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144 comments

American Hero. (5, Interesting)

Whiney Mac Fanboy (963289) | more than 8 years ago | (#15138002)

Ed Felten is a true American hero - he's:
* Convinced the Music Industry watermarking is unworkable (saving us from poor quality files)

* Testified against predatory monopolists as a witness for the US govt.

* Exposed holes in Sony's "fix" for XCP malware CDs (that turned out to be almost as dangerous as the original rootkit)

* Given us the memorable quote Given a choice between dancing pigs and security, users will pick dancing pigs every time.
(gleaned from wikipedia) [wikipedia.org]

Also - anyone thinking the 40 'conspiring' devices makes it impractical to break HDCP/HDMI - think again. It just means 40 (or less) like minded hackers have to get together - not particularly hard to imagine these days.

Re:American Hero. (1)

GeorgeMonroy (784609) | more than 8 years ago | (#15138071)

Who doesn't like pigs? Of course they would choose pigs over security.

I would do it (3, Funny)

eclectro (227083) | more than 8 years ago | (#15138003)


But I don't have room for the forty big-screen TVs.

Re:I would do it (4, Funny)

gEvil (beta) (945888) | more than 8 years ago | (#15138011)

That's okay. You can store them here at my place.

Re:I would do it (1)

cortana (588495) | more than 8 years ago | (#15138047)

The Internet does though.

Half a building connected to the same set-top-box (0)

Anonymous Coward | more than 8 years ago | (#15138189)

It wasnt rare to find a building with half of the neighbors connected to the same set-top-box when PPV arrived. ;))

Re:I would do it (0)

Anonymous Coward | more than 8 years ago | (#15139014)

But they're only a few inches thick.

Where did you get 40? (-1)

Murphy Murph (833008) | more than 8 years ago | (#15138012)

From the article:
This sounds pretty cool. But it has a very large problem: if any four devices conspire, they can break the security of the system.

Re:Where did you get 40? (1)

Murphy Murph (833008) | more than 8 years ago | (#15138022)

NVFM

Re:Where did you get 40? (-1, Redundant)

Anonymous Coward | more than 8 years ago | (#15138030)

In the real system, where the secret vectors have forty entries, not four, it takes a conspiracy of about forty devices, with known private vectors, to break HDCP completely. But that is eminently doable, and it's only a matter of time before someone does it. I'll talk next time about the implications of that fact.

Re:Where did you get 40? (2, Informative)

Anonymous Coward | more than 8 years ago | (#15138041)

From TFA:

In the real system, where the secret vectors have forty entries, not four, it takes a conspiracy of about forty devices, with known private vectors, to break HDCP completely. But that is eminently doable, and it's only a matter of time before someone does it. I'll talk next time about the implications of that fact.

 
Four was an example for the article.

Re:Where did you get 40? (-1, Redundant)

Anonymous Coward | more than 8 years ago | (#15138043)

In the real system, where the secret vectors have forty entries, not four, it takes a conspiracy of about forty devices, with known private vectors, to break HDCP completely.
I managed to decypher this information by reading the article all the way through. Now did I already break the DMCA?

Re:Where did you get 40? (1)

sholden (12227) | more than 8 years ago | (#15138045)

Try reading the rest of the article.

A little tougher than that... (5, Interesting)

weetjerm (637949) | more than 8 years ago | (#15138049)

His attack methodology is correct, but it will take more than 40 devices to break the system. The chances are very low that all 40 devices being linearly independent, and therefore each one offering non-duplicate information about the system. If you read the comments, he actually inadvertantly ran into this problem with his small example of 4 keys.

However, in writing this, I realize that I do not know how many keys you would need to present a good probability of solving the system of equations. Anyone want to run a simulation?

Re:A little tougher than that... (1)

bperkins (12056) | more than 8 years ago | (#15138227)

You're right, I get 80 devices to get a 50/50 chance.

OTOH, since the addition rules are public, you can target your cracking to devices that have the types of keys you want.

Re:A little tougher than that... (4, Informative)

Maljin Jolt (746064) | more than 8 years ago | (#15138229)

Anyone want to run a simulation?

No funny simulation is needed, a math paper refered by TFA contains the info you want: 50 KSV's have probability 0.999, by the properties of linear algebra over Z/2exp56Z.

Exactly. Ed's math is borked. (2, Insightful)

goombah99 (560566) | more than 8 years ago | (#15138293)

I had exactly the same thought. I think this attack may fail. Or rather not be as immediately successful as imagined. Ironically, the fatal flaw is contained in the same algebra mistake made in the orginal post.

In order to prevent this attack from being done easily, the central authority could deliberately hand out linearly dependent addition vectors to any company that applies. For example, suppose a company applies for 10,000 keys. The central authority gives them 10,000 keys and 10,000 addition vectors. But the addition vectors are all crammed into the first 14 or 15 bits of the 40 bit addition vector. (that is bits 16 to 40 are zero). This would assure that the addition vectors are linearly dependent and the code cannot be cracked.

In effect the 10,000 keys are hobbled to representing no more than 15 independent keys, not the requisite 40 to crack this.

Thinking even more globally, the central authority could reserve say the last 10 bits of the addition vector, so that all devices manufactured from 2008 to 2010 never used the last 10 bits. then all devices manufactured from 2010 to 2012 always used the 31st bit but none of the last 9. Then in 2013-2014, all devices always use the 32nd bit but none of the last 8. and so on.

thus they can prevent anyone from collecting all 40 so far into the future that they can assure that any crack that works this year will fail on all new devices.

Of course, the hackers only need to stay on the ball and update their hacks as they can. But it's going to take a very large consipiracy among multiple companies to collect large enough set of addition vectors to crack this.

Re:Exactly. Ed's math is borked. (1)

Mattcelt (454751) | more than 8 years ago | (#15138480)

Ok, so help me out here. Doesn't that reduce the effective keyspace by an order of 2^16? Seems to me that would make a brute-force attack much more practical. (It doesn't matter if you set the first 16, last 16, or any arbitrary (but consistent) combination of bits to zero, it will still reduce the keyspace for all devices by the same amount.)

Of course, I don't know much about the algorithm itself, but from the blog's example, it should be simple to test the validity of any arbitrary key with any device.

Increasing the keyspace as you have suggested would actually make the codes more secure as time went on - but given that there are always going to be those initial devices with 24 instead of 40 bits, those will always be the most attractive target. And it only takes one key, one time, to build a device which can output an unencrypted stream, breaking the whole system catastrophically.

