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I have support set of length 120 of a bent function over $GF(256)$. $GF(256)=<\beta>$ is generated by the polynomial $x^8+ x^4 + x^3 + x^2 + 1$. The support set is of the form {$\alpha 0 1 0$, $\alpha^2 0 \alpha 0$, $1 0 \alpha^2 0$, $0 \alpha 0 1$, $0 \alpha^2 0 \alpha$, ...} in which each element of a 4_tuple comes from $GF(4)$ and $\alpha$ is generator of $GF(4)$ s.t. $\alpha^2 + \alpha + 1=0$.

Is there a way that I can calculate the Walsh transform by using these 4_tuples? or

Or is there any way to write each 4_tuple element of the form $\beta$. If it is possible, I can calculate the Walsh transform of this function.

Kenan123
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  • Bent functions are defined here. – Jean Marie Feb 18 '23 at 19:48
  • An interesting answer here about a useful representation of GF(256). – Jean Marie Feb 18 '23 at 19:51
  • Are those 4-tuples the coordinate sequences of elements of $GF(256)$ with respect to a monomial basis like $1,\beta,\beta^2,\beta^3$? If so, we need more information because that degree eight polynomial splits into a product of two quartics over $GF(4)$. Or are they something else? – Jyrki Lahtonen Feb 19 '23 at 04:47
  • I wrote each 4-tuple as elements of $\mathbb{F}{2}^8$ so that $\alpha^2=11$, $\alpha=10$, $1=01$ and $0=00$ . e.g. $(\alpha 1 \alpha^2 0)$ = $(10011100)$. Then I used the 120 elements as a support set of a Boolean function $\mathbb{F}{2}^8\rightarrow \mathbb{F}{2}$ and I calculated Walsh transform. I wonder whether there is a way of calculating Walsh transform by using 4-tuples. Or how can I change each 4-tuple to elements of $\mathbb{F}{2^8}$. – Kenan123 Feb 19 '23 at 11:35

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