Why was A doubled in size

Tarick Welling

3/22/24, 4:22 PM

Why was the dimension of A doubled in kyber?

LWE encryption uses a public matrix A of dimension K but kyber uses a double matrix A resulting in $A ^{ k * k * n }$

When deriving the results of the definition of gen, enc en dec this results in:

$$ RA_0 S + RA_1 S + RE + E_3 + \frac{q}{2}m - RA_0 S + RA_1 S + ES \\ $$

Which reduces into: $$ RE + E_3 + \frac{q}{2}m - ES $$

Which is equivalent to the $A^k$ scheme. Why was the size of A increased by $k$?

0 + 3

lattice-crypto

kyber

yoyo

3/22/24, 8:44 PM

For some module dimension $k$ (a security parameter), $A\in R_q^{k\times k}$, a $k\times k$ matrix with entries in $R_q=\mathbb{Z}[x]/(q,x^{256}+1)$ with $q=3329$. In the specification, it is actually $\hat{A}$, the number theoretic transform of $A$ that is expanded from seed (or its transpose). For practical purposes, $A$ or $\hat{A}$ is a $k\times k$ matrix with entries that are 256-dimensional vectors over $\mathbb{Z}/q\mathbb{Z}$, or more simply a $k\times k \times256$ array with integer entries (say in the range $[0,q)$).

Tarick Welling

3/22/24, 9:23 PM

@yoyo exactly, but the theory already works for a $k \times 256$ array, so why was it made to be $k \times k \times 256$?

yoyo

3/22/24, 9:40 PM

You may want to compare different learning with errors key exchanges: LWE (like FRODO), RLWE (like NewHope), and MLWE (like Kyber). They're all very similar, but use algebraic structures of increasing complexity.

Elon Musk

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