Mixing
Fast inversion over large finite fields
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I was wondering if there is a "fastest" way to compute inversions over finite fields, especially if they are very large. I know that the standard way is the extended Euclidean algorithm, which runs in $O(log^2p)$, with $p$ being the characteristic of the finite field. In my case, $p=2^u3^vpm 1$ a very large prime. Is there a way to compute $a^{-1}text{ mod } p$ faster than with the Euclidean algorithm, in general, or maybe by somehow using the structure of the characteristic to our advantage?
Thank you for any help!
finite-fields computational-complexity
asked Jan 21 at 11:00
GemeisGemeis
133
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add a comment |
$begingroup$
I was wondering if there is a "fastest" way to compute inversions over finite fields, especially if they are very large. I know that the standard way is the extended Euclidean algorithm, which runs in $O(log^2p)$, with $p$ being the characteristic of the finite field. In my case, $p=2^u3^vpm 1$ a very large prime. Is there a way to compute $a^{-1}text{ mod } p$ faster than with the Euclidean algorithm, in general, or maybe by somehow using the structure of the characteristic to our advantage?
Thank you for any help!
finite-fields computational-complexity
asked Jan 21 at 11:00
GemeisGemeis
133
$endgroup$
1
$begingroup$
I'm not an expert on this so I cannot say for sure, but... have you checked out the potential of using Montgomery representatives? One of the first hits to the buzzword Montgomery inverse.
$endgroup$
– Jyrki Lahtonen
Jan 21 at 14:52
add a comment |
$begingroup$
I was wondering if there is a "fastest" way to compute inversions over finite fields, especially if they are very large. I know that the standard way is the extended Euclidean algorithm, which runs in $O(log^2p)$, with $p$ being the characteristic of the finite field. In my case, $p=2^u3^vpm 1$ a very large prime. Is there a way to compute $a^{-1}text{ mod } p$ faster than with the Euclidean algorithm, in general, or maybe by somehow using the structure of the characteristic to our advantage?
Thank you for any help!
finite-fields computational-complexity
asked Jan 21 at 11:00
GemeisGemeis
133
$endgroup$
I was wondering if there is a "fastest" way to compute inversions over finite fields, especially if they are very large. I know that the standard way is the extended Euclidean algorithm, which runs in $O(log^2p)$, with $p$ being the characteristic of the finite field. In my case, $p=2^u3^vpm 1$ a very large prime. Is there a way to compute $a^{-1}text{ mod } p$ faster than with the Euclidean algorithm, in general, or maybe by somehow using the structure of the characteristic to our advantage?
Thank you for any help!
finite-fields computational-complexity
finite-fields computational-complexity
asked Jan 21 at 11:00
GemeisGemeis
133
asked Jan 21 at 11:00
GemeisGemeis
133
asked Jan 21 at 11:00
GemeisGemeis
133
asked Jan 21 at 11:00
GemeisGemeis
133
asked Jan 21 at 11:00
GemeisGemeis
133
133
1
$begingroup$
I'm not an expert on this so I cannot say for sure, but... have you checked out the potential of using Montgomery representatives? One of the first hits to the buzzword Montgomery inverse.
$endgroup$
– Jyrki Lahtonen
Jan 21 at 14:52
add a comment |
1
$begingroup$
I'm not an expert on this so I cannot say for sure, but... have you checked out the potential of using Montgomery representatives? One of the first hits to the buzzword Montgomery inverse.
$endgroup$
– Jyrki Lahtonen
Jan 21 at 14:52
1
1
$begingroup$
I'm not an expert on this so I cannot say for sure, but... have you checked out the potential of using Montgomery representatives? One of the first hits to the buzzword Montgomery inverse.
$endgroup$
– Jyrki Lahtonen
Jan 21 at 14:52
$begingroup$
I'm not an expert on this so I cannot say for sure, but... have you checked out the potential of using Montgomery representatives? One of the first hits to the buzzword Montgomery inverse.
$endgroup$
– Jyrki Lahtonen
Jan 21 at 14:52
add a comment |
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$begingroup$
I'm not an expert on this so I cannot say for sure, but... have you checked out the potential of using Montgomery representatives? One of the first hits to the buzzword Montgomery inverse.
$endgroup$
– Jyrki Lahtonen
Jan 21 at 14:52