2017 Day 20 Complete!

master
Brian Buller 5 years ago
parent ef98db2967
commit 3d7ac83d3f
  1. 215
      2017/day20/day20.go
  2. 1000
      2017/day20/input
  3. 121
      2017/day20/problem
  4. 2
      2017/day20/testinput

@ -0,0 +1,215 @@
package main
import (
"bufio"
"fmt"
"log"
"math"
"os"
"strconv"
"strings"
)
var particles map[int]*Particle
var confidenceThreshold int
func main() {
inp := StdinToStrings()
particles = make(map[int]*Particle)
for i := range inp {
particles[i] = NewParticle(inp[i], i)
}
confidenceThreshold = 1000
doPart := 2
if len(os.Args) > 1 {
if os.Args[1] == "-1" {
doPart = 1
}
}
switch doPart {
case 1:
part1(inp)
case 2:
part2(inp)
}
}
func part1(inp []string) {
var ticks, confidence, last int
cPart := -1
for {
cDist := 10000000000
ticks++
for i := range particles {
particles[i].Tick()
if particles[i].GetDistance() < cDist {
cPart = i
cDist = particles[i].GetDistance()
}
}
if cPart == last {
confidence++
} else {
confidence = 0
last = cPart
}
if confidence >= confidenceThreshold {
break
}
}
fmt.Println("After", ticks, "the closest is", cPart)
}
func part2(inp []string) {
var ticks, lastNum, confidence int
for {
ticks++
for i := range particles {
particles[i].Tick()
}
var collisions []int
for k1, v1 := range particles {
for k2, v2 := range particles {
if k1 == k2 {
continue
}
if v1.Collides(v2) {
collisions = AddToSlice(collisions, k1)
collisions = AddToSlice(collisions, k2)
break
}
}
}
for i := len(collisions) - 1; i >= 0; i-- {
delete(particles, collisions[i])
}
if len(particles) == lastNum {
confidence++
} else {
confidence = 0
lastNum = len(particles)
}
if confidence >= confidenceThreshold {
break
}
}
fmt.Println("After", ticks, "there are", len(particles), "particles left")
}
func AddToSlice(sl []int, v int) []int {
for i := range sl {
if sl[i] == v {
return sl
}
}
sl = append(sl, v)
return sl
}
type Coordinate struct {
X, Y, Z int
}
func NewCoordinate(inp string) Coordinate {
inp = inp[1 : len(inp)-1]
pts := strings.Split(inp, ",")
px, py, pz := Atoi(pts[0]), Atoi(pts[1]), Atoi(pts[2])
return Coordinate{X: px, Y: py, Z: pz}
}
func (c *Coordinate) toString() string {
return fmt.Sprintf("<%d,%d,%d>", c.X, c.Y, c.Z)
}
type Particle struct {
Index int
Position Coordinate
Velocity Coordinate
Acceleration Coordinate
LastPosition Coordinate
}
func NewParticle(inp string, idx int) *Particle {
ret := new(Particle)
ret.Index = idx
pts := strings.Split(inp, " ")
for i := range pts {
keyVal := strings.Split(pts[i], "=")
if keyVal[1][len(keyVal[1])-1] == ',' {
keyVal[1] = keyVal[1][:len(keyVal[1])-1]
}
switch keyVal[0] {
case "p":
ret.Position = NewCoordinate(keyVal[1])
case "v":
ret.Velocity = NewCoordinate(keyVal[1])
case "a":
ret.Acceleration = NewCoordinate(keyVal[1])
}
}
ret.LastPosition = NewCoordinate(ret.Position.toString())
return ret
}
func (p *Particle) Collides(tst *Particle) bool {
if p.Position.X == tst.Position.X &&
p.Position.Y == tst.Position.Y &&
p.Position.Z == tst.Position.Z {
return true
}
return false
}
func (p *Particle) String() string {
return fmt.Sprintf(
"p=<%d,%d,%d>, v=<%d,%d,%d>, a=<%d,%d,%d>",
p.Position.X, p.Position.Y, p.Position.Z,
p.Velocity.X, p.Velocity.Y, p.Velocity.Z,
p.Acceleration.X, p.Acceleration.Y, p.Acceleration.Z,
)
}
func (p *Particle) Tick() {
p.LastPosition = NewCoordinate(p.Position.toString())
p.Velocity.X += p.Acceleration.X
p.Velocity.Y += p.Acceleration.Y
p.Velocity.Z += p.Acceleration.Z
p.Position.X += p.Velocity.X
p.Position.Y += p.Velocity.Y
p.Position.Z += p.Velocity.Z
}
func (p *Particle) IsDone() bool {
return p.GetDistance() == p.GetLastDistance()
}
func (p *Particle) GetDistance() int {
return GetAbs(p.Position.X) + GetAbs(p.Position.Y) + GetAbs(p.Position.Z)
}
func (p *Particle) GetLastDistance() int {
return int(math.Abs(float64(p.LastPosition.X + p.LastPosition.Y + p.LastPosition.Z)))
}
func StdinToStrings() []string {
var input []string
scanner := bufio.NewScanner(os.Stdin)
for scanner.Scan() {
input = append(input, scanner.Text())
}
return input
}
func GetAbs(v int) int {
return int(math.Abs(float64(v)))
}
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
}

