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Message no. 1
From: Adam Getchell <acgetche@****.UCDAVIS.EDU>
Subject: Re: Quantum Cryptography
Date: Wed, 3 Aug 1994 09:24:28 -0700
The public key is sent *one* photon at a time. This means that
an eavesdropper cannot sample the message without fixing the polarization
of the photon, so that the intended recipient of the key will
automatically know that it has been tapped.
I didn't get into it, but there is a method of "privacy
amplification" wherein sender/recievers knowing that someone is sampling
their data can distill from the original transmission a shorter one that
the sampler is unlikely to have sampled even one photon.
If the sender/reciever sends/gets 800 good photons out of 1000
sent, they can use an algorithm based upon a 20% suspected sampling rate
to construct a new message which has much less than an even 1% chance.
If you want more details, try finding the article, or if there is
sufficient interest I will post a longer summary.

+-------------+---------------------------------------------------------------+
|Adam Getchell|acgetche@****.engr.ucdavis.edu | ez000270@*******.ucdavis.edu |
| acgetchell |"Invincibility is in oneself, vulnerability is in the opponent"|
+-------------+---------------------------------------------------------------+
Message no. 2
From: Adam Getchell <acgetche@****.UCDAVIS.EDU>
Subject: Re: Quantum Cryptography
Date: Wed, 3 Aug 1994 09:30:10 -0700
Gian-Paolo wrote:

> I'll bet the NSA can come up with a way to break it.

The point is not that they can't break it, but that quantum
mechanics guarantees that the original sender and reciever will know that
this is being done.
And by the way, there do exist mathematically unbreakable
encryption algorithms. It's just that the key is longer than the
original message (required) and the key can only be used once, as
multiple uses would give the mathematicians enough information to
compromise it.

+-------------+---------------------------------------------------------------+
|Adam Getchell|acgetche@****.engr.ucdavis.edu | ez000270@*******.ucdavis.edu |
| acgetchell |"Invincibility is in oneself, vulnerability is in the opponent"|
+-------------+---------------------------------------------------------------+
Message no. 3
From: Gian-Paolo Musumeci <musumeci@***.LIS.UIUC.EDU>
Subject: Re: Quantum Cryptography
Date: Wed, 3 Aug 1994 13:14:22 -0500
I bet NSA can find a way around it.
Message no. 4
From: Stainless Steel Rat <ratinox@***.NEU.EDU>
Subject: Re: Quantum Cryptography
Date: Wed, 3 Aug 1994 16:39:50 -0400
>>>>> "GP" == Gian-Paolo Musumeci
<musumeci@***.LIS.UIUC.EDU> writes:

GP> I bet NSA can find a way around it.

ANY encryption scheme can be broken, given enough time and effort. The
trick is to make cracking take so long that it's not worth while.

--
Rat <ratinox@***.neu.edu> |"Odds don't matter when Humans want
http://www.ccs.neu.edu/home/ratinox|something bad enough." --UHED series 507
PGP Public Key: Ask for one today! |
Message no. 5
From: Christopher Gottbrath <chrisg@****.EDU>
Subject: Re: Quantum Cryptography
Date: Wed, 3 Aug 1994 15:23:28 MST
Rat <ratinox@***.neu.edu> wrote:
"ANY encryption scheme can be broken, given enough time and effort."

agreed for encryption, however i doubt that without the benifit of magic
anyone is going to be "breaking" a fundamental law of Quantum Mechanics. :^)

now since magic is a factor in SR ....


chris

research student at Kitt Peak National Observatory: " the eyes of the
human race "
chrisg@****.edu
Message no. 6
From: Stainless Steel Rat <ratinox@***.NEU.EDU>
Subject: Re: Quantum Cryptography
Date: Wed, 3 Aug 1994 19:24:52 -0400
>>>>> "Christopher" == Christopher Gottbrath
<chrisg@****.EDU> writes:

"ANY encryption scheme can be broken, given enough time and effort."

Christopher> agreed for encryption, however i doubt that without
Christopher> the benifit of magic anyone is going to be "breaking" a
Christopher> fundamental law of Quantum Mechanics. :^)

Um, no. If you can decrypt it mundanely, it /can/ be broken mundanely.
Accept this very important fact about encryption: there is no such thing
as an unbreakable encryption algorithm.

What was described is something similar to OTP (one time pad), an
encryption technique that modifies the cleartext based on pseudo-random
numbers; decryption is accomplished by generating the same pseudo-random
sequence and reversing the modification. Depending on the encryption
engine, OTPs can be very resistant to standard cryptanalytical attacks. But
as I said, ANY encryption scheme can be broken, given enough time and effort.

--
Rat <ratinox@***.neu.edu> |Guns cause crime and cars cause vehicular
http://www.ccs.neu.edu/home/ratinox|homicide.
PGP Public Key: Ask for one today! |
Message no. 7
From: Luke Kendall <luke@********.CANON.OZ.AU>
Subject: Re: Quantum Cryptography
Date: Thu, 4 Aug 1994 10:08:02 +1000
Stainless Steel Rat writes:

> ANY encryption scheme can be broken, given enough time and effort.

True, but that's not being argued. There is _no_ way to intercept it
without the intended recipient getting garbage, instead of the
encrypted message. Hence, they know that the message was intercepted.

luke
Message no. 8
From: Stainless Steel Rat <ratinox@***.NEU.EDU>
Subject: Re: Quantum Cryptography
Date: Wed, 3 Aug 1994 21:17:20 -0400
>>>>> "Luke" == Luke Kendall <luke@********.CANON.OZ.AU>
writes:

Luke> True, but that's not being argued. There is _no_ way to intercept it
Luke> without the intended recipient getting garbage, instead of the
Luke> encrypted message. Hence, they know that the message was intercepted.

Never say never; nothing is impossible, just highly improbable. We'll have
a solution for you tomorow.

Another point, even assuming decryption is improbably difficult, such a
scheme wouldn't enter widespread use, simply because such a means of
communication is woefully unreliable. It would be damaging to you for me to
obtain the information you are sending--that is why you encrypt it. It
would be devestating for you if I prevent the intended recipient from ever
getting it--that is why you use a reliable transmission scheme. The purpose
of secure commo systems is to make it as difficult as possible for the
"enemy" to tamper with your lines of communication. If I can completely
hose your commo by splicing in a piece of fibre, you're system is not
secure. It's the trade off between security and usability; the ultimate in
security at the cost of an unusable system.

--
Rat <ratinox@***.neu.edu> |"One likes to believe in the freedom of
http://www.ccs.neu.edu/home/ratinox|baseball."; --Geddy Lee
PGP Public Key: Ask for one today! |
Message no. 9
From: Gian-Paolo Musumeci <musumeci@***.LIS.UIUC.EDU>
Subject: Re: Quantum Cryptography
Date: Wed, 3 Aug 1994 20:51:45 -0500
Rat> ANY encryption scheme can be broken, given enough time and effort.

