HE/20
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.
from typing import List, Tuple
def find_closest_elements(numbers: List[float]) -> Tuple[float, float]:
""" From a supplied list of numbers (of length at least two) select and return two that are the closest to each
other and return them in order (smaller number, larger number).
>>> find_closest_elements([1.0, 2.0, 3.0, 4.0, 5.0, 2.2])
(2.0, 2.2)
>>> find_closest_elements([1.0, 2.0, 3.0, 4.0, 5.0, 2.0])
(2.0, 2.0)
"""
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.
from typing import List, Tuple
def find_closest_elements(numbers: List[float]) -> Tuple[float, float]:
""" From a supplied list of numbers (of length at least two) select and return two that are the closest to each
other and return them in order (smaller number, larger number).
>>> find_closest_elements([1.0, 2.0, 3.0, 4.0, 5.0, 2.2])
(2.0, 2.2)
>>> find_closest_elements([1.0, 2.0, 3.0, 4.0, 5.0, 2.0])
(2.0, 2.0)
"""
closest_pair = None
distance = None
for idx, elem in enumerate(numbers):
for idx2, elem2 in enumerate(numbers):
if idx != idx2:
if distance is None:
distance = abs(elem - elem2)
closest_pair = tuple(sorted([elem, elem2]))
else:
new_distance = abs(elem - elem2)
if new_distance < distance:
distance = new_distance
closest_pair = tuple(sorted([elem, elem2]))
return closest_pair
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.
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.
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).
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 solves the stated task and passes both provided examples. The algorithm uses a nested loop to compare all pairs of elements, tracking the minimum distance and corresponding pair. The logic is sound: it correctly excludes self-comparisons (idx != idx2), properly computes absolute differences, and returns the pair in sorted order as required. However, there are two minor issues: (1) The algorithm has O(n²) time complexity with redundant comparisons (each pair is checked twice), whereas sorting the list first and comparing adjacent elements would achieve O(n log n) with cleaner logic. (2) The code initializes distance and closest_pair to None and relies on the first comparison to set them, which works but is slightly inelegant compared to initializing distance to infinity. The solution is correct and idiomatic enough for the task, but the inefficiency and minor style issues prevent a perfect score.
The solution correctly finds the two closest elements and returns them sorted (smaller, larger). It handles both provided examples correctly. The logic is sound: it iterates all pairs (using index comparison to avoid self-pairing), tracks minimum distance, and sorts the result. Minor issues: (1) It's O(n²) when O(n log n) is possible via sorting first, but correctness is not affected. (2) It compares each pair twice (i,j) and (j,i), which is redundant but harmless. (3) Edge cases like a two-element list work fine. The solution is correct and handles the stated task well.
The solution correctly identifies the closest pair of numbers and returns them in sorted order. It passes the provided test cases, but it has inefficiencies due to the nested loops, leading to a time complexity of O(n^2). Additionally, it does not handle edge cases such as lists with duplicate values or very close floating-point numbers effectively, which could lead to incorrect results in some scenarios.