1. Networks and Communication Systems Department1
NET 311D
INFORMATION SECURITY
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TA. Anfal AlHazzaa
TUTORIAL 3 (How DES Works in Detail )

2. Recalling 1
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3. Cryptography Categories3
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4. Symmetric Key Symmetric Key Traditional Ciphers Symmetric Key4
Symmetric Key
Symmetric Key
Modern Round Ciphers
Traditional Ciphers
Simple Modern Cipher
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5. Modern Round Ciphers (bit orinted)5
Symmetric Key
Modern Round Ciphers
Data Encryption Standard (DES)
Advanced Encryption Standard (AES)
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6. Example: How DES Works in Detail6
Example:
Let M be the plain text message
M = ABCDEF
After convert it to binary (Base 16)
M =
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7. Example: How DES Works in Detail cont.7
Let K be the hexadecimal key
 K = BBCDFF1.
After convert it to binary (Base 16)
 K =
DES operates on the 64-bit blocks using key sizes of bits.
The keys are actually stored as being 64 bits long, but every 8th bit in the key is not used (i.e. bits numbered 8, 16, 24, 32, 40, 48, 56, and 64).
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8. Data Encryption Standard (DES)8
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9. Step 1: Create 16 subkeys, each of which is 48-bits long.9
The 64-bit key is permuted according to the following table, PC-1 ( to make it 56 bits)
From the original 64-bit key
K =
we get the 56-bit permutation
K+ =
001111
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10. How to perform the permutation ?
The 64-bit key is permuted according to the following table, PC-1 ( to make it 56 bits) K= First : we make a table of the bits , Second : we number these bits 1 2 3 4 5 6 7 8
The same way apply on slide 14,18,21,28,30 the only differences is the permutation table 1 9 16 17 24 25 32 33 40 41 48 49 56 57 64

11. Con.
Third : we rearrange the bits according to given table here the table is PC-1 : for example the bit number 57 will move to bit number 1 , bit number 49 will move to bit number 2 , bit number 41 will be in number 3 and so on .

12. Con.
Forth : we will make new table called K+ based on these new arrangement
Original key K+ 1
FINALLY The key will be 56 bit
K+ =

13. Step 1: Create 16 subkeys, each of which is 48-bits long.10
Next, split this key into left and right halves, C0 and D0, where each half has 28 bits.
Example: From the permuted key K+, we get
K+ = |
C0 =
D0 =
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14. Step 1: Create 16 subkeys, each of which is 48-bits long.11
With C0 and D0 defined, we now create sixteen blocks Cn and Dn, 1<=n<=16.
Each pair of blocks Cn and Dn is formed from the previous pair Cn-1 and Dn- 1, respectively, for n = 1, 2, ..., 16,
Using the following schedule of "left shifts" of the previous block. To do a left shift, move each bit one place to the left, except for the first bit, which is cycled to the end of the block.
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15. Step 1: Create 16 subkeys, each of which is 48-bits long.12
Example: From original pair C0 and D0 we obtain:
C0 =
D0 =
C1 =
D1 =
C2 =
D2 =
C3 =
D3 =
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16. Step 1: Create 16 subkeys, each of which is 48-bits long.13
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17. Step 1: Create 16 subkeys, each of which is 48-bits long.14
We now form the keys Kn, for 1<=n<=16, by applying the following permutation table to each of the concatenated pairs CnDn.
Each pair has 56 bits, but PC-2 only uses 48 of these.
Example: For the first key we have C1D1 =
which, after we apply the permutation PC-2, becomes
K1 =
110010 
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18. Step 1: Create 16 subkeys, each of which is 48-bits long.15
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19. Review Step 1 1 Converting the key from hexadecimal to binary 2
Permuting the 64-bit key to make it 56 bits 3
Splitting the key into left and right halves C0 and D0, 4
Creating sixteen blocks Cn and Dn by shifting 5
Permuting each of the concatenated pairs Cn Dn

20. (Review) Step 1: Create 16 subkeys, each of which is 48-bits long
Convert the key to binary (16 base)  64 bits
The 64-bit key is permuted according to the PC-1 table ( to make it 56 bits)
Split this key into left and right halves, C0 and D0, where each half has 28 bits.
Using the schedule of "left shifts“, we now create sixteen blocks from C0
and D0
We now form the keys Kn, for 1<=n<=16, by applying the permutation table (PC-2) to each of the concatenated pairs CnDn.
Now : we have 16 subkeys (48 bits)
TA. Anfal AlHazzaa -Networks and Communication Systems Department

