# AoC Day 19: Go With the Flow

On Day 19, we're headed back to the time machine's inner workings to play with more opcodes, and this time, our opcodes can fiddle with the program's instruction pointer!

It seemed like people mostly enjoyed the opcode challenge a few days ago (me included!), so this should be fun! Let's see your solutions.

### Discussion

For part1 I translated the code directly into C with a program in python:

import os
import string

TEMPLATE = string.Template("""
#include <stdio.h>

int regs[] = $registers; void addr(int a, int b, int c) { regs[c] = regs[a] + regs[b]; } void addi(int a, int b, int c) { regs[c] = regs[a] + b; } void mulr(int a, int b, int c) { regs[c] = regs[a] * regs[b]; } void muli(int a, int b, int c) { regs[c] = regs[a] * b; } void banr(int a, int b, int c) { regs[c] = regs[a] & regs[b]; } void bani(int a, int b, int c) { regs[c] = regs[a] & b; } void borr(int a, int b, int c) { regs[c] = regs[a] | regs[b]; } void bori(int a, int b, int c) { regs[c] = regs[a] | b; } void setr(int a, int b, int c) { regs[c] = regs[a]; } void seti(int a, int b, int c) { regs[c] = a; } void gtir(int a, int b, int c) { regs[c] = (a > regs[b]) ? 1 : 0; } void gtri(int a, int b, int c) { regs[c] = (regs[a] > b) ? 1 : 0; } void gtrr(int a, int b, int c) { regs[c] = (regs[a] > regs[b]) ? 1 : 0; } void eqir(int a, int b, int c) { regs[c] = (a == regs[b]) ? 1 : 0; } void eqri(int a, int b, int c) { regs[c] = (regs[a] == b) ? 1 : 0; } void eqrr(int a, int b, int c) { regs[c] = (regs[a] == regs[b]) ? 1 : 0; } int main() { while (1) { switch(regs[$ipreg]) {
$cases default: printf("Execution: %d\\n", regs[0]); return 0; } regs[$ipreg] += 1;
}
}
""")

def compile_line(line):
opname = line[:4]
a, b, c = line[4:].split()
return '%s(%s, %s, %s);' % (opname, a, b, c)

def compile_lines(lines):
cases = []
for lineno, line in enumerate(lines):
cases.append('      case %d: %s break;' % (lineno, compile_line(line)))
return '\n'.join(cases)

def compile_file(fname, registers):
with open(fname, 'r') as program:
registers = str(registers).replace('[', '{').replace(']', '}')
cases = compile_lines(line.strip() for line in program)
return TEMPLATE.substitute(ipreg=ipreg, cases=cases, registers=registers)

def part1(fname):
name, _ = os.path.splitext(fname)
with open(name + '_part1.c', 'w') as compiled:
program = compile_file(fname, [0, 0, 0, 0, 0, 0])
compiled.write(program)

def part2(fname):
name, _ = os.path.splitext(fname)
with open(name + '_part2.c', 'w') as compiled:
program = compile_file(fname, [1, 0, 0, 0, 0, 0])
compiled.write(program)

if __name__ == "__main__":
part1('test_input.txt')
part2('test_input.txt')
part1('input.txt')
part2('input.txt')

The program was able to execute part1 but part2 was taking forever.

Then, after some failed ideas, I did a translator into "assembler" to study the code:

import os

OPS = {'addi': 'r[{c}] = r[{a}] + {b}',
'addr': 'r[{c}] = r[{a}] + r[{b}]',
'muli': 'r[{c}] = r[{a}] * {b}',
'mulr': 'r[{c}] = r[{a}] * r[{b}]',
'seti': 'r[{c}] = {a}',
'setr': 'r[{c}] = r[{a}]',
'eqrr': 'r[{c}] = 1 if r[{a}] == r[{b}] else 0',
'gtrr': 'r[{c}] = 1 if r[{a}] > r[{b}] else 0'
}

