One can compute the result of the function without applying xor \\(n – 1\\) times. ### Theorem \\begin{align} 2m \\oplus (2m + 1) = 1 \\ \\ \\forall m \\in \\mathbb{N}\_{0} \\end{align} By this fact, \\begin{align} 4m \\oplus (4m+1) \\oplus (4m+2) \\...