prize18-6  Exercise 3: Truncate Without Splitting a Cluster
===========================================================

Plan: split into clusters, keep limit - 1 of them and add the ellipsis when the text is too long (so the result has exactly limit clusters),
and return the text unchanged when it already fits. The naive slice s[:limit] counts code points, so it can cut a cluster in two.

ex3.py (needs the third-party regex module):

  import regex

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

  def fits(s, limit):
      return len(graphemes(s)) <= limit

  def truncate(s, limit, marker="\u2026"):
      g = graphemes(s)
      if len(g) <= limit:
          return s
      return "".join(g[:limit - 1]) + marker

  for s in ["hello world", "cafe\u0301 au lait", "go \U0001F468\u200D\U0001F469\u200D\U0001F467 now"]:
      t = truncate(s, 6)
      print(ascii(t), len(graphemes(t)), len(t))
  print(ascii("go \U0001F468\u200D\U0001F469\u200D\U0001F467 now"[:6]))

Real output (Python 3.14):

  'hello\u2026' 6 6
  'cafe\u0301 \u2026' 6 7
  'go \U0001f468\u200d\U0001f469\u200d\U0001f467 \u2026' 6 10
  'go \U0001f468\u200d\U0001f469'

Each result has exactly 6 clusters, though 6, 7 and 10 code points: the accent and the family stay whole. The naive slice (last line) cuts the
family emoji after two members, so only two of the three remain.