def compile_line(line, lineno, ipreg):
opname = line[:4]
a, b, c = map(int, line[4:].split())
if opname == 'addi' and c == ipreg and a == ipreg:
return 'jump #%d' % (b+lineno+1,)
elif opname == 'addr' and c == ipreg and a == ipreg:
return 'jump r[%d] + %d' % (b, lineno+1)
elif opname == 'addr' and c == ipreg and b == ipreg:
return 'jump r[%d] + %d' % (a, lineno+1)
elif opname == 'seti' and c == ipreg:
return 'jump #%d' % (a+1,)
elif opname == 'mulr' and a == ipreg and b == ipreg and c == ipreg:
return 'jump #%d' %(lineno * lineno + 1,)
elif opname == 'addr' and b == ipreg:
return 'r[%d] = r[%d] + %d' %(c, a, lineno)
elif opname == 'addr' and a == ipreg:
return 'r[%d] = r[%d] + %d' % (c, b, lineno)
elif opname == 'mulr' and b == ipreg:
return 'r[%d] = r[%d] * %d' %(c, a, lineno)
elif opname == 'mulr' and a == ipreg:
return 'r[%d] = r[%d] * %d' % (c, b, lineno)
elif opname == 'setr' and a == ipreg:
return 'r[%d] = %d' % (c, lineno)
else:
return OPS[opname].format(a=a, b=b, c=c)

def compile_lines(lines, ipreg):
cases = []
for lineno, line in enumerate(lines):
cases.append("%2d# %s" % (lineno, compile_line(line, lineno, ipreg)))
return '\n'.join(cases)

def compile_file(fname, registers):
with open(fname, 'r') as program:
registers = 'r = ' + str(registers) + '\n'
instructions = compile_lines((line.strip() for line in program), ipreg)
return registers, instructions

def part2(fname):
name, _ = os.path.splitext(fname)
with open(name + '_disassembled.txt', 'w') as compiled:
registers, instructions = compile_file(fname, [1, 0, 0, 0, 0, 0])
compiled.write(registers)
compiled.write(instructions)

if __name__ == "__main__":
part2('test_input.txt')
part2('input.txt')

Then I studied the result:

0# jump #17
1# r[3] = 1
2# r[5] = 1
3# r[4] = r[3] * r[5]
4# r[4] = 1 if r[4] == r[1] else 0
5# jump r[4] + 6
6# jump #8
7# r[0] = r[3] + r[0]
8# r[5] = r[5] + 1
9# r[4] = 1 if r[5] > r[1] else 0
10# jump r[4] + 11
11# jump #3
12# r[3] = r[3] + 1
13# r[4] = 1 if r[3] > r[1] else 0
14# jump r[4] + 15
15# jump #2
16# jump #257
17# r[1] = r[1] + 2
18# r[1] = r[1] * r[1]
19# r[1] = r[1] * 19
20# r[1] = r[1] * 11
21# r[4] = r[4] + 3
22# r[4] = r[4] * 22
23# r[4] = r[4] + 13
24# r[1] = r[1] + r[4]
25# jump r[0] + 26
26# jump #1
27# r[4] = 27
28# r[4] = r[4] * 28
29# r[4] = r[4] + 29
30# r[4] = r[4] * 30
31# r[4] = r[4] * 14
32# r[4] = r[4] * 32
33# r[1] = r[1] + r[4]
34# r[0] = 0
35# jump #1

And, after drawing a flowchart, I reassembled the code into:

if (r0 == 0)
r1 = 915;
else
r1 = 10551315;
r3 = 1;
do {
r5 = 1;
do {
r4 = r3 * r5;
if (r4 == r1)
r0 = r3 + r0;
r5 = r5 + 1;
} while (r3 <= r1);
r3 = r3 + 1;
} while (r5 <= r1);
cout << r0;

The code sums all factors of the number stored in r1 !!!!

So, I factored 10551315 = 3*5*31*22691 and then sum all the divisors, as the program does.

For me it has been the best problem so far !!!

Part 1 was straight forward execution of the program.

For Part 2 I wasn't really sure how we were expected to solve, but I went with rewriting asm code into php and then looked for some possible optimization (early break from the loop in my case).

