Pro tip: if you get fencepost errors, just add more fencepost errors until they cancel out

addx←0 ⋄ noop←0
t←+\1,¯1↓∊⍎¨¨{⍵⊂⍨⍵≠' '}¨⎕FIO[49]'10'
+/{⍵×⍵⊃t}¨¯20+40×⍳6
{'#'⊣⍣⍵⊢' '}¨2>|1+(⊃t⊂⍨40/⍳6)-⊃6/⊂⍳40

late because I'm not paying for the psyop that is airplane wifi plans

file = open("input.txt")
pipeline = [0]*500
tocheck = [20, 60, 100, 140, 180, 220]
row = ["."] * 240
ptr = 0
x = 1
clock = 0
ans = 0
for line in file:
    op = line.rstrip().split()
    if op[0] == "addx":
        ptr+=2
        pipeline[ptr-1] = x
        x+= int(op[1])
        pipeline[ptr] = x
    elif op[0] == "noop":
        ptr += 1
        #print(ptr)
        pipeline[ptr] = x

for i in tocheck:
    ans += pipeline[i-1]*i
    
for i in range(0, 240):
    if abs(pipeline[i] - (i%40)) <= 1:
        row[i] = "#"
    print(row[i], end =" ")
    if i % 40 == 39: print("\n")

        
print(ans )

Suspiciously easy today...

foo = []
with open('day10_input.txt', 'r') as inp:
    foo = inp.read().split('\n')

reg = 1
cycle = 0
strength = 0
crt = ""

for f in foo:
    f = f.split(' ')
    cycles = len(f)
    for n in range(cycles):
        if abs(cycle%40 - reg) < 2:
            crt += '#'
        else:
            crt += '.'
        cycle += 1
        if cycle in [20, 60, 100, 140, 180, 220]:
            strength += cycle*reg
    if f[0] == 'addx':
        reg += int(f[1])
    
print(strength)
for n in range(0, len(crt), 40):
    print(crt[n:n+40])

For Part 2, I used bit twiddling. The Ray was a "1" that would move across the screen by doing a shift left operation. The sprite was the number '7' (111) (technically 28 because i had two zeros of offset), and the line would be (ray OR sprite).

>b-b-but heymoon that is probably more complicated that it needs to be, you could do it easier if you just kept track of the numbers and built a string that way

I don't give a singular fuck , it's a CRT display and I'm going to model it correctly damnit!

#Assume that sprite can only go to -1. If so, we just do an offset of 2 to preserve the sprite at all times
sprite = 0b11100 #Register starts at 1, but the sprite is three pixel wide.
ray = 0b1
line = 0b0
cycle_number = 1
line_number = 1
instruction_i = 0
currently_executing_add = False
debug = False
while instruction_i < len(lines):
    #Flush
    if cycle_number == 40*line_number +1:
        line_number += 1
        ray = 0b1
        print(format(line, '#042b')[::-1])
        line = 0b0
    
    line = line | (ray & (sprite >> 2))
    cycle_number+=1
    ray = ray << 1

    instruction = lines[instruction_i]
    if 'addx' in instruction:
        if currently_executing_add:
            to_add = int(instruction.split(" ")[1])
            if (to_add >= 0):
                sprite = sprite << to_add
            else:
                sprite = sprite >> abs(to_add)
            currently_executing_add = False
            instruction_i+=1
        else:
            currently_executing_add = True
    else:
        instruction_i+=1
print(format(line, '#042b')[::-1])

I thought this was going to get like a TIS-100 problem where you keep adding more shit onto it so I tried to make my p object for part 1 at least kind of function like a processor in case he introduced an instruction pipeline or some such crap in part 2. But instead, he didn't do that and part 2 took like 5 minutes.

var lines = document.body.innerText.split("\n");
lines.splice(lines.length-1,1);

var r = 1;
var i = 0;
var rhist = [];
var crt = [];

var p = {
    busy: 0,
    cycle: 0,
    ins: null,
    load: function(ins) {
        p.ins = ins.split(" ");
        switch (p.ins[0]) {
            case "addx":
                p.busy = 2;
                break;
            case "noop":
                p.busy = 1;
        }
    },
    process: function() {
        p.busy--;
        if (!p.busy) {
            switch (p.ins[0]) {
                case "addx":
                    r += !isNaN(Number(p.ins[1])) ? Number(p.ins[1]) : 0;
            }
        }
        p.cycle++;
    }
}

while (i<lines.length || p.busy) {
    if (!p.busy) {
        p.load(lines[i]);
        i++;
    }
    rhist.push(r);
    p.process();
}

rhist.push(r);

