Another Almost-Cartesian Consonant Inventory

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eldin raigmore
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Another Almost-Cartesian Consonant Inventory

Post by eldin raigmore »

Is the 38-consonant inventory described below a reasonable consonant inventory for a conlang that’s supposed to be naturalistic and realistic? Or at least learnable, speakable, and understandable?
It has
7 voiceless fricatives
7 voiced fricatives
7 nasals
6 voiceless plosives
6 voiced plosives
5 approximants.
::::::::::
It has
6 alveolars
6 retroflexes
6 palatals
6 velars
5 bilabials
5 uvulars
4 labiodentals
....................
It doesn’t have any labiodental plosives, neither voiceless nor voiced.
It doesn’t have any bilabial approximant nor any uvular approximant.
....................
I didn’t remember to say so at first; but all of its consonants are pulmonic egressive.
..........


Critique?
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Re: Another Almost-Cartesian Consonant Inventory

Post by Pabappa »

well you didnt post the inventory itself, so Im not sure .... im curious how you got 4 labiodentals if none of them is a stop .... im guessing one nasal and one approximant? that by itself puts you on the edge, because although a labiodental nasal is attested, its quite rare and seems to pattern almost as if it were an allophone of /mʷ/. likewise im curious what the five uvulars are.
I'll take the theses, and you can have the thoses.
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Re: Another Almost-Cartesian Consonant Inventory

Post by Creyeditor »

This is the best guess, I can come up with. It has some hardly natural parts in the labials and the palatals, but maybe someone can fix it.

/m ɱ n ɳ ɲ ŋ ɴ/
/p b t d ʈ ɖ c ɟ k g q ɢ/
/ɸ β f v s z ʂ ʐ ç ʝ x ɣ X ʁ/
/ʋ l ɻ j ʟ/
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Re: Another Almost-Cartesian Consonant Inventory

Post by eldin raigmore »

Creyeditor wrote: 28 May 2020 16:24 This is the best guess, I can come up with. It has some hardly natural parts in the labials and the palatals, but maybe someone can fix it.

/m ɱ n ɳ ɲ ŋ ɴ/
/p b t d ʈ ɖ c ɟ k g q ɢ/
/ɸ β f v s z ʂ ʐ ç ʝ x ɣ X ʁ/
/ʋ l ɻ j ʟ/
Mostly right.

12 stops/plosives: /p b t d ʈ ɖ c ɟ k g q ɢ/
7 nasals: /m ɱ n ɳ ɲ ŋ ɴ/
14 fricatives: /ɸ β f v s z ʂ ʐ ç ʝ x ɣ x̩ ʁ/
5 approximants: /ʋ ɹ ɻ j ɰ /

:::::

5 bilabials: /p b m ɸ β/
4 labiodentals: /ɱ f v ʋ/
6 alveolars: /t d n s z ɹ /
6 retroflexes: /ʈ ɖ ɳ ʂ ʐ ɻ/
6 palatals: /c ɟ ɲ ç ʝ j/
6 velars: /k g ŋ x ɣ ɰ /
5 uvulars: /q ɢ ɴ x̩ ʁ/

..........

It doesn’t have any laterals. So no /l/ and no /ʟ/. (Not having any laterals is a minority thing; but having more than one would also have been a minority thing.)
Also the voiceless uvular fricative is a lowercase Greek letter chi, rather than an uppercase one.

I don’t know how you typed all those IPA letters! I put in the Unicode hex for the three corrections; I hope you can read them!
&#x0270; turned lowercase Latin letter m with long right leg ɰ
&#x0279; turned lowercase Latin letter r ɹ
&#x03C7; lowercase Greek letter chi. Maybe I should use x̩ instead.

====================

@Pabappa:
I also worry about having both a bilabial consonant and a labiodental consonant with the same manner of articulation.
The labiodental approximant is IIANM the newest consonant phoneme symbol in the 2018 edition of the IPA.
But apparently some languages do have both phonemes of each such pair with a minimal pair of words attesting they can make accepted words with different meanings.

I also worry that it might crowd the space to articulate different consonants to have all four of alveolar, palatal, velar, and uvular places of articulation for the same manner of articulation. Each two such consonants are attested together in some natlang, but I am not certain all four are.
Actually it might be tough to have alveolar and palatal and velar at every MoA; and/or to have palatal and velar and uvular at every MoA.

==========

This is my second “almost-Cartesian” consonant inventory.
In my first, I included every consonant that either UPSID or PHOIBLE said was attested in >50% of the languages in their databases. Then I tried to include every phoneme that was the same place-of-articulation of one (or more) such consonant(s) and also the same voicing and manner-of-articulation of one (or more) such consonant(s); unless it was impossible to pronounce.
The only one left out was an affricate of a co-articulated bilabial-and-velar; I don’t think it can be done.
if i recall correctly there were 7 places (counting that co-articulation as a “place”) and 5 manners? I’ll have to look it back up.
Edit: viewtopic.php?f=6&t=711&p=152499&hilit= ... an#p152499
That table has 14 columns and 5 rows. Each combination of PoA and voicing has a column. Labio-velar is counted as a PoA; and affricate is counted as an MoA.
The thing is, every row (MoA) and every column (PoA) contained one (or two) consonant(s) that appears in half or more of the world’s natlangs; but all of the other consonants appear in around 1% or 2% of the world’s natlangs, if i recall correctly.

