WiiU`s compression strength continues from Wii at about 3 to1 per TEV

#51AceMosPosted 9/3/2013 4:05:35 AM
TC using big words dont make you smart

and using them out of context just makes you foolish
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3 things 1. i am female 2. i havea msucle probelm its hard for me to typ well 3.*does her janpuu dance*
#52oq7sterPosted 9/4/2013 12:33:48 AM
Something like?

Y
044
0334
02234
012344
000000X

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3DS: 1805-3478-2204; NNID: oq7ster
NDF Superbus membrum
#53SkyCrackersPosted 9/4/2013 5:56:52 PM
You sound like you really do know something about the Wii/Wii U's TEV unit, and real-time graphics in general, but this mostly sounds like meaningless babble. I have a massive nerd obsession with real-time graphics algorithms, so If I can't make sense of it, chances are no one else can, either. You're either horrible at expressing your thoughts, or you're just posting random strings of jargon... why do you do this? Is it to troll? Just trying to look smart? I must know! D:
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"Darwin gave us sensory organs for a purpose" -my friend, demonstrating his understanding of evolution
#54Infinity8378(Topic Creator)Posted 9/5/2013 8:02:00 PM
[This message was deleted at the request of the original poster]
#55Infinity8378(Topic Creator)Posted 9/5/2013 8:04:18 PM
The way you get compression is you break the bits into constants of rounded increments derived from the equation.

(2^(n-1))/(2^(e^((n(n-1)/2 + 1)/5))))

stored in an array only for needed values

where n is the number of bits
#56Golden MavenPosted 9/5/2013 8:30:11 PM
http://devkitpro.org/viewtopic.php?f=3&t=6755&view=previous

He's just copy-pasting forum posts from developer help sites.
#57Infinity8378(Topic Creator)Posted 9/8/2013 10:23:31 AM
[This message was deleted at the request of the original poster]
#58Infinity8378(Topic Creator)Posted 9/8/2013 2:04:10 PM(edited)
(2^(8))/2^(Q+(1+sqrt(5)/2)^(10-n)-(1-sqrt(5)/2)^(10-n)))

Where Q is 2 or 3
Where it's needed and bits>4bits works best
Then where there is repetition in the 16-32 combination range with the constants group the values together so you only have 1 6bit or 7bit constant instead of 2.
This is only used where it's needed most

Though you could represent bits other ways.

Basically

2^(5 , 4,3 or 1)
#59Infinity8378(Topic Creator)Posted 9/10/2013 11:37:23 PM
[This message was deleted at the request of the original poster]
#60Infinity8378(Topic Creator)Posted 9/11/2013 7:59:35 PM(edited)
This may work better

(2^(1+(1+sqrt(5)/2)^(n-4)-(1-sqrt(5)/2)^(n-4)))

Where it's needed and bits>4bits works best
Then where there is repetition in the 16-32 combination range with the constants group the values together so you only have 1 6bit or 7bit constant instead of 2.
This is only used where it's needed most

Though you could represent bits other ways.

Basically

2^(4 , 3,2or 2)

filled every 2^18 efficient at 48+16+8 72*96 bytes about 1 KB.