Kaelygon

joined 1 year ago
[–] Kaelygon 4 points 2 weeks ago* (last edited 2 weeks ago)

It's condensed content with simpler terms and plain English, which is helpful for those who aren't native speakers, like Gamba said.
Simple wiki also comes in handy in topics like biology, which can have very specialized vocabulary.

But in this context, the people who unironically believe in things like the moon not being a reflector can't be reasoned with. They won't change their mind no matter how simple English you explain the fact.

[–] Kaelygon 5 points 2 weeks ago

Finnish bank osuuspankki logo comes close to qp outline, but it has extra shaft at the top

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memory_DEBUG.c [OC] (lemmy.world)
submitted 1 month ago* (last edited 1 month ago) by Kaelygon to c/poetry
 

Here's what a successful free message looks like

I was testing some unconventional C methods and got bit off track with the error messages. Source code is in pastebin.

There's others small details in the sources, such as the addresses needed to correctly free the memory is invalidated right after the "regret" verse (allockStack = NULL), if the "Worry not" verse doesn't come up.

Moral of the story: free your allocated memory.

[–] Kaelygon 11 points 1 month ago* (last edited 1 month ago) (1 children)

Google search results are often completely unrelated so it's not any better. If the thing I'm looking for is obscure, AI often finds some thread that I can follow, but I always double check that information.
Know your tool limits, after hundreds of prompts I've learned pretty well when the AI is spitting bullshit answers. Real people on the internet can be just as wrong and biased, so it's best to find multiple independent sources

[–] Kaelygon 9 points 1 month ago* (last edited 1 month ago) (11 children)

thankfully modern ones like molten salt reactors have passive safety, where they stop the reaction if overheating occurs.
edit: My mistake, there's no active commercial molten salt reactors.
But nuclear power is very safe nowadays because of the multiple fail-safes, which some can still be passive like emergency cooling.
I much rather get electricity from magic rocks than destroying rain forest in developing countries drilling oil, gas or mining coal.
The biggest risk in nuclear is environmental disasters like in Fukushima's case, which is the last significant nuclear incident in past 13 years

[–] Kaelygon 1 points 1 month ago

Thank you! :3
I think the very limited palette and dithering does it. Even gameboy color could display 32768 colors, but the sprites were limited to 4 colors per block due to vram limitation.

[–] Kaelygon 2 points 1 month ago

I loved every bit of Rain World! But I ended up quitting it mid play through when it became too hard. I found a way to gather stacks of berries to have enough reattempts for the hard parts, but then got lost where I was even supposed to go and gave up after ~25 hours playtime

[–] Kaelygon 2 points 1 month ago

Whoop, I mixed up dark souls 3 with Elden ring. Though, the same applies. I did like the gritty atmosphere and lore. The main issue I had was the learning curve and when trying to playing co-op there was no way to turn off strangers joining what I recall. But I bet by now there's mods for all of that like you said.

[–] Kaelygon 26 points 1 month ago (15 children)

I once made the mistake googling easy mode for Elden ring that someone gifted to me. Once I saw the gatekeeping on Reddit, I decided it's not a game for me and uninstalled. I'm sorry that I suck at video games

[–] Kaelygon 1 points 1 month ago* (last edited 1 month ago)

It appears to always run in ~30 milliseconds regardless of the tested number, so this might be O(1) until some bottleneck kicks in. Though I have yet to verify the complexity as the quality of division rule depends on a,b and c ranges.
Edit: after some testing it's some logarithmic complexity when P is bigger than 10^2000

P size, time seconds
10^3000, 3.11
10^4000, 6.43
10^5000, 11.27
10^6000, 17.69
10^7000, 26.31
10^8000, 37.09

Plotting these gave about O(log(P)^2.5)
The bRange, math.gcd() and reciprocal scale with P digit count but rest of the calculations are O(1).
I have no idea why you would need 10^8000 divisibility rule designed hand calculations, but you can get one under a minute and this isn't even multithreaded!

