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For PKBDF2-HMAC-SHA256 on my Intel NUC i3 from 2018 I was getting 200000 iteration calculated roughly in 20ms and 400000 iters in 40ms. On my Mac M1 2021 results were about 10% slower. Here are more details and test script in Ruby and Crystal: https://lukas.zapletalovi.com/2021/05/finding-the-right-cost-for-bcryptpbkdf2.html


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Build a simple calibration function that will calculate how much it takes to calculate X iterations, run it during app startup if that's feasible and set iteration count accordingly but not smaller than X (minimum count - a safe margin). This works really well for on-premise web applications where you can set the target execution time to be 30 or 50ms. This ...


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Wow, as I typed this out, it became a much longer answer than I expected. Summary: Yes they all happen at the same time. In some contexts you may be able to do with fewer layers, but generally speaking, each layer is addressing a different threat model -- either the information you're protecting or the attacker you're protecting it from are different at each ...


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The answer is that they all protect against different kinds of eavesdropping, so they are not at all in use at the same time, and are not always relevant. For instance: WPA2 protects the physical link layer, but only in the case of WIFI communication. WPA2 is what stops someone else on your WIFI network from listening to the traffic between your computer ...


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This isn't a question of cryptography, but of auditing. Bank A and Bank B both keep records of the transactions they've performed. Eventually, they're going to compare notes and realize that the two sets of records don't match. Specifically, Bank B has a collection of transactions that transfer money into Oscar's account without counterpart transactions ...


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But how do I know which hash (of all the other hashes in the database) to compare to the user input? Each user has a row in your database with entries for columns: (1) username; (2) bcrypt output. Also, if bcrypt could check if it matches without a salt, couldn't anyone do that? bcrypt can not check without the salt. However, how do I know which salt to ...


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Strictly speaking, TLS already supports what you're asking about. There are "null cipher suites" - that is, TLS cipher suites that have no ("null") encryption - that would do what you want. Using one will allow you to use existing TLS libraries and take advantage of all the work and hard thought and lessons learned that has gone into the ...


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Private key storage by user would be a nice feature, but it could open you up to a world of issues as you would be unable to access the web based version without uploading the key--thus traversing networks that are public. I trust proton more than I trust starbucks


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If you want to keep this simple you can just make the password a big long number. Then create several smaller numbers that together add up to the big number. If a calculator that can handle such a long number won't be available, then the numbers can be made easier add up by hand by selecting smaller numerals which don't overflow in to the next number's spot ...


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This sounds like Secret Sharing or Secret Splitting. Secret sharing (also called secret splitting) refers to methods for distributing a secret among a group of participants, each of whom is allocated a share of the secret. The secret can be reconstructed only when a sufficient number, of possibly different types, of shares are combined together; individual ...


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A password with 256 bits of entropy is more than enough to withstand even the most well-resourced brute-force attacks. See https://crypto.stackexchange.com/questions/1145/how-much-would-it-cost-in-u-s-dollars-to-brute-force-a-256-bit-key-in-a-year for some interesting reading on this subject. The command below will produce a random password with 256 bits of ...


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Given that your cyphertext is all As, I am going to assume that the key is of the same length as the plaintext (there are edge cases where this is technically untrue, but for decryption purposes it's probably true). When your Vigenère key is as long as your plaintext, you are effectively using a One-time pad, which is impossible to decrypt.


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