phc-discussions - Re: [PHC] Schvrch is broken

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Message-ID: <54181820.5010408@ciphershed.org>
Date: Tue, 16 Sep 2014 06:59:44 -0400
From: Bill Cox <waywardgeek@...hershed.org>
To: discussions@...sword-hashing.net
Subject: Re: [PHC] Schvrch is broken

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On 09/15/2014 11:29 PM, Rade Vuckovac wrote:
> case 1 carry = b xor c xor d
> 
> case 2 carry = b xor c xor ~d
> 
> case 3 carry = b xor ~c xor d
> 
> case 4 carry = b xor ~c xor ~d
> 
> case 5 carry = ~b xor c xor d
> 
> case 6 carry = ~b xor c xor ~d
> 
> case 7 carry = ~b xor ~c xor d
> 
> case 8 carry = ~b xor ~c xor ~d
> 
> a’ = a xor (can be any of the cases above*) carry


The value of carry is equal in cases 1, 4, 6, and 7.  The value of
carry is also equal in case 2, 3, 5, and 8.

There are only 2 outcomes here, not 8.  These are basic XOR identities:

a xor b == ~a xor ~b
~a xor b == a xor ~b == ~(a xor b)
~a = a xor 1

I like to add a few more so that I can convert any Boolean equation to
Read Muller form by hand easily.  I'll use + for xor, because xor is
addition, and it makes it easier to see:

a(b + c) = ab + bc (distributive)
a*a = a (power)
a + a = 0 (like terms cancel)
a*!a = 0
a or b = !(!a!b) = !a!b + 1 (convert OR to AND)

So, for example, when I try to simplify Wolfram's Rule 30, I do it
this way:

a2 = a1 + (b1 or c1) = a1 + !b1!c1 + 1
a3 = a2 + !b2!c2 + 1
   = (a1 + !b1!c1 + 1) + !(b1 + !c1!d1 + 1)!(c1 + !d1!e1 + 1) + 1
   = a1 + !b1!c1 + (b1 + !c1!d1)(c1 + !d1!e1)
   = a1 + !b1!c1 + b1c1 + b1!d1!e1 + !c1!d1!e1
       b1c1 = (!b1 + 1)(!c1 + 1) = !b1!c1 + !b1 + !c1 + 1
   = a1 + !b1 + !c1 + b1!d1!e1 + !c1!d1!e1 + 1
   = a1 + !b1 + !c1 + !d1!e1(b1 + !c1) + 1
   = a1 + !b1 + !c1 + (!d1!e1 + 1)(b1 + !c1) + b1 + !c1 + 1
   = a1 + (!d1!e1 + 1)(b1 + !c1)
   = a1 + (d1 or e1)(b1 + !c1)

I probably made a mistake in there...

Anyway, in silicon, if I did not make a mistake, this is 2 XOR gates,
and OR gate, and an AND gate.  I think I can get this running about
50% faster than the straight forward computation.

Bill
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