[EMAIL PROTECTED] wrote:
>> From: [EMAIL PROTECTED]
>> [mailto:[EMAIL PROTECTED] Behalf Of Dave Howe
>>
>> Peter Fairbrother may well be in possession of a break for the QC hard
>> problem - his last post stated there was a way to "clone" photons with
>> high accuracy in retention of their polari
Peter Fairbrother wrote:
> I promised some links about the 5/6 cloning figure. You've had a few
> experimental ones, here are some theory ones.
has anyone with better number theory / probability skills than me taken a
stab at exactly *how* accurate cloning would have to be (and how many
clones you
I promised some links about the 5/6 cloning figure. You've had a few
experimental ones, here are some theory ones.
Cloning machines:
http://www.fi.muni.cz/usr/buzek/mypapers/96pra1844.pdf
Theoretically optimal cloning machines:
http://www.gap-optique.unige.ch/Publications/Pdf/PRL02153.pdf
1/6 d
I'm always stuck on that little step where Alice tells Bob what basis
she used for each photon sent. Tells him how? They need integrity
protection and endpoint authentication for N bits of basis. Is the
quantum trick converting those N bits to N/2 privacy-protected bits
really as excit
John S. Denker wrote:
>After the key exchange has taken place, Alice
>and Bob can use the key to set up a tunnel to
>keep their discussions private. Probably one
>of the first things they will do is exchange
>authentication messages through the newly
>created tunnel. Thereby Alice can decide
>whe
Matt Crawford wrote:
>> BTW, you can decrease the wavelength of a photon by bouncing it off
>> moving
>> mirrors.
>
> Sure. To double the energy (halve the wavelength), move the mirror at
> 70% of the speed of light. And since you don't know exactly when the
> photon is coming, keep it moving a
BTW, you can decrease the wavelength of a photon by bouncing it off
moving
mirrors.
Sure. To double the energy (halve the wavelength), move the mirror at
70% of the speed of light. And since you don't know exactly when the
photon is coming, keep it moving at that speed ...
---
[EMAIL PROTECTED] wrote:
>> From: [EMAIL PROTECTED]
>> [mailto:[EMAIL PROTECTED] Behalf Of Dave Howe
>>
>> Peter Fairbrother may well be in possession of a break for the QC hard
>> problem - his last post stated there was a way to "clone" photons with
>> high accuracy in retention of their polari
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] Behalf Of Dave Howe
>
> Peter Fairbrother may well be in possession of a break for the QC hard
> problem - his last post stated there was a way to "clone" photons with
> high accuracy in retention of their polarization
> [SNIP]
>
Not a break at
I always understood that QKD is based on a hard problem of which the theory of
physics says it is impossible to find a solution (if not, then i'd like to
know). Then if QKD breaks, the current theory of physics was wrong.
On the other hand, if DH or RSA breaks, factoring or the discrete log turn
On Sun, Sep 21, 2003 at 01:37:21PM +0100, Peter Fairbrother wrote:
[cloning photons]
> There is also another less noisy cloning technique which has recently been
> done in laboratories, though it doubles the photon's wavelength, which would
> be noticeable,
To get rid of the wavelength change it s
Peter Fairbrother wrote:
> If the channel is authentic then a MitM is hard - but not impossible. The
> "no-cloning" theorem is all very well, but physics actually allows imperfect
> cloning of up to 5/6 of the photons while retaining polarisation, and this
> should be allowed for as well as the n
>> no. its the "underlieing hard problem" for QC. If there is
>> a solution to any of the Hard Problems, nobody knows about them.
>right, so it's no better than the arguable hard problem of
>factoring a 2048 bit number.
Peter Fairbrother may well be in possession of a break for the QC hard
problem
Again, replying to all.
also sprach John S. Denker <[EMAIL PROTECTED]> [2003.09.19.0038 +0200]:
> Other key-exchange methods such as DH are comparably
> incapable of solving the DoS problem. So why bring up
> the issue?
