/**
* EdDSA-Java by str4d * * To the extent possible under law, the person who associated CC0 with * EdDSA-Java has waived all copyright and related or neighboring rights * to EdDSA-Java. * * You should have received a copy of the CC0 legalcode along with this * work. If not, see <https://creativecommons.org/publicdomain/zero/1.0/>. * */
package net.i2p.crypto.eddsa.math.ed25519;
import net.i2p.crypto.eddsa.math.*;
/**
* Helper class for encoding/decoding from/to the 32 byte representation. * <p> * Reviewed/commented by Bloody Rookie (nemproject@gmx.de) */
public class Ed25519LittleEndianEncoding extends Encoding {
/**
* Encodes a given field element in its 32 byte representation. This is done in two steps:
* <ol>
* <li>Reduce the value of the field element modulo $p$.
* <li>Convert the field element to the 32 byte representation.
* </ol><p>
* The idea for the modulo $p$ reduction algorithm is as follows:
* </p>
* <h2>Assumption:</h2>
* <ul>
* <li>$p = 2^{255} - 19$
* <li>$h = h_0 + 2^{25} * h_1 + 2^{(26+25)} * h_2 + \dots + 2^{230} * h_9$ where $0 \le |h_i| \lt 2^{27}$ for all $i=0,\dots,9$.
* <li>$h \cong r \mod p$, i.e. $h = r + q * p$ for some suitable $0 \le r \lt p$ and an integer $q$.
* </ul><p>
* Then $q = [2^{-255} * (h + 19 * 2^{-25} * h_9 + 1/2)]$ where $[x] = floor(x)$.
* </p>
* <h2>Proof:</h2>
* <p>
* We begin with some very raw estimation for the bounds of some expressions:
* <p>
* $$
* \begin{equation}
* |h| \lt 2^{230} * 2^{30} = 2^{260} \Rightarrow |r + q * p| \lt 2^{260} \Rightarrow |q| \lt 2^{10}. \\
* \Rightarrow -1/4 \le a := 19^2 * 2^{-255} * q \lt 1/4. \\
* |h - 2^{230} * h_9| = |h_0 + \dots + 2^{204} * h_8| \lt 2^{204} * 2^{30} = 2^{234}. \\
* \Rightarrow -1/4 \le b := 19 * 2^{-255} * (h - 2^{230} * h_9) \lt 1/4
* \end{equation}
* $$
* <p>
* Therefore $0 \lt 1/2 - a - b \lt 1$.
* <p>
* Set $x := r + 19 * 2^{-255} * r + 1/2 - a - b$. Then:
* <p>
* $$
* 0 \le x \lt 255 - 20 + 19 + 1 = 2^{255} \\
* \Rightarrow 0 \le 2^{-255} * x \lt 1.
* $$
* <p>
* Since $q$ is an integer we have
* <p>
* $$
* [q + 2^{-255} * x] = q \quad (1)
* $$
* <p>
* Have a closer look at $x$:
* <p>
* $$
* \begin{align}
* x &= h - q * (2^{255} - 19) + 19 * 2^{-255} * (h - q * (2^{255} - 19)) + 1/2 - 19^2 * 2^{-255} * q - 19 * 2^{-255} * (h - 2^{230} * h_9) \\
* &= h - q * 2^{255} + 19 * q + 19 * 2^{-255} * h - 19 * q + 19^2 * 2^{-255} * q + 1/2 - 19^2 * 2^{-255} * q - 19 * 2^{-255} * h + 19 * 2^{-25} * h_9 \\
* &= h + 19 * 2^{-25} * h_9 + 1/2 - q^{255}.
* \end{align}
* $$
* <p>
* Inserting the expression for $x$ into $(1)$ we get the desired expression for $q$.
*/
public byte[] encode(FieldElement x) {
int[] h = ((Ed25519FieldElement)x).t;
int h0 = h[0];
int h1 = h[1];
int h2 = h[2];
int h3 = h[3];
int h4 = h[4];
int h5 = h[5];
int h6 = h[6];
int h7 = h[7];
int h8 = h[8];
int h9 = h[9];
int q;
int carry0;
int carry1;
int carry2;
int carry3;
int carry4;
int carry5;
int carry6;
int carry7;
int carry8;
int carry9;
// Step 1:
// Calculate q
q = (19 * h9 + (1 << 24)) >> 25;
q = (h0 + q) >> 26;
q = (h1 + q) >> 25;
q = (h2 + q) >> 26;
q = (h3 + q) >> 25;
q = (h4 + q) >> 26;
q = (h5 + q) >> 25;
q = (h6 + q) >> 26;
q = (h7 + q) >> 25;
q = (h8 + q) >> 26;
q = (h9 + q) >> 25;
// r = h - q * p = h - 2^255 * q + 19 * q
// First add 19 * q then discard the bit 255
h0 += 19 * q;
carry0 = h0 >> 26; h1 += carry0; h0 -= carry0 << 26;
carry1 = h1 >> 25; h2 += carry1; h1 -= carry1 << 25;
carry2 = h2 >> 26; h3 += carry2; h2 -= carry2 << 26;
carry3 = h3 >> 25; h4 += carry3; h3 -= carry3 << 25;
carry4 = h4 >> 26; h5 += carry4; h4 -= carry4 << 26;
