PANOPTES
panoptes-d636a93f

HE/26

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
0.900
inter-judge spread
0.200
judges polled
2
sampling draws
10
1. the task

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

from typing import List


def remove_duplicates(numbers: List[int]) -> List[int]:
    """ From a list of integers, remove all elements that occur more than once.
    Keep order of elements left the same as in the input.
    >>> remove_duplicates([1, 2, 3, 2, 4])
    [1, 3, 4]
    """
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 remove_duplicates(numbers: List[int]) -> List[int]:
    """ From a list of integers, remove all elements that occur more than once.
    Keep order of elements left the same as in the input.
    >>> remove_duplicates([1, 2, 3, 2, 4])
    [1, 3, 4]
    """
    import collections
    c = collections.Counter(numbers)
    return [n for n in numbers if c[n] <= 1]
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
1.000
$0.0027
gpt-4o-mini
0.800
$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 = 1.68e-2
total predictive variance, 2 judges
24% aleatoric76% epistemic
aleatoric (irreducible)
4.00e-3
epistemic (reducible)
1.28e-2
self-consistency
gpt-4o-mini
0.840[0.803, 0.903]
posterior mean · 90% Bayesian-bootstrap CI
variance: 8.00e-3
n samples: 5
self-consistency
claude-haiku
1.000[1.000, 1.000]
posterior mean · 90% Bayesian-bootstrap CI
variance: 0.00e+0
n samples: 5
semantic-entropy
gpt-4o-mini
H = 1.332/ log 51.609
4 semantic clusters via bidirectional NLI
entropy as fraction of max
cluster sizes
c1=1 · c2=1 · c3=2 · c4=1
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

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 1.000 · 1843 tokens · 2193ms

The solution correctly implements the task. It uses `collections.Counter` to count occurrences of each element, then filters the list to keep only elements that appear exactly once (count <= 1). This preserves the original order since it iterates through the input list in sequence. The provided doctest example [1, 2, 3, 2, 4] → [1, 3, 4] is handled correctly: 1 appears once (kept), 2 appears twice (removed), 3 appears once (kept), 4 appears once (kept). Edge cases are handled well: empty list returns empty list, single-element list returns that element, all duplicates returns empty list. The code is idiomatic, clear, and efficient (O(n) time complexity).

gpt-4o-mini
score 0.800 · 911 tokens · 2494ms

The solution correctly removes duplicates and maintains the order of elements as specified in the task. It passes the provided example and handles edge cases like empty lists. However, it could be more efficient by using a single pass to count and filter instead of creating a Counter object, which adds unnecessary overhead.