Why this is O(2ⁿ)
Why O(2ⁿ). Each call branches into two — the recursion tree has roughly 2^n leaves. (The exact ratio is the golden-ratio-based φ^n ≈ 1.618^n, but in Big-O we round up to the nearest standard class — exponential.)
Fix. Memoise it and the recursion collapses to O(n). This is the canonical "why memoisation matters" snippet.
How to recognize exponential complexity in the wild
O(2ⁿ) is the cost of exploring every subset or every yes/no branch without pruning. Naive recursive Fibonacci, brute-force subset-sum, generating the power set.
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 7 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.