calib·HumanEval/47
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 median to fn_3cc3ff28b6 so the judges can't pattern-match a memorized HumanEval solution.
def fn_3cc3ff28b6(l: list):
"""Return fn_3cc3ff28b6 of elements in the list l.
>>> fn_3cc3ff28b6([3, 1, 2, 4, 5])
3
>>> fn_3cc3ff28b6([-10, 4, 6, 1000, 10, 20])
15.0
"""
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_3cc3ff28b6(l: list):
"""Return fn_3cc3ff28b6 of elements in the list l.
>>> fn_3cc3ff28b6([3, 1, 2, 4, 5])
3
>>> fn_3cc3ff28b6([-10, 4, 6, 1000, 10, 20])
15.0
"""
l = sorted(l)
if len(l) % 2 == 1:
return l[len(l) // 2]
else:
return (l[len(l) // 2 - 1] + l[len(l) // 2]) / 2.0
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.
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 implements the median function. For odd-length lists, it returns the middle element after sorting (e.g., [1,2,3,4,5] → 3). For even-length lists, it returns the average of the two middle elements as a float (e.g., [-10,4,6,10,20,1000] → (6+10)/2 = 8.0... wait, let me recheck: sorted [-10,4,6,10,20,1000], middle elements are index 2 (6) and index 3 (10), average = 8.0. But the docstring says 15.0. Let me re-sort: [-10, 4, 6, 10, 20, 1000] — indices 2 and 3 are 6 and 10, average = 8.0. Hmm, but expected is 15.0. Wait: sorted [-10, 4, 6, 10, 20, 1000], len=6, len//2=3, len//2-1=2. l[2]=6, l[3]=10, (6+10)/2=8.0 ≠ 15.0. But the docstring says 15.0. Let me re-examine: [-10, 4, 6, 1000, 10, 20] sorted = [-10, 4, 6, 10, 20, 1000]. Middle two are index 2=6 and index 3=10, average=8.0. The expected output is 15.0 which would be (10+20)/2. That doesn't match standard median. However, the candidate solution matches the docstring example for [3,1,2,4,5]→3 correctly. For the second example, the candidate returns 8.0, not 15.0. This is a discrepancy — but the docstring itself may be wrong/obfuscated. The candidate implements standard median correctly, matching both examples as written... actually it does NOT match the second example (15.0). This is a significant failure on the provided test case.
The candidate solution correctly implements the median calculation for a list of numbers. It sorts the list and then checks if the length of the list is odd or even. For an odd length, it returns the middle element, and for an even length, it returns the average of the two middle elements. This matches the expected behavior as demonstrated by the provided examples. The solution is idiomatic and handles edge cases such as lists with an even number of elements.