prize18-3  Exercise 2: Latin-1 Against Windows-1252
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Prediction: the two tables agree everywhere except 0x80 to 0x9F (32 values). In that block Latin-1 has control characters and Windows-1252 has printable
characters, except for five bytes that Windows-1252 leaves undefined (0x81, 0x8D, 0x8F, 0x90, 0x9D). So 0x9D should print "(undefined)" and the rest should differ.

ex2.py:

  def compare(b):
      one = bytes([b]).decode("latin-1")
      try:
          two = bytes([b]).decode("cp1252")
      except UnicodeDecodeError:
          two = "(undefined)"
      return one, two

  for b in [0x80, 0x85, 0x92, 0x93, 0x99, 0x9D, 0xA3, 0xE9]:
      one, two = compare(b)
      print(f"0x{b:02X}  latin-1 U+{ord(one):04X}  cp1252 {two!r}")

  diff = [b for b in range(256) if compare(b)[0] != compare(b)[1]]
  print(len(diff), "differ, all inside", hex(min(diff)), "-", hex(max(diff)))

Real output (Python 3.14):

  0x80  latin-1 U+0080  cp1252 '€'
  0x85  latin-1 U+0085  cp1252 '…'
  0x92  latin-1 U+0092  cp1252 '’'
  0x93  latin-1 U+0093  cp1252 '“'
  0x99  latin-1 U+0099  cp1252 '™'
  0x9D  latin-1 U+009D  cp1252 '(undefined)'
  0xA3  latin-1 U+00A3  cp1252 '£'
  0xE9  latin-1 U+00E9  cp1252 'é'
  32 differ, all inside 0x80 - 0x9f

Note that the five undefined bytes count as "differ" because the string "(undefined)" is not the Latin-1 control character.