8. (a) (1 point) The following is the encryption part of AES in ECB mode. Draw the decryption part of the mode. (b) (1 point) Prove the correctness of the ECB mode.
Added by Charles L.
Close
Step 1
Each message block ($m_i$) is independently encrypted using the same key ($k$) and encryption function ($F_k$) to produce ciphertext blocks ($c_i$). The ECB mode stands for Electronic Codebook mode. Show more…
Show all steps
Your feedback will help us improve your experience
Adi S and 66 other AP CS educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Consider a simple symmetric block encryption algorithm in which 64-bit blocks of plaintext are encrypted using a 128-bit key. The plaintext is divided into two 32-bit blocks (Lo, Ro), and the key is divided into 4 32-bit blocks (Ko, K1, K2, K3). Encryption involves repeated application of a pair of rounds, defined as follows for rounds i and i+1: Ri = Li-1 Li = Ri-1 ⊕ F(Li-1, Ko, K1) Ri+1 = Li Li+1 = Ri ⊕ F(Li, K2, K3) Where F is defined as: F(M, Ki, Kj) = ((M ≪ 3) ⊞ Ki) ⊕ ((M ≫ 2) ⊞ Kj) If only one pair of rounds is used, then the ciphertext consists of the 64-bit block (L2, R2). For this case, answer the following questions: a) Illustrate the operation of the above encryption using a block diagram. (6 pts) b) Express the decryption algorithm in terms of equations. (8 pts) c) Illustrate the operation of the decryption using a block diagram. (6 pts)
Adi S.
This question concerns block cipher modes. We will use a simple affine cipher, which can be expressed in C as follows. char cipher(unsigned char block, char key){ return (key+11*block)%256;}The inverse of this cipher is shown below.char inv_cipher(unsigned 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 0x08. We now decrypt various ciphertexts using modes for this cipher. In every case in which the mode requires an IV, the IV will be 0xAA. In the case of CTR mode, we use a (nonce || counter) arrangement in which the nonce is the left 5 bits of 0xAA 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 given in hexadecimal. a) Decrypt the ciphertext "2227370922273709" using CTR mode. Please enter your answer in ASCII characters (aka words). b) Decrypt the ciphertext "3EAC33803EAC3380" using ECB mode. Please enter your answer in ASCII characters (aka words). c) Decrypt the ciphertext "314C23FAA3314C23FAA3" using CFB mode. Please enter your answer in ASCII characters (aka words). d) Decrypt the ciphertext "8ADF9821F48ADF9821F4" using CBC mode. Please enter your answer in ASCII characters (aka words). e) Decrypt the ciphertext "3BD56939533BD5693953" using OFB mode. Please enter your answer in ASCII characters (aka words)
Akash M.
Which of the following best describes the decryption in Triple DES? (A) Plain text $=D_{K_{1}}\left(E_{K_{2}}\left(D_{k_{1}}\right.\right.$ (ciper text))) (B) Plain text $=D_{K_{1}}\left(E_{K_{2}}\left(D_{k_{3}}\right.\right.$ (ciper text))) (C) Plain text $=E_{K_{1}}\left(D_{K_{2}}\left(E_{k_{1}}\right.\right.$ (ciper text))) (D) Plain text $=E_{K_{1}}\left(D_{K_{2}}\left(E_{k_{1}}\right.\right.$ (ciper text)))
NETWORKS, INFORMATION SYSTEMS, SOFTWARE ENGINEERING AND WEB TECHNOLOGY
NETWORK SECURITY
Recommended Textbooks
Computer Science and Information Technology
Introduction to Programming Using Python
Computer Science - An Overview
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD