I suppose I already suggested that one computes a^{-1} mod b
as a^{b-1} mod b, using a plain old modexp.
I realise that this will be asymptotically slower, in this setting
O(n^3) vs O(n^2), but it ought have a much lower constant factor.
Torbjörn
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Ciao,
Il Ven, 27 Dicembre 2013 12:53 am, Torbjorn Granlund ha scritto:
I realise that this will be asymptotically slower, in this setting
O(n^3) vs O(n^2), but it ought have a much lower constant factor.
We will introduce a side-channel silent threshold...
Regards,
m
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