prize18-1  Exercise 1: Code Points and Byte Counts
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Predictions: ASCII letters need 1 byte; U+0080-U+07FF need 2; U+0800-U+FFFF need 3; U+10000 and above need 4.
So: A=1, ss (U+00DF)=2, Omega (U+03A9)=2, euro (U+20AC)=3, the CJK character (U+8A9E)=3, party popper (U+1F389)=4.

ex1.py:

  for ch in ["A", "\u00df", "\u03a9", "\u20ac", "\u8a9e", "\U0001F389"]:
      cp = ord(ch)
      print(f"{ch}  U+{cp:04X}  decimal={cp:<7} utf-8 bytes={len(ch.encode('utf-8'))}")

Real output (Python 3.14, run with py -X utf8 ex1.py):

  A  U+0041  decimal=65      utf-8 bytes=1
  ß  U+00DF  decimal=223     utf-8 bytes=2
  Ω  U+03A9  decimal=937     utf-8 bytes=2
  €  U+20AC  decimal=8364    utf-8 bytes=3
  語  U+8A9E  decimal=35486   utf-8 bytes=3
  🎉  U+1F389  decimal=127881  utf-8 bytes=4

The byte count grows with the size of the code point, matching the ranges in the chapter.
Note: the \u escapes above are written as plain escapes so the file itself stays ASCII.