In this second attempt, I am just trying to make the inventory really be almost Cartesian.
Every combination of MoA and voicing is represented in all 7, or all but one (I.e. 6), or all but two (I.e. 5), PoAs;
7 7 7 6 6 5
And every PoA is represented in all 6, or all but one (I.e. 5), or all but two (I.e. 4), combinations of MoA and voicing;
6 6 6 6 5 5 4

I understand that most natlangs’ consonant inventories are more symmetrical than they would be if they were purely random, but less symmetrical than a Cartesian (or as close as possible to Cartesian) inventory would be.
But, barring that bit of unnaturalness and irrealism, I’m hoping this is (otherwise) naturalistic and/or realistic; or, at least, learnable, understandable, and speakable.

Obviously, so far, this language is destined to be an engelang!

——————————

Thanks for your comments, both of you!

......

Should I call this quasi-Cartesian? (Almost Cartesian?)
Or meta-Cartesian? (Next to Cartesian?)
Or semi-Cartesian? (Half Cartesian?)
Last edited by eldin raigmore on 31 May 2020 23:19, edited 4 times in total.
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Re: Another Almost-Cartesian Consonant Inventory

Post by Creyeditor »

How about Para-Cartesian?
Edit: Also, why not a smaller inventory cartesian language? Just stops vs. non-stops and labial vs. coronal vs. dorsal?
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Re: Another Almost-Cartesian Consonant Inventory

Post by eldin raigmore »

Creyeditor wrote: 29 May 2020 22:38 How about Para-Cartesian?
Maybe!
Or maybe “sub-Cartesian” or “pre-Cartesian”?
Edit: Also, why not a smaller inventory cartesian language? Just stops vs. non-stops and labial vs. coronal vs. dorsal?
Why not, indeed?
.....
This second inventory was for a hypothetical 5Cons* in which no PoA nor MoA could be used twice in the same root; and PoAs and MoAs could be permuted independently of each other to mutate or derive or inflect the root.
I already know that such a system would be impractical (too few (126) roots and too many (14400) derivations/inflections), but I went ahead with the consonant inventory anyway.
*quinquiliteral roots
.....
Your proposed six-consonant inventory could be the third in the {quasi|meta|para|semi}-Cartesian “family”(?) of consonant inventories. What other features of a conlang would it help or hinder?
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Re: Another Almost-Cartesian Consonant Inventory

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eldin raigmore wrote: 31 May 2020 14:19 This second inventory was for a hypothetical 5Cons* in which no PoA nor MoA could be used twice in the same root; and PoAs and MoAs could be permuted independently of each other to mutate or derive or inflect the root. I already know that such a system would be impractical (too few (126) roots and too many (14400) derivations/inflections), but I went ahead with the consonant inventory anyway.
Interesting eng-y idea. Have you tried fleshing it out a bit more. Would these numbers be significantly improved if you allowed for fully identical consonants? That's what some natlangs do.
eldin raigmore wrote: 31 May 2020 14:19 Your proposed six-consonant inventory could be the third in the {quasi|meta|para|semi}-Cartesian “family”(?) of consonant inventories. What other features of a conlang would it help or hinder?
So, I guess you could have a lot of allophony, if we only keep the explicit features constant. There are lots of labial non-stops for example. Also if we don't allow for phonemic sonorancy plateaus, we could get a maximal syllable structure that is basically CCVCC. This would give us 4*4*4*4=256 consonantal syllable if onsets and codas are optional. If we allow for trisyllabic roots these should be 256^3=16777216 possible combinations. These should be enough, I guess. So, we would have complex syllables and long words, but only 6 consonants.
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Re: Another Almost-Cartesian Consonant Inventory

Post by eldin raigmore »

Creyeditor wrote: 01 Jun 2020 23:57 Interesting eng-y idea. Have you tried fleshing it out a bit more?
There are 13 PoAs in the 2018 IPA chart of pulmonic-egressive consonants.*
And, 13 combinations of MoA-and-voicing.*
(For purposes of this post I’ll temporarily include voicing as part of “MoA”.)
Edit: *No, there are 11. I must have meant to pick up two more from “off-chart”.

Consider a 4Cons in which the PoAs and MoAs can be permuted independently to derive and/or inflect the roots.

There are 715 subsets of 4 things out of 13 things;
there are 495 subsets of 4 things out of 12 things;
and there are 330 subsets of 4 things out of 11 things.

I consider 330*330 = 108,900 too few roots. At least, it’s fewer roots than the OED says English has.
But 495*330 = 330*495 = 163,350 is nearly enough;
and 715*330 = 330*715 = 235,950 is probably enough,
as is 495*495 = 245,025.
715*495 = 495*715 = 353,925 is more than enough.

So we’d want 13 PoAs and 11 MoAs, or 11 PoAs and 13 MoAs, or 12 PoAs and 12 MoAs, at least.

