This question concerns block cipher modes. We will use a simple affine cipher, which can be expressed in C as follows. char cipherfunsigned char block, char key) return (key+11'block)%256; The inverse...


This question concerns block cipher modes. We will use a simple affine cipher, which can be expressed in C as follows.<br>char cipherfunsigned char block, char key)<br>return (key+11'block)%256;<br>The inverse of this cipher is shown below.<br>char inv_cipherfunsigned char block, char key)<br>{/ 163 is the inverse of 11 mod 256<br>return (163 block-key+256)%256;<br>Note that the block size is 8 bits, which is one byte (and one ASCII character). We will work with the fixed key Ox08.<br>We now decrypt various ciphertexts using modes for this cipher. In every case in which the mode requires an IV, the IV will be OXAA. In the case of CTR<br>mode, we use a (nonce || counter) arrangement in which the nonce is the left 5 bits of OKAA and the counter is a 3 bit counter that begins at 0. In all of the<br>problems given below, one character is one block. Each character of the plaintext should be regarded as its corresponding ASCII code. The ciphertext is<br>n hexadecimal.<br>a) Decrypt the ciphertext

Extracted text: This question concerns block cipher modes. We will use a simple affine cipher, which can be expressed in C as follows. char cipherfunsigned char block, char key) return (key+11'block)%256; The inverse of this cipher is shown below. char inv_cipherfunsigned char block, char key) {/ 163 is the inverse of 11 mod 256 return (163 block-key+256)%256; Note that the block size is 8 bits, which is one byte (and one ASCII character). We will work with the fixed key Ox08. We now decrypt various ciphertexts using modes for this cipher. In every case in which the mode requires an IV, the IV will be OXAA. In the case of CTR mode, we use a (nonce || counter) arrangement in which the nonce is the left 5 bits of OKAA and the counter is a 3 bit counter that begins at 0. In all of the problems given below, one character is one block. Each character of the plaintext should be regarded as its corresponding ASCII code. The ciphertext is n hexadecimal. a) Decrypt the ciphertext "212C330F184EBB* using CTR mode. Please enter your answer in ASCII characters (aka words). b) Decrypt the ciphertext "ACCDF904' using ECB mode. Please enter your answer in ASCIl characters (aka words). c) Decrypt the ciphertext *324 F1 F363623' using CFB mode. Please enter your answer in ASCil characters (aka words). d) Decrypt the ciphertext '5BSA91C9274170* using CBC mode. Piease enter your answer in ASCII charactors (aka words). e) Decrypt the ciphertext "3EDF673842* using OF B mode. Please enter your answer in ASCII characters (aka words). Note: You can earn partial credit on this pmhiem
Jun 11, 2022
SOLUTION.PDF

Get Answer To This Question

Related Questions & Answers

More Questions »

Submit New Assignment

Copy and Paste Your Assignment Here