I've heard that quantum computers may soon be powerful enough to crack standard encryption. If this is true, could we potentially harness them to end ransomware? Instead of paying the ransom or coming up with some alternate solution, could someone who was hit with ransomware just pay a company to break their computer's encryption?
1 Answer
... quantum computers may soon be powerful enough to crack standard encryption.
I don't know what "soon" means for you, but surely not in the next few years.
And, quantum computers might be in some time be powerful enough to crack what is considered today "standard encryption", but encryption moves on too. For one, one can for some time just increase the length of keys, but there are also already algorithms for post quantum cryptography which are hopefully resistant against quantum computers.
I mean hopefully, since what ransomware uses to take away data from their rightful owners are basically the same algorithms which are used to protect data from unauthorized access. So any major advantages in cracking ransomware in general also means loss of rightful data protection.
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Excellent answer, as usual. WRT, 'quantum computers might be in some time be powerful enough to crack what is considered today "standard encryption"' - are you referring to quantum computers finding a weakness in today's encryption algorithms? Or, are you referring to quantum computers brute-forcing the encryption key?– mti2935Oct 27, 2022 at 18:13
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1@mti2935: Asymmetric algorithms used in signatures, encryption and key exchange based on RSA, ECC or DH are at risk to be broken with quantum computer specific methods like Shor's algorithm, i.e. it is not simply brute force. Symmetric algorithms like AES are less at risk - here it is considered to be sufficient to double the key size. Oct 27, 2022 at 18:41
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1To add to the above, Grover's algorithm is the most commonly referenced quantum computing algorithm for performing search operations against black-box functions (i.e. finding a function inversion), as it operates in square root time complexity, or O(n^½). This means that it reduces the time complexity of cracking a 128-bit symmetric key from 2¹²⁸ operations down to 2⁶⁴ operations. Newer work indicates that QCs might be able to achieve O(n^⅓) for some classes of hidden variable problems. Oct 27, 2022 at 19:00
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Don't forget that Grover's algorithm needs 2⁶⁴ sequential evaluation for 128-bit key-sized block ciphers, and it is not clear how long each setup will take time. And, Grover can be paralleled, though, a
t
time parallelization can only speed upsqrt(t)
– kelalakaOct 27, 2022 at 22:17