You are to encrypt a ciphertext using the permutation p : {0,1}4 → {0,1}4 defined as p(x) = ~x, ie, toggle each bit (O to 1 and 1 to 0). If you need an IV use 1010. If you need a nonce use 01. If the mode uses padding to handle arbitrary plaintext lengths, add 10* padding. If you need a counter, begin at 1. Encrypt the ciphertext 1110 1110 111 using ECB mode. Write four bits per box, with the final box possibly having fewer bits. Encrypt the ciphertext 1110 1110 111 using CBC mode. Write four bits per box, with the final box possibly having fewer bits. Encrypt the ciphertext 1110 1110 111 using CTR mode. Write four bits per box, with the final box possibly having fewer bits.

You are to encrypt a ciphertext using the permutation p : {0,1}4 → {0,1}4 defined as p(x) = ~x, ie, toggle each bit (O to 1 and 1 to 0). If you need an IV use 1010. If you need a nonce use 01. If the mode uses padding to handle arbitrary plaintext lengths, add 10* padding. If you need a counter, begin at 1. Encrypt the ciphertext 1110 1110 111 using ECB mode. Write four bits per box, with the final box possibly having fewer bits. Encrypt the ciphertext 1110 1110 111 using CBC mode. Write four bits per box, with the final box possibly having fewer bits. Encrypt the ciphertext 1110 1110 111 using CTR mode. Write four bits per box, with the final box possibly having fewer bits.

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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To prevent the tapping and use of information on a wireless network which has brought untold mistrust amongst the directors of Okonko Systems, you suggested that every message that is transmitted should go through the following: A systematic two left bit shift; After “a” above, the algorithm f[(6m +2) + k] mode 13 should be applied to the cipher text, where m is the message and k =9 is a key; After “b” above, perform a systematic right bit shift to the new cipher; The message should then be transmitted. If on a trial basis, the message is “genomics” Use the mono – alphabetic letters where a = 0, b = 1, ….z = 25 to find the cipher that would be transmitted
Applications of Cryptography:
provide your code link form repl.it.comThe ciphertext below was encrypted using a substitution cipher. In your favorite language (in C++ or Python) crack encrypted text below using frequency analysis lrvmnir bpr sumvbwvr jx bpr lmiwv yjeryrkbi jx qmbm wi bpr xjvni mkd ymibrut jx irhx wi bpr riirkvr jx ymbinlmtmipw utn qmumbr dj w ipmhh but bj rhnvwdmbr bpr yjeryrkbi jx bpr qmbm mvvjudwko bj yt wkbrusurbmbwjk lmird jk xjubt trmui jx ibndt wb wi kjb mk rmit bmiq bj rashmwk rmvp yjeryrkb mkd wbi iwokwxwvmkvr mkd ijyr ynib urymwk nkrashmwkrd bj ower m vjyshrbr rashmkmbwjk jkr cjnhd pmer bj lr fnmhwxwrd mkd wkiswurd bj invp mk rabrkb bpmb pr vjnhd urmvp bpr ibmbr jx rkhwopbrkrd ywkd vmsmlhr jx urvjokwgwko ijnkdhrii ijnkd mkd ipmsrhrii ipmsr w dj kjb drry ytirhx bpr xwkmh mnbpjuwbt lnb yt rasruwrkvr cwbp qmbm pmi hrxb kj djnlb bpmb bpr xjhhjcwko wi bpr sujsru msshwvmbwjk mkd wkbrusurbmbwjk w jxxru yt bprjuwri wk bpr pjsr bpmb bpr riirkvr jx jqwkmcmk qmumbr cwhh urymwk wkbmvb…
This question concems 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)%6256, 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 encrypt various plaintexts 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 OXAA 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 a) Encrypt the plaintext "cheery using CTR mode. Please enter your answer in hex. (Please do *"not"" enter an 0x, as this has been done) Ox b) Encrypt the plaintext "chirper" using ECB mode Please enter your answer in hex Ox C) Encrypt the plaintext…
Description: You are required to implement a server program that responds to both TCP and UDP requests from the corresponding clients. The service provided by the server is simply a substitute cipher/cryptosystem. The cipher replaces one character to another based on some key. The input message (plaintext) can be of any size but composed only of small-letter (abcdefghijklmnopqrstuvwxyz). The output (ciphertext) has the same size and composed also of small-letter characters. The key is limited to 7 characters (also small-letter). You need to treat each digit as a number according to its position in the alphabet. For example, a=0, b=1, …,z=26. To encrypt a message, you add a plaintext character with the corresponding key character (mod 26). The key might be repeated to match the length of the plaintext message. In the decryption, you subtract the corresponding key character from the ciphertext character (mod 26). Again the key might be repeated as necessary. For example, if the…
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Suppose we have an alphabet of 30 characters, labelled as the integers modulo 30. Suppose that a 2 × 2 affine shift cipher is used to encode a message and it is known that the translation vector B is (0,0). The cipher text comes along as "21, 2, 13, 3, 14, 7, 0, 15,..." and you figure out through some other means that the plaintext starts "15, 16, 1, 1, plaintext? ." What is the encryption key? What are the next four digits of the ....
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The problem at the center of RSA is finding the e’th root of the ciphertext c modulo N where N = pq is a product of (distinct) primes. This is hard to do for large N if you don’t know the factors p and q, but once you know p and q it becomes easy! Write a finction findRoot(c,e,p,q) which solves the equation xe ≡ c mod N for x where N = pq