EBK COMPUTER NETWORKING
7th Edition
ISBN: 8220102955479
Author: Ross
Publisher: PEARSON
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Chapter 8, Problem P5P
Program Plan Intro
Given Information:
For a given “key” Alice and Bob would need to keep eight tables, where each table is of 8 bits by 8 bits.
Block Cipher:
A Block Cipher is a method of encrypting text in which a cryptographic key and
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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
01001000 = rgx1L2X3X4X5X6X7
when encrypted by the LFSR produced the ciphertext
10000011 = 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, Po = 1).
Say you obtain a ciphertext of an English sentence encrypted with an unknown key using this scheme. For a brute-force attack, what is the maximum number of keys would you have to test?
Alice and Bob are using the EIGamal cipher with
the parameters p = 89 and a = 3.
Suppose Alice is observed sending public ephemeral
key KE 7 and Bob sends public key ß = 44.
=
Their private keys are i = 81 and d = 28, respectively.
a) What is the masking key k?
b) What is the encryption of x = 31?
Chapter 8 Solutions
EBK COMPUTER NETWORKING
Ch. 8 - Prob. R1RQCh. 8 - Prob. R2RQCh. 8 - Prob. R3RQCh. 8 - Prob. R4RQCh. 8 - Prob. R5RQCh. 8 - Prob. R6RQCh. 8 - Prob. R7RQCh. 8 - Prob. R8RQCh. 8 - Prob. R9RQCh. 8 - Prob. R10RQ
Ch. 8 - Prob. R11RQCh. 8 - Prob. R12RQCh. 8 - Prob. R13RQCh. 8 - Prob. R14RQCh. 8 - Prob. R15RQCh. 8 - Prob. R16RQCh. 8 - Prob. R17RQCh. 8 - Prob. R18RQCh. 8 - Prob. R19RQCh. 8 - Prob. R20RQCh. 8 - Prob. R21RQCh. 8 - Prob. R22RQCh. 8 - Prob. R23RQCh. 8 - Prob. R24RQCh. 8 - Prob. R25RQCh. 8 - Prob. R26RQCh. 8 - Prob. R27RQCh. 8 - Prob. R28RQCh. 8 - Prob. R29RQCh. 8 - Prob. R30RQCh. 8 - Prob. R31RQCh. 8 - Prob. R32RQCh. 8 - Prob. R33RQCh. 8 - Prob. P1PCh. 8 - Prob. P2PCh. 8 - Prob. P3PCh. 8 - Prob. P4PCh. 8 - Prob. P5PCh. 8 - Prob. P6PCh. 8 - Prob. P8PCh. 8 - Prob. P12PCh. 8 - Prob. P13PCh. 8 - Prob. P14PCh. 8 - Prob. P18PCh. 8 - Prob. P20PCh. 8 - Prob. P21PCh. 8 - Prob. P22PCh. 8 - Prob. P23P
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Similar questions
- Alice and Bob are using the ElGamal cipher with the parameters p = 199 and a = 3. Their parameters are small and we decide to crack the cipher. Suppose Alice is observed sending public ephemeral key k = 103 and Bob sends public key = 136. ß Alice is observed transmitting the ciphertext y = 68. a) What is the masking key km? b) What is the plaintext?arrow_forwardSuppose 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.arrow_forwardSuppose 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 ....arrow_forward
- 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). 0010arrow_forwardConsider the following scenario: a lengthy string of text is encrypted using a Vigen'ere cypher, and the same string is discovered in the ciphertext beginning at various locations. Demonstrate how this information may be utilised to assist calculate the length of the key by explaining how it is obtained.arrow_forwardAttacks on a cipher that use letter frequency can be made more difficult by assigning multiple codes to the most frequently used letters in the alphabet. Such a cipher is called a homophonic cipher. Explain why multiple codes cause no increase in the time to decipher a message if the key is known.arrow_forward
- A Man-in-the-Middle attack on the Diffie-Hellman key exchange protocol may be effectively carried out if the adversary produces two public-private key pairs for the assault. Could the same assault be carried out with a single pair? Justify and describe the event in full.arrow_forwardA secret agent successfully intercepts an encrypted message below: Ciphertext 1 Y C P B R I M R E P J T M P V It is found that the encrypted message is encrypted using a Vigenère cipher and can be decrypted into Plaintext 1 M E E T M E A T T H E P A R K What is the key for the above cipher text and if the next cipher text is encrypted using the same key as the previous message, what is the plaintext? Ciphertext 2 A L W G F J F C C N T Y D N Xarrow_forwardConsider a plaintext consisting 1000 elements of {0,1,2,3}. The text is encrypted using a poly-alphabetic cipher with a key of length 250 and the resulting ciphertext is a sequence of 1000 threes. Explain why this implies the sum of all the elements of the plaintext is zero modulo 4.arrow_forward
- Suppose a plaintext message x is encrypted using the Caesar cipher b = a +26 10 where each letter of the alphabet is assigned a numerical value (A = 0, B = 1, C = 2, ..., Z = 25), the variable a represents the numerical value of a letter in x, and b represents the numerical value of the corresponding letter in the ciphertext message y. If y = WKDR SC MYYV, then decrypt y to obtain the plaintext message x. Enter the decryption of each letter of y in the boxes below. NOTE: All answers should be capitalized and do not insert extra blank spaces. W → K →→ D→ R→ S→ C → M→ Y → Y → RemakingMIN... V →arrow_forwardAlice and Bob are using the ElGamal cipher with the parameters ?=19p=19 and ?=2α=2. Their parameters are small and we decide to crack the cipher. Suppose Alice is observed sending public ephemeral key ??=7kE=7 and Bob sends public key ?=17β=17. Alice is observed transmitting the ciphertext ?=5y=5. a) What is the masking key ??kM? b) What is the plaintext?arrow_forwardAlice and Bob are using the ElGamal cipher with = : 89 and a = 3. the parameters p Suppose Alice is observed sending public ephemeral key KE 7 and Bob sends public key = 44. = Their private keys are i a) What is the masking key km? = 81 and d = b) What is the encryption of x = 31? 28, respectively.arrow_forward
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