This question concerns block cipher padding. Suppose the block cipher has block size of 18 bytes. A certain message ends with a partial block which is 5 bytes. We must reversibly pad out the message to use some block cipher modes, such as CBC (even if the last block is full). Consider the following padding strategy. Append a single byte with value 128, and then as many zero bytes as required to make the overall length a multiple of 18. Suppose the final (possibly partial) block of the message is "Ox0001020304" in hexadecimal. Write out the complete final block in hexadecimal. Please do not put a leading Ox as this has been written already. Ox

This question concerns block cipher padding. Suppose the block cipher has block size of 18 bytes. A certain message ends with a partial block which is 5 bytes. We must reversibly pad out the message to use some block cipher modes, such as CBC (even if the last block is full). Consider the following padding strategy. Append a single byte with value 128, and then as many zero bytes as required to make the overall length a multiple of 18. Suppose the final (possibly partial) block of the message is "Ox0001020304" in hexadecimal. Write out the complete final block in hexadecimal. Please do not put a leading Ox as this has been written already. Ox

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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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…
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Do not copy other's work. Show your own
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