It might be that not all combinations of a set of four PoAs with a set of four MoAs have a pronounceable instantantiation.
Spoiler:
For instance, suppose the PoAs were
{bilabial, uvular, labiodental, velar} or
{bilabial, uvular, labiodental, retroflex};
and the MoAs were
{trill, lateral approximant, approximant, nasal} or
{trill, lateral approximant, approximant, voiced plosive} or
{tap-or-flap, lateral approximant, approximant, nasal} or
{tap-or-flap, lateral approximant, approximant, voiced plosive}.
For instance, if the PoAs were
{dental, postalveolar, pharyngeal, glottal}
and the MoAs were
{voiceless lateral fricative, voiced lateral fricative, trill, tap-or-flap};
there’d be no way.

—————

If each of the four MoAs can be pronounced at each of the four PoAs,
there’d be 4! = 24 ways to permute the MoAs and 4! =24 ways to permute the PoAs.
So, 24*24 = 576 ways to derive or inflect the root by just permuting the PoAs and MoAs.
I don’t know of any natlang that has much more than 100 basic word forms per root.
So that’s kind of too many.

However for some pairs of a set of four MoAs and a set of four PoAs, there’s only one way to match them up.
So you’d have four consonants you could permute in 24 ways.
Or, there might be two ways to match the MoAs with the PoAs, and thus 2*24 = 48 variations;
or four matches, so 4*24 = 96 variations;
or six matches, so 6*24 = 144 variations.

—————
Would these numbers be significantly improved if you allowed for fully identical consonants? That's what some natlangs do.
I have considered using sets of four consonants, no two of them homorganic.
And allowing permuting the individual consonants, but not the PoAs independently of the MoAs.
That would allow each root to be derived/inflected into 24 tetraliteral variants just by permuting the consonants.
I don’t know how many sets of four mutually-non-homorganic pulmonic egressive consonants there might be. But I bet there’re a lot more than 108,900.

That would just abandon the most-fun part; permuting the MoAs independently from the PoAs. Oh, well.

Another idea would be to allow a root to consist of any set of four distinct consonants as long as no three of them are homorganic; and allow any permutation that doesn’t put two homorganic consonants next to each other.
That would allow at least eight permutations per root.

For instance suppose we want to use /s n x g/.
We could allow
g-n-x-s
g-s-x-n
x-n-g-s
x-s-g-n
n-g-s-x
n-x-s-g
s-g-n-x
s-x-n-g
.
Eight variations.

————

Some, even many, languages share a strong preference, or possibly-violable constraint, against having the same phoneme (or, at least, the same consonant) appear twice in the same morpheme. I was going to see how far I could get making that constraint inviolable.

The reason for avoiding consecutive homorganic consonants, is that I wanted to avoid sound changes such as affrication, pre-nasalization, and pre-affrication.

It’s my understanding that for 3Cons natlangs, either the natlang itself or some ancestor thereof, had, for many of its roots
such as C1-C2-C3,
a form C1C2VC3
and a form C1VC2C3.
(Not necessarily the same vowel both times. And not necessarily the same vowel for two different roots!)

So I didn’t want two consecutive consonants of one root to be at the same PoA.
If the consonants can be permuted freely, that means no two of them can be at the same PoA.

—————
So, I guess you could have a lot of allophony, if we only keep the explicit features constant. There are lots of labial non-stops for example. Also if we don't allow for phonemic sonorancy plateaus, we could get a maximal syllable structure that is basically CCVCC. This would give us 4*4*4*4=256 consonantal syllable if onsets and codas are optional. If we allow for trisyllabic roots these should be 256^3=16777216 possible combinations. These should be enough, I guess. So, we would have complex syllables and long words, but only 6 consonants.
If we have optional onsets up to two-consonant clusters, and optional codas up to two-consonant clusters, and don’t allow the same consonant twice in the same syllable; and have no phonotactic constraints on which clusters can occur in either syllable margin; then we’d have
however many vowels there are (call it #V) of syllables of shape V
6*#V syllables shaped like CV
6*#V syllables shaped like VC
6*5*#V = 30*#V syllables shaped like CCV
6*5*#V = 30*#V syllables shaped like CVC
6*5*#V = 30*#V syllables shaped like VCC
6*5*4*#V = 120*#V syllables shaped like CCVC
6*5*4*#V = 120*#V syllables shaped like CVCC
6*5*4*3*#V = 360*#V syllables shaped like CCVCC

Totaling (1 + 2*6 + 3*30 + 2*120 + 360)*#V
= (1 + 12 + 90 + 240 + 360)*#V
= 703*#V
possible syllables.

In bisyllabic words or morphemes, there’d have to be some way to locate the internal syllable-boundary in shapes like
CCVCVCC or CCVCCVCC or CCVCCCVCC.
We wouldn’t really have (703^2)*(#V^2) distinct two-syllable sequences.

So something fewer than 494,209 consonant-sequences (including “no consonants”) with slots reserved for two as-yet-to-be-specified vowels.

That’s more than the OED says English has roots. And if I understand correctly only one other natlang is known to have more roots than English has!

—————

I’m not sure I followed why you decided to calculate 4*4*4*4=256 CCVCC patterns per vowel.
256*256 = 65,536 may be way the heck more roots than any conlang is likely to have, but still not much more than half as many roots as some of the more voluble natlangs have.
256*256*256 = 16,777,216 is indeed more than 16 times as many possibilities as I ever contemplated for a conlang!

....

Why, again, did you choose to calculate 4*4*4*4?

