You have 8 identical-looking balls. One is slightly heavier than the rest. Using a balance scale, what is the minimum number of weighings needed to guarantee you find the heavy ball? Explain your strategy.

Judge: Correct answer (2 weighings) with a complete, thorough strategy covering all branches. Explains the information-theoretic lower bound (3^W >= 8), details the 3-3-2 split, and walks through every outcome for both weighings. Exceptionally clear presentation with all cases covered.

Judge: States the answer is 2 weighings and gives a complete strategy: 3v3 first weighing with both branches (balanced -> 7v8; unbalanced -> 1v2 with balance implying 3) fully covered. Also explains why 1 weighing is insufficient using the 3-outcome argument. Clear and well-structured.

Judge: States the correct answer of 2 weighings with a valid 3-3-2 strategy covering every branch of both weighings. It also gives the information-theoretic argument (3^2 = 9 >= 8) for why 2 suffices and why 1 weighing cannot. Complete, rigorous, and clearly laid out.

Judge: States the correct answer of 2 weighings with a complete 3-3-2 strategy covering every branch of the decision tree, and includes the information-theoretic argument (3 outcomes per weighing, 3 < 8 <= 9) for why 1 weighing is insufficient. All hard constraints are met and the presentation is clean. This is essentially a model answer.

Judge: Correct answer (2 weighings) with a complete strategy covering all branches: 3v3 with 2 aside, then second weighing handles both the group-of-3 and group-of-2 cases. Includes the information-theoretic justification (3^2=9 >= 8). The strategy is clearly explained with all decision paths enumerated. The explanation is concise and elegant.

Judge: Correct answer of 2 weighings with a complete strategy covering all branches. Includes the information-theoretic bound (3^k >= 8) as justification for why 1 weighing is insufficient. The strategy clearly covers both outcomes of the first weighing (balanced vs unbalanced) and all sub-cases of the second weighing. Exceptionally clear presentation.

Judge: Correct answer of 2 with a valid, complete strategy (3/3/2 split, all branches handled) and the 3^2=9>8 argument for why 2 suffices. Slightly less explicit than ideal on why 1 weighing alone is insufficient, but otherwise thorough and clear.

Judge: Correct answer of 2 weighings with a complete strategy covering all decision branches. The 3-3-2 split is optimal and all outcomes are traced. Minor gap: doesn't explicitly argue why 1 weighing is insufficient.

Judge: Correct answer (2 weighings) with a complete, well-explained strategy. Uses information-theoretic reasoning (3 outcomes per weighing, 3^2=9 >= 8) to establish the lower bound, then provides the constructive strategy (3-3-2 split). Covers all branches including the 2-suspect case. Clear formatting and logical flow. A near-perfect response.

Judge: Correct answer of 2 with both the sufficiency (3^2 = 9 >= 8) and necessity (1 weighing only 3 outcomes) arguments. The heavier-side branch is illustrated with a 'suppose 1,2,3 is heavier' example rather than stating the symmetric case explicitly, and the intermediate 'among those 6 balls' phrasing is briefly loose before correcting to 3, so the branch coverage is a notch below complete. Clear overall.

Judge: Correct answer (2 weighings) with a complete strategy covering all branches. The 3-3-2 split is optimal, and both outcomes (3 suspects and 2 suspects) are explained. Includes the information-theoretic justification (3^2=9 > 8). Well-structured and clear.

Judge: Correct answer (2 weighings) with a complete strategy covering all branches. The divide-into-thirds approach is optimal and well-explained. Covers all three outcomes of the first weighing and all sub-cases of the second. The information theory justification (3^2 = 9 >= 8) is a nice bonus that explains why 2 is sufficient and 1 is not. Clear, well-structured presentation.

Judge: Correct answer (2 weighings) with a complete strategy covering all branches. The 3-3-2 split is the standard optimal approach. All cases are covered: balanced first weighing leads to comparing the 2-ball group, unbalanced leads to comparing 2 of 3 from the heavy group. Clear formatting with step-by-step analysis of each branch.

Judge: Correct answer of 2 weighings, but the strategy explanation has gaps. The first weighing (3 vs 3) is correctly described, but Step 5 and beyond are vague about the second weighing -- it says 'take half of those identified balls' without explicitly walking through both branches (3 balls case: weigh 1 vs 1; 2 remaining balls case: weigh 1 vs 1). The 'remaining 4' scenario from a balanced first weighing is wrong (should be 2 remaining, not 4).