<?php
$input = require_once 'readFile.php';$ip = explode(' ', array_shift($input))[1][0];$programm = array_values(array_map(function($str) { return explode(' ', preg_replace("/\r|\n/", "",$str));
}, $input)); function execute($r, $programm,$ip) {
$asm = require_once 'opcodes.php'; while (true) { [$opcode, $a,$b, $c] =$programm[$r[$ip]];
$r =$asm[$opcode]($r, $a,$b, $c);$r[$ip]++; if (empty($programm[$r[$ip]])) {
return $r[0]; } } } function executeOptimized($r) {
[$r0,$r1, $r2,$r3, $r4,$r5] = $r;$r1 = 6 * 22 + 6;
$r3 = 2 * 2 * 19 * 11 +$r1;

if ($r0 == 1) {$r1 = (27 * 28 + 29) * 30 * 14 * 32;
$r3 +=$r1;
$r0 = 0; } while($r4 <= $r3) {$r5 = 1;

while ($r5 <=$r3) {
$r1 =$r4 * $r5; if ($r1 > $r3) { break; } if($r1 == $r3) {$r0 += $r4; }$r5++;
}
$r4++; } return$r0;
}

$r = [0, 0, 0, 0, 0, 0]; // echo "Part 1: " . execute($r, $programm,$ip) . "\n";
echo "Part 1: " . executeOptimized($r) . "\n";$r[0] = 1;
echo "Part 2: " . executeOptimized($r) . "\n"; ?> opcodes.php <?php return [ 'addr' => function($r, $a,$b, $c) {$r[$c] =$r[$a] +$r[$b]; return$r;
},
'addi' => function($r,$a, $b,$c) {
$r[$c] = $r[$a] + $b; return$r;
},
'mulr' => function($r,$a, $b,$c) {
$r[$c] = $r[$a] * $r[$b];
return $r; }, 'muli' => function($r, $a,$b, $c) {$r[$c] =$r[$a] *$b;
return $r; }, 'banr' => function($r, $a,$b, $c) {$r[$c] =$r[$a] &$r[$b]; return$r;
},
'bani' => function($r,$a, $b,$c) {
$r[$c] = $r[$a] & $b; return$r;
},
'borr' => function($r,$a, $b,$c) {
$r[$c] = $r[$a] | $r[$b];
return $r; }, 'bori' => function($r, $a,$b, $c) {$r[$c] =$r[$a] |$b;
return $r; }, 'setr' => function($r, $a,$b, $c) {$r[$c] =$r[$a]; return$r;
},
'seti' => function($r,$a, $b,$c) {
$r[$c] = $a; return$r;
},
'gtir' => function($r,$a, $b,$c) {
$r[$c] = $a >$r[$b] ? 1 : 0; return$r;
},
'gtri' => function($r,$a, $b,$c) {
$r[$c] = $r[$a] > $b ? 1 : 0; return$r;
},
'gtrr' => function($r,$a, $b,$c) {
$r[$c] = $r[$a] > $r[$b] ? 1 : 0;
return $r; }, 'eqir' => function($r, $a,$b, $c) {$r[$c] =$a == $r[$b] ? 1 : 0;
return $r; }, 'eqri' => function($r, $a,$b, $c) {$r[$c] =$r[$a] ==$b ? 1 : 0;
return $r; }, 'eqrr' => function($r, $a,$b, $c) {$r[$c] =$r[$a] ==$r[$b] ? 1 : 0; return$r;
},
];
?>

<?php
$file = fopen("input.txt", "r") or exit("Unable to open file"); while(!feof($file)) {
$array[] = fgets($file);
}

fclose($file); return array_filter($array);
?>

Part 2 of this one stumped me. I'm posting my code for part 1 here so I don't feel like a total failure, but I'm pretty bummed. I knew that you had to manually step through the instructions or convert them to some sort of higher level thing so that you could see what they were trying to do slowly and shortcut it. But, when I went through all that work and came up with a wrong answer, I looked here. I was thinking that we were summing all of the numbers up until a certain large number, but I was on the wrong track. After looking here at the answer, it seemed like cheating to finish a solution and get the star.

That's one point for you, Advent.

