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Science Forum Index » Cryptography Forum » Idea for algo.
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| David Eather |
 Posted: Sat Jan 03, 2004 6:47 pm |
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Guest
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Comments for Peter from a (relatively) crypto newbe:
It would appear that your cipher can't be strengthened enough to be
secure in today's environment.
But there are several reasons to take heart -
- You have definitely moved out of the pen and pencil strength ciphers -
congratulations
VB is not the ideal language for measuring speed with - it has to drag
windows around with it
- your cipher is probably faster than you imagine. You could check by not
encrypting a file but rather a variable in memory (encrypt it as many times
as you need to for a measurable time - also make the variable small so it
all fits in the CPU cache. It sounds like cheating but isn't. In the end
you will get a figure that tells (in ideal circumstances) how long each
byte/block takes to encrypt without the need to wait for disk i/o, memory
access etc. That figure should be heartening.
- You re-invented something that was first described by someone else who
had more experience in crypto than you. So your thinking is not far off as
to be "out of the ball park" - perhaps John Savard can tell you more about
the effects of the swap function you reinvented.
Most people with an interest in crypto will remain enthusiastic amateurs,
some will become aficionados (hope the spelling is correct) and a very few
will become world class crypto gurus. (or if you prefer, most hobbyists,
some tradespeople and a very few artists). Read the faq sheet for sci.crypt
and try to get an introductory book on modern cryptography, "Applied
Cryptography" by Schneier is probably the easiest to get into, but it will
still be a steep learning curve.
Even if you never become a guru or artist with published ciphers and
"breaks" of other ciphers that need not stop you from experimenting in, and
enjoying cryptography.
"Peter" <peter_rabbit@shaw.ca> wrote in message
news:LPKIb.894773$9l5.45015@pd7tw2no...
Quote: quite a simple exchange (swap) of the byte involved with 2 others.
a= a_key(x) 'the involved Byte
a_key(x)=a_key(y)
a_key(y)=a_key(z)
a_key(z)=a
it is actually a bit more involved, since you must shuffle 2 arrays, the
a_key and the ref. array. I don't want to bore everybody with the
details and wear out my already thin welcome here, but to anyone
interested, I'll gladly mail the whole thing. (be forewarned, though, it
is in VB6 and not in God's language C)
Peter
Michael Amling wrote:
Peter wrote:
you do not need to shuffle the whole array! Simply shuffle the byte...
Uh, how is it again that you choose the byte with which you're going
to shuffle the byte just used? (I presume that "shuffle" means
"exchange" in this context.)
--Mike Amling
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