phc-discussions - Re: [PHC] Re: BlaMka loses entropy

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Message-ID: <CAOLP8p7QSYpvFC-vHS3=c049HCL0-NYhFzfp=f74OwiKb6Lt-A@mail.gmail.com>
Date: Mon, 19 Oct 2015 12:12:25 -0700
From: Bill Cox <waywardgeek@...il.com>
To: "discussions@...sword-hashing.net" <discussions@...sword-hashing.net>
Subject: Re: [PHC] Re: BlaMka loses entropy

On Mon, Oct 19, 2015 at 12:01 PM, Marcos Simplicio <mjunior@...c.usp.br>
wrote:

> Hi, Bill.
>
> I was about to point out the mistake with a Java snippet we developed to
> check if BlaMka is a permutation for several bit-lengths.
>
> You were faster than me, though :)
>
> BR,
>
> Marcos Simplicio.
>
>
Yeah, I should have better checked my work before posting.  Sorry for
making so many mistakes.  Here's a simple proof that BlaMka preserves
entropy:

Let a and b be two 64-bit integers, and let ah and bh be the high 32 bis of
a and b, while al and bl are the lower 32 bits.  Then the BlaMka function
inner multiplication step is:

    a' = (a + b + 2*al*bl) % 2^64
    a' = (ah + al + bh + bl + 2*al*bl) % 2^64
    a' = (ah + b + al*(1 + 2*bl) ) % 2^64

Given a' and b, we can compute a:

    (ah + b + al*(1 + 2*bl) ) % 2^64

First, subtract b:

    (ah + al*(1 + 2*bl) ) % 2^64

The factor (1 + 2*bl) is an odd integer.  Therefore, it has an inverse mod
2^64.  Call this inverse bi, and multiply by it:

    (ah*bi + al) % 2^64

We now know al, since ah*bi has 0's for the lower 32-bits, while al has 0's
for the upper 32 bits.  So, subtract al, and multiply by the inverse of bi,
which is (1 + 2*bl), to reveal ah.

Bill

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