Day 10 - Advent of Code 2022

--- Day 10: Cathode-Ray Tube ---

You= avoid the ropes, plunge into the river, and swim to shore.

The Elves yell something about meeting back up with them upriver, but th= e river is too loud to tell exactly what they're saying. They finish crossi= ng the bridge and disappear from view.

Situations like this must be why the Elves prioritized getting the commu= nication system on your handheld device working. You pull it out of your pa= ck, but the amount of water slowly draining from a big crack in its screen = tells you it probably won't be of much immediate use.

Unless, that is, you can design a replacement for the device's = video system! It seems to be some kind of cathode-ray tube screen and simple CPU that are both driven= by a precise clock circuit. The clock circuit ticks at a constant= rate; each tick is called a cycle.

Start by figuring out the signal being sent by the CPU. The CPU has a si= ngle register, X, which starts with the value 1. = It supports only two instructions:

addx V takes two cycles to complete. After two cycles, the X register is increased by the value = V. (V can be negative.)

noop takes one cycle to complete. It has no other= effect.

The CPU uses these instructions in a program (your puzzle input) to, som= ehow, tell the screen what to draw.

Consider the following small program:

noop addx 3 addx -5

Execution of this program proceeds as follows:

At the start of the first cycle, the noop instruction begi= ns execution. During the first cycle, X is 1. Aft= er the first cycle, the noop instruction finishes execution, d= oing nothing.

At the start of the second cycle, the addx 3 instruction b= egins execution. During the second cycle, X is still 1.
During the third cycle, X is still 1. After t= he third cycle, the addx 3 instruction finishes execution, set= ting X to 4. At the start of the fourth cycle, the addx -5 instruction = begins execution. During the fourth cycle, X is still 4<= /code>. During the fifth cycle, X is still 4. After t= he fifth cycle, the addx -5 instruction finishes execution, se= tting X to -1.
Maybe you can learn something by looking at the value of the X register throughout execution. For now, consider the signal strengt= h (the cycle number multiplied by the value of the X regi= ster) during the 20th cycle and every 40 cycles after that (that i= s, during the 20th, 60th, 100th, 140th, 180th, and 220th cycles). For example, consider this larger program: addx 15 addx -11 addx 6 addx -3 addx 5 addx -1 addx -8 addx 13 addx 4 noop addx -1 addx 5 addx -1 addx 5 addx -1 addx 5 addx -1 addx 5 addx -1 addx -35 addx 1 addx 24 addx -19 addx 1 addx 16 addx -11 noop noop addx 21 addx -15 noop noop addx -3 addx 9 addx 1 addx -3 addx 8 addx 1 addx 5 noop noop noop noop noop addx -36 noop addx 1 addx 7 noop noop noop addx 2 addx 6 noop noop noop noop noop addx 1 noop noop addx 7 addx 1 noop addx -13 addx 13 addx 7 noop addx 1 addx -33 noop noop noop addx 2 noop noop noop addx 8 noop addx -1 addx 2 addx 1 noop addx 17 addx -9 addx 1 addx 1 addx -3 addx 11 noop noop addx 1 noop addx 1 noop noop addx -13 addx -19 addx 1 addx 3 addx 26 addx -30 addx 12 addx -1 addx 3 addx 1 noop noop noop addx -9 addx 18 addx 1 addx 2 noop noop addx 9 noop noop noop addx -1 addx 2 addx -37 addx 1 addx 3 noop addx 15 addx -21 addx 22 addx -6 addx 1 noop addx 2 addx 1 noop addx -10 noop noop addx 20 addx 1 addx 2 addx 2 addx -6 addx -11 noop noop noop The interesting signal strengths can be determined as follows: During the 20th cycle, register X has the value 21, so the signal strength is 20 * 21 =3D 420. (The 20th cycle = occurs in the middle of the second addx -1, so the value of re= gister X is the starting value, 1, plus all of th= e other addx values up to that point: 1 + 15 - 11 + 6 - 3 + 5 = - 1 - 8 + 13 + 4 =3D 21.) During the 60th cycle, register X has the value 19, so the signal strength is 60 * 19 =3D 1140. During the 100th cycle, register X has the value 18<= /code>, so the signal strength is 100 * 18 =3D 1800.<= /li> During the 140th cycle, register X has the value 21<= /code>, so the signal strength is 140 * 21 =3D 2940.<= /li> During the 180th cycle, register X has the value 16<= /code>, so the signal strength is 180 * 16 =3D 2880.<= /li> During the 220th cycle, register X has the value 18<= /code>, so the signal strength is 220 * 18 =3D 3960.<= /li> 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 15140.--- Part Two --- It seems like the X= register controls the horizontal position of a sprite. Specifically, the sprite is 3 pixel= s 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 w= here 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 dra= ws 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 rig= ht-most pixel in each row is in position 39. Like the CPU, the CRT is tied closely to the clock circuit: the CRT draw= s 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: Cycle 1 -> #######################################= # <- Cycle 40 Cycle 41 -> ######################################## = <- Cycle 80 Cycle 81 -> ######################################## = <- Cycle 120 Cycle 121 -> ######################################## = <- Cycle 160 Cycle 161 -> ######################################## = <- Cycle 200 Cycle 201 -> ######################################## = <- Cycle 240 So, by carefully 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 foll= ows: 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: ...................###.................. Allowing the program to run to completion causes the CRT to produce the = following image: ##..##..##..##..##..##..##..##..##..##.. ###...###...###...###...###...###...###. ####....####....####....####....####.... #####.....#####.....#####.....#####..... ######......######......######......#### #######.......#######.......#######..... Render the image given by your program. What eight capital letters a= ppear on your CRT? Your puzzle answer was BPJAZGAP. Both parts of this puzzle are complete! They provide two gold stars: ** At this point, you should return to your Advent calendar and try another puzz= le. If you still want to see it, you can get your puzz= le input. You can also [Shareo= n Twitter Mastodon<= /a>] this puzzle.