Mixing
Any Efficient Multiplication with a Primitive Root over Prime Field?
2
$begingroup$
Description: to multiply the "complex unit" $w^{N/4}$ over a prime field, i.e., $w^N equiv 1 bmod (,p)$ (suppose $p, N$ do provide such primitive root).
- I am implementing the radix-4 Number Theoretic Transform which involves the multiplications with the complex unit "$j$".
- It seems that in the complex domain, multiplication with the complex unit is "almost" free, i.e., $a + bj to -b + aj$. That is one of the reasons that radix-4 FFT can save more number of multiplications compared with the radix-2 implementation.
- However, over the prime field, I have no idea whether this free multiplication is possible or not.
Thanks.
fast-fourier-transform
asked Jan 21 at 5:03
FionserFionser
114
$endgroup$
add a comment |
2
$begingroup$
Description: to multiply the "complex unit" $w^{N/4}$ over a prime field, i.e., $w^N equiv 1 bmod (,p)$ (suppose $p, N$ do provide such primitive root).
- I am implementing the radix-4 Number Theoretic Transform which involves the multiplications with the complex unit "$j$".
- It seems that in the complex domain, multiplication with the complex unit is "almost" free, i.e., $a + bj to -b + aj$. That is one of the reasons that radix-4 FFT can save more number of multiplications compared with the radix-2 implementation.
- However, over the prime field, I have no idea whether this free multiplication is possible or not.
Thanks.
fast-fourier-transform
asked Jan 21 at 5:03
FionserFionser
114
$endgroup$
1
$begingroup$
Finite field multiplication is always almost free when representing elements as powers of a primitive root. Also, you don't always have a "complex unit" (i.e., -1 is not always a square). So maybe you can clarify what you are asking?
$endgroup$
– Morgan Rodgers
Jan 21 at 5:40
1
$begingroup$
@MorganRodgers Suppose that a such complex unit do exist, such as when $p = 1 bmod 2N$. Thank your for the answer.
$endgroup$
– Fionser
Jan 21 at 5:51
1
$begingroup$
Then yes, if $alpha$ is primitive element of $mathbb{F}_{q}$ and $j = alpha^{frac{q-1}{4}}$ then $j^{2} = -1$ and so $j(a+bj) = -b+aj$. Is this what you are asking?
$endgroup$
– Morgan Rodgers
Jan 21 at 6:20
1
$begingroup$
No, I mean for complex numbers, we only need to switch its real part and image part for multiplying with $j$, which is almost free. However, for a prime field, the $j$ is given as an integer $alpha^{q-1/4}$. To multiply $alpha^{q-1/4}$ with any element in the prime field, it seems we need to do the expensive integer multiplication followed with a modulus correction step.
$endgroup$
– Fionser
Jan 21 at 9:59
1
$begingroup$
Usually finite field multiplication is done using a table lookup (see Zech's logarithms).
$endgroup$
– Morgan Rodgers
Jan 21 at 17:23
add a comment |
2
2
2
$begingroup$
Description: to multiply the "complex unit" $w^{N/4}$ over a prime field, i.e., $w^N equiv 1 bmod (,p)$ (suppose $p, N$ do provide such primitive root).
- I am implementing the radix-4 Number Theoretic Transform which involves the multiplications with the complex unit "$j$".
- It seems that in the complex domain, multiplication with the complex unit is "almost" free, i.e., $a + bj to -b + aj$. That is one of the reasons that radix-4 FFT can save more number of multiplications compared with the radix-2 implementation.
- However, over the prime field, I have no idea whether this free multiplication is possible or not.
Thanks.
fast-fourier-transform
asked Jan 21 at 5:03
FionserFionser
114
$endgroup$
Description: to multiply the "complex unit" $w^{N/4}$ over a prime field, i.e., $w^N equiv 1 bmod (,p)$ (suppose $p, N$ do provide such primitive root).
