https://adventofcode.com/2021/day/17
Day 17's puzzle was a change of pace from previous puzzles. Instead of
a multiline chunk of text representing the puzzle, we got a single
line, which looked like this:
target area: x=20..30, y=-10..-5
All that would vary would be the numbers.
The puzzle itself was about a quasi-physics simulation process, with
integer positions, velocities and accelerations.
The target area represents a rectangular region on a two dimensional
map. We get to "launch a probe" with some initial x velocity and y
velocity. It starts at time=0 at location x=0, y=0.
Additionally, after each step, x velocity is reduced by 1 in magnitude
until it reaches 0. And, after each step, y velocity is reduced by 1
and will can reach arbitrary negative values.
Additionally, since this is an integer based simulation, our probe
would fall or fly right past the target area if probe's velocity is
sufficient to cross that target area, in one step. For the probe to
hit the target area, it has to be in the target area at the end of one
of it's steps.
For part A, we were asked: how high can our probe go while still
landing in the target area.
The key here, for our puzzle input (which had a negative value range
for y where ymax-ymin was significantly smaller in magnitude than
ymin) was to realize that "maximum height" corresponded to a velocity
moved the probe from position 0 (which is where it would fall to after
being launched upwards) to the bottom of our puzzle input.
Our launch velocity would be 1 unit less than that (since each step
increases the proble's negative velocity by 1).
In other words, given trian=: -:@* >: for our puzzle input the height
for part A would be trian >./|targetY
My implementation for part B was nothing nearly as elegant. My problem
was visualizing the possibilities. Also, I did not document my
thinking adequately, and I mostly used a trial and error approach to
quickly simulate the possibilities. And, I included a simplification
(that Y would be significantly negative) which means that my part B
implementation does not work on the sample data (which would be
sample=: 20 30,.10 _5 ).
Anyways... here's what I had for part B
simy=: {{
Y0=. {:<./m
Y1=. {:>./m
Y=.0
k=.0
y=.-1+y
while. Y >: Y0 do.
Y=.Y+y
if. (Y>:Y0) * Y <: Y1 do. 1 return. end.
y=.y-1
end.
0
}}
simX=: {{
k=.0
X=.0
r=.i.0
X0=: {.<./m
X1=: {.>./m
while.(X <: X1)*y>0 do.
k=.k+1
X=. X+y
if. (X>:X0)*X<:X1 do.
r=.r,k
end.
y=.y-*y
end.
critical=: I. (X0&<: * X1&>:) trian i.X0
if. y=0 do. ~.r,(*#r)#(#~ k&<:) critical return. end.
r
}}
b17=:{{
N=. 0
'X0 Yp0'=. <./|y
'X1 Yp1'=. >./|y
Ypos=. I.y simy"0 i.1+Yp1
Xref=: y simX"0 i.{.1+>./|y
V=: i.0
for_s. i.>./critical do.
V=:V,<I.Xref+./ .= s
end.
Yneg=. 1+Ypos
N=. N+(#Ypos)*#critical
for_Y. -Yneg do.
py=. +/\(|Y)+i.20
steps=. 1+I.(Yp0 <: py)*Yp1>:py
if. (>./critical) > <./steps do.
Vs=. i.0
for_s.steps do.
Vs=.~.Vs,(s {::V)-.0
end.
N=. N+#Vs NB. r=.r,>{Y;Vs
else.
N=.N+#critical
end.
end.
N
}}
That's not a very good approach, though it did work for me...
If someone has an elegant solution to part B, that would be great to see.
Thanks,
--
Raul
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