—————

Thank you for a very interesting response!
Last edited by eldin raigmore on 09 Jun 2020 18:03, edited 3 times in total.
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Re: Another Almost-Cartesian Consonant Inventory

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eldin raigmore wrote: 02 Jun 2020 03:52 Why, again, did you choose to calculate 4*4*4*4?
I though that we have four consonant slots in a CCVCC syllable and each of them can have any of three consonants or be empty. I figured it is three consonants and not six, because I did not want tautosyllabic consonant clusters of the type stop-stop or nonstop-nonstop. That's what I meant by sonorancy plateaus. But there might still be a logical mistake, this is all just midnight babbling [:D]
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Re: Another Almost-Cartesian Consonant Inventory

Post by eldin raigmore »

Creyeditor wrote: 02 Jun 2020 23:22I though that we have four consonant slots in a CCVCC syllable and each of them can have any of three consonants or be empty. I figured it is three consonants and not six, because I did not want tautosyllabic consonant clusters of the type stop-stop or nonstop-nonstop. That's what I meant by sonorancy plateaus. But there might still be a logical mistake, this is all just midnight babbling [:D]
I see! I misunderstood you about the sonority, then.
You’re suggesting, maybe,
(T)(N)V(N)(T)
Where (T) represents an optional sTop and (N) represents an optional Non-stop.
Right?
I do think that would give 256 syllables per nucleus. Just as you calculated.

———

If a bisyllabic string were like
(T)(N)VNTNV(N)(T)
we’d still need a way to decide whether that middle T was part of the coda of the first syllable or part of the onset of the second syllable.
(T)(N)VNT + NV(N)(T)
or
(T)(N)VN + TNV(N)(T)

We might have a convention that would settle it, perhaps like the Maximal Onset Principle.
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Re: Another Almost-Cartesian Consonant Inventory

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eldin raigmore wrote: 02 Jun 2020 23:48 If a bisyllabic string were like
(T)(N)VNTNV(N)(T)
we’d still need a way to decide whether that middle T was part of the coda of the first syllable or part of the onset of the second syllable.
(T)(N)VNT + NV(N)(T)
or
(T)(N)VN + TNV(N)(T)

We might have a convention that would settle it, perhaps like the Maximal Onset Principle.
Right, that's what I was looking for. I assumed that sonorancy plateaus, e.g. stop-stop sequences can occur across syllable boundaries. I imagined a bisyllabic string as following:

(T)(N)V(N)(T).(T)(N)V(N)(T)

The problem you mentioned is more severe. A string with a single intervocalic T or N can be generated in two ways, because there are two possible slots. This would reduce the number of possible roots again.

Still, I think including vowels (and thereby syllables) in roots makes it easier to reach a higher number of possible roots. Also, of course, I did not include any root consonant coocurrence restrictions like you did, but maybe including vowels in roots (and more complex syllable structure) allows you to solve your problem too?
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Re: Another Almost-Cartesian Consonant Inventory

Post by eldin raigmore »

Creyeditor wrote: 03 Jun 2020 17:29 Right, that's what I was looking for. I assumed that sonorancy plateaus, e.g. stop-stop sequences can occur across syllable boundaries. I imagined a bisyllabic string as following:

(T)(N)V(N)(T).(T)(N)V(N)(T)

The problem you mentioned is more severe. A string with a single intervocalic T or N can be generated in two ways, because there are two possible slots. This would reduce the number of possible roots again.

Still, I think including vowels (and thereby syllables) in roots makes it easier to reach a higher number of possible roots. Also, of course, I did not include any root consonant coocurrence restrictions like you did, but maybe including vowels in roots (and more complex syllable structure) allows you to solve your problem too?
I estimate there are almost 256^2 = 65536 two-syllable sequences per vowel-pair.
65536 is more than most conlangers would ever actually create; the usual limit is probably 50,000 or fewer.
But a natlang might require some multiple of that; maybe 3 to 5 times as many.

If we have two vowels and either syllable’s nucleus can be either vowel independently of what’s happening with the other,
there might be up to 2^2 * 256^2 = 512^2 = 256K = 262,144 two-syllable strings.

If we have three vowels, and the only restriction is each syllable’s nucleus must be a vowel different from the other syllable’s nucleus, we might have something a bit shy of 3*2 * 65,536 = 393,216 two-syllable morphemes (roots?). I’m sure that’s plenty even for a natlang.

....

The shapes of the two-syllable utterances are:
(T)(N)VV(N)(T) (256 times however many VV are permitted);
(T)(N)VCV(N)(T) (1536 times however many (V,V) pairs are permitted);
(T)(N)VCCV(N)(T) (9216 times however many (V,V) pairs are permitted);
(T)(N)VNTNV(N)(T) (6912 times however many (V,V) pairs are permitted);
(T)(N)VNTTV(N)(T) (6912 times however many (V,V) pairs are permitted);
(T)(N)VTTNV(N)(T) (6912 times however many (V,V) pairs are permitted);
and
(T)(N)VNTTNV(N)(T) (20,736 times however many (V,V) pairs are permitted).

That’s 52,480 times the number of permitted (V,V) pairs — 205/256 (about 80.08%) of my former estimate of 65,536.
With 2*2=4 permitted (V,V) pairs, there could be 209,920 two-syllable roots; which might be enough, or might not.
If there are 3*2 = 6 permitted (V,V) pairs, there could be 314,880 two-syllable roots; that would be comfortably more than enough IMHO.
But if there are 3*3 = 9 permitted (V,V) pairs, there could be 472,320 two-syllable roots, comfortably more than enough IMHO.

