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Researchers Simplify Quantum Cryptography

kdawson posted more than 6 years ago | from the again-with-the-bob-and-alice dept.

Security 106

Stony Stevenson writes "Quantum cryptography, the most secure method of transmitting data, has taken a step closer to mainstream viability with a technique that simplifies the distribution of keys. Researchers at NIST claim that the new 'quantum key distribution' method minimizes the required number of detectors, the most costly components in quantum crypto. Four single-photon detectors are usually required (these cost $20K to $50K each) to send and decode cryptography keys. In the new method, the researchers designed an optical component that reduces the required number of detectors to two. (The article mentions that in later refinements to the published work, they have reduced the requirement to one detector.) The researchers concede that their minimum-detector arrangement cuts transmission rates but point out that the system still works at broadband speeds."

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Quantum Post! (5, Funny)

Majik Sheff (930627) | more than 6 years ago | (#23634813)

Either this post is first or it isn't. I won't know until I press submit.

Re:Quantum Post! (-1, Offtopic)

Anonymous Coward | more than 6 years ago | (#23635059)

Either your mom is fucking me or she isn't. I won't know until I turn on the lights.

Re:Quantum Post! (1, Funny)

Anonymous Coward | more than 6 years ago | (#23636973)

And then you find out it actually was your cat, and thus the counselling begins.

Re:Quantum Post! (5, Funny)

FrYGuY101 (770432) | more than 6 years ago | (#23635121)

Either this post is first or it isn't. I won't know until I press submit.
No fair! You changed the outcome by posting it!

Re:Quantum Post! (1)

mrsteveman1 (1010381) | more than 6 years ago | (#23635145)

Only Bruce Schneier knows for sure

Re:Quantum Post! (0)

Anonymous Coward | more than 6 years ago | (#23636171)

Bruce Schneier is Chuck Norris with a beard!

Re:Quantum Post! (2, Funny)

calmond (1284812) | more than 6 years ago | (#23636101)

Hey, has anyone seen my cat...

Re:Quantum Post! (1, Funny)

Anonymous Coward | more than 6 years ago | (#23636839)

Heisenberg, leave Schrodinger alone!

Actually... (1)

RudeIota (1131331) | more than 6 years ago | (#23635499)

Actually... This could have been the first post, not the first post, or both. And even though you had pressed submit, you probably wouldn't be able to find out anyway.

Re:Actually... (2, Interesting)

Impy the Impiuos Imp (442658) | more than 6 years ago | (#23638329)

Not immediately, I suppose. It's interesting to note that a proper database system would discrete-ize the posting numbers to arrive at a definite first post, even though relativity makes a mockery of first posts.

Hey wait, that shouldn't be possible.

> Quantum cryptography, the most secure method of transmitting data,

Technically it would only be tied at best with a one-time pad, and, at worst, slightly less secure. I wonder if it has codes that could be cracked by social engineering, as one time pad's could, or if you must physically have the proper connection device.

But I hear it's also possible to do a quantum simulation of the entire universe using a quantum device. Hence it may be trivial to crac...hey, waitaminnit!

Maybe this whole universe is someone's attempt to quantum crack some encoded pr0n. DAMMIT!

DAMMIT! My life is just being a cog in someone's un-encoding of some pr0n! >:(

Actually, I feel my life is more valuable doing that then it turning out this universe was Yahweh's twice-patched fuckup (Noah, and Jesus) wise and perfect plan all along.

Entangled (-1, Offtopic)

Anonymous Coward | more than 6 years ago | (#23634815)

Fist Pole!

Re:Entangled (1)

datapharmer (1099455) | more than 6 years ago | (#23636465)

I see someone didn't read the instructions and looked at the light while posting.

what's the big deal (0)

ILuvRamen (1026668) | more than 6 years ago | (#23634843)

Anyone care to explain why anyone should use this? So it's 100% random...why cares? So equation and clock based pseudo random based encrptions can technically be predicted cuz they're not 100% random. But nobody ever knows the exact equation AND exact millisecond it was calculated to generate the key so noobdy can predict it. I don't think it's any better. It's just a product they're pushing with some ridiculous statement like "but it's better cuz it's completely random!" and never back it up with any facts about it truly being harder to crack.

Re:what's the big deal (2, Informative)

Anonymous Coward | more than 6 years ago | (#23634905)

It is impossible to crack because there is no way to decode it without the right key. Algorithms like RSA or DES can be brute forced with enough horse power, for instance, when the quantum computer is invented it could make short work of them. Quantum cryptology will be the only defense.

Re:what's the big deal (0)

mrbluze (1034940) | more than 6 years ago | (#23635043)

Quantum cryptology will be the only defense.

Maybe not. IANAC (I am not a cryptographer) but my limited understanding of things is that quantom computing is not the be-all-and-end-all of encryption. It is possible to develop encryption algorithms that run on conventional processors which are resistant to what quantum computers are probably going to be capable of doing with numbers.

I don't have the vocabulary to go beyond that at this time of day but a while back while almost-blind-drunk I had one of those moments of clarity and it all made sense.

Re:what's the big deal (2, Informative)

khellendros1984 (792761) | more than 6 years ago | (#23635459)

From what I understand, quantum computing will basically allow the equivalent of massively parallel computation, so you can find the key that solves the message easily. In RSA, it means that it could factor the large prime numbers that make up the public key, and mathematically generate the private key from those.

Re:what's the big deal (3, Informative)

arotenbe (1203922) | more than 6 years ago | (#23635467)

"Conventional" encryption algorithms can be brute forced even without the correct key - it will just take a really long time. As I understand it, the point of quantum cryptography is that it is completely impossible to break, because the transmission would be scrambled the moment someone tries to tap the connection.

Don't expect the above to be completely correct, though - I'm hardly a cryptography expert (which doesn't stop me from putting a reference in my sig [wikipedia.org] ).

Re:what's the big deal (1)

ark1 (873448) | more than 6 years ago | (#23639557)

Actually, breaking quantum cryptography should be fairly simply. All you have to do is break the laws of quantum mechanics. A good lawyer should be enough to accomplish this feat.

Re:what's the big deal (1)

Super Jamie (779597) | more than 6 years ago | (#23635103)

when the quantum computer is invented
But will it run Crysis?