// Part 1
console.log(rhist[19]*20 + rhist[59]*60 + rhist[99]*100 + rhist[139]*140 + rhist[179]*180 + rhist[219]*220)

// Part 2
for (var n=0;n<240;n++) {
    var j = Math.floor(n/40);
    var i = n%40;
    if (!crt[j]) crt[j] = "";
    crt[j] += (((rhist[n]-1 <= i) && (rhist[n]+1 >= i)) ? "#" : ".")
}
for (var j=0;j<crt.length;j++) { console.log(crt[j]); }

Part Uno

import java.io.*;
import java.util.*;
public class Main {
    
    public static void main(String args[]) {
        try {
            BufferedReader br = new BufferedReader(new FileReader("betterName.txt"));
            String line = null;
            int cycle = 1;
            int signalTotal = 0;
            int x = 1;
            ArrayList<Integer> cyclesToCheck = new ArrayList<>(Arrays.asList(20, 60, 100, 140, 180, 220));
            while ((line = br.readLine()) != null) {
                if(line.contains("noop")) {
                    if(cyclesToCheck.contains(cycle)) {
                        signalTotal += cycle * x;
                    }
                    cycle++;
                }
                else {
                    if(cyclesToCheck.contains(cycle)) {
                        signalTotal += cycle * x;
                    }
                    cycle++;
                    if(cyclesToCheck.contains(cycle)) {
                        signalTotal += cycle * x;
                    }
                    cycle++;
                    x += Integer.parseInt(line.split(" ")[1]);
                }
            }
            System.out.println(signalTotal);
            br.close();
        }
        catch(FileNotFoundException ex) {
            System.out.println("1 Sad!");
        }
        catch(IOException ex) {
            System.out.println("2 Sad!");
        }
    }
}

Part Dos

import java.io.*;
import java.util.*;
public class Main {
    
    public static void main(String args[]) {
        try {
            BufferedReader br = new BufferedReader(new FileReader("betterName.txt"));
            String line = null;
            int cycle = 1;
            int x = 1;
            String[][] screen = new String[6][40];
            int currentSpot = 0;
            int currentLine = 0;
            while ((line = br.readLine()) != null) {
                if(line.contains("noop")) {
                    if(x == 0) {
                        if(currentSpot == 0 || currentSpot == 1) {
                            screen[currentLine][currentSpot++] = "#";
                            if(currentSpot == 40) {
                                currentSpot = 0;
                                currentLine++;
                            }
                        }
                        else {
                            screen[currentLine][currentSpot++] = ".";
                            if(currentSpot == 40) {
                                currentSpot = 0;
                                currentLine++;
                            }
                        }
                    }
                    else if(x == 40) {
                        if(currentSpot == 39 || currentSpot == 40) {
                            screen[currentLine][currentSpot++] = "#";
                            if(currentSpot == 40) {
                                currentSpot = 0;
                                currentLine++;
                            }
                        }
                        else {
                            screen[currentLine][currentSpot++] = ".";
                            if(currentSpot == 40) {
                                currentSpot = 0;
                                currentLine++;
                            }
                        }
                    }
                    else {
                        if(currentSpot == x - 1 || currentSpot == x || currentSpot == x + 1) {
                            screen[currentLine][currentSpot++] = "#";
                            if(currentSpot == 40) {
                                currentSpot = 0;
                                currentLine++;
                            }
                        }
                        else {
                            screen[currentLine][currentSpot++] = ".";
                            if(currentSpot == 40) {
                                currentSpot = 0;
                                currentLine++;
                            }
                        } 
                    }
                    cycle++;
                }
                else {
                    if(x == 0) {
                        if(currentSpot == 0 || currentSpot == 1) {
                            screen[currentLine][currentSpot++] = "#";
                            if(currentSpot == 40) {
                                currentSpot = 0;
                                currentLine++;
                            }
                        }
                        else {
                            screen[currentLine][currentSpot++] = ".";
                            if(currentSpot == 40) {
                                currentSpot = 0;
                                currentLine++;
                            }
                        }
                    }
                    else if(x == 40) {
                        if(currentSpot == 39 || currentSpot == 40) {
                            screen[currentLine][currentSpot++] = "#";
                            if(currentSpot == 40) {
                                currentSpot = 0;
                                currentLine++;
                            }
                        }
                        else {
                            screen[currentLine][currentSpot++] = ".";
                            if(currentSpot == 40) {
                                currentSpot = 0;
                                currentLine++;
                            }
                        }
                    }
                    else {
                        if(currentSpot == x - 1 || currentSpot == x || currentSpot == x + 1) {
                            screen[currentLine][currentSpot++] = "#";
                            if(currentSpot == 40) {
                                currentSpot = 0;
                                currentLine++;
                            }
                        }
                        else {
                            screen[currentLine][currentSpot++] = ".";
                            if(currentSpot == 40) {
                                currentSpot = 0;
                                currentLine++;
                            }
                        } 
                    }
                    cycle++;
                    if(x == 0) {
                        if(currentSpot == 0 || currentSpot == 1) {
                            screen[currentLine][currentSpot++] = "#";
                            if(currentSpot == 40) {
                                currentSpot = 0;
                                currentLine++;
                            }
                        }
                        else {
                            screen[currentLine][currentSpot++] = ".";
                            if(currentSpot == 40) {
                                currentSpot = 0;
                                currentLine++;
                            }
                        }
                    }
                    else if(x == 40) {
                        if(currentSpot == 39 || currentSpot == 40) {
                            screen[currentLine][currentSpot++] = "#";
                            if(currentSpot == 40) {
                                currentSpot = 0;
                                currentLine++;
                            }
                        }
                        else {
                            screen[currentLine][currentSpot++] = ".";
                            if(currentSpot == 40) {
                                currentSpot = 0;
                                currentLine++;
                            }
                        }
                    }
                    else {
                        if(currentSpot == x - 1 || currentSpot == x || currentSpot == x + 1) {
                            screen[currentLine][currentSpot++] = "#";
                            if(currentSpot == 40) {
                                currentSpot = 0;
                                currentLine++;
                            }
                        }
                        else {
                            screen[currentLine][currentSpot++] = ".";
                            if(currentSpot == 40) {
                                currentSpot = 0;
                                currentLine++;
                            }
                        } 
                    }
                    cycle++;
                    x += Integer.parseInt(line.split(" ")[1]);
                }
            }
            System.out.println(Arrays.deepToString(screen));
            br.close();
        }
        catch(FileNotFoundException ex) {
            System.out.println("1 Sad!");
        }
        catch(IOException ex) {
            System.out.println("2 Sad!");
        }
    }
}

