# Find the Two Heaviest Balls with Minimum Comparisons There are 32 balls with distinct weights. A comparison reveals which of two balls is heavier. Determine the minimum number of comparisons needed to identify both the heaviest and second-heaviest balls, give an algorithm that achieves it, and prove that fewer comparisons cannot always succeed. ### Constraints & Assumptions - All weights are distinct. - Only pairwise comparisons are allowed. - The algorithm may choose later comparisons based on earlier outcomes. - Both identities, not their numeric weights, must be returned. ### Clarifying Questions to Ask - Are comparisons deterministic and error-free? - Must the lower bound apply to every adaptive comparison strategy? - How should the argument generalize when the item count is not a power of two? ```hint Preserve comparison evidence For every ball excluded from either returned position, be able to point to observed comparison outcomes that justify the exclusion. ``` ```hint Separate construction from lower bound An algorithm that achieves a count does not by itself prove that every adaptive strategy needs that many comparisons. ``` ### What a Strong Answer Covers - A valid adaptive comparison algorithm that returns both requested identities. - Enough retained comparison evidence to restrict which balls can still be second-heaviest. - A derivation of the exact comparison count without revealing it in the prompt. - A worst-case lower bound that applies to arbitrary adaptive comparison strategies, not only a balanced construction. ### Follow-up Questions - What is the comparison count for n balls when n is a power of two? - How would the algorithm track the third-heaviest ball?