Re:what's the big deal (1)

mattmcm (1143125) | more than 6 years ago | (#23636977)

Maybe.

Re:what's the big deal (1)

Tanktalus (794810) | more than 6 years ago | (#23638117)

Who cares? I want to know if it will run Duke Nukem Forever!

Re:what's the big deal (5, Informative)

BadAnalogyGuy (945258) | more than 6 years ago | (#23634917)

The big deal is that the cracking time for non-quantum algorithms reduces to O(n) for length n keys. OTOH, for quantum encryption, the cracking time minimum threshold is O(n^n) for length n keys. Hyperbolically, the linear analog is also true in that with quantum decryption, it is possible to crack non-quantum algorithms in O(n) time (again for length n keys), but quantum algorithms require O(n^n) to decrypt. Note that without the correct key, the quantum algorithm requires O(n^n) regardless of whether the cracker is employing spherical numerical analysis techniques or advanced quantum distribution array matrices.

The fact of the matter is that quantum encryption provides much greater security than standard algorithmic encryption.

Re:what's the big deal (5, Interesting)

Anonymous Coward | more than 6 years ago | (#23635029)

You also failed to mention that it is impossible to eavesdrop on the communication of the keys. This is probably the most important part because it can make one time pad encryption useful on computer networks. Without quantum cryptography, your one time pad is only as safe as how you send it (RSA encryption, chaos encryption, snail mail). Additionally, quantum cryptography can't be reverse engineered to find the algorithm for your one time pad.

This is all nice, but it is going to be tricky to implement it in the future. How do you send a photon from one computer to another a long distance away without using repeaters or branches? It will be a little tricky. Would this require a fiber optic connection between every computer that wants to communicate with quantum encryption? Or can you adjust the medium so that photons are transmitted and branched undisturbed?

Re:what's the big deal (3, Informative)

mapsjanhere (1130359) | more than 6 years ago | (#23637649)

People are mixing up two different things here - quantum transmission, the one you can't read unnoticed, and encryption/decryption using quantum computers and algorithms.
The first one has been demonstrated, and works over limited distances.
The second is an "advanced concept", right next to fusion reactors.

Re:what's the big deal (1)

kmac06 (608921) | more than 6 years ago | (#23644169)

How do you send a photon from one computer to another a long distance away without using repeaters or branches?
You don't. You just use branches and repeaters! It sounds counter-intuitive, but you can do something called entanglement swapping [wikipedia.org] to send the quantum state from Alice to Bob through Charlie, without Charlie changing (or even being able to detect) the state Alice sent.

Also you can control/change the path of the photon using electro-optical controls without changing it's state (or at least without changing the part of the state that carries the information), though this would likely be more difficult/expensive to implement on a wide scale.

Re:what's the big deal (2, Funny)

SeekerDarksteel (896422) | more than 6 years ago | (#23635357)

Plus with quantum encryption you can utilize lunar wainshafts to feed the unilatral phase dectractors and Karnot-Graham meters.

...

Re:what's the big deal (1)

NormalVisual (565491) | more than 6 years ago | (#23636909)

Well, that's how the Federation does it anyway.

Re:what's the big deal (2, Insightful)

spazdor (902907) | more than 6 years ago | (#23635373)

spherical numerical analysis techniques or advanced quantum distribution array matrices.


I was with you up until about there. It occurs to me that there are any number of mathematical terms that could be combined at random to induce the same effect in me, and I wonder if this is true of all the people who modded you up.

I think i'm just gonna take your word for it. :)

Re:what's the big deal (3, Funny)

MrMr (219533) | more than 6 years ago | (#23635411)

It's not so hard, let me explain:
spherical numerical analysis techniques: That is standard maths; If you need to compute something involving for instance a cow, you start with "Assume a spherical cow with radius R".
advanced quantum distribution array matrices: That just your normal quantum distribution array matrices but with the new icons and toolbar.

Re:what's the big deal (1)

redxxx (1194349) | more than 6 years ago | (#23636679)

You left out the new 'quantum bar'. It's really cool because when you click the little star, it creates a quantum super state where the address is simultaneously bookmarked and not bookmarked. A lot of people have been complaining about all the extra clutter, but that's just because they are observing the results.

Re:what's the big deal (1)

beckerist (985855) | more than 6 years ago | (#23638323)

hmmm.... that sounds a lot like coolwebsearch...

Re:what's the big deal (1)

jonaskoelker (922170) | more than 6 years ago | (#23637011)

"Assume a spherical cow with radius R".
You're leaving out the most crucial part:

Uniform density!

Re:what's the big deal (0)

Anonymous Coward | more than 6 years ago | (#23641381)

David Jatt: "Is it because they don't have legs, that makes spherical cows so bad, or is it because there's some kind of cruelty involved, I mean what what what what what what what what?"

Oliver Skeete: "I, I'm not actually sure what they do to the cows, right."

DJ: "Well they breed them in a particular direction, interfere with their genes so they're just a big ball of meat."

OS: "No, well you see I I wouldn't like to eat that, definitely if I knew I was eating that I would, I would go absolutely mad, it's like somebody putting... puh, I don't know - a bit of snot in a bit of bread and giving it to you, it's the same thing -"

DJ: "Really, that sort of thing?"

OS: "Yeah."

DJ: "Somebody's at a restaurant, to give you this scenario right, somebody out there, and they get what they think is a piece of spherical cow on their plate, what do they do?"

OS: "They just eat it don't they."

DJ: "Yeah but what _should_ they do for god's sake, yeah yeah yeah yeah yeah!"

OS: "I mean what they should do really is make a stink about it, they should start phoning up people and start complaining about, you know the meat that they're eating."

DJ: "This came from is a spherical cow, how dare you sell this to me?"

OS: "That's right, of course."

Re:what's the big deal (1)

gweihir (88907) | more than 6 years ago | (#23636315)

The fact of the matter is that quantum encryption provides much greater security than standard algorithmic encryption.

In your dreams. No quantum computer exists that can break encryption used today by a very, very large margin. It seems doubtful whether researchers can get beyond a few bits at all, let alone scale up to a few thousands. Presently this is all hype to get research money. There have been much more similar things that failed and only very few that deliverd on their claims.

