Re: Unbreakable code using XOR for one time pad?




Unruh wrote:

Guy Macon <http://www.GuyMacon.com/> writes:

As for the "far more secure", Given the basic scheme of sending
multiple OTP keys and XORing them at the end, sending at least
one of the keys using public key encryption would make the key
distribution mechanism at least as secure as simply using public
key encryption would be, and certainly would make the chances of
an attacker getting that key far lower that the one in a million
used in the above calculation.

Sure. It would make the total probability essentially the chance
of breaking the public key cypher. So why not send the message
that way?

You appear to have missed the point of what I wrote. Nowhere
did I say that the method described is better than just using
PGP. My claim was narrow and specific: claims of insecurity
and inconvenience in distributing a large OTP key rather than
a small shared secret cipher key are overblown. I wrote that the
task of distributing a 128 bit AES key pretty much has the same
degree of security and convenience that sending an 8 gigabyte
OTP key has. I stand by that statement.




Unruh wrote:

Guy Macon <http://www.GuyMacon.com/> writes:

Unruh wrote:

He then argues that the probability of anyone getting all 6 is low and you
are OK. But the probability of someone getting all 6 is surely far far
greater than the probability of someone guessing a private key for a
symmetric cypher, or getting say an AES key via exhaustive search.

Hmmm. Is it really?

Let's assume a 1 in a million chance of getting each key. That's
6 times 20 bits, or 120 bits. Not too far from the AES-128 keysize.
Even if we assume a one in 16,384 chance for each, the more-secure
9-key scheme that I discussed in the same post would be 9 times 14
bits -- 126 bits.

This reminds me of the Drake equation. Multiply a bunch of totally unknown
probabilities together and then declare the answer as cast in stone and
correct.("I have no idea what the probability of life arising on a planet
the same temperature distance from a star as the earth is, so because I
have no idea, that probability must be about 1/2--- OOOh look how probably
life on other planets is ")

You are confusing things that have totally unknown probabilities
with things where we can set a lower and upper bound on the odds.

I would estimate the probability of getting each key for a
determined attacker as 1/10 not 1/1000000. What does that
do to your estimate?

It calls into question your estimating ability. Consider:

I send key 1 through a library computer using PGP
encrypted email.

I send key 2 through a computer at an Internet cafe.
This time I use stego to hide it in a series of porns
that I put up in a binaries Usenet newsgroup.

I send key 3 through my home computer using TOR, and I
run a TOR node from my house.

I send key 4 by US mail by walking it into a post office
20 miles from where I live and sending it to PO box at
at the destination city. It is a USB thumb drive that
I concealed inside a RC race car toy. For the odds of
interception to be 1 in 10, someone would have to intercept
10% of all US mail and tear apart every toy they find.

Key 5 goes by FedEx from another nearby city to a private
"mailboxes R us" box and the key is hidden in the slack
space of a laptop computer. For the odds of interception
to be 1 in 10, someone would have to intercept 10% of all
FedEx shipments *and* figure out that the data is there.

Key 6 goes by UPS, to general pickup at a UPS store in
the destination city and is in the unused portions
of some DVDs with family photos on them.

Key 7 goes by a courier service on a DVD labeled "Patent
Infringement lawsuit data"

My brother hand-delivers key 8, which I put inside a cake.

Key 9 is delivered by one of my employees embedded in a
prototype.

For the odds of interception to be 1 in 10, someone would
have to open 10% of all US mail, FedEx, and UPS shipments,
have a 10% chance of compromising a courier service employee,
my employee and my brother, be able to intercept 10% of all
PGP emails and TOR connections, and watch 10% of all Internet
cafes in Los Angeles 24/7 to catch me going in to one. Or maybe
all Starbucks Wi-Fi points -- I might use one of those.

And even if the attacker did achieve a 1 in 10 chance for
each key, he has a one in a million chance of getting all nine.
Far easier to simply grab me off the street and beat my
secrets out of me.

--
Guy Macon
<http://www.GuyMacon.com/>

.

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