Anyways, here's my code for part 1 in Python:

"""Day 19: Go with the flow

Figure out control flow with time machine opcodes.
"""

from day16_ops import NAMED_OPERATIONS

def execute(instructions, ip_register, initial_registers=None):
"""Executes a bunch of instructions against 6 registers"""
ip = 0
if initial_registers is None:
registers = [0, 0, 0, 0, 0, 0]
else:
registers = initial_registers[:]
while 0 <= ip < len(instructions):
op, a, b, c = instructions[ip]
registers[ip_register] = ip
registers = op(registers, a, b, c)
ip = registers[ip_register]
ip += 1
return registers

def parse_instructions(text):
"""Parses a text file of instructions into actual instructions"""
lines = text.splitlines()
ip_register = int(lines[0][-1])
instructions = []
for line in lines[1:]:
op_name, a, b, c = line.split()
instructions.append([NAMED_OPERATIONS[op_name], int(a), int(b), int(c)])
return ip_register, instructions

if __name__ == "__main__":
# Part 1
with open("python/data/day19.txt", "r") as f:

ip_register, instructions = parse_instructions(puzzle_input)
results = execute(instructions, ip_register)
print("Value of register 0: ", results[0])

# Part 2

# results = execute(instructions, ip_register, [1, 0, 0, 0, 0, 0])
# print("New value of register 0: ", results[0])
# TOO SLOW

## JavaScript solution

I have a generic solution for Part 1 and a customized solution to my input for Part 2.

For Part 2, because it was taking forever, I decided to print the output and let it run for a few minutes and try to understand how the registers vary and why.

Then I followed the suggestions from @jmgimeno and analyzed my program to understand what should happen and what does the loop try to accomplish, and after some deep thoughts I concluded the register 0 is the sum of the dividers of the number in register 2! So I did that algorithm myself, manually.

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

#### 19-common.js

const {
operations
} = require('./16-common');

const readInput = lines => {
const ipRegex = /^#ip (?<ipRegister>\d)$/; const instructionRegex = /^(?<op>\w+) (?<a>\d+) (?<b>\d+) (?<c>\d+)$/;

let ipRegister;
const program = [];
for (const line of lines) {
if (ipRegister === undefined) {
const match = line.match(ipRegex);
if (match) {
ipRegister = +match.groups.ipRegister;
}
}
else {
const match = line.match(instructionRegex);
if (match) {
const { op, a, b, c } = match.groups;
program.push([op, +a, +b, +c]);
}
}
}
return { ipRegister, program };
};

const runProgram = (ipRegister, program, registers) => {
const n = program.length;
let ip = registers[ipRegister];
let i = 0;
while (ip >= 0 && ip < n) {
const instruction = program[ip];
const op = operations[instruction[0]];
registers = op(instruction)(registers);
ip = ++registers[ipRegister];
i++;
i % 1000000 === 0 && console.log(${i},${registers.join(',')});
}

return registers;
};

module.exports = {
runProgram
};

#### 19a.js

const {
runProgram
} = require ('./19-common');

(async () => {

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

const result = runProgram(ipRegister, program, [0, 0, 0, 0, 0, 0]);

console.log(The state of the registers after executing the test program is ${result}); })(); #### 19b.js const { readFile } = require('./reader'); const { readInput, runProgram } = require ('./19-common'); const analyzeProgram = (ipRegister, program) => { for (let i = 0; i < program.length; i++) { const [op, a, b, c] = program[i]; let analysis; const target = c === ipRegister ? 'jumps ip to ' : [${c}] = ;
if (op === 'addi') { //result[c] = registers[a] + b;
analysis = target + [${a}] +${b};
}
else if (op === 'addr') { //result[c] = registers[a] + registers[b];
analysis = target + [${a}] + [${b}];
}
else if (op === 'seti') { //result[c] = a;
analysis = target + ${a}; } else if (op === 'setr') { //result[c] = registers[a]; analysis = target + [${a}];
}
else if (op === 'muli') { //result[c] = registers[a] * b;
analysis = target + [${a}] *${b};
}
else if (op === 'mulr') { //result[c] = registers[a] * registers[b];
analysis = target + [${a}] * [${b}];
}
else if (op === 'eqrr') { //result[c] = registers[a] === registers[b] ? 1 : 0;
analysis = target + 1 if [${a}] === [${b}] or 0 otherwise;
}
else if (op === 'gtrr') { //result[c] = registers[a] > registers[b] ? 1 : 0;
analysis = target + 1 if [${a}] > [${b}] or 0 otherwise;
}
console.log(${i}:${analysis});
}
};

const runOptimizedProgram = () => {
const number = 10551298;
let sumDivisors = 0;
for (let i = 1; i <= number; i++) {
if (number % i === 0) {
sumDivisors += i;
}
}
return sumDivisors;
}

(async () => {