Re:what's the big deal (2, Insightful)

devman (1163205) | more than 6 years ago | (#23636655)

The reason governments throw money at it is because whoever can build the first working one would theoretically have the keys to the kingdom. If a working quantum computer existed (and worked as theorized) suddenly secure communication over the internet would be a liability, thats assuming if one ever gets built (and works) and the public knows about it.

Re:what's the big deal (1)

Tolkien (664315) | more than 6 years ago | (#23637033)

Where's the car, Mr. BadAnalogyGuy?

Re:what's the big deal (1)

kmac06 (608921) | more than 6 years ago | (#23644261)

Nonsense. Quantum encryption, when properly implemented, cannot be cracked. Period. If you have a perfectly good and fast quantum computer, you still could not crack quantum encryption. RSA can be broken using Shor's algorithm, but that is O(log(n)^3), not O(n). And if you don't have a quantum computer lying around, then it's a lot worse than that.

Parent shouldn't be modded informative, just about everything he said was wrong.

Re:what's the big deal (0)

Anonymous Coward | more than 6 years ago | (#23634925)

Thank goodness the guy before me said that...my response would have been a little more rude.

Re:what's the big deal (4, Informative)

Inf0phreak (627499) | more than 6 years ago | (#23634935)

I think you've misunderstood something. "Quantum encryption" is something of a misnomer. It's actually a physical process that can be used by Alice and Bob to establish a commonly shared secret that is random (and unknown to even Alice and Bob before the process starts). This secret is then typically used as a one-time pad [wikipedia.org] .

Re:what's the big deal (1)

menace3society (768451) | more than 6 years ago | (#23634973)

Here's something I've never understood. Alice prepares a one-time pad and sends it along using this quantum dealie. Eve intercepts it. Now supposedly this thing changes every time someone observes it, but could Eve just generate a new one based on the data she acquired? Alice created one 'from scratch', why can't Eve?

Re:what's the big deal (1)

Firehed (942385) | more than 6 years ago | (#23634997)

Presumably because the one-time pad is your decryption key. Encryption wouldn't be especially useful if you could just put in a password (not "the" password, but "a" password) and unlock the secrets, would it?

That's my best guess, I've never really understood the theory either. It IS quantum physics, after all.

Re:what's the big deal (5, Informative)

SeekerDarksteel (896422) | more than 6 years ago | (#23635111)

The reason Eve can't just generate a new pad is because there are two methods of generating a photon and two methods of measuring a photon. Each method of generating a photon has a matched way of measuring it. If you use the matched measurement method you correctly get the bit Alice sent. If you use the incorrect method you get a random 0 or 1 no matter what bit Alice sent. Eve (and Bob too) has no way of telling which method Alice used. In quantum key distribution, after sending the photons, Alice would contact Bob over a different channel. They would tell which method they used, and if they used matching methods keep that bit. If they used different methods they would throw out the bit. If Eve regenerated the bits, she could not have used the same methods as Alice since she doesnt know which methods were used. So Alice and Bob's keys won't match up which will result in any information passed between them to be undecodable and they will know someone eavesdropped.

Quantum Key Distribution is, in its most naive form, still vulnerable to man in the middle attacks. It makes it a little more difficult because you must be able to intercept information on two different channels (the quantum channel and the classical electronic channel), but it is still doable. (There are, however, cryptographic methods of detecting man in the middle attacks, but thats a subject for another time).

Re:what's the big deal (1)

devman (1163205) | more than 6 years ago | (#23636697)

There is a way to do the key exchange without using the classical electronic channel. Bob must retransmit his guesses back to Alice, and Alice tells Bob which ones were incorrect because Alice reads the response with her generated key. Eve has no clue because she can't know which methods for reading the photons Bob used.

Re:what's the big deal (1)

Urkki (668283) | more than 6 years ago | (#23635203)

This is how I understand it: In normal cryptography, you have to worry about "the man in the middle" intercepting the message and then cracking it at their leisure. In quantum crypto, "the man in the middle" can't do this. They need the keys beforehand to even record the message. And another thing is, they can't just eavesdrop passively, they must do actual "man in the middle", ie. intercept the message and re-send it in real time.

Somebody correct me if I'm badly mistaken...

Re:what's the big deal (2, Interesting)

locofungus (179280) | more than 6 years ago | (#23635849)

Here's something I've never understood. Alice prepares a one-time pad and sends it along using this quantum dealie. Eve intercepts it. Now supposedly this thing changes every time someone observes it, but could Eve just generate a new one based on the data she acquired? Alice created one 'from scratch', why can't Eve?

Lookup quantum cloning and the "no cloning theorem".

But basically (and this is a naive implementation that won't actually work), Alice transmits to Bob using linearly polarized photons. Now, if you remember from your school days, if you shine a light though a polarization filter and then through another filter at the same angle, all the light that gets through the first filter gets through the second filter as well.

So, let Alice transmit a horizontally polarized photon |H> if she wants to send a 1 and |V> if she wants to send a 0.

Bob uses a horizontally polarized filter = 1 (the photon gets through and he detects it, = 0. The photon gets stopped and he doesn't detect it.

So far, so good but... Eve does exactly what you suggested and measures the photon and then regenerates it - so Bob doesn't see any difference.

Now it starts getting clever ... Again, hopefully you remember from your school days that, if you send that polarized light though a second filter at an angle to the first, a proportion of the light gets through (cos^2 theta). But, QM says that you cannot predict which photons will make it through the second filter, it's entirely probabilistic.

So as well as using |H>,|V> to transmit 1 and 0, Alice also uses |+>,|-> where these are 45 degree polarizations. Alice uses one or the other completely at random.

Bob, when he measures at his end also choses whether to measure the horizontal polarization = Alice and Bob use the same polarization angle so Bob detects the photon
  = Alice and Bob use crossed polarization filters so Bob doesn't detect the photon
  = Alice and Bob's filters are at 45 degrees so Bob may or may not detect a photon
  = ditto
  = ditto
  = ditto
  = Alice and Bob use the same polarization angle so Bob detects the photon
  = Alice and Bob use crossed polarization filters so Bob doesn't detect the photon

Once Alice and Bob have done this, Bob tells Alice which measurement he's done (over a classical channel, they don't care who might eavesdrop.) If Alice and Bob have used the same basis - i.e. Alice used |H>,|V> to transmit her bit and Bob used ,|-> to transmit her bit and Bob used or |V>. She can retransmit that value but, if Alice sent |+> or |-> instead then she'll have corrupted the bit. If she measures or |-> but if Alice sent |H> or |V> then she'll corrupt that bit instead. Infact, on average, regardless of which measurement she makes, she'll end up corrupting 1/2 of the values that Alice and Bob have "successfully" exchanged.

