/* * Copyright © 2016 Southern Storm Software, Pty Ltd. * Copyright © 2017-2021 WireGuard LLC. All Rights Reserved. * SPDX-License-Identifier: Apache-2.0 */ package com.wireguard.crypto; import com.wireguard.util.NonNullForAll; import java.util.Arrays; import androidx.annotation.Nullable; /** * Implementation of Curve25519 ECDH. *
* This implementation was imported to WireGuard from noise-java: * https://github.com/rweather/noise-java *
* This implementation is based on that from arduinolibs: * https://github.com/rweather/arduinolibs *
* Differences in this version are due to using 26-bit limbs for the * representation instead of the 8/16/32-bit limbs in the original. *
* References: http://cr.yp.to/ecdh.html, RFC 7748 */ @SuppressWarnings({"MagicNumber", "NonConstantFieldWithUpperCaseName", "SuspiciousNameCombination"}) @NonNullForAll public final class Curve25519 { // Numbers modulo 2^255 - 19 are broken up into ten 26-bit words. private static final int NUM_LIMBS_255BIT = 10; private static final int NUM_LIMBS_510BIT = 20; private final int[] A; private final int[] AA; private final int[] B; private final int[] BB; private final int[] C; private final int[] CB; private final int[] D; private final int[] DA; private final int[] E; private final long[] t1; private final int[] t2; private final int[] x_1; private final int[] x_2; private final int[] x_3; private final int[] z_2; private final int[] z_3; /** * Constructs the temporary state holder for Curve25519 evaluation. */ private Curve25519() { // Allocate memory for all of the temporary variables we will need. x_1 = new int[NUM_LIMBS_255BIT]; x_2 = new int[NUM_LIMBS_255BIT]; x_3 = new int[NUM_LIMBS_255BIT]; z_2 = new int[NUM_LIMBS_255BIT]; z_3 = new int[NUM_LIMBS_255BIT]; A = new int[NUM_LIMBS_255BIT]; B = new int[NUM_LIMBS_255BIT]; C = new int[NUM_LIMBS_255BIT]; D = new int[NUM_LIMBS_255BIT]; E = new int[NUM_LIMBS_255BIT]; AA = new int[NUM_LIMBS_255BIT]; BB = new int[NUM_LIMBS_255BIT]; DA = new int[NUM_LIMBS_255BIT]; CB = new int[NUM_LIMBS_255BIT]; t1 = new long[NUM_LIMBS_510BIT]; t2 = new int[NUM_LIMBS_510BIT]; } /** * Conditional swap of two values. * * @param select Set to 1 to swap, 0 to leave as-is. * @param x The first value. * @param y The second value. */ private static void cswap(int select, final int[] x, final int[] y) { select = -select; for (int index = 0; index < NUM_LIMBS_255BIT; ++index) { final int dummy = select & (x[index] ^ y[index]); x[index] ^= dummy; y[index] ^= dummy; } } /** * Evaluates the Curve25519 curve. * * @param result Buffer to place the result of the evaluation into. * @param offset Offset into the result buffer. * @param privateKey The private key to use in the evaluation. * @param publicKey The public key to use in the evaluation, or null * if the base point of the curve should be used. */ public static void eval(final byte[] result, final int offset, final byte[] privateKey, @Nullable final byte[] publicKey) { final Curve25519 state = new Curve25519(); try { // Unpack the public key value. If null, use 9 as the base point. Arrays.fill(state.x_1, 0); if (publicKey != null) { // Convert the input value from little-endian into 26-bit limbs. for (int index = 0; index < 32; ++index) { final int bit = (index * 8) % 26; final int word = (index * 8) / 26; final int value = publicKey[index] & 0xFF; if (bit <= (26 - 8)) { state.x_1[word] |= value << bit; } else { state.x_1[word] |= value << bit; state.x_1[word] &= 0x03FFFFFF; state.x_1[word + 1] |= value >> (26 - bit); } } // Just in case, we reduce the number modulo 2^255 - 19 to // make sure that it is in range of the field before