"Functional Collection Programming with Semi-Ring Dictionaries"

[dmm]

arxiv.org/abs/2103.06376

page 2: "Semi-ring dictionaries realize the well-known connection between relations and tensors" (from "In-Database Learning with Sparse Tensors" 2016-2018 paper)

Comments

[dmm]

I am using her to help me, both here and in many other situations (in this case, I can't closely read 40 pages fast enough, and I can't allocate enough time without knowing if the payoff is likely to be high).

In this case, I can double check their definition and note that they indeed do consider the generality of non-commutative rings (section 2.1, page 3). So her remark is correct.

(This is, actually, very good, because doing this with rings of matrices might be quite interesting to try.)

The point is, I want to make progress without diving deeply. Later, if diving deeply turns out to be warranted, I would do that.

***

Other than that, I know my paper really well, I am trying to get a quick feel whether the techniques from this one are likely to help me make progress with some open problems with my approach. If one does not know either paper, I don't quite see why one would be interested in comparing these two papers. But I have some very particular needs in mind (the main reason why my neural machines are not very practical compared to Transformers is that my machines are not GPU-friendly and not highly optimized in their full generality, so rectifying that is fairly central to me).

[chaource]

Look again at this phrase:

Multiplication is defined in a way that is not commutative by default, where multiplying a dictionary with a scalar results in each value of the dictionary being multiplied by the scalar

This phrase is confusing: it says that multiplication is not commutative by default (what does "default" mean here?) and then it talks about multiplying dictionaries by scalars as if to illustrate the non-commutativity.

To understand what this means, you need to actually look into the paper and find what is commutative and what is non-commutative.

A reasonable rewrite would be: "Multiplication by scalars is in general non-commutative. (full stop, it's going to be a different topic now!) Multiplying a dictionary with a scalar results in ..."

[dmm]

Да, мелких погрешностей полно...

(Как и в том, что я пишу, особенно, если я решаю не исправлять, а оставляю, как получилось (это - некоторая особенность текущей парадигмы, что она может думать только вслух, а не про себя; это можно чинить, но до какой степени надо торопиться это чинить, это не очень понятно).)