PANOPTES
panoptes-e86ef9e3

calib·HumanEval/0

humaneval-calibprobe · code

This is one (task, candidate response) pair flowing through the full PANOPTES pipeline. Each section below is a stage of the analysis: the task itself, the candidate solution being evaluated, every judge's score + rationale, the sampling-pass dispersion that captures within-judge noise, and the uncertainty-quantification metrics computed on top.

mean score
0.875
inter-judge spread
0.150
judges polled
2
sampling draws
20
1. the task

The function signature + docstring presented to both the model under test and to every judge. The entry-point name was obfuscated from has_close_elements to fn_de004141fa so the judges can't pattern-match a memorized HumanEval solution.

from typing import List


def fn_de004141fa(numbers: List[float], threshold: float) -> bool:
    """ Check if in given list of numbers, are any two numbers closer to each other than
    given threshold.
    >>> fn_de004141fa([1.0, 2.0, 3.0], 0.5)
    False
    >>> fn_de004141fa([1.0, 2.8, 3.0, 4.0, 5.0, 2.0], 0.3)
    True
    """
2. the candidate response judged

The full solution every judge is grading. For these runs the candidate is the reference solution prepended with the task signature so it parses as a complete program.

from typing import List


def fn_de004141fa(numbers: List[float], threshold: float) -> bool:
    """ Check if in given list of numbers, are any two numbers closer to each other than
    given threshold.
    >>> fn_de004141fa([1.0, 2.0, 3.0], 0.5)
    False
    >>> fn_de004141fa([1.0, 2.8, 3.0, 4.0, 5.0, 2.0], 0.3)
    True
    """
    for idx, elem in enumerate(numbers):
        for idx2, elem2 in enumerate(numbers):
            if idx != idx2:
                distance = abs(elem - elem2)
                if distance < threshold:
                    return True

    return False
3a. point-pass scores (temperature 0)

One call per judge at temperature=0. This is the "best single guess" each judge has. Disagreement here is structural. The judges literally see this candidate differently.

claude-sonnet
0.950
$0.0082
gpt-4o
0.800
$0.0036
3b. sampling-pass dispersion (temperature 1)

n draws per judge at temperature=1. Dots are individual draws; vertical bar is the mean; shaded band is ±1σ. Wide band = the judge is uncertain even with itself; narrow band = it's consistent.

4. uncertainty quantification

Three statistical methods computed on top of the raw scores above. Each one tells you something different about how much to trust the headline number.

semantic-entropy
claude-sonnet
H = 0.000/ log 102.303
1 semantic cluster via bidirectional NLI
entropy as fraction of max
cluster sizes
c1=10
semantic-entropy
gpt-4o
H = 0.611/ log 102.303
2 semantic clusters via bidirectional NLI
entropy as fraction of max
cluster sizes
c1=7 · c2=3

How to read these: self-consistency = does the judge agree with itself when resampled? Narrower CI means more consistent. Semantic entropy = do the judge's rationales cluster into one meaning, or several? Higher entropy means the judge is internally conflicted about why. Decomposition = how much of the total variance comes from within-judge sampling noise (aleatoric, hard to fix) vs between-judge disagreement (epistemic, fixable by calling more judges).

5. judge rationales

The natural-language explanation each judge gave alongside its score. These are produced via tool-use structured output, so the score and the rationale are guaranteed to come from the same forward pass.

claude-sonnet
score 0.950 · 1935 tokens · 4561ms

The solution correctly implements the task: it checks all pairs of distinct elements (using index comparison to avoid self-comparison) and returns True if any two numbers are closer than the threshold. Both provided examples would pass. The logic is sound and handles edge cases like empty lists (returns False) and single-element lists (returns False). Minor inefficiency: it checks each pair twice (i,j) and (j,i), but this doesn't affect correctness. A slightly more idiomatic approach would use itertools.combinations or a nested loop with idx2 starting at idx+1, but the current approach is correct and clear.

gpt-4o
score 0.800 · 1078 tokens · 1901ms

The candidate solution correctly implements the task of checking if any two numbers in the list are closer to each other than the given threshold. It iterates over all pairs of numbers and checks their distance, returning True if any pair is closer than the threshold. The solution passes the provided examples and handles edge cases like empty lists and lists with a single element correctly. However, the solution is not optimal in terms of efficiency, as it uses a nested loop resulting in O(n^2) complexity, which could be improved by sorting the list first and then checking adjacent pairs.