# Advent of Code: Day 7

Source

Part 1: Given a list of definitions of the form 123 -> x, NOT e -> f, and x AND y -> z, with possible operations NOT, AND, OR, LSHIFT, and RSHIFT, find the value of a. Assume all values are 16-bit integers.

This one is actually really cool. It’s basically a full declarative programming language.

monops = {
'NOT': lambda x : ~x & 0xFFFF,
}

binops = {
'AND': operator.and_,
'OR': operator.or_,
'LSHIFT': operator.lshift,
'RSHIFT': operator.rshift,
}

machine = {}

for line in sys.stdin:
line = line.strip()

m = (
re.match(r'(\w+) -> (\w+)', line)
or re.match(r'(\w+) (\w+) (\w+) -> (\w+)', line)
or re.match(r'(\w+) (\w+) -> (\w+)', line)
).groups()

machine[m[-1]] = m[:-1]

def evaluate(register_or_value):
try:
return int(register_or_value)
except:
return run(register_or_value)

def run(register, state = {}):
if not register in state:
command = machine[register]

if len(command) == 1:
value, = command
state[register] = evaluate(value)

elif len(command) == 2:
monop, input = command
state[register] = monops[monop](evaluate(input))

elif len(command) == 3:
input_1, binop, input_2 = command
state[register] = binops[binop](evaluate(input_1), evaluate(input_2))

return state[register]

print(run('a'))

Basically, we have two interesting functions: evaluate and run. Each of those will be applied to any parameters. evaulate will check first if the parameter is an integer, if so return it directly. If not, fall back to run, which is a memoized virtual machine.

If we have the values of any inputs (either because they are numeric or because we’ve already calculated them), it will directly calculate the value for that gate and cache it. If not, it will calculate any recursive outputs it needs (caching them as well), and then calculate it’s own. Through the power of dynamic programming, this will naturally resolve the order the gates need to be run in while still running in O(n) time to the number of gates. Very cool, in my opinion1.

The only other oddity is the definition of NOT. Since Python integers are not 16-bits, doing a bitwise and with 0xFFFF (the maximum 16-bit value) will lock the result into that range.

Part 2: Take the value of a after running part 1 and assign it to b. Run the simulation again.

Due to how I load in the instructions, this is as easy as adding a line of the form 14710 -> b to the end of my input before running it. That will replace the previous command for b. You can do it in a zsh one liner2:

{cat input.txt; (echo "\n" cat input.txt | python part-1.py "-> b")} | python3 part-1.py

1. which would be why I have a blog [return]
2. I’m sure other, inferior3shells would work just as well [return]
3. Just kidding4 [return]
4. Or am I? [return]