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Yes, Poly1305-AES can safely be modified to use AES-256 rather than AES-128; but if AES is implemented in software beware of not introducing a timing vulnerability in the implementation. Change of the cipher in Poly1305-AES is explicitly endorsed; quoting D. J. Bernstein's The Poly1305-AES message-authentication code There is nothing special about AES ...


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I add my whitebox AES implementation on GitHub in: C++ Java C++ version implements both Chow's (mixing bijections, input/output encodings, external encodings) and Karroumi's (dual AES in each column) whitebox AES scheme plus Billet's key recovery attack on both schemes. Java implements Chow's scheme only. PS: Due to low reputation I post links to ...


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You can use your HardwareID as basis for the encryption key. If the ID provides enough entropy it'll work. However, if anyone can somehow obtain the ID (which might be quite easy to do) one can decrypt the file. For CFB-Mode the IV must indeed be unpredictable (but need not be secret), so random is just fine, but DO NOT REUSE AN IV. Encryption large ...


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Without Mix Columns, every byte of the ciphertext would not depend on every byte of the plaintext, but only on the one byte at the same position. This would fully break AES, because we don't have an avalanche effect on the whole text anymore. Without the Row shifting, every column of plaintext would only affect the same column of the ciphertext. This would ...


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If the MAC'ing is done right (after the encryption) and if you pad the CBC data correctly there's no security risk. Howver changing to CTR or even better to GCM would be better, as both don't operate on whole blocks and GCM even provides authentication.


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The ZFS file system uses AES in CCM or GCM modes. This works because in ZFS the data and file system metadata is encrypted but the block pointers are in the clear, the AuthTag (MAC) is stored in the block pointer. ZFS also has a SHA256 based merkle tree based on the block pointers that is used for data integrity for resilvering and navigation purposes. ...


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AES sub-bytes is defined over the finite field $GF(2^8).$ This field has 256 elements. Its multiplicative group $GF(2^8)^{\ast}$ has 255 elements. In any multiplicative group $a^N=identity$ for all $a$ in the group. This means the multiplicative order of all the group elements (the smallest power they can be raised to and obtain 1) divides $N.$ Here, by ...


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It depends on the mode of operation. If you use an AEAD mode (which contains a message authentication code, to ensure that the message wasn't tampered with en route to you), then decryption will fail because the MAC is invalid for that message with that key (the whole point of AEAD is that someone without the key can't create a valid ciphertext, so a ...



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