view 2017/day23/problem @ 35:1d99d733cf13 tip @

day08: replace static foreach with workaround
author Jordi Gutiérrez Hermoso <>
date Tue, 16 Jan 2018 11:28:55 -0500
parents 049fb8e56025
line wrap: on
line source

--- Day 23: Coprocessor Conflagration ---

You decide to head directly to the CPU and fix the printer from there.
As you get close, you find an experimental coprocessor doing so much
work that the local programs are afraid it will halt and catch fire.
This would cause serious issues for the rest of the computer, so you
head in and see what you can do.

The code it's running seems to be a variant of the kind you saw
recently on that tablet. The general functionality seems very similar,
but some of the instructions are different:

    set X Y sets register X to the value of Y.

    sub X Y decreases register X by the value of Y.

    mul X Y sets register X to the result of multiplying the value
    contained in register X by the value of Y.

    jnz X Y jumps with an offset of the value of Y, but only if the
    value of X is not zero. (An offset of 2 skips the next
    instruction, an offset of -1 jumps to the previous instruction,
    and so on.)

    Only the instructions listed above are used. The eight registers
    here, named a through h, all start at 0.

The coprocessor is currently set to some kind of debug mode, which
allows for testing, but prevents it from doing any meaningful work.

If you run the program (your puzzle input), how many times is the mul
instruction invoked?

Your puzzle answer was 3025.

--- Part Two ---

Now, it's time to fix the problem.

The debug mode switch is wired directly to register a. You flip the
switch, which makes register a now start at 1 when the program is

Immediately, the coprocessor begins to overheat. Whoever wrote this
program obviously didn't choose a very efficient implementation.
You'll need to optimize the program if it has any hope of completing
before Santa needs that printer working.

The coprocessor's ultimate goal is to determine the final value left
in register h once the program completes. Technically, if it had
that... it wouldn't even need to run the program.

After setting register a to 1, if the program were to run to
completion, what value would be left in register h?

Your puzzle answer was 915.

Both parts of this puzzle are complete! They provide two gold stars: **