...and NSA has both the time and the effort if they want to. Trust me. =)
Message no. 10
From: Gian-Paolo Musumeci <musumeci@***.LIS.UIUC.EDU>
Subject: Re: Quantum Cryptography
Date: Wed, 3 Aug 1994 20:52:29 -0500
> There is _no_ way to intercept it without the intended recipient getting

La de da de da. Don't intercept it. Divert it. Then mirror-echo off a
copy right back down the line. Next victim?
Message no. 11
From: Luke Kendall <luke@********.CANON.OZ.AU>
Subject: Re: Quantum Cryptography
Date: Thu, 4 Aug 1994 12:33:04 +1000
Gian-Paolo Musumeci:

> > There is _no_ way to intercept it without the intended recipient getting
>

GPM> La de da de da. Don't intercept it. Divert it. Then mirror-echo off a
GPM> copy right back down the line. Next victim?

Rat> Never say never; nothing is impossible, just highly improbable. We'll have
Rat> a solution for you tomorow.

We're not talking about decryption, we're talking about _fundamental_
laws of physics.

Look, I'm not trying to be insulting, but it really seems to me that you
(and Rat) simply don't know what you're talking about - for once. :-)

This is a quantum effect. The message can only be observed once, ever,
and then it's randomised. (You're disputing the Uncertainty Principle!!!)

There's even good arguments for saying that magic couldn't get around this
one, for god's sake!

And I can imagine a communication strategy, where knowing that a message
was intercepted was of more worth than actually getting the message.

luke
Message no. 12
From: Janne Jalkanen <jalkanen@*********.CERN.CH>
Subject: Re: Quantum Cryptography
Date: Thu, 4 Aug 1994 09:30:11 +0200
On Wed, 3 Aug 1994, Gian-Paolo Musumeci wrote:

> > There is _no_ way to intercept it without the intended recipient getting
>
> La de da de da. Don't intercept it. Divert it. Then mirror-echo off a
> copy right back down the line. Next victim?

Since the speed of light is constant, the recipient will notice a delay
in the photons, and will thus know that the message has been intercepted.
(I played also with the theorem that you could measure the quanta and then
send exact copies down the line. Unfortunately it does not work that way ;)

Also, again - as Adam correctly stated - if your encryption key is longer
than your message, all you really can do anymore is trial and error -
analysis to the message.

Janne Jalkanen ///! For those who have to fight for it
jalkanen@******.cern.ch /// ! life has a flavor
Janne.Jalkanen@***.fi \\\/// ! the protected will never understand
-'Keep on going...' \XX/ ! (anonymous, Viet Nam, 1968)
Message no. 13
From: Neil Smith <NSMITH@***.AC.UK>
Subject: Re: Quantum Cryptography
Date: Thu, 4 Aug 1994 09:30:00 GMT
I get the feeling that quantum cryptography wouldn't work over the
matrix anyway; there are all sorts of signal switchers, boosters and
telephone exchanges that would be liable to mess up the
polarisations. You'd need a dedicated monofilament.

Neil.

(Who knows next to nothing about opto-electronics)
Message no. 14
From: Damion Milliken <u9467882@******.UOW.EDU.AU>
Subject: Re: Quantum Cryptography
Date: Thu, 4 Aug 1994 18:37:20 +0000
Janne writes:

> Since the speed of light is constant, the recipient will notice a delay
> in the photons, and will thus know that the message has been intercepted.
> (I played also with the theorem that you could measure the quanta and then
> send exact copies down the line. Unfortunately it does not work that way ;)

Don't optical fibre cables need relays to boost or re-send the signals every
couple of hundred meters? This would make the delay impossible to predict
unless you knew the spacing and operating perameters of the relays. BTW, would
the relays not stuff up the quanta the same way intercepting and re-sending the
messgae would?

> Also, again - as Adam correctly stated - if your encryption key is longer
> than your message, all you really can do anymore is trial and error -
> analysis to the message.

Not according to Rat :-)

--
Damion Milliken University of Wollongong E-Mail: u9467882@******.uow.edu.au

(Geek Code 2.1) GE d@ H s++:-- !g p? !au a18 w+ v C+ U P? !L !3 E? N K- W+ M
!V po@ Y t(+) !5 !j r+(++) G(+) !tv(--) b++ D+ B? e+ u@ h+(*)
f+@ !r n--(----)@ !y+
Message no. 15
From: Janne Jalkanen <jalkanen@*********.CERN.CH>
Subject: Re: Quantum Cryptography
Date: Thu, 4 Aug 1994 11:05:57 +0200
On Thu, 4 Aug 1994, Damion Milliken wrote:

> Don't optical fibre cables need relays to boost or re-send the signals every
> couple of hundred meters? This would make the delay impossible to predict
> unless you knew the spacing and operating perameters of the relays. BTW, would

I'm not in my own domain here, but I think reading that the *current*
optical fibre cables need boosting every few kilometers, not hundred
meters. Also, when using satellite-to-satellite communication, optical
means could be _very_ effective, especially when coded with this sort of
a system. And no relay's needed.

I think that the quanta coding is very effective, except it is still too
cumbersome to be used as a reliable means of communication. Maybe in 20
years.

> the relays not stuff up the quanta the same way intercepting and re-sending
the
> messgae would?

Probably. Which again restricts the usability of this system.

> > Also, again - as Adam correctly stated - if your encryption key is longer
> > than your message, all you really can do anymore is trial and error -
> > analysis to the message.
>
> Not according to Rat :-)

I really, really would like to know if there are any better methods in
this. Not just 'yes there are, trust me' - type messages. If anyone has
an algorithm or an idea, I'd like to hear it. However, I think that
according to US law, discussion about encoding/decoding algorithms with
foreigners is a felony, I doubt that I won't get too many answers... :(

Janne Jalkanen ///! For those who have to fight for it
jalkanen@******.cern.ch /// ! life has a flavor
Janne.Jalkanen@***.fi \\\/// ! the protected will never understand
-'Keep on going...' \XX/ ! (anonymous, Viet Nam, 1968)
Message no. 16
From: Alexander Borghgraef <Alexander.Borghgraef@***.AC.BE>
Subject: Re: Quantum Cryptography
Date: Thu, 4 Aug 1994 11:17:59 --100
>Since the speed of light is constant, the recipient will notice a delay
>in the photons, and will thus know that the message has been intercepted.

Oh dear, that transmission is 2 microseconds late,I suppose the line is tapped!
Let me ask you a question:do you know exactly how longthe telephone line is to
every person you call?
Message no. 17
From: Janne Jalkanen <jalkanen@*********.CERN.CH>
Subject: Re: Quantum Cryptography
Date: Thu, 4 Aug 1994 11:54:35 +0200
On Thu, 4 Aug 1994, Alexander Borghgraef wrote:

> >Since the speed of light is constant, the recipient will notice a delay
> >in the photons, and will thus know that the message has been intercepted.
>
> Oh dear, that transmission is 2 microseconds late,I suppose the line is
tapped!
> Let me ask you a question:do you know exactly how longthe telephone line is to
> every person you call?

