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Robert Mion
Robert Mion

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Coprocessor Conflagration

Advent of Code 2017 Day 23

Try the simulator to walk through your puzzle input

  • I built a simulator
  • It doesn't generate the correct answer for me for Part 2
  • But it does show how a valid program runs

Part 1

  1. Solve for X where X =...
  2. mul instruction?
  3. Referring to Day 18 to ensure full understanding
  4. Copy-paste-refactor
  5. After troubleshooting clumsy mistakes, success!

Solve for X where X =...

the amount of times the mul instruction is invoked

mul instruction?

  • My input is a program containing instructions
  • There are four types of instructions: set, sub, mul and jnz
  • Each instruction works with two operands: a register and a value
  • The pattern of each instruction resembles: instruction X Y
  • The registers are one of 8 variables, labeled a-h
  • All eight registers initially store the value 0

Working through Advent of Code backwards seems to present me with an advantage here, since this feels very similar to all the Intcode puzzles from later years.

For that reason, I feel confident I'll be able to solve at least Part 1.

However, I sense I need to refer to Day 18's puzzle. The instructions mention it as the first encounter of this type of puzzle.

Referring to Day 18 to ensure full understanding

  • Flash forward: I solved Part 1 for Day 18
  • The missing context was expected: the program runs until an address pointer refers outside of the bounds of the instruction list
  • Good news: I wrote an algorithm that accounts for eight instructions...and can now just copy it, remove some instructions, and update where appropriate!


  • I'll start by reusing my algorithm from Day 18 Part 1
  • Then I'll remove several unused instructions
  • And finally, adjust the jump instruction based on today's puzzle's definition

After troubleshooting clumsy mistakes, success!

  • My code from Day 18 Part 1 was missing some important possibilities that my code from Part 2 incorporated

Logic like:

  • Accounting for strings or numbers as either X or Y operands
  • The case where a key may not be in registers at the moment a jnz instruction runs
  • There is no longer a case where an instruction could receive only one operand

Alas, after resolving these issues... algorithm generated the correct answer!

Part 2

  1. Solve for X where X =...
  2. What's so unique about a and h?
  3. Running the program with a set to 1
  4. Building a simulator to step through at my own pace

Solve for X where X =...

the value left in register h if the program were to run to completion after setting register a to 1

What's so unique about a and h?

  • Both a and h are featured in only one instruction
  • a is featured in a jnz instruction - which acts depending on whether a is a value other than 0
  • h is featured in a sub instruction - which subtracts some amount from the value stored in register h

Running the program with a set to 1


  • Setting a to 1 initially causes the program to process four important instructions that set registers b thru f
  • b becomes a six-digit positive integer
  • g becomes nearly the same integer, but negative
  • The program cycles through a subset of instructions until g climbs its way back up to 0
  • Only to be reset to that six-digit negative integer by way of a jump back a few instructions later
  • f must be 0 for the program to avoid skipping the only line that adjusts h
  • g must be 0 for the program to reach that f jump instruction
  • But g will never be 0 as long as b causes g to reset to some non-0 number in the instruction prior to the jump

Building a simulator to step through at my own pace

  • I reused the code from my Day 18 Part 2 simulator
  • After a lot of deletions, I had it setup to run Day 23 Part 2, with a starting as 1

I quickly saw the parts that looped endlessly:

  1. jnz g -8
  2. jnz g -13

Due to this earlier instruction:
sub b -100000

And several instances of this instruction:
set g b

Both jnz instructions initialized loops that took hundreds of thousands of times to bring g to 0.

However, I remain baffled.

Why isn't the answer 1?

The last few instructions of my puzzle input are as follows:

jnz g -13
jnz f 2
sub h -1
set g b
sub g c
jnz g 2
jnz 1 3
sub b -17
jnz 1 -23
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  • If g is 0, it would jump to the next line
  • If f is 0, it would jump to the next line
  • I see no other way to arrive at this instruction that subtracts -1 from h
  • Since no other line sets h, it must start at and be 0 when this instruction runs
  • Causing it to store the value 1
  • Assuming the next two instructions run
  • The next jnz would need to be skipped in order to arrive at the next instruction
  • Because jnz 1 3 seems like the only way to cause this program to halt

And once it halts, h would have the value 1.

Sadly, that's not the correct answer.

So, as in Day 18 Part 2:

  • I'm stumped

Celebrating my accomplishments

  • I solved Part 1!
  • I built a simulator to see how the program runs...specifically to understand Part 2!


  • I didn't understand what the instructions meant by optimize the program: did it mean to change the input?
  • I was wrong about what I felt was the only possible value that h could store once the program terminated
  • Thus, I did not solve Part 2

I sense that for Part 2 of Days 18 and 23 I am missing some hidden trick to make the program arrive at a particular instruction with the right value stored in a register.

Maybe Reddit's Solution Megathread can reveal the answer.

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