File diff suppressed because it is too large Load Diff

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Advent of Code
--- Day 20: Particle Swarm ---
Suddenly, the GPU contacts you, asking for help. Someone has
asked it to simulate too many particles, and it won't be able
to finish them all in time to render the next frame at this
rate.
It transmits to you a buffer (your puzzle input) listing each
particle in order (starting with particle 0, then particle 1,
particle 2, and so on). For each particle, it provides the X,
Y, and Z coordinates for the particle's position (p), velocity
(v), and acceleration (a), each in the format <X,Y,Z>.
Each tick, all particles are updated simultaneously. A
particle's properties are updated in the following order:
* Increase the X velocity by the X acceleration.
* Increase the Y velocity by the Y acceleration.
* Increase the Z velocity by the Z acceleration.
* Increase the X position by the X velocity.
* Increase the Y position by the Y velocity.
* Increase the Z position by the Z velocity.
Because of seemingly tenuous rationale involving z-buffering,
the GPU would like to know which particle will stay closest to
position <0,0,0> in the long term. Measure this using the
Manhattan distance, which in this situation is simply the sum
of the absolute values of a particle's X, Y, and Z position.
For example, suppose you are only given two particles, both of
which stay entirely on the X-axis (for simplicity). Drawing
the current states of particles 0 and 1 (in that order) with
an adjacent a number line and diagram of current X positions
(marked in parenthesis), the following would take place:
p=< 3,0,0>, v=< 2,0,0>, a=<-1,0,0> -4 -3 -2 -1 0 1 2 3 4
p=< 4,0,0>, v=< 0,0,0>, a=<-2,0,0> (0)(1)
p=< 4,0,0>, v=< 1,0,0>, a=<-1,0,0> -4 -3 -2 -1 0 1 2 3 4
p=< 2,0,0>, v=<-2,0,0>, a=<-2,0,0> (1) (0)
p=< 4,0,0>, v=< 0,0,0>, a=<-1,0,0> -4 -3 -2 -1 0 1 2 3 4
p=<-2,0,0>, v=<-4,0,0>, a=<-2,0,0> (1) (0)
p=< 3,0,0>, v=<-1,0,0>, a=<-1,0,0> -4 -3 -2 -1 0 1 2 3 4
p=<-8,0,0>, v=<-6,0,0>, a=<-2,0,0> (0)
At this point, particle 1 will never be closer to <0,0,0> than
particle 0, and so, in the long run, particle 0 will stay
closest.
Which particle will stay closest to position <0,0,0> in the
long term?
Your puzzle answer was _______.
The first half of this puzzle is complete! It provides one
gold star: *
--- Part Two ---
To simplify the problem further, the GPU would like to remove
any particles that collide. Particles collide if their
positions ever exactly match. Because particles are updated
simultaneously, more than two particles can collide at the
same time and place. Once particles collide, they are removed
and cannot collide with anything else after that tick.
For example:
p=<-6,0,0>, v=< 3,0,0>, a=< 0,0,0>
p=<-4,0,0>, v=< 2,0,0>, a=< 0,0,0> -6 -5 -4 -3 -2 -1 0 1 2 3
p=<-2,0,0>, v=< 1,0,0>, a=< 0,0,0> (0) (1) (2) (3)
p=< 3,0,0>, v=<-1,0,0>, a=< 0,0,0>
p=<-3,0,0>, v=< 3,0,0>, a=< 0,0,0>
p=<-2,0,0>, v=< 2,0,0>, a=< 0,0,0> -6 -5 -4 -3 -2 -1 0 1 2 3
p=<-1,0,0>, v=< 1,0,0>, a=< 0,0,0> (0)(1)(2) (3)
p=< 2,0,0>, v=<-1,0,0>, a=< 0,0,0>
p=< 0,0,0>, v=< 3,0,0>, a=< 0,0,0>
p=< 0,0,0>, v=< 2,0,0>, a=< 0,0,0> -6 -5 -4 -3 -2 -1 0 1 2 3
p=< 0,0,0>, v=< 1,0,0>, a=< 0,0,0> X (3)
p=< 1,0,0>, v=<-1,0,0>, a=< 0,0,0>
------destroyed by collision------
------destroyed by collision------ -6 -5 -4 -3 -2 -1 0 1 2 3
------destroyed by collision------ (3)
p=< 0,0,0>, v=<-1,0,0>, a=< 0,0,0>
In this example, particles 0, 1, and 2 are simultaneously
destroyed at the time and place marked X. On the next tick,
particle 3 passes through unharmed.
How many particles are left after all collisions are resolved?
Although it hasn't changed, you can still get your puzzle
input.
Answer: _____________________
References
Visible links
. http://adventofcode.com/
. http://adventofcode.com/2017/about
. http://adventofcode.com/2017/support
. http://adventofcode.com/2017/events
. http://adventofcode.com/2017/settings
. http://adventofcode.com/2017/auth/logout
. http://adventofcode.com/2017
. http://adventofcode.com/2017
. http://adventofcode.com/2017/leaderboard
. http://adventofcode.com/2017/stats
. http://adventofcode.com/2017/sponsors
. http://adventofcode.com/2017/sponsors
. https://en.wikipedia.org/wiki/Z-buffering
. https://en.wikipedia.org/wiki/Taxicab_geometry
. http://adventofcode.com/2017/day/20/input

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p=<3,0,0>, v=<2,0,0>, a=<-1,0,0>
p=<4,0,0>, v=<0,0,0>, a=<-2,0,0>
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