HE/10
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.
def is_palindrome(string: str) -> bool:
""" Test if given string is a palindrome """
return string == string[::-1]
def make_palindrome(string: str) -> str:
""" Find the shortest palindrome that begins with a supplied string.
Algorithm idea is simple:
- Find the longest postfix of supplied string that is a palindrome.
- Append to the end of the string reverse of a string prefix that comes before the palindromic suffix.
>>> make_palindrome('')
''
>>> make_palindrome('cat')
'catac'
>>> make_palindrome('cata')
'catac'
"""
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 is_palindrome(string: str) -> bool:
""" Test if given string is a palindrome """
return string == string[::-1]
def make_palindrome(string: str) -> str:
""" Find the shortest palindrome that begins with a supplied string.
Algorithm idea is simple:
- Find the longest postfix of supplied string that is a palindrome.
- Append to the end of the string reverse of a string prefix that comes before the palindromic suffix.
>>> make_palindrome('')
''
>>> make_palindrome('cat')
'catac'
>>> make_palindrome('cata')
'catac'
"""
if not string:
return ''
beginning_of_suffix = 0
while not is_palindrome(string[beginning_of_suffix:]):
beginning_of_suffix += 1
return string + string[:beginning_of_suffix][::-1]
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 implements the algorithm described in the docstring. It finds the longest palindromic suffix by iterating from the start of the string, checking if each suffix is a palindrome. Once found, it appends the reverse of the prefix that precedes that suffix. Testing against the provided examples: (1) empty string returns ''; (2) 'cat' → finds 'cat' is not palindrome, 'at' is not, 't' is → appends reverse of 'ca' → 'catac'; (3) 'cata' → finds 'cata' is not, 'ata' is → appends reverse of 'c' → 'catac'. The logic is sound, handles edge cases (empty input), and the code is clear and idiomatic.
The solution correctly implements the described algorithm: it finds the longest palindromic suffix by incrementing `beginning_of_suffix` until `string[beginning_of_suffix:]` is a palindrome, then appends the reverse of the prefix before that suffix. All three doctests pass: `make_palindrome('')` → `''`, `make_palindrome('cat')` → `'catac'`, `make_palindrome('cata')` → `'catac'`. Edge cases like an already-palindromic string (e.g., `'aba'`) correctly return the string unchanged since `beginning_of_suffix` stays 0 and nothing is appended. The code is clean, idiomatic, and matches the stated algorithm exactly.
The candidate solution correctly implements the `make_palindrome` function according to the specified algorithm. It handles edge cases, such as an empty string, and produces the expected results for the provided examples. The logic is clear and idiomatic, and it efficiently finds the shortest palindrome by appending the necessary characters. All edge cases are covered, including the empty input and the provided test cases.