we start. // This eliminates values between 2^255 - 19 and 2^256 - 1. state.reduceQuick(state.x_1); state.reduceQuick(state.x_1); } else { state.x_1[0] = 9; } // Initialize the other temporary variables. Arrays.fill(state.x_2, 0); // x_2 = 1 state.x_2[0] = 1; Arrays.fill(state.z_2, 0); // z_2 = 0 System.arraycopy(state.x_1, 0, state.x_3, 0, state.x_1.length); // x_3 = x_1 Arrays.fill(state.z_3, 0); // z_3 = 1 state.z_3[0] = 1; // Evaluate the curve for every bit of the private key. state.evalCurve(privateKey); // Compute x_2 * (z_2 ^ (p - 2)) where p = 2^255 - 19. state.recip(state.z_3, state.z_2); state.mul(state.x_2, state.x_2, state.z_3); // Convert x_2 into little-endian in the result buffer. for (int index = 0; index < 32; ++index) { final int bit = (index * 8) % 26; final int word = (index * 8) / 26; if (bit <= (26 - 8)) result[offset + index] = (byte) (state.x_2[word] >> bit); else result[offset + index] = (byte) ((state.x_2[word] >> bit) | (state.x_2[word + 1] << (26 - bit))); } } finally { // Clean up all temporary state before we exit. state.destroy(); } } /** * Subtracts two numbers modulo 2^255 - 19. * * @param result The result. * @param x The first number to subtract. * @param y The second number to subtract. */ private static void sub(final int[] result, final int[] x, final int[] y) { int index; int borrow; // Subtract y from x to generate the intermediate result. borrow = 0; for (index = 0; index < NUM_LIMBS_255BIT; ++index) { borrow = x[index] - y[index] - ((borrow >> 26) & 0x01); result[index] = borrow & 0x03FFFFFF; } // If we had a borrow, then the result has gone negative and we // have to add 2^255 - 19 to the result to make it positive again. // The top bits of "borrow" will be all 1's if there is a borrow // or it will be all 0's if there was no borrow. Easiest is to // conditionally subtract 19 and then mask off the high bits. borrow = result[0] - ((-((borrow >> 26) & 0x01)) & 19); result[0] = borrow & 0x03FFFFFF; for (index = 1; index < NUM_LIMBS_255BIT; ++index) { borrow = result[index] - ((borrow >> 26) & 0x01); result[index] = borrow & 0x03FFFFFF; } result[NUM_LIMBS_255BIT - 1] &= 0x001FFFFF; } /** * Adds two numbers modulo 2^255 - 19. * * @param result The result. * @param x The first number to add. * @param y The second number to add. */ private void add(final int[] result, final int[] x, final int[] y) { int carry = x[0] + y[0]; result[0] = carry & 0x03FFFFFF; for (int index = 1; index < NUM_LIMBS_255BIT; ++index) { carry = (carry >> 26) + x[index] + y[index]; result[index] = carry & 0x03FFFFFF; } reduceQuick(result); } /** * Destroy all sensitive data in this object. */ private void destroy() { // Destroy all temporary variables. Arrays.fill(x_1, 0); Arrays.fill(x_2, 0); Arrays.fill(x_3, 0); Arrays.fill(z_2, 0); Arrays.fill(z_3, 0); Arrays.fill(A, 0); Arrays.fill(B, 0); Arrays.fill(C, 0); Arrays.fill(D, 0); Arrays.fill(E, 0); Arrays.fill(AA, 0); Arrays.fill(BB, 0); Arrays.fill(DA, 0); Arrays.fill(CB, 0); Arrays.fill(t1, 0L); Arrays.fill(t2, 0); } /** * Evaluates the curve for every bit in a secret key. * * @param s The 32-byte secret key. */ private void evalCurve(final byte[] s) { int sposn = 31; int sbit = 6; int svalue = s[sposn] | 0x40; int swap = 0; // Iterate over all 255 bits of "s" from the highest to the lowest. // We ignore the high bit of the 256-bit representation of "s". while (true) { // Conditional swaps on entry to this bit but only if we // didn't swap on the previous bit. final int select = (svalue >> sbit) & 0x01; swap ^= select; cswap(swap, x_2, x_3); cswap(swap, z_2, z_3); swap = select; // Evaluate the