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+--- 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 150.
+
+--- 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?
+
+Your puzzle answer was 657.
+
+Both parts of this puzzle are complete! They provide two gold stars: **