3
$\begingroup$

What do you think about the combination of NTRU Prime and X25519 ECDH in OpenSSH 9.0?

Why is the combination of NTRU Prime and X25519 ECDH in OpenSSH believed to be secure against quantum attacks in the future?

when I talk about one-way encryption in quantum computing would reflect would i be able to generate 100% solution that would prevent me from causing a quantum computing collision? We are currently at 99.9% with the current one-way encryption methods, in my opinion?

the current standard password methods have already lost out against the quantum attacks from quantum computing compared to asynchronous encryption methods?

Can i prevent against the 'harvest now, decrypt later' in theory with quantum computing?

I'm going to start a bounty because i need your opinions from

https://quantumcomputing.stackexchange.com

Report from heise.de:

The encryption of user data using a symmetrical method such as AES with 256 or almost any number of bits is not the real problem in attacks by quantum computers. It's "only" about cracking the exchange of the random key pair generated for this purpose, which has so far been done with a conventional Diffie-Hellman key exchange based on elliptic curves (X25519 ECDH).

This method is considered to be susceptible to quantum cryptographic attacks, even if the current (known) performance – i.e. the number of qubits – of the first quantum computer is far from sufficient.

There are many estimates of how many qubits are needed to crack asymmetric encryption in a timely manner, probably in the range of several thousand qubits.

Where is the current research? Google last year showed off its Sycamore processor with 54 qubits.

The OpenSSH developers, as part of the OpenBSD community, are aware that new implementations may contain bugs. Instead of just using a new algorithm, the OpenSSH developers are combining NTRU with the X25519 ECD, which has been the standard up to now. This is supposed to form a kind of safety net, because the old protection still applies even in the case of unknown security gaps or the secret backdoors of the secret services.

What does that mean in theory? For example, if I have in performance, the number of 10,000 or 100,000 of qubits (that's just the theory!!!) would have could I crack the asymmetric encryption?

I don't want detailed explanations about these 2 procedures but why the combination should make it safe.

And now back to my question:

Why is the combination of NTRU PRIME and X25519 ECDH in OpenSSH believed to be secure against quantum attacks in the future?

Some links on the topic:

OpenSSH now defaults to protecting against quantum computer attacks

NATO cybersecurity center finishes tests of quantum-proof network/

cryptsetup(8) — Linux manual page

OpenSSH 9.0/9.0p1 (2022-04-08)

secret keys that protect the traffic

$\endgroup$
8
  • 4
    $\begingroup$ Nothing is secure against unknown bugs, gaps, or backdoors. But NTRU is believed to be quantum safe. $\endgroup$
    – Danylo Y
    Commented Apr 16, 2022 at 10:54
  • 3
    $\begingroup$ Also this question may be better suited at crypto.stackexchange.com $\endgroup$ Commented Apr 16, 2022 at 13:25
  • $\begingroup$ is there a relation between the question in the title and the body of the post under "update"? Please note that StackExchange is not suitable for opinion-based discussions, and that each post should contain a single, laser-focused question $\endgroup$
    – glS
    Commented Apr 18, 2022 at 16:07
  • $\begingroup$ It's is one question. sry if i wrote to much you can edit the question. $\endgroup$
    – Z0OM
    Commented Apr 18, 2022 at 16:09
  • 2
    $\begingroup$ @BlockchainOffice it certainly improved it $\endgroup$
    – glS
    Commented Apr 20, 2022 at 18:08

1 Answer 1

2
+50
$\begingroup$

Before I address your question, I'll mention that the security of a cryptosystem in theory is not really the same as the security of a cryptosystem in practice. The specific reasons why NTRU is considered secure as a whole depends on a lot of implementation details not really relevant to what makes quantum attacks difficult. However, there are still potentially (classical) attacks when you have a faulty implementation of a good idea.

The security of a (classical) cryptosystem in theory is essentially guaranteed by hardness assumptions of certain problems. One might prove mathematically that "cracking" a key reduces to solving an instance of a problem. RSA for example relies on the hardness of factoring large integers, Diffie-Hellman relies on the hardness of discrete log. To break RSA for any key, you must be able to factor any large number quickly.

In the case of factoring, however, we have quantum algorithms that provide speedups over best known classical algorithms. The assumption of that factoring is a hard problem turned out to be wrong if you consider quantum algorithms.

NTRU is a lattice-based cryptosystem, and the problem that it is based on is the Shortest Vector Problem (SVP). It is known that SVP is NP-hard under randomized reductions, so it is not expected that there is a fast algorithm for solving SVP.

$\endgroup$
2
  • $\begingroup$ RSA hasn't been reduced the the hardness of factoring integers, it's just generally assumed to be equivalent. $\endgroup$
    – forest
    Commented Jun 14, 2022 at 19:31
  • 1
    $\begingroup$ It's known that the ability to quickly factor very large integers is enough to break RSA. However, we don't yet known if RSA could be broken even faster with some other (yet unknown) method without factoring any integers. From mathematical standpoint, it seems plausible that RSA cannot be broken without factoring so that's why it's assumed to be equivalent. This is because ability to break RSA "without factoring" would provide a new faster method to factor (some subset of) large integers in general which would make it equal to factoring at least for that subset of integers. $\endgroup$ Commented Nov 12, 2022 at 10:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.