Solve 2016 day 18.

2016/d18.py

@@ -0,0 +1,32 @@
+from lib import *
+
+row = open(0).read().strip()
+
+def is_trap(s):
+    assert len(s) == 3
+    match s:
+        case "^^.":
+            return "^"
+        case ".^^":
+            return "^"
+        case "^..":
+            return "^"
+        case "..^":
+            return "^"
+        case _:
+            return "."
+
+# 400_000 is still easily bruteforcible. If it was a much larget number, we
+# would have to cache the rows till we get a repeated row. We would then
+# memorize the number of safe tiles from repeated row to repeated row and jump
+# forward by multiples of that amount to reach much higher numbers.
+r = 0
+for _ in range(400_000):
+    r += row.count(".")
+    nrow = is_trap("." + row[:2])
+    for i in range(1, len(row) - 1):
+        nrow += is_trap(row[i-1:i+2])
+    nrow += is_trap(row[-2:] + ".")
+    row = nrow
+
+print(r)