Have I got that right?

Re:Exactly. Ed's math is borked. (1)

goombah99 (560566) | more than 8 years ago | (#15138570)

You have it partly right and partly wrong.
First, HDCP does not require super security. It's not how the media is encoded it's just the transport from the player to the viewer that is being encoded. There's a whole nother more secure code for the media encryption. I think what they want to avoid is some gizmho you could put inline that would decode it. SO if they can create a situtation where there is no universal gizmho for every player/viewer combination or one that breaks every year when a new device is released it accomplishes a lot of their purposes.

One supposes that the point of attack has to be outside the media player (dvd) since otherwise there is no need to attack the transport layer and you already have everything you need to decode the video if you are in controll of the inner workings of the player.

So in trying to attack the transport layer there's no reduction of the complexity of the key by restricting the addition vector to a subset of the possible bits. In general it's always going to be about half the bits (half on half off).

By restricting the addition key space its sort of like restricting the space of challenge codes to a challenge response algorithm. The main effect of this is to prevent a challenge code from being seen previously and thus the response learned.

Of course it does, as you surmise, reduce the brute force number of challenge codes one might try to learn every possible attack for that series of player. But I suspect the set is still so large it matters not. And moreover, as I said, that still wont let you build a universal decoded gizmho, just one that works for that particular player for that model year.

Of course for some folks that's all the want. e.g. if it becomes known that there is a gizmho that can be attached to a 2007 sony model XXXXX that can then be spoofed with a certain addition vector then all the hackers will go out and buy that 2007 model which will then work indefintely. But one guesses that maybe the media will then come with something that recognizes that model number and refuses to play in high def. Not sure if they could get away with that as it would piss off some consumers.

Re:Exactly. Ed's math is borked. (0)

Anonymous Coward | more than 8 years ago | (#15138624)

a whole nother ???

Re:Exactly. Ed's math is borked. (1)

Don_dumb (927108) | more than 8 years ago | (#15138762)

But one guesses that maybe the media will then come with something that recognizes that model number and refuses to play in high def. Not sure if they could get away with that as it would piss off some consumers.
Not to mention the manufacturer, I cant imagine Sony being too happy when Fox puts a "cannot be played on Sony xxxxxx players" on its media, as consumers may buy another player instead. If this was to be attempted then we could see a wonderful end to the HDCP madness as Sony (or another player maker, of course) would send in their lawyers to stop anyone "blacklisting specifically their equipment".

And of course, I can only hope this happens as the only way to prevent all erosion of our consumer rights is to let the big corps fight amongst themselves. The battle between HD-DVD and Blu-Ray is one example (I hope they both lose).

My math is borked, too. (1)

Mattcelt (454751) | more than 8 years ago | (#15139101)

I see what you mean, that makes more sense.

So let's say, for the sake of argument, that the whole keyspace is tested; i.e., that for an arbitrary key that you create you have gathered the entire range of challenge responses from a particular device and stored each. Is an addition vector an NP problem that wouldn't give up the secrets of the key itself even if all the challenge responses were known?

It would seem that it must be to serve the intended purpose. It's much more damaging to be able to spoof a particular device to other devices than to spoof a response sequence with a single devices, yes? That way you could sell a device to any user that emulates a "2007 sony model XXXXX" to any other device to decrypt the stream in real-time, versus having to buy a "2007 sony model XXXXX" to work with the theoretical gizmo. But all of that would rest on the ability of the addition vector to be reverse-engineered, which I must confess I'm ignorant about.

Re:My math is borked, too. (1)

goombah99 (560566) | more than 8 years ago | (#15139576)

close but not quite.

Here's how spoofing would fail. Suppose I tell a new device I'm a a sony xxxx and my addition key is 1,4,7, ... etc and it omitts the last ten bits. Okay that half of the process works. but then the player replys, I'm a panasonic yyyy and my addition key is 1,3,15,...39,40.

Now you're screwed because your spoof device does not know what the keys for 39 and 40 are.

Thus you can't work with the new device. You CAN work with any old device whose subspace of addition keys you have mapped, but not any new device.

Finally just for completeness note that when I say certain bits are held back, that's a simplification. What I mean is that certain basis vectors are held back. Thus to make the point. if every time 39 appears, 40 were also to appear in the addition vector then you can never reverse engineer what 39 and 40 are in the key. you can only figure out what 39+40 are. Thus this talk of certain bits being held back is just for pedagogical simplification.

Here's what will happen (2, Insightful)

Omaze (952134) | more than 8 years ago | (#15138615)

Someone will connect an oscilloscope to the wire(s) that connect(s) the devices and reverse engineer the communications signal. They will then construct a custom breadboard able to talk to any HDCP device while being able to impersonate a device with a programmable HDCP vector/rule. With a link (ethernet or serial) to any modern day PC they'll just brute force it.

It won't be difficult.

Re:Here's what will happen (1)

MoonBuggy (611105) | more than 8 years ago | (#15138837)

Someone's already mentioned this scenario in the comments on the blog, it seems plausible in theory but there's also very little reason for the HDCP chips not to limit handshake attempts to (say) one per second - you're not going to get more attempts than that legitimately anyway. Since the keys are 56 bit numbers and you're adding them together you've got a fair amount of ground to cover - it's going to take a hell of a lot of time going through x1+x2=1; fail; x1+x2=2; fail; ... x1+x2=379654; pass; x1+x3=1; fail; and so on. If my calculations are correct (which they are quite possibly not) you're talking anything up to centuries to brute force 40 56-bit keys at that speed.

Re:Here's what will happen (1)

name773 (696972) | more than 8 years ago | (#15138979)

each device only has one key vector and addition rule. why would it handshake more than once per device?

Re:Here's what will happen (2, Insightful)

tadmas (770287) | more than 8 years ago | (#15138970)

Someone will connect an oscilloscope to the wire(s) that connect(s) the devices and reverse engineer the communications signal.

There is no need to do this -- the signal itself would have to be according to some kind of standard or else a brand X DVD player couldn't work with a brand Y television. Just look up the communications protocol.

With a link (ethernet or serial) to any modern day PC they'll just brute force it.

Riiiiight. The DVD's addition rule is [1]+[3] and the TV's is [6]+[17]. What's our secret key? It could be 24 (7+17 and 9+15) or 57 (17+40 and 56+1) or 29387 (12412+16975 and 19280+10107).... Each is equally likely, so yes you could brute force it, but if the actual keys are big enough, it would take a Really Long Time to do it. This is the idea behind just about all forms of modern encryption; they can be broken by brute force, but it takes so long it's not worth it.

