified dragon curve.
Start with an appropriate initial state (your puzzle input). Then, so long as you don't have enough data
yet to fill the disk, repeat the following steps:
 Call the data you have at this point "a".
 Make a copy of "a"; call this copy "b".
 Reverse the order of the characters in "b".
 In "b", replace all instances of 0 with 1 and all 1s with 0.
 The resulting data is "a", then a single 0, then "b".
For example, after a single step of this process,
 1 becomes 100.
 0 becomes 001.
 11111 becomes 11111000000.
 111100001010 becomes 1111000010100101011110000.
Repeat these steps until you have enough data to fill the desired disk.
Once the data has been generated, you also need to create a checksum of that data. Calculate the checksum
only for the data that fits on the disk, even if you generated more data than that in the previous step.
The checksum for some given data is created by considering each non-overlapping pair of characters in the
input data. If the two characters match (00 or 11), the next checksum character is a 1. If the characters
do not match (01 or 10), the next checksum character is a 0. This should produce a new string which is
exactly half as long as the original. If the length of the checksum is even, repeat the process until you
end up with a checksum with an odd length.
For example, suppose we want to fill a disk of length 12, and when we finally generate a string of at
least length 12, the first 12 characters are 110010110100. To generate its checksum:
 Consider each pair: 11, 00, 10, 11, 01, 00.
 These are same, same, different, same, different, same, producing 110101.
 The resulting string has length 6, which is even, so we repeat the process.
 The pairs are 11 (same), 01 (different), 01 (different).
 This produces the checksum 100, which has an odd length, so we stop.
Therefore, the checksum for 110010110100 is 100.
Combining all of these steps together, suppose you want to fill a disk of length 20 using an initial state
of 10000:
 Because 10000 is too short, we first use the modified dragon curve to make it longer.
 After one round, it becomes 10000011110 (11 characters), still too short.
 After two rounds, it becomes 10000011110010000111110 (23 characters), which is enough.
 Since we only need 20, but we have 23, we get rid of all but the first 20 characters:
10000011110010000111.
 Next, we start calculating the checksum; after one round, we have 0111110101, which 10 characters long
(even), so we continue.
 After two rounds, we have 01100, which is 5 characters long (odd), so we are done.
In this example, the correct checksum would therefore be 01100.
The first disk you have to fill has length 272. Using the initial state in your puzzle input, what is the
correct checksum?
Your puzzle answer _____________________.
The first half of this puzzle is complete! It provides one gold star: *
--- Part Two ---
The second disk you have to fill has length 35651584. Again using the initial state in your puzzle input,
what is the correct checksum for this disk?
Your puzzle input is still 00111101111101000.
Answer: _____________________
References
Visible links
. http://adventofcode.com/
. http://adventofcode.com/2016/about
. http://adventofcode.com/2016/support
. http://adventofcode.com/2016/events
. http://adventofcode.com/2016/settings
. http://adventofcode.com/2016/auth/logout
. http://adventofcode.com/2016
. http://adventofcode.com/2016
. http://adventofcode.com/2016/leaderboard
. http://adventofcode.com/2016/stats
. http://adventofcode.com/2016/sponsors
. http://adventofcode.com/2016/sponsors
. https://en.wikipedia.org/wiki/Dragon_curve