prize18-5  Exercise 2: A Matching Key for User Names
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Prediction: the first three names (composed, decomposed, upper case) and the fullwidth-letter one should share one key; "Zoe" (no accent) and "zoes" (a
different word) stay separate. So six raw names should reduce to three keys.

ex2.py:

  import unicodedata as ud

  def key(name):
      return ud.normalize("NFKC", name).casefold()

  names = ["Zo\u00eb", "Zoe\u0308", "ZO\u00cb", "\uff3a\uff4f\u00eb", "Zoe", "zo\u00ebs"]
  seen = {}
  for n in names:
      seen.setdefault(key(n), []).append(n)
  for k, group in seen.items():
      print(f"{k!r:8} <- {group}")
  print(len(names), "names,", len(seen), "distinct")
  print(len({n for n in names}), "distinct if compared raw")

Real output (Python 3.14):

  'zoë'    <- ['Zoë', 'Zoë', 'ZOË', 'Ｚｏë']
  'zoe'    <- ['Zoe']
  'zoës'   <- ['zoës']
  6 names, 3 distinct
  6 distinct if compared raw

Note that "Zoë" and "Zoe" stay different: normalization never removes an accent. Whether they should count as the same name is a policy decision for your
application, not a Unicode one. Also note that a key like this is for comparison only; display and store the name the user typed.