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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)

Date: 2023-12-04 09:04 am (UTC)
chaource: (Default)
From: [personal profile] chaource
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

That seems incorrect to me. Multiplication by scalars is obviously commutative.

Since I'm not familiar with technical details of any of this stuff, it's really useless for me to read what chatgpt said. I can spot an error only in a domain where I already know the technical details. In domains where I am just trying to learn new stuff, chatgpt's output will mislead without me noticing.

For example, it says "Yes, this is closely related" and actually it could be unrelated, how do we know? Chatgpt didn't say why they were closely related other than superficially (dictionaries with numeric leaves vs. nested dictionaries with semiring-valued leaves).
Edited Date: 2023-12-04 09:43 am (UTC)

Date: 2023-12-05 10:57 am (UTC)
chaource: (Default)
From: [personal profile] 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 ..."

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