In block ciphers, if we use AES for encryption, it does not matter whether the cipher is CBC-based or ECB-based (both will have equally strong results). true or false
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In block ciphers, if we use AES for encryption, it does not matter whether the cipher is CBC-based or ECB-based (both will have equally strong results).
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- Question 1 Study the scenario and complete the question(s) that follow: Ceasar Cipher The Caesar Cipher technique is one of the earliest and simplest method of encryption technique. It's simply a type of substitution cipher, i.e., each letter of a given text is replaced by a letter some fixed number of positions down the alphabet. For example with a shift of 1, A would be replaced by B, B would become C, and so on. The method is apparently named after Julius Caesar, who apparently used it to communicate with his officials. 1.1 Write a Ceasar cipher algorithm in such a way that a character D is changed to N. Derive the encryption of the other characters accordingly. 1.2 Based on your algorithm, what will be the encrypted code of the message “my mother is not home". End of Question 1In an affine cipher, if k = (a, b) = (3, 7), then what is the encryption of x=9. Answer by typing just the resulting number in the box.A block cipher has block length 3. When the key is k, the results of encrypting a block with this cipher are given in the table below: m E(m, k) m E(m, k) 000 011 100 101 001 111 101 110 010 001 110 010 011 000 111 100 (i) Decrypt the ciphertext 010011011 given that it was created by using this block cipher in CBC mode, with key k and IV 111. (ii) Decrypt the ciphertext 010011011 given that it was created by using this block cipher in ECB mode, with key k. (iii) Decrypt the ciphertext 010011011 given that it was created by using this block cipher in CTR mode, with key k and initial counter value 100.
- 1. An affine cipher has the form c = (am+b) mod n. Suppose m is an integer between 0 and 25, each integer representing a letter. (a) Let n = 26, a = 3, and b = 123. What is the ciphertext corresponding to the phrase THIS IS A CIPHER MESSAGE. (b) A requirement for a cipher is that every plaintext letter correspond to a different ciphertext letter. If either a and b is not relatively prime to n, does the affine cipher meet this property? Either prove it does or present a counterexample.A symmetric block encryption algorithm is shown below. 16-bit blocks of plaintext P are encrypted using a 32-bit key. Encryption is defined as: C = (P EX_OR Ko) + K1 C is the ciphertext, K is the secret key, Ko is the leftmost 16 bits of K, K1 is the rightmost 16 bits of K, EX_OR is bitwise exclusive OR, and + is binary addition. a. The ciphertext C must be the same size as the plaintext P, that is, it must not be larger than 16 bits. How can this be achieved? b. Show the decryption equation. How will the encrypted message be decrypted?I need help with the following problem: A ciphertext was generated using an affine cipher. The most frequent letter in the ciphertext is “C” and the second most frequent letter is “V”. Break the code. Assume that the most frequent letter in English is “E” and the second most frequent letter is “T”.
- The answer above is NOT correct. Note: The notation from this problem is from Understanding Cryptography by Paar and Pelzl. We conduct a known-plaintext attack against an LFSR. Through trial and error we have determined that the number of states is m = 4. The plaintext given by 11100010 — ХоXјX2XҙX4XsX6X7 when encrypted by the LFSR produced the ciphertext 11010110 — Уo У1 У2 Уз Уз У5 У6 Ут. What are the tap bits of the LFSR? Please enter your answer as unspaced binary digits (e.g. 0101 to represent p3 = 0, p2 = 1, p1 = 0, po = 1). 00106. A good block cipher has a 64-bit block size and uses 50-bit keys. How can we convert that in to a strong block cipher with brute force strength of 2^150?Let k=( 4 1 3 2) be the encryption and key k = (2 4 3 1) be the decryption key used in a transposition cipher. Show all the steps to decrypt the following ciphertext using the keyed approach: “XYSEESRU”
- Use the two prime numbers p 5 and q =13 in the first step to give ONE integrated example to show how the five steps in the basic process of RSA Cryptography works. Based on your example, demonstrate how the number 60 as an original message is encrypted into ciphertext and decrypted back correctly to the original message. If you need to choose a number in any step, you must choose it from the set {r e R: 8 Sx< 12}. You must show all detailed steps involved.In a substitution cipher, we “encrypt” (i.e., conceal in a reversible way) a message by replacing every letter with another letter. To do so, we use a key: in this case, a mapping of each of the letters of the alphabet to the letter it should correspond to when we encrypt it. To “decrypt” the message, the receiver of the message would need to know the key, so that they can reverse the process: translating the encrypt text (generally called ciphertext) back into the original message (generally called plaintext). A key, for example, might be the string NQXPOMAFTRHLZGECYJIUWSKDVB. This 26-character key means that A (the first letter of the alphabet) should be converted into N (the first character of the key), B (the second letter of the alphabet) should be converted into Q (the second character of the key), and so forth. A message like HELLO, then, would be encrypted as FOLLE, replacing each of the letters according to the mapping determined by the key. Let’s write a program called hw6.py…The hacker group Desdeia tries to break our old weak cipher which we used for the communication with the Duchess of Ligovia. It's generally known that an actual breaking of the cipher needs the reviewing a huge amount of variants which probably lasts a couple of months. Additionally we know that the reviewing of different variants needs always the same server resource. Using our intelligence sources we tried to get some information about the actual progress of breaking, but we got only some fragments: • The hacker group Desdeia uses for breaking their powerful four-core server which resources are fully dedicated to breaking process. . The actual breaking process began exactly at noon, but we don't know, on which day. . By the noon of 2nd of April, already 648,467,673 variants were successfully reviewed. Exactly at this noon, when the breaking process was already lasted for 19 days, it was attempted to double the server speed. The doubling itself was failed, but instead of it there was…