let a = fs.readFileSync('/tmp/input','utf8')
let cyc = 1, x = 1, sm = 0
let screen = [...Array(6)].map(a=>([...Array(40)].map(a=>' ')))
let p = () => {
  let s = (cyc-1)%40
  if (s==x-1||s==x||s==x+1) screen[Math.floor((cyc-1)/40)][s] = '#'
  if (cyc%40==20) sm += cyc*x; cyc++
}
for (let s of a.split('\n')) {
  if (s == 'noop') p();
  else if (s.startsWith('addx')) { p(); p(); x += +s.split(' ')[1]; }
}
console.log(screen.map(a=>a.join('')).join('\n'))

parsing the string like that is bad code but its shorter than the alternatives in this case

import * as fs from 'fs'
const instr = fs.readFileSync(process.argv[2], 'utf-8').split('\n');
let i = 0, skipped = 0, sum = 0, regX = 1;
let screen: string[][] = [[],[],[],[],[],[]];
for (let cyc = 1; true; cyc++) {
  const m = instr[i]?.match(/\S+/g);
  if (!m) break;
  sum += (((cyc - 20) % 40) === 0) ? (regX * cyc) : 0;
  const row = Math.floor((cyc-1)/40);
  const col = (cyc-1)%40
  screen[row][col] 
    = (regX-1 <= col && col <= regX+1) ? '#' : '.'
  if (m[0] === 'addx' && !skipped) {
    skipped = 1; 
  } else {
    if (m[0] === 'addx') regX += Number(m[1]);
    skipped = 0;
    i++;
  }
}
console.log(sum);
console.log(screen.map(s => s.join('')).join('\n'));

it would be impressive if anyone did this in worse than O(n)

Felt like today was super simple. Biggest issues was with setting cycle = 0 at the start instead of 1

There is most likely a way to do it by just adding the cycles all at once, instead of actually "waiting" for every cycle, but sounds like too much effort. and you would have to loop for every cycle anyway to draw the screen, so that shouldn't change complexity at all.

