On Wed, 26 Jun 2013 22:49:48 -0400, Herman Miller <[log in to unmask]> wrote:

>Here's a variation on an idea Alex Fink was talking about back in 2009
>("Unambiguous prosody for trees"). The nice thing is that in the simple
>cases, it looks like something that might be found in a reasonable
>spoken language. 
>So far we've got:
>N -> A N
>A -> A D
>D -> C D
>Now here's the cool part. To expand a C, the left-hand side is inflected
>like a C, and the right-hand side is like an N.
>C -> C N

This is beautiful, well done!  I agree that it's nearly certainly an improvement in usability over the 2009 scheme, given that (as you were pointing out below) it has the advantage that the expansion of each category X contains an X in it, and thus the expansions can be reasonably thought of as having _heads_.   The 2009 scheme had the one wacky rule b -> a B without this property.  

But it's not _completely_ head-final: the categories N and D are head-final, but the categories A and C are head-initial.  It's only the choice of N as the top-level category that makes it head-final!  If you had chosen A or C as the top-level category, you'd get exactly the same system left-right reflected with the names of the categories changed; if you had chosen D, you'd get the same system unreflected with the names of the categories changed.  So really it's quite symmetric until you pick out N for special treatment.  

Here's, accordingly, another way to look at it.  Draw your binary syntax tree, and give it a fictive parent S of which it's the right child.  For each word (i.e. leaf node) W, descend along the tree from S to W, and write down on W *the number of times you changed direction* during the descent.  (For example, since you always begin by going right, the rightmost word always gets a 0, and is the only word that can get a 0.)  This numeric labelling is the same one described by the following system of expansion rules, starting from a 0:
0 -> 1 0
1 -> 1 2
2 -> 3 2
3 -> 3 4
4 -> 5 4
5 -> 5 6

But now your new scheme is just what you get from this numeric scheme by making these replacements:
numbers 0 mod 4 become N
numbers 1 mod 4 become A
numbers 2 mod 4 become D
numbers 3 mod 4 become C

To put all those numbers in a more linguistically palatable cloak, let's suppose our language has nouns but no adjectives as bases, plus one adjectivising suffix (case?) .A and one nominalizing suffix (case?) .N, the two of which participate in suffixaufnahme with each other.  Suppose noun phrases are head-final but adjective phrases are head-initial.  Then the occurring binary expansions are these, using B to mean a base:
B -> B.A B
B.A -> B.A B.A.N
B.A.N -> B.A.N.A B.A.N
B.A.N.A -> B.A.N.A B.A.N.A.N
B.A.N.A.N -> B.A.N.A.N.A B.A.N.A.N
B.A.N.A.N.A -> B.A.N.A.N.A B.A.N.A.N.A.N
and I'm sure you're going bananas at this point lining up the suffixes.  Anyhow. 

The number in my numeric scheme counts the total number of .As and .Ns.  You recover your scheme by first cancelling all copies of the string .A.N.A.N and then making these replacements on what's left:
B becomes N
B.A becomes A
B.A.N becomes D
B.A.N.A becomes C

On Thu, 27 Jun 2013 20:59:28 -0400, Herman Miller <[log in to unmask]> wrote:

>On 6/27/2013 4:45 AM, And Rosta wrote:
>> Did the Miller--Fink scheme of 2009, which I admired but never got my head
>> around, also involve four inflections? I dimly recall it had two.
>There were four, but I notated them as a, b, A, B. 

Yup, and given the asymptotics of the Catalan numbers, four is the fewest that can be hoped for.  Trying it with only two inflections you'll find that there are just barely too many trees on a seven-word string (132 of them) to be able to represent them with seven bits (2^7 = 128).  

>[...] but I made an exception for a final (a
>b), which is left as is without reducing it to (a). This was so the
>scheme could represent both (a (a (a (a b)))) and ((((a b) b) b) b) type
>sequences without preference.
>The new system has no exceptions to the rules,

Well, that exception is an artificial thing you're bolting on.  It's not a difference that exists at the level of the sets of four binary expansions, taken alone, and this is the level at which I'd rather compare first.  

That said, your current scheme is objectively more symmetric than the 2009 one.  (So there was no symmetry to rescue, exception schmexception!)  Even if you were to forget which kinds of nodes are "marked" and change all the labels, the 2009 system is not the same as its left-right reflection: since b is the only non-headed category, it can be picked out, and then you can tell whether you're in the original system or the reflected one by whether b is a right child or a left child of its parents.