calib·HumanEval/40
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 triples_sum_to_zero to fn_cd672a055a so the judges can't pattern-match a memorized HumanEval solution.
def fn_cd672a055a(l: list):
"""
fn_cd672a055a takes a list of integers as an input.
it returns True if there are three distinct elements in the list that
sum to zero, and False otherwise.
>>> fn_cd672a055a([1, 3, 5, 0])
False
>>> fn_cd672a055a([1, 3, -2, 1])
True
>>> fn_cd672a055a([1, 2, 3, 7])
False
>>> fn_cd672a055a([2, 4, -5, 3, 9, 7])
True
>>> fn_cd672a055a([1])
False
"""
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_cd672a055a(l: list):
"""
fn_cd672a055a takes a list of integers as an input.
it returns True if there are three distinct elements in the list that
sum to zero, and False otherwise.
>>> fn_cd672a055a([1, 3, 5, 0])
False
>>> fn_cd672a055a([1, 3, -2, 1])
True
>>> fn_cd672a055a([1, 2, 3, 7])
False
>>> fn_cd672a055a([2, 4, -5, 3, 9, 7])
True
>>> fn_cd672a055a([1])
False
"""
for i in range(len(l)):
for j in range(i + 1, len(l)):
for k in range(j + 1, len(l)):
if l[i] + l[j] + l[k] == 0:
return True
return False
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 a three-sum check using three nested loops with distinct indices (i < j < k), ensuring the three elements are at distinct positions. It passes all provided doctests. However, the problem says "three distinct elements" — this could mean distinct values rather than distinct indices. For example, [0, 0, 0] would return True with this implementation (three positions summing to zero), but whether [0, 0, 0] should return True depends on interpretation of "distinct elements." The doctest [1, 3, -2, 1] returns True (using indices for 1, -2, 1 which are distinct positions but repeated value 1), suggesting position-based distinctness is intended, which the solution handles correctly. The solution is O(n³) which is acceptable for this type of problem. Minor concern: the "distinct elements" wording is slightly ambiguous, but the solution aligns with the provided examples.
The candidate solution correctly implements the task of finding three distinct elements in a list that sum to zero. It uses a triple nested loop to iterate over all possible combinations of three elements, checking if their sum is zero. This approach is correct and matches the problem requirements. The solution passes all provided examples and handles edge cases such as lists with fewer than three elements. The code is clear and idiomatic for the problem at hand.