2022 day 10: Bash parts 1 & 2

2022/RESULTS.txt

@@ -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

2022/day10/README.org

@@ -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: **

2022/day10/aoc.bash

@@ -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