PANOPTES
panoptes-696da4d5

HE/18

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
0.983
inter-judge spread
0.050
judges polled
3
sampling draws
15
1. the task

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



def how_many_times(string: str, substring: str) -> int:
    """ Find how many times a given substring can be found in the original string. Count overlaping cases.
    >>> how_many_times('', 'a')
    0
    >>> how_many_times('aaa', 'a')
    3
    >>> how_many_times('aaaa', 'aa')
    3
    """
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.



def how_many_times(string: str, substring: str) -> int:
    """ Find how many times a given substring can be found in the original string. Count overlaping cases.
    >>> how_many_times('', 'a')
    0
    >>> how_many_times('aaa', 'a')
    3
    >>> how_many_times('aaaa', 'aa')
    3
    """
    times = 0

    for i in range(len(string) - len(substring) + 1):
        if string[i:i+len(substring)] == substring:
            times += 1

    return times
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
0.950
$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 = 1.37e-3
total predictive variance, 3 judges
12% aleatoric88% epistemic
aleatoric (irreducible)
1.67e-4
epistemic (reducible)
1.20e-3
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
0.940[0.923, 0.949]
posterior mean · 90% Bayesian-bootstrap CI
variance: 5.00e-4
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 = 0.500/ log 51.609
2 semantic clusters via bidirectional NLI
entropy as fraction of max
cluster sizes
c1=4 · c2=1

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 · 1912 tokens · 3204ms

The solution correctly implements the required functionality. It iterates through the string with a sliding window of size len(substring), comparing each window to the substring and counting matches. The loop range `len(string) - len(substring) + 1` is correct and handles overlapping cases properly. Testing against the provided examples: (1) empty string returns 0 (loop doesn't execute), (2) 'aaa' with 'a' returns 3 (matches at indices 0, 1, 2), (3) 'aaaa' with 'aa' returns 3 (matches at indices 0, 1, 2). Edge cases are handled correctly: empty substring would cause division issues but is not tested; empty string works correctly; single character matches work. The code is idiomatic, clear, and efficient with O(n*m) time complexity where n is string length and m is substring length.

claude-sonnet
score 0.950 · 1839 tokens · 5185ms

The solution correctly implements overlapping substring counting using a sliding window approach. It passes all three provided examples: empty string returns 0, 'aaa'/'a' returns 3, 'aaaa'/'aa' returns 3. Edge cases like empty substring would cause issues (range(len(string)+1) would iterate and always match, returning len(string)+1), but that's an ambiguous edge case not covered by the spec. The logic is sound and idiomatic for the stated task.

gpt-4o-mini
score 1.000 · 909 tokens · 1238ms

The solution correctly counts overlapping occurrences of the substring in the string, passing all provided test cases and handling edge cases like empty strings. The implementation is clear and idiomatic.