2022 day 10: Bash parts 1 & 2
This commit is contained in:
@@ -195,3 +195,23 @@ aoc.bash: res=2376
|
||||
aoc-c: res=2376
|
||||
time: 0:00.00 real, 0.00 user, 0.00 sys
|
||||
context-switch: 0+1, page-faults: 0+131
|
||||
|
||||
=========================================
|
||||
================= day10 =================
|
||||
=========================================
|
||||
|
||||
+++++++++++++++++ part 1
|
||||
aoc.bash: res=13220
|
||||
time: 0:00.00 real, 0.00 user, 0.00 sys
|
||||
context-switch: 1+1, page-faults: 0+261
|
||||
|
||||
+++++++++++++++++ part 2
|
||||
aoc.bash: res=
|
||||
###..#..#..##..#..#.#..#.###..####.#..#.
|
||||
#..#.#..#.#..#.#.#..#..#.#..#.#....#.#..
|
||||
#..#.#..#.#..#.##...####.###..###..##...
|
||||
###..#..#.####.#.#..#..#.#..#.#....#.#..
|
||||
#.#..#..#.#..#.#.#..#..#.#..#.#....#.#..
|
||||
#..#..##..#..#.#..#.#..#.###..####.#..#.
|
||||
time: 0:00.01 real, 0.00 user, 0.00 sys
|
||||
context-switch: 0+1, page-faults: 0+263
|
||||
|
||||
@@ -229,3 +229,161 @@ The sum of these signal strengths is =13140=.
|
||||
|
||||
Find the signal strength during the 20th, 60th, 100th, 140th, 180th, and
|
||||
220th cycles. /What is the sum of these six signal strengths?/
|
||||
|
||||
Your puzzle answer was =13220=.
|
||||
|
||||
** --- Part Two ---
|
||||
It seems like the =X= register controls the horizontal position of a
|
||||
[[https://en.wikipedia.org/wiki/Sprite_(computer_graphics)][sprite]].
|
||||
Specifically, the sprite is 3 pixels wide, and the =X= register sets the
|
||||
horizontal position of the /middle/ of that sprite. (In this system,
|
||||
there is no such thing as "vertical position": if the sprite's
|
||||
horizontal position puts its pixels where the CRT is currently drawing,
|
||||
then those pixels will be drawn.)
|
||||
|
||||
You count the pixels on the CRT: 40 wide and 6 high. This CRT screen
|
||||
draws the top row of pixels left-to-right, then the row below that, and
|
||||
so on. The left-most pixel in each row is in position =0=, and the
|
||||
right-most pixel in each row is in position =39=.
|
||||
|
||||
Like the CPU, the CRT is tied closely to the clock circuit: the CRT
|
||||
draws /a single pixel during each cycle/. Representing each pixel of the
|
||||
screen as a =#=, here are the cycles during which the first and last
|
||||
pixel in each row are drawn:
|
||||
|
||||
#+begin_example
|
||||
Cycle 1 -> ######################################## <- Cycle 40
|
||||
Cycle 41 -> ######################################## <- Cycle 80
|
||||
Cycle 81 -> ######################################## <- Cycle 120
|
||||
Cycle 121 -> ######################################## <- Cycle 160
|
||||
Cycle 161 -> ######################################## <- Cycle 200
|
||||
Cycle 201 -> ######################################## <- Cycle 240
|
||||
#+end_example
|
||||
|
||||
So, by [[https://en.wikipedia.org/wiki/Racing_the_Beam][carefully]]
|
||||
[[https://www.youtube.com/watch?v=sJFnWZH5FXc][timing]] the CPU
|
||||
instructions and the CRT drawing operations, you should be able to
|
||||
determine whether the sprite is visible the instant each pixel is drawn.
|
||||
If the sprite is positioned such that one of its three pixels is the
|
||||
pixel currently being drawn, the screen produces a /lit/ pixel (=#=);
|
||||
otherwise, the screen leaves the pixel /dark/ (=.=).