Now Eve can get really clever. Instead of measuring the photon, she can clone it and then measure her clone. Now it turns out that there is a limit to how good her cloning machine can be so, although it won't corrupt half of the bits that Alice and Bob transmit, it will corrupt at least 1/6.

(Actually, in the naive scheme outlined above I think Eve can do:
a|H> + b|V> => a|HH> + b|VV>, store her photon, wait for Bob to measure, eavesdrop the message from Bob to Alice and then make the same measurement on her stored photon. But this only works because the only possible values for a,b in the naive scheme are (0,1), (1,0), (1/sqrt2,1/sqrt2), (1/sqrt2, -1/sqrt2) but I'm right on the limits of my understanding of QM and entangled photons now so I could be completely wrong)

Tim.

Re:what's the big deal (1)

ArsenneLupin (766289) | more than 6 years ago | (#23637221)

Actually, in the naive scheme outlined above I think Eve can do: a|H> + b|V> => a|HH> + b|VV>, store her photon, wait for Bob to measure, eavesdrop the message from Bob to Alice and then make the same measurement on her stored photon.
Actually, the message telling which bits were send and/or measured at a 45% angle is only exchanged after Bob has measured the quantum bits. So, even can't just store them, measure them and re-inject them after the fact.

So, rather than just eavesdropping the message from Bob to Alice, she would actually need to destructively intercept it and change it.

Re:what's the big deal (1)

locofungus (179280) | more than 6 years ago | (#23642549)

Actually, the message telling which bits were send and/or measured at a 45% angle is only exchanged after Bob has measured the quantum bits. So, even can't just store them, measure them and re-inject them after the fact.

So, rather than just eavesdropping the message from Bob to Alice, she would actually need to destructively intercept it and change it.


Eve isn't intercepting the bits. She's creating a pair of correlated photons without actually making any measurement. She sends one on to Bob and stores the other for measuring later.

a|H> + b|V> => a|HH> + b|VV>. I'm pretty sure this transformation is allowed for all values of a and b. Note that she is NOT cloning (a|H> + b|V>)(a|H> + b|V>) which would be prohibited.

Having said that, I think my quick thoughts were incorrect. I was assuming
|HH> = k|++> + k|--> when I think it should have been 1/2|++> + 1/2|+-> + 1/2|-+> + 1/2|-->
(k=1/sqrt2) now that I think about it properly.

Tim.

Re:what's the big deal (1)

locofungus (179280) | more than 6 years ago | (#23635871)

Grrr. < needed :-(

Here's something I've never understood. Alice prepares a one-time pad and sends it along using this quantum dealie. Eve intercepts it. Now supposedly this thing changes every time someone observes it, but could Eve just generate a new one based on the data she acquired? Alice created one 'from scratch', why can't Eve?

Lookup quantum cloning and the "no cloning theorem".

But basically (and this is a naive implementation that won't actually work), Alice transmits to Bob using linearly polarized photons. Now, if you remember from your school days, if you shine a light though a polarization filter and then through another filter at the same angle, all the light that gets through the first filter gets through the second filter as well.

So, let Alice transmit a horizontally polarized photon |H> if she wants to send a 1 and |V> if she wants to send a 0.

Bob uses a horizontally polarized filter <H| to measure Alice's photon. <H|H> = 1 (the photon gets through and he detects it, <H|V> = 0. The photon gets stopped and he doesn't detect it.

So far, so good but... Eve does exactly what you suggested and measures the photon and then regenerates it - so Bob doesn't see any difference.

Now it starts getting clever ... Again, hopefully you remember from your school days that, if you send that polarized light though a second filter at an angle to the first, a proportion of the light gets through (cos^2 theta). But, QM says that you cannot predict which photons will make it through the second filter, it's entirely probabilistic.

So as well as using |H>,|V> to transmit 1 and 0, Alice also uses |+>,|-> where these are 45 degree polarizations. Alice uses one or the other completely at random.

Bob, when he measures at his end also choses whether to measure the horizontal polarization <H| or diagonal polarization <+| completely at random. There are eight cases:

<H|H> = Alice and Bob use the same polarization angle so Bob detects the photon
<H|V> = Alice and Bob use crossed polarization filters so Bob doesn't detect the photon
<H|+> = Alice and Bob's filters are at 45 degrees so Bob may or may not detect a photon
<H|-> = ditto
<+|H> = ditto
<+|V> = ditto
<+|+> = Alice and Bob use the same polarization angle so Bob detects the photon
<+|-> = Alice and Bob use crossed polarization filters so Bob doesn't detect the photon

Once Alice and Bob have done this, Bob tells Alice which measurement he's done (over a classical channel, they don't care who might eavesdrop.) If Alice and Bob have used the same basis - i.e. Alice used |H>,|V> to transmit her bit and Bob used <H| to measure it or Alice used |+>,|-> to transmit her bit and Bob used <+| to measure it then Alice says "correct" and Bob knows what bit Alice sent. If Bob used the wrong measurement then Alice says "incorrect" and they both throw that bit away.

So, on average, for each two bits that Alice tries to send to Bob, one will be thrown away and the other will be a good value.

Now Eve tries to eavesdrop. If she measures <H| then she'll detect |H> or |V>. She can retransmit that value but, if Alice sent |+> or |-> instead then she'll have corrupted the bit. If she measures <+| instead then she can retransmit |+> or |-> but if Alice sent |H> or |V> then she'll corrupt that bit instead. Infact, on average, regardless of which measurement she makes, she'll end up corrupting 1/2 of the values that Alice and Bob have "successfully" exchanged.

Now Eve can get really clever. Instead of measuring the photon, she can clone it and then measure her clone. Now it turns out that there is a limit to how good her cloning machine can be so, although it won't corrupt half of the bits that Alice and Bob transmit, it will corrupt at least 1/6.