Could this be broken on a modern PC? Assuming you could easily verify that you got the unencrypted form and the secret keys are 17 decimal digits, then on average it would take you 5e17 guesses to brute force it. If you assume checking 1,000,000 per second, that's 5e11 seconds > 15844 years. Don't hold your breath.

This is why the attack in TFA is useful. Instead of having to try billions of possible keys, you can algebraically figure out a secret vector, so then cracking the encryption is a simple elementary school addition problem. Solving a set of linear equations to get the secret vector can be done in slightly less than thousands of years.

It won't be difficult.

Yes, it will. That's just like saying "cracking RSA is super-easy because it's just finding the prime factors of a number!!!!!!!11!!1one" So, why can't anyone with a modern PC bring RSA to its knees? After all, when you publish your public key, you're also publishing your private key, too.... if someone can figure out the factors of your modulus. You can just brute force it -- it won't be difficult.

Re:Here's what will happen (1)

Omaze (952134) | more than 8 years ago | (#15138993)

If they can't get to it from the case connector they'll open the box and find a different set of wires on the circuit board to tap into. Yes, it'll take more research into the chips on the board but eventually a weak point will be found.

It's been going on for centuries. Keep arguing. Unless you're willing to bet that HDCP will be the be all and end all of encryption methods and no other better method will ever be needed then you'd best just pack up and shut up now. If you are willing to bet on it then I'll be more than happy to take you up on your offer of free money.

Re:Here's what will happen (1)

tadmas (770287) | more than 8 years ago | (#15139075)

If they can't get to it from the case connector they'll open the box and find a different set of wires on the circuit board to tap into. Yes, it'll take more research into the chips on the board but eventually a weak point will be found.

Well, duh. The point is to prevent a descrambling device in the middle that end users can use, such as the cable descramblers that are used today. If you could descramble at will, you can copy the HD content all you want. However, most end users won't take apart their new high definition DVD player and start hooking up wires.

It's much easier to stop a handful of people that mass-produce illegal copies than it is to stop millions of end users from making just a few copies.

I never said this was a particularly good encryption system, either. I just pointed out that it's naive to think "we'll just brute force it, and it'll be easy".

Nope (1)

goombah99 (560566) | more than 8 years ago | (#15139645)

No this scheme won't work. Here's why.

the keys are never transmeitted only the addition rules. So here's a hypthetical exchange

device 1: my addition rule is 17+13
device 2: my addition rule is 24+5
device 1: okay I computed the secret= key[24]+key[5] (which I alone know)
device 2: okay I computed the secret = key[17]+key[13] (which I alone know)

at this point both secrets are the same but neither secret has appeared on any tapable wire.

now dev1 says:
dev1: youre challenge is to encrypt this number: rand = 1380912
dev2: my resonpsne is theat encrypting 1380912 with my secret key gives 478120181
dev1: hey that's right, I was able to check that using my secret
dev2: youre challenge is to encrypt this number: eand = 18171710 ...

and so on.

now each device has poven to each other they share the same secret key but they have never transmitted it.
You cna't memorize the transaction pattern for two reasons. 1) the random challenge will vary even if the addition keys dont

and any time you connect a new device the addition keys will change.

Re:Exactly. Ed's math is borked. (1)

doormat (63648) | more than 8 years ago | (#15139254)

But the addition vectors are all crammed into the first 14 or 15 bits of the 40 bit addition vector. (that is bits 16 to 40 are zero). This would assure that the addition vectors are linearly dependent and the code cannot be cracked.

Didnt the article say that the vectors always have 20 1's and 20 zeros? Doesnt that limit the permutability of the vector?

Also, if you were to hand vectors out 10,000 keys like that to one manufacturer, woudln't you only need 14 or 15 of those types of devices to conspire to break the system? You could essentially break any device of that manufacturer (or whoever made the internals) with a fewer number of devices.

Re:Exactly. Ed's math is borked. (1)

goombah99 (560566) | more than 8 years ago | (#15139552)

okay then 20 not 15. whatever. they just don't release the full basis to any vendor. then you cant universally reverse emgineer it.

and no. you are confusing devices with dimensionality. a 20 dimensional spaces spans much more then 10,000 devices.

The problem with any security system is ... (0)

Anonymous Coward | more than 8 years ago | (#15138050)

security vs. ease of use. You can make something so secure that it is almost impossible to break. The trouble is that it becomes so cumbersome that no one wants to use it.

The example that comes quickly to mind is copy protection on software. At some point it drives away paying customers and doesn't deter the pirates.

Personally, I think I will continue to use the analog hole because there isn't that much stuff that really needs high definition to be enjoyed.

Cool, but nor practical (1, Insightful)

pla (258480) | more than 8 years ago | (#15138065)

if any 40 devices conspire together, they can break the security of the system

From TFA:
it takes a conspiracy of about forty devices, with known private vectors, to break HDCP completely. But that is eminently doable, and it's only a matter of time before someone does it.
Apparently Mr. Felten has a somewhat twisted idea of "eminently doable".


The HDCP CA will certainly only give out keys to people who sign very very scary agreements not to engage in exactly the sort of activities described. While a few of them might "accidentally" leak their keys, I find it exceedingly unlikely that 40 such companies will pay for a key vector, just to take the risk of getting sued out of existence.


Though I have to wonder about the actual security of these keys under the condition of physical access. That point might make Felten's proposed crack viable, if we just need to find a weaknedd in 40 devices out of the thousands that will eventually hit the market - ESPECIALLY if player software needs to have a valid key as well.



I also wonder why we need to "know" even one, much less 40, secret keys beforehand, however... It doesn't sound like you need to come up with the correct answer to get a single response. If you faked 40 devices, couldn't you still get the target device to respond at least once to each, thereby getting the necessary 40 unknowns? Sure, this would reduce to 40 instances of cracking a 56(?) bit key, but a modern PC can brute-force that in under a day.

Re:Cool, but nor practical (1)

Midnight Thunder (17205) | more than 8 years ago | (#15138161)

Most things are doable, though not necessarily in a lifetime. I am sure you could insert a sniffer device to monitor the data going through the cable. Also, apparently this technology will only prevent you access from the HD content. Maybe like aeroglass, the low quality content will be enough for many people.