21. Data Encryption Standard (DES)17
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22. Step 2: Encode each 64-bit block of data18
There is an initial permutation IP of the 64 bits of the message data M. This rearranges the bits according to the following table
Example: Applying the initial permutation to the block of text M, given previously, we get
M =
IP =
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23. Step 2: Encode each 64-bit block of data19
Next divide the permuted block IP into a left half L0 of 32 bits, and a right half R0 of 32 bits.
Example: From IP, we get L0 and R0
IP = |
L0 =
R0 =
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24. Step 2: Encode each 64-bit block of data20
Then for n going from 1 to 16 we calculate
Ln = Rn-1
Rn = Ln f(Rn-1,Kn)
Example: For n = 1, we have
L1 = R0 =
R1 = L0 + f(R0,K1)
K1 =
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25. Step 2: Encode each 64-bit block of data21
To Calculate function f:
1. We first expand each block Rn-1 from 32 bits to 48 bits(This is done by using a selection table that repeats some of the bits in Rn-1. We'll call the use of this selection table the function E ,Thus E(Rn-1) has a 32 bit input block, and a 48 bit output block)
Example: We calculate E(R0) from R0 as follows:
R0 =
E(R0) =
010101
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26. Step 2: Encode each 64-bit block of data22
To Calculate function f:
2. Next in the f calculation, we XOR the output E(Rn-1) with the key Kn:
Kn E(Rn-1).
Example: For K1 , E(R0), we have
K1 =
E(R0) =
K1+E(R0) =
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27. Step 2: Encode each 64-bit block of data23
To Calculate function f:
3. We now have 48 bits, or eight groups of six bits.
- We now have 48 bits, or eight groups of six bits.
Write the previous result, which is 48 bits, in the form:
Kn E(Rn-1) =B1B2B3B4B5B6B7B8,
where each Bi is a group of six bits. We now calculate
S1(B1)S2(B2)S3(B3)S4(B4)S5(B5)S6(B6)S7(B7)S8(B8)
where Si(Bi) referres to the output of the i-th S box.
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29. Step 2: Encode each 64-bit block of data25
 S1(011000) 
 Answer: 0101  Hence S1(011011) = 0101.
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30. Step 2: Encode each 64-bit block of data26
 S2(010001) S2 
 Answer: 1100  Hence S1(010001) = 1100
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31. Step 2: Encode each 64-bit block of data27
Example:
For the first round, we obtain as the output of the eight S boxes:
K1 + E(R0) =
S1(B1)S2(B2)S3(B3)S4(B4)S5(B5)S6(B6)S7(B7)S8(B8) =
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32. Step 2: Encode each 64-bit block of data28
To Calculate function f:
4. The final stage in the calculation of f is to do a permutation P of the S-box output to obtain the final value of f:
f = P(S1(B1)S2(B2)...S8(B8))
Example: From the output of the eight S boxes:
 S1(B1)S2(B2)S3(B3)S4(B4)S5(B5)S6(B6)S7(B7)S8(B8) =
 f =
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33. Step 2: Encode each 64-bit block of data29
 R1 = L0 + f(R0 , K1 ) = =
For round 1 (n=1)
L1 = R0 =
R1 = L0 + f(R0,K1) =
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34. Step 2: Encode each 64-bit block of data30
In the next round(n-2) , we will have L2 = R1, which is the block we just calculated, and then we must calculate R2 =L1 + f(R1, K2), and so on for 16 rounds.
At the end of the sixteenth round we have the blocks L16 and R16.
We then reverse the order of the two blocks into the 64-bit block
R16L16
Apply a final permutation IP-1 as defined by the following table: -
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35. Step 2: Encode each 64-bit block of data31
Example:
If we process all 16 blocks using the method defined previously, we get, on the 16th round,
L16 =
R16 =
We reverse the order of these two blocks and apply the final permutation to
 R16L16 =
 IP-1 =
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36. Step 2: Encode each 64-bit block of data32
which in hexadecimal format is
85E813540F0AB405.
This is the encrypted form of
M = ABCDEF  C = 85E813540F0AB405.
Decryption is simply the inverse of encryption, following the same steps as
above, but reversing the order in which the subkeys are applied.
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37. Review step 2 1 Converting the message from hexadecimal to binary 2
Initial permutation IP 3
Dividing the permuted block IP into a left half L0 of 32 bits, and a right half R0 of 32 bits.

38. Con. Review The Feistel (F) function1
Calculating
Ln = Rn-1 Rn = Ln-1 + f(Rn-1,Kn) 2
Looping 16 times 3
Expanding each block Rn-1 4
XORing the output E(Rn-1) with the key Kn. 5
CalculateingS1(B1)S2(B2)S3(B3)S4(B4)S5(B5)S6(B6)S7(B7)S8(B8)using S-box 6
Permutating the S-box output to obtain the final value of f. 7
Reversing the order of the two blocks R16L16 8
Final permutation IP-1 9
Converting the IP-1 from binary to hexadecimal to get the cipher text

39. (Review) Step 2: Encode each 64-bit block of data33
Convert message to binary (base 16)  64 bits
Apply the initial permutation to the block of text M.
Divide the permuted block IP into a left half L0 of 32 bits, and a right half R0 of 32 bits.
Then for n going from 1 to 16 we calculate
Ln = Rn-1
Rn = Ln-1 + f(Rn-1,Kn)
To calculate f:
Expand each block Rn-1 from 32 bits to 48 bits(using selection table E)
Kn E(Rn-1)  48 bits
Where each Bi is a group of six bits, calculate
S1(B1)S2(B2)S3(B3)S4(B4)S5(B5)S6(B6)S7(B7)S8(B8)
Do a permutation P of the S-box output to obtain the final value of f
Repeat these steps 16
times
5. At the end , Reverse the order of these two blocks ( L16, R16 ) and apply the final permutation.
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40. References 34
The DES Algorithm Illustrated. (n.d.). Retrieved 2015, from
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