(Actually, in the naive scheme outlined above I think Eve can do:
a|H> + b|V> => a|HH> + b|VV>, store her photon, wait for Bob to measure, eavesdrop the message from Bob to Alice and then make the same measurement on her stored photon. But this only works because the only possible values for a,b in the naive scheme are (0,1), (1,0), (1/sqrt2,1/sqrt2), (1/sqrt2, -1/sqrt2) but I'm right on the limits of my understanding of QM and entangled photons now so I could be completely wrong)

Tim.

Re:what's the big deal (1)

ILuvRamen (1026668) | more than 6 years ago | (#23635005)

yeah, so it's completely random and comes out of nowhere. But a gigantic equation based on an exact millisecond on the computer's clock might as well have come from nowhere too cuz nobody can record or measure that. But the next couple replies actually make sense (almost) about how it prevents eavesdropping. I still don't buy the completely made up, cat in a box, quantum flux until someone "measures it" even though measures it doesn't make sense in the tradition sense and is unproven in the absolute molecular sense but I'll just take their word for it and assume it means it can only be decrypted once or something. In that case I guess a test packet could be sent first and if it arrives unencrypted, someone else decrypted it first and they can stop transmission. But wait, it would have to contain a previously agreed upon message to be able to tell if it was encrypted or not so the man in the middle could fake it. I'm no quantum physicist but there's a way to beat everything.

Re:what's the big deal (1)

locofungus (179280) | more than 6 years ago | (#23635925)

You don't transmit the message using QC, you transmit a OTP. So if Eve does intercept it then all she gets is a bunch of random bits, Alice and Bob detect the interception, throw the OTP away and start again.

The best that Eve can do is a DOS attack.

Tim.

Re:what's the big deal (1)

amrik98 (1214484) | more than 6 years ago | (#23634953)

Quantum cryptography is supposed to provide secrecy by detecting eavesdroppers, since other parties observing your conversation should disturb some physical properties that you can subsequently measure.

Impossible to eavesdrop, otherwise, a big yawn (5, Interesting)

Mathinker (909784) | more than 6 years ago | (#23634955)

The sexy part is that if there is a third party who tries to eavesdrop, the attempt will both fail and can be detected by the two communicating parties, and that the security of quantum cryptography has nothing to do with the lack of ability to factor large numbers, but is instead based on physical principles (quantum mechanics). Of course, the sensitivity to eavesdropping means that the system is probably vulnerable to a denial of service attack, depending on how the two communicating parties relate to eavesdropping.

Otherwise, you are perfectly correct. Many cryptographers, including Bruce Schneier, believe that quantum cryptography is a solution to the wrong problem. Nowadays, most probably, the least secure part of your communication system isn't in your key distribution scheme, but is somewhere else --- like in social engineering, or the computer systems which deal with the decrypted cleartext.

Re:Impossible to eavesdrop, otherwise, a big yawn (0)

Anonymous Coward | more than 6 years ago | (#23635239)

If you're able to phyiscally access the channel, you might as well mount a man in the middle attack [wikipedia.org] instead of trying to eavesdrop.

Re:Impossible to eavesdrop, otherwise, a big yawn (1)

Tom (822) | more than 6 years ago | (#23636693)

Otherwise, you are perfectly correct. Many cryptographers, including Bruce Schneier, believe that quantum cryptography is a solution to the wrong problem. Nowadays, most probably, the least secure part of your communication system isn't in your key distribution scheme, but is somewhere else --- like in social engineering, or the computer systems which deal with the decrypted cleartext.
I disagree with Bruce on this, no matter how much I respect him. Once we have a reliable crypto scheme, people will stop looking at the technology when there's a failure, and start looking at the real issues. And I'm pessimist enough to believe that nothing else, no amount of education, training or awareness, will make them do that.

Re:what's the big deal (1)

gweihir (88907) | more than 6 years ago | (#23636293)

There is an even bigger problem with quantum ''encryption'': The pysical models are only ever exact to some degree. It is quite possible that some minor, not yet discoverd, effect exists that completely breaks security. If you really want to be secure, definitely stick to the stuff we understand.

What are you doing here (-1, Flamebait)

Anonymous Coward | more than 6 years ago | (#23634847)

It's summer time, world. Get out and enjoy it. Only dickweeds and geeks hang out on slashdot in the summer time.

Re:What are you doing here (-1, Offtopic)

Anonymous Coward | more than 6 years ago | (#23634857)

Looks like a case of the pot calling the kettle dickweed.

Re:What are you doing here (2, Funny)

BlockedThreads (870266) | more than 6 years ago | (#23635017)

It's a summertime Northern Hemisphere and a wintertime Southern Hemisphere. Slice the world the other way and its daytime in one hemisphere and nighttime in the other. And its always dark down here in my parents' basement.

Real world usage? (1, Offtopic)

BadAnalogyGuy (945258) | more than 6 years ago | (#23634873)

Is there anywhere in the world actually using this sort of technology? Is it used in the military at all?

Not much (3, Informative)

Mathinker (909784) | more than 6 years ago | (#23635015)

From Wikipedia:

Quantum encryption technology provided by the Swiss company Id Quantique was used in the Swiss canton (state) of Geneva to transmit ballot results to the capitol in the national election occurring on Oct. 21, 2007.[8]

In 2004, the world's first bank transfer using quantum cryptography was carried in Vienna. An important cheque, which needed absolute security, was transmitted from the Mayor of the city to an Austrian bank.[9]
Both of these look like special uses set up for publicity by vendors.

Re:Not much (1)

gweihir (88907) | more than 6 years ago | (#23636335)

Both of these look like special uses set up for publicity by vendors.

They are. Nobody competent would use quantum techniques for things that really need to be secure. The physics is not that well established, for one thing, leading to an unknown risks. Sure, the properties look nice, but when was the last time a fundamental physical theory turned out to be not quite accurate? Yes, that is true, when the current theory replaced the last one. That has happened so far to every theory except the respective current last ones. Why should there not be additional theory refinements? And the current theory was emphatically not made to predict any security properties. It is more a result of some physics people trying their hand at philosophy (and failing IMO).

I doubt it. (0, Redundant)

jd (1658) | more than 6 years ago | (#23635045)

You can achieve comparable security using Byzantine methods to split a one-time encryption pad into relatively secure fragments. Since the fragments can be reordered and then randomly embedded in the data, you can achieve everything boasted for quantum encryption but using methods tried-and-tested.