|
||||
|
||||
The first few pixels from the larger example above are drawn as follows:
|
||||
|
||||
#+begin_example
|
||||
Sprite position: ###.....................................
|
||||
|
||||
Start cycle 1: begin executing addx 15
|
||||
During cycle 1: CRT draws pixel in position 0
|
||||
Current CRT row: #
|
||||
|
||||
During cycle 2: CRT draws pixel in position 1
|
||||
Current CRT row: ##
|
||||
End of cycle 2: finish executing addx 15 (Register X is now 16)
|
||||
Sprite position: ...............###......................
|
||||
|
||||
Start cycle 3: begin executing addx -11
|
||||
During cycle 3: CRT draws pixel in position 2
|
||||
Current CRT row: ##.
|
||||
|
||||
During cycle 4: CRT draws pixel in position 3
|
||||
Current CRT row: ##..
|
||||
End of cycle 4: finish executing addx -11 (Register X is now 5)
|
||||
Sprite position: ....###.................................
|
||||
|
||||
Start cycle 5: begin executing addx 6
|
||||
During cycle 5: CRT draws pixel in position 4
|
||||
Current CRT row: ##..#
|
||||
|
||||
During cycle 6: CRT draws pixel in position 5
|
||||
Current CRT row: ##..##
|
||||
End of cycle 6: finish executing addx 6 (Register X is now 11)
|
||||
Sprite position: ..........###...........................
|
||||
|
||||
Start cycle 7: begin executing addx -3
|
||||
During cycle 7: CRT draws pixel in position 6
|
||||
Current CRT row: ##..##.
|
||||
|
||||
During cycle 8: CRT draws pixel in position 7
|
||||
Current CRT row: ##..##..
|
||||
End of cycle 8: finish executing addx -3 (Register X is now 8)
|
||||
Sprite position: .......###..............................
|
||||
|
||||
Start cycle 9: begin executing addx 5
|
||||
During cycle 9: CRT draws pixel in position 8
|
||||
Current CRT row: ##..##..#
|
||||
|
||||
During cycle 10: CRT draws pixel in position 9
|
||||
Current CRT row: ##..##..##
|
||||
End of cycle 10: finish executing addx 5 (Register X is now 13)
|
||||
Sprite position: ............###.........................
|
||||
|
||||
Start cycle 11: begin executing addx -1
|
||||
During cycle 11: CRT draws pixel in position 10
|
||||
Current CRT row: ##..##..##.
|
||||
|
||||
During cycle 12: CRT draws pixel in position 11
|
||||
Current CRT row: ##..##..##..
|
||||
End of cycle 12: finish executing addx -1 (Register X is now 12)
|
||||
Sprite position: ...........###..........................
|
||||
|
||||
Start cycle 13: begin executing addx -8
|
||||
During cycle 13: CRT draws pixel in position 12
|
||||
Current CRT row: ##..##..##..#
|
||||
|
||||
During cycle 14: CRT draws pixel in position 13
|
||||
Current CRT row: ##..##..##..##
|
||||
End of cycle 14: finish executing addx -8 (Register X is now 4)
|
||||
Sprite position: ...###..................................
|
||||
|
||||
Start cycle 15: begin executing addx 13
|
||||
During cycle 15: CRT draws pixel in position 14
|
||||
Current CRT row: ##..##..##..##.
|
||||
|
||||
During cycle 16: CRT draws pixel in position 15
|
||||
Current CRT row: ##..##..##..##..
|
||||
End of cycle 16: finish executing addx 13 (Register X is now 17)
|
||||
Sprite position: ................###.....................
|
||||
|
||||
Start cycle 17: begin executing addx 4
|
||||
During cycle 17: CRT draws pixel in position 16
|
||||
Current CRT row: ##..##..##..##..#
|
||||
|
||||
During cycle 18: CRT draws pixel in position 17
|
||||
Current CRT row: ##..##..##..##..##
|
||||
End of cycle 18: finish executing addx 4 (Register X is now 21)
|
||||
Sprite position: ....................###.................