That is not the issue. Quanta coding is too cumbersome to be utilized in
standard telephone conversations, but I would imagine for instance the
military/governments would use it for extremely high security crypting.
And I am pretty sure THEY know the length of their lines ;-)

Also, you can break down your transmission into 800 different lines and
if they arrive asynchronously, you know someone is wiretapping. (Better
yet, put them in different lines altogether and reroute around the
world.) Breaking down your message / adding deliberate noise is an old,
but effective method. For instance, you can send your recipient a 200K
message, out of which only ~1K is true data and the rest is just garbage.
The recipient has an algorithm that will tell which of the data is real
and which is fake. If the algorithm is well designed (for instance, true
random data) all you can really do is to go and steal the decryption key.
Which, in turn makes shadowrunners necessary and thus links nicely back
to the SR world. ;-)


Janne Jalkanen ///! For those who have to fight for it
jalkanen@******.cern.ch /// ! life has a flavor
Janne.Jalkanen@***.fi \\\/// ! the protected will never understand
-'Keep on going...' \XX/ ! (anonymous, Viet Nam, 1968)
Message no. 18
From: Damion Milliken <u9467882@******.UOW.EDU.AU>
Subject: Re: Quantum Cryptography
Date: Thu, 4 Aug 1994 20:04:25 +0000
Janne writes:

> and which is fake. If the algorithm is well designed (for instance, true
> random data) all you can really do is to go and steal the decryption key.

I thought there was no such thing as "true randomness" when it came to
computers. I would still vouch for Rats idea with the "extra garbage" method
though. If you were willing to spend enough time you could crack it. Also if
you got hold of multiple different transmisions then you might have it a bit
easier. But I know nothing about data encryption or cryptology (sp?) so I'm
probably completely wrong here.

--
Damion Milliken University of Wollongong E-Mail: u9467882@******.uow.edu.au

(Geek Code 2.1) GE d@ H s++:-- !g p? !au a18 w+ v C+ U P? !L !3 E? N K- W+ M
!V po@ Y t(+) !5 !j r+(++) G(+) !tv(--) b++ D+ B? e+ u@ h+(*)
f+@ !r n--(----)@ !y+
Message no. 19
From: Janne Jalkanen <jalkanen@*********.CERN.CH>
Subject: Re: Quantum Cryptography
Date: Thu, 4 Aug 1994 12:36:12 +0200
On Thu, 4 Aug 1994, Damion Milliken wrote:

> I thought there was no such thing as "true randomness" when it came to
> computers. I would still vouch for Rats idea with the "extra garbage"
method
> though. If you were willing to spend enough time you could crack it. Also if
> you got hold of multiple different transmisions then you might have it a bit
> easier. But I know nothing about data encryption or cryptology (sp?) so I'm

When using an algorithmic method of creating random numbers, all you will
get are so-called 'pseudo'-random numbers, which are not truly random.
There are some pretty good algorithms that will give a good distribution,
but what I was talking about was true random data, for instance from
observing radioactive decaying... You put a CDROM full of that random
data and send it by a courier to your recipient, then you will have 600
Megabytes of secret passcode to play with ;)

The trouble when you add random data is that you can construct also false
messages from it. I'll demonstrate this with a simple example:

Our pass key is very simple, it is using the 9 first prime numbers
(1,2,3,5,7,11,13,17,19) and the coded message is:

shagdaorygwyrgaxuandnd

Now, if you get the wrong algorithm, you can get for instance
'garygygaxadnd' instead of 'shadowrun' - which is of course the right
message. Or you might easily get 'shawn' or 'go wyrd' or 'do as you are'
or whatever. As a matter of fact, there's nothing to stop you from
reusing the same letters over and over again, you can traverse the
sentence backwards and forwards, you can scramble it so that it resembles
random data... There are ways to produce encryption schemes that are
*extremely* hard to break. If 99% of your message is garbage, (and you
even iterate the method a couple of times) you'll get an extremely
bloated message with little or no chance of decrypting without the proper
key. And shadowrunners are called again. Naturally, the backside of this
is that your messages are gonna be big, and since they consist of random
data, they cannot even be compressed.

Also, the good, old book-techique (which is just an variation of this) is
also extremely hard to break.

Incidentally, both methods were already invented several hundred years
ago, but they are not very widely used mainly due to the increased
bandwidth... I could imagine that as these are the techniques (or
variations thereof) used on the most secure comm lines, since they can
afford it ;)

Yes, it is true that the more messages you have, the easier the crypting
is to crack. By using a different key each time, you will maximise your
security.

Ah yes, and there's even a more ultimate crypting technique (for short
messages) which was invented a long time ago (cannot remember who)... The
idea of it is similar, except that you use NO algorithm WHATSOEVER to
pick up the coding key. And you code phrases, not letters. Increases the
time to break considerably.

And BTW, if you Americans don't get this message, you know that NSA has
been listening ;-)

Janne Jalkanen ///! For those who have to fight for it
jalkanen@******.cern.ch /// ! life has a flavor
Janne.Jalkanen@***.fi \\\/// ! the protected will never understand
-'Keep on going...' \XX/ ! (anonymous, Viet Nam, 1968)
Message no. 20
From: Stainless Steel Rat <ratinox@***.NEU.EDU>
Subject: Re: Quantum Cryptography
Date: Thu, 4 Aug 1994 09:14:49 -0400
>>>>> "Janne" == Janne Jalkanen
<jalkanen@*********.cern.ch> writes:

[Lots of good info]

Janne> Ah yes, and there's even a more ultimate crypting technique (for
Janne> short messages) which was invented a long time ago (cannot remember
Janne> who)... The idea of it is similar, except that you use NO algorithm
Janne> WHATSOEVER to pick up the coding key. And you code phrases, not
Janne> letters. Increases the time to break considerably.

That's technically not encryption but encoding with a cipher. The
techniques for breaking a cypher are simiiar to breaking encryption, but
there are differences as well. Ciphers tend to be resistant to brute force
attacks, but are vulnerable to conventional cryptanalysis.

The technique you describe comes under the "security by obscurity" axiom,
hiding your secrets in plain sight.

