calib·HumanEval/37
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.
The function signature + docstring presented to both the model under test and to every judge. The entry-point name was obfuscated from sort_even to fn_4d58b82570 so the judges can't pattern-match a memorized HumanEval solution.
def fn_4d58b82570(l: list):
"""This function takes a list l and returns a list l' such that
l' is identical to l in the odd indicies, while its values at the even indicies are equal
to the values of the even indicies of l, but sorted.
>>> fn_4d58b82570([1, 2, 3])
[1, 2, 3]
>>> fn_4d58b82570([5, 6, 3, 4])
[3, 6, 5, 4]
"""
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 fn_4d58b82570(l: list):
"""This function takes a list l and returns a list l' such that
l' is identical to l in the odd indicies, while its values at the even indicies are equal
to the values of the even indicies of l, but sorted.
>>> fn_4d58b82570([1, 2, 3])
[1, 2, 3]
>>> fn_4d58b82570([5, 6, 3, 4])
[3, 6, 5, 4]
"""
evens = l[::2]
odds = l[1::2]
evens.sort()
ans = []
for e, o in zip(evens, odds):
ans.extend([e, o])
if len(evens) > len(odds):
ans.append(evens[-1])
return ans
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.
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.
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.
The solution correctly extracts even-indexed elements, sorts them, keeps odd-indexed elements unchanged, and interleaves them back. It handles the edge case where the list has an odd length (more even-indexed elements than odd-indexed). Both provided doctests pass: [1,2,3]→[1,2,3] and [5,6,3,4]→[3,6,5,4]. Edge cases like empty list and single-element list also work correctly.
The candidate solution correctly implements the task by sorting the even-indexed elements while keeping the odd-indexed elements unchanged. It handles the example cases correctly and accounts for lists with an odd number of elements by appending the last even-indexed element if necessary. However, the solution could be slightly improved in terms of efficiency by using a more idiomatic approach to combine the sorted even and unchanged odd elements, such as using list comprehension or itertools. Overall, the solution is correct but has minor style issues.