phc-discussions - Re: [PHC] Compute time hardness

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Message-ID: <20150403101949.GA24805@openwall.com>
Date: Fri, 3 Apr 2015 13:19:49 +0300
From: Solar Designer <solar@...nwall.com>
To: discussions@...sword-hashing.net
Subject: Re: [PHC] Compute time hardness

On Thu, Apr 02, 2015 at 02:55:46PM +0300, Solar Designer wrote:
> Can we estimate the number of transistors needed for a naive multiplier
> (minimizing solely the circuit depth)?  60,000 / 1024 partial product
> bits gives us 60 transistors per bit.  Perhaps half of them would be
> spent on the first set of additions (since 1024 is close to
> 512+256+128+64 = 960 inputs to further additions).  If so, we have 30
> transistors to spend on one full adder, which is more than we need -
> which is 8 to 22, from what I could find.  Of course, we wouldn't use
> ripple-carry adders, but 30 transistors per bit feels about right for a
> carry look-ahead adder.

I made an error here.  1024 bits need only 512 full adders for the
first set of additions, so we have 60 transistors per full adder there.

> So the circuit size appears on par with the naive one (and their fastest
> one is probably larger than the naive one).  Clearly, so much effort was
> put into this not just for the sake of weirdness.

With the above correction, my estimate is that the 60,000 transistor
circuit is maybe twice larger than a naive one would have been, and
the fastest circuit from this PhD thesis is 3x larger than a naive one.

Alexander