--
Rat <ratinox@***.neu.edu> |"Wot we gonna do tonight, Brain?"
"The same
http://www.ccs.neu.edu/home/ratinox|thing we do every night, Pinky, try to take
PGP Public Key: Ask for one today! |over the world!"
Message no. 21
From: Gian-Paolo Musumeci <musumeci@***.LIS.UIUC.EDU>
Subject: Re: Quantum Cryptography
Date: Thu, 4 Aug 1994 08:30:42 -0500
It is a quantum effect, yes, and keep in mind that quantum physi/mechanics
is still a highly theoetical field. Also, this des not stop my method of
intercept and copyback.
Message no. 22
From: Hamish Laws <h_laws@**********.UTAS.EDU.AU>
Subject: Re: Quantum Cryptography
Date: Fri, 5 Aug 1994 08:50:08 +1000
Janne Jalkanen writes

SNIP

>Since the speed of light is constant, the recipient will notice a delay
>in the photons, and will thus know that the message has been intercepted.
>(I played also with the theorem that you could measure the quanta and then
>send exact copies down the line. Unfortunately it does not work that way ;)
>
Only if he knows exactly when the message was sent as the copying
process would presumably be extremely rapid.


*************************************************
There has to be an optimist around here somewhere
*************************************************

Hamish Laws
Message no. 23
From: Adam Getchell <acgetche@****.UCDAVIS.EDU>
Subject: Re: Quantum Cryptography
Date: Thu, 4 Aug 1994 15:56:08 -0700
On Wed, 3 Aug 1994, Stainless Steel Rat wrote:

> Um, no. If you can decrypt it mundanely, it /can/ be broken mundanely.
> Accept this very important fact about encryption: there is no such thing
> as an unbreakable encryption algorithm.

My reading of the literature, in particular the Quantum
Cryptography article on page 50 of the Oct. 1992 Scientific American
seems to disagree with the above.
They specifically talk about mathematical constructs which are
impossible to crack without repeated exposure; specifically, as long as
the key is longer than the message decryption is impossible. This is
guaranteed by using the code only once, hence the one-time pad use. It
is of note that computers can churn these codes out relatively rapidly;
the disadvantage in this approach lies only in its awkwardness. That is,
James Bond must have a separate code/key for each message he plans to
send/recieve.
The quantum cryptography method uses public-key encryption, and
the notable thing about the technique is not that it cannot be broken,
but that the users will know about the attempt.
After all, in order to break the code you must either 1)Intercept
code samples or 2)Steal the code key. QC ensures that 1) is impossible
without the sender/reciever knowing about it. There is also a technique,
using QM, that makes 2) impossible without knowledge of the attempt, but
it is currently impractical (it would require storing photons for long
periods of time).
As another interesting note, I believe a certain mathematician
has come up with a theory that calls into question the randomness of the
numbers we generate; it seems that almost all of them have a particular
type of underlying pattern that repeats quite rapidly, far sooner than we
would desire.

> Rat <ratinox@***.neu.edu> |Guns cause crime and cars cause vehicular
> http://www.ccs.neu.edu/home/ratinox|homicide.
> PGP Public Key: Ask for one today! |
>

+-------------+---------------------------------------------------------------+
|Adam Getchell|acgetche@****.engr.ucdavis.edu | ez000270@*******.ucdavis.edu |
| acgetchell |"Invincibility is in oneself, vulnerability is in the opponent"|
+-------------+---------------------------------------------------------------+
Message no. 24
From: Adam Getchell <acgetche@****.UCDAVIS.EDU>
Subject: Re: Quantum Cryptography
Date: Thu, 4 Aug 1994 16:02:52 -0700
On Wed, 3 Aug 1994, Stainless Steel Rat wrote:

> Never say never; nothing is impossible, just highly improbable. We'll have
> a solution for you tomorow.

As long as our current understanding of QM is unflawed (and we
have no reason to think so since Quantum-Chromodynamics by R. Feynman et.
al has held up well) it _IS_ impossible for observation of an event to
not alter the event.
Just like it's impossible for you to jump up, flap your arms, and
fly (without mechanical contrivances, and assuming you're on the Earth).
Notice, though, that physics _ALWAYS_ works under a set of
conditions. It's just that the conditions for QM to apply happens to be
the entire range of the current Universe.

> Rat <ratinox@***.neu.edu> |"One likes to believe in the freedom
of

+-------------+---------------------------------------------------------------+
|Adam Getchell|acgetche@****.engr.ucdavis.edu | ez000270@*******.ucdavis.edu |
| acgetchell |"Invincibility is in oneself, vulnerability is in the opponent"|
+-------------+---------------------------------------------------------------+
Message no. 25
From: Adam Getchell <acgetche@****.UCDAVIS.EDU>
Subject: Re: Quantum Cryptography
Date: Thu, 4 Aug 1994 16:05:24 -0700
On Wed, 3 Aug 1994, Gian-Paolo Musumeci wrote:

> La de da de da. Don't intercept it. Divert it. Then mirror-echo off a
> copy right back down the line. Next victim?

You didn't get the point of there only being *one* photon. You
*cannot* divert it and mirror-echo it, because you don't know the
information content or polarization of the photon, and can't without
examining it in which case you change it....
Got it?

+-------------+---------------------------------------------------------------+
|Adam Getchell|acgetche@****.engr.ucdavis.edu | ez000270@*******.ucdavis.edu |
| acgetchell |"Invincibility is in oneself, vulnerability is in the opponent"|
+-------------+---------------------------------------------------------------+
Message no. 26
From: Adam Getchell <acgetche@****.UCDAVIS.EDU>
Subject: Re: Quantum Cryptography
Date: Thu, 4 Aug 1994 16:20:36 -0700
You can send signals in packets called phonons (they occur
naturally in crystals) which cancels the tendencies for wave packets to
spread out and randomize and average their frequencies (i.e. dissipate).
That way, losses only come from media.
And don't they have 1-way matrix lines already? This might be why.

+-------------+---------------------------------------------------------------+
|Adam Getchell|acgetche@****.engr.ucdavis.edu | ez000270@*******.ucdavis.edu |
| acgetchell |"Invincibility is in oneself, vulnerability is in the opponent"|
+-------------+---------------------------------------------------------------+
Message no. 27
From: Gian-Paolo Musumeci <musumeci@***.LIS.UIUC.EDU>
Subject: Re: Quantum Cryptography
Date: Thu, 4 Aug 1994 19:44:27 -0500
Uh, wrong.Divert the photon; I mean splice your recorder thingy into the
fiberop. Then have your recordr thingy observe the photon. Ooh, look, we
know the goodies on the photon. Send out an IDENTICAL photon down the line, so
the reciever gets it just fine. Now you still have a copy of the first photon,
so you can decrypt it at your leisure.
Got it?
Message no. 28
From: Gian-Paolo Musumeci <musumeci@***.LIS.UIUC.EDU>
Subject: Re: Quantum Cryptography
Date: Thu, 4 Aug 1994 20:54:21 -0500
Ew, I just read my own message. 'sorry about the typos, my phone line seems to
be disintegrating.
Message no. 29
From: Luke Kendall <luke@********.CANON.OZ.AU>
Subject: Re: Quantum Cryptography
Date: Fri, 5 Aug 1994 12:21:02 +1000
Gian-Paolo Musumeci writes:

> Uh, wrong.Divert the photon; I mean splice your recorder thingy into the
> fiberop. Then have your recordr thingy observe the photon. Ooh, look, we
> know the goodies on the photon. Send out an IDENTICAL photon down the line,
so
> the reciever gets it just fine. Now you still have a copy of the first
photon,
> so you can decrypt it at your leisure.
> Got it?