carry5 = h5 >> 25; h6 += carry5; h5 -= carry5 << 25;
carry6 = h6 >> 26; h7 += carry6; h6 -= carry6 << 26;
carry7 = h7 >> 25; h8 += carry7; h7 -= carry7 << 25;
carry8 = h8 >> 26; h9 += carry8; h8 -= carry8 << 26;
carry9 = h9 >> 25; h9 -= carry9 << 25;
// Step 2 (straight forward conversion):
byte[] s = new byte[32];
s[0] = (byte) h0;
s[1] = (byte) (h0 >> 8);
s[2] = (byte) (h0 >> 16);
s[3] = (byte) ((h0 >> 24) | (h1 << 2));
s[4] = (byte) (h1 >> 6);
s[5] = (byte) (h1 >> 14);
s[6] = (byte) ((h1 >> 22) | (h2 << 3));
s[7] = (byte) (h2 >> 5);
s[8] = (byte) (h2 >> 13);
s[9] = (byte) ((h2 >> 21) | (h3 << 5));
s[10] = (byte) (h3 >> 3);
s[11] = (byte) (h3 >> 11);
s[12] = (byte) ((h3 >> 19) | (h4 << 6));
s[13] = (byte) (h4 >> 2);
s[14] = (byte) (h4 >> 10);
s[15] = (byte) (h4 >> 18);
s[16] = (byte) h5;
s[17] = (byte) (h5 >> 8);
s[18] = (byte) (h5 >> 16);
s[19] = (byte) ((h5 >> 24) | (h6 << 1));
s[20] = (byte) (h6 >> 7);
s[21] = (byte) (h6 >> 15);
s[22] = (byte) ((h6 >> 23) | (h7 << 3));
s[23] = (byte) (h7 >> 5);
s[24] = (byte) (h7 >> 13);
s[25] = (byte) ((h7 >> 21) | (h8 << 4));
s[26] = (byte) (h8 >> 4);
s[27] = (byte) (h8 >> 12);
s[28] = (byte) ((h8 >> 20) | (h9 << 6));
s[29] = (byte) (h9 >> 2);
s[30] = (byte) (h9 >> 10);
s[31] = (byte) (h9 >> 18);
return s;
}
static int load_3(byte[] in, int offset) {
int result = in[offset++] & 0xff;
result |= (in[offset++] & 0xff) << 8;
result |= (in[offset] & 0xff) << 16;
return result;
}
static long load_4(byte[] in, int offset) {
int result = in[offset++] & 0xff;
result |= (in[offset++] & 0xff) << 8;
result |= (in[offset++] & 0xff) << 16;
result |= in[offset] << 24;
return ((long)result) & 0xffffffffL;
}
/**
* 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 $2^{25.5}$ bit representation.
*/
public FieldElement decode(byte[] in) {
long h0 = load_4(in, 0);
long h1 = load_3(in, 4) << 6;
long h2 = load_3(in, 7) << 5;
long h3 = load_3(in, 10) << 3;
long h4 = load_3(in, 13) << 2;
long h5 = load_4(in, 16);
long h6 = load_3(in, 20) << 7;
long h7 = load_3(in, 23) << 5;
long h8 = load_3(in, 26) << 4;
long h9 = (load_3(in, 29) & 0x7FFFFF) << 2;
long carry0;
long carry1;
long carry2;
long carry3;
long carry4;
long carry5;
long carry6;
long carry7;
long carry8;
long carry9;
// Remember: 2^255 congruent 19 modulo p
carry9 = (h9 + (long) (1<<24)) >> 25; h0 += carry9 * 19; h9 -= carry9 << 25;
carry1 = (h1 + (long) (1<<24)) >> 25; h2 += carry1; h1 -= carry1 << 25;
carry3 = (h3 + (long) (1<<24)) >> 25; h4 += carry3; h3 -= carry3 << 25;
carry5 = (h5 + (long) (1<<24)) >> 25; h6 += carry5; h5 -= carry5 << 25;
carry7 = (h7 + (long) (1<<24)) >> 25; h8 += carry7; h7 -= carry7 << 25;
carry0 = (h0 + (long) (1<<25)) >> 26; h1 += carry0; h0 -= carry0 << 26;
carry2 = (h2 + (long) (1<<25)) >> 26; h3 += carry2; h2 -= carry2 << 26;
carry4 = (h4 + (long) (1<<25)) >> 26; h5 += carry4; h4 -= carry4 << 26;
carry6 = (h6 + (long) (1<<25)) >> 26; h7 += carry6; h6 -= carry6 << 26;
carry8 = (h8 + (long) (1<<25)) >> 26; h9 += carry8; h8 -= carry8 << 26;
int[] h = new int[10];
h[0] = (int) h0;
h[1] = (int) h1;
h[2] = (int) h2;
h[3] = (int) h3;
h[4] = (int) h4;
h[5] = (int) h5;
h[6] = (int) h6;
h[7] = (int) h7;
h[8] = (int) h8;
h[9] = (int) h9;
return new Ed25519FieldElement(f, h);
}
/**
* 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 $\{0,2,4,\dots,q-1\}$
* <p>
* Preconditions:
* </p><ul>
* <li>$|x|$ bounded by $1.1*2^{26},1.1*2^{25},1.1*2^{26},1.1*2^{25}$, etc.
* </ul>
*
* @return true if $x$ is in $\{1,3,5,\dots,q-2\}$, false otherwise.
*/
public boolean isNegative(FieldElement x) {
byte[] s = encode(x);
return (s[0] & 1) != 0;
}
}