[Cryptography] [PHC] [OT] Improvement to Momentum PoW

[Bill Cox]

On Tue, Jul 28, 2015 at 4:36 AM, Dmitry Khovratovich <khovratovich at gmail.com
> wrote:

> I am not sure what you're calling by "parallel rho", but the parallel
> collision search by van Oorschot-Wiener deals exactly with this problem.
> They call it "golden collision" search in Section 4 of
> http://people.scs.carleton.ca/~paulv/papers/JoC97.pdf .
>
> The tradeoff in that paper is very favourable to the attacker: when the
> memory is reduced by P^2, the time grows by the factor of P only, so it is
> the sublinear growth. Storing only some small range of Hb(i) gives  a much
> worse, linear tradeoff only.
>
> Dmitry
>

This is incorrect.  This is covered in the paper in the section "Run-Time
Analysis for Finding a Large Number of Collisions" under finding golden
collisions on page 9.

The hash table algorithm uses O(sqrt(n)) memory, for n points, and runs in
time O(n) to generate all collisions.  Decreasing it by a factor of T,
causes it to run T times longer to generate all collisions, so memory*time
= O(n^1.5), regardless of T.  The paper's algorithm runs with w
distinguished points, using O(w) memory, and runs in time O(sqrt(n^3/w)).
Memory*time is O(n^1.5*sqrt(w)).  This is far worse than the hash table,
for all values of w >> 1.

For constant memory, both algorithms converge to O(n^2) runtime, no better
than comparing random points.  This isn't in the paper, but it's clear from
the algorithm if you imagine storing 1 distinguished point.  You have to
take O(n) steps to hit it once, and the paper's algorithm requires hitting
it twice.  Since random points collide with probability 1/n, you only need
O(n) random comparisons to find a collision.  The same goes for the hash
table algorithm.

So, distinguished point algorithms fail at every memory size compared to
using a hash table.

Parallel Pollard Rho is where we share the distinguished points among a
large number of workers.  This algorithm fails vs a shared hash table for
the same reasons as above.

Anyway, thanks for taking some time to examine it.  For now, I still think
this algorithm works better than regular Momentum.

Bill
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