## \-\-- 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 ``. 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](https://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/Taxicab_geometry), 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 parentheses), 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 `258`. ## \-\-- Part Two \-\-- {#part2} 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? Your puzzle answer was `707`. Both parts of this puzzle are complete! They provide two gold stars: \*\* At this point, you should [return to your Advent calendar](/2017) and try another puzzle. If you still want to see it, you can [get your puzzle input](20/input).