Why this is O(n!)
Why O(n!). permutations(cities[1:]) enumerates (n−1)! orderings. For each, we sum an O(n) path cost. Total: O((n−1)! × n) = O(n!) — the factorial dominates.
The Big-O-dle lesson. When you see itertools.permutations over a non-trivial input, alarm bells should ring. There's almost always a smarter approach — TSP itself drops to 2ⁿ with bitmask DP.
How to recognize factorial complexity in the wild
O(n!) is permutation territory. Generating every ordering, brute-force travelling salesman, anything that asks 'in how many orders can I arrange n things.' Even n=12 is already a billion permutations.
The eight rungs of the ladder
Big-O classifies the asymptotic growth of a function, not the wall-clock time. The ladder Bugdle uses has eight rungs ordered from cheapest to most ruinous: O(1), O(log n), O(n), O(n log n), O(n²), O(n³), O(2ⁿ), O(n!). The puzzle's answer sits at rung 8 of 8. The four-guess budget plus higher/lower hints means the puzzle is solvable in at most ⌈log₂(8)⌉ = 3 perfectly-read guesses; the fourth attempt absorbs misreads of the snippet.
Common confusables
Nested loops aren't automatically quadratic — the inner loop has to range over n, not a constant. A loop that always runs ten iterations is still O(1) work per outer iteration. Similarly, a recursive function isn't automatically exponential just because it calls itself twice — if the recursion has overlapping subproblems and you memoise, you collapse back to polynomial. Always ask: what does n really mean here, and how many distinct subproblems are there?
External reference: Big O notation — Wikipedia.