TomHChappell wrote: ↑Fri Aug 02, 2019 2:19 am
Using Case-Endings and Postpositions and Prepositions to Mark Case-Like Stuff
Imagine a language with a few case-endings and a few postpositions and a few prepositions.
Imagine the following are true:
* any noun-phrase can be used with no case-ending and no postposition and no preposition.
* any noun-phrase can be used with any case-ending and no postposition and no preposition.
* any noun-phrase can be used with any postposition and no case-ending and no preposition.
* any noun-phrase can be used with any preposition and no case-ending and no postposition.
Assume all of those marking possibilities indicate different syntactic roles and/or different semantic roles; ad-verbal or ad-nominal.
More assumptions:
* any noun-phrase can be used with any case-ending and
nearly any postposition and no preposition.
* any noun-phrase can be used with any case-ending and
nearly any preposition and no postposition.
* any noun-phrase can be used with any postposition and
nearly any case-ending and no preposition.
* any NP can be used with any postposition and
nearly any preposition and no case-ending.
* any NP can be used with any preposition and
nearly any case-ending and no postposition.
* any NP can be used with any preposition and
nearly any postposition and no case-ending.
Assume all of the marking possibilities mentioned so far indicate either different syntactic roles or different semantic roles, whether ad-verbal or ad-nominal.
One more assumption:
* For any case-ending and any postposition and any preposition, if that case-ending can be used with each of the adpositions separately, and those adpositions can be used together with no case-ending, then any NP can be used with that case-ending and that postposition and that preposition all together at once.
And, with different semantic or syntactic implications than any other of the markings mentioned so far.
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I’m about to get ready to figure out how many different marking combinations there are.
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If for each postposition there is one and only one case-ending with which it is incompatible (ie they can’t be used together), then there aren’t more postpositions than case-endings.
If for each preposition there is one and only one case-ending it is incompatible with, there aren’t more prepositions than case-endings.
If for each case-ending there’s one and only one postposition it’s incompatible with, there aren’t more case-endings than postpositions.
If for each case-ending there’s exactly one preposition it’s incompatible with, there aren’t more case-endings than prepositions.
If for each preposition there’s just one postposition it can’t be used with, there aren’t more prepositions than postpositions.
If for every postposition there’s just one preposition it can’t be used with, there aren’t more postpositions than prepositions.
I’m going to assume all six of the above hypotheses. That will mean the number of case-endings and the number of postpositions and the number of prepositions are all equal. And that will make the following algebra a lot simpler.
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Let n be the number of case-endings and the number of postpositions and the number of prepositions.
Then, there are
1 way to “mark” a NP with no case-ending and no adposition,
n ways to mark a NP with a case-marking but no adposition,
n ways to mark a NP with a postposition but no case-ending and no preposition,
n ways to mark a NP with a preposition but no case-ending and no postposition,
n(n-1) ways to mark a NP with a case-ending and a postposition but no preposition,
n(n-1) ways to mark a NP with a case-ending and a preposition but no postposition,
n(n-1) ways to mark a NP with a postposition and a preposition but no case-ending.
So far that’s 1+3n+3n(n-1) = 1 + 3n + 3(n^2) - 3n = 1 + 3(n^2) ways to mark a NP with none ore one or two, but not all three, of a case-ending and/or a postposition and/or a preposition.
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How many marking-combinations there are, when all three types — a case-ending, a postposition, and a preposition — are used, depends on how the incompatibilities line up with each other.
I’m going to calculate under two assumptions, which I’m guessing are the extremes. I doubt they’re the only possibilities unless n is small.
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First; assume:
* if a case-ending is incompatible with both a postposition and a preposition, then that postposition and that preposition are incompatible with each other.
* if a postposition is incompatible with both a case-ending and a preposition, then that case-ending and that preposition are incompatible with each other.
* if a preposition is incompatible with both a case-ending and a postposition, then that case-ending and that postposition are incompatible with each other.
If we assume those, then the number of ways to mark a noun-phrase with all three of a case-ending and a postposition and a preposition, each compatible with each of the others, is n(n-1)(n-2) = (n^3) - 3(n^2) + 2n.
So the total number of grammatical ways to mark up a NP in this language is
(n^3) - 3(n^2) + 2n + 3(n^2) + 1 = (n^3) + 2n + 1
If n is 0 this is 1. Boring.
If n is 1 this is 4, because you can’t use the case-ending with either adposition, and you can’t use both adpositions together. Still kinda boring.
If n is 2 this is 13. Now maybe we’re getting somewhere.
If n is 3 this is 34. That is at least as many as in Stanley Starosta’s “The Case for Lexicase”, unless my memory is completely misleading me, which gets ever likelier year by year.
If n is 4 this is 73. Four is my favorite value for n.
If n is 5 this is 136. Nearly as many as Tsez’s cases.
If n is 6 this is 229. More than Tsez has cases. I think that’s enough and I’m not going for bigger n under this assumption.
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The other assumption I plan to check out, about how the incompatibilities line up with each other, is the following threefold statement:
If a case-ending is incompatible with each of a postposition and a preposition, then that postposition and that preposition can be used together with any other case-ending, and also with no case-ending.
If a postposition is incompatible with each of a case-ending and a preposition, then that case-ending and that preposition can be used together with any other postposition, and also with no postposition.
If a preposition is incompatible with each of a case-ending and a postposition, then that case-ending and that postpositions can be used together with any other preposition, and also with no preposition.
Under this assumption, the number of grammatical ways this language has to markup case-wise and adposition-wise a NP with each of a case-ending and a postposition and a preposition all together at once, is
n(((n-1)^2) - 2) = n((n^2) - 2n + 1 - 2) = n((n^2) - 2n - 1) = (n^3) -2(n^2) - n .
So the total number of grammatical marking possibilities is
(n^3) - 2(n^2) - n + 3(n^2) + 1 = (n^3) + (n^2) - n + 1.
(This will usually exceed (n^3) + 2n + 1, as long as n-1 > 2.)
So for each value of n from 0 to 6, how many is (n^3) + (n^2) - n + 1 ?
If n is 0, this is 1. Of course. Boring.
If n is 1, this is 2. That doesn’t make sense. I don’t think all my other assumptions are consistent together with n=2.
If n is 2, this is 11. Not as many, yet, as the previous assumption.
If n is 3, this is 34. Same number as the previous assumption.
If n is 4, this is 77. A few more than under the previous assumption.
If n is 5, this is 146. Several more than the previous assumption.
If n is 6, this is 247. 28 more than under the previous assumption.
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So what does anyone think?