Ryan Palo

Posted on Dec 21, 2018

AoC Day 21: Chronal Conversion

#adventofcode #discuss

...

Honestly? Your guess is as good as mine. It's Day 21, and we're exploiting an integer underflow bug in the space-time continuum? In order to save Christmas? I guess?

I think I'm going to have to spend some time on this one before I even understand exactly what it is that we have to do. All I know is that we are fiddling with the registers in our time machine again so that we don't go back in time indefinitely forever.

Ho. Ho. Ho.

Good luck! 🤶

Top comments (4)

Juan Manuel Gimeno • Dec 21 '18

First I have translated the input into "assembler" using the disassembler I did for day 19, and then I reassembled the solution into C++.

In the code, the program exits when register 5 is equal to register 0.

For part1 the value of register 5 the first time the condition is evaluated, is the solution.
For part2 you have to check for a cycle in the values of register 5 at the condiction. The last one before the cycle is the solution.

So I've added the code to check those two conditions.

#include <iostream>
#include <set>

using namespace std;

int main() {
  set<int> seen;
  int last_r5 = 0;
  int r0, r1, r3, r4, r5;

  l6: 
  r1 = r5 | 65536;
  r5 = 10678677;

  l8:
  r4 = r1 & 255;
  r5 += r4;
  r5 &= 16777215;
  r5 *= 65899;
  r5 &= 16777215;
  if (256 > r1) goto l28;
  r4 = 0;

  l18:
  r3 = r4 + 1;
  r3 *= 256;
  if (r3 > r1) goto l26;
  r4 += 1;
  goto l18;

  l26:
  r1 = r4;
  goto l8;

  l28:
  if (last_r5 == 0) {
    cout << "Part1: " << r5 << endl;
  }
  if (seen.find(r5) == seen.end()) {
    seen.insert(r5);
    last_r5 = r5;
  } else {
    cout << "Part2: " << last_r5 << endl;
    return 0;
  }
  goto l6;
}

E. Choroba • Dec 22 '18

For part 1, I used the same interpreter as in the previous days. I had to convert all registers to numbers, then I just waited until the final condition could be checked.

while ($r[$ip] <= $#code) {
    $_ = 0 + $_ for @r;
    my $line = $code[ $r[$ip] ];
    my ($instruction, @args) = split ' ', $line;
    if ($r[$ip] == 28) {
        say $r[2];
        exit;
    }
    $opcodes{$instruction}->(@args);
    ++$r[$ip];
}

For part 2, the interpreter was too slow, so I rewrote the whole program by hand. Running it then took 2.5 minutes.

#!/usr/bin/perl
use strict;
use feature qw{ say };

my $prev;
my ($r0, $ip, $r2, $r3, $r4, $r5) = (0) x 6;
my %seen;

$r2 = 123;
l1:
$r2 &= 456;

$r2 = $r2 == 72;
goto l1 unless $r2;

$r2 = 0;

l6:
$r5 = 0+$r2 | 65536;
$r2 = 4843319;

l8:
$r4 = 0+$r5 & 255;
$r2 += $r4;
$r2 &= 16777215;
$r2 *= 65899;
$r2 &= 16777215;
$r4 = 256 > $r5;
goto l28 if $r4;

$r4 = 0;

l18:
$r3 = $r4 + 1;
$r3 *= 256;
$r3 = $r3 > $r5;
goto l26 if $r3;

++$r4;
goto l18;

l26:
$r5 = $r4;
goto l8;

l28:
$r4 = $r2 == $r0;
$seen{$r2}++ and end();
$prev = $r2;
end() if $r4;
goto l6;

sub end {
    say $prev;
    exit
}

Tiago Romero • Dec 27 '18

JavaScript solution

My case was exactly the same as @jmgimeno , I used the same analyzer from day 19 and I came down to the same conclusions as him:

for part 1, the solution is the first time the code reaches the check if register[5] is the same as register0
for part 2, I decided that I had to check for loops and the last solution before the 1st loop is the solution.

As usual, I'm gonna omit reader.js which is the same as the other solutions and jump to the point:

21a.js

const { readFile } = require('./reader');
const {
    operations
} = require('./16-common');
const {
    readInput,
    analyzeProgram
} = require ('./19-common');

const runProgram = (ipRegister, program) => {
    let registers = [0, 0, 0, 0, 0, 0];
    const n = program.length;
    let ip = registers[ipRegister];
    while (ip >= 0 && ip < n) {
        const instruction = program[ip];
        const op = operations[instruction[0]];
        registers = op(instruction)(registers);
        ip = ++registers[ipRegister];
        if (ip === 28) {            
            return registers[5];
        }
    }

    return registers[5];
};

(async () => {
    const lines = await readFile('21-input.txt');

    const { ipRegister, program } = readInput(lines);

    //analyzeProgram(ipRegister, program);

    const solution = runProgram(ipRegister, program);

    console.log(`The lowest non-negative integer value for register 0 that causes the program to halt after executing the fewest instructions is ${solution}.`);
})();

21b.js

const { readFile } = require('./reader');
const {
    operations
} = require('./16-common');
const {
    readInput,
    analyzeProgram
} = require ('./19-common');

const runProgram = (ipRegister, program) => {
    let registers = [0, 0, 0, 0, 0, 0];
    const n = program.length;
    let ip = registers[ipRegister];
    const frequency = new Set();
    let lastSolution;
    while (ip >= 0 && ip < n) {
        const instruction = program[ip];
        const op = operations[instruction[0]];
        registers = op(instruction)(registers);
        ip = ++registers[ipRegister];
        if (ip === 28) {            
            if (frequency.has(registers[5])) {
                return lastSolution;
            }
            else  {
                lastSolution = registers[5];
                frequency.add(lastSolution);
            }
        }
    }

    return registers[5];
};

(async () => {
    const lines = await readFile('21-input.txt');

    const { ipRegister, program } = readInput(lines);

    //analyzeProgram(ipRegister, program);

    const solution = runProgram(ipRegister, program);

    console.log(`The lowest non-negative integer value for register 0 that causes the program to halt after executing the most instructions is ${solution}.`);
})();

Grégoire Willmann • Dec 21 '18

Don't know if I'm alone here but seeing all your posts starting with "AoC" over the past few days triggered a reaction in my brain that made me want to play AoE!!!

DEV Community

AoC Day 21: Chronal Conversion

Top comments (4)

JavaScript solution

21a.js

21b.js

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