Re:Cool, but nor practical (1)

Firehed (942385) | more than 8 years ago | (#15138429)

Maybe like aeroglass, the low quality content will be enough for many people.
But Aero Glass is the fully pretty one - you must mean plain Aero. Anyways, the whole purpose of buying HD media is for the HD. If it's then downscaled right back to just-slight-above DVD quality, I think people are going to be, pardon my French, pretty fucking pissed. Especially the early adopters who have the highest chance of getting screwed over.

Re:Cool, but nor practical (1)

jamesshuang (598784) | more than 8 years ago | (#15138168)

The cipher is probably based on matricies (maybe even some sort of advanced hill cipher?). With 40 known matricies, it's merely a matter of multiplying them with the cipher text (or however it's encoded), and the main key pops out. That's why exactly 40 are needed - it's mathematics.

Re:Cool, but nor practical (0)

Anonymous Coward | more than 8 years ago | (#15138211)

You misunderstood the method of attack. The secret vector, which is stored in the device, is not used directly in this crack, but is calculated. The secret vector turns out to be a solution to a set of linear equations as snooped from HDMI handshakes. To get a single secret vector, 40+ HDMI device handshakes need to be snooped, with atleast 40 of those having unique non-linear-combination secret keys. This is why 40 devices are needed, though they don't need to be in a chain to crack the encryption as suggested in the story.

Re:Cool, but nor practical (2, Insightful)

Anonymous Coward | more than 8 years ago | (#15138221)

I find it exceedingly unlikely that 40 such companies will pay for a key vector, just to take the risk of getting sued out of existence.
According to the article, keys are being sold in quantities of 10000, which makes it sound like each physical device has its own unique key. If this is the case, then one not-quite-tamper-proof production run of some player will yield more than enough keys for the attack to be practical.

Re:Cool, but nor practical (1, Informative)

Anonymous Coward | more than 8 years ago | (#15138251)

You don't need a license to obtain the secret keys. You can create your own thus making the approach extremely doable. Please read the article to see how this is done.

Re:Cool, but nor practical (1)

johndoe42 (179131) | more than 8 years ago | (#15138436)

First, the HDCP CA gives a lot of keys to each company, I think. So you'd only need one crooked company.

About your other idea: From the paper referenced in the article, it looks like the device sends a hash of the sum over the wire. So you'd have to invert a hash on each try (which may still be doable -- the input space isn't all that huge). But the attacker can cleverly choose a basis for the KSV space, thereby recovering the target's private key in exactly 40 tries. This attack would probably take a week or two on computation, so it wouldn't be all that great on its own. But... an attacker could run this attack 40-50 times and use the results to run the attack in the article, breaking the entire system. No broken devices required. Eenteresting.

Re:Cool, but nor practical (1)

pkhuong (686673) | more than 8 years ago | (#15138892)

Well, the hash is lossy (56 -> 16 bits, iirc), so you'd probably need ~4 attempts of the same challenge with 4 different seeds to recover the sum. Still very much in the realm of the doable.

Re:Cool, but nor practical (1)

johndoe42 (179131) | more than 8 years ago | (#15139372)

I think you mean 2^(56-16) = a lot of attempts. Unless there's corresponding weakness in the hash.

(I didn't realize it was a hash that short. But 16 bits sounds absurd -- the hash gives the shared secret and 16 bits is way too short.)

Re:Cool, but nor practical (5, Informative)

quentin_quayle (868719) | more than 8 years ago | (#15138676)

Did the moderators Read The Fine Article before giving the parent points?

Felten in talking about "a conspiracy of about forty devices" is not saying that (defectors at) forty device makers have to reveal secret keys. What he's saying is that you just need to the 40 devices themselves, or rather (as post above pointed out) enough to get 40 different key sets (and some math and programming ability). Then the crack is done by analysing the bit streams between the devices (between player and display, or whatevre).

The expense is the cost of all those tvs and players. Bribing the device makers is a *different* kind of attack which Felten rules out as impractical.

Re:Cool, but nor practical (1)

pla (258480) | more than 8 years ago | (#15139076)

Did the moderators Read The Fine Article before giving the parent points?

Did you? Or did we somehow read entirely different articles?


Felten in talking about "a conspiracy of about forty devices" is not saying that (defectors at) forty device makers have to reveal secret keys.

The linked article specifically says exactly that! The described attack requires knowing the key vector of each of the 40 devices used in the attack:
There are two things to notice about this process. First, in order to do it, you need to know either Alice's or Bob's secret vector.
[...]
In the real system, where the secret vectors have forty entries, not four, it takes a conspiracy of about forty devices, with known private vectors, to break HDCP completely.



Then the crack is done by analysing the bit streams between the devices (between player and display, or whatevre).

Really, now? Perhaps you could quote where he says that? Because, I can't help but notice that it says NOTHING about analyzing the conversation itself. In my last paragraph, I hypothesized that a brute-force attack on the actual conversation might suffice, but Felton said nothing at all about that. In fact, to apply the method he describes, you don't even need to ever build the devices - You just need to know their keys and the victim-device's addition rules.

From that, you can solve a 40-variable linear equasion to produce arbitrary valid keys, which comes as close to a full crack as matters for any practical application.


Now, I did not know, as one or two others pointed out, that anyone can obtain huge numbers of keys without significant expense or contractual restrictions. That would seem to make Felten's attack trivial, and if true, I look forward to buying a black-market HDCP-stripping dongle in the very near future. I admit my lack of information on that point. But the points you take issue with don't even seem to come from the current topic!

Re:Cool, but nor practical (1)

imaginieus (897756) | more than 8 years ago | (#15139731)

All it would take to crack HDCP is a single person at one of these companies to obtain 40 keys and then run the crack himself. He would then be able to produce 50 new, untraceable keys that could be released publicly.

this seems unwise (-1, Troll)

Anonymous Coward | more than 8 years ago | (#15138073)

Someone needs to take this information down before the terrorists see it.

Sites like this one are literally handing the enemy weapons to use against us. This site needs to come down - if they won't do it voluntarily, then perhaps law enforcement needs to get involved.

Re:this seems unwise (0)

Anonymous Coward | more than 8 years ago | (#15138153)

If the terrorists want to crack HDCP, I'm all for it as long as they release it to the general public... which they probably wouldnt being terrorists and all. They would reprogram their pre-hdcp complient HDTV's to get full high def channels. those bastards.

Why Reveal this Now? (2, Interesting)

PingXao (153057) | more than 8 years ago | (#15138074)

As a poster said at TFA, why did they reveal this attack so soon? It would have been much better to wait another few months until HDCP displays and video cards were shipping in larger numbers. That being said, who's comes up with these lame cryptosystems anyway? First CSS, which was a joke, now this, and you know the Advanced CSS will have holes in it big enough to drive a truck through. The bad news is that some day they will start hiring people who know what they're doing with cryptosystems and then we're all screwed.