curve. add(A, x_2, z_2); // A = x_2 + z_2 square(AA, A); // AA = A^2 sub(B, x_2, z_2); // B = x_2 - z_2 square(BB, B); // BB = B^2 sub(E, AA, BB); // E = AA - BB add(C, x_3, z_3); // C = x_3 + z_3 sub(D, x_3, z_3); // D = x_3 - z_3 mul(DA, D, A); // DA = D * A mul(CB, C, B); // CB = C * B add(x_3, DA, CB); // x_3 = (DA + CB)^2 square(x_3, x_3); sub(z_3, DA, CB); // z_3 = x_1 * (DA - CB)^2 square(z_3, z_3); mul(z_3, z_3, x_1); mul(x_2, AA, BB); // x_2 = AA * BB mulA24(z_2, E); // z_2 = E * (AA + a24 * E) add(z_2, z_2, AA); mul(z_2, z_2, E); // Move onto the next lower bit of "s". if (sbit > 0) { --sbit; } else if (sposn == 0) { break; } else if (sposn == 1) { --sposn; svalue = s[sposn] & 0xF8; sbit = 7; } else { --sposn; svalue = s[sposn]; sbit = 7; } } // Final conditional swaps. cswap(swap, x_2, x_3); cswap(swap, z_2, z_3); } /** * Multiplies two numbers modulo 2^255 - 19. * * @param result The result. * @param x The first number to multiply. * @param y The second number to multiply. */ private void mul(final int[] result, final int[] x, final int[] y) { // Multiply the two numbers to create the intermediate result. long v = x[0]; for (int i = 0; i < NUM_LIMBS_255BIT; ++i) { t1[i] = v * y[i]; } for (int i = 1; i < NUM_LIMBS_255BIT; ++i) { v = x[i]; for (int j = 0; j < (NUM_LIMBS_255BIT - 1); ++j) { t1[i + j] += v * y[j]; } t1[i + NUM_LIMBS_255BIT - 1] = v * y[NUM_LIMBS_255BIT - 1]; } // Propagate carries and convert back into 26-bit words. v = t1[0]; t2[0] = ((int) v) & 0x03FFFFFF; for (int i = 1; i < NUM_LIMBS_510BIT; ++i) { v = (v >> 26) + t1[i]; t2[i] = ((int) v) & 0x03FFFFFF; } // Reduce the result modulo 2^255 - 19. reduce(result, t2, NUM_LIMBS_255BIT); } /** * Multiplies a number by the a24 constant, modulo 2^255 - 19. * * @param result The result. * @param x The number to multiply by a24. */ private void mulA24(final int[] result, final int[] x) { final long a24 = 121665; long carry = 0; for (int index = 0; index < NUM_LIMBS_255BIT; ++index) { carry += a24 * x[index]; t2[index] = ((int) carry) & 0x03FFFFFF; carry >>= 26; } t2[NUM_LIMBS_255BIT] = ((int) carry) & 0x03FFFFFF; reduce(result, t2, 1); } /** * Raise x to the power of (2^250 - 1). * * @param result The result. Must not overlap with x. * @param x The argument. */ private void pow250(final int[] result, final int[] x) { // The big-endian hexadecimal expansion of (2^250 - 1) is: // 03FFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF // // The naive implementation needs to do 2 multiplications per 1 bit and // 1 multiplication per 0 bit. We can improve upon this by creating a // pattern 0000000001 ... 0000000001. If we square and multiply the // pattern by itself we can turn the pattern into the partial results // 0000000011 ... 0000000011, 0000000111 ... 0000000111, etc. // This averages out to about 1.1 multiplications per 1 bit instead of 2. // Build a pattern of 250 bits in length of repeated copies of 0000000001. square(A, x); for (int j = 0; j < 9; ++j) square(A, A); mul(result, A, x); for (int i = 0; i < 23; ++i) { for (int j = 0; j < 10; ++j) square(A, A); mul(result, result, A); } // Multiply bit-shifted versions of the 0000000001 pattern into // the result to "fill in" the gaps in the pattern. square(A, result); mul(result, result, A); for (int j = 0; j < 8; ++j) { square(A, A); mul(result, result, A); } } /** * Computes the reciprocal of a number modulo 2^255 - 19. * * @param result The result. Must not overlap with x. * @param x The argument. */ private void recip(final int[] result, final int[] x) { // The reciprocal is the same as x ^ (p - 2) where p = 2^255 - 19. // The big-endian hexadecimal