- I am implementing the radix-4 Number Theoretic Transform which involves the multiplications with the complex unit "$j$".
- It seems that in the complex domain, multiplication with the complex unit is "almost" free, i.e., $a + bj to -b + aj$. That is one of the reasons that radix-4 FFT can save more number of multiplications compared with the radix-2 implementation.
- However, over the prime field, I have no idea whether this free multiplication is possible or not.
Thanks.
fast-fourier-transform
fast-fourier-transform
asked Jan 21 at 5:03
FionserFionser
114
asked Jan 21 at 5:03
FionserFionser
114
edited Jan 21 at 5:49
Fionser
asked Jan 21 at 5:03
FionserFionser
114
asked Jan 21 at 5:03
FionserFionser
114
asked Jan 21 at 5:03
FionserFionser
114
114
1
$begingroup$
Finite field multiplication is always almost free when representing elements as powers of a primitive root. Also, you don't always have a "complex unit" (i.e., -1 is not always a square). So maybe you can clarify what you are asking?
$endgroup$
– Morgan Rodgers
Jan 21 at 5:40
1
$begingroup$
@MorganRodgers Suppose that a such complex unit do exist, such as when $p = 1 bmod 2N$. Thank your for the answer.
$endgroup$
– Fionser
Jan 21 at 5:51
1
$begingroup$
Then yes, if $alpha$ is primitive element of $mathbb{F}_{q}$ and $j = alpha^{frac{q-1}{4}}$ then $j^{2} = -1$ and so $j(a+bj) = -b+aj$. Is this what you are asking?
$endgroup$
– Morgan Rodgers
Jan 21 at 6:20
1
$begingroup$
No, I mean for complex numbers, we only need to switch its real part and image part for multiplying with $j$, which is almost free. However, for a prime field, the $j$ is given as an integer $alpha^{q-1/4}$. To multiply $alpha^{q-1/4}$ with any element in the prime field, it seems we need to do the expensive integer multiplication followed with a modulus correction step.
$endgroup$
– Fionser
Jan 21 at 9:59
1
$begingroup$
Usually finite field multiplication is done using a table lookup (see Zech's logarithms).
$endgroup$
– Morgan Rodgers
Jan 21 at 17:23
add a comment |
1
$begingroup$
Finite field multiplication is always almost free when representing elements as powers of a primitive root. Also, you don't always have a "complex unit" (i.e., -1 is not always a square). So maybe you can clarify what you are asking?
$endgroup$
– Morgan Rodgers
Jan 21 at 5:40
1
$begingroup$
@MorganRodgers Suppose that a such complex unit do exist, such as when $p = 1 bmod 2N$. Thank your for the answer.
$endgroup$
– Fionser
Jan 21 at 5:51
1
$begingroup$
Then yes, if $alpha$ is primitive element of $mathbb{F}_{q}$ and $j = alpha^{frac{q-1}{4}}$ then $j^{2} = -1$ and so $j(a+bj) = -b+aj$. Is this what you are asking?
$endgroup$
– Morgan Rodgers
Jan 21 at 6:20
1
$begingroup$
No, I mean for complex numbers, we only need to switch its real part and image part for multiplying with $j$, which is almost free. However, for a prime field, the $j$ is given as an integer $alpha^{q-1/4}$. To multiply $alpha^{q-1/4}$ with any element in the prime field, it seems we need to do the expensive integer multiplication followed with a modulus correction step.
$endgroup$
– Fionser
Jan 21 at 9:59
1
$begingroup$
Usually finite field multiplication is done using a table lookup (see Zech's logarithms).
$endgroup$
– Morgan Rodgers
Jan 21 at 17:23
1
1
$begingroup$
Finite field multiplication is always almost free when representing elements as powers of a primitive root. Also, you don't always have a "complex unit" (i.e., -1 is not always a square). So maybe you can clarify what you are asking?
