calib·HumanEval/32
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 find_zero to fn_6324b7c6ab so the judges can't pattern-match a memorized HumanEval solution.
import math
def poly(xs: list, x: float):
"""
Evaluates polynomial with coefficients xs at point x.
return xs[0] + xs[1] * x + xs[1] * x^2 + .... xs[n] * x^n
"""
return sum([coeff * math.pow(x, i) for i, coeff in enumerate(xs)])
def fn_6324b7c6ab(xs: list):
""" xs are coefficients of a polynomial.
fn_6324b7c6ab find x such that poly(x) = 0.
fn_6324b7c6ab returns only only zero point, even if there are many.
Moreover, fn_6324b7c6ab only takes list xs having even number of coefficients
and largest non zero coefficient as it guarantees
a solution.
>>> round(fn_6324b7c6ab([1, 2]), 2) # f(x) = 1 + 2x
-0.5
>>> round(fn_6324b7c6ab([-6, 11, -6, 1]), 2) # (x - 1) * (x - 2) * (x - 3) = -6 + 11x - 6x^2 + x^3
1.0
"""
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.
import math
def poly(xs: list, x: float):
"""
Evaluates polynomial with coefficients xs at point x.
return xs[0] + xs[1] * x + xs[1] * x^2 + .... xs[n] * x^n
"""
return sum([coeff * math.pow(x, i) for i, coeff in enumerate(xs)])
def fn_6324b7c6ab(xs: list):
""" xs are coefficients of a polynomial.
fn_6324b7c6ab find x such that poly(x) = 0.
fn_6324b7c6ab returns only only zero point, even if there are many.
Moreover, fn_6324b7c6ab only takes list xs having even number of coefficients
and largest non zero coefficient as it guarantees
a solution.
>>> round(fn_6324b7c6ab([1, 2]), 2) # f(x) = 1 + 2x
-0.5
>>> round(fn_6324b7c6ab([-6, 11, -6, 1]), 2) # (x - 1) * (x - 2) * (x - 3) = -6 + 11x - 6x^2 + x^3
1.0
"""
begin, end = -1., 1.
while poly(xs, begin) * poly(xs, end) > 0:
begin *= 2.0
end *= 2.0
while end - begin > 1e-10:
center = (begin + end) / 2.0
if poly(xs, center) * poly(xs, begin) > 0:
begin = center
else:
end = center
return begin
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 uses bisection search to find a root. The bracket-expansion loop correctly doubles the interval until poly(begin) and poly(end) have opposite signs. The bisection loop then narrows to 1e-10 precision. Both doctests pass: for [1,2] it returns ~-0.5, and for [-6,11,-6,1] it returns ~1.0. The approach is sound for the stated constraints (even number of coefficients, largest non-zero coefficient guaranteed, implying a sign change exists). Minor issues: (1) the expansion loop could infinite-loop if the polynomial has no real root (but the problem guarantees a solution), (2) returning `begin` instead of `center` at the end introduces a tiny asymmetry (up to 1e-10 error, acceptable), (3) the expansion strategy of doubling both begin and end symmetrically works for most cases but could theoretically miss a root bracketed asymmetrically — however given the guarantee it's practically fine. Overall correct and clean.
The candidate solution correctly implements a method to find a root of a polynomial using the bisection method. It handles the task requirements by ensuring that the polynomial has an even number of coefficients and that the largest non-zero coefficient guarantees a solution. The solution passes the provided examples, demonstrating correctness for those cases. However, there are minor issues: the solution does not explicitly check for the even number of coefficients or the largest non-zero coefficient condition, which could lead to incorrect assumptions about the input. Additionally, the solution could be more efficient by reducing redundant calculations of the polynomial at the same points. Despite these minor issues, the solution is mostly correct and idiomatic.