INFA 640 Cryptology and Data Protection Part 1: Hash Codes. Provide the expected numerical answer along with your explanation. 1-a. A certain communications system uses messages that are 1024 bit...

INFA 640
Cryptology and Data Protection
Part 1: Hash Codes. Provide the expected numerical answer along with your
explanation.

1-a. A certain communications system uses messages that are 1024 bit
long. How many different messages can be supported my such system?
Solution:
2
1024
= 179 769 313 486 231 590 772 930 519 078 902 473 361 797 697 894 230 657 273
430 081 157 732 675 805 500 963 132 708 477 322 407 536 021 120 113 879 871 393
357 658 789 768 814 416 622 492 847 430 639 474 124 377 767 893 424 865 485 276
302 219 601 246 094 119 453 082 952 085 005 768 838 150 682 342 462 881 473 913
110 540 827 237 163 350 510 684 586 298 239 947 245 938 479 716 304 835 356 329
624 224 137 216 different types of message.

1-b. Often messages are accompanied by hash codes. If the bit length of the
hash code is 2 to the sixth power, how many such hash codes can be
produced?
Solution:
2
6
= 64 different types of hash.

1-c. A certain communications system uses messages that are 2 to the
tenth power bits long. The accompanying hash code is 64 bits long. How
many messages could produce the same hash code?
Solution:
Since we use 64 bits long hash code then only 2
64
hash messages can be generated, after
that the message have to have same structure in order to match the same hash code for two
different messages. This is also known as collision resistance of the hashes.


Part 2. Alpha Encryption. Provide the answer along with a brief explanation

2-c. An encryption algorithm turned

GOOD MORNING AMERICA into FNNC LNQMHMF ZLDQHJZ

Identify the encryption algorithm and express it in a mathematical form.

Solution:
The encryption algorithm used here is the simple Caeser Cipher with key as -1.
Calculation:
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

Encrypting letter ‘G’:

G: (G + key) mod 26 = (7+(-1)) mod 26 = 6 hence it will be F. For a case when letter ‘A’
is used it will come ‘0’ hence, it will reverse and go to ‘Z’.

Part 3. PGP Desktop.

3-a. In the PGP Desktop encryption system there is (or is not) a green dot under a
field labeled Validity. Describe the significance of that dot.
Solution:
A green circle for valid keys that you do not own

3-b. In the PGP Desktop, describe the steps necessary to...