Re:Real world usage? (0)

Anonymous Coward | more than 6 years ago | (#23635931)

This might be a baseless rumor, but I've heard that some kind of quantum thingy is used in nuclear missile launch systems.

Like between the control bunker and the launch pad.

Presumably to prevent hot wiring.

Yay! Saved! (1, Funny)

SEWilco (27983) | more than 6 years ago | (#23634919)

With this simplification, thousands of cats are saved from having to deliver code keys.

Oblig XKCD (3, Funny)

azakem (924479) | more than 6 years ago | (#23634947)

Every time I hear about Alice and Bob, I now think of this [xkcd.com]

Broadband Speeds (1, Interesting)

enoz (1181117) | more than 6 years ago | (#23634991)

Describing the rate as "Broadband Speeds" is about as useful as describing the performance of a supercar as "roadworthy" (there's your car analogy).

For reference, in Australia not only does the incumbent Telco consider 256/64kbps to be broadband, but they also describe it as "Fast [bigpond.com] ".

Re:Broadband Speeds (1)

TubeSteak (669689) | more than 6 years ago | (#23635291)

Describing the rate as "Broadband Speeds" is about as useful as describing the performance of a supercar as "roadworthy" (there's your car analogy).
Quantum cryptography doesn't need to be all that fast.
The whole point is to be able to securely pass an encryption key.
Then you can encrypt and use any method you like for transporting the encrypted data.
Whether it's Australian 'fast', Internet2 fast, or a stationwagon full of Terabyte hard drives fast.

Broadband? (1, Insightful)

Tubal-Cain (1289912) | more than 6 years ago | (#23635013)

What speeds are they calling broadband? 200Kbps?

Re:Broadband? (1)

Tubal-Cain (1289912) | more than 6 years ago | (#23635225)

TROLL? The FCC defines broadband [fcc.gov] as greater than 200 Kbps. I was simply wondering if they were using the same metric or a better one.

Re:Broadband? (1)

spazdor (902907) | more than 6 years ago | (#23635399)

I would mod you funny but I've already posted further up. :(

Not the most secure (2, Insightful)

BlueParrot (965239) | more than 6 years ago | (#23635061)

There is only one cryptography scheme with proven secrecy, and that is the one time pad. Even if you assume no errors occur in its implementation, no physicist can guarantee there will never be discovered a way to eavesdrop on transmissions that use Quantum Cryptography. In contrast with the one time pad a Mathematician can more or less prove, at least to the extent you can prove anything at all, that eavesdropping is only possible if the implementation is flawed.

In practice none of this is relevant since the hassles associated with correctly implementing either QC or a OTP are sufficiently large that for most applications they are both inferior to public key cryptography and symmetric ciphers. There are some exceptions, but the only way you could possibly justify describing quantum cryptography as "the most secure way to transmit data" would be by ignoring so many aspects of information security that it will have no relevance to practical applications.

Re:Not the most secure (2, Informative)

Cairnarvon (901868) | more than 6 years ago | (#23635147)

-1 Failed Attempt at Sounding Insightful.
Quantum cryptography schemes are guaranteed to inform both Alice and Bob if their communication is intercepted. That's the entire point, and what has everyone so excited about quantum cryptography in the first place. Secrecy in the sense of undecryptability isn't the point of quantum cryptography (as data isn't even *encrypted* in the classical sense), just certainty that there are no eavesdroppers.
Your post just suggests that you haven't actually read anything about quantum cryptography, you've just heard something about one-time pads and thought this would be a good time to misapply your knowledge.

Quantum cryptography isn't a cipher. It's a method of transmitting data, which does one specific thing, which is guarantee that you'll be able to tell if people have attempted to eavesdrop.
It's not a complete cryptosystem; it's not meant to be. It's meant to be just one component of cryptosystems, and in doing what it does, it's provably secure in the sense that secure is being used here.

(Incidentally, mathematical proofs aren't like scientific proofs; it *is* possible to prove with absolute certainty in mathematics.)

Re:Not the most secure (2, Interesting)

Creepy Crawler (680178) | more than 6 years ago | (#23635351)

Ok. What is an observer?

Or better yet, what would happen if some new device could record without observing?

---Quantum cryptography isn't a cipher. It's a method of transmitting data, which does one specific thing, which is guarantee that you'll be able to tell if people have attempted to eavesdrop. It's not a complete cryptosystem; it's not meant to be. It's meant to be just one component of cryptosystems, and in doing what it does, it's provably secure in the sense that secure is being used here.

Of course not. quantum crypto solves the problem with OTP's: secure transmission of the OTP lookup sheet itself. I forget the bit rate of the machine, but it's something damned slow (100 bits/s?). The ultimate problem with any crypto system is still the people though. Too bad it doesnt fix that ^_^

---(Incidentally, mathematical proofs aren't like scientific proofs; it *is* possible to prove with absolute certainty in mathematics.)

Also, unfortunately, QC math isnt a math proof. It's a proof on physics that attempts to estimate the real math of our universe. Until we know the "real" math, there will be holes in our knowledge, and therefore unprovable. It's probably pretty accurate though, but is it accurate to trust?

Re:Not the most secure (1)

Cairnarvon (901868) | more than 6 years ago | (#23635687)

Ok. What is an observer?

Or better yet, what would happen if some new device could record without observing?

I spend most of my time debating creationists, not laypersons who misunderstand quantum physics, but I bet physicists get as tired of shouting "IT DOESN'T WORK LIKE THAT" at people like you as I do at creationists.

It's probably pretty accurate though, but is it accurate to trust?
You'll remember Feynman once compared our understanding of quantum physics to measuring the distance between New York and Los Angeles to within the width of a human hair. I'd say anything with that kind of predictive power is probably accurate enough to trust your life with.

Re:Not the most secure (1)

blueg3 (192743) | more than 6 years ago | (#23637219)

I spend most of my time debating creationists, not laypersons who misunderstand quantum physics, but I bet physicists get as tired of shouting "IT DOESN'T WORK LIKE THAT" at people like you as I do at creationists.
All the time.

