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Date: Tue, 13 Mar 2018 10:31:22 +0100 (CET)
From: Stefan.Lucks@...-weimar.de
To: discussions@...sword-hashing.net
Subject: Re: [PHC] How to: A super-efficient cryptographic accumulator?

Hey Denis,

you have a nice idea, but it won't work for information-theoretical 
reasons. I'll explain below.

> [...] the best measure for password safety is probably to prevent users 
> from using any of billions of known revealed passwords, but the 
> compressed database is 5 GB in size:
>
> https://github.com/berzerk0/Probable-Wordlists/blob/master/Downloads.md

Let us denote this list of bad passwords as "L", and the universe of 
passwords not in L (up to a fixed length) as "U".

> Now. I sort of have this intuition that it’s possible to compress this crap
> into 20 bytes. That's not a typo – 20 bytes, not megabytes or something.
> But I don’t know how.
>
> If we have an infinitely fast computer, then we can simply find X such 
> that:
>
> TruncateTo32Bits ( SHA256 ( X | P1 ) ) == 0x00000000
> TruncateTo32Bits ( SHA256 ( X | P2 ) ) == 0x00000000
> TruncateTo32Bits ( SHA256 ( X | anything else ) ) == whatever

What you mean is

   (1)  Trunc32 ( SHA256 ( X | P ) ) == 0x00000000
        for all P in L

and ideally

   (2)  Trunc32 ( SHA256 ( X | Q ) ) != 0x00000000
        for all Q in U

or, for a

   (2') Pr [Trunc32 ( SHA256 ( X  | Q ) ) == 0x00000000] < 0.01
        for Q in U chosen at random

> If we crank the infinitely fast computer long enough, we should be able to
> find this X,

If we consider SHA256 as a "random oracle", such an X exists, and, 
assuming an unlimited amount of computer power, we can find it.

> and it should be small in size.

Why do you think it would be small in size? That is the whole point: X 
will be HUGE! If we want to meet both (1) and (2), then we can 
statistically expect X to be as big as L. The reason is simple: Whenever 
we change L, we need another X.

Changing from (2) to (2') doesn't safe the day: X can be slightly smaller, 
but only by a few bits.

> The only reason we should not be able to find this X is if SHA256 has 
> some structural issue that prohibits an X like that from existing.

Yes, the iterated structure of SHA256 (and essentially all other hash 
functions ever published) makes the existence of X extremely unlikely.
Neither SHA512 nor SHA3 will do, nor Blake, JH, Groestl, Skein, ...

BTW, if X where actually small in size, the iterated structure would not 
pose a problem, and even SHA256 would do.

> In that case we use a better hash function.

Good luck with designing an entirely new type of cryptographic hash 
function!

> Now, the reason we can’t do that is, the way SHA256 is designed, it
> requires us to have an infinitely fast computer.

There are three obstacles to doing that:

1. The way SHA256 and essentially all other hash functions are designed, 
such an X is unlikely to exist.

2. Given an ideal hash function (a "random oracle") X should exist, but we 
would need an absurdly fast computer to find X.

3. Given an ideal hash function and an absurdly fast computer, X would 
still be about as large as L. So even then, there is not much of an 
advantage over using L directly.

Finally, you asked about a quantum computer. If s is the expected size of 
X (in bits), the search for X will take about 2^(2^s) steps on a classical 
computer. Using a quantum computer, we can apply Grover's algorithm to 
reduce this value to its square-root: 2^((2^s)-1). If L is a few 
Gigabytes, then s is 30 or larger. Even using a quantum computer, the sun 
will long be cold before your quantum computer can output X.


Stefan

--------  I  love  the  taste  of  Cryptanalysis  in  the morning!  --------
www.uni-weimar.de/de/medien/professuren/mediensicherheit/people/stefan-lucks
----Stefan.Lucks (at) uni-weimar.de, Bauhaus-Universität Weimar, Germany----

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