一串字绕成圆，怎样选同一个开头？提前停下为什么会选错

"""Booth's least circular shift, WITHOUT the invalid early-return shortcut.

Algorithm reference: https://www.cs.ubc.ca/~ksbooth/PUB/LCS.shtml
Offsets and comparisons use Python Unicode code points, without normalization.
"""
from itertools import product


def require(ok, message):
    if not ok:
        raise AssertionError(message)


def rotate(text, start):
    return text[start:] + text[:start]


def least_rotation(text, shortcut=False, emit=None):
    if type(text) is not str:
        raise TypeError("text must be str")
    if type(shortcut) is not bool:
        raise TypeError("shortcut must be bool")
    n = len(text)
    if not n:
        return 0
    failure = [-1] * (2*n)
    k = 0
    for j in range(1, 2*n):
        # Intentionally faulty mode reproduces the author's documented shortcut.
        if shortcut and j-k > n:
            if emit:
                emit(j, k, [], True)
            return k
        i = failure[j-k-1]
        updates = []
        while i != -1 and text[j % n] != text[(k+i+1) % n]:
            if text[j % n] < text[(k+i+1) % n]:
                previous = k
                k = j-i-1
                updates.append((previous, k))
            i = failure[i]
        if i == -1 and text[j % n] != text[(k+i+1) % n]:
            if text[j % n] < text[(k+i+1) % n]:
                previous = k
                k = j
                updates.append((previous, k))
            failure[j-k] = -1
        else:
            failure[j-k] = i+1
        if emit:
            emit(j, k, updates, False)
    return k


def brute(text):
    # Full enumeration shares neither failure links nor candidate updates.
    return min((rotate(text, k), k) for k in range(len(text))) if text else ("", 0)


def main():
    for text in ("aaba", "0010", "alfa", "baba", "aaaa", ""):
        good = least_rotation(text)
        bad = least_rotation(text, True)
        require(rotate(text, good) == brute(text)[0], "fixed example")
        print(f"text={text!r}; start={good}; minimum={rotate(text, good)!r}; shortcut_start={bad}; shortcut_result={rotate(text,bad)!r}")
    require(least_rotation("aaba", True) == 0, "reproduce faulty shortcut")
    print("trace aaba without shortcut:")
    least_rotation("aaba", emit=lambda j,k,changes,early: print(f"j={j}; candidate={k}; updates={changes}; early={early}"))
    counts = 0
    for alphabet, maximum in (("ab", 11), ("abc", 7)):
        for n in range(maximum + 1):
            for letters in product(alphabet, repeat=n):
                text = "".join(letters)
                index = least_rotation(text)
                minimum, _ = brute(text)
                require(rotate(text,index) == minimum, "rotation oracle")
                require(index == 0 if n == 0 else 0 <= index < n, "index range")
                counts += 1
    cyclic_checks = 0
    for text in ("caba", "aaba", "baba", "aaaa", "你呀你", "😀a😀"):
        for k in range(len(text)):
            shifted = rotate(text,k)
            require(rotate(shifted, least_rotation(shifted)) == brute(text)[0], "rotation invariant")
            cyclic_checks += 1
    for n in range(2, 101):
        text = "a" * n + "ba"
        require(rotate(text,least_rotation(text)) == brute(text)[0], "published regression family")
    for invalid in (None, 123, b"ab", ["a"]):
        try:
            least_rotation(invalid)
        except TypeError:
            pass
        else:
            raise AssertionError("invalid input accepted")
    print(f"exhaustive_strings={counts}; cyclic_invariance={cyclic_checks}; regression_family=99; invalid_inputs=4")
    print("PASS: minimum rotation string agrees with enumeration; tied positions may coexist")


if __name__ == "__main__":
    main()