You'll remember Feynman once compared our understanding of quantum physics to measuring the distance between New York and Los Angeles to within the width of a human hair. I'd say anything with that kind of predictive power is probably accurate enough to trust your life with.
That's actually quantum electrodynamics. QED was "solved" by Feynman using approximations that involve shady cancellations of infinite quantities. It's accepted as an approximate solution, but the quality of the approximation is good. Simple quantum mechanics, like what quantum cryptography is based on, is much better-understood.

Re:Not the most secure (1)

locofungus (179280) | more than 6 years ago | (#23635965)

Or better yet, what would happen if some new device could record without observing?

"record without observing" doesn't make sense. In the words of Pauli, "That's not right. That's not even wrong." But trying to guess that you mean "observe without affecting in any way".

Then you can violate causality. Give me a device that can "record without observing" and I'll build you a device to communicate faster than the speed of light.

Tim.

Re:Not the most secure (0)

Anonymous Coward | more than 6 years ago | (#23635153)

... except that Quantum Cryptography is used in practice as a secure method of transmitting One Time Pads.

So. Uh. What were you saying again?

Apples and oranges (2, Interesting)

Chuck Chunder (21021) | more than 6 years ago | (#23635227)

There is only one cryptography scheme with proven secrecy, and that is the one time pad. Even if you assume no errors occur in its implementation, no physicist can guarantee there will never be discovered a way to eavesdrop on transmissions that use Quantum Cryptography. In contrast with the one time pad a Mathematician can more or less prove, at least to the extent you can prove anything at all, that eavesdropping is only possible if the implementation is flawed.


You are comparing apples with oranges. The bit your mathematician can "prove" is only part of the problem quantum encryption aims to solve. Ie quantum encryption also includes key exchange (and in fact typically uses a one time pad for the data transfer).

You can't simply ignore the key exchange problems on the mathematicians side.

Perhaps the laws of physics that are supposed to protect quantum encryption will turn out to be false but based on our current understanding there is no better way to do it.

How is your mathemetician going to distribute his one time pad?

Re:Apples and oranges (2, Interesting)

Creepy Crawler (680178) | more than 6 years ago | (#23635323)

---How is your mathemetician going to distribute his one time pad?

A one time pad guarantees perfect secrecy. A QC channel allows secrecy as any "listening" devices become in part with the system, thereby allowing detection.

I do think this is a bit excessive by stating.. Data is always time-dependent. Therefore, we only need protect data for X amount of years.

What combination of encryption technologies can we use to make the data physically hard to crack? We need a multi-tiered encryption setup that uses multiple algorithms and multiple layers. Assuming mathematical proof of said encryption and no holes in implementation, we can calculate CPU years in brute-forcing each layer. Also, assuming that MIPS/s is increasing exponentially, we can calculate a "cracked by" date, and wrap said data in the date we need.

Having "perfect" data integrity and "perfect" communication seems... not right. It's just a gut feeling.

Re:Apples and oranges (0)

Anonymous Coward | more than 6 years ago | (#23635567)

We need a multi-tiered encryption setup that uses multiple algorithms and multiple layers.

Having "perfect" data integrity and "perfect" communication seems... not right. It's just a gut feeling.
Don't you have an mba class you're supposed to be at or something?

Re:Apples and oranges (1)

Creepy Crawler (680178) | more than 6 years ago | (#23637875)

Do YOU want to be the one to wager that quantum theory is 100% correct and possibly have your data compromised?

Re:Apples and oranges (1, Insightful)

Anonymous Coward | more than 6 years ago | (#23635545)

How is your mathemetician going to distribute his one time pad?


The same way you are going to tell the receiver of the QC message to use Quantum cryptography in the first place.

Lots of people seem to have this confused. Quantum cryptography does NOT give you a way to do secure key exchange without meeting the person you are going to communicate with. If you think about it for a second you will realize that this CANNOT be done no matter what encryption scheme, and no matter what "spooky action at a distance" you come up with. No matter what encryption scheme I use to communicate with you, I first need to agree with you to use that scheme, or for that matter, I have to get to know you for the concept of "you" to even make sense. If we have agree to an encryption scheme, we could just as well have exchanged one time pads, or RSA public keys, while doing it.

To put it this way, I made this post, but there is NO way for you to EVER determine who I was. Quantum cryptography won't do it because you don't know who you should tell to be at the receiving end. Asking Slashdot what is in their server log doesn't do it, because somebody could be doing a man in the middle between me and Slashdot right now. If 3 people came up to you tomorrow, each claiming to have made this post, there is fuck all you can do to determine who is telling the truth.

As a consequence, since you cannot even know who I am, you cannot possibly communicate securely with me in the future. After exchanging keys you can communicate securely with some person who may or may not be me, but you will never be able to know for sure it was that person who made this post.

Finally it is also worth mentioning here that QC, like the OTP, is limited in the amount of data you can send before meeting again to exchange entangled particles to be used for future communications. If it was not it would have an advantage over the OTP in that the OTP only lets you send so much data before you have exhausted your pad. As it happens, however, each entangled particle pair can be used only once, since any measurement destroys the entanglement.

Re:Apples and oranges (1)

locofungus (179280) | more than 6 years ago | (#23635753)

QC allows you to securely exchange a OTP. And it doesn't depend on using entangled particles at all.

Until Mrs. Tenney becomes sloppy. (1)

Mateorabi (108522) | more than 6 years ago | (#23635229)

You do realize that QC is just a method of securly distributing a one-time-pad between two endpoints, right? They don't use the photons to send the message data, that gets XORed later and sent via normal channels. So if everyone is wrong about quantum mechanics translates directly to "the OTP implementation is flawed". While OTPs are hard to implement (Where did I put that onionpaper again?) the whole point of developing QC is to get to the point someday where it IS practical/hasslefree to distribute the "pad". Ever faster and over longer distances.

Mod parent up... (1)

argent (18001) | more than 6 years ago | (#23636351)

... for making the comment I was about to make. :)

... at this level the weakest link (1)

Fallen Andy (795676) | more than 6 years ago | (#23635379)

is people. Either reusing an OTP or failing to RTFM for the QC equipment. It doesn't really matter that it would take e.g. the NSA longer than the time to the heat death of the universe to "crack" a cipher if $100k in a suitcase can "crack" Alice.

