The DMZ

A: VPN - Tor: can VPN provider see the data?

8:15 AM

Between your Tor client and the guard node, the communications are encrypted with three layers of AES128. The first layer is a key shared between you and the guard node. The second layer is a key shared between you and the middle node. The third layer is a key shared between you and the exit node...

forest does some "simple math" here, but I'm dumb enough to not undertand. Can someone explain where the 2^128*2 comes from and how it equals 2^130?

1 hour later…

ThoriumBR

9:44 AM

I would like to have an explanation too... I believe 2^128*4 would be 2^130...

2 hours later…

11:24 AM

@nobody (2^128)*2 = 2^129

It's a simple mistake, nothing more

Where does (2^128)*2 come from in the first place?

There are three 128 bit keys

So that's equivalent to a 384 bit key

If you take in account the meet in the middle attack

then its 256 bits

I think that's a question for the sidechannel

I have a weird feeling in my throat. I think I got the coof

coof == cough?

the rona, my dude

With only a 98% chance of survival, I'm as good as dead

RIP

11:38 AM

I can already see the funeral service held in the DMZ

"Here lies MechMK1. He may not always have been the nicest person, and he had some very objectional political views, crude humor and generally lowered the quality of discussion in the DMZ for years... the sentence ends here."

11:52 AM

Atleast you get a funeral service here. I wouldn't even get that.

Makes sense though

No funeral service if nobody dies.

Wait, that means... if I die because of covid, then everybody can stop worrying about covid, because nobody was killed by it

Exactly

Nobody got infected, nobody died. Pandemic over.

whelp! where do I hide?

Luckily, nobody can hide from the virus

12:14 PM

@nobody exponents don't work that way. If you multiply two numbers you add their exponents

Anonymous

Morning.

Anonymous

Afternoon, nevermind.

Anonymous

What is time anyway?

@ConorMancone Isn't that what I did? (2^128)*(2^128)*(2^128)=2^384

ThoriumBR

@J-- it's 15 past something...

12:18 PM

It is a time

Take 2^3 for example (aka 8). 2*8=16, aka 2^1 * 2^3 = 2^4

When you multiply numbers you add their exponents

(sorry, I am thinking out loud)

Ah yes: so, if you had 2 keys with 128 bits of entropy, this is not the same thing as a single key with 256 bits of entropy

2 keys with 128 bits is "double" a single 128 but key, so you have 2^1 * 2^128 = 2 ^ 129 bits

I knew something like that was going on

But fuck I am so tired

Anonymous

15 past tired.

Anonymous

You could ask me what color the sky was and I'd say "Whatever, green or such"

12:23 PM

@ConorMancone How so? The number of different possible key combinations are still 2^256

(2^128)*(2^128) doesn't mean that you have 2 keys with 128 bits: it would mean that you have 2^128 keys with 2^128 bits!

Anonymous

Today I am bypassing ASLR.

Anonymous

Fun.

@nobody no, a 256 bit key is not just double a 128 but key - it is far, far, far stronger

@ConorMancone Pretty sure you're wrong here. 3DES has 112 bit strength after the meet in the middle attack

If you were right, it would be only 58 bits

12:25 PM

true

fuck, I'm retarded

@MechMK1 your chance of survival is actually only like 99.99%, assuming you are under 60 and not in a high risk category

It's been fun dude, I'll miss you!

Anonymous

That's not what Tom said :D

Anonymous

I would miss Mech dearly, however.

The DMZ would be dead without him

Anonymous

Literally.

Anonymous

12:28 PM

I wouldn't even be bothered to post.

Anonymous

Unless forest was here. Then I might.

@nobody try it with small numbers. I have 2 keys with 4 bits each. That gives me 8 bits of entropy total

Anonymous

No offense to anyone else I'm just...

Anonymous

Lazy.

Anonymous

Anti social.

Anonymous

12:28 PM

Mechs biggest fan.

Anonymous