For one, I can un-DoS with QC at any point in time. This may be
relevant for
There are lots of types of QC. I'll just mention two.
In "classic" QC Alice generates polarised photons at randomly chosen either
"+" or "x" polarisations. Bob measures the received photons using a randomly
chosen polarisation, and tells Alice whether the measurement polarisation he
chose was "+"
At 6:38 PM -0400 9/18/03, John S. Denker wrote:
Yes, Mallory can DoS the setup by reading (and thereby
trashing) every bit. But Mallory can DoS the setup by
chopping out a piece of the cable. The two are equally
effective and equally detectable. Chopping is cheaper and
easier.
Other key-exchange
> Date: Fri, 19 Sep 2003 11:57:22 -0400
> From: Ian Grigg <[EMAIL PROTECTED]>
> If I understand this correctly, this is both
> an eavesdropping scenario and an MITM scenario.
>
> In the above, Eve is acting as Mallory, as she
> is by definition intercepting the bits and re-
> sending them on?
As
On 09/19/2003 12:07 PM, Matt Crawford wrote:
I'm always stuck on that little step where Alice tells Bob what basis
she used for each photon sent. Tells him how?
That's a fair question. Here's an outline of
the answer.
We choose an eps << 1.
We ask how many people accurately received a
fractio
I'm always stuck on that little step where Alice tells Bob what basis
she used for each photon sent. Tells him how? They need integrity
protection and endpoint authentication for N bits of basis. Is the
quantum trick converting those N bits to N/2 privacy-protected bits
really as exciting as
Ian Grigg wrote:
> If I understand this correctly, this is both
> an eavesdropping scenario and an MITM scenario.
>
> In the above, Eve is acting as Mallory, as she
> is by definition intercepting the bits and re-
> sending them on?
I think it is more a question of style - a classic "passive" Eve c
"R. Hirschfeld" wrote on QKD:
> The eavesdropper Eve doesn't know with which basis to measure the
> polarity of the each intercepted photon. When she guesses right, she
> gets the correct information and can send it on undetectably. When
> she guesses wrong, she gets a zero or one with equal pro
On Fri, 19 Sep 2003, martin f krafft wrote:
> But Newton gets more wrong the faster you go. So it's not F = m.a,
> that theory was only a good approximation, nothing more.
Actually it still is F = m.a, but the numbers depend on the observer.
F=m.a is a fundamental consequence of the conservation
also sprach [EMAIL PROTECTED] <[EMAIL PROTECTED]> [2003.09.19.1115 +0200]:
> The sender sends RANDOM BITS to the receiver. Those that don't get
> eavesdropped can then be concatenated at both ends to produce an
> identical string of random bits. Since this is known to both
> endpoint parties, and n
martin f krafft wrote:
>This is what I don't buy. If Mallory sees the data, it must be
>detected, because otherwise the approach is flawed.
As I understand it, there are four possible "rotations" for the photon
( call them '\' '|' '/' and '-' ) so two choices for a filter (straight or
slant). a s
> Date: Thu, 18 Sep 2003 18:02:50 +0200
> From: martin f krafft <[EMAIL PROTECTED]>
>
>
I don't know a lot about QKD, but I believe the following is true:
The eavesdropper Eve doesn't know with which basis to measure the
polarity of the each intercepted photon. When she guesses right, she
gets
I wrote:
>>
>> *) In each block, Mallory has a 50/50 chance of being able to
>> copy a bit without being detected.
On 09/18/2003 12:02 PM, martin f krafft wrote:
>
> This is what I don't buy. If Mallory sees the data, it must be
> detected, because otherwise the approach is flawed. But in any cas
It took me a while. I would herewith like to reply to all posts on
this I received so far:
also sprach John S. Denker <[EMAIL PROTECTED]> [2003.09.13.2343 +0200]:
> *) In each block, Mallory has a 50/50 chance of being able to
> copy a bit without being detected.
