NWS620S Tutorial 1: Symmetric Encryption - DES Encryption is the translation of data into a secret code so that only authorised entities can read it. Encrypting data is considered a very effective way of achieving data security. To access encrypted data, you must have access to a secret key that enables you to decrypt it. Unencrypted data is called plain text; encrypted data is referred to as cipher text. There are two types of encryption: • Symmetric encryption • Asymmetric encryption/public key encryption Data Encryption Standard is the most widely used encryption scheme. It was issued in 1977 as Federal Information Processing Standard 46 (FIPS 46) by the National Institute of Standards and Technology (NIST). The algorithm itself is referred to as the Data Encryption Algorithm (DEA). You can visit the following for more information on DES • http://page.math.tu-berlin.de/~kant/teaching/hess/krypto-ws2006/des.htm • https://ccm.net/contents/134-introduction-to-encryption-with-des OR Read the document DES found in Network Security course on E-Leaning Questions/problems
1. How many keys are required for two people to communicate via a symmetric cipher? ___________________________________________________________________________
2. How can symmetric cyphertext be created? ___________________________________________________________________________ ___________________________________________________________________________
3. The paragraph following is a description of DES algorithm. Fill in the blank spaces with words below:
64 bits, 16, Feistel network, encryption , 56 bits The plaintext is _______________ in length and the key is __________________in length; longer plaintext amounts are processed in 64-bit blocks. The DES structure is a minor variation of the ____________________. There are 16 rounds of processing. From the original 56-bit key, __________________________subkeys are generated, one of which is used for each round. The process of decryption with DES is essentially the same as the ____________process. The rule is as follows: Use the ciphertext as input to the DES algorithm, but use the subkeys Ki in reverse order. That is, use K16 on the first iteration, K15 on the second iteration, and so on until K1 is used on the 16th and last iteration.
4. Consider the condition imposed on one-time pad: One-time pad - requires the use of a one-time pre-shared key the same size as, or longer than, the message being sent. Select in the table below whether it is true that the key given would be suitable to be used as a pre-shared key for the plaintext to be encrypted with one-time pad. Plaintext size Pre-shared key size True/False 64 56 102 123 89 89
5. DES decryption rule is as follows: Use the ciphertext as input to the DES algorithm, but use the subkeys Kn in reverse order. That is, use K16 on the first iteration, K15 on the second iteration, and so on until K1 is used on the 16th and last iteration
a. Which key is used on the third iteration? ___________________________________________________________________________
b. Which key is used on the last iteration? ___________________________________________________________________________
6. Look at Figure 1 on How DES works. Given: the plaintext: 0123456789ABCDEF
a. Convert it to binary ______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________
b. The first step of DES enciphering plaintext is the initial permutation (IP) given by: 58 50 42 34 26 18 10 2 60 52 44 36 28 20 12 4 62 54 46 38 30 22 14 6 64 56 48 40 32 24 16 8 57 49 41 33 25 17 9 1 59 51 43 35 27 19 11 3 61 53 45 37 29 21 13 5 63 55 47 39 31 23 15 7 What is the result of applying IP on your plaintext you got in (6a)? ________________________________________________________________________ ________________________________________________________________________
c. IP you got in (6b) is then divided into two equal sizes block; left half (L) and right half (R). What is the size of each half? ________________________________________________________________________
d. Write down left half (L0). ________________________________________________________________________
e. Write down right half (R0). ________________________________________________________________________
f. Let K be ABCD1234DEFC5678. Follow the example in class to derive K1, K2, K3, K4, K5, K6 Write down the values of the keys. ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________
g. We know Rn = Ln-1 + f(Rn-1, Kn). Given that f(R0; K1) is 11111010100001011110110001011111. + is bitwise XOR . What is R1? ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________
h. What is the value of L1? ___________________________________________________________________________ ___________________________________________________________________________
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