From: Matthias Blume (find@my.address.elsewhere) Subject: Re: [OT] Finns and education Newsgroups: comp.lang.scheme Date: 2004-09-03 18:45:40 PST [ I know I shouldn't do this, but here we go again... ] richter@math.northwestern.edu (Bill Richter) writes: > There is absolutely no reason to think that Will thought his "add a > new start symbol" semantics is compositional. There is every reason. Why else would he even have brought it up? > You didn't explain what else DS means besides > compositionality, nor did Will, and we haven't been arguing about it. I don't want to explain, given that this is pretty much independent of the problem at hand. I already told you (several times, in fact) that compositionality is merely one desirable property of a DS. Just being compositional does not make anything a DS. > You can define whatever you want, as long as you make a good > well-defined mathematical definition. Yes. But if you want to talk to people in a field where there already is an established defintition, you better stick to that. Otherwise no meaningful communication is possible (as you can observe in this very thread). > C-F's already made my kinda def; they defined compositionality as a > property of functions: I lost track of what "C-F" means. But whoever it is (Cartwright/Felleisen, presumably?), I bet they didn't. My guess is that they might have said/written something that under an anal-retentive interpretation sounds like Bill-compositionality, but really meant the established notion. You know, it is much easier to say "the function is compositional" than "the definition of the function we just gave is compositional". It is also safe as long as everybody knows what this is supposed to mean. > No. Only the second is a definition of compositionality. > > I suppose you mean, "only the 2nd is is the accepted def of comp." Yes. > And if we take the liberty of changing the grammar of the language, > the first definition is also meaningless as it makes all functions > on syntax trees compositional. > > OK, but that goes for the 2nd def as well. Nonsense. If you change the grammar, you subsequently have to change the definition of the function. Once you chance the definition, anything you say about the new definition has no bearing on the original one. > But if you change the grammar enough, you can get all > kinds of silliness. I only changed the grammar in the ways that's > accepted in OpS. "Accepted in OpS"?!? Indeed, if you change the grammar (strike the "enough"!), you can get all kind sof silliness. You have given a perfect example of such silliness. (And, as I have demonstrated to you, when you work hard enough, you can even be equally silly without changing the grammar.) > Have /you/ understood /my/ version? Last time you said you can't > read the ML code... > > No I didn't, sorry. And you have the NERVE to claim I didn't understand yours?!? > Please, dear god! In your framework this is a "fact" about *every* > function. It hasn't been missed. It has merely not considered > worth any attention because it is such a triviality, because it > buys absolutely nothing. > > That indicates you didn't in fact understand my semantics, so we're > each ignorant of each others work. To spell it out for you: In your framework, a function on programs is compositional if one can find a grammar for the function such that there exists a corresponding compositional definition of the function in question. My claim was that this is true for every function, and the proof for this is trivially given by Will's "add a new start symbol" trick. If you don't understand that, you demonstrate THAT YOU DON'T UNDERSTAND YOUR OWN SEMANTICS WHICH IS MERELY AN OVERLY COMPLICATED VERSION OF THE SAME TRICK. > In Will's "add a new start symbol" > semantics, yes, it's trivial, and buys you nothing. In my framework, > it amounts to the Scheme fact that in evaluating a combination, we > first evaluate the arguments, and then apply the first value (better > be a procedure) to the 2nd. So you are saying that certain changes of the grammar are legal, and others are not (or "meaningless")? Could you make this idea a bit more precise? What distinguishes Will's approach from yours in such a way that yours is non-trivial and buys something while Will's isn't and doesn't? (Rhetorical question -- you don't have to answer.) [By the way, since ultimately both your and Will's function are the same -- namely the original OpS, everything you say about your function applies to his. Moreover, what you say about your function definition is arguably wrong -- but I am too tired of this to go into any detail as to why.] > Now previously, you've answered this > point of mine with "But other DS's do this better", so perhaps you did > in fact understand my semantics, but just spaced it out here. It has by now become clear to me that I understand your semantics better than you do. The same is probably true for Will. > No, I'm really confident in my pure Math skills, and this is just a > pure Math problem. The whole problem with the world is that fools and fanatics are always so certain of themselves, but wiser people so full of doubts. -Russell Matthias