You really don't know the theory behind this, do you? Each photon in question
is one of a pair; if you `touch' one, then the other one changes (its
quantum state changes from indeterminate to determinate), and this is an
irreversible change. So you can't really produce an identical photon.

Got it?

luke
Message no. 30
From: "J.D. Falk" <jdfalk@****.CAIS.COM>
Subject: Re: Quantum Cryptography
Date: Thu, 4 Aug 1994 23:42:57 -0400
On Thu, 4 Aug 1994, Alexander Borghgraef wrote:

> >Since the speed of light is constant, the recipient will notice a delay
> >in the photons, and will thus know that the message has been intercepted.
>
> Oh dear, that transmission is 2 microseconds late,I suppose the line is tapped!
> Let me ask you a question:do you know exactly how longthe telephone line is to
> every person you call?

Alex, this idea would be totally useless except in special
situations, for that exact reason. However, it would make sense for
corps to use it between Corp holdings, because they have the resources to
find out how long the lines are.
Message no. 31
From: Hamish Laws <h_laws@**********.UTAS.EDU.AU>
Subject: Re: Quantum Cryptography
Date: Fri, 5 Aug 1994 14:39:26 +1000
luke Kendell writes
>Gian-Paolo Musumeci writes:
>
>You really don't know the theory behind this, do you? Each photon in question
>is one of a pair; if you `touch' one, then the other one changes (its
>quantum state changes from indeterminate to determinate), and this is an
>irreversible change. So you can't really produce an identical photon.
>
>Got it?
>
Um, no. Seeing as to detect it the receiver must also change its
state from indeterminate to determinate what's the hassle with rebeaming a
proton like the photon you have intercepted?
I don't know much about quantum physics but if the state is changed
by one person 'reading' it surely it must be changed by anyone else reading
it so that the information must be there in the determinate state.


*************************************************
There has to be an optimist around here somewhere
*************************************************

Hamish Laws
Message no. 32
From: Hamish Laws <h_laws@**********.UTAS.EDU.AU>
Subject: Re: Quantum Cryptography
Date: Fri, 5 Aug 1994 14:41:41 +1000
J.D.Falk wrote
>On Thu, 4 Aug 1994, Alexander Borghgraef wrote:
>
SNIP
>
> Alex, this idea would be totally useless except in special
>situations, for that exact reason. However, it would make sense for
>corps to use it between Corp holdings, because they have the resources to
>find out how long the lines are.

Um, yes but how accurate is the timing system being used, and how
would they go about checking back, for every photon?


*************************************************
There has to be an optimist around here somewhere
*************************************************

Hamish Laws
Message no. 33
From: Adam Getchell <acgetche@****.UCDAVIS.EDU>
Subject: Re: Quantum Cryptography
Date: Fri, 5 Aug 1994 00:27:05 -0700
On Thu, 4 Aug 1994, Gian-Paolo Musumeci wrote:

> Uh, wrong.Divert the photon; I mean splice your recorder thingy into the
> fiberop. Then have your recordr thingy observe the photon. Ooh, look, we
> know the goodies on the photon. Send out an IDENTICAL photon down the line, so

Uh, wrong. You are not understanding the quantum mechanics
behind the scheme. I will repost a short section of the article covering
this very topic; Eve is the eavesdropper, Alice the sender and Bob the
reciever:
"Because of the uncertainty principle, Eve cannot measure both
rectilinear and diagonal polarizations of the same photon. If, for a
particular photon, she makes the wrong measurement, then, even if she
resends Bob a photon consistent with the result of her measurement, she
will have irretrievable randomized the polarization originally sent by
Alice. The net effect is to cause errors in one quarter of the bits in
Bob's data that have been subjected to eavesdropping."
This is because there are _four_ polarizations, 0, 45, 90 and 135.
Got it?

+-------------+---------------------------------------------------------------+
|Adam Getchell|acgetche@****.engr.ucdavis.edu | ez000270@*******.ucdavis.edu |
| acgetchell |"Invincibility is in oneself, vulnerability is in the opponent"|
+-------------+---------------------------------------------------------------+
Message no. 34
From: Adam Getchell <acgetche@****.UCDAVIS.EDU>
Subject: Re: Quantum Cryptography
Date: Fri, 5 Aug 1994 00:35:28 -0700
On Fri, 5 Aug 1994, Luke Kendall wrote:

> You really don't know the theory behind this, do you? Each photon in question
> is one of a pair; if you `touch' one, then the other one changes (its
> quantum state changes from indeterminate to determinate), and this is an
> irreversible change. So you can't really produce an identical photon.

This is absolutely correct. This is how to deal with the second
of the two problems in cryptography, interception of data and theft of
the encryption key.
By storing the encryption key as a pattern of photon pairs,
stealing the key automatically randomizes half of it, and the key owners
will automatically know that it's compromised.
This is the part that's hard to do with modern technology,
because it involves long term storage of photons.

> luke

+-------------+---------------------------------------------------------------+
|Adam Getchell|acgetche@****.engr.ucdavis.edu | ez000270@*******.ucdavis.edu |
| acgetchell |"Invincibility is in oneself, vulnerability is in the opponent"|
+-------------+---------------------------------------------------------------+
Message no. 35
From: Adam Getchell <acgetche@****.UCDAVIS.EDU>
Subject: Re: Quantum Cryptography
Date: Fri, 5 Aug 1994 00:44:40 -0700
On Thu, 4 Aug 1994, J.D. Falk wrote:
>
> Alex, this idea would be totally useless except in special
> situations, for that exact reason. However, it would make sense for
> corps to use it between Corp holdings, because they have the resources to
> find out how long the lines are.

It is also totally unnecessary for reasons heretofore explained;
regardless of whether or not the eavesdroppers can successfully decrypt
the data, it is impossible for them to send exactly the same message that
they recieved. The BB84 scheme described would produce errors in 1/4 of
the encrypted, intercepted bits.
Also, the BB84 scheme described in the article uses the Vernam
cipher, which has a key exactly as long as the message and was proven by
Claude E. Shannon in the 40's to be able to completely and
unconditionally immune to decryption, as long as the key is random and
used only once.
Janne suggested an easy way to get good random data, and BB84
scheme provides an easy way to continually change keys.
I should note for reference that what Shannon actually proved was
that any encryption key shorter than its message automatically "leaks"
information about what is being encrypted; he then went on to prove that
the inverse was also true, namely that any key equally, or longer than
its message is unbreakable under the necessary conditions of randomness
and one-time use.

