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Home/ Questions/Q 4019146
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Editorial Team
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Editorial Team
Asked: May 20, 20262026-05-20T10:04:11+00:00 2026-05-20T10:04:11+00:00

So we all agree keys are a fixed-length of 128bits or 192bits or 256bits.

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So we all agree keys are a fixed-length of 128bits or 192bits or 256bits. If our context was 50 characters in size (bytes) % 16 = 2 bytes. So we encrypt the context in 3 times, but the remaining two bytes how will they be stored in the State block. Should I pad them, the standard doesn’t specify how to handle such conditions.

MixColumns stage is the most complicated aspect in the AES, however I have been unable to understand the mathematical representation. I have an understanding of the matrix multiplication, but I’m surprised of the mathematical results. Multiplying a value by 2, shift left for little endian 1 position and shift right for big endian. If we had the most significant bit was set as 1 (0x80) then we should XOR the shifted result with 0x1B. I thought by multiplying by 3 it would mean to shift the value 2 positions.

I’ve checked the various sources on Wikipedia, even the tutorial that provides a C implementation. But I’m more interested to complete my own implementation! Thank you for any possible input.

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  1. Editorial Team
    Editorial Team
    2026-05-20T10:04:11+00:00Added an answer on May 20, 2026 at 10:04 am

    In the mix columns stage the exponents are being multiplied.

    take this example
    AA*3
    10101010*00000011
    is
    x^7+x^5+x^3+x^1*x^1+x^0
    x^1+x^0 is 3 represented in polynomial form
    x^7+x^5+x^3+x^1 is AA represented in polynomial form
    first take x^1 and dot multiply it by the polynomial for AA.
    that results in...
    x^8+x^6+x^4+x^2 ... adding one to each exponent
    then reduce this to 8 bits by XoRing by 11B
    11B is x^8+x^4+x^3+x^1+x^0 in polynomial form.
    so...
    x^8+x6+x^4+    x^2
    x^8+   x^4+x^3+    x^1+x^0
    leaves
    x^6+x^3+x^2+x^1+x^0 which is AA*2
    now take AA and dot multiply by x^0 (basically AA*1)
    that gives you
    x^7+x^5+x^3+x^1 ... a duplicate of the original value.
    then exclusive or AA*2 with AA*1
    x^7+    x^5+x^3+    x^1
        x^6+    x^3+x^2+x^1+x^0
    which leaves
    x^7+x^6+x^5+x^2+x^0 or 11100101 or E5
    I hope that helps.
    here also is a document detailing the specifics of how mix columns works.
    

    mix_columns.pdf

    EDIT:Normal matrix multiplication does not apply to this ..so forget about normal matrices.

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