Ontopic Random Computer-Electronics Thread

[Valve1138]

Anyone here have an ultra wide monitor that they use when playing video games on PC?

I asked about this recently.

Apparently you’re a rich bourgeoisie if you have one.

 

[OzSTEEZ]

I asked about this recently.

Apparently you’re a rich bourgeoisie if you have one.

Ugh... This shit here on UF is getting stupid. i.e. Want better sound? Spending a bazillion dallors on this receiver system. What a better TV? If you spend more than 5 dollars you're an idiot.

 

[Valve1138]

Ugh... This shit here on UF is getting stupid. i.e. Want better sound? Spending a bazillion dallors on this receiver system. What a better TV? If you spend more than 5 dollars you're an idiot.

You spent 8 cents on that post.

 

[MadDog]

yeah i have https://www.asus.com/us/Monitors/ROG-SWIFT-PG348Q/

its p nice

 

[OzSTEEZ]

yeah i have https://www.asus.com/us/Monitors/ROG-SWIFT-PG348Q/

its p nice

Nice, that's one of the ones I've been looking into. Did your frame rate drop much?

 

[MadDog]

Nice, that's one of the ones I've been looking into. Did your frame rate drop much?

nah i have a 1080ti

 

[fly]

OK, fine. Let's assume an integer byte, a 6 drive array, and 2 way parity (P and Q). So, we're going to store 4 bytes and the two parity bytes:
D0 = 8, D1 = 12, D2 = 3, and D3 = 7 .

P = 8 + 12 + 3 + 7 = 30
Q = 1*8 + 2*12 + 3*3 + 4*7 = 69

If we lose D0, we can always recalculate:
30 = D0 + 22
8 = D0

If we lose D0 and D1, we have a slightly harder problem, but let's simplify using the knowns and collecting like terms:
D0 + D1 + 3 + 7 = 30
D0 + D1 = 20

69 = D0 + 2*D1 + 9 + 28
69 = D0 + 2*D1 + 37
D0 + 2*D1 = 32

From there, it's just a system of equations. Let's try substitution after solving for D0 in the first equation (D0 = 20 - D1) :
(20 - D1) + 2*D1 = 32
20 + D1 = 32
D1 = 12

Then we'll back substitute:
D0 = 20 - 12
D0 = 8

See? Easy peasy. The actual math uses XOR and bitshifts, but this should be easier for your apparently tiny brain to understand.

yeah, my brain is still too tiny. No idea how that works. I'll just call it magic.

 

[Mr. Argumentor]

yeah, my brain is still too tiny. No idea how that works. I'll just call it magic.

At that level it's just algebra. You're smart enough to do that. Look up "solving for x and y given two equations" for a basic version, then reread his post.

 

[Darth Handsome]

 

[wetwillie]

Your attention span is too short to explain . . nvm

 

[Jehannum]

yeah, my brain is still too tiny. No idea how that works. I'll just call it magic.

So, adding and multiplying are too hard?

 

[fly]

So, adding and multiplying are too hard?

It seriously might as well be in Aramaic. That's okay though, I can't be smart at *everything*. Thank you for trying tho.

 

[Duke]

It seriously might as well be in Aramaic. That's okay though, I can't be smart at *everything*. Thank you for trying tho.

Dummy

 

[gee]

Any text written in Comic Sans can be safely disregarded.

 

[Darth Handsome]

Any text written in Comic Sans can be safely disregarded.

 

[Jehannum]

It seriously might as well be in Aramaic. That's okay though, I can't be smart at *everything*. Thank you for trying tho.

Hmm.

Maybe I didn't explain the equations well enough?

let P = the first parity value
let Q = the second parity value

P = D0 + D1 + D2 + D3
Q = 1*D0 + 2*D1 + 3*D2 + 4*D3

So, if we lose any one value, you can solve using the first equation to get that value back.

If we lose any two values, you can solve the first equation to isolate one of the unknown values, then solve the second, back-substitute from the first you solved to get the value for the second lost value, and you'll be back to the first scenario (one lost value).

 

[fly]

Hmm.

Maybe I didn't explain the equations well enough?

let P = the first parity value
let Q = the second parity value

P = D0 + D1 + D2 + D3
Q = 1*D0 + 2*D1 + 3*D2 + 4*D3

So, if we lose any one value, you can solve using the first equation to get that value back.

If we lose any two values, you can solve the first equation to isolate one of the unknown values, then solve the second, back-substitute from the first you solved to get the value for the second lost value, and you'll be back to the first scenario (one lost value).

AHHHHHHHHH now I get it!

Thanks!

 

[Jehannum]

AHHHHHHHHH now I get it!

Thanks!

Cool. Sometimes I suck at teaching, but I honestly thought the first one was a fairly clear explanation. I can see now that someone without my background wouldn't necessarily connect the multiplication with the bit shifting.

 

[fly]

Cool. Sometimes I suck at teaching, but I honestly thought the first one was a fairly clear explanation. I can see now that someone without my background wouldn't necessarily connect the multiplication with the bit shifting.

Yep, I have no doubt that your first description was spot on.

 

[OzSTEEZ]

A couple days ago I came across this code of a colleague

var entity = entity != null ? entity : null;