From the math.stackexchange question: The homophonic group: a mathematical diversion:

By definition, English words have the same pronunciation if their phonetic spellings in the dictionary are the same. The homophonic group H is generated by the letters of the alphabet, subject to the following relations: English words with the same pronunciation represent equal elements of the group. Thus be=bee, and since H is a group, we can conclude that e=1 (why?). Try to determine the group H.

This is an exercise from Michael Artin's Algebra on, well, abstract algebra.

In this exercise for the English language, words are equal if they are homophones, kind of like a formalisation of the joke that sin(x)/n=6. So in English:

  • bee=be → This implies e=1 by cancellation of b and e.

  • buy=by → This implies u=1 by cancellation of b and y.

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

  • canvass = canvas → This implies s=1 by cancellation of c,a,n,v,a and s. By canvass=canvas and rase=raze, we have s=z=1.

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

What then is the analagous group for Spanish? Equivalently, what Spanish letters won't equal 1?

  • 5
    I think i would upvote this if i could understand the question – Mike Aug 16 '18 at 19:40
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    @Mike The game is considering that each letter is a number, and words are "numbers put together in multiplication. So buy=by actually means "b·u·y=b·y, so u must equal 1. – FGSUZ Aug 16 '18 at 20:07
  • 1
    so. "copa" = "paco" ? – Mike Aug 16 '18 at 20:47
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    @Mike assuming the group is commutative, yes. – brazofuerte Aug 16 '18 at 21:58
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    @BCLC nope, but in English neither do till and lilt, despite being equal elements of the group in English (since all letters are the identity). The set of homophone words is the equality relation we start with, but this implies equality between many more words which aren't homophones. E.g. e=1 → bat = bat•1 = bat•e = bate which is not pronounced the same as bat. – brazofuerte Aug 17 '18 at 14:38

This question may have different answers depending on a number of factors, including:

  • the dialect(s) of Spanish you consider
  • which words you include
  • which letters you consider distinct
  • how narrow a phonemic/phonetic transcription you use for equivalence

Abusing a combination of interpretations of these rules, we may be able to get Spanish down to the trivial group {e}, but let's see how far we can get with a minimal number of assumptions.

Note: all rules appendixed with dup duplicate other equivalences and aren't strictly necessary. The overlap is maintained here in case an earlier rule is disputed.

A. The Castilian Case


To start off, let's assume:

  • distinción
  • no yeísmo
  • letters with diacritics are distinct
  • only consider words which appear in the DLE
  • only consider phonetic transcriptions or equivalent pronunciations described/implied by the RAE in its publications (DLE, DPD etc)


When it doesn't appear in the digraph ch, h is silent. Thus, for example:

0. Silent letters

  1. ha = a ⇒ h=1

Some letters are pronounced identically (in certain contexts), for example:

  1. Identical sounds
    1. bote = vote ⇒ b=v
    2. agito = ajito ⇒ g=j
    3. encima = enzima ⇒ c=z
    4. cappa = kappa ⇒ c=k
    5. samurái = samuray ⇒ i=y (1.6)
    6. Diacritics
      • cual=cuál ⇒ a=á
      • el=él ⇒ e=é
      • si=sí ⇒ i=í
      • como=cómo ⇒ o=ó
      • tu=tú ⇒ u=ú
  1. Old orthographies
    1. quilo = kilo ⇒ qu=k ⇒ cu=k (3.1) ⇒ u=1 (1.4)
    2. México = Méjico ⇒ x=j
  1. Latinisms
    1. cuórum = quorum ⇒ c=q
    2. sub iudice = sub judice ⇒ i=j
  1. Unnativised orthographies
    1. huincha = wincha ⇒ hu=w ⇒ u=w (0.1) ⇒ w=1 (2.1)
    2. wolframio = volframio ⇒ w=v ⇒ v=1 (4.1)
    3. detall = detal ⇒ ll=l ⇒ l=1
    4. sunní = suní ⇒ nn=n ⇒ n=1
    5. judo = yudo ⇒ j=y dup 1.5, 3.2
  1. Greek consonant clusters
    1. gneis = neis ⇒ gn=n ⇒ g=1
    2. psicología = sicología ⇒ ps=s ⇒ p=1
    3. cneoráceo = neoráceo ⇒ cn=n ⇒ c=1
    4. mnemónica = nemónica ⇒ mn=n ⇒ m=1
  1. Reduced consonant clusters (prefixes)
    1. substancia = sustancia ⇒ subs=sus ⇒ b=1 dup
    2. transalpino = trasalpino ⇒ trans=tras ⇒ n=1 dup
    3. consciencia = conciencia ⇒ cons=con ⇒ s=1
    4. postmoderno = posmoderno ⇒ post=pos ⇒ t=1
  1. Alophones
    1. huaca = guaca ⇒ hu=gu ⇒ h=g ⇒ g=1 (0.1) dup
    2. huemul = güemul ⇒ hu=gü ⇒ hu=ü (5.1) ⇒ u=ü (0.1) ⇒ ü=1 (2.1)
    3. excusa = escusa ⇒ xc=sc ⇒ x=s
    4. envasar = embasar ⇒ nv=mb ⇒ n=m (1.1) dup
  1. Synalepha
    1. contraalmirante = contralmirante ⇒ aa=a ⇒ a=1 (see also bezaar > bezar etc)

