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Implement Best-of-N (rejection) sampling for LLM generation.

Given N candidate responses with their reward scores, select the best one. Also compute the KL cost of this selection strategy.

The expected KL divergence cost of Best-of-N sampling is approximately: KLBoN≈log⁡(N)−N−1N\text{KL}_{\text{BoN}} \approx \log(N) - \frac{N-1}{N}

Input:

  • Line 1: N (number of candidates)
  • Next N lines: reward_score response_text

Output:

  • Line 1: The best response text (highest reward)
  • Line 2: The best reward score, rounded to 4 decimal places
  • Line 3: The KL cost approximation, rounded to 4 decimal places

Example:

Input:
3
1.5 Hello world
2.3 Hi there
0.8 Hey
Output:
Hi there
2.3000
0.4319
Reasoning:
  • We read the input values: N=3N = 3, and the reward scores with their corresponding responses: (1.5,Hello world)(1.5, \text{Hello world}), (2.3,Hi there)(2.3, \text{Hi there}), and (0.8,Hey)(0.8, \text{Hey}).
  • We select the best response by finding the highest reward score, which is 2.32.3 for the response Hi there\text{Hi there}.
  • We calculate the KL cost approximation using the formula: KLBoN≈log⁡(N)−N−1N=log⁡(3)−3−13≈0.4319\text{KL}_{\text{BoN}} \approx \log(N) - \frac{N-1}{N} = \log(3) - \frac{3-1}{3} \approx 0.4319.
  • The final output is the best response text, the best reward score rounded to 4 decimal places (2.30002.3000), and the KL cost approximation rounded to 4 decimal places (0.43190.4319).

Constraints:

  • 1 <= N <= 100
  • If tied, pick the first one encountered
  • Round to 4 decimal places
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solution.py

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Best-of-N Sampling - Hard | PixelBank