This is what I don't buy. If
QC is currently a one-time pad distribution mechanism - or at lower rates a
key establishment mechanism most suitable for symmetric algorithms.
You are correct that authentication is not inherent. Then again, this is
also true for "classical" symmetric and PKI schemes. To be usable, all
crypto r
On Sat, Sep 13, 2003 at 09:06:56PM +, David Wagner wrote:
>
> You're absolutely right. Quantum cryptography *assumes* that you
> have an authentic, untamperable channel between sender and receiver.
So as a result, Quantum cryptography depends on the known
methods to provide authenticity and
martin f krafft wrote:
> So MagiQ and others claim that the technology is theoretically
> unbreakable. How so? If I have 20 bytes of data to send, and someone
> reads the photon stream before the recipient, that someone will have
> access to the 20 bytes before the recipient can look at the 20
> b
Arnold G. Reinhold wrote:
>I think there is another problem with quantum cryptography. Putting
>aside the question of the physical channel, there is the black box at
>either end that does all this magical quantum stuff. One has to trust
>that black box.
>
>- Its design has to thoroughly audited
David Wagner wrote:
> One could reasonably ask how often it is in practice that we have a
> physical channel whose authenticity we trust, but where eavesdropping
> is a threat. I don't know.
The only answer that I have come across - to which I
ascribe no view on accuracy - is "undersea fibre" [1
martin f krafft wrote:
and the general hype about quantum cryptography, I am bugged by
a question that I can't really solve. I understand the quantum
theory and how it makes it impossible for two parties to read the
same stream. However, what I don't understand is how that adds to
security.
It's ve
At 10:18 PM + 9/13/03, David Wagner wrote:
...
One could reasonably ask how often it is in practice that we have a
physical channel whose authenticity we trust, but where eavesdropping
is a threat. I don't know.
I think there is another problem with quantum cryptography. Putting
aside the que
Martin F Krafft asked:
> So MagiQ and others claim that the technology is theoretically
> unbreakable. How so? If I have 20 bytes of data to send, and someone
> reads the photon stream before the recipient, that someone will have
> access to the 20 bytes before the recipient can look at the 20
> b
On 09/13/2003 05:43 PM, David Wagner wrote:
>
> I believe the following is an accurate characterization:
> Quantum provides confidentiality (protection against eavesdropping),
> but only if you've already established authenticity (protection
> against man-in-the-middle attacks) some other way.
I
martin f krafft wrote:
>David Wagner <[EMAIL PROTECTED]> writes:
>> You're absolutely right. Quantum cryptography *assumes* that you
>> have an authentic, untamperable channel between sender and
>> receiver. The standard quantum key-exchange protocols are only
>> applicable when there is some oth
> On 09/13/2003 05:06 PM, David Wagner wrote:
> > Quantum cryptography *assumes* that you
> > have an authentic, untamperable channel between sender and receiver.
>
> Not true. The signal is continually checked for
> tampering; no assumption need be made.
Quantum crypto only helps me exchange
On 09/13/2003 03:52 PM, martin f krafft wrote:
> ... any observation of the quantum stream is immediately
> detectable -- but at the recipient's side, and only if checksums are
> being employed, which are not disturbed by continual or sporadic
> photon flips.
>
> someone will have
> access to the 2
also sprach David Wagner <[EMAIL PROTECTED]> [2003.09.13.2306 +0200]:
> You're absolutely right. Quantum cryptography *assumes* that you
> have an authentic, untamperable channel between sender and
> receiver. The standard quantum key-exchange protocols are only
> applicable when there is some oth
martin f krafft wrote:
>So MagiQ and others claim that the technology is theoretically
>unbreakable. How so? If I have 20 bytes of data to send, and someone
>reads the photon stream before the recipient, that someone will have
>access to the 20 bytes before the recipient can look at the 20
>bytes,
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