PANOPTES
panoptes-696da4d5

HE/0

humaneval · 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
0.917
inter-judge spread
0.200
judges polled
3
sampling draws
15
1. the task

The function signature + docstring presented to both the model under test and to every judge.

from typing import List


def has_close_elements(numbers: List[float], threshold: float) -> bool:
    """ Check if in given list of numbers, are any two numbers closer to each other than
    given threshold.
    >>> has_close_elements([1.0, 2.0, 3.0], 0.5)
    False
    >>> has_close_elements([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 has_close_elements(numbers: List[float], threshold: float) -> bool:
    """ Check if in given list of numbers, are any two numbers closer to each other than
    given threshold.
    >>> has_close_elements([1.0, 2.0, 3.0], 0.5)
    False
    >>> has_close_elements([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-haiku
0.800
$0.0030
claude-sonnet
0.950
$0.0079
gpt-4o-mini
1.000
$0.0002
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.

decomposition
aggregate (all judges)
Var = 1.09e-2
total predictive variance, 3 judges
29% aleatoric71% epistemic
aleatoric (irreducible)
3.19e-3
epistemic (reducible)
7.73e-3
self-consistency
claude-haiku
0.830[0.785, 0.899]
posterior mean · 90% Bayesian-bootstrap CI
variance: 9.50e-3
n samples: 5
self-consistency
gpt-4o-mini
1.000[1.000, 1.000]
posterior mean · 90% Bayesian-bootstrap CI
variance: 0.00e+0
n samples: 5
self-consistency
claude-sonnet
0.954[0.950, 0.960]
posterior mean · 90% Bayesian-bootstrap CI
variance: 8.00e-5
n samples: 5
semantic-entropy
claude-haiku
H = 0.000/ log 51.609
1 semantic cluster via bidirectional NLI
entropy as fraction of max
cluster sizes
c1=5
semantic-entropy
gpt-4o-mini
H = 0.500/ log 51.609
2 semantic clusters via bidirectional NLI
entropy as fraction of max
cluster sizes
c1=4 · c2=1
semantic-entropy
claude-sonnet
H = 0.000/ log 51.609
1 semantic cluster via bidirectional NLI
entropy as fraction of max
cluster sizes
c1=5

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-haiku
score 0.800 · 1978 tokens · 2937ms

The solution correctly implements the required logic and passes both provided test cases. The algorithm checks all pairs of distinct elements and returns True if any pair has a distance less than the threshold. The logic is sound: it properly uses enumerate to get indices, correctly checks idx != idx2 to avoid comparing an element with itself, and correctly uses abs() for distance calculation. However, there is a minor efficiency issue: the nested loop structure is O(n²) and redundantly checks each pair twice (once as (i,j) and again as (j,i)). A more efficient approach would be to use `for idx2 in range(idx+1, len(numbers))` to avoid duplicate comparisons. Additionally, the solution could be more Pythonic using itertools.combinations. These are minor style/efficiency concerns that don't affect correctness on the stated task.

claude-sonnet
score 0.950 · 1910 tokens · 4426ms

The solution correctly implements the task using a brute-force O(n²) double loop. It checks all pairs (i≠j), computes absolute difference, and returns True if any pair is closer than the threshold. Both doctest examples pass. Edge cases like empty list or single-element list return False correctly. The only minor issue is slight inefficiency (checks each pair twice, e.g., (i,j) and (j,i)), but correctness is not affected. Style is clear and idiomatic enough.

gpt-4o-mini
score 1.000 · 1012 tokens · 2788ms

The solution correctly implements the logic to check if any two numbers in the list are closer than the given threshold. It passes the provided test cases and handles edge cases, such as lists with fewer than two elements. The code is clear and idiomatic, using nested loops to compare each pair of elements effectively.