calib·HumanEval/49
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 modp to fn_4fccd624ee so the judges can't pattern-match a memorized HumanEval solution.
def fn_4fccd624ee(n: int, p: int):
"""Return 2^n modulo p (be aware of numerics).
>>> fn_4fccd624ee(3, 5)
3
>>> fn_4fccd624ee(1101, 101)
2
>>> fn_4fccd624ee(0, 101)
1
>>> fn_4fccd624ee(3, 11)
8
>>> fn_4fccd624ee(100, 101)
1
"""
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_4fccd624ee(n: int, p: int):
"""Return 2^n modulo p (be aware of numerics).
>>> fn_4fccd624ee(3, 5)
3
>>> fn_4fccd624ee(1101, 101)
2
>>> fn_4fccd624ee(0, 101)
1
>>> fn_4fccd624ee(3, 11)
8
>>> fn_4fccd624ee(100, 101)
1
"""
ret = 1
for i in range(n):
ret = (2 * ret) % p
return ret
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 computes 2^n mod p and passes all provided test cases. It uses iterative multiplication with modular reduction, which avoids large integer overflow issues. However, it runs in O(n) time rather than O(log n) — Python's built-in `pow(2, n, p)` or binary exponentiation would be far more efficient. For very large n (e.g., n=10^18), this solution would be impractically slow. The "be aware of numerics" hint in the docstring likely refers to using modular exponentiation for large n, which this solution doesn't address. Since correctness is maintained for all given examples but efficiency is a real concern for large inputs, this lands at 0.80.
The candidate solution correctly implements the task of computing 2^n modulo p using an iterative approach. It passes all the provided test cases, including edge cases like n=0. The use of a loop to compute the power is clear and idiomatic, though not the most efficient method for large n, where a more efficient approach like exponentiation by squaring could be used. However, the solution is correct and clear for the given task.