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This is inspired by an exercise from Michael Artin's Algebra on, well, abstract algebra. This exercise for the English language is kind of like a formalisation of the joke that sin(x)/n = 6. Letters are interpreted as elements of a non-commutative group¹, a word is interpreted as the product of its letters, and two words are equal if they are homophones.

Using examples from English:

  • bee and be are homophones. Therefore we have bee = be and therefore e = 1 (by cancellation of b and e).

  • rase and raze are homophones. This implies s = z by cancellation of r, a, and e.

In English all 26 letters will eventually equal 1. Apparently, this was done for French and Czech.

My question is: What letters are not 1 when applying these rules to the German language?


¹ If you do not know what a group is, think of normal multiplication, just that you are not allowed to use a·b = b·a.

  • I'm voting to close this question as off-topic because it asks for a list of letters (examples). Hence, it does not really fit the Q&A format. – Arsak Aug 21 '18 at 3:58
  • Comments are not for extended discussion; this conversation has been moved to chat. – Takkat Aug 21 '18 at 20:00
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I will deliberately try to avoid proper names, and (recent) loan words as far and long as possible. I will not use alternative spellings of the same word or non-standard pronunciation at all.

Letter Duplication

In German orthography, duplication of a consonant indicates that the preceding vowel is short, while duplication of the vowels a, e, and o indicates that they are long. However, length of a vowel may also be given through what syllable in a word of how many syllables we have (or not at all), so duplication of letters may be unnecessary. Sometimes this duplication is then used to distinguish homophones in spelling or for etymological reasons. This allows us to reduce some common letters to 1:

  • Waagen = Wagena = 1
  • Heer = here = 1
  • schafft = Schaftf = 1
  • schallte = schaltel = 1
  • Mann = mann = 1
  • Boot = boto = 1
  • tappst = tapstp = 1
  • wirrst = wirstr = 1
  • dass = dass = 1
  • Spot = Spottt = 1 (alternatively via Kit = Kitt or Mathe = Matte)

Letter Quasiduplication

In some case, instead of letter duplication, another combination is used (kk → ck, zz → tz, ßß → ss, ii → ie) and the letter h indicates length of the preceding vowel in general. This allows to make similar reductions:

  • strickt = striktc = 1
  • wahr = warh = 1
  • Lied = Lide = 1 (not new)
  • Hertz = Herzt = 1 (not new)

Terminal Devoicing

As a next step, German pronunciation has a property called terminal devoicing that allows us to equate several consonants to each other (some of which we already know to be 1):

  • Mob = Moppb = pp = 1
  • Rad = Ratd = t = 1
  • Bug = bukg = k
  • reist = reißtß = s = 1

Homophonic letters and combinations

For the last systematic step, there are some letters or combinations of letters that can have the same pronunciation:

  • Lerche = Lärcheä = e = 1
  • heute = häuteä = e (not new)
  • Saite = Seitea = e = 1 (not new)
  • lax = Lachs = Lacksx = chs = 1 and g = k = cks = chs = 1
  • fiel = vielv = f = 1
  • Willen = Villenw = v = 1
  • Stadt = statt ⇒ d = t = 1 (not new)

Multiple Mechanisms

Some homophones which feature multiple of the above mechanisms or new ones help us to further weed out some letters:

  • Hemd = hämmtm = mm ⇒ m = 1 (thanks to npst)
  • Mythen = mühtenyth = ühty = ü
  • Trotz = Trottsz = ts = 1

Intermediate Result

At this point, we are left with i, j, q, u, y = ü, and ö.

Loanwords and Proper Names

By including recent loanwords, proper names, and similar, we can make further progress:

  • pulen = poolenu = oo = 1
  • Beat = biet ⇒ i = 1
  • Haydn = Heideny = ü = 1
  • Hoechst = höchstö = oe = 1.
  • Jens = Yensj = y = 1

After that, we are only left with q. Until further notice, the German Homophonic Group is the infinite cyclic group generated by the German letter q: {…, q⁻², q⁻¹, 1, q, q², q³, …}.

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    Ich finde nicht, dass Spot=Spott. – Carsten S Aug 17 '18 at 12:55
  • @CarstenS: Wenn es Dir um [spɔt] vs. [ʃpɔt] geht, so sind zumindest gemäß meiner Erfahrung und dem Duden beide Aussprachen geläufig. So oder so, es gibt auch Alternativen, die allerdings auch Eindeutschungen oder von Eigennamen Abgeleitetes bemühen (siehe meine Änderung). – Wrzlprmft Aug 17 '18 at 13:06
  • Die Gleichsetzung von Rad = Rat und von Bug = buk und auch von Hemd = hemmt funktioniert nur in jenen Regionen, in denen die Auslautverhärtung ausgeführt wird. Das ist im Süden (namentlich in Österreich) nicht der Fall. – Hubert Schölnast Aug 17 '18 at 13:50

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