Re:Why Reveal this Now? (0)

Anonymous Coward | more than 8 years ago | (#15138091)

If you read the entire set of blog comments, you'll see that this attack has been known for over 15 years - see http://www.freedom-to-tinker.com/?p=1005#comment-2 6675 [freedom-to-tinker.com]

Re:Why Reveal this Now? (0)

Anonymous Coward | more than 8 years ago | (#15138092)

I dont think it is a case that they do not know what they are doing. The problem is virtually impossible to solve with todays technology (which is why there is a push for Trusted Computing).

Unless you can point to another solution which stops someone copying media that they have physical access to.

Even if the copy protection is really really strong, it is only a matter of a few years before those systems can be brut forced.

Re:Why Reveal this Now? (0)

Anonymous Coward | more than 8 years ago | (#15139562)

Unless you can point to another solution which stops someone copying media that they have physical access to.

This ones easy but you wont like it: Make blank media cost more than the actual product meaning that there is no incentive to copy it.

Re:Why Reveal this Now? (0)

Anonymous Coward | more than 8 years ago | (#15138104)

Read the replies to that post as well. It's been "revealed" for several years now, yet the industry went ahead and implemented this.

Re:Why Reveal this Now? (5, Interesting)

Anonymous Coward | more than 8 years ago | (#15138197)

The bad news is that some day they will start hiring people who know what they're doing with cryptosystems and then we're all screwed.


Rather unlikely. The whole concept of DRM is bankrupt as a cryptographic concept because you are handing over the ciphertext, the plaintext and last but not least the key over to your adversary (usually called "consumer" or "hacker"). Sure you can try to make it hard for him to actually get them but you already handed them over and it just remains a question of time until they are recovered.
Meanwhile, a single break is a class break for at least all the content released up to the point of the break (even with "revokable" keys). Also, once a broke the system once, the content is freed forever and can be distributed at leisure (darknet hypothesis), which means even some small quality loss may be acceptable to the attacker since that loss would only occure once.

In short, DRM is a DReaM indeed.

Re:Why Reveal this Now? (1)

PingXao (153057) | more than 8 years ago | (#15138747)

Interesting point. Why make it as AC? I wouldn't have even seen it except I wanted to read the replies to my post.

Re:Why Reveal this Now? (0)

Anonymous Coward | more than 8 years ago | (#15138343)

It was revealed to Intel by a Dutch researcher, but Intel dismissed it as only theoretical. He didn't go public because he didn't want to have to worry about being arrested under the DMCA when visiting the US.

Re:Why Reveal this Now? (1, Insightful)

Anonymous Coward | more than 8 years ago | (#15138698)

As others have pointed out, the attack is not new. What HDCP does is *not* protect content (at least, not seriously)... it forces the makers of consumer electronics to sign legal agreements with Intel, and more critically with the MPAA... and these legal agreements dictate what features the manufacturers can add. If you want to sell players legally, you have to make them they way you are told... not the way the consumer wants.

It's about control, not copy protection (can't fast forward through adverts etc etc)... and getting your sticky royalty grabbing fingers into the equipment pie.

Re:Why Reveal this Now? (1)

TwilightSentry (956837) | more than 8 years ago | (#15139627)

It doesn't matter how strong the crypto is; the real purpose is to allow the content industry to sue the heck out of anyone (In the US) who tries to excersize fair use. The DMCA doesn't care whether the crypto is strong or weak...

It's 4 not 40 (0)

symphara (225088) | more than 8 years ago | (#15138083)

From the article:

"This sounds pretty cool. But it has a very large problem: if any four devices conspire, they can break the security of the system.

To see how, let's do an example. Suppose that Alice, Bob, Charlie, and Diane conspire, and that the conspiracy wants to figure out the secret vector of some innocent victim, Ed. Ed's addition rule is "[1]+[4]", and his secret vector is, of course, a secret."

Re:It's 4 not 40 (0)

Anonymous Coward | more than 8 years ago | (#15138096)

RTFA, fucktard. "In the real system, where the secret vectors have forty entries, not four, it takes a conspiracy of about forty devices, with known private vectors, to break HDCP completely."

Re:It's 4 not 40 (0)

Anonymous Coward | more than 8 years ago | (#15138108)

His example had vectors with four entries, real devices have vectors with forty entries.

Re:It's 4 not 40 (1)

Maru Dubshinki (804451) | more than 8 years ago | (#15138116)

"An example will help to make this clear. In the example, we'll save space by pretending that the vectors have four secret numbers rather than forty, but the idea will be the same."

'Nuff said.

No, it's 40, not 4 (4, Informative)

Space cowboy (13680) | more than 8 years ago | (#15138124)

In real life the devices have a vector of 40 secret numbers, he's using a vector of 4 to illustrate withour bogging down the reader.

The key is that with N variables (the number of different numbers in the vector), you need N equations to solve the set of equations for all of those variables - it's simple linear algebra.

When you purchase a licence, you get a bunch of 10000 keys for $16000, so S.O.Mebody could use this within an organisation to analyse the generation matrix, and actually produce 40 new keys and release them to the wild. No comeback.

Simon

Re:No, it's 40, not 4 (1)

kanweg (771128) | more than 8 years ago | (#15138266)

In another post, Weetjerm wrote "His attack methodology is correct, but it will take more than 40 devices to break the system. The chances are very low that all 40 devices being linearly independent, and therefore each one offering non-duplicate information about the system. If you read the comments, he actually inadvertantly ran into this problem with his small example of 4 keys."

So, what they could do is sell you 10.000 linearly dependent keys.

Bert
A patent lawyer who detests software patents and DRM that punishes honest people only

Re:No, it's 40, not 4 (1)

phoenix.bam! (642635) | more than 8 years ago | (#15138715)

If they are linearly dependent keys you can still crack the subspace those keys span and access any media coming from those 10,000 devices.

Re:It's 4 not 40 (1)

run4ever79 (949047) | more than 8 years ago | (#15138147)

His example was 4, he goes on to say that the actaul vector size is 40.

Re:It's 4 not 40 (0)

Anonymous Coward | more than 8 years ago | (#15139376)

Read the article, imbecile!

Wait wait wait (1)

neonenergy (888041) | more than 8 years ago | (#15138193)

He assumes that all the keys are different.

