The Nineteenth Byte

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att

00:17

@NewPosts i feel like this is basically just "find the primes < some bitlength"

1 hour later…

Unrelated String

01:25

@att ...especially considering it's not required to be in any particular format, so you can literally just translate a list of primes naively to DNF

@Fmbalbuena yeah sorry I didn't actually check this 😭 I kinda just automatically don't bother looking at fastest-code stuff usually

Unrelated String

01:37

...unless the output encodes primality testing instead :3

New Posts

02:17

Q: Decide Equality of Closed Surfaces

Objective Given two closed surfaces (a.k.a. closed 2-manifolds), decide whether they're homeomorphic. Introduction In layman's terms, a closed surface is a shape that resembles a flat plane everywhere (say, when magnified very much). Familiar examples of a closed surface include a sphere and a to...

Dannyu NDos

We need more challenges involving topology.

5 hours later…

mousetail

07:18

I found this weird code:

    let mut to_surface = |r, g, b, a| {
        let tr = a * r + 0x80;
        let tg = a * g + 0x80;
        let tb = a * b + 0x80;
        surface[i + 0] = (((tb >> 8) + tb) >> 8) as u8;
        surface[i + 1] = (((tg >> 8) + tg) >> 8) as u8;
        surface[i + 2] = (((tr >> 8) + tr) >> 8) as u8;
        surface[i + 3] = a as u8;

        i += SURFACE_CHANNELS;
    };

I think it's abusing bitwise operators to do rounding

It's.... beautiful

Also fuck the concept of alpha premultiplication

1 hour later…

mousetail

08:27

Also updated version:

2 hours later…

l4m2

09:59

(1/5)^^inf mod 9
= (1/5)^(1/5)^^inf mod 6
= (1/5)^(1/5)^(1/5)^^inf mod 2
= (1/5)^(1/5)^(1/5)^(1/5)^^inf mod 1
= (1/5)^(1/5)^(1/5)^0 mod 2
= (1/5)^(1/5)^1 mod 6
= (1/5)^5 mod 9
= 2^5 mod 9
= 5

Can such thing get defined for (1/5)^^inf mod 7?

4 hours later…

Mukundan314

13:37

isn't the answer 7 for mod 9? my reasoning was:

(1/5)^^inf mod 9
= 2^^inf mod 9
= 2^(2^^inf mod 6) mod 9
= 2^(2^((2^^inf mod 2) + 2) mod 6) mod 9
= 2^(2^2 mod 6) mod 9
= 2^4 mod 9
= 7

the + 2 is because the cycle 2^x mod 6 doesn't include 1

similar logic for mod 7:

(1/5)^^inf mod 7
= 3^^inf mod 7
= 3^(3^^inf mod 6) mod 7
= 3^(3^((3^^inf mod 1) + 1) mod 6) mod 7
= 3^(3^1 mod 6) mod 7
= 3^3 mod 7
= 6

CowperKettle

14:28

DeepMind AI crushes tough maths problems on par with top human solvers

l4m2

15:23

@Mukundan314 Nope (x^y) mod 5 ≠ (x^(y mod 5)) mod 5

Mukundan314

I used phi(mod)

l4m2

so (1/5)^^inf mod 9 ≠ 2^^inf mod 9

Mukundan314

interesting

can't believe how I forgot that

any reason why we can't use same reasoning for mod 7?

(1/5)^^inf mod 7
= (1/5)^^inf mod 7
= 3^((1/5)^^inf mod 6) mod 7
= 3^(5^((1/5)^^inf mod 2) mod 6) mod 7
= 3^(5^(1^^inf mod 2) mod 6) mod 7
= 3^(5^1 mod 6) mod 7
= 3^5 mod 7
= 5

l4m2

I thought it would meet 1/5 mod 5, seems not here but other num would

Mukundan314

15:53

ig we will have problems if modulo is 11 since then exponent would be under modulo 10

6 hours later…