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2018 Day 19 Complete!

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Advent of Code
--- Day 19: Go With The Flow ---
With the Elves well on their way constructing the North Pole base, you turn
your attention back to understanding the inner workings of programming the
device.
You can't help but notice that the device's opcodes don't contain any flow
control like jump instructions. The device's manual goes on to explain:
"In programs where flow control is required, the instruction pointer can be
bound to a register so that it can be manipulated directly. This way,
setr/seti can function as absolute jumps, addr/addi can function as relative
jumps, and other opcodes can cause truly fascinating effects."
This mechanism is achieved through a declaration like #ip 1, which would
modify register 1 so that accesses to it let the program indirectly access the
instruction pointer itself. To compensate for this kind of binding, there are
now six registers (numbered 0 through 5); the five not bound to the
instruction pointer behave as normal. Otherwise, the same rules apply as the
last time you worked with this device.
When the instruction pointer is bound to a register, its value is written to
that register just before each instruction is executed, and the value of that
register is written back to the instruction pointer immediately after each
instruction finishes execution. Afterward, move to the next instruction by
adding one to the instruction pointer, even if the value in the instruction
pointer was just updated by an instruction. (Because of this, instructions
must effectively set the instruction pointer to the instruction before the one
they want executed next.)
The instruction pointer is 0 during the first instruction, 1 during the
second, and so on. If the instruction pointer ever causes the device to
attempt to load an instruction outside the instructions defined in the
program, the program instead immediately halts. The instruction pointer starts
at 0.
It turns out that this new information is already proving useful: the CPU in
the device is not very powerful, and a background process is occupying most of
its time. You dump the background process' declarations and instructions to a
file (your puzzle input), making sure to use the names of the opcodes rather
than the numbers.
For example, suppose you have the following program:
#ip 0
seti 5 0 1
seti 6 0 2
addi 0 1 0
addr 1 2 3
setr 1 0 0
seti 8 0 4
seti 9 0 5
When executed, the following instructions are executed. Each line contains the
value of the instruction pointer at the time the instruction started, the
values of the six registers before executing the instructions (in square
brackets), the instruction itself, and the values of the six registers after
executing the instruction (also in square brackets).
ip=0 [0, 0, 0, 0, 0, 0] seti 5 0 1 [0, 5, 0, 0, 0, 0]
ip=1 [1, 5, 0, 0, 0, 0] seti 6 0 2 [1, 5, 6, 0, 0, 0]
ip=2 [2, 5, 6, 0, 0, 0] addi 0 1 0 [3, 5, 6, 0, 0, 0]
ip=4 [4, 5, 6, 0, 0, 0] setr 1 0 0 [5, 5, 6, 0, 0, 0]
ip=6 [6, 5, 6, 0, 0, 0] seti 9 0 5 [6, 5, 6, 0, 0, 9]
In detail, when running this program, the following events occur:
• The first line (#ip 0) indicates that the instruction pointer should be
bound to register 0 in this program. This is not an instruction, and so
the value of the instruction pointer does not change during the processing
of this line.
• The instruction pointer contains 0, and so the first instruction is
executed (seti 5 0 1). It updates register 0 to the current instruction
pointer value (0), sets register 1 to 5, sets the instruction pointer to
the value of register 0 (which has no effect, as the instruction did not
modify register 0), and then adds one to the instruction pointer.
• The instruction pointer contains 1, and so the second instruction, seti 6
0 2, is executed. This is very similar to the instruction before it: 6 is
stored in register 2, and the instruction pointer is left with the value
2.
• The instruction pointer is 2, which points at the instruction addi 0 1 0.
This is like a relative jump: the value of the instruction pointer, 2, is
loaded into register 0. Then, addi finds the result of adding the value in
register 0 and the value 1, storing the result, 3, back in register 0.
Register 0 is then copied back to the instruction pointer, which will
cause it to end up 1 larger than it would have otherwise and skip the next
instruction (addr 1 2 3) entirely. Finally, 1 is added to the instruction
pointer.
• The instruction pointer is 4, so the instruction setr 1 0 0 is run. This
is like an absolute jump: it copies the value contained in register 1, 5,
into register 0, which causes it to end up in the instruction pointer. The
instruction pointer is then incremented, leaving it at 6.
• The instruction pointer is 6, so the instruction seti 9 0 5 stores 9 into
register 5. The instruction pointer is incremented, causing it to point
outside the program, and so the program ends.
What value is left in register 0 when the background process halts?
Your puzzle answer was 888.
--- Part Two ---
A new background process immediately spins up in its place. It appears
identical, but on closer inspection, you notice that this time, register 0
started with the value 1.
What value is left in register 0 when this new background process halts?
Your puzzle answer was 10708992.
Both parts of this puzzle are complete! They provide two gold stars: **
References
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