Part 1
Task
You avoid the ropes, plunge into the river, and swim to shore.
The Elves yell something about meeting back up with them upriver, but the river is too loud to tell exactly what they're saying. They finish crossing the bridge and disappear from view.
Situations like this must be why the Elves prioritized getting the communication system on your handheld device working. You pull it out of your pack, 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 single register,
X, which starts with the value1. It supports only two instructions:
addx Vtakes two cycles to complete. After two cycles, theXregister is increased by the valueV. (Vcan be negative.)
nooptakes one cycle to complete. It has no other effect.The CPU uses these instructions in a program (your puzzle input) to, somehow, tell the screen what to draw.
Consider the following small program:
noop addx 3 addx -5Execution of this program proceeds as follows:
At the start of the first cycle, the
noopinstruction begins execution. During the first cycle,Xis1. After the first cycle, thenoopinstruction finishes execution, doing nothing.At the start of the second cycle, the
addx 3instruction begins execution. During the second cycle,Xis still1.During the third cycle,
Xis still1. After the third cycle, theaddx 3instruction finishes execution, settingXto4.At the start of the fourth cycle, the
addx -5instruction begins execution. During the fourth cycle,Xis still4.During the fifth cycle,
Xis still4. After the fifth cycle, theaddx -5instruction finishes execution, settingXto-1.Maybe you can learn something by looking at the value of the
Xregister throughout execution. For now, consider the signal strength (the cycle number multiplied by the value of theXregister) during the 20th cycle and every 40 cycles after that (that is, 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 noopThe interesting signal strengths can be determined as follows:
During the 20th cycle, register
Xhas the value21, so the signal strength is 20 * 21 = 420. (The 20th cycle occurs in the middle of the secondaddx -1, so the value of registerXis the starting value,1, plus all of the otheraddxvalues up to that point: 1 + 15 - 11 + 6 - 3 + 5 - 1 - 8 + 13 + 4 = 21.)During the 60th cycle, register
Xhas the value19, so the signal strength is 60 * 19 =1140.During the 100th cycle, register
Xhas the value18, so the signal strength is 100 * 18 =1800.During the 140th cycle, register
Xhas the value21, so the signal strength is 140 * 21 =2940.During the 180th cycle, register
Xhas the value16, so the signal strength is 180 * 16 =2880.During the 220th cycle, register
Xhas the value18, so the signal strength is 220 * 18 =3960.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?
Solution
For today's solution, I tried to take the simplest solution. Since I was sick maniple days and needed to catch up on the puzzles.
Our solution would look as follow:
Initialize 2 variables that will represent our register and cycle state.
Map each instruction to its signal strength
sum all strengths
Keep in mind that when calculating the strength we need to know whether to include the new value or not.
Let's start by adding an extension function. This function would return us the cycle signal strength:
private fun Int.getCycleSignal(register: Int) =
if (this % 40 == 20) this * register
else 0
Now, we will map each instruction in our input into its strength. To do so, we will follow the following instructions:
Increase the cycle by 1
get the current signal strength
if this is a
addxoperation:
In the end, we will just sum all signals and receive our results:
fun solvePart1(): Int {
var register = 1
var cycle = 1
return input
.filter { it.isNotBlank() }
.map { instruction ->
cycle++
var signal = cycle.getCycleSignal(register)
if (instruction.startsWith("addx")) {
// addx operation takes 2 cycles
cycle++
val (_, value) = instruction.split(" ")
register += value.toInt()
// Update signal if needed for addx operation
signal += cycle.getCycleSignal(register)
}
signal
}
.sumOf { it }
}
We're done. Let's run our tests:
@Test
fun `Part 1 - Example`() {
val day10 = Day10(exampleInput)
assertEquals(13140, day10.solvePart1())
}
@Test
fun `Part 1 - Real Input`() {
val day10 = Day10(resourceAsList("2022/day10.txt"))
assertEquals(15260, day10.solvePart1())
}
Part 2
Task
It seems like the
Xregister controls the horizontal position of a sprite. Specifically, the sprite is 3 pixels wide, and theXregister 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 position39.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:Cycle 1 -> ######################################## <- Cycle 40 Cycle 41 -> ######################################## <- Cycle 80 Cycle 81 -> ######################################## <- Cycle 120 Cycle 121 -> ######################################## <- Cycle 160 Cycle 161 -> ######################################## <- Cycle 200 Cycle 201 -> ######################################## <- Cycle 240So, 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 follows:
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 appear on your CRT?
Solution
To be honest, I had to read the instructions roughly 10 times until I manage to understand the task. So to clarify that we're all understanding it I will simplify it:
Print the rows column by column
For each column:
I would start by introducing a new function that would return the character to print on the CRT screen. Note that this function also breaks the line after 40 characters.
private fun getCRTCharacter(index: Int, register: Int): String {
val adjustedIndex = index % 40
return when (adjustedIndex) {
in register - 1..register + 1 -> "#"
else -> "."
}.let {
// If that's the 1st character of the new line break line
if (adjustedIndex == 0) System.lineSeparator() + it
else it
}
}
Now, all we need to do is:
iterate over the instructions.
For each instruction, append the relevant character(s)
We will use a StringBuilder object to improve our performance a bit:
fun solvePart2(): String {
val crt = StringBuilder()
var register = 1
var cycle = 1
input.filter { it.isNotBlank() }
.forEach { instruction ->
crt.append(getCRTCharacter(cycle - 1, register))
cycle++
if (instruction.startsWith("addx")) {
crt.append(getCRTCharacter(cycle - 1, register))
cycle++
register += instruction.substringAfter("addx").toInt()
}
}
return crt
.toString()
// Remove the 1st break line
.removePrefix(System.lineSeparator())
}
Note that we need to remove our 1st breaking line just for the tests as otherwise, the response will not match our expected results.
Ok, we're almost done, let's run our tests:
@Test
fun `Part 2 - Example`() {
val exampleOutput = """
##..##..##..##..##..##..##..##..##..##..
###...###...###...###...###...###...###.
####....####....####....####....####....
#####.....#####.....#####.....#####.....
######......######......######......####
#######.......#######.......#######.....
""".trimIndent()
val day10 = Day10(exampleInput)
assertEquals(exampleOutput, day10.solvePart2())
}
@Test
fun `Part 2 - Real Input`() {
val realOutput = """
###...##..#..#.####..##..#....#..#..##..
#..#.#..#.#..#.#....#..#.#....#..#.#..#.
#..#.#....####.###..#....#....#..#.#....
###..#.##.#..#.#....#.##.#....#..#.#.##.
#....#..#.#..#.#....#..#.#....#..#.#..#.
#.....###.#..#.#.....###.####..##...###.
""".trimIndent()
val day10 = Day10(resourceAsList("2022/day10.txt"))
assertEquals(realOutput, day10.solvePart2())
}
All of the code that was presented in the article, along with the utility classes that are used are available in my GitHub account.
See you on the next challenge!
Links
- The code of this post is available here.
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