[wfunction]

> It still doesn't make a difference in my argument. My point is that you gain no added strength via a stacked hash implementation because it's as weak as the first hash in the sequence

God, I wish really I could downvote replies.

Nobody said you should be applying the hash functions in sequence. There are at last 3 obvious approaches: (1) applying the functions sequentially, (2) concatenating their outputs, (3) XORing their outputs. None of these takes rocket science to figure out, and some 5 seconds of thinking would easily rule out #1 and #2 as inferior to #3.

Honest question: did you even spend 5 seconds actually thinking about what I wrote before deciding I must be wrong? I'm not sure if you realize this, but when you reply so confidently without thinking, you (and many others) active harm the whole field of infosec. I'm so frustrated and fed up with you and so many other people's overconfidence and lack of willingness to think for 5 seconds when it comes to cryptography.

>> I feel like you should already realize this (in which case I don't get why you're posting the comment), but while that's a cute mathematical existence proof, it's totally irrelevant as it's not something that can just happen out of the blue.

> Fundamentally, all modern crypto relies heavily on math. I made a "cute mathematical existence proof" to make it obvious how stacking ciphers can weaken an encryption system

Again: are you reading and thinking? Or are you just writing?

You're simultaneously literally claiming that two secure ciphers can be combined to result in an insecure cipher when their keys are generated independently. This is far more astonishing than the claim that the ciphers you're using are actually secure in the first place. You're already accepting the latter despite any sort of proof, yet you're bothered by the former? Hell, you haven't even shown shown this is possible for any pair of secure ciphers; your "example" was missing the most crucial part of the cipher -- the key. The whole argument is so crazy it's just utterly ridiculous.

[electronvolt]

> here are at last 3 obvious approaches: (1) applying the functions sequentially, (2) concatenating their outputs, (3) XORing their outputs. None of these takes rocket science to figure out, and some 5 seconds of thinking would easily rule out #1 and #2 as inferior to #3.

This is wrong. Concatenation would be harder to attack than XOR. Finding two things which hash to to two particular values in two separate hash functions is necessarily harder than finding two things which will hash to values which will XOR to the same value--almost a priori. You replace a double collision (across two hashes) which is very unlikely with an XOR collision, which is going to be exponentially easier.

> You're simultaneously literally claiming that two secure ciphers can be combined to result in an insecure cipher when their keys are generated independently. This is far more astonishing than the claim that the ciphers you're using are actually secure in the first place.

Since you seem to want practical examples on recent crypto: consider meet in the middle attacks on 2DES as an example of why combined cryptosystems are not necessarily as strong as you'd imagine. It's admittedly a weak example--still stronger than 1DES, and an old system. Fundamentally, combining cryptosystems, even with separate keys, gives you a new cryptosystem which requires separate analysis.

> Hell, you haven't even shown shown this is possible for any pair of secure ciphers; your "example" was missing the most crucial part of the cipher -- the key. The whole argument is so crazy it's just utterly ridiculous.

If I had a good attack on RSA + ECC, I'd be writing a paper about it. I'm gonna posit that if that's the kind of proof you want to believe you're "wrong", you'll remain happily "correct" in this scenario.

[wfunction]

> This is wrong. Concatenation would be harder to attack than XOR. Finding two things which hash to to two particular values in two separate hash functions is necessarily harder than finding two things which will hash to values which will XOR to the same value

No, you're the one who's wrong. You're assuming the hash is secure and then trying to brute-force it. But the entire discussion is not about brute force; it's about when an adversary breaks the hash, i.e. one who is able to produce multiple inputs with the same hash much more quickly than with brute force. This means they can attack the hashes independently, whereas if you XOR, they can't do that. Heck, if you XOR, the probability that they'll be able to tell which algorithms you used already becomes astronomically low, let alone them breaking it.

> Since you seem to want practical examples on recent crypto: consider meet in the middle attacks on 2DES as an example of why combined cryptosystems are not necessarily as strong as you'd imagine.

Again, you're wrong. You said you were "making it obvious how stacking ciphers can WEAKEN an encryption system". All you proved is that it isn't twice as strong. I never claimed nor even imagined that it was twice as strong. I merely claimed that the probability of it being WEAKER is astonishingly lower than the probability of the crypto layers being strong in the first place. Why do you keep changing your arguments?

>> Hell, you haven't even shown shown this is possible for any pair of secure ciphers; your "example" was missing the most crucial part of the cipher -- the key. The whole argument is so crazy it's just utterly ridiculous.

> If I had a good attack on RSA + ECC, I'd be writing a paper about it. I'm gonna posit that if that's the kind of proof you want to believe you're "wrong", you'll remain happily "correct" in this scenario.

Way to keep changing the topic just to win the argument. I just pointed out that your counterexample ciphers didn't even have independent keys, for God's sake!! Instead of accepting that you made silly mistake, you're spreading meaningless FUD. Why can't you just accept you made an error instead of giving me this nonsense? Are you just a troll? If you keep trolling don't expect me to respond.

[Natanael_L]

If you have a combined hash that is XYZ = weak_hash(m) XOR strong_hash(m), then you still have a birthday collision attack available. Just keep permutating the message in various ways, and in 2^(hash length / 2) operations you have two identical hashes with different inputs.

Edit: yes, this isn't new, but it is strictly weaker than to append the two hashes. It increases the difficulty as you have two hard targets you must hit with the same input, vs one target, which may even be weakened.

You're also assuming the hashes are fully uncorrelated. If the designs are similar, there could be a correlation between the two which biases the output, such that some bits often will be the same or different in certain ways. This can drastically reduce the number of possible outputs in known ways, and could even enable cryptanalysis to break it faster than bruteforce if part of the weaker hash counteracts parts of the other.

You also forgot timing attacks and other sidechannels in layered encryption.

[wfunction]

> If you have a combined hash that is XYZ = weak_hash(m) XOR strong_hash(m), then you still have a birthday collision attack available. Just keep permutating the message in various ways, and in 2^(hash length / 2) operations you have two identical hashes with different inputs.

You understand what "good enough" means, right? Brute force is irrelevant if your hashes are long enough (e.g. say, 256 bits). Again, like the other guy, you're confused: we're trying to guard against broken hashes, not brute force attacks. Brute force attacks are trivial to guard against just by changing the length.

> Edit: yes, this isn't new, but it is strictly weaker than to append the two hashes. It increases the difficulty as you have two hard targets you must hit with the same input, vs one target, which may even be weakened.

I thought I already explained this, but I'll explain it again. For a brute force attack, sure, it's weaker. But like I said, that's already trivial to guard against already, so that's irrelevant. For a broken hash, it's MUCH STRONGER, since in the concatenation case, both hashes can leak information about the input (meaning breaking one can help break the other), whereas in this case it's the opposite: the adversary won't get any information about the input unless he breaks both hashes simultaneously.

You're thinking about the problem wrong.

> You also forgot timing attacks and other sidechannels in layered encryption.

Somebody already mentioned this in another comment chain. I initially thought of it but then forgot it, yeah. But it's just something to watch out for, not an argument for not doing it in the first place. It's also trivial to guard against if you apply a standard layer first (which doesn't preclude applying another standard one last, and putting your own layer in between).