Well lets say a company had 40 keys... but they all have the same addition formula. What now? Everything would come out the same.

In a related question... (2, Interesting)

dpilot (134227) | more than 8 years ago | (#15138213)

I was checking the Sunday advertising fliers this morning, and see that many of the new TVs are advertising HDMI as well as PC connections. Can someone please explain my limitations?

1: Can I hook up my current VGA or DVI to one of these, and display the content I can currently display?

2: Is the only limitation/constraint the new HD/BlueRay DVDs with "double-plus-good super-duper copy-protection, put there to protect me AND the children"?

3: Related to both, assume I have MythTV running with an HD capture card. (I don't yet, but plan to, before they become illegal. What's the latest status?) Can I run my captured content out through one of these new displays?

Re:In a related question... (1)

The Jonas (623192) | more than 8 years ago | (#15138307)

1: Can I hook up my current VGA or DVI to one of these, and display the content I can currently display?
I can only help answer your first question. I bought a 32" LCD with multiple inputs including HDMI for for my PC's. I have yet to find a graphics card that is HDMI compliant. Therefore, at this time I can not use the 1920 x 1080i @ 60Hz that the display can handle. I am using the RGB-PC inputs. There may be a card, but I have not found it yet.

Re:In a related question... (1)

makomk (752139) | more than 8 years ago | (#15138443)

I can only help answer your first question. I bought a 32" LCD with multiple inputs including HDMI for for my PC's. I have yet to find a graphics card that is HDMI compliant. Therefore, at this time I can not use the 1920 x 1080i @ 60Hz that the display can handle. I am using the RGB-PC inputs. There may be a card, but I have not found it yet.

Try a graphics card with a DVI out - you should generally be able to connect a DVI out to a HDMI in. However, you can only connect a HDMI output to a DVI input if the video isn't copy-protected or the device you're using supports HDCP on its DVI input...

Re:In a related question... (3, Interesting)

frzndrag (252873) | more than 8 years ago | (#15138481)

HDMI compliance is not required, you just need a DVI to HDMI is just a rework of the DVI cable to allow for easier consumer connections and include audio.
from http://www.ramelectronics.net/ [ramelectronics.net] "HDMI - Digital connection for Video and 8-channels of Digital Audio as well as device control features. Electronically better potential for supporting longer cable lengths than DVI for digital video.
Specification supports up to 12 bit Y-Pr-Pb video (rarely implemented on equipment) as opposed to 8 bit limit of DVI RGB."
I've used them before for other AV media conversion products and they make pretty good stuff.

also see the HDMI FAQ at http://www.hdmi.org/about/faq.asp [hdmi.org]
which states "Is HDMI backward-compatible with DVI (Digital Visual Interface)?
Yes, HDMI is fully backward-compatible with DVI using the CEA-861 profile for DTVs. HDMI DTVs will display video received from existing DVI-equipped products, and DVI-equipped TVs will display video from HDMI sources."

Re:In a related question... (0)

Anonymous Coward | more than 8 years ago | (#15138419)

1: Can I hook up my current VGA or DVI to one of these, and display the content I can currently display?

Check the specs on the display to see if it has a VGA, Composite or Component connectors. HDCP doesn't come into play here.

HDCP is used to protect the higher resolution digital video. If you connect a monitor or recorder with DVI/HDMI that doesn't support HDCP, the video source is supposed to just give you a lower relsolution version over your DVI interface. So you may get SD instead of the HD the source is capable of.

2: Is the only limitation/constraint the new HD/BlueRay DVDs with [HDCP]

HDCP could also be (is?) used by your cable or satellite receiver. But again, HDCP would only apply to the Higher Def. DVI/HDMI output of any of those devices, and they should have a lower def. stream available to DVI/HDMI devices w/o HDCP. They may or may not have lower def. analog outputs that are unprotected.

3: Related to both, assume I have MythTV running with an HD capture card.

The problem would be that the HD capture card would be unlikely to run HDCP, so you'd only capture a lower def. video stream. I think any HD capture card manufacturer that released a card that did HDCP on an "untrusted" (read no DRM) OS, would find their card revoked rather quickly.

Re:In a related question... (4, Informative)

nsayer (86181) | more than 8 years ago | (#15138425)

1. There are HDMI to DVI cables. The only question mark is the type of DVI your card uses. There are 3 types, depending on which sets of signals the jack has: DVI-A, DVI-D and DVI-I. HDMI is all digital, but its backwards compatible with DVI-D (DVI-I is a combination of both A and D - analog and digital). So unless your card is DVI-A, you should be able to use a DVI-to-HDMI cable to hook up your display. You will need to make separate arrangements for audio, however, since DVI (unlike HDMI) has no provisions for it.

This does presume that the card is able to put out a mode/timing that's compatible with the set, of course.

2. What you're probably talking about is the requirement that non HDCP-hardened outputs from HD players are supposed to be down-resed to 480p (or whatever). I don't know for certain, but I'm willing to bet that this is not an absolute requirement, but that there's a bit that the disk can set to require this behavior. Not all studios or titles will make the decision to flip that bit on on their content, and I'd certainly expect them not to bother until/unless the technology to take DVI-B and rip it to MPEG4 becomes widespread. Unlike macrovision on analog outputs, which largely went unnoticed with DVDs, this bit does threaten to have a real impact on folks, so I would expect a site to pop up relatively shortly with a list of disks "not to buy" unless you have HDCP. The industry might even respond with a standardized icon on the box whose meaning is "HDCP required for full resolution."

The other obvious restriction is that the HD media is itself encrypted, so when HD-DVD-ROM drives come out, you won't be able to read the data off of them (except in the context of an HD-DVD movie player app), at least not until it's reverse engineered and cracked like DVDs were.

3. I may be wrong, but I am unaware of any HD video capture cards. There are HD tuner cards/boxes out there that will do HDTV, but they're decoding the RF from a TV station and getting MPEG2 streams. That's not the same thing as ripping 1080i from a DVI connector and turning THAT into MPEG2. Even if that were possible, the original source (HDTV, HD-DVD, DVD, whatever) was probably compressed in the first place, so you'll be recompressing it, which will degrade the picture some (more).

ripping HD from DVI (1)

mccoyspace (590866) | more than 8 years ago | (#15138775)

This is an interesting device:
http://www.doremilabs.com/products/XDVI-20.htm [doremilabs.com]
It converts a DVI signal into an SDI-HD signal.
Then with a card like this -- http://www.blackmagic-design.com/products/hd/ [blackmagic-design.com]
and a disk array that could handle about 1.5 gbits/sec you could record the high-def signal in an accessible form.
With the drives we're in the $1500 range for all the gear, so it's not cheap, but it is 'prosumer' level.