......

It’s clear how to calculate the number of three-syllable strings; it’s just too tedious to actually do it at the moment.
I guesstimate there’ll be more than 10,760,000 of them. (Rounding off 256*205*205.)
I could be way off.

————

Thoughts?
Edit:
Also, of course, I did not include any root consonant coocurrence restrictions like you did, but maybe including vowels in roots (and more complex syllable structure) allows you to solve your problem too?
I meant the constraint against two uses of the same phoneme (or, at least, the same consonant) in the same morpheme, to be part of my motivation for the consonant inventory I proposed at the head of this thread.
I don’t think that constraint, or any other phonotactic constraint than the ones you proposed, are really going to be compatible with such a small (6) consonant inventory as you’ve proposed.
I did exclude any NT onset-clusters and any TN coda-clusters. I thought you would want that. Is I wrong?
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Re: Another Almost-Cartesian Consonant Inventory

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eldin raigmore wrote: 03 Jun 2020 18:13 I did exclude any NT onset-clusters and any TN coda-clusters. I thought you would want that. Is I wrong?
No you is totally right [:D]
eldin raigmore wrote: 03 Jun 2020 18:13 I estimate there are almost 256^2 = 65536 two-syllable sequences per vowel-pair.
65536 is more than most conlangers would ever actually create; the usual limit is probably 50,000 or fewer.
But a natlang might require some multiple of that; maybe 3 to 5 times as many.

If we have two vowels and either syllable’s nucleus can be either vowel independently of what’s happening with the other,
there might be up to 2^2 * 256^2 = 512^2 = 256K = 262,144 two-syllable strings.

If we have three vowels, and the only restriction is each syllable’s nucleus must be a vowel different from the other syllable’s nucleus, we might have something a bit shy of 3*2 * 65,536 = 393,216 two-syllable morphemes (roots?). I’m sure that’s plenty even for a natlang.

....

The shapes of the two-syllable utterances are:
(T)(N)VV(N)(T) (256 times however many VV are permitted);
(T)(N)VCV(N)(T) (1536 times however many (V,V) pairs are permitted);
(T)(N)VCCV(N)(T) (9216 times however many (V,V) pairs are permitted);
(T)(N)VNTNV(N)(T) (6912 times however many (V,V) pairs are permitted);
(T)(N)VNTTV(N)(T) (6912 times however many (V,V) pairs are permitted);
(T)(N)VTTNV(N)(T) (6912 times however many (V,V) pairs are permitted);
and
(T)(N)VNTTNV(N)(T) (20,736 times however many (V,V) pairs are permitted).

That’s 52,480 times the number of permitted (V,V) pairs — 205/256 (about 80.08%) of my former estimate of 65,536.
With 2*2=4 permitted (V,V) pairs, there could be 209,920 two-syllable roots; which might be enough, or might not.
If there are 3*2 = 6 permitted (V,V) pairs, there could be 314,880 two-syllable roots; that would be comfortably more than enough IMHO.
But if there are 3*3 = 9 permitted (V,V) pairs, there could be 472,320 two-syllable roots, comfortably more than enough IMHO.

......

It’s clear how to calculate the number of three-syllable strings; it’s just too tedious to actually do it at the moment.
I guesstimate there’ll be more than 10,760,000 of them. (Rounding off 256*205*205.)
I could be way off.
Thank you for calculating (and guesstimate) this. So a language with TNVNT syllable structure, 6 consonants, three vowels and bisyllabic roots CAN have enough roots. That I find very comforting in some way.
eldin raigmore wrote: 03 Jun 2020 18:13
Edit: Also, of course, I did not include any root consonant coocurrence restrictions like you did, but maybe including vowels in roots (and more complex syllable structure) allows you to solve your problem too?
I meant the constraint against two uses of the same phoneme (or, at least, the same consonant) in the same morpheme, to be part of my motivation for the consonant inventory I proposed at the head of this thread.
I was unclear about what I meant I think. Imagine your quasi-cartesian consonant inventory plus quinquiliteral root, but modified. Instead of having only one root CCCCC for a given five-consonant combination you could have different syllable structures. This would make sense, since you only want to PoA and MoA permutation for inflection and not syllable structure. For our CCCCC root and (C)(C)V(C)(C) syllables and maximally bisyllabic roots (and assuming some Maximum Onset Principle) this gives 8 possible syllabifications.
CV.CCVCC
VC.CCVCC
VCC.CCVC
CVC.CCVC
CCV.CVCC
CCV.CCVC
CVCC.CCV
CCVC.CCV

If we have 10 vowels that would give you 126*8*10 roots, which is still clearly not enough. So maybe syllable structure does not help so much after all.
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Re: Another Almost-Cartesian Consonant Inventory

Post by eldin raigmore »

10,080 roots isn’t enough, but it’s still quite a few.
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Re: Another Almost-Cartesian Consonant Inventory

Post by eldin raigmore »

