Solve 2024:17 p1-2 "Chronospatial Computer"

Slow burner.

First part was a straightforward implementation
of a runner, the code is extracted to `org_version()`.

Part 2 was way more tricky. Like many others, trying
to understand the program felt necessary. This
intepretation is this:

    """Python version of puzzle input program."""
    a = <unknown>
    b, c = 0, 0
    while a:
        b = (a % 8) ^ 1
        c = a // (2**b)
        b = (b ^ 5) ^ c
        a = a // (2**3)
        print(b % 8)

Some clarification here:

- the value is octagonal. 1-8 is the key.
- The value of reg a is _high_. It is not meant to
be brute forced.

The idea, which is not originally mine, is to go
from right to left. This code can be used to try
out some patterns:

    while True:
        python_version(input("Provide A:"))

Here, it was apparent a=4 gives the last digit of
my puzzle input. a=37 (4 * 8 + 3) gives the last
2 digits. a=222 (37 * 8 + 6) gives the last 3
digits, and so on.

Knowing the program could be reconstructed like
this, the first code halted at wrong values. Turns
out some steps give more than 1 possible scenario.

The code was therefore in need of a queue.
This commit is contained in:
Anders Englöf Ytterström 2024-12-21 01:20:13 +01:00
parent ded5c4f28c
commit 14520a4625

View file

@ -0,0 +1,75 @@
from collections import deque
from output import ints
def solve(data):
R, program = data.split("\n\n")
R = {k: int(v) for k, v in zip("abc", ints(R))}
program = ints(program)
P = ",".join(map(str, program))
p1 = org_version(R["a"], program)
p2 = float("inf")
Q = deque([1])
while Q:
x = Q.popleft()
for i in range(8):
o = org_version(x + i, program)
if P.endswith(o):
if o == P:
p2 = min(p2, x + i)
else:
Q.append((x + i) * 8)
return p1, p2
def org_version(A, program):
R = {"a": A, "b": 0, "c": 0}
def combo(v):
return v if v < 4 else R[chr(93 + v)]
i = 0
out = []
while i < len(program):
cl = program[i + 1]
match program[i]:
case 0:
R["a"] = R["a"] // (2 ** combo(cl))
i += 2
case 1:
a, b = R["b"], cl
R["b"] = a ^ b
i += 2
case 2:
R["b"] = combo(cl) % 8
i += 2
case 3:
i = (i + 2) if R["a"] == 0 else cl
case 4:
a, b = R["b"], R["c"]
R["b"] = a ^ b
i += 2
case 5:
out.append(combo(cl) % 8)
i += 2
case 6:
R["b"] = R["a"] // (2 ** combo(cl))
i += 2
case 7:
R["c"] = R["a"] // (2 ** combo(cl))
i += 2
return ",".join(map(str, out))
if __name__ == "__main__":
with open("./input/17.txt", "r") as f:
inp = f.read().strip()
p1, p2 = solve(inp)
print(p1)
print(p2)