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Theorem Let $A:{GF(2^m)}^n \to {GF(2^m)}^n$ be an $n\times n \{0,1\}$-matrix over $GF(2^m)$. Then the branch number of $A$ is at most $\frac{2n+4}{3}$. Let $A$ be the almost-MDS matrix. So its branch number is $n$. Using above theorem we have $$n \leq \frac{2n+4}{3} $$ So $n=2,3$ or $4$. The size of matrix is a tool for finding upper bound of branch ...


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The matrix is not MDS over $GF(2)$; No binary MDS codes exist and non nonbinary (over $GF(2^n)$ MDS codes would have this generator whose scalar entries are in the field $GF(2)$). Over $GF(2^n)$ The branch number, which is the minimum weight of the corresponding linear code is 4, in $GF(2^n)$ for all $n$. This covers all possible fields of interest for ...



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