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Date: Mon, 27 Jan 2014 14:34:13 +0100
From: Dmitry Khovratovich <khovratovich@...il.com>
To: discussions@...sword-hashing.net
Subject: Re: [PHC] An argument for Catena-2

Hi Bill,

it seems that your 10X estimate is a bit pessimistic. If you have only half
of memory available, you can split it in half and store every 4th point
(v_i) from the first two rows in memory. Hence while computing the second
row, you need on average 1.5 steps on the first row from a distinguished
point to the immediate predecessor, and 2.5 steps on average in total.
Having spent G steps for the first row, you need 2.5*G for the second row.

The same holds for the third row, where for each new node you need on
average 1.5 steps on the second row. As they cost 2.5 on average, you will
need 2.5*1.5+1 = 4.75 steps on average for the third row, or 4.75*G in
total. Overall, you need about (4.75+2.5+1)G=8.25G steps, which should be
compared to 3G with the full memory available. Thus the penalty is 8.25/3 =
2.75 .

I have applied this algorithm to the example graph with g=3 from the Catena
presentation, and obtained 66 steps, which perfectly matches the above
estimate.


In general, if you have 1/q of memory available, this algorithm gives G for
the first row, ((2q-1)/2+1)G for the second row, and
(((2q-1)/2+1)(2q-1)/2+1) for the third row, or (q^2+3q+2.25)G in total,
which should be compared to 3G. The overall penalty is quadratic of q, but
grows not so fast. At q=10 the penalty is 44X. The second row calculation
can be further optimized.

Best regards,
Dmitry


On Sun, Jan 26, 2014 at 9:32 PM, Bill Cox <waywardgeek@...il.com> wrote:

> I've done some more work on my pebbling program, and tested Catena-2
> against Catena-3.  Catena-3 does have a much higher penalty for
> attackers using a TMTO, however, I think Catena-2 may be good enough.
>
> When I save 1 memory location for Catena-2, I get a 50% recalculation
> penalty.  No one is going to try and save 1 memory location against
> Catena-2.  When I try something reasonable, like half, then I get a
> recomputation penalty of over 10X.  Compute cores are cheap compared
> to memory, so maybe I can run parallel cores and reduce the runtime
> penalty, but I doubt I can get that 10X to run enough in parallel to
> get the recomputation time to below 2X.  I should write some more code
> to determine how much parallelism is available in the recomputations.
> My guess is it is not enough to make it worthwhile.
>
> I benchmarked Catena-2, and it is running 3.4X faster than Scrypt,
> compared to 2.6X faster for Catena-3.  It's still 2.6X slower than
> NoelKDF for the same memory, which is right in the middle of the
> theoretical range of 2X to 3X.  It does 2 reads and 2 writes to every
> memory location, compared to 1 read and 1 write for NoelKDF, which is
> where the 2X lower bound comes from.  When memory bandwidth limited,
> we should see a 2X difference.  For compute limited situations,
> Catena-2 does 3X more computation than NoelKDF for the same memory
> size, which is where the 3X upper bound comes from.
>
> Performance does matter.  I'll do some more analysis on the chances of
> a parallel attack against Catena-2 succeeding, but for now I'm leaning
> towards Catena-2 for a cache-timing attack resistant algorithm.
>
> Bill
>



-- 
Best regards,
Dmitry Khovratovich

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