which one of the following is correct the Encrypt of "class is over felows" using Caesar cipher with 4 shifts?sziv jipsaw gpeww mw1gpeww mw sziv jipsawupp ibse up cf ibse jipsawclassAisAoverAfelowsAThe "Playfair Cipher" uses a matrix of sizeО бхб3х3O 4x 4О 5x5

In Caesar cipher technique, each letter is replaced by number of positions said in number of shifts.…

which one of the following is correct the Encrypt of "class is over felows" using Caesar cipher with 4 shifts? sziv jipsaw gpeww mw gpeww mw sziv jipsaw upp ibse up cf ibse jipsaw classAisAoverAfelowsA The "Playfair Cipher" uses a matrix of size бхб O 3 x 3 O 4x 4 O 5x 5

which one of the following is correct the Encrypt of "class is over felows" using Caesar cipher with 4 shifts? sziv jipsaw gpeww mw gpeww mw sziv jipsaw upp ibse up cf ibse jipsaw classAisAoverAfelowsA The "Playfair Cipher" uses a matrix of size бхб O 3 x 3 O 4x 4 O 5x 5

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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Decrypt Using these two prime numbers p: 11472448678961361675809388014248341 q: 46838031327322626556194693766252933 an RSA public key was constructed (these are same prime numbers we used in class) and the following message (an encoded palindromic sentence) has been sent to you: s: 206852398229333288394377911072663657292972237656591756203408205688278 What is that palindromic sentence?
Consider the simple “Caesar” cipher C = (L + 3) mod 26 giving Letter: ABCDEFGHIJKLMNOPQRSTUVWXYZ (0 1 2 24 25) Cipher: DEFGHIJKLMNOPQRSTUVWXYZABC (0 1 2 24 25) Encode the string “HELLO” What is the deciphering formula?
Q1: Based on Vigenere Encryption Cipher, if the plain text is "THE LENGTH OF BLOCK DEPENDS"" and the corresponding cipher text is " HWGKJGTV HBSZJCNGG PT IMRX". Find the key, in the case we have used uppercase and space.
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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 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 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. The ciphertext is given in hexadecimal. a) Decrypt the ciphertext "303…
use python please
What will be the ciphered text corresponding to “ALGORITHM” if bifid cipher is used for encryption with key as “KEY” with a period as 5? a) SBPISZTKZH b) PLNSOWGKQM c) SELFQEXBHM d) YFSGWNFBW
4. Encryption is used to secure data messages in computer networks. One possible way of encrypting text consists of rewriting the text by including first the characters found at even indexes (starting from 0) followed by the characters stored at odd indexes, while ignoring whitespace characters. For instance, the String "Exam is fun" can be encoded using the previously explained technique as "Easfnxmiu". To be able to recover the original message based on the encrypted version of the message, some additional information should be added to the encrypted message, namely, the number of characters found at even indexes, the number of characters found at odd indexes, and the indexes at which the white space characters existed in the original message. In this way, reconsidering our previous example, the string "Exam is fun" can be encoded using the previously explained technique as "5 4 4 7 Easfnxmiu". This is particularly true since the original message had 5 characters at even indexes…
Affine cipher is an example of? 1. Mono alphabetic cipher 2. Poly alphabetic cipher 3. Transposition cipher 4. Asymmetric cipher
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; } 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 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 "spider" 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 "spelling" using ECB mode. Please enter your answer in hex. Ox c) Encrypt the…
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; } 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 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 "lippo" using CTR mode. Please enter your answer in hex. (Please do **not** enter an Ox, as this has been done.) Ox b) Encrypt the plaintext "lippi" using ECB mode. Please enter your answer in hex. Ox c) Encrypt the…
A very different kind of mapping is achieved by performing some sort of permutation on the plaintext letters. This technique is referred to as a transposition cipher. Apply Row transposition on below given data. Plain Text: WE ARE DISCOVERDED SAVE YOUR-SELF. Key: 5444