+-------------+---------------------------------------------------------------+
|Adam Getchell|acgetche@****.engr.ucdavis.edu | ez000270@*******.ucdavis.edu |
| acgetchell |"Invincibility is in oneself, vulnerability is in the opponent"|
+-------------+---------------------------------------------------------------+
Message no. 36
From: Adam Getchell <acgetche@****.UCDAVIS.EDU>
Subject: Re: Quantum Cryptography
Date: Fri, 5 Aug 1994 00:48:20 -0700
On Fri, 5 Aug 1994, Hamish Laws wrote:

> I don't know much about quantum physics but if the state is changed
> by one person 'reading' it surely it must be changed by anyone else reading
> it so that the information must be there in the determinate state.

Nope. Not true.
It's okay. Quantum mechanics is tricky stuff, and the nastiest
physics class I ever took (well, okay Statistical Mechanics was a close
second).

> Hamish Laws

+-------------+---------------------------------------------------------------+
|Adam Getchell|acgetche@****.engr.ucdavis.edu | ez000270@*******.ucdavis.edu |
| acgetchell |"Invincibility is in oneself, vulnerability is in the opponent"|
+-------------+---------------------------------------------------------------+
Message no. 37
From: Alexander Borghgraef <Alexander.Borghgraef@***.AC.BE>
Subject: Re: Quantum Cryptography
Date: Fri, 5 Aug 1994 11:26:41 --100
>You really don't know the theory behind this, do you? Each photon in question
>is one of a pair; if you `touch' one, then the other one changes (its
>quantum state changes from indeterminate to determinate), and this is an
>irreversible change. So you can't really produce an identical photon.

Great, the EPR experiment!But how does the sender know what kind of photon
he has sent?If he knows the photon is determined.
Message no. 38
From: Alexander Borghgraef <Alexander.Borghgraef@***.AC.BE>
Subject: Re: Quantum Cryptography
Date: Fri, 5 Aug 1994 11:30:05 --100
> Alex, this idea would be totally useless except in special
>situations, for that exact reason. However, it would make sense for
>corps to use it between Corp holdings, because they have the resources to
>find out how long the lines are.

How would the receiver know exactly when the photon has been sent?There must be an error
on that.
Message no. 39
From: Damion Milliken <u9467882@******.UOW.EDU.AU>
Subject: Re: Quantum Cryptography
Date: Fri, 5 Aug 1994 19:38:13 +0000
Adam writes:

> They specifically talk about mathematical constructs which are
> impossible to crack without repeated exposure; specifically, as long as
> the key is longer than the message decryption is impossible. This is
> guaranteed by using the code only once, hence the one-time pad use. It
> is of note that computers can churn these codes out relatively rapidly;
> the disadvantage in this approach lies only in its awkwardness. That is,
> James Bond must have a separate code/key for each message he plans to
> send/recieve.

I assume both partys have a "book" or whatever, and they publicly broadcast
what code/key to use?

> As another interesting note, I believe a certain mathematician
> has come up with a theory that calls into question the randomness of the
> numbers we generate; it seems that almost all of them have a particular
> type of underlying pattern that repeats quite rapidly, far sooner than we
> would desire.

No doubt we do. But the numbers we come up with could only be predicted by
careful and lengthy observation. If you were using a computer algorithm, then
inspecting the algorithm may help somewhat.

luke writes:

> You really don't know the theory behind this, do you? Each photon in question
> is one of a pair; if you `touch' one, then the other one changes (its
> quantum state changes from indeterminate to determinate), and this is an
> irreversible change. So you can't really produce an identical photon.

I will admit I do not understand the principles, but it seems this way to me:
A person can produce a photon (pair if thats the way the come) the way they
want it, to represent a particular piece of data right? Likewise a person can
read these photons to get the message out of them, altering them when they do
this. Then why can't the person who just read the photons send out an
identical message to the one they just received? The person who sent it did.

> Got it?

Nope :-)

Adam writes:

> Uh, wrong. You are not understanding the quantum mechanics
> behind the scheme. I will repost a short section of the article covering
> this very topic; Eve is the eavesdropper, Alice the sender and Bob the
> reciever:
> "Because of the uncertainty principle, Eve cannot measure both
> rectilinear and diagonal polarizations of the same photon. If, for a
> particular photon, she makes the wrong measurement, then, even if she
> resends Bob a photon consistent with the result of her measurement, she
> will have irretrievable randomized the polarization originally sent by
> Alice. The net effect is to cause errors in one quarter of the bits in
> Bob's data that have been subjected to eavesdropping."

What if she doesn't make a wrong measurement?

> Got it?

Again, nope. Must be thick eh? :-)

> This is absolutely correct. This is how to deal with the second
> of the two problems in cryptography, interception of data and theft of
> the encryption key.
> By storing the encryption key as a pattern of photon pairs,
> stealing the key automatically randomizes half of it, and the key owners
> will automatically know that it's compromised.

Well, I assume somebody put the photons there in the fasion that they wanted to
in the first place, so whats to stop me coming along, reading the encryption
key, and then putting a bunch of photons back in there just the same way the
person who originally put them there did?


> It is also totally unnecessary for reasons heretofore explained;
> regardless of whether or not the eavesdroppers can successfully decrypt
> the data, it is impossible for them to send exactly the same message that
> they recieved. The BB84 scheme described would produce errors in 1/4 of
> the encrypted, intercepted bits.

Why is it impossible to send the same message that you received? And what is
the BB84 scheme?

> Also, the BB84 scheme described in the article uses the Vernam
> cipher, which has a key exactly as long as the message and was proven by
> Claude E. Shannon in the 40's to be able to completely and
> unconditionally immune to decryption, as long as the key is random and
> used only once.

If this scheme is so good, then why do we need anything like the photons? All
they really do is allow us to tell someone has tapped our line, with this
super scheme, who would give two hoots if someone tapped your line? If its
totally uncrackable, it wouldn't matter.

--
Damion Milliken University of Wollongong E-Mail: u9467882@******.uow.edu.au

(Geek Code 2.1) GE d@ H s++:-- !g p? !au a18 w+ v C+ U P? !L !3 E? N K- W+ M
!V po@ Y t(+) !5 !j r+(++) G(+) !tv(--) b++ D+ B? e+ u@ h+(*)
f+@ !r n--(----)@ !y+
Message no. 40
From: Janne Jalkanen <jalkanen@*********.CERN.CH>
Subject: Re: Quantum Cryptography
Date: Fri, 5 Aug 1994 11:40:48 +0200
On Fri, 5 Aug 1994, Alexander Borghgraef wrote:

> How would the receiver know exactly when the photon has been sent?There must
be an error on that.

