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>
   $$
   \begin{align}
   x &amp;= h - q * (2^{255} - 19) + 19 * 2^{-255} * (h - q * (2^{255} - 19)) + 1/2 - 19^2 * 2^{-255} * q - 19 * 2^{-255} * (h - 2^{230} * h_9) \\
     &amp;= h - q * 2^{255} + 19 * q + 19 * 2^{-255} * h - 19 * q + 19^2 * 2^{-255} * q + 1/2 - 19^2 * 2^{-255} * q - 19 * 2^{-255} * h + 19 * 2^{-25} * h_9 \\
     &amp;= h + 19 * 2^{-25} * h_9 + 1/2 - q^{255}.
   \end{align}
   $$
   <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;
}

}