Re:ripping HD from DVI (1)

Stealth210 (447350) | more than 8 years ago | (#15139115)

Correct me if I'm wrong, but you would need to recompress the data on the fly before writing to disk. 1.5gb/sec would be approx 187MB/sec (1500/8) which I do not believe any consumer disk array can achieve. Also, is 1.5gb the standard data rate for 1280x720 + 5.1 audio?

Re:In a related question... (1)

Wesley Felter (138342) | more than 8 years ago | (#15139068)

A recent Ask Slashdot thread revealed several DVI capture cards on the market, but they're in the $3,000 range; and you'd need a pretty hefty computer to record uncompressed HD (and then recompress it).

One thing I hate worse ... (2, Interesting)

Midnight Thunder (17205) | more than 8 years ago | (#15138231)

There is one thing I hate worse than this DRM (Draconian Rights Management) crap: region encoding. DRM only effects me if I want to make a backup or play a disk I bought with Linux. Now if I buy a disk in Europe and want to play it in Canada it is not doable, officially. Unofficially I have to get a DVD player with a backdoor, or a PC DVD player with the Firmware hacked or rip the DVD - all this for a DVD I bought legitimately!?

And then there is something that scares me: how unaware of this many people I speak to are, even some people working in IT!

Region Coding vs. Fair Use (0)

Anonymous Coward | more than 8 years ago | (#15138473)

IANAIPL, but....

You're arguing your fair use rights against the copyright holders rights to regionally control distribution. I don't think this is quite as cut and dried.

On the other hand, region encoding doesn't seem to make as much sense in this day of global economies. Do they still even have staggered releases of movies?

Re:Region Coding vs. Fair Use (2, Interesting)

ClamIAm (926466) | more than 8 years ago | (#15139374)

Sorry, but in the age of global trade, nobody has a "right" to the type of region-controlling the media cartels do. In fact, this type of collusion is most likely illegal under lots of treaties and jurisdictions.

This is what.... (1)

Firehed (942385) | more than 8 years ago | (#15138404)

This is what the guy who originally said he could easily crack HDCP said. And the only reason he didn't release specifics (which could have allowed them to fix it before it went 'public') is because he'd have been in some boiling legal water thanks to the DMCA. As it is, the publisher of this story probably will be, but the system will still be cracked *very* quickly, and we'll all have AnyHDCP running in our trays so our computers are stupid-proof.

Re:This is what.... (1)

sconeu (64226) | more than 8 years ago | (#15138930)

because he'd have been in some boiling legal water thanks to the DMCA. As it is, the publisher of this story probably will be

Ed Felten has gone toe to toe with the xxAA before.

Re:This is what.... (0)

Anonymous Coward | more than 8 years ago | (#15139128)

The full details for cracking HDCP were publicly released years ago; Felten hasn't added anything new. Even though the details have been known, no one has yet to actually crack HDCP.

The comment about "AnyHDCP" shows that you don't have any clue how HDCP works. You can't upgrade a Fast Ethernet card to Gigabit using software and you sure can't crack HDCP in software. At a minimum you need an FPGA.

not as easy as it seems, or am I misunderstanding? (1)

dioscaido (541037) | more than 8 years ago | (#15138413)

I may be totally misunderstanding, but won't the 40 devices need to have their private numbers assigned from the central authority as well (and presumably have to pay $$$$$ for it)? Otherwise, when they send [1]+[2] to the device they are cracking, and get back [3]+[4], it will be meaningless unless the hacker's internal numbers' 3+4 addition equals 1+2 of the remote device.

What!? Hasn't he heard of the /. Effect? (1)

kadathseeker (937789) | more than 8 years ago | (#15138438)

Oh, I see, breaking the security of the systems. Right. Didn't see that the first time. Sorry.

How does this stop pirates? (0)

Anonymous Coward | more than 8 years ago | (#15138561)

[Note: I define pirate as someone who infringes copyright on a large scale for profit. That doesn't mean others aren't infringing.]

All this anti-piracy encryption is still missing the point. Long before CSS was cracked, pirates were bit copying DVDs.

AFAIKT, the new disc formats don't have any magic that prevents a pirate from making physcial (i.e. analog) copies of a purchased source disc. Players have to be able to read the bits off the disc. A recorder can write the same bits back on a disc. No decryption necessary for a perfect copy.

Maybe they'll bring back the "bad sector" copy protection schemes. Remember when you couldn't play a game w/o its install floppy that had deliberate bad sectors on it? And you couldn't copy it as the PC would helpfully correct the bad sectors. Yes, I've keyed in boot vectors in binary from the front panel, why do you ask?

One attack in many (4, Interesting)

bhima (46039) | more than 8 years ago | (#15138596)

Wow so many folks sort of missed the point here...

Felton's description of the weaknesses of DHCP handshakes is of only one potential attack. Combined with other attacks and it's entirely possible that a group effort could crank out new secret vectors faster than the M.A.F.I.A.A. could revoke known compromised ones.

For example: If more was known (than I know) about the encryption algorithm used (AKA "the hdcpRngCipher") work could be started on creating dense & smart Time-Memory Trade-Off tables. This is a non-trivial task involving tens of thousands of CPU hours... a perfect thing for a validating distributed computing application (oh. this. has. so. been. done. before).

Also a HDMI repeater or splitter isn't very far from being a sniffer... I think all it lacks is a little I2C to USB help. This, the tables above, & a HDCP device will net you all the vectors you need to employ Felton's attack. Once one set has been compromised and the methodology worked out it's just a matter of turning the crank to get more and potentially very, very quickly.

The utility of these attacks goes well beyond being able to view 1080p on a non DHCP device... one could render revocation useless be attacking high-end components sold by M.A.F.I.A.A. members (i.e. Sony). This eventually must lead hardware devices running out of un-revoked vectors and becoming inoperable... an untenable situation for the M.A.F.I.A.A.

Now, if such a concerted attack is organized on the hi-def media... I feel that we will be right where we are now... a reasonably astute person can watch any DVD wherever they want and they can retain a backup of that media in a format of their choosing.

Re:One attack in many (1)

arodland (127775) | more than 8 years ago | (#15139668)

Not to be rude, but his name still isn't Felton, just like it hasn't been any other time some slashdotter misspells it.