By using high-precision clocks you can determine the time to nanoseconds
and faster. All you really have to do is to agree on a protocol (like
photons are sent on at every 1/100,000 seconds or something like that).
And in the beginning of your message you can tag the exact time of
sending the first photon.

Janne Jalkanen ///! For those who have to fight for it
jalkanen@******.cern.ch /// ! life has a flavor
Janne.Jalkanen@***.fi \\\/// ! the protected will never understand
-'Keep on going...' \XX/ ! (anonymous, Viet Nam, 1968)
Message no. 41
From: Damion Milliken <u9467882@******.UOW.EDU.AU>
Subject: Re: Quantum Cryptography
Date: Fri, 5 Aug 1994 20:10:51 +0000
Janne writes:

> > How would the receiver know exactly when the photon has been sent?There must
> be an error on that.
>
> By using high-precision clocks you can determine the time to nanoseconds
> and faster. All you really have to do is to agree on a protocol (like
> photons are sent on at every 1/100,000 seconds or something like that).
> And in the beginning of your message you can tag the exact time of
> sending the first photon.

Could you not intercept the message and re-send it with a modified tag time?

--
Damion Milliken University of Wollongong E-Mail: u9467882@******.uow.edu.au

(Geek Code 2.1) GE d@ H s++:-- !g p? !au a18 w+ v C+ U P? !L !3 E? N K- W+ M
!V po@ Y t(+) !5 !j r+(++) G(+) !tv(--) b++ D+ B? e+ u@ h+(*)
f+@ !r n--(----)@ !y+
Message no. 42
From: Janne Jalkanen <jalkanen@*********.CERN.CH>
Subject: Re: Quantum Cryptography
Date: Fri, 5 Aug 1994 12:14:15 +0200
On Fri, 5 Aug 1994, Damion Milliken wrote:

> this. Then why can't the person who just read the photons send out an
> identical message to the one they just received? The person who sent it did.

Because he should know what sort of measurements he should do. That's the
passkey, knowing the measurements. (If I understood this correctly)

> What if she doesn't make a wrong measurement?

If there are two possibilities (measure either horizontal/vertical
polarization), the chances are 50 % that she can guess correctly. But,
since the message is probably longer than 1 bit, the chances of getting
them all correctly go down the drain. I believe that the chances are
(.5)^N that she'll guess it correctly. For instance, if you have a
message of 64 bits, the probability of making correct measurements is
.5^64 = 1 against 18,446,744,073,709,551,616. If your message is 10,000
bits long...

> Well, I assume somebody put the photons there in the fasion that they wanted
to
> in the first place, so whats to stop me coming along, reading the encryption
> key, and then putting a bunch of photons back in there just the same way the
> person who originally put them there did?

Umm... Those 400 armed security guards? ;)

Actually, you'd again need to know the correct measurements. Which are
probably locked away for good somewhere.

The idea behind this system is the same than in all encryption: put your
message into two bits, so that the other one is completely useless
without the other (passkey/message). However, this system guarantees that
if you try to eavesdrop the message, the recipient will get to know of
it, immediately, which is something current encryption messages don't do.

> Why is it impossible to send the same message that you received? And what is
> the BB84 scheme?

Because 1/4 of the data you resend would be utter garbage. You see,
you have 4 polarizations. You can observe them in two different ways,
we'll call the polarizations A(0), B(45), C(90) and D(135). Vertical
polarization will yield an 1 on A, B and D, and 0 on C, whereas
horizontal polarization will yield an 1 on B, C and D (and 0 on A).

Thus, if you observe a 1 with H polarization, you cannot really tell
which one it was. If you were supposed to measure it in H polarization,
it could be B C or D, but if you were supposed to measure it in V
polarization, it could be A B or D. And no, you can't measure it both
ways (this is forbidden by the uncertainty principle).

And if it is A, it is very possible that it in reality should've been a
0, instead of a 1 (or vice versa). So, he really cannot tell which
polarization to send down the line again... You see, if he sees an one,
he'll have different choices depending on his measurement. With a zero,
there's no problem. If he sends down the wrong polarization, the
recipient will know of it immediately (or right after checking the
CRC/whatever they must be using).

So, the poor wiretapper is confused. Now, how does the recipient know
what to do? He knows that this photon should be measured in a H fashion,
so he knows that it is supposed to be one and writes it down as an one.

> If this scheme is so good, then why do we need anything like the photons? All
> they really do is allow us to tell someone has tapped our line, with this
> super scheme, who would give two hoots if someone tapped your line? If its
> totally uncrackable, it wouldn't matter.

You might want to use another method of encryption, because Vernam key is
a one-time pad. Also, that'd also mean that now it is time to dispatch
troopers to kill the wiretappers, who have penetrated the complex ;)

Janne Jalkanen ///! For those who have to fight for it
jalkanen@******.cern.ch /// ! life has a flavor
Janne.Jalkanen@***.fi \\\/// ! the protected will never understand
-'Keep on going...' \XX/ ! (anonymous, Viet Nam, 1968)
Message no. 43
From: Janne Jalkanen <jalkanen@*********.CERN.CH>
Subject: Re: Quantum Cryptography
Date: Fri, 5 Aug 1994 12:15:52 +0200
On Fri, 5 Aug 1994, Damion Milliken wrote:

> Could you not intercept the message and re-send it with a modified tag time?

Yes, if you can decode/recode it immediately. For observing two-hour
delays you don't really need a clock anymore...

Janne Jalkanen ///! For those who have to fight for it
jalkanen@******.cern.ch /// ! life has a flavor
Janne.Jalkanen@***.fi \\\/// ! the protected will never understand
-'Keep on going...' \XX/ ! (anonymous, Viet Nam, 1968)
Message no. 44
From: Damion Milliken <u9467882@******.UOW.EDU.AU>
Subject: Re: Quantum Cryptography
Date: Fri, 5 Aug 1994 21:01:30 +0000
Janne writes:

> > Well, I assume somebody put the photons there in the fasion that they wanted
> to
> > in the first place, so whats to stop me coming along, reading the encryption
> > key, and then putting a bunch of photons back in there just the same way the
> > person who originally put them there did?
>
> Umm... Those 400 armed security guards? ;)

Same as any crypto key really.

> Actually, you'd again need to know the correct measurements. Which are
> probably locked away for good somewhere.

Someone must know them, otherwise the messages could not be read. In fact
anyone sending/receiving messages would need to know them.

> The idea behind this system is the same than in all encryption: put your
> message into two bits, so that the other one is completely useless
> without the other (passkey/message). However, this system guarantees that
> if you try to eavesdrop the message, the recipient will get to know of
> it, immediately, which is something current encryption messages don't do.