“Creyeditor” wrote: I was unclear about what I meant I think. Imagine your quasi-cartesian consonant inventory plus quinquiliteral root, but modified. Instead of having only one root CCCCC for a given five-consonant combination you could have different syllable structures. This would make sense, since you only want to PoA and MoA permutation for inflection and not syllable structure. For our CCCCC root and (C)(C)V(C)(C) syllables and maximally bisyllabic roots (and assuming some Maximum Onset Principle) this gives 8 possible syllabifications.
CV.CCVCC
VC.CCVCC
VCC.CCVC
CVC.CCVC
CCV.CVCC
CCV.CCVC
CVCC.CCV
CCVC.CCV

If we have 10 vowels that would give you 126*8*10 roots, which is still clearly not enough. So maybe syllable structure does not help so much after all.
I think the main problem — the main reason I came up with only 126=6*21 quinquiliteral roots in a system where PoAs and MoAs can be permuted independently under the same head-lexeme, and can’t be re-used within the same root —— is down to the fact that CHOOSE(A,B) = ((B!)/((A!)((B-A)!))) goes DOWN once A rises higher than B/2.
If A is 5, we really need B >= 9, or we start losing numbers.
With 6 MoAs and 7 PoAs we have CHOOSE(5,6) = CHOOSE(1,6) = 6 and CHOOSE(5,7) = CHOOSE(2,7) = (7*6)/2 = 21; so only 6*21 = 126 roots.

I could expand the system to 51 consonants.
9 MoAs
—— voiceless fricatives, voiced fricatives, nasals, voiceless plosives, voiced plosives, approximants, lateral approximants, trills, and taps-or-flaps ——
and 8 PoAs —
— alveolar, retroflex, palatal, velar, bilabial, uvular, labiodental, and glottal —
— and get 126 sets of 5 MoAs and 56 sets of 5 PoAs, for most of 126 * 56 = 7056 roots — way too few, I think.

If I use tetraliteral roots instead, I get 126 sets of 4 PoAs and 70 sets of 4 MoAs for most of 8820 roots — still too few in my opinion.

Thing is, it’s only “most”, because the number of consonants in each MoA are
8 8 7 7 6 5 4 3 3
and the number in each PoA are
9 8 7 7 6 6 5 3
.
So if one of the MoAs chosen for a given root is trills, or taps-or-flaps, or one of the PoAs chosen is glottal,
it’s likely to be true that not every one of the chosen MoAs occurs at every one of the chosen PoAs;
so the MoA set-of-4 and the PoA set-of-4 may be incompatible with each other.

.....

I think I might only use metathesis of the first two consonants’ MoAs and PoAs independently of each other, and metathesis of the last two consonants’ MoAs and PoAs independently of each other, as derivational and/or inflectional morphology for the roots.
Edit: That means the number of wordforms per root would be at most only 4*4 = 16 times as many as there are ways of inserting vowels.
If there are 4 or more consonants per root, that would mean neither of the first two consonants would ever switch places with either of the last two consonants, producing another wordform of the same root.
If I still want to avoid consecutive consonants having the same PoA, then at 4 consonants per root I’d still need 4 distinct PoAs.
But with 5 or more consonants-per-root, I’d need the first 3 PoAs to be distinct, and also need the last 3 PoAs to be distinct. But it might be OK to have one of the first two PoAs to also be one of the last two PoAs.
I could also get more roots by just deleting the constraint that consecutive consonants can’t have the same MoA.

Note though that I’d still want to not have the same consonant occur twice in the same morpheme, even if it were a root morpheme.
If a quinquiliteral root were to have five distinct MoAs, then in this new scheme either or both of the first two PoAs could also be one of the last two PoAs, and not violate any of the remaining constraints.
But if the first two MoAs were also the last two MoAs, then the root would need five distinct PoAs, to avoid any possibility of having the same consonant be one of the first two consonants and also one of the last two consonants.
....
In other words; shit could get complicated.
I’ll never figure it out in my head before dawn; it’s already midnight!
....
I hope you enjoy some of this!
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Re: Another Almost-Cartesian Consonant Inventory

Post by eldin raigmore »

I have a new “scheme” worked out for tetraconsonantal roots with the first two PoAs and the first two MoAs being metathesizable independently, and also the last two PoAs and the last two MoAs being metathesizable independently.

A “root” will consist of;
* an unordered pair of PoAs to be the PoAs of the first two consonants;
* an unordered pair of MoAs to be the MoAs of the first two consonants;
* independently, an unordered pair of PoAs for the last two consonants; and
* independently, an unordered pair of MoAs for the last two consonants.

Either the consonant-inventory must contain both of the first two MoAs at both of the first two PoAs;
or it must contain both of the last two MoAs at both of the last two PoAs;
or both.
(Also, I’m just kinda hoping it goes without saying, each MoA in each pair occurs at some PoA in that pair, and vice-versa.)

If there are 8 PoAs and 7 MoAs, or 8 MoAs and 7 PoAs, in the consonant-inventory,
there would be up to (and probably most of)
(8*7)/2 = 28 pairs of PoAs and (7*6)/2 = 21 pairs of MoAs, or 28 pairs of MoAs and 21 pairs of MoAs.
So there would be most of, and up to, 28*21 = 588 ways to choose a pair of PoAs and a pair of MoAs for the first two consonants;
and the same number of ways to choose a pair of last two PoAs and a pair of last two MoAs.

