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 119.

   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: 471

References

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   . https://en.wikipedia.org/wiki/Z-buffering
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