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Dimitri
Dimitri
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In the equation  $ a^x = a^y + 1$ over a finite field $\text{GF}(2^n)$, where '$a$' is the primitive element, can one calculate $x$ as a function of $y$ without having to resort to taking the logarithm of $(a^y + 1)$?

I.e. calculate $x$ as a function of $y$, without using "$\text{log}(k)$" as part of this function.

I need to implement this in hardware, and using the $\text{log}()$ function is very costly, so I was thinking if there is a smarter way to do it.

Regards,

-Dimitri

In the equation$ a^x = a^y + 1$ over a finite field $\text{GF}(2^n)$, where '$a$' is the primitive element, can one calculate $x$ as a function of $y$ without having to resort to taking the logarithm of $(a^y + 1)$?

I.e. calculate $x$ as a function of $y$, without using "$\text{log}(k)$" as part of this function.

I need to implement this in hardware, and using the $\text{log}()$ function is very costly, so I was thinking if there is a smarter way to do it.

Regards,

-Dimitri

In the equation  $ a^x = a^y + 1$ over a finite field $\text{GF}(2^n)$, where '$a$' is the primitive element, can one calculate $x$ as a function of $y$ without having to resort to taking the logarithm of $(a^y + 1)$?

I.e. calculate $x$ as a function of $y$, without using "$\text{log}(k)$" as part of this function.

I need to implement this in hardware, and using the $\text{log}()$ function is very costly, so I was thinking if there is a smarter way to do it.

Regards,

-Dimitri

I edited in the format of latex
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edit approved May 30, 2014 at 10:58
user21574
edit approved May 30, 2014 at 10:58
user21574

In the equation a^x = a^y + 1$ a^x = a^y + 1$ over a finite field GF(2^n)$\text{GF}(2^n)$, where 'a''$a$' is the primitive element, can one calculate x$x$ as a function of y$y$ without having to resort to taking the logarithm of (a^y + 1)$(a^y + 1)$?

I.e. calculate x$x$ as a function of y$y$, without using "log(k)"$\text{log}(k)$" as part of this function.

I need to implement this in hardware, and using the log()$\text{log}()$ function is very costly, so I was thinking if there is a smarter way to do it.

Regards,

-Dimitri

In the equation a^x = a^y + 1 over a finite field GF(2^n), where 'a' is the primitive element, can one calculate x as a function of y without having to resort to taking the logarithm of (a^y + 1)?

I.e. calculate x as a function of y, without using "log(k)" as part of this function.

I need to implement this in hardware, and using the log() function is very costly, so I was thinking if there is a smarter way to do it.

Regards,

-Dimitri

In the equation$ a^x = a^y + 1$ over a finite field $\text{GF}(2^n)$, where '$a$' is the primitive element, can one calculate $x$ as a function of $y$ without having to resort to taking the logarithm of $(a^y + 1)$?

I.e. calculate $x$ as a function of $y$, without using "$\text{log}(k)$" as part of this function.

I need to implement this in hardware, and using the $\text{log}()$ function is very costly, so I was thinking if there is a smarter way to do it.

Regards,

-Dimitri

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Dimitri
Dimitri
  • 87
  • 5

In a finite field with characteristic 2, can I calculate the log(K+1) based on the log(K)?

In the equation a^x = a^y + 1 over a finite field GF(2^n), where 'a' is the primitive element, can one calculate x as a function of y without having to resort to taking the logarithm of (a^y + 1)?

I.e. calculate x as a function of y, without using "log(k)" as part of this function.

I need to implement this in hardware, and using the log() function is very costly, so I was thinking if there is a smarter way to do it.

Regards,

-Dimitri