[–] Kaelygon 1 points 1 month ago (1 children)

Funnily enough, I just sped up my own solution by 25000% without compromising anything.
https://pastebin.com/Dkbq2chV

I realized that multiplying the divisor P by its non-zero reciprocal digits, gets you near 10^n which are ideal numbers for the divisibility rules. Which should have been obvious since n * (1/n) = 1, and cutting off the reciprocal results in approximation of 1, which can be scaled by 10^b.
e.g. finding divisibility rules for 7

1/7=0.14285...
7*14=98
7*143=1001
7*1429=10003

The first script was very naive brute force approach.
So instead of searching every combination of a, b and c, I can just check the near multiples of P*reciprocal.
The variables can be solved by P*N = a*10^b + c when b is given and a is 1 to 9
7*1429=10003 would expand to P*N=1*10^4+3

[–] Kaelygon 3 points 2 months ago* (last edited 1 month ago) (3 children)

I wrote some terrible python code to search divisibility rules for a given number and it tests example product divisibility

Edit2: https://pastebin.com/Dkbq2chV Yet another revision, I got caught up in this project but I think it has enough features now. I added few command line options and details you can edit in the script.
I need to stop before I add more features. Here's example output:

$ python ./findDivRules.py -h

python ./findDivRules.py [Integer or "(Start,End)"] [Show example? (0,1)] 
   [Example is divisible? (0,1)] [Parker style? (0,1)] [Rule count] [Rule index]
Default command : python ./findDivRules.py 313 True False True 10 0
Range example   : python ./findDivRules.py "[1,11]" 0 0 1 2 0

$ python ./findDivRules.py 313 True True True           

Found 3 rules for 313, showing first 10:
  P    N    a    b    c     P*N
313   16    5    3    8    5008
313   32    1    4   16   10016
313  639    2    5    7  200007

313 has following divisibility rule using B*a-A*c
Split the tested number into A and B after 3rd digit.
Multiply A by 8 and multiply B by 5
Subtract A from B = B*5-A*8

Example:
Using rules P=313 a=5 b=3 c=8
Testing 700807 divisibility by 313

A|B      B*a-A*c        Intermed
700|807  807*5-700*8       -1565
-2|435   435*5+2*8          2191
2|191    191*5-2*8           939
0|939    939*5+0*8          4695

Smallest iteration 939 = 313*3
700807 is divisible by 313
8
submitted 2 months ago* (last edited 2 months ago) by Kaelygon to c/math
 

I really hoped Matt Parker would have shown how to construct these divisibility rules, so I came up with my own method. Find prime P and natural number N such that P*N = a *10^b + c
The smaller a and c are, the better. b determines where you split the number.

Example: P=313 N=16 a=5 b=3 c=8

313*16=5008  
313*16=5 *10^3 + 8  

Now we can test for 223795=313*13*11*5
Split 223795 after the b:th number, 3rd in this case.
223795/10^3=223.795 decimal point separates the components A, B

Multiply the A=223 by c and subtract from the rest B=795 multiplied by a

B*a-A*c=  
795*5-223*8=2191  

Repeat if needed till you get to small enough number.

2191/10^3=2.191  
191*5-2*8=939  

which is easy to see that's 3*313
Some bad combinations don't reduce the starting number but they are at least always divisible by P. Those cases could be called Parker divisibility rules.

You can also see that when c is negative, A*c is added rather than subtracted, which explains why some method add or subtract like Vsauce vs James.
This is just a funny trick to simulate division and modulus by 10^b to get smaller number, while preserving the congruence.

It's possible I made mistake somewhere, but was able to get correct answers with other few examples.

[–] Kaelygon 3 points 2 months ago (1 children)

Thank you so much! ^^
Pixel art has very different techniques to other art but I still spent lot of time tweaking the proportions and line gradients pixel by pixel. Gen 2 pokemon sprites are an other impressive example what you can do with just 4 colors and 56x56 sprite.

 

60x60px 12 colors RGB15, I always love how much you can do with so little in pixelart.
This is upscaled to 360 because no website uses neighbor nearest scaling. The original image is just 1.4 kb

Kaelygon is my first furry OC and to this date I use it as a artist- and nickname.
Back in 2016 I stumbled upon Dutch Angel Dragons and it was the first furry community that I was active.

 

I converted my speculative species into a thematically fitting normal/poison Pokemon.

Venom Bite is the only custom move.
Physical, poison, 120 power, 70% accuracy and 30% chance to poison the target.
It's supposed to be poison variant of the Gen 2 Thunder.

No actual rom hack was done. This was a cheeky video edit, but all of the stats, sprite and flavor texts are within the original game constraints. Gen 2 Pokemon have 10 character limit so "Maned dragon" had to be abbreviated.

I scaled the IVs and EVs referring to my L73 Vaporeon, so every stat should be plausible and moves in the video is what Manedragon would have in wild.
They're not optimized for anything.


