Why this is O(n³)
Why O(n³). All-pairs shortest paths via dynamic programming. Three nested loops over the n vertices — each combination of (intermediate, source, destination) is considered once.
The constant factor matters here. Floyd–Warshall's n³ is a *tight* inner loop with no branching cost — it's faster in practice than running Dijkstra n times (O(n × (n log n + e))) for dense graphs.
How to recognize cubic complexity in the wild
O(n³) shows up in triple-nested loops (matrix multiplication, certain dynamic-programming tables, Floyd–Warshall). Rare in idiomatic application code but common in numerical kernels.
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 6 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.