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back to article Boffins confirm quantum crypto can keep a secret

Over recent years, the gap between theoretical security of quantum crytography and practical implementation has provided plenty of fun for super-geniuses the world over. Yes, quantum cryptography is supposed to be unbreakable. After all, if anybody even observes the state of a qubit that Alice has prepared, entangled with …

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Bronze badge

Great Idea...

...lets bring another single point of failure (charlie) into the mix!

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Facepalm

Confused?

Was Eve having an affair with Charlie? was Bob upset because Alice was jealous? What was the relationship between Alice and Eve? Did Charlie ever have a relationship with Bob? Was Alice related to Eve?

What about Rita, Sue and Bob Too?

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Anonymous Coward

Re: Great Idea...

Dunno why this got a downvote. It's goes right to the essence of the situation.

we're now being told that you can't have provably secure comms between two parties. You need a third.

That is an almost insurmountable additional complexity.

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Holmes

Re: Great Idea...

But Charlie may just be a device in Bob's phone....

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Re: Great Idea...

> we're now being told that you can't have provably secure comms between two parties. You need a third.

What I am given to understand is that the third party is the evesdropper.

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Mushroom

Re: Great Idea...

Simplicity by Complexity - or it takes another right Charlie to Foxtrot the system.

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Alice, Bob, Charlie... and Eve. What happened to Dave? Or Desdemona?

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Boffin

Alice & Bob

Traditionally, Eve refers to an eavesdropper (Eve... geddit?). Charlie is a 3rd (non-malicious) participant. It's usually Dan or Dave for the 4th. They're all derived by Bruce Schneier I think.

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Dave's not here, man ...

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Pint

Re: Alice & Bob

They're all derived by Bruce Schneier I think.

WP says the names Alice and Bob were first used by Ron Rivest (the 'R' of "RSA") in 1978, and that other names have been added as need arose. The popularity of the naming scheme may have increased as a result of John Gordon's oft-quoted after-dinner speech in 1984. Bruce scheier has certainly used them, not least in Applied Cryptography (1994).

http://en.wikipedia.org/wiki/Alice_and_Bob

Beer, because there's no port icon and a good after-dinner speech deserves a full glass.

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Silver badge

Dave ...

... for endless repeating signals from the past.

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Terminator

I'm sorry Dave, I can't do that.

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Headmaster

Re: Alice & Bob

"They're all derived by Bruce Schneier I think."

This sentence could be:

"They're all devised by Bruce Schneier I think." or:

"They're all derived from Bruce Schneier I think."

Which is it?

Personally I think that the first makes more sense though both are reasonably correct, grammatically.

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Joke

@ Frank Haney

We all know what happened to Desdemona - Shakespeare's still collecting royalties!

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Coat

This isn't really going to help is it ?

Because when Charlie sends his yes or no to Bob, then Bob will know the polarisation state of the photon he has sent to Alice. Therefore a measurement has taken place and the quantum state of the other photon

(the one Alice gets) will collapse to the opposite state. So instead of needing to intercept the signal at Bob's end you only need to intercept it at 'Charlie's place' to intercept the key.

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FAIL

Re: This isn't really going to help is it ?

You are reading it wrong.

"the Charlie detector doesn't report on Alice's / Bob's polarisation – only the difference between their polarisations"

So no-one knows the polarisation yet (in fact, it hasn't even been "set" at all). Only their difference!

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FAIL

Re: This isn't really going to help is it ?

Charlie has to MEASURE the polarisation of the signal coming from Alice AND Bob to be able to compare them right ?

So the NSA tap can be placed at Charlie's...

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Silver badge
Holmes

Re: This isn't really going to help is it ?

You are not supposing for a second these guys haven't thought about this, do you?

We are not in classical space here.

I will have to read the paper later but I suppose that in this case you "measure" whether two qubits are equal or not without measuring the qbits...

So, your original state is one of |00>, |01>, |10>, |11>

Then you ask the question: is the state one of (|01>, |10>) or one of (|11>, |00>)

This gives you the classical Yes/No bit, at which point your photon pair is still either

(|01>, |10>)

or

(|11>, |00>)

But you don't know which, and A and B will still have to measure to see whether the 0 or the 1 is on their end. Charlie doesn't know anything helpful about that.

We need the "are you serious, commenter" icon ASAP-

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Anonymous Coward

Re: This isn't really going to help is it ?

And Charlie can't possibly have malicious intent, instead of comparing Alice' polarization with Bobs polarization to his own polarization for each, thus create false keys and enable a suitably equipped third party, lets call him Mallory, to be the man in the middle with his new keys, which are viable to transceive signals to Alice and Bob?

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Holmes

Re: This isn't really going to help is it ?

Pretty sure that no, but for that you have to go over the math. Unfortunatley I am currently busy shifting decidedly classical bits from the database to user program and back.

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Bronze badge

So, not only have these boffins done some fantastic work, but they have also finally proved that it _IS_ possible to prove a negative...

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25 kbit secure keys

Oh. My. God.

I wonder what key length we will need in the 4th millennium.

Then again, we might be more interested by a log of wood by then.

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Meh

Dave - Aka - Tremendous Knowlege

Dave, the keeper of the spare keys.

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Facepalm

Re: Dave - Aka - Tremendous Knowlege

No

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Coffee/keyboard

This sounds like a decent idea for a porno involving secret agents from various nations.

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Ok I've got a hard enough time getting my head around regular encryption. My brain now hurts, a lot.

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