So, with our initial conditions we can reduce the alphabet down to the free group on 6 generators:

A:    {1,d,e,f,ñ,o,r}

B. Crude loanwords and abbreviations/acronyms

Including crude loanwords (those that appear in italics in the DLE) which have nativised doublets, we can gain a few more relations:

  1. Further loans
    1. sioux = siux ⇒ o=1
    2. soufflé = suflé ⇒ ouf=u (9.1) ⇒ uf=u ⇒ f=1
    3. toffee = tofe ⇒ fe=1 ⇒ e=1 (9.2)
    4. caddie = cadi ⇒ caddi=cadi (9.3) ⇒ dd=d ⇒ d=1

Now, finally we come to rr. First, the following lemma on hyphens:

  1. Hyphens
    1. fino-ugrio = finoúgrio > -=1


  1. Acronyms and abbreviations
    1. CD-ROM = cederrón (10.1) ⇒ CDROM=cederrón ⇒ r=rr ⇒ r=1
    2. Letters
      1. r = erre ⇒ r=rr ⇒ r=1 dup 11.1
      2. c = ce ⇒ e=1 dup 9.2

Which leaves us with:

A,B:    {1,ñ}

The free group on one generator.

Assuming a different dialect and relaxing our restriction on RAE-sanctioned pronunciations (i.e. including pronunciations it recognises, but admonishes as not belonging to 'habla culta'), we can reduce the group without relying on italicised loanwords or abbreviations (B):

C. Andalusia

We revise our assumptions. Note the following changes only add further equivalences, and do not negate existing ones:

  1. ...
    1. distinción seseo
      • seda = ceda = zeda ⇒ s=c=z dup
    2. no yeísmo yeísmo
      • arrollo = arroyo ⇒ ll=y dup
    3. elision of intervocal d:
      • cantador = cantaór ⇒ d=1 (1.6)
    4. elision of terminal r and d:
      • comer = [ko'me] = comed ⇒ r=d ⇒ r=1 (12.3)

We thus achieve the free group on 4 generators:

A,C:    {1,e,f,ñ,o}

D. South America

Although the RAE proscribes pronouncing formal examples of hiatus as diphthongs in 'habla esmerada', it does note that this occurs even in educated speech in Mexico and other South American countries. Thus we might also assume the following equivalences:

  1. Hiatus > diphthong
    1. noroeste = norueste ⇒ o=u (e.g. toalla > [ˈtwaja]) ⇒ o=1
    2. óleo = olio ⇒ i=e (e.g. beatitud > [bʝatiˈtuð]) ⇒ e=1

  1. Hypercorrection
    1. buganvilla = buganvilia ⇒ ll=li ⇒ l=i ⇒ l=1 dup

Thus we could alternatively reduce the group to:

A,D:    {1,d,f,ñ,r}

E. Other alophones

As opposed to using italicised loanwords or a specific dialect, we could consider other alophonic equivalences not explicitly stated by the RAE:

  1. Further Alophones
    1. icnita = ignita ⇒ c=g (/k/ voiced approximant [ɣ] before a voiced consonant) dup
    2. yezgo = yedgo⇒ z=d (/θ/ voiced [ð] before a voiced consonant) ⇒ d=1
    3. zafra = [ˈθavɾa] ~ [ˈθaβɾa] = zabra ⇒ f=b (/f/ voiced [v] before a voiced consonant, [v] allophone of /b/) (e.g. afgano = [avˈgano]) ⇒ f=1
    4. desrabar = derrabar1 2 (p.50) ⇒ sr=rr ⇒ s=r ⇒ r=1 Elision of 's' in consonant cluster 'sr' and fricative realisation of 'r' i.e. 'r' > 'rr' (fricative interpreted by native speakers as allophonic to trill, never to tap)
      (Note: the RAE itself claims sr=srr)

This would leave us with:

A,E:    {1,e,o,ñ}

Note: if we assume a seseo dialect we must omit rule 14.2 since the voiced seseo realisation of /θ/ is [z], not [ð].