Unless both the first-two pair of pairs, and the last-two pair of pairs, are “locally non-Cartesian” — that is, the language doesn’t contain both of the first two MoAs at both of the first two PoAs, and also doesn’t contain both of the last two MoAs at both of the last two PoAs — the two halves of the root are compatible with each other, and can be chosen independently of each other.

So the maximum possible number of “roots” would be (21^2)*(28^2) = 441*784 = 588^2 = 345,744. That’s plenty.

....

In metathesizing the PoAs, if a PoA occurs in both the first pair and the last pair, one is constrained not to make the same PoA occur both in second place and in third place.
And, in metathesizing the MoAs, if an MoA occurs in both the first pair and the last pair, one is constrained from matching the same MoA with the same PoA twice.
Because consecutive consonants in a root can’t have the same PoA; and a root can’t contain two instances of one consonant.
But even if both of the first two PoAs were both of the last two PoAs, and both of the first two MoAs were both of the last two MoAs, there would still be four sequences of consonants that could be considered derivational or inflectional forms of one and the same lexeme.
If OTOH just one of the first two PoAs were one of the last two PoAs, and one of the first two MoAs were one of the last two MoAs, then, provided both pairs-of-pairs were “locally Cartesian”, there would be nine ways to arrange the PoAs and MoAs to make wordbases that were derived or inflected from the same lexeme.
If there were four distinct PoAs and four distinct MoAs, there would be 16 derived wordbases.
All before inserting any vowels!
....
Speaking of vowels:
Suppose there are five vowels /a e i o u/. Suppose they have to be inserted to make CVCVCVC wordforms. Suppose the same vowel can’t be used twice.
Suppose also the same vowel-height can’t be used twice in a row.
That is, if the middle vowel is /i/ or /u/ then neither the first nor the last vowel can be /i/ or /u/;
and if the middle vowel is /e/ or /o/ then neither the first nor the last vowel can be /e/ or /o/;
and (unnecessary to say) if the middle vowel is /a/ then neither the first nor the last vowel can be/a/.

Then the allowable vowel sequences would be the following 3536, if I did it right:
/-a-e-i-/
/-a-e-u-/
/-a-i-e-/
/-a-i-o-/
/-a-o-i-/
/-a-o-u-/
/-a-u-e-/
/-a-u-o-/
/-e-a-i-/
/-e-a-o-/
/-e-a-u-/
/-e-i-a-/
/-e-i-o-/
/-e-u-a-/
/-e-u-o-/
/-i-a-e-/
/-i-a-o-/
/-i-a-u-/
/-i-e-a-/
/-i-e-u-/
/-i-o-a-/
/-i-o-u-/
/-o-a-e-/
/-o-a-i-/
/-o-a-u-/
/-o-i-a-/
/-o-i-e-/
/-o-u-a-/
/-o-u-e-/
/-u-a-e-/
/-u-a-i-/
/-u-a-o-/
/-u-e-a-/
/-u-e-i-/
/-u-o-a-/
/-u-o-i-/
Edit: If the middle vowel is /a/, there are four other vowels that are not also low/open. So there are 4*3 = 12 allowed vowel-sequences with /-a-/ in the middle.
Otherwise, there are three other vowels not the same height as the middle vowel; and so 3*2 = 6 allowed vowel-sequences for each of the 4 other possible middle vowels.
That’s 12 + 6*4 = 12+24 = 36 vowel sequences; not 35. I miscounted. By one.
So, there might be as many as 35*4 = 140 up to 35*16 = 560 ways to inflect each root. No prefixes nor suffixes required!
Edit: correction: 36*4 = 144 up to 36*16 = 576.

..........

Whatcha think?
Edit: If we also didn’t allow two consecutive front vowels nor two consecutive back vowels, there would be 20 allowable vowel-sequences instead of 36. We’d leave out the ones that start with /e-i-/ or /i-e-/ or /o-u-/ or /u-o-/, and the ones that end with /-e-i/ or /-i-e/ or /-o-u/ or /-u-o/. There are two of each of those, so we’d leave out 8*2 = 16 of the 36 we had. There might be as many as 20*4=80 to 20*16=320 ways to inflect each root without prefixes or suffixes.
Last edited by eldin raigmore on 08 Jun 2020 07:11, edited 2 times in total.
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Re: Another Almost-Cartesian Consonant Inventory

Post by eldin raigmore »

Forbidden combinations, and conditionally-limited combinations.
____________________
Forbidden Combinations
____________________
If one of the MoAs is taps-or-flaps, then the PoAs cannot be two of:
Glottal
Uvulars
Bilabials
Velars
Palatals
...
If one of the MoAs is trills, then the PoAs cannot be two of:
Glottal
Labiodentals
Velars
Palatals
Retroflexes
...
If one of the MoAs is lateral approximants, then the PoAs cannot be two of:
Glottal
Labiodentals
Uvulars
Bilabials
...
If one of the MoAs is approximants, then the PoAs cannot be two of:
Glottal
Uvulars
Bilabials
...
If one of the MoAs is voiced plosives, then the PoAs cannot be glottal and labiodentals.
__ __ __ __ __
If one of the PoAs is glottal, then the MoAs cannot be two of:
Taps-or-flaps
Trills
Lateral approximants
Approximants
Voiced plosives
Nasals
...
If one of the PoAs is labiodentals, then the MoAs cannot be two of:
Trills
Lateral approximants
Voiced plosives
Voiceless plosives
...
If one of the PoAs is uvulars, then the MoAs cannot be two of:
Taps-or-flaps
Lateral approximants
Approximants
...
If one of the PoAs is bilabials, then the MoAs cannot be two of:
Taps-or-flaps
Lateral approximants
Approximants
...
If one of the PoAs is velars, then the MoAs cannot be taps-or-flaps and trills.
...
If one of the PoAs is palatals, then the MoAs cannot be taps-or-flaps and trills.

