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I was wondering in the Chaskey, when the message m is split into fixed size blocks such as m1,m2,m3....ml, what would be the size of each message block?

Is it 128 bits each or it can be of any variable size?

m=m1||m2||m3||...||ml

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Looking at the original paper, the block sizes can be anything but all blocks except for last have to be of same size.

3.1 Mode of Operation
Chaskey uses an n-bit key K to process a message m of arbitrary size into a tag τ of t ≤ n bits.

For every key K, two subkeys K1, K2 are generated as shown in Algorithm 2. The message m is split into ℓ blocks m1, m2, . . . , mℓ of n bits each, except for the lastblock mℓ which may be incomplete.

But according to the author it is specially designed for 128 bit security so my guess is that it is better to use 128 bit keys.

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  • Thank You for your response. According to the research paper the key size and the indivisual message block should be the same. So if key is 128 bit, then eventually the each message block should be 128 bit as well.
    – Ni09
    Commented Jan 23, 2017 at 20:12
  • @Ni09 the question had no mention of key sizes so I didnt mention that in my answer as well. But in general, why the key size is equal to the block size in block cipher
    – Limit
    Commented Jan 23, 2017 at 22:10

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