The existence of the following pair:

pergeño, pergenio

And given /n/ has a palatalized alophone [ɲ] ("ñ") before palatals ([ʎ], [j], [ʝ], [dʒ]), makes it very tempting to try and find a way to reduce ñ to 1, but so far I haven't been able to find a convincing example.

Related questions:

- Exception to the Phonetic Rule
- "Guion" vs "Guión" - Are there other words which could be written in multiple ways?

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  • 1
    But are all these actually homophones? Aren't some indeed variations of pronunciations, and as such, all but homophones? – Rafael Aug 17 '18 at 14:13
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    @Rafael I've reworked my answer to more clearly delineate which assumptions on equivalences I'm making at each step. Hopefully this is clearer. – brazofuerte Feb 18 '19 at 17:10
  • Can possibly remove 'e' also: abc = abecé ⇒ eé=1 ⇒ ee=1 etc – brazofuerte Apr 29 '19 at 12:26
  • Can remove d=g: lema.rae.es/dpd/srv/search?key=s%E1ndwich – brazofuerte Jan 1 at 10:03

It may be easier to consider this in the reverse. Which letters could be 1?

Let's begin with the vowels:

A, E, O, and U can only be written as such, and are always pronounced (except for the u in the sequence [gq][ue][iy]). Removing any of them will change the pronunciation of the word. While I next to another vowel and Y are indeed generally equivalent, they are always pronounced, and again removing them will alter the pronunciation. In practice, it may be possible to consider words like cree to give E a score of 1, but in even minimally formal speech, each E is distinguished and thus cree is distinct from cre. So no vowels can have a score of 1.

For consonants, we'd need to find sequences with equivalent pronunciations. That would require a doubled letter or in someway finding letter sequences with equivalent pronunciations.

The only doubled consonants are cc, ll, nn, rr, each of which represent different sounds than an equivalently positioned c, l, n, r). Imported words like pizza may have special rules, but I'm not sure we can derive much from them, as there's no way to predict the rule that the reduced version would follow. Thus C/L/N/R cannot score a 1.

The only letter sequences with equivalent pronunciations in all dialects are B/V but since they are never found next to each other that doesn't change anything (and if they were, like with nn they'd be split between syllables), and N/M and X/J in certain positions. Dialectically, there are other possibilities (S/Z/C, LL/Y, D/Z, D/T, H/J, X/S, S/J) but they will all be beholden to the same issues and won't allow a score of 1.

H can be removed without varying the pronunciation in words in which it is not preceded by a C. Thus it can score a 1. Dialectically, an H and and a J could potentially have the same pronunciation, so in some areas, if I'm understanding the examples on the other page correctly, J could then equal 1 by extension of J=H, but if and only if those letters appear in separate distinct words. But that requires relying on a specific dialect, and isn't generalizable. But using that one and other dialects, we could maybe extend it to a few more letters, as sometimes S can sound like J, and thence to Z and C. But I think that'd be the extent of it.

All other consonants will only appear once in words, with no phonetic substitution possible, and thus their removal necessitates scores distinct from 1.

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  • Just a note, the combination 'bv' does occasionally appear: obvención, obvencional, obviable, obviamente, obviar, obviedad, obvio, subvención, subvencionable, subvencionar, subvenio, subvenir, subversión, subversivo, subversor, subvertir – brazofuerte Aug 16 '18 at 16:03
  • @ukemi ah, yes, I didn't think of prefixation. Here the removal of one will modify the pronunciation but with even less of a distinction as cree -> cre, although technically both /b/ are pronounced (consider how tiny the difference between the sequences obio, obvio and obpio. There is a difference, but in rapid speech it may be imperceptible). In that case, we may consider B=V=1 and possibly even P=1, but I don't think further extension is possible. – user0721090601 Aug 16 '18 at 16:17
  • Guifa do you agree with ukemi? – BCLC Aug 17 '18 at 11:09

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