PokeDex:

No. ???
Manedragon
Venomous
Type / Normal Poison
HT 3'11"
WT 99.2lb

Stat Value
HP 85
Attack 110
Defense 70
Special Attack 80
Special Defense 85
Speed 105
Total 535

Flavor texts:
Silver

It prefers to hunt
alone and is ready
to strike with its
venom. Yet it
forms unique bonds
with its own kind.

Gold:

Its venomous bite
can defeat larger
prey than itself.
Despite its name
and looks it's not
a true dragon.

Rest of the Manedragon movepool, stats and details

Based on speculative species Maned dragon

Cry editor Github

56x56 Sprite

10
submitted 2 months ago* (last edited 2 months ago) by Kaelygon to c/math
 

I have been exploring this particular prime 13238717 which is sum of two squares and has Pythagorean triple.
I found this interesting property and so far I haven't found any texts about what I wrote below.

This is just my conjecture, I have no formal proof and I have only tested few small primes.
I haven't found any counter examples yet, but I have checked only few dozen primes and couple composites by hand.

I modified ChatGPT script which lists numbers that have both forms P^2=a^2+b^2 and P=c^2+d^2 and it appears to generate the exact same sequence as: A004431
5 10 13 17 20 25 26 29 34 37 40 41 45 50...

Numbers P seem to always have both following forms Hypotenuse numbers (Pythagorean triples) A009003: P^2=a^2+b^2
Numbers that are the sum of 2 squares A001481: P=c^2+d^2

Wolfram notes, "-- one side of every Pythagorean triple is divisible by 3, another by 4, and another by 5."
I noticed if P is prime and have both forms: one of the Pythagorean sides (a or b) whichever is divisible by 4 has the exact factors that construct both of the square sum components 'c' and 'd', with the exception of extra factor '2'.

Here's the conjecture put out more formally based solely on my observations:
Pythagorean sides: a,b and square sum componentsc,d are natural numbers and n#, m# are prime factors.

a^2 + b^2 = P^2 (pyth. triple)  
c^2 + d^2 = P (sum of two squares) 

(a or b) mod 4 = 0
(a or b) factors are = 2* (n1*n2*n3...) * (m1*m2*m3...)
c = n1*n2*n3...
d = m1*m2*m3...
(a or b) = 2*(c*d)

Here's couple examples:
primes 2 and 5 are trivial exceptions as 1 isn't a prime factor.

1^2+2^2=5  
1^2+1^2=2

Prime: 13 (first non-trivial prime case)

5^2 + 12^2 = 13^2 (pyth. triple) 
2^2 + 3^2 = 13 (sum of two squares)  

12 factors are 2 2 3
12 mod 4 = 0
c=2
d=3
12 = 2* (2*3)

Prime: 821

429^2 + 700^2 = 821^2 (pyth. triple)  
14^2 + 25^2 = 821 (sum of two squares)  
  
700 mod 4 = 0
700 factors are 2 (5 5) (2 7)
c = 5*5 = 25
d = 2*7 = 14
700 = 2* (25*14)

prime: 13238717

1315508^2 + 13173195^2 = 13238717^2 (pyth. triple)
181^2 + 3634^2 = 13238717 (sum of two squares)  
  
1315508 mod 4 = 0
1315508 factors are 2 (2 23 79) (181)
c=181
d=2*23*79=3634
1315508 = 2* (181*3634)

Some composites have multiple ways to write sum of two squares, which each have different (a or b) counterpart, but not necessarily divisible by 4. composite: 260

(four valid pythagorean side pairs)  
132^2+224^2 = 64^2+252^2 = 100^2+240^2 = 156^2+208^2 = 260^2 
(two valid square sums)
8^2+14^2 = 2^2+16^2 = 260

8^2+14^2:
224 mod 4 = 0
224 factors 2 (2 2 2) (2 7)
c= 2*2*2 = 8
d= 2*7 = 14
224/(8*14) = 2
  
2^2+16^2:
64 mod 4 = 0
64 factors 2 (2) (2 2 2 2)
c = 2
d = 2^4 = 16
64 = 2* (2*16)

composite: 58

40^2+42^2=58^2 (pyth. triple)  
3^2+7^2=58  (sum of two squares) 

42 mod 4 = 2 
Not 0 mod 4 congruent, unlike primes. 
Might be result of both c, d being odd.
42 factors 2 3 7
c=3
d=7
58 = 2* (3*7)

Generally it seems that there's always at least one Pythagorean component where (a or b) = c*d*2, but I haven't quite figured why this is the case.

