Page 202 - eProceeding - IRSTC & RESPEX 2017

Robbi Rahim / JOJAPS – JOURNAL ONLINE JARINGAN COT POLIPD

KE2 (Round 3) = C2DC
KE3 (Round 3) = CE86
KE4 (Round 3) = DEDC

3. Third Round
RotateLeft (CCCAE4CADCC6CA9AC2D8C2DCCE86DEDC, 25) = 95B98D953585B185B99D0DBDB99995C9

Split into 8 parts
KE5 (Round 3) = 95B9
KE6 (Round 3) = 8D95
KE1 (Round 4) = 3585
KE2 (Round 4) = B185
KE3 (Round 4) = B99D
KE4 (Round 4) = 0DBD
KE5 (Round 4) = B999
KE6 (Round 4) = 95C9

Do the rotation until the 8th round so that the results obtained are as follows:

Rotate Left
(E86DEDCCCCAE4CADCC6CA9AC2D8C2DCC, 25) = 99995C995B98D953585B185B99D0DBDB

Split into 8 parts (last 4 parts are not used):
KE1 (Transformation Output) = 9999
KE2 (Transformation Output) = 5C99
KE3 (Transformation Output) = 5B98
KE4 (Transformation Output) = D953

After determining the key used, the next is to encrypt the message process, below are the process:
Message = ICEEIEOK
Key = ConferenceMalang

1. First Round
01) L#1 = (X1 * K1) mod (2^16 + 1) = 4943 * 436F mod (2^16 + 1) = 39C1
02) L#2 = (X2 + K2) mod 2^16 = 4545 + 6E66 mod 2^16 = B3AB
03) L#3 = (X3 + K3) mod 2^16 = 4945 + 6572 mod 2^16 = AEB7
04) L#4 = (X4 * K4) mod (2^16 + 1) = 4F4B * 656E mod (2^16 + 1) = 89D0
05) L#5 = L#1 XOR L#3 = 39C1 XOR AEB7 = 9776
06) L#6 = L#2 XOR L#4 = B3AB XOR 89D0 = 3A7B
07) L#7 = (L#5 * K5) mod (2^16 + 1) = 9776 * 6365 mod (2^16 + 1) = 28C0
08) L#8 = (L#6 + L#7)) mod 2^16 = 3A7B + 28C0 mod 2^16 = 633B
09) L#9 = (L#8 * K6) mod (2^16 + 1) = 633B * 4D61 mod (2^16 + 1) = 3A5D
10) L#10 = (L#7 + L#9)) mod 2^16 = 28C0 + 3A5D mod 2^16 = 631D
11) L#11 = L#1 XOR L#9 = 39C1 XOR 3A5D = 039C
12) L#12 = L#3 XOR L#9 = AEB7 XOR 3A5D = 94EA
13) L#13 = L#2 XOR L#10 = B3AB XOR 631D = D0B6
14) L#14 = L#4 XOR L#10 = 89D0 XOR 631D = EACD

For the next round the round key is used:
X1 = L#11 = 039C
X2 = L#12 = 94EA
X3 = L#13 = D0B6
X4 = L#14 = EACD

2. Second Round
01) L#1 = (X1 * K1) mod (2^16 + 1) = 039C * 6C61 mod (2^16 + 1) = 2C95
02) L#2 = (X2 + K2) mod 2^16 = 94EA + 6E67 mod 2^16 = 0351

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