____________________
Conditionally Limited Combinations
At most one of the two pairs-of-pairs (either the initial pair, or the final pair, or neither, but not both), can contain any of the following combinations of a PoA and an MoA.
Glottal and tap-or-flap
Glottal and trills
Glottal and lateral approximants
Glottal and approximants
Glottal and voiced plosives
Glottal and nasals
Labiodentals and trills
Labiodentals and lateral approximants
Labiodentals and voiced plosives
Labiodentals and voiceless plosives
Uvulars and taps-or-flaps
Uvulars and lateral approximants
Uvulars and approximants
Bilabials and taps-or-flaps
Bilabials and lateral approximants
Bilabials and approximants
Velars and taps-or-flaps
Velars and trills
Palatals and taps-or-flaps
Palatals and trills
Retroflexes and trills
.......... .......... .......... .......... ..........
===== ===== ===== ===== =====
From these nine MoAs and eight PoAs I have not yet calculated how many pairs of pairs are not forbidden, nor how many of those are also not conditionally limited; and so I haven’t yet calculated how many possible “tetra-consonantal roots” there might be.

(BTW the nine MoAs are:
Voiceless fricatives
Voiced fricatives
Nasals
Voiceless plosives
Voiced plosives
Approximants
Lateral approximants
Trills
Taps-or-flaps )

(And the eight PoAs are
Alveolars
Retroflexes
Palatals
Velars
Bilabials
Uvulars
Labiodentals
Glottal .)
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Re: Another Almost-Cartesian Consonant Inventory

Post by eldin raigmore »

I’m going to say any two nominal roots that begin with the same MoA are likely to have some grammatical feature in common; e.g. case and/or pragmatic status (such as definiteness) etc.
Inspired by initial consonant mutation in modern Celtic languages.

I’m going to say any two verbal roots that end with the same MoA are likely to have some grammatical feature in common; e.g. tense and/or aspect and/or voice.
Inspired by liquid verbs of classical languages.

—————

Maybe two nominal roots with the same first PoA have the same seinsart?
Maybe two verbal roots with the same last PoA have the same aktionsart?
Or some other probabilistic lexical semantic feature?

And for roots belonging to some other part-of-speech there should be some other statistical correlation between similarity of sound and similarity of grammatical or semantico-pragmatic features?

All just “maybes” so far.
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Re: Another Almost-Cartesian Consonant Inventory

Post by Creyeditor »

I really like your solution. I think it is more complicated than before, but it seems to work. Actually, in order to be easier to understand, could you give a concrete example of two possible roots with their meaning and two possible inflections? No need to commit yourself, just a concrete example.

Also, what is "seinsart"? [:D]
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Re: Another Almost-Cartesian Consonant Inventory

Post by eldin raigmore »

“Seinsart” is German for “mode of being”.
“Sart ” is cognate with “sort” and means something like “type”.

“Aktionsart” is “sort of action” or “type of action” most literally; in linguistics it is used to mean “lexically-inherent aspectual class”, usually shortened to “lexical aspect” or “inherent aspect” or “aspectual class”.
The actual, precise meaning of a given aspect marking on a given verb, is a function of the marking and of the verb’s aktionsart.

“Seinsart” is to nouns what “aktionsart” is to verbs. I’m afraid I have not found an exact or explicit definition of it as used in linguistics. It is not in the Oxford Concise Dictionary of Linguistics; at least not in the online-searchable version of the 2nd edition.
It should be some lexically-inherent class of a noun’s meaning that, together with some inflectional marking on the noun, determines the precise meaning of that particular marking on that particular noun.

I mentioned it in my post because I wanted the first PoA of a nominal root to tend to indicate some lexically-inherent thing about the noun, independent of whatever is communicated by the first MoA, that would determine the precise meaning of a particular initial mutation for that particular noun.
It might include some things like common vs proper, and/or count vs mass-or-measure, or possessible vs non-possessible, or obligatorily-possessed vs independent.

See https://books.google.com/books?id=E02SS ... rt&f=false

“Lexical properties denoted by nouns”
“Inherent semantic features of terms”
+-human +-animate +-concrete
Edit: upon further thought I think seinsart might be the things I have been calling “gender” in my conlang Adpihi. It’s several different mostly-two-valued inherent semantic features of nouns’ referents that influence how likely they are to occupy the grammatical or syntactic relations or functions in clauses and phrases and speech-acts. They include the three above, and also others such as +-living, +-rational, +-sentient. However in Adpihi animacy is a three-valued “scalar”(?) feature; the animate referents are divided between those which are translocomotively mobile and those which are sessile. In the earlier language the various seinsart features will be less logically independent of each other than in the science-fiction-technology part of the language’s history.
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