prize18-6  Exercise 1: Code Points Against Clusters
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Predictions: every input is one thing to a reader, so every cluster count is 1: the thumbs-up (1 code point), the thumbs-up with a tone modifier (2),
the person-with-tone joined to a laptop (4: person, tone, ZWJ, laptop), the flag (2 regional indicators), the plain heart (1) and the heart with
variation selector 16 (2).

ex1.py (needs the third-party regex module: pip install regex):

  import regex

  def count(s):
      return len(regex.findall(r"\X", s))

  def report(s):
      cps = " ".join(f"U+{ord(c):04X}" for c in s)
      print(f"{len(s)} code points, {count(s)} clusters: {cps}")

  report("\U0001F44D")
  report("\U0001F44D\U0001F3FD")
  report("\U0001F469\U0001F3FB\u200D\U0001F4BB")
  report("\U0001F1E9\U0001F1EA")
  report("\u2764")
  report("\u2764\uFE0F")

Real output (Python 3.14):

  1 code points, 1 clusters: U+1F44D
  2 code points, 1 clusters: U+1F44D U+1F3FD
  4 code points, 1 clusters: U+1F469 U+1F3FB U+200D U+1F4BB
  2 code points, 1 clusters: U+1F1E9 U+1F1EA
  1 code points, 1 clusters: U+2764
  2 code points, 1 clusters: U+2764 U+FE0F

All six clusters equal 1, as predicted. The code point count ranges from 1 to 4, so len() would give a wrong answer to "how many characters does the user see?".