[119004] in Cypherpunks
Re: Multiplying Binary Matrices
daemon@ATHENA.MIT.EDU (Jonathan Stafford)
Tue Oct 12 14:38:35 1999
Date: Tue, 12 Oct 1999 14:21:11 -0400 (EDT)
From: Jonathan Stafford <jestaff2@unity.ncsu.edu>
To: cypherpunks@einstein.ssz.com
In-Reply-To: <19991012110709.A21576@einstein.ssz.com>
Message-ID: <Pine.SOL.4.05.9910121356510.19409-100000@eos00du.eos.ncsu.edu>
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Reply-To: Jonathan Stafford <jestaff2@unity.ncsu.edu>
On Tue, 12 Oct 1999, Jim Choate wrote:
>
> Excuse me, but what the hell are you talking about? In your first post
> you multiply the matrices as if they contain normal numbers. In your next
> post you (supposedly) treat them like binary matrices, but I'm unsure
> where you get 0 | 0 | 1 = 1 from (except perhaps the (3,1) element, but
> why the hell would you be using that anyway).
>
> -------------------------------------------------------------------------
>
> I figured somebody would be more interested in proving me wrong than actualy
> checking Bill's work. You and one other fell for it. I can read titles of
Definitely easier to point out your mistakes than to go to the trouble of
actually checking Bill's work.
Actually, it's not that much trouble, so here it is:
abc abc 010
def * def = 001
ghi ghi 110
From this we create a system of equations:
1) aa + bd + cg = 0
2) ab + be + ch = 1
3) ac + bf + ci = 0
4) da + ed + fg = 0
5) db + ee + fh = 0
6) dc + ef + fi = 1
7) ga + hd + ig = 1
8) gb + he + ih = 1
9) gc + hf + ii = 0
From 1, 5, and 9, we see that a, e, and i must be 0.
After removing all terms containing a, e, or i:
1) bd + cg = 0
2) ch = 0
3) bf = 0
4) fg = 0
5) db + fh = 0
6) dc = 1
7) hd = 1
8) gb = 1
9) gc + hf = 0
Examining 6 and 8, we see that b, c, d, and g must be 1.
This however contradicts 9, therefore, there is no (binary) square root.
Paying attention Bill?
> posts, it was a pre-meditated trick to draw folks out. I was hoping Bill
> would toot his horn.
>
> I get 0 + 0 + 1 = 1
>
> as the basic definition of the OR operator. It's simple addition in base
> two. 0 + 0 = 0, 0 + 1 = 1 in ALL bases. Therefore,
>
> 0 + 0 + 1 = (0 + 0 ) + 1 = 0 + 1 = 1
Thanks for the review of how OR works, I almost forgot, especially since
it is the simplest binary operator there is.
> As to XOR, you brought that up, not me.
Yes, as an explanation of what Bill did, not claiming that he did it
correctly.
> Bottem line, Bill Payne's math is wrong the matris is unitary (ie. all
> elements are a 1).
Yes, all the elements should be one, because he should be using OR and not
XOR.
Jonathan
--
carry on underling