Well, I still don't see why, but I'm getting ready to give up on it. :-)

> Because 1/4 of the data you resend would be utter garbage. You see,
> you have 4 polarizations. You can observe them in two different ways,
> we'll call the polarizations A(0), B(45), C(90) and D(135). Vertical
> polarization will yield an 1 on A, B and D, and 0 on C, whereas
> horizontal polarization will yield an 1 on B, C and D (and 0 on A).
>
> Thus, if you observe a 1 with H polarization, you cannot really tell
> which one it was. If you were supposed to measure it in H polarization,
> it could be B C or D, but if you were supposed to measure it in V
> polarization, it could be A B or D. And no, you can't measure it both
> ways (this is forbidden by the uncertainty principle).
>
> And if it is A, it is very possible that it in reality should've been a
> 0, instead of a 1 (or vice versa). So, he really cannot tell which
> polarization to send down the line again... You see, if he sees an one,
> he'll have different choices depending on his measurement. With a zero,
> there's no problem. If he sends down the wrong polarization, the
> recipient will know of it immediately (or right after checking the
> CRC/whatever they must be using).
>
> So, the poor wiretapper is confused.

You think the poor wiretapper is confused! :-) I didn't quite catch that.
Perhaps I should give up and just accept what you all say, wait till I do
quantum physics and learn for myself.

> Now, how does the recipient know
> what to do? He knows that this photon should be measured in a H fashion,
> so he knows that it is supposed to be one and writes it down as an one.

Just how does he know it is to be measured in a H fashion?

--
Damion Milliken University of Wollongong E-Mail: u9467882@******.uow.edu.au

(Geek Code 2.1) GE d@ H s++:-- !g p? !au a18 w+ v C+ U P? !L !3 E? N K- W+ M
!V po@ Y t(+) !5 !j r+(++) G(+) !tv(--) b++ D+ B? e+ u@ h+(*)
f+@ !r n--(----)@ !y+
Message no. 45
From: Janne Jalkanen <jalkanen@*********.CERN.CH>
Subject: Re: Quantum Cryptography
Date: Fri, 5 Aug 1994 13:24:31 +0200
On Fri, 5 Aug 1994, Damion Milliken wrote:

> Someone must know them, otherwise the messages could not be read. In fact
> anyone sending/receiving messages would need to know them.
>
> Just how does he know it is to be measured in a H fashion?

Because that is the passkey. For instance, you could agree with the
sender that every other measurement is to be done in a H and every other
in a V fashion. This, obviously, is not a very good passkey, but with
some suitable random data, you could make up really long (Vernam)
passkeys.

Janne Jalkanen ///! For those who have to fight for it
jalkanen@******.cern.ch /// ! life has a flavor
Janne.Jalkanen@***.fi \\\/// ! the protected will never understand
-'Keep on going...' \XX/ ! (anonymous, Viet Nam, 1968)
Message no. 46
From: Skrub <skrub@******.SSNET.COM>
Subject: Quantum Cryptography
Date: Fri, 5 Aug 1994 16:23:56 -0400
I don't know a whole lot bout optical physics etc, BUT. If the idea is to
have one photon indicating whether or not there is a big hole. one end,
A, would send it. I will assume there is a way yo generate it which will
control all aspects which are important, and send it to B. You (don't
know who so I'll say "you") say it cannot be intercepted w/ out B knowing
because the interception will change the vital statistics of the photon
used to detect this sort of thing (Yes, I am not an Engineer :). Sounds
good.

How does B know if it's correct, won't B looking at it change it?
Therefore it would ALWAYS appear that someone tampered? If B can measure
the photon than so can the interceptor. The interceptor may then change
it, but can re emit the whole package and no one will know unless it is
being timed.

My 2 $ (it was a big thought)

-Skrub
Message no. 47
From: Damion Milliken <u9467882@******.UOW.EDU.AU>
Subject: Re: Quantum Cryptography
Date: Sat, 6 Aug 1994 14:59:41 +0000
Skrub writes:

[stuff on receiving and resending photons]

This is what I tried to ask a while back, and as yet no-one have given me an
[understandable] explanation for it.

--
Damion Milliken University of Wollongong E-Mail: u9467882@******.uow.edu.au

(Geek Code 2.1) GE d@ H s++:-- !g p? !au a18 w+ v C+ U P? !L !3 E? N K- W+ M
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f+@ !r n--(----)@ !y+
Message no. 48
From: Ivy Ryan <ivyryan@***.ORG>
Subject: Re: Quantum Cryptography
Date: Mon, 8 Aug 1994 13:18:01 -0700
Two points here, folks,

On Wed, 3 Aug 1994, Gian-Paolo Musumeci wrote:

> Rat> ANY encryption scheme can be broken, given enough time and effort.

There IS a system that hasn't been broken as yet, and NSA started trying
in 1956.

> ...and NSA has both the time and the effort if they want to. Trust me. =)
>
NSA can, and will, break anything else. They have over 2,000 people and
unimaginable computation power. Code breaking and message interception
are their jobs. I worked there for a while, and they are very good at it.

The unbroken encryption system involves unbroken transmission of
encrypted information in whict the start and end points of the actual
message are unknown and undiscoverable. Anything less than that they
will break.

Ivy
Message no. 49
From: Stainless Steel Rat <ratinox@***.NEU.EDU>
Subject: Re: Quantum Cryptography
Date: Mon, 8 Aug 1994 17:37:30 -0400
>>>>> "Ivy" == Ivy Ryan <ivyryan@***.ORG> writes:

Rat> ANY encryption scheme can be broken, given enough time and effort.
Ivy> There IS a system that hasn't been broken as yet, and NSA started
Ivy> trying in 1956.

It's a variant of the one time pad (OTP). As long as the keys remain truely
random, it works. In the case of this example an additional level of
randomness is added by denying the knowledge of the start of the message,
thus you can't tell how to do your bit splitting; or else it's using an XOR
style system, using the previous chunk text to assist in the generation of
the key for the next chunk of text. Either way, if you don't know where the
breaks are it becomes a royal pain to analyze.

The most obvious weakness with OTPs, or any other single-key cryptosystem
is distributing the keys. If your distribution channel is secure, why do
you need encryption? If it's not secure, why are you trusting it for key
disemination?

BTW, just because they haven't broken it doesn't mean they never will. And
besides, the NSA has a wondeful track record of not telling people things
about itself (officially the NSA didn't even exist in 1956 :).

--
Rat <ratinox@***.neu.edu> |fnord fnord fnord fnord fnord fnord fnord f
http://www.ccs.neu.edu/home/ratinox|nord fnord fnord fnord fnord fnord fnord fn
PGP Public Key: Ask for one today! |ord fnord fnord fnord fnord fnord fnord fno

Further Reading

If you enjoyed reading about Quantum Cryptography, you may also be interested in:

Disclaimer

These messages were posted a long time ago on a mailing list far, far away. The copyright to their contents probably lies with the original authors of the individual messages, but since they were published in an electronic forum that anyone could subscribe to, and the logs were available to subscribers and most likely non-subscribers as well, it's felt that re-publishing them here is a kind of public service.

Thwap!