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 determinedthat the number of states is m = 4.The plaintext given by01110010 = 0X1X2X3X4X5X6X7when encrypted by the LFSR produced the ciphertext11000011 = YOY1Y2Y3Y3Y5Y6Y7-What are the tap bits of the LFSR? Please enter your answer as unspaced binary digits (e.g. 0101 torepresent p3 = 0, p2 = 1, p1 = 0, po = 1).

Answered: Image /qna-images/answer/9979e61a-f4e8-4950-ac33-8d1469b33214.jpg

Answered: Note: The notation from this problem is…

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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Solve the Encryption and Decryption process using Cipher Block Channing (CBC) for the given Plaintext M=1100101111111000. Key= 3214. IV=1011.
Consider a homophonic cipher in which H(A) = {000, 001, 101, 011} and H(B) = {111, 110, 010, 100}. What is the plaintext that results from the decryption of 110000100011? Answer:
Note: The notation from this problem is from Understanding Cryptography by Paar and Pelzl. Consider the LFSR represented by the polynomial x^4+x^3+x^2+1. What are the tap bits of the LFSR? Please enter your answer as unspaced binary digits (e.g. 010101 to represent p5=0,p4=1,p3=0,p2=1,p1=0,p0=1).
ONLY THE TAP BITS. 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=4m=4.The plaintext given by 01001011=x0x1x2x3x4x5x6x701001011=x0x1x2x3x4x5x6x7 when encrypted by the LFSR produced the ciphertext 11010110=y0y1y2y3y3y5y6y711010110=y0y1y2y3y3y5y6y7. 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,p0=1p3=0,p2=1,p1=0,p0=1).
Fast and correct solution please.
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.
Suppose your RSA public key is PK: {n, e} = {13861, 37}. Your friend sends you a ciphertext C = 9908. But unfortunately you have forgotten your private key, now you have to crack it yourself.a) Write down a possible condition of factors p and q.p= q=b) What is your private key SK: {d}?d=c) What is the plaintext of your friend’s message?The plaintext M =d) Suppose the plaintext M is a 12-digit number consisting of a prefix “19” and 1234567890. What is the corresponding ciphertext? Since M is greater than n, you only need to encrypt four digits at a time. Ignore redundant zeros. e.g., 0001 = 1. The result should contain three integers.The ciphertext C0 =
Consider the case of Alice sending a message, m, to Bob. Both Alice and Bob use public key cryptography and each has a public and private key as described in the text. The figure attached below shows the operations that Alice must perform to provide confidentiality, authentication, and integrity when sending a message to Bob over the network. We can use either symmetric key or public key cryptography to encrypt a message. For our purposes, either technique will encrypt the message, and applying both doesn't make it "more secure". Also, we can assume that the session key would remain a secret so the fact that it is discarded does not make it "more secure". Why do we use a session key, Ks, instead of relying only on public key Cryptography? In other words, why do we use both public key and symmetric key cryptography?
Please explain the problem in detail so that I can understand.
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?
Suppose a public-key cryptosystem encrypts integers (say, modulo another integer n). Show that if this scheme has the property that Dec(Enc(x) · Enc(y)) = x + y for any messages x, y, then the scheme is necessarily vulnerable to a CCA2 attack.
In RSA, suppose p = 3, q = 11, e = 7, and d = 3. Given the message m = 5, show the encryption process to obtain the ciphertext c from m and the decryption process to derive m back from c.

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