$endgroup$
– Morgan Rodgers
Jan 21 at 5:40
$begingroup$
Finite field multiplication is always almost free when representing elements as powers of a primitive root. Also, you don't always have a "complex unit" (i.e., -1 is not always a square). So maybe you can clarify what you are asking?
$endgroup$
– Morgan Rodgers
Jan 21 at 5:40
1
1
$begingroup$
@MorganRodgers Suppose that a such complex unit do exist, such as when $p = 1 bmod 2N$. Thank your for the answer.
$endgroup$
– Fionser
Jan 21 at 5:51
$begingroup$
@MorganRodgers Suppose that a such complex unit do exist, such as when $p = 1 bmod 2N$. Thank your for the answer.
$endgroup$
– Fionser
Jan 21 at 5:51
1
1
$begingroup$
Then yes, if $alpha$ is primitive element of $mathbb{F}_{q}$ and $j = alpha^{frac{q-1}{4}}$ then $j^{2} = -1$ and so $j(a+bj) = -b+aj$. Is this what you are asking?
$endgroup$
– Morgan Rodgers
Jan 21 at 6:20
$begingroup$
Then yes, if $alpha$ is primitive element of $mathbb{F}_{q}$ and $j = alpha^{frac{q-1}{4}}$ then $j^{2} = -1$ and so $j(a+bj) = -b+aj$. Is this what you are asking?
$endgroup$
– Morgan Rodgers
Jan 21 at 6:20
1
1
$begingroup$
No, I mean for complex numbers, we only need to switch its real part and image part for multiplying with $j$, which is almost free. However, for a prime field, the $j$ is given as an integer $alpha^{q-1/4}$. To multiply $alpha^{q-1/4}$ with any element in the prime field, it seems we need to do the expensive integer multiplication followed with a modulus correction step.
$endgroup$
– Fionser
Jan 21 at 9:59
$begingroup$
No, I mean for complex numbers, we only need to switch its real part and image part for multiplying with $j$, which is almost free. However, for a prime field, the $j$ is given as an integer $alpha^{q-1/4}$. To multiply $alpha^{q-1/4}$ with any element in the prime field, it seems we need to do the expensive integer multiplication followed with a modulus correction step.
$endgroup$
– Fionser
Jan 21 at 9:59
1
1
$begingroup$
Usually finite field multiplication is done using a table lookup (see Zech's logarithms).
$endgroup$
– Morgan Rodgers
Jan 21 at 17:23
$begingroup$
Usually finite field multiplication is done using a table lookup (see Zech's logarithms).
$endgroup$
– Morgan Rodgers
Jan 21 at 17:23
add a comment |
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1
$begingroup$
Finite field multiplication is always almost free when representing elements as powers of a primitive root. Also, you don't always have a "complex unit" (i.e., -1 is not always a square). So maybe you can clarify what you are asking?
$endgroup$
– Morgan Rodgers
Jan 21 at 5:40
1
$begingroup$
@MorganRodgers Suppose that a such complex unit do exist, such as when $p = 1 bmod 2N$. Thank your for the answer.
$endgroup$
– Fionser
Jan 21 at 5:51
1
$begingroup$
Then yes, if $alpha$ is primitive element of $mathbb{F}_{q}$ and $j = alpha^{frac{q-1}{4}}$ then $j^{2} = -1$ and so $j(a+bj) = -b+aj$. Is this what you are asking?
$endgroup$
– Morgan Rodgers
Jan 21 at 6:20
1
$begingroup$
No, I mean for complex numbers, we only need to switch its real part and image part for multiplying with $j$, which is almost free. However, for a prime field, the $j$ is given as an integer $alpha^{q-1/4}$. To multiply $alpha^{q-1/4}$ with any element in the prime field, it seems we need to do the expensive integer multiplication followed with a modulus correction step.
$endgroup$
– Fionser
Jan 21 at 9:59
1
$begingroup$
Usually finite field multiplication is done using a table lookup (see Zech's logarithms).
$endgroup$
– Morgan Rodgers
Jan 21 at 17:23