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day08: replace static foreach with workaround
author Jordi Gutiérrez Hermoso <jordigh@octave.org>
date Tue, 16 Jan 2018 11:28:55 -0500
parents 049fb8e56025
children
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--- Day 18: Duet ---

You discover a tablet containing some strange assembly code labeled
simply "Duet". Rather than bother the sound card with it, you decide
to run the code yourself. Unfortunately, you don't see any
documentation, so you're left to figure out what the instructions mean
on your own.

It seems like the assembly is meant to operate on a set of registers
that are each named with a single letter and that can each hold a
single integer. You suppose each register should start with a value of
0.

There aren't that many instructions, so it shouldn't be hard to figure
out what they do. Here's what you determine:

    snd X plays a sound with a frequency equal to the value of X.

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

    add X Y increases 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.

    mod X Y sets register X to the remainder of dividing the value
    contained in register X by the value of Y (that is, it sets X to
    the result of X modulo Y).

    rcv X recovers the frequency of the last sound played, but only
    when the value of X is not zero. (If it is zero, the command does
    nothing.)

    jgz X Y jumps with an offset of the value of Y, but only if the
    value of X is greater than zero. (An offset of 2 skips the next
    instruction, an offset of -1 jumps to the previous instruction,
    and so on.)


Many of the instructions can take either a register (a single letter)
or a number. The value of a register is the integer it contains; the
value of a number is that number.

After each jump instruction, the program continues with the
instruction to which the jump jumped. After any other instruction, the
program continues with the next instruction. Continuing (or jumping)
off either end of the program terminates it.

For example:

set a 1
add a 2
mul a a
mod a 5
snd a
set a 0
rcv a
jgz a -1
set a 1
jgz a -2

    The first four instructions set a to 1, add 2 to it, square it,
    and then set it to itself modulo 5, resulting in a value of 4.

    Then, a sound with frequency 4 (the value of a) is played.

    After that, a is set to 0, causing the subsequent rcv and jgz
    instructions to both be skipped (rcv because a is 0, and jgz
    because a is not greater than 0).

    Finally, a is set to 1, causing the next jgz instruction to
    activate, jumping back two instructions to another jump, which
    jumps again to the rcv, which ultimately triggers the recover
    operation.


At the time the recover operation is executed, the frequency of the
last sound played is 4.

What is the value of the recovered frequency (the value of the most
recently played sound) the first time a rcv instruction is executed
with a non-zero value?

Your puzzle answer was 7071.

--- Part Two ---

As you congratulate yourself for a job well done, you notice that the
documentation has been on the back of the tablet this entire time.
While you actually got most of the instructions correct, there are a
few key differences. This assembly code isn't about sound at all -
it's meant to be run twice at the same time.

Each running copy of the program has its own set of registers and
follows the code independently - in fact, the programs don't even
necessarily run at the same speed. To coordinate, they use the send
(snd) and receive (rcv) instructions:

    snd X sends the value of X to the other program. These values wait
    in a queue until that program is ready to receive them. Each
    program has its own message queue, so a program can never receive
    a message it sent.

    rcv X receives the next value and stores it in register X. If no
    values are in the queue, the program waits for a value to be sent
    to it. Programs do not continue to the next instruction until they
    have received a value. Values are received in the order they are
    sent.


Each program also has its own program ID (one 0 and the other 1); the
register p should begin with this value.

For example:

snd 1
snd 2
snd p
rcv a
rcv b
rcv c
rcv d

Both programs begin by sending three values to the other. Program 0
sends 1, 2, 0; program 1 sends 1, 2, 1. Then, each program receives a
value (both 1) and stores it in a, receives another value (both 2) and
stores it in b, and then each receives the program ID of the other
program (program 0 receives 1; program 1 receives 0) and stores it in
c. Each program now sees a different value in its own copy of register
c.

Finally, both programs try to rcv a fourth time, but no data is
waiting for either of them, and they reach a deadlock. When this
happens, both programs terminate.

It should be noted that it would be equally valid for the programs to
run at different speeds; for example, program 0 might have sent all
three values and then stopped at the first rcv before program 1
executed even its first instruction.

Once both of your programs have terminated (regardless of what caused
them to do so), how many times did program 1 send a value?

Your puzzle answer was 8001.

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