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Cryptography And Network Security Principles And Practice 6th Edition William-Stallings Solutions Manual
Full Download: http://alibabadownload.com/product/cryptography-and-network-security-principles-and-practice-6th-edition-wi




CHAPTER 2 CLASSICAL ENCRYPTION
TECHNIQUES

ANSWERS TO QUESTIONS
2.1 Plaintext, encryption algorithm, secret key, ciphertext, decryption
algorithm.

2.2 Permutation and substitution.

2.3 One key for symmetric ciphers, two keys for asymmetric ciphers.

2.4 A stream cipher is one that encrypts a digital data stream one bit or
one byte at a time. A block cipher is one in which a block of plaintext
is treated as a whole and used to produce a ciphertext block of equal
length.

2.5 Cryptanalysis and brute force.

2.6 Ciphertext only. One possible attack under these circumstances is the
brute-force approach of trying all possible keys. If the key space is very
large, this becomes impractical. Thus, the opponent must rely on an
analysis of the ciphertext itself, generally applying various statistical
tests to it. Known plaintext. The analyst may be able to capture one
or more plaintext messages as well as their encryptions. With this
knowledge, the analyst may be able to deduce the key on the basis of
the way in which the known plaintext is transformed. Chosen
plaintext. If the analyst is able to choose the messages to encrypt, the
analyst may deliberately pick patterns that can be expected to reveal
the structure of the key.

2.7 An encryption scheme is unconditionally secure if the ciphertext
generated by the scheme does not contain enough information to
determine uniquely the corresponding plaintext, no matter how much
ciphertext is available. An encryption scheme is said to be
computationally secure if: (1) the cost of breaking the cipher exceeds
the value of the encrypted information, and (2) the time required to
break the cipher exceeds the useful lifetime of the information.

-9-

This sample only, Download all chapters at: alibabadownload.com

, 2.8 The Caesar cipher involves replacing each letter of the alphabet with
the letter standing k places further down the alphabet, for k in the
range 1 through 25.

2.9 A monoalphabetic substitution cipher maps a plaintext alphabet to a
ciphertext alphabet, so that each letter of the plaintext alphabet maps
to a single unique letter of the ciphertext alphabet.

2.10 The Playfair algorithm is based on the use of a 5  5 matrix of
letters constructed using a keyword. Plaintext is encrypted two letters at
a time using this matrix.

2.11 A polyalphabetic substitution cipher uses a separate
monoalphabetic substitution cipher for each successive letter of
plaintext, depending on a key.

2.12 1. There is the practical problem of making large quantities of random
keys. Any heavily used system might require millions of random
characters on a regular basis. Supplying truly random characters in
this volume is a significant task.
2. Even more daunting is the problem of key distribution and
protection. For every message to be sent, a key of equal length is
needed by both sender and receiver. Thus, a mammoth key
distribution problem exists.

2.13 A transposition cipher involves a permutation of the plaintext
letters.

2.14 Steganography involves concealing the existence of a message.


ANSWERS TO PROBLEMS
2.1 a. No. A change in the value of b shifts the relationship between
plaintext letters and ciphertext letters to the left or right uniformly,
so that if the mapping is one-to-one it remains one-to-one.
b. 2, 4, 6, 8, 10, 12, 13, 14, 16, 18, 20, 22, 24. Any value of a larger
than 25 is equivalent to a mod 26.
c. The values of a and 26 must have no common positive integer factor
other than 1. This is equivalent to saying that a and 26 are relatively
prime, or that the greatest common divisor of a and 26 is 1. To see
this, first note that E(a, p) = E(a, q) (0 ≤ p ≤ q < 26) if and only if
a(p – q) is divisible by 26. 1. Suppose that a and 26 are relatively
prime. Then, a(p – q) is not divisible by 26, because there is no way
to reduce the fraction a/26 and (p – q) is less than 26. 2. Suppose


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