I'm wondering whether or not there is an encryption algorithm, or at least the possibility of one, which takes a lot of work to encrypt, yet can be decrypted very easily?

I have come across this question Are there encryption schemes with which it takes (significantly) longer to encrypt than to decrypt? and https://stackoverflow.com/questions/9766579/high-cost-encryption-but-less-cost-decryption, but neither answer my question.

I am NOT looking for a hash of the data, I specifically need to encrypt it in a way which requires lot of work, and then be able to decrypt it very easily.

Whether the algorithm is symmetric or asymmetric is irrelevant.


  • $\begingroup$ Having an encryption scheme that is generally hard isn't that useful. For instance, if you have a block cipher then the time to encrypt would be linear with the size of the plaintext message. That's not commonly a requirement. As the question has already been asked, can you explain why a proof-of-work scheme does not suffice? Why is it that symmetric or asymmetric is irrelevant? Key management requirements are likely rather different for symmetric and asymmetric schemes, so they are unlikely to be a good fit for the same usage scenario. $\endgroup$
    – Maarten Bodewes
    Commented Jan 18, 2018 at 12:01
  • $\begingroup$ I have a situation where i need to prove that the same data is stored in multiple places(or at least multiple times, I'll live with it being stored multiple times on the same drive, even though this is pointless), and the best way I can think to do this is to encrypt it uniquely and then have it stored. If the encryption cost is greater than the storage cost, then naturally any bad players would opt to store it multiple times, rather than try to spoof the system. The encryption is not there to hide the data, therefore symmetric/ assymetric doesn't matter. $\endgroup$
    – Irontiga
    Commented Jan 18, 2018 at 12:27
  • $\begingroup$ You may want to have a look at proof of storage and proofs of retrievability systems. $\endgroup$
    – SEJPM
    Commented Jan 18, 2018 at 21:26
  • $\begingroup$ These algorithms don't address the issue of redundancy :/ $\endgroup$
    – Irontiga
    Commented Jan 20, 2018 at 7:57


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