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
Assumptions
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)
Relations
When it doesn't appear in the digraph ch, h is silent. Thus, for example:
0. Silent letters
- ha = a ⇒ h=1
{a,á,b,c,d,e,é,f,g,[h,1],i,í,j,k,l,m,n,ñ,o,ó,p,q,r,s,t,u,ú,ü,v,w,x,y,z}
Some letters are pronounced identically (in certain contexts), for example:
- Identical sounds
- bote = vote ⇒ b=v
- agito = ajito ⇒ g=j
- encima = enzima ⇒ c=z
- cappa = kappa ⇒ c=k
- samurái = samuray ⇒ i=y (1.6)
- Diacritics
- cual=cuál ⇒ a=á
- el=él ⇒ e=é
- si=sí ⇒ i=í
- como=cómo ⇒ o=ó
- tu=tú ⇒ u=ú
{[a,á],[b,v],[z,c,k],d,[e,é],f,[g,j],[h,1],[i,í,y],l,m,n,ñ,[o,ó],p,q,r,s,t,[u,ú],ü,w,x}
- Old orthographies
- quilo = kilo ⇒ qu=k ⇒ cu=k (3.1) ⇒ u=1 (1.4)
- México = Méjico ⇒ x=j
{[a,á],[b,v],[z,c,k,qu],d,[e,é],f,[g,j,x],[h,1],[i,í,y],l,m,n,ñ,[o,ó],p,q,r,s,t,[u,ú],ü,w}
- Latinisms
- cuórum = quorum ⇒ c=q
- sub iudice = sub judice ⇒ i=j
{[a,á],[b,v],[z,c,k,q],d,[e,é],f,[g,x,j,i,í,y],[h,u,ú,1],l,m,n,ñ,[o,ó],p,r,s,t,ü,w}
- Unnativised orthographies
- huincha = wincha ⇒ hu=w ⇒ u=w (0.1) ⇒ w=1 (2.1)
- wolframio = volframio ⇒ w=v ⇒ v=1 (4.1)
- detall = detal ⇒ ll=l ⇒ l=1
- sunní = suní ⇒ nn=n ⇒ n=1
- judo = yudo ⇒ j=y
dup 1.5, 3.2
{[a,á],[z,c,k,q],d,[e,é],f,[g,x,j,i,í,y],[l,h,b,v,w,u,ú,n,1],m,ñ,[o,ó],p,r,s,t,ü}
- Greek consonant clusters
- gneis = neis ⇒ gn=n ⇒ g=1
- psicología = sicología ⇒ ps=s ⇒ p=1
- cneoráceo = neoráceo ⇒ cn=n ⇒ c=1
- mnemónica = nemónica ⇒ mn=n ⇒ m=1
{[a,á],d,[e,é],f,[z,c,k,q,p,m,g,x,j,i,í,y,l,h,b,v,w,u,ú,n,1],ñ,[o,ó],r,s,t,ü}
- Reduced consonant clusters (prefixes)
- substancia = sustancia ⇒ subs=sus ⇒ b=1
dup
- transalpino = trasalpino ⇒ trans=tras ⇒ n=1
dup
- consciencia = conciencia ⇒ cons=con ⇒ s=1
- postmoderno = posmoderno ⇒ post=pos ⇒ t=1
{[a,á],d,[e,é],f,[z,c,k,q,p,m,g,x,j,i,í,y,l,h,b,v,w,u,ú,t,n,1],ñ,[o,ó],r,s,ü}
- Alophones
- huaca = guaca ⇒ hu=gu ⇒ h=g ⇒ g=1 (0.1)
dup
- huemul = güemul ⇒ hu=gü ⇒ hu=ü (5.1) ⇒ u=ü (0.1) ⇒ ü=1 (2.1)
- excusa = escusa ⇒ xc=sc ⇒ x=s
- envasar = embasar ⇒ nv=mb ⇒ n=m (1.1)
dup
{[a,á],d,[e,é],f,[z,c,k,q,p,m,g,x,s,j,i,í,y,l,h,b,v,w,u,ú,ü,t,n,1],ñ,[o,ó],r}
- Synalepha
- contraalmirante = contralmirante ⇒ aa=a ⇒ a=1 (see also bezaar > bezar etc)
{d,[e,é],f,[a,á,z,c,k,q,p,m,g,x,s,j,i,í,y,l,h,b,v,w,u,ú,ü,t,n,1],ñ,[o,ó],r}
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:
- Further loans
- sioux = siux ⇒ o=1
- soufflé = suflé ⇒ ouf=u (9.1) ⇒ uf=u ⇒ f=1
- toffee = tofe ⇒ fe=1 ⇒ e=1 (9.2)
- caddie = cadi ⇒ caddi=cadi (9.3) ⇒ dd=d ⇒ d=1
{[e,é,o,ó,f,a,á,z,c,k,q,p,m,g,x,s,j,i,í,y,l,h,b,v,w,u,ú,ü,t,n,d,1],ñ,r}
Now, finally we come to rr. First, the following lemma on hyphens:
- Hyphens
- fino-ugrio = finoúgrio > -=1
Now:
- Acronyms and abbreviations
- CD-ROM = cederrón (10.1) ⇒ CDROM=cederrón ⇒ r=rr ⇒ r=1
- Letters
- r = erre ⇒ r=rr ⇒ r=1
dup 11.1
- 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:
- ...
distinción seseo
- seda = ceda = zeda ⇒ s=c=z
dup
no yeísmo yeísmo
- arrollo = arroyo ⇒ ll=y
dup
- elision of intervocal d:
- cantador = cantaór ⇒ d=1 (1.6)
- 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:
- Hiatus > diphthong
- noroeste = norueste ⇒ o=u (e.g. toalla > [ˈtwaja]) ⇒ o=1
- óleo = olio ⇒ i=e (e.g. beatitud > [bʝatiˈtuð]) ⇒ e=1
- Hypercorrection
- 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:
- Further Alophones
- icnita = ignita ⇒ c=g (/k/ voiced approximant [ɣ] before a voiced consonant)
dup
- yezgo = yedgo⇒ z=d (/θ/ voiced [ð] before a voiced consonant) ⇒ d=1
- zafra =
[ˈθavɾa]
~ [ˈθaβɾa]
= zabra ⇒ f=b (/f/ voiced [v] before a voiced consonant, [v] allophone of /b/) (e.g. afgano = [avˈgano]) ⇒ f=1
- 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]
.
Ñ
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?
e=1 → bat = bat•1 = bat•e = bate
which is not pronounced the same as bat.