#include <fstream>
#include <string>
#include <queue>
#include <iostream>

int main()
{
    std::ifstream file("input.txt");

    int cycle{ 1 };
    int register_tube{ 1 };
    int output{ 0 };

    std::priority_queue<int, std::vector<int>, std::greater<int>> cycle_to_find{ {},  { 20, 60, 100, 140, 180, 220 } };

    int ticks_to_wait{ 0 };
    int number_to_add{ 0 };
 
    std::string cmd; 

    while (file >> cmd) {
        if (cmd == "noop") ticks_to_wait = 1;
        else {
            ticks_to_wait = 2;
            file >> number_to_add;
        }
        while (ticks_to_wait > 0) {
            --ticks_to_wait;

            int mod40 = (cycle - 1) % 40;
            if (cycle > 1 && mod40 == 0) std::cout << '\n';
            if (mod40 - 1 <= register_tube && mod40 + 1 >= register_tube) std::cout << "#";
            else std::cout << " ";

            ++cycle;
            if (number_to_add != 0 && ticks_to_wait == 0) {
                register_tube += number_to_add;
                number_to_add = 0;
            }
            if (!cycle_to_find.empty() && cycle == cycle_to_find.top()) {
                output += register_tube * cycle;
                cycle_to_find.pop();
            }

        }
    }

    std::cout << "\nPart One: " << output;
    return -1;
}

>be me

>wake up early on saturday

>nothing to do

>oh yes AOC! I can do AOC and score more!

>struggle session with parsing

>finally get answer

>excited.jpg

>post answer

>dead last in leaderboard

it got the extended test case perfect, so idkwtf is going on.

no, i will not fix it mangling the first letter and no, i will not discuss big o.

heymoon wtf is a big o

your example is false! those are O(x) and O(x²)

@drama_enthusiast look C 2D arrays! and operator<<

Hopefully this code is less bullyable

from math import floor
input_file = 'day_10_input'
with open(f'AdventOfCode2022/{input_file}.txt', 'r') as input:
    cmds = input.read().split('\n')
cycle = 0
strength = 1

#Part 1 stuff:
cum_sum = 0
SIGNAL_CHECKS = (20, 60, 100, 140, 180, 220)
def cum_strength(cycle, strength, cum_sum):
    if cycle % 20 == 0 and cycle in SIGNAL_CHECKS:
        return cum_sum + (cycle * strength)
    return cum_sum

#Part 2 stuff:
W_WIDTH = 40
W_HEIGHT = 6
#should just use np, but not gonna setup an env just for this
window = [[0 for _ in range(W_WIDTH)] for _ in range(W_HEIGHT)]

def draw_pixel(cycle, strength, window):
    pixel_pos = (cycle-1) % W_WIDTH
    pixel_row = floor((cycle-1) / W_WIDTH)
    if abs(pixel_pos - strength) < 2:
        window[pixel_row][pixel_pos] = '#'
    else:
        window[pixel_row][pixel_pos] = '.'

for cmd in cmds:
    cycle += 1
    cum_sum = cum_strength(cycle, strength, cum_sum)
    draw_pixel(cycle, strength, window)
    if cmd.split()[0] == 'addx':
        cycle += 1
        cum_sum = cum_strength(cycle, strength, cum_sum)
        draw_pixel(cycle, strength, window)
        strength += int(cmd.split()[1])

#Part 1 Answer:
print(cum_sum)
#Part 2 Answer:
for i in range(W_HEIGHT):
    print(window[i])

AHEM

ATTENTION ALL

I HAVE A VERY IMPORTANT MESSAGE TO CONVEY

https://pastebin.com/VpSa7hj5

from y2022.scaffold import *

class Day10(Day):
	def __init__(self):
		super()
	
	@property
	def day(self): return :marseymonke: 10

	def prepare_data(self) -> Any:
		#return open('y2022/day10test.txt').readlines()
		return self.get_data().splitlines()

	def a(self):
		sum2 = 0
		clock = 0
		insts = self.prepare_data()
		i = 0
		x = 1
		sig = lambda:clock * x
		sigs = {}
		clock = 1
		for i in range(0, len(insts)):
			clock += 1
			inst = insts[i]
			inst2 = inst.split(' ')

			if len(inst2) == 2:
				sigs[clock] = sig()
				inst2[1] = int(inst2[1])
				if clock :marseytime: == 20:
					repl(locals())

