Exercise 3 solution

    from datetime import datetime, timezone
    from zoneinfo import ZoneInfo

    def classify(dt):
        first = dt.replace(fold=0).utcoffset()
        second = dt.replace(fold=1).utcoffset()
        if first == second:
            return "normal"
        return "repeated" if first > second else "gap"

    ny = ZoneInfo("America/New_York")
    lon = ZoneInfo("Europe/London")
    tests = [
        ("New York", datetime(2026, 3, 8, 2, 30, tzinfo=ny)),
        ("New York", datetime(2026, 11, 1, 1, 30, tzinfo=ny)),
        ("New York", datetime(2026, 11, 1, 2, 30, tzinfo=ny)),
        ("London", datetime(2026, 3, 29, 1, 30, tzinfo=lon)),
        ("London", datetime(2026, 10, 25, 1, 30, tzinfo=lon)),
        ("London", datetime(2026, 7, 4, 12, 0, tzinfo=lon)),
    ]
    for city, dt in tests:
        print(f"{city:<9} {dt:%Y-%m-%d %H:%M}  {classify(dt)}")

    rep = datetime(2026, 11, 1, 1, 30, tzinfo=ny)
    for f in (0, 1):
        r = rep.replace(fold=f)
        print(f"fold={f}: {r.tzname()}  UTC {r.astimezone(timezone.utc):%H:%M}")

Output (run, and exactly as shown):

    New York  2026-03-08 02:30  gap
    New York  2026-11-01 01:30  repeated
    New York  2026-11-01 02:30  normal
    London    2026-03-29 01:30  gap
    London    2026-10-25 01:30  repeated
    London    2026-07-04 12:00  normal
    fold=0: EDT  UTC 05:30
    fold=1: EST  UTC 06:30

Explanation: the same wall reading is asked for its offset with fold=0 and
fold=1. If the two answers agree the time is ordinary. In a repeated hour the
first pass has the larger offset (summer time), so first > second. In a gap
Python's convention gives the pre-jump offset for fold=0 and the post-jump
offset for fold=1, so first < second. New York 02:30 on 1 November is ordinary
because the repeated hour there is 01:00-02:00. The final lines show the two
passes through the repeated hour are one real hour apart in UTC.