expansion of (p - 2) is: // 7FFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFEB // Start with the 250 upper bits of the expansion of (p - 2). pow250(result, x); // Deal with the 5 lowest bits of (p - 2), 01011, from highest to lowest. square(result, result); square(result, result); mul(result, result, x); square(result, result); square(result, result); mul(result, result, x); square(result, result); mul(result, result, x); } /** * Reduce a number modulo 2^255 - 19. * * @param result The result. * @param x The value to be reduced. This array will be * modified during the reduction. * @param size The number of limbs in the high order half of x. */ private void reduce(final int[] result, final int[] x, final int size) { // Calculate (x mod 2^255) + ((x / 2^255) * 19) which will // either produce the answer we want or it will produce a // value of the form "answer + j * (2^255 - 19)". There are // 5 left-over bits in the top-most limb of the bottom half. int carry = 0; int limb = x[NUM_LIMBS_255BIT - 1] >> 21; x[NUM_LIMBS_255BIT - 1] &= 0x001FFFFF; for (int index = 0; index < size; ++index) { limb += x[NUM_LIMBS_255BIT + index] << 5; carry += (limb & 0x03FFFFFF) * 19 + x[index]; x[index] = carry & 0x03FFFFFF; limb >>= 26; carry >>= 26; } if (size < NUM_LIMBS_255BIT) { // The high order half of the number is short; e.g. for mulA24(). // Propagate the carry through the rest of the low order part. for (int index = size; index < NUM_LIMBS_255BIT; ++index) { carry += x[index]; x[index] = carry & 0x03FFFFFF; carry >>= 26; } } // The "j" value may still be too large due to the final carry-out. // We must repeat the reduction. If we already have the answer, // then this won't do any harm but we must still do the calculation // to preserve the overall timing. The "j" value will be between // 0 and 19, which means that the carry we care about is in the // top 5 bits of the highest limb of the bottom half. carry = (x[NUM_LIMBS_255BIT - 1] >> 21) * 19; x[NUM_LIMBS_255BIT - 1] &= 0x001FFFFF; for (int index = 0; index < NUM_LIMBS_255BIT; ++index) { carry += x[index]; result[index] = carry & 0x03FFFFFF; carry >>= 26; } // At this point "x" will either be the answer or it will be the // answer plus (2^255 - 19). Perform a trial subtraction to // complete the reduction process. reduceQuick(result); } /** * Reduces a number modulo 2^255 - 19 where it is known that the * number can be reduced with only 1 trial subtraction. * * @param x The number to reduce, and the result. */ private void reduceQuick(final int[] x) { // Perform a trial subtraction of (2^255 - 19) from "x" which is // equivalent to adding 19 and subtracting 2^255. We add 19 here; // the subtraction of 2^255 occurs in the next step. int carry = 19; for (int index = 0; index < NUM_LIMBS_255BIT; ++index) { carry += x[index]; t2[index] = carry & 0x03FFFFFF; carry >>= 26; } // If there was a borrow, then the original "x" is the correct answer. // If there was no borrow, then "t2" is the correct answer. Select the // correct answer but do it in a way that instruction timing will not // reveal which value was selected. Borrow will occur if bit 21 of // "t2" is zero. Turn the bit into a selection mask. final int mask = -((t2[NUM_LIMBS_255BIT - 1] >> 21) & 0x01); final int nmask = ~mask; t2[NUM_LIMBS_255BIT - 1] &= 0x001FFFFF; for (int index = 0; index < NUM_LIMBS_255BIT; ++index) x[index] = (x[index] & nmask) | (t2[index] & mask); } /** * Squares a number modulo 2^255 - 19. * * @param result The result. * @param x The number to square. */ private void square(final int[] result, final int[] x) { mul(result, x, x); } }