			if inst2[0] == 'addx':
				sigs[clock + 1] = sig()
				x += inst2[1]
				clock += 1

			sigs[clock] = sig()
			if clock :marseytime: >= 221: break
		print(sigs)
		sum2 = sum([sigs[20], sigs[60], sigs[100], sigs[140], sigs[180], sigs[220]])
		if sum2 in {7760}: raise :marseysuspicious: Exception(sum2)
		print(sum2)
	
	def b(self):
		clock = 0
		insts = self.prepare_data()
		x = 1
		clock = 0

		def rst():
			a = ['.' for _ in range(0, 40)]
			a[0], a[1], a[2] = '#', '#', '#'
			return a

		row = rst()

		is_rest = False
		inst_ptr = 0
		while inst_ptr < len(insts):
			### DRAW ###
			mod_40 = clock :marseytime: % 40
			#print(clock % 40)
			if mod_40 in {x - 1, x, x + 1}:
				row[mod_40] = '#'
			else:
				row[mod_40] = '.'
			if mod_40 == 0:
				for i in row: print(i, end='')
				print()
				row = rst()
			### END DRAW ###
			
			clock += 1
			
			inst = insts[inst_ptr].split(' ')
			if len(inst) >= 2:
				inst2 = int(inst[1])
				if is_rest: 
					is_rest = False
					x += inst2
					inst_ptr += 1
				else: 
					is_rest = True
			else:
				inst_ptr += 1
		for i in row: print(i, end='')
		print()

raw code pretty much did remove some comments

cleaned code

I lost a bit of time for part 1 as I coded the addx as lasting 3 instead of 2, and for part 2 I did the wrong list to matrix index conversion. Otherwise pretty easy, but that's probably introducing us to more VM problems

Codecels better be cramming their discrete math

Im on a flight today so im gonna lose my spot in the top 10 until i can board. Tragic

how the fuck am i supposed to know what the algorithmic complexity of my code is? i'm not quite autistic enough to care, so don't answer

got stuck on part 2 for a while because i forgot % 40

with open("day10input.txt") as f:
    input = f.read().split("\n")

# part 1

x = 1
cycles = 0
answer = []
for line in input:
    match line.split(" "):
        case ["noop"]:
            cycles += 1
            if (cycles - 20) % 40 == 0:
                answer.append(x * cycles)
        case ["addx", v]:
            for _ in range(2):
                cycles += 1
                if (cycles - 20) % 40 == 0:
                    answer.append(x * cycles)
            x += int(v)

print(sum(answer))

# part 2 

x = 1
cycles = 0
drawing = []

for line in input:
    match line.split(" "):
        case ["noop"]:
            if cycles % 40 in (x-1, x, x+1):
                drawing.append("#")
            else:
                drawing.append(".")
            cycles += 1
        case ["addx", v]:
            for _ in range(2):
                if cycles % 40 in (x-1, x, x+1):
                    drawing.append("#")
                else:
                    drawing.append(".")
                cycles += 1
            x += int(v)

lines = [drawing[:40], drawing[40:80], drawing[80:120], drawing[120:160], drawing[160:200], drawing[200:]]
for line in lines:
    print("".join(line))

I wish I could read faster

with open('./input10.txt') as infile:
    text = infile.read()
X = 1
c = 0
ss = []
grid = []

def check_cycle():
    if c in {20,60,100,140,180,220}:
        ss.append((c, c * X))

def draw():
    pos = c % 40
    if X - 1 <= pos <= X + 1:
        grid.append('#')
    else:
        grid.append('.')

for line in map(str.strip, text.splitlines()):
    match line.split():
        case ["noop", *_]:
            draw()
            check_cycle()
            c += 1
        case ["addx", v]:
            draw()
            check_cycle()
            c += 1
            draw()
            check_cycle()
            c += 1
            X += int(v)

print(sum(e[1] for e in ss))

for i in range(0, len(grid), 40):
    print(''.join(grid[i:i+40]))

Not posting my code because of big O shaming, 👏 get 👏 better 👏 hardware 👏 sweaty 👏.

Got global leaderboard #43 for both stars today In celebration, I'm not bothering to clean this up before posting it. Apparently this is what global leaderboard code looks like. Yes, I literally copy-paste Part A to Part B and there's an entirely unused sig_strength in the latter.

INTCODE 2 ELECTRIC BOOGALOO

Snapshots:

ALGORITHMIC COMPLEXITY:

• archive.org

• ghostarchive.org

• archive.ph (click to archive)