I reckon it has something to do with the fact that mod 4 congruence of 4k+1 doesn't change when you square it: (4k+1)^2 = 8k*(2k+1)+1
Additionally the fact that when sum of two squares is prime or odd, exactly one of the components is always odd, which may explain why (a or b) isn't always divisible by 4 with composites.

ChatGPT wrote a counterexample finder I checked up to 100 000.

To my knowledge there isn't straight up equation that would spit out a Pythagorean triple or sum of square solutions for any integer.
There might be some other way to prove or disprove that a or b = c*d*2 when P is prime, but so far I couldn't think of any.
It might be something obvious that I am missing, or it's simply all about congruence rules of additions and multiplications.

Again, this is just what I've found from my few tests and I don't have any formal proof. I couldn't find any papers or posts specifically about this. This is nothing too important, but I found it interesting enough to share.
I just write bad python code out of interest in number theory without an university degree.


Thanks for the solution goes to: @0v0
Here's summary how I understood this:
Brahmagupta–Fibonacci identity

P=c²+d²
Squaring 'P' results in:
P²=(c²-d²)² + (2cd)²

These 'P²' sum components are equivalent to the Pythagorean legs 'a' and 'b':
a=c²-d²
b=2*c*d

Hence, '(a or b)' always contains the factors '2' and all factors of 'c' and 'd'. 

Additionally, the 2*c*d divisibility by 4 is result of P being odd.
2*c*d being divisible by 4 is true for any odd number as one of the sum of square components 'c' or 'd' must be even.

c=2*k
b = 2*c*d = 4*k*d
 

Steps to reproduce:

  1. Be logged out
  2. Search Communities
  3. Paste the full community link to search -> No results.
  4. Login
  5. Two results appear and if you edit the text field, a duplicate result shows up each key press.
117
50 baguettes [fishtrouts] (64.media.tumblr.com)
submitted 3 months ago* (last edited 3 months ago) by Kaelygon to c/[email protected]
 

image
image

I really love fishtrouts comics.
You can find the original post here!

 

Not a dragon in traditional sense, but I took some inspiration from furred dragons and drakes.
The original concept is mix of canines and lizards.

 

Is it a lizard or just a funny looking dog?
I made cover art for my own species!

It sure has been a while since I drew ferals, but I think this ended up pretty alright.
I barely draw nowadays as I mostly focus in 3D art.
Feathers are bit wacky and shading isn't very accurate as I mostly winged it, but I just liked the process of drawing shades.

595
submitted 3 months ago* (last edited 3 months ago) by Kaelygon to c/politicalmemes
 

It's totally fine if you believe that life starts at conception.
The thing that actually baffles me are the states that passed anti-abortion laws, but struggle to provide adequate health care, especially for those who are not financially stable.

I found this article, "States with more abortion restrictions have higher maternal and infant mortality", but feel free to correct or educate me on the topic.

Edit: removed "this article" appearing twice and tried to fix preview

 

Furred dragons have become a wide range of creatures, and classifying them is vague at best, so this will be my subjective view.
My fursona Tiena recently had an identity crisis as I asked the important question What is a dragon?

The most common feature among furred dragons is the snout in the picture, which is usually similar to reptiles and the nostrils are on top of it. But not exclusively, but this is by far the most common trait. I believe that the only requirement for a furred dragon is to have fur or feathers.

Traditional dragons usually are mythological, and one or more of these features: breath attack, large size, wings, scales, whiskers, reptilian features, lays eggs, horns, etc.
There are also dragon types such as, Eastern, Eastern, Wyrm, Serpent, Wyvern, and so on.

Furred dragons don't have strict rules about what they are other than fuzzy dragons.
They may lack wings, have ears, be mammals, and require no supernatural abilities or relation to mythology. Which begs the question of whether they are dragons at all.
Although I have yet to see a furred dragon that doesn't have any draconic features, you may add that as the second required trait.
Of course a creator of their character can ignore the semantics and have any artistic freedom on what makes them a dragon. Not necessarily in a traditional sense.

Originally around 2018 my sona took inspiration from komodo dragons and maned wolves, then 1-2 years later I found about furred dragons, which shared many similarities due to the reptilian aspects. From that point on, I called my sona a furred dragon until now.

I concluded that my sona isn't an actual dragon, as his only dragon-like features are the snout and tail profile, but even those are based on komodo dragons.

So, I made a new species based on maned wolves and komodo dragons and I came up with the very original name feather maned dragon. But since the species is based on early real-life mammals, the name is a misnomer as he's not an actual dragon. Which is rather poetic, and this all comes back to a full circle.

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