Cryptography-Data Communication Systems-Assignment Solution, Exercises of Data Communication Systems and Computer Networks

This file contains solution to problems related Data Communication Systems. Mr. Prajin Ahuja assigned task at Birla Institute of Technology and Science. Its main points are: Cryptography, Exercise, Session, Symmetric, Encryption, NBCM, Encryption, Plaintext, Cipehertext, Alice

Typology: Exercises

2011/2012

Uploaded on 07/26/2012

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CHAPTER 30
Cryptography
Solutions to Odd-Numbered Review Questions and Exercises
Review Questions
1. Only one key (the shared secret key) is needed for two-way communication. How-
ever, for more security, it is recommended that a different key be used for each
direction.
3. Each person in the first group needs to have 10 keys to communicate with all peo-
ple in the second group. This means we need at least 10 ร— 10 = 100 keys. Note that
the same keys can be used for communication in the reverse direction. However,
note that we are not considering the communication between the people in the
same group. For this purpose, we would need more keys.
5. For two-way communication, 4 keys are needed. Alice needs a private key and a
public key; Bob needs a private key and a public key.
7. For two-way communication, the people in the first group need 10 pairs of keys,
and the people in the second group need a separate 10 pairs of keys. In other
words, for two-way communication 40 keys are needed.
Exercises
9. If the two persons have two pairs of asymmetric keys, then they can send messages
using these keys to create a session symmetric key, a key which is valid for one
session and should not be used again. Another solution is to use a trusted center
that creates and send symmetric keys to both of them using the symmetric key or
asymmetric key that has been already established between each person and the
trusted center. We will discuss this mechanism in Chapter 31.
11.
a. We can show the encryption character by character. We encode characters A to
Z as 0 to 25. To wrap, we subtract 26.
T 1 9 + 2 0 = 3 9 โˆ’ 26 = 13 โ†’ N
H 0 7 + 2 0 = 2 7 โˆ’ 26 = 01 โ†’ B
I 08 + 2 0 = 2 8 โˆ’ 26 = 02 โ†’ C
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CHAPTER 30

Cryptography

Solutions to Odd-Numbered Review Questions and Exercises

Review Questions

  1. Only one key (the shared secret key) is needed for two-way communication. How- ever, for more security, it is recommended that a different key be used for each direction.
  2. Each person in the first group needs to have 10 keys to communicate with all peo- ple in the second group. This means we need at least 10 ร— 10 = 100 keys. Note that the same keys can be used for communication in the reverse direction. However, note that we are not considering the communication between the people in the same group. For this purpose, we would need more keys.
  3. For two-way communication, 4 keys are needed. Alice needs a private key and a public key; Bob needs a private key and a public key.
  4. For two-way communication, the people in the first group need 10 pairs of keys, and the people in the second group need a separate 10 pairs of keys. In other words, for two-way communication 40 keys are needed.

Exercises

  1. If the two persons have two pairs of asymmetric keys, then they can send messages using these keys to create a session symmetric key , a key which is valid for one session and should not be used again. Another solution is to use a trusted center that creates and send symmetric keys to both of them using the symmetric key or asymmetric key that has been already established between each person and the trusted center. We will discuss this mechanism in Chapter 31.

a. We can show the encryption character by character. We encode characters A to Z as 0 to 25. To wrap, we subtract 26.

T 19 + 20 = 39 โˆ’ 26 = 13 โ†’ N H 07 + 20 = 27 โˆ’ 26 = 01 โ†’ B I 08 + 20 = 28 โˆ’ 26 = 02 โ†’ C

The encrypted message is NBCM CM UH YRYLWCMY.

b. We can show the decryption character by character. We encode characters A to Z as 0 to 25. To wrap the negative numbers, we add 26.

The decrypted message is THIS IS AN EXERCISE.

  1. We can, but it is not safe at all. The best we can do is to change a 0 sometimes to 0 and sometimes to 1 and to change a 1 sometimes to 0 and sometimes to 1. It can be easily broken using trial and error.

S 18 + 20 = 38 โˆ’ 26 = 12 โ†’ M

I 08 + 20 = 28 โˆ’ 26 = 02 โ†’ C

S 18 + 20 = 38 โˆ’ 26 = 12 โ†’ M

A 00 + 20 = 20 โ†’ U

N 13 + 20 = 33 โˆ’ 26 = 07 โ†’ H

E 04 + 20 = 24 โ†’ Y

X 23 + 20 = 43 โˆ’ 26 = 17 โ†’ R

E 04 + 20 = 24 โ†’ Y

R 17 + 20 = 37 โˆ’ 26 = 11 โ†’ L

C 02 + 20 = 22 โ†’ W

I 08 + 20 = 28 โˆ’ 26 = 02 โ†’ C

S 18 + 20 = 38 โˆ’ 26 = 12 โ†’ M

E 04 + 20 = 24 โ†’ Y

N 13 โˆ’ 20 = โˆ’07 + 26 = 19 โ†’ T

B 01 โˆ’ 20 = โˆ’19 + 26 = 07 โ†’ H

C 02 โˆ’ 20 = โˆ’18 + 26 = 08 โ†’ I

M 12 โˆ’ 20 = โˆ’08 + 26 = 18 โ†’ S

C 02 โˆ’ 20 = โˆ’18 + 26 = 08 โ†’ I

M 12 โˆ’ 20 = โˆ’08 + 26 = 18 โ†’ S

U 20 โˆ’ 20 = 00 โ†’ A

H 07 โˆ’ 20 = โˆ’13 + 26 = 13 โ†’ N

Y 24 โˆ’ 20 = 04 โ†’ E

R 17 โˆ’ 20 = โˆ’03 + 26 = 23 โ†’ X

Y 24 โˆ’ 20 = 04 โ†’ E

L 11 โˆ’ 20 = โˆ’09 + 26 = 17 โ†’ R

W 22 โˆ’ 20 = 02 โ†’ C

C 02 โˆ’ 20 = โˆ’18 + 26 = 08 โ†’ I

M 12 โˆ’ 20 = โˆ’08 + 26 = 18 โ†’ S

Y 24 โˆ’ 20 = 04 โ†’ E