|
||||
|
||||
Start cycle 19: begin executing noop
|
||||
During cycle 19: CRT draws pixel in position 18
|
||||
Current CRT row: ##..##..##..##..##.
|
||||
End of cycle 19: finish executing noop
|
||||
|
||||
Start cycle 20: begin executing addx -1
|
||||
During cycle 20: CRT draws pixel in position 19
|
||||
Current CRT row: ##..##..##..##..##..
|
||||
|
||||
During cycle 21: CRT draws pixel in position 20
|
||||
Current CRT row: ##..##..##..##..##..#
|
||||
End of cycle 21: finish executing addx -1 (Register X is now 20)
|
||||
Sprite position: ...................###..................
|
||||
#+end_example
|
||||
|
||||
Allowing the program to run to completion causes the CRT to produce the
|
||||
following image:
|
||||
|
||||
#+begin_example
|
||||
##..##..##..##..##..##..##..##..##..##..
|
||||
###...###...###...###...###...###...###.
|
||||
####....####....####....####....####....
|
||||
#####.....#####.....#####.....#####.....
|
||||
######......######......######......####
|
||||
#######.......#######.......#######.....
|
||||
#+end_example
|
||||
|
||||
Render the image given by your program. /What eight capital letters
|
||||
appear on your CRT?/
|
||||
|
||||
Your puzzle answer was =RUAKHBEK=.
|
||||
|
||||
Both parts of this puzzle are complete! They provide two gold stars: **
|
||||
|
||||
68
2022/day10/aoc.bash
Executable file
68
2022/day10/aoc.bash
Executable file
@@ -0,0 +1,68 @@
|
||||
#!/usr/bin/env bash
|
||||
#
|
||||
# aoc.bash: Advent of Code 2022, day 10
|
||||
#
|
||||
# Copyright (C) 2022 Bruno Raoult ("br")
|
||||
# Licensed under the GNU General Public License v3.0 or later.
|
||||
# Some rights reserved. See COPYING.
|
||||
#
|
||||
# You should have received a copy of the GNU General Public License along with this
|
||||
# program. If not, see <https://www.gnu.org/licenses/gpl-3.0-standalone.html>.
|
||||
#
|
||||
# SPDX-License-Identifier: GPL-3.0-or-later <https://spdx.org/licenses/GPL-3.0-or-later.html>
|
||||
|
||||
. common.bash
|
||||
|
||||
declare -i cycle=0 regx=1
|
||||
|
||||
draw() {
|
||||
local -i pos
|
||||
|
||||
if (( cycle % 40 )); then
|
||||
(( pos = cycle % 40 - regx - 1, pos = pos < 0? -pos: pos ))
|
||||
(( pos < 2 )) && res+="#" || res+="."
|
||||
else
|
||||
res+=.$'\n'
|
||||
fi
|
||||
}
|
||||
|
||||
tick() {
|
||||
(( cycle ++ ))
|
||||
if (( part == 1 )); then
|
||||
(( (cycle + 20) % 40 )) || (( res += cycle * regx ))
|
||||
else
|
||||
draw
|
||||
fi
|
||||
}
|
||||
|
||||
do_noop() {
|
||||
tick
|
||||
}
|
||||
|
||||
do_add() {
|
||||
tick
|
||||
tick
|
||||
(( regx += $1 ))
|
||||
}
|
||||
|
||||
parse() {
|
||||
while read -r instr value; do
|
||||
# not too hacky, to prepare for next puzzles ;-)
|
||||
case "$instr" in
|
||||
"addx")
|
||||
do_add "$value"
|
||||
;;
|
||||
"noop")
|
||||
do_noop
|
||||
;;
|
||||
esac
|
||||
done
|
||||
}
|
||||
|
||||
solve() {
|
||||
# remove last '\n', add starting '\n'
|
||||
(( part == 2 )) && res=$'\n'${res::-1}
|
||||
}
|
||||
|
||||
main "$@"
|
||||
exit 0
|
||||
Reference in New Issue
Block a user