AES Encryption (cryptography) Question to solve .

I’m studying and need help with a Probability question to help me learn.

Question Part: (i am posting “WHAT” example in the example part to get and idea to solve AES)

Use this message:

“UCUMBERLAND IS A+”

Fill the message in the 4X4 MATRIX

and NOW Multiply by the AES MIX ROW – IN GALOIS FIELD (Below)

2 3 1 1
1 2 3 1
1 1 2 3
3 1 1 2

Remember the 2 is actually is actually 0010

3 is 0011

1 is 0001

YOU ARE PERFORMING MATRIX MULTIPLICATION (DOT PRODUCT) AND REDUCE IF NEEDED. THE RESULT HAS TO FIT IN ONE BYTE

THIS IS DIFFICULT AND TAKES TIME UNLESS YOU KNOW HOW TO USE THE LOOK UP TABLE IN WIKEPEDIA (MUL2 and MUL3)

For Reference posting WHAT example which is mentioned in the question :

When encrypting with AES, we only need to multiply by the Galois fields for 1, 2 and 3.

A few things to remember –

1. Multiply a Galois field by a Galois field for 1 – you get the same thing. It is like multiplying a number by 1. You get the same number back.
2. Multiply by two or three are different. The original values des change. We can multiply or we can use the look up tables.
3. There are two tables that are available MUL2 and MUL3.
4. Remember the resulting values have to be MOD 100011011  Do you remember why? This is the reducer to make the result fit in a byte

Let’s look at the word ‘what’ and apply AES to it

W is a 57 in HEX – represented by 0x57
h is a 68 in HEX – represented by 0x68
a is a 61 in HEX – represented by 0x61
t is a 74 in HEX – – represented by 0x74

Now we have to take the DOT PRODUCT and it is complex.
0x57
0x68 [ 0x02 0x03 0x01 0x01 ]
0x61
0x74

5 = 0101 (in bits)
7 = 0111 (in bits)
57 = 01010111 (Or use your scientific calculator)

NOW
How about the 02? -> (10)

Now multiple the 01010111 by 10 (0x57 times 0x02 – HEX) (remember you did this before)

X^6 + x^4 + x^2 + x^1 + x^0 multiply by X^1

0 1
1 2
2 3
4 5
6 7

Now we count
How many

7 – 1

6 — 0
5 — 1
4 — 0
3 — 1
2 — 1
1 – 0

So 0x57 DOT 0x02 = 1 0 1 0 1 1 0

Now do you need to reduce 

THE ANSWER IS NO.

Why  because the bits fit in a BYTE

NO REDUCER NEEDED.

****

Note: For the question posted the RUDUCER needed to solve

course: cryptography

AES Encryption Topic

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