For a really really good look at security, try to track down the earliest black+white TV series of Mission Impossible - (almost no gadgets, lots of neat social engineering).

Andy

Actually, Quantum Cryptography Compliments OTP (1)

AlfredR (1095283) | more than 6 years ago | (#23636525)

The one-time pad requires a shared-secret key in order to be able to encode and decode encrypted messages. Sharing the key securely becomes a huge logistical problem.

Quantum cryptography promises, through quantum theory, that anyone trying to skim data from a secure channel ultimately corrupts it. So by measuring the noise level in the channel you can detect an eavesdropper.

A typical Quantum cryptography scheme requires two channels. One of the channels is a classical channel, like the internet, which is used to exchange the encrypted message. The other is a low noise quantum channel, which is capable of exchanging some kind of physical entity with information about the key encoded in its configuration. An example of such a physical entity would be a collection of polarized photons.

The rough idea is that you exchange the key over the quantum channel. If while doing that your "noise level" rises beyond a certain threshold, you abort the transmission. Otherwise the key to your OTP is now shared and you transmit your encrypted message over the classical channel.

I didn't miss your point, science isn't provably true and quantum mechanics may someday turn out to be wrong, exposing a loophole which allows for eavesdropping. But quantum cryptography isn't as much of a cryptography scheme as it is a transmission vehicle.

Still, if you want to find a flaw with quantum cryptography, you don't have to look very hard. Quantum cryptography assumes that your man in middle just wants to read data out of the channel without breaking the link. It is easily defeated if someone can make themselves into a relay.

i.e. Quantum cryptography is defeated if an eavesdropper cuts both the quantum and classical channels and inserts herself into the middle:

Sender ===>>=== Attacker ===>>== Receiver

where she pretends to be the receiver to the sender and rebroadcasts the message as if she were the sender to the receiver.

Re:Not the most secure (1)

blueg3 (192743) | more than 6 years ago | (#23637191)

Of course, with encrypted data, you need to ensure that nobody in the future will be able to defeat the cryptography.

The process of quantum "cryptography" is such that you need to ensure that nobody *now* (when you are transmitting the data) will be able to eavesdrop successfully.

Quantum hardware DRM... (0)

rastoboy29 (807168) | more than 6 years ago | (#23635093)

...would suck.

Re:Quantum hardware DRM... (1)

spazdor (902907) | more than 6 years ago | (#23635423)

Whoa boy, you are right about that.

Why is this practical? (3, Insightful)

Johnno74 (252399) | more than 6 years ago | (#23635161)

I'm all for R&D into pure science, and I'm not bagging the concept of quantum cryptography, but why does this need to be a commercial product?

Is there really anyone out there paranoid enough to need/want this besides various three-letter agencies? Maybe this is proveably secure, we think, but what is more likely - Someone finds a loophole in the very weird world of quantum mechanics that makes quantum cryptography as we know it obsolite, or someone figures out a way to find prime factors of obsenely large numbers in a reasonable time.

This article is about how it may be possible have a quantum crypto setup with a bandwidth of maybe 1024kbps by spending only $20k-$50k on one component to the system. I bet there is a lot of other components.
Compare this with a basic commodity PC, which can could encrypt 1024kbps using AES with ridiculous ease.

Re:Why is this practical? (2, Insightful)

mrcaseyj (902945) | more than 6 years ago | (#23635333)

I was thinking that it would be foolhardy to trust that some discovery in quantum physics would not render quantum cryptography insecure. But then I realized two things. First, you can always just use QC to wrap conventional cryptography so you get the security of both. Second, conventional cryptography can be eavesdropped on and recorded to be broken someday when weaknesses or more computational power is available. With QC your communications can't even be recorded for future breaking until some new physics comes along.

Re:Why is this practical? (1)

Chuck Chunder (21021) | more than 6 years ago | (#23635355)

but what is more likely - Someone finds a loophole in the very weird world of quantum mechanics that makes quantum cryptography as we know it obsolite, or someone figures out a way to find prime factors of obsenely large numbers in a reasonable time.


Given that we don't know it makes sense to have both.

Advances in quantum computing may make the factoring problem an easy one.

Of course the commercial applications are rather niche right now and cost is no small part of that, but how many things often start that way, including your commodity PC?

Re:Why is this practical? (1)

MobyDisk (75490) | more than 6 years ago | (#23639635)

The part tht scares me isn't the power of quantum cryptography. It's the power of a quantum computer to decrypt classical cryptography. It really does look like the end to privacy might be just around the corner.

Researchers Simplify Quantum Cryptography (1)

dgun (1056422) | more than 6 years ago | (#23636049)

That's a relief. I was worried there for a while.

th9is FP f0r GNAA (-1, Troll)

Anonymous Coward | more than 6 years ago | (#23636203)

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This isn't news for QKD researchers (0)

Anonymous Coward | more than 6 years ago | (#23636285)

As someone who has did several years of research in QKD (aka Quantum Cryptography), implementations of QKD that only use one detector have been around since the 90s, and the awareness that it is possible to implement them have probably been around since the 80's. I'm too lazy to get a source if someone else wants to get the karma. So this article isn't entirely news and is there is probably some artificial hype added to it.

QKD works by making measurements of the Quantum basis that a sender sent, and generally in optical schemes you need two or four photon detectors on an optical interferometer to detect each basis. If you reduce the number of photon detectors to one, you can only detect one basis. However, the mechanisms involved allow you to still recover data and keep the key exchange secure. It isn't a fundamentally novel concept to anyone who knows the basic theory of implementing a real world QKD system.

I haven't worked on QKD in a few years, so my knowledge of the theory might be rusty, so feel free to correct me if anyone knows better.

Thanks to this new breakthrough... (5, Funny)

Ihmhi (1206036) | more than 6 years ago | (#23636535)

...quantum cryptography now requires 30% less cats and 46% fewer radioactive isotopes.

50k? no. 5k! (1)

Briden (1003105) | more than 6 years ago | (#23638897)

Article Summary states "Four single-photon detectors are usually required (these cost $20K to $50K each)"

sounds expensive.

except, the real article states: "Bob uses four single-photon detectors, costing approximately $5,000-$20,000 each."

still pretty expensive, but it sounds like you could have a working one of these for only 10k in detectors!

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