60 lines
2.5 KiB
Markdown
60 lines
2.5 KiB
Markdown
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## \-\-- Day 20: Particle Swarm \-\--
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Suddenly, the GPU contacts you, asking for
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help.
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Someone has asked it to simulate *too many particles*, and it won\'t be
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able to finish them all in time to render the next frame at this rate.
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It transmits to you a buffer (your puzzle input) listing each particle
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in order (starting with particle `0`, then particle `1`, particle `2`,
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and so on). For each particle, it provides the `X`, `Y`, and `Z`
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coordinates for the particle\'s position (`p`), velocity (`v`), and
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acceleration (`a`), each in the format `<X,Y,Z>`.
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Each tick, all particles are updated simultaneously. A particle\'s
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properties are updated in the following order:
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- Increase the `X` velocity by the `X` acceleration.
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- Increase the `Y` velocity by the `Y` acceleration.
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- Increase the `Z` velocity by the `Z` acceleration.
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- Increase the `X` position by the `X` velocity.
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- Increase the `Y` position by the `Y` velocity.
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- Increase the `Z` position by the `Z` velocity.
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Because of seemingly tenuous rationale involving
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[z-buffering](https://en.wikipedia.org/wiki/Z-buffering), the GPU would
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like to know which particle will stay closest to position `<0,0,0>` in
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the long term. Measure this using the [Manhattan
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distance](https://en.wikipedia.org/wiki/Taxicab_geometry), which in this
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situation is simply the sum of the absolute values of a particle\'s `X`,
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`Y`, and `Z` position.
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For example, suppose you are only given two particles, both of which
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stay entirely on the X-axis (for simplicity). Drawing the current states
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of particles `0` and `1` (in that order) with an adjacent a number line
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and diagram of current `X` positions (marked in parentheses), the
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following would take place:
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p=< 3,0,0>, v=< 2,0,0>, a=<-1,0,0> -4 -3 -2 -1 0 1 2 3 4
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p=< 4,0,0>, v=< 0,0,0>, a=<-2,0,0> (0)(1)
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p=< 4,0,0>, v=< 1,0,0>, a=<-1,0,0> -4 -3 -2 -1 0 1 2 3 4
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p=< 2,0,0>, v=<-2,0,0>, a=<-2,0,0> (1) (0)
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p=< 4,0,0>, v=< 0,0,0>, a=<-1,0,0> -4 -3 -2 -1 0 1 2 3 4
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p=<-2,0,0>, v=<-4,0,0>, a=<-2,0,0> (1) (0)
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p=< 3,0,0>, v=<-1,0,0>, a=<-1,0,0> -4 -3 -2 -1 0 1 2 3 4
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p=<-8,0,0>, v=<-6,0,0>, a=<-2,0,0> (0)
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At this point, particle `1` will never be closer to `<0,0,0>` than
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particle `0`, and so, in the long run, particle `0` will stay closest.
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*Which particle will stay closest to position `<0,0,0>`* in the long
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term?
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To begin, [get your puzzle input](20/input).
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Answer:
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