> - Chinese EVs are dominating every market outside the US.

Not so in Europe, where the top 10 EVs sold in 2025 [1] has not a single Chinese car.

[1] https://autorai.nl/tesla-van-de-troon-gestoten-als-populairs...

I see sources saying BYD sold 187,657 vehicles in Europe in 2025, and MG sold 307,812. Interesting.

Your source only says: “Top 5 best-selling EV brands in Europe in 2025 Volkswagen – 274.417 (+56%) Tesla – 238.765 (-27%) BMW – 193.186 (+15%) Skoda – 172.100 (+117%) Audi – 153.848 (+51%)”

BYD sells a lot of hybrids rather than BEVs fyi. I believe its the majority of their sales maybe?

Of course, because the agreement between the European Union and Chinese limits battery electric vehicles, but does not limit hybrids the same way… which I think speaks again to the ability of the different manufacturers to compete on a “fair” basis

Correction: the European Union imposed additional tariffs (up to 35%) on Chinese-made battery-electric vehicles, while conventional hybrids and plug-in hybrids were exempt from those additional tariffs, giving Chinese manufacturers an incentive to shift toward hybrid sales.

Chinese EV have heavy import taxes in the EU. Also in France, (I don’t know in other countries), you get more subvention when buying an EU-made EV. This heavily distorts the market to try to save VW and friends.

How long until they are devoured by the Chinese? Geely already owns Volvo.

0.5 GB of SRAM is exactly what the Cuckatoo32 [1] Proof-of-Work scheme needs for optimal solving efficiency. Solutions are 42-cycles in random bipartite graphs of 2^32 edges on 2^32 + 2^32 nodes, which can be found by repeatedly removing edges that have only one adjacent edge. Tracking which nodes have incident edges takes 2^32 (completely randomly accessed) bits which is exactly 0.5GB. Another 0.5 GB is needed to track alive edges but those can be accessed serially for which DRAM suffices.

[1] https://github.com/tromp/cuckoo

The only way he could stay in the White House is if Melania got elected (the Clinton route).

> The only way he could stay in the White House is if Melania got elected

Melania is not a natural born US citizen so that would be impossible

One of his ridiculous children, perhaps.

My favorite is the (partly analog) PRW-7000FC-1JF, as seen on https://cdn11.bigcommerce.com/s-e31c8/images/stencil/1280x12...

I love that, but have dainty little bird wrists. Men's watches look OBNOXIOUSLY large on me.

I wish they made something like that, that wasn't 500mm.

> I wish they made something like that, that wasn't 500mm.

That doesn't sound dainty in the least ;)

A language that was designed to make Kolmogorov complexity concrete is Binary Lambda Calculus, a self interpreter of which takes only 168 bits (21 bytes). Unlike the Lisp self interpreter in the article, it operates on a bitstream and thus has to include a parser.

I recommend parent's ioccc gem https://www.ioccc.org/2012/tromp/

> Overall, I built a Graph Reduction engine

I did the same for my performant implementation of pure functional programming language BLC/BLC2, which in 400+ lines contains a graph reduction engine for combinatory logic, to which the lambda calculus programs are converted by Kiselyov's bracket abstraction algorithm.

[1] https://github.com/tromp/AIT/blob/master/uni.c

The function named clapp was likely a beast to debug.

I'm more interested in the converse question: how well does an LLM perform as a compressor, compared to gzip (ignoring its insanely lower speed)?

Top contestant in the Hutter Prize uses a neural network for compression. So fair to say, LLMs would perform pretty well compared to gzip.

Even ignoring speed per GP, the Hutter Prize's metric includes the size of the decompressor. LLMs would be disqualified for being larger than 1GB.

Hutter prize does have speed restrictions. If it did not, LLMs would win even with counting the size of the model (which is the most reasonable choice imo) as per the main benchmark: https://www.mattmahoney.net/dc/text.html.

Huh? The main benchmark is compressed size + program size. The uncompressed size of enwik9 is defined as 1GB, so no "L"LM qualifies.

And the hutter prize disallows GPU's. If you allow use of a powerful GPU, you can do quite a bit better.

hallucinations are lossy compression artefacts

can they be considered to have compressed the entirety of their training data into their weights?

If you consider them lossy compression, then yes.

The goal would be to find the minimum model that, with a fixed seed, would exactly reproduce your text.

So I could hide information in a model basically

I think this would be the exact opposite. This model would be optimized to only output your information.

thats what I meant. For a given input give me the output I want without anyone knowing why that is there

lossless vs lossy is the question.

How lossy? Because I can lossy compress anything into 0 bits.

0 bits is just a very small bloom filter.

Much better

Not every bitcoin tx output.

With taproot (P2TR), scripts are optional, and outputs can be based solely on Schnorr signatures.

P2TR outputs have a script that always starts with OP_1. That script may or may not commit to a tapscript.

When code is a maze describes my 1988 IOCCC entry [1]

    char*M,A,Z,E=40,J[40],T[40];main(C){for(*J=A=scanf(M="%d",&C);
    --            E;             J[              E]             =T
    [E   ]=  E)   printf("._");  for(;(A-=Z=!Z)  ||  (printf("\n|"
    )    ,   A    =              39              ,C             --
    )    ;   Z    ||    printf   (M   ))M[Z]=Z[A-(E   =A[J-Z])&&!C
    &    A   ==             T[                                  A]
    |6<<27<rand()||!C&!Z?J[T[E]=T[A]]=E,J[T[A]=A-Z]=A,"_.":" |"];}

[1] https://tromp.github.io/pearls.html#maze

Very cool, thanks for sharing!

If, like me, you're trying to compile it, don't forget to set the

    -std=gnu89

flag in gcc, so:

    gcc -std=gnu89 -w -o maze maze.c

Then run it as:

    echo 10 | ./maze

Got seg fault (gcc 15.2.0 on nixos 26.05) when writing to

    M[Z] = ...

Had to add:

    char fmt[] = "%d";

and change to:

    scanf(M=fmt,&C);

Works now:

    char*M,A,Z,E=40,J[40],T[40];main(C){char fmt[]="%d";for(*J=A=scanf(M=fmt,&C);
    --            E;             J[              E]             =T
    [E   ]=  E)   printf("._");  for(;(A-=Z=!Z)  ||  (printf("\n|"
    )    ,   A    =              39              ,C             --
    )    ;   Z    ||    printf   (M   ))M[Z]=Z[A-(E   =A[J-Z])&&!C
    &    A   ==             T[                                  A]
    |6<<27<rand()||!C&!Z?J[T[E]=T[A]]=E,J[T[A]=A-Z]=A,"_.":" |"];}