Decode a FieldElement from its $(b-1)$-bit encoding. The highest bit is masked out. @param in the $(b-1)$-bit encoding of a FieldElement. @return

Decodes a given field element in its 10 byte $2^{25.5}$ representation. @param in The 32 byte representation. @return The field element in its $2^{25

@deprecated

@throws InvalidAlgorithmParameterException if spec is ONE_SHOT_MODE and update() already called @see #ONE_SHOT_MODE

@throws SignatureException if in one-shot mode

@return the public key

@return the hash

Gets the representation of the group element. @return The representation.

@return will be null if constructed directly from the private key

Gets the $T$ value of the group element. This is for most representation the projective $T$ coordinate. @return The $T$ value.

Gets the $X$ value of the group element. This is for most representation the projective $X$ coordinate. @return The $X$ value.

Gets the $Y$ value of the group element. This is for most representation the projective $Y$ coordinate. @return The $Y$ value.

Gets the $Z$ value of the group element. This is for most representation the projective $Z$ coordinate. @return The $Z$ value.

@return the private key

Create a new Ed25519::SigningKey from the given seed value @param seed [String] 32-byte seed value from which the key should be derived

String inspection that does not leak secret values

Show hex representation of serialized coordinate in string inspection

Invert this field element. <p> The inverse is found via Fermat's little theorem:<br> $a^p \cong a \mod p$ and therefore $a^{(p-2)} \cong a^{-1} \mod p

From the Ed25519 paper:<br> $x$ is negative if the $(b-1)$-bit encoding of $x$ is lexicographically larger than the $(b-1)$-bit encoding of -x. If $q$

From the Ed25519 paper:<br> $x$ is negative if the $(b-1)$-bit encoding of $x$ is lexicographically larger than the $(b-1)$-bit encoding of $-x$. If $

Is the FieldElement negative in this encoding? <p> Return true if $x$ is in $\{1,3,5,\dots,q-2\}$<br> Return false if $x$ is in $\{0,2,4,\dots,q-1\}$

Gets a value indicating whether or not the field element is non-zero. @return 1 if it is non-zero, 0 otherwise.

Verify that a point is on its curve. @return true if the point lies on its curve.

$(ab+c) \bmod q$ <p> Input: </p><ul> <li>$a[0]+256 a[1]+\dots+256^{31} a[31] = a$ <li>$b[0]+256 b[1]+\dots+256^{31} b[31] = b$ <li>$c[0]+256 c[1]+\dot

$h = -f$ <p> TODO-CR BR: see above. <p> Preconditions: </p><ul> <li>$|f|$ bounded by $1.1 2^{25},1.1 2^{24},1.1 2^{25},1.1 2^{24},$ etc. </ul><p> Post

Gets this field element to the power of $(2^{252} - 3)$. This is a helper function for calculating the square root. <p> TODO-CR BR: I think it makes s

Reduction modulo the group order $q$. <p> Input: $s[0]+256 s[1]+\dots+256^{63} s[63] = s$ <p> Output: $s[0]+256 s[1]+\dots+256^{31} s[31] = s \bmo

To efficiently sign all the data in one shot, if it is available, use this method, which will avoid copying the data. Same as: <pre> setParameter(O

$h = 2 f f$ <p> Can overlap $h$ with $f$. <p> Preconditions: </p><ul> <li>$|f|$ bounded by $1.65 2^{26},1.65 2^{25},1.65 2^{26},1.65 2^{25},$ etc. </u

$h = f - g$ <p> Can overlap $h$ with $f$ or $g$. <p> TODO-CR BR: See above. <p> Preconditions: </p><ul> <li>$|f|$ bounded by $1.1 2^{25},1.1 2^{24},1.

To efficiently verify all the data in one shot, if it is available, use this method, which will avoid copying the data. Same as: <pre> setParameter