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Cryptography

Questions tagged [ecb]

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Electronic Code Book (ECB) is a mode of operation for a block cipher, with the characteristic that each possible block of plaintext has a defined corresponding ciphertext value and vice versa.

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I'm writing an application that encrypts data with a shared key using AES in ECB mode. The level of security for this application isn't extremely high. But I do want to scramble the blocks before ...
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I am wondering if it is possible to find a K that decrypts a chosen ciphertext C into a chosen plaintext P in Camellia 128 ECB-mode? We can assume that there is no IV. There will exist some K that ...
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I wrote my oracle so any identical blocks will cause it to classify as ECB, and no collisions means CBC. If I encrypt text that is highly repetitive (will cause obvious collisions in ECB mode), my ...
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I'm reading "Cryptography Made Simple" Theorem 13.4, which states that "ECB mode is OW-CPA secure assuming the underlying block cipher ek acts like a pseudo-random permutation." I ...
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Given two ciphertexts encrypting the same (natural language) plaintext, where we know that one of the ciphertexts was encrypted with AES-CBC (and unknown IV) and one was encrypted with AES in EBC mode,...
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Forgive my ignorance, but I've run into some confusion in an AES implementation I'm testing. To the best of my understanding, ECB provides no diffusion, the same plaintext with the same key will yield ...
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I'm working on a network where nodes communicate using AES-CCM encryption, within a context of tight bandwidth limitations, making every bit count. The setup uses a nonce that combines a 16-bit node ...
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Known conditions: The block size of this algorithm is 16 bytes. If there are multiple encryption modes, then it is limited to ECB mode. Always use only one immutable key. Since only ECB encryption ...
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I got two audio files one is encrypted and the other file is a part of the plaintext. I want to extract all the plaintext. I am using the AES-ECB mode, the size of the block is 128 bits and the start ...
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I have $C_1 \oplus C_2$ where $C_1$ = $AES128_k(P_1)$ and $C_2$ = $AES128_k(P_2)$. I know the k, it is 0 (i.e. 16 bytes of zeros). I know the piece of $P_1$ (7 ...
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I now understand the initial key from all the round keys will be the original 4x4 block for 128-bit keys, but I do not know how it would work for something else, like AES-256 or 192. Would it be that ...
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I'm trying to pad some plaintext to be a multiple of 16 bytes to use it as input in AES ECB block cipher. ...
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I was going to encrypt with AES, But I noticed ECB mode is not safe for CPA. So I thought about preventing CPA by padding input text to a multiple of 16 bytes. First, perform custom padding system: ...
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Given this plaintext: 101010101010, and the block is 3, with a IV = 000 and the key as k = |1 2 3| |2 1 3| What are the required steps to get the same result like these in ECB mode: c1 = Ek(m1) = 011, ...
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For example, if we run this openssl command: openssl aes-128-ecb -a -in <(echo -n "AAAAAAAAAAAAAAAA") -K "deadbeefdeadbeefdeadbeefdeadbeef" ...
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