PANOPTES
panoptes-696da4d5

HE/9

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
1.000
inter-judge spread
0.000
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, Tuple


def rolling_max(numbers: List[int]) -> List[int]:
    """ From a given list of integers, generate a list of rolling maximum element found until given moment
    in the sequence.
    >>> rolling_max([1, 2, 3, 2, 3, 4, 2])
    [1, 2, 3, 3, 3, 4, 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, Tuple


def rolling_max(numbers: List[int]) -> List[int]:
    """ From a given list of integers, generate a list of rolling maximum element found until given moment
    in the sequence.
    >>> rolling_max([1, 2, 3, 2, 3, 4, 2])
    [1, 2, 3, 3, 3, 4, 4]
    """
    running_max = None
    result = []

    for n in numbers:
        if running_max is None:
            running_max = n
        else:
            running_max = max(running_max, n)

        result.append(running_max)

    return result
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
$0.0029
claude-sonnet
1.000
$0.0076
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 = 0.00e+0
total predictive variance, 3 judges
aleatoric (irreducible)
0.00e+0
epistemic (reducible)
0.00e+0
self-consistency
claude-haiku
1.000[1.000, 1.000]
posterior mean · 90% Bayesian-bootstrap CI
variance: 0.00e+0
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
1.000[1.000, 1.000]
posterior mean · 90% Bayesian-bootstrap CI
variance: 0.00e+0
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 1.000 · 1923 tokens · 2129ms

The solution correctly implements the rolling maximum function. It iterates through the input list, maintaining a running maximum and appending it to the result list at each step. The logic is sound: it initializes running_max to None, sets it to the first element, then updates it with max(running_max, n) for subsequent elements. Testing against the provided example: [1, 2, 3, 2, 3, 4, 2] produces [1, 2, 3, 3, 3, 4, 4] as expected. Edge cases are handled correctly: empty list returns empty list, single element returns that element, and negative numbers work properly. The code is idiomatic, clear, and efficient (O(n) time, O(n) space).

claude-sonnet
score 1.000 · 1856 tokens · 5402ms

The solution correctly implements rolling maximum. It initializes `running_max` to `None` and updates it with `max(running_max, n)` for each element, appending to the result list. This handles the canonical example ([1,2,3,2,3,4,2] → [1,2,3,3,3,4,4]) correctly. Edge cases are also handled: empty list returns [], single element returns that element, negative numbers work correctly since `None` check initializes properly. The logic is clean and idiomatic.

gpt-4o-mini
score 1.000 · 957 tokens · 1615ms

The solution correctly implements the rolling maximum functionality as specified in the task. It handles the provided example and will also work for edge cases such as empty lists. The code is clear and idiomatic, using a straightforward loop and max function to maintain the running maximum.