PANOPTES
panoptes-696da4d5

HE/14

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


def all_prefixes(string: str) -> List[str]:
    """ Return list of all prefixes from shortest to longest of the input string
    >>> all_prefixes('abc')
    ['a', 'ab', 'abc']
    """
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 all_prefixes(string: str) -> List[str]:
    """ Return list of all prefixes from shortest to longest of the input string
    >>> all_prefixes('abc')
    ['a', 'ab', 'abc']
    """
    result = []

    for i in range(len(string)):
        result.append(string[:i+1])
    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
$0.0027
claude-sonnet
1.000
$0.0070
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 = 0.000/ log 51.609
1 semantic cluster via bidirectional NLI
entropy as fraction of max
cluster sizes
c1=5
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 · 1792 tokens · 2543ms

The solution correctly implements the all_prefixes function. It iterates through the string length, building prefixes from index 0 to i+1 for each iteration, which produces all prefixes from shortest to longest. The logic is sound: for a string of length n, the loop runs n times (i from 0 to n-1), generating prefixes of lengths 1 through n. The example case 'abc' correctly produces ['a', 'ab', 'abc']. Edge cases are handled properly: an empty string returns an empty list (loop doesn't execute), and single-character strings return a list with that character. The code is idiomatic, clear, and efficient with O(n²) time complexity (unavoidable for this task since the output itself is O(n²)).

claude-sonnet
score 1.000 · 1719 tokens · 3253ms

The solution correctly iterates from index 0 to len(string)-1, appending string[:i+1] each time, producing all prefixes from shortest to longest. It handles the canonical example ('abc' → ['a', 'ab', 'abc']), edge cases like empty string (returns []), and single-character strings correctly. The logic is clean and idiomatic.

gpt-4o-mini
score 1.000 · 853 tokens · 2164ms

The solution correctly generates all prefixes of the input string from shortest to longest, as demonstrated by the provided example. It handles edge cases, such as an empty string, by returning an empty list, which is the expected behavior. The code is clear and idiomatic, following Python conventions.