Ok, fine, but where do you get the info? (1)

Opportunist (166417) | more than 8 years ago | (#15138610)

First of all, let me admit that I'm not big into electronics. Best I can do is hook a cable onto my computer and pray that it gets the signal across. So please educate me.

How is he going to find out what the device "wants to hear"? Is he going to sniff into the communication between two "legit" devices? Or is he going to try to "talk" with one of them and brute force through try and error (because it's unlikely the device will send him the "right" answer to the question as well)?

How's he getting the information?

Knowing the vectors is only half the deal (1)

Opportunist (166417) | more than 8 years ago | (#15138680)

When you know the vectors of a machine, you only know what it can send you, but not what it expects from you. When the machine tells you to add [1] and [3], you have to know the index of this rule in its ruleset as well, so you know first of all what it wants to hear from you, and second which indexes it wants to get asked from you so it adds up to the same number.

Technically you could of course go ahead and implement the same vectors and keys, which would of course yield the same results. But you need the ruleset, too, or at the very least the same keys the machine has.

A true generic hack that works against any machine would require vectors and math rules, so you could tell the correct answer without knowing the asking machine's ruleset.

If any 40 devices conspire together... (1)

archeopterix (594938) | more than 8 years ago | (#15138856)

"If any 40 devices conspire together, they can break the security of the system."

Ah, that explains the 40 suspicious looking toasters gathered in my basement whispering to each other.

Re:If any 40 devices conspire together... (1)

Gothmolly (148874) | more than 8 years ago | (#15139799)

Were they running *BSD?

Easier? (1)

Spazmania (174582) | more than 8 years ago | (#15139094)

Couldn't you get this without first gaining the secret vectors for 40 devices? Suppose you only knew the secret vector for just one device. Borrowing from the article's example, couldn't you do something like the following:

Alice is a device whose secret vector has been obtained through means not addressed here. Bob is a commercially purchased device with an unknown secret vector.

Known: Alice secret vector is (26,19,12,7)
Known: Alice addition rule is [1]+[2]
Known: Bob's addition rule is [2]+[4]
Unknown: Bob's secret vector (b1,b2,b3,b4)

Hacker impersonating Alice receives data from Bob and decrypts it into DATA.

Hacker now knows that b1+b2 = a2+a4 = 19+7 = 26

Hacker changes his addition rule [1]+[3] and tries again.
Hacker receives encrypted data from Bob. [1]+[3] is some Keysize number (2^56?). Hacker performs a brute force attack against the encrypted data until he finds key K that produces the same decrypted DATA as before. Hacker now knows that b1+b3 = K.
26-K = (b1+b2)-(b1+b3) = b2-b3.

Repeat a couple times and you have enough equations to solve for the individual vector values. This gives you Bob's secret vector.

Repeat against 38 more devices and you have the requisite number to break the whole algorithm.

Someone better at math than I am, please feel free to jump in and tear holes in the argument.

IT'S NOT ABOUT PIRACY! (5, Insightful)

nagora (177841) | more than 8 years ago | (#15139233)

This stuff, just like region encoding, is about price-fixing. That's why the security is crap: its only purpose is to prevent the 99.99% of consumers who will never crack even a trivial encryption from recording a TV programme instead of going out and buying the HDDVD of the series later in the year. That keeps the price of those DVD's up and that's all this is about.

It used to be called "a cartel" and it used to be illegal.

TWW

Re:IT'S NOT ABOUT PIRACY! (1)

ClamIAm (926466) | more than 8 years ago | (#15139400)

You are correct, but this principle is relevant in a much more general sense, this being that greed is nearly always the underlying factor in witch-hunts, business decisions, and government policy. People scream about things like "piracy", "corporate restructuring", and "terrorism" (to name a few), yet the underlying reasons are almost always love of money and power. Instead of debating whether or not the evil du jour is legitimate, we should instead be asking whether or not greed is a good enough explanation. Unsurprisingly, it most usually is.

Apparently this is easy. (2, Insightful)

mozu (862682) | more than 8 years ago | (#15139404)

The solution is easy according to an anonymous physicist. I showed him the problem and it took him 2 min to do this. He laughed when I told him this is a multi-billion dollar cipher system.

If (no. of eqns.) >= (no. of variables), the equations are solvable.

Given

x1 + x2 = 33 - (1)
x2 + x4 = 18 - (2)
x1 + x3 = 41 - (3)
x2 + x3 = 24 - (4)

Rearrange (4)
--> x2 = 24 - x3

Sub (5) into (1)
x1 + ( 24 - x3 ) = 33
x1 - x3 = 33 - 24
x1 - x3 = 9


(6) + (3) --> 2(x1) + 0 = 9 + 41
2(x1) = 50
x1 = 25

Sub (7) into (1) --> 25 + x2 = 33
x2 = 8

Sub (8) into (2) 8 + x4 = 18
x4 = 10

Sub (8) into (4) 8 + x3 = 24
x3 = 16

Summary
-->
x1 = 25
x2 = 8
x3 = 16
x4 = 10

Apparently any 1st year maths student can do this. This is not the best method however and using a matrix to solve for lambda is the best way, so he says. By the way it took me about 2 hours brute forcing it by logical trial and error using pen and paper.

Re:Apparently this is easy. (0)

Anonymous Coward | more than 8 years ago | (#15139628)

The "theorem" you gave is most definitely false. Its definitely not true for non-linear equations. Furthermore, its not always true for linear equations either. It doesn't sound like you are terribly mathematically inclined, so I will not bore you with details. If you are actually curious, pick up just about any Linear Algebra book.

However, while I'm not certain, we can probably count on the system being linear in this case. And, we can probably count on the existence of a solution as well. So, you probably are right in this case, but I just cringe as a mathematician when someone says an incorrect theorem.

I'd also like to point out that the method you gave for solving the system is essentially equivalent to using a matrix and row-reducing. The only difference is notation

Its all pointless anyway (1)

stewwy (687854) | more than 8 years ago | (#15139800)

I think everyone is getting things too complicated! in ANY system that can be used by humans (ie viewed and/or heard) there comes a point whereby whatever data is used becomes 'human' readable, at that point all security becomes useless. I can read from just before the output device, why not just unplug the LCD screen and read the signals direct? All HDPC does is try to stop me from reading the signal PC to LCD electronics, as far as I can make out I can read the internal signal to the actual crystal matrix with no problem. Just as for any audio I can plug the speaker output into an input and read that. All any DRM does is make pirated copies MORE attractive.
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