From 7498e792262c690de3c07ab4a0bd57c06fe24c33 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?Anders=20Engl=C3=B6f=20Ytterstr=C3=B6m?= <yttan@fastmail.se>
Date: Sat, 29 Nov 2025 12:38:32 +0100
Subject: [PATCH] Solve 2022 Day 22 pt 1-2

This is how far I got the Dec 2022 run in Elixir.

Had no idea what so ever for pt 2, so I had to
solve it by the solutions thread in the subreddit.

At first, I wondered if circular values would be the
case. It was not.

Apparently, modular arithmetics are what I lack
knowledge about.

Not entirely sure what to call the operator, it is
either a greatest common divisor or a least common
nominator.

What it actually does is to reduce the numbers by
modulo for pt 2, which will keep them low and faster
to compute.

My math is rusty.
---
 2022-python/output/day_11.py | 74 ++++++++++++++++++++++++++++++++++++
 1 file changed, 74 insertions(+)
 create mode 100644 2022-python/output/day_11.py

diff --git a/2022-python/output/day_11.py b/2022-python/output/day_11.py
new file mode 100644
index 0000000..197a2d5
--- /dev/null
+++ b/2022-python/output/day_11.py
@@ -0,0 +1,74 @@
+from copy import deepcopy
+from collections import defaultdict
+from output import ints
+
+
+def solve(data):
+    M = []
+    for md in data.split("\n\n"):
+        m = {}
+        starting_items, operation, divisor, if_true, if_false = md.splitlines()[1:]
+        m["items"] = ints(starting_items)
+        m["if_false"] = ints(if_false)[0]
+        m["if_true"] = ints(if_true)[0]
+        m["divisor"] = ints(divisor)[0]
+        m["operation_values"] = operation.split()[-3:]
+        M.append(m)
+    p12 = []
+    # Part 2 resolver.
+    #
+    # Not sure what to call this, so I go for Greatest Common
+    # Divisor. Had to look for hints at subreddit to find it out.
+    # It basically fasten up the computation speed by keeping
+    # the numbers low.
+    #
+    # Apparently, this is common Modular arithmetic:
+    # https://en.wikipedia.org/wiki/Modular_arithmetic
+    gcd = 1
+    for d in [m["divisor"] for m in M]:
+        gcd *= d
+    for rounds, wld in [(20, True), (10_000, False)]:
+        B = defaultdict(int)
+        MM = deepcopy(M)
+        for _ in range(rounds):
+            for i, m in enumerate(MM):
+                while m["items"]:
+                    wl = m["items"].pop(0)
+                    B[i] += 1
+                    x, y, z = m["operation_values"]
+                    x = int(wl) if x == "old" else int(x)
+                    z = int(wl) if z == "old" else int(z)
+                    match y:
+                        case "+":
+                            wl = x + z
+                        case "*":
+                            wl = x * z
+                    if wld:
+                        wl = wl // 3
+                    else:
+                        wl = wl % gcd
+                    if wl % m["divisor"] == 0:
+                        MM[m["if_true"]]["items"].append(wl)
+                    else:
+                        MM[m["if_false"]]["items"].append(wl)
+        a, b, *_ = sorted(B.values(), reverse=True)
+        p12.append(a * b)
+    p1, p2 = p12
+    return p1, p2
+
+
+def _state(M):
+    return tuple([(tuple(m["items"])) for m in M])
+
+
+if __name__ == "__main__":
+    with open("./input/11.txt", "r") as f:
+        inp = f.read().strip()
+
+    p1, p2 = solve(inp)
+
+    print(p1)
+    print(p2)
+
+    assert p1 == 50616
+    assert p2 == 11309046332