int log2 = DoubleMath.log2(d, FLOOR);
      assertTrue(StrictMath.pow(2.0, log2) <= d);
      assertTrue(StrictMath.pow(2.0, log2 + 1) > d);
    }
  }

  @GwtIncompatible // DoubleMath.log2(double, RoundingMode), StrictMath
  public void testRoundLog2Ceiling() {
    for (double d : POSITIVE_FINITE_DOUBLE_CANDIDATES) {
      int log2 = DoubleMath.log2(d, CEILING);

      int log2 = DoubleMath.log2(d, FLOOR);
      assertTrue(StrictMath.pow(2.0, log2) <= d);
      assertTrue(StrictMath.pow(2.0, log2 + 1) > d);
    }
  }

  @GwtIncompatible // DoubleMath.log2(double, RoundingMode), StrictMath
  public void testRoundLog2Ceiling() {
    for (double d : POSITIVE_FINITE_DOUBLE_CANDIDATES) {
      int log2 = DoubleMath.log2(d, CEILING);

    return result;
  }

  /**
   * Generates a number in [0, 2^numBits) with an exponential distribution. The floor of the log2 of
   * the result is chosen uniformly at random in [0, numBits), and then the result is chosen in that
   * range uniformly at random. Zero is treated as having log2 == 0.
   */
  static BigInteger randomNonNegativeBigInteger(int numBits) {
    int digits = RANDOM_SOURCE.nextInt(numBits);

    return result;
  }

  /**
   * Generates a number in [0, 2^numBits) with an exponential distribution. The floor of the log2 of
   * the result is chosen uniformly at random in [0, numBits), and then the result is chosen in that
   * range uniformly at random. Zero is treated as having log2 == 0.
   */
  static BigInteger randomNonNegativeBigInteger(int numBits) {
    int digits = RANDOM_SOURCE.nextInt(numBits);

     * definitely >= floor(sqrt(x)), and then continue the iteration until we reach a fixed point.
     */
    BigInteger sqrt0;
    int log2 = log2(x, FLOOR);
    if (log2 < Double.MAX_EXPONENT) {
      sqrt0 = sqrtApproxWithDoubles(x);
    } else {
      int shift = (log2 - DoubleUtils.SIGNIFICAND_BITS) & ~1; // even!
      /*
       * We have that x / 2^shift < 2^54. Our initial approximation to sqrtFloor(x) will be

   * </ul>
   *
   * <p>The computed result is within 1 ulp of the exact result.
   *
   * <p>If the result of this method will be immediately rounded to an {@code int}, {@link
   * #log2(double, RoundingMode)} is faster.
   */
  public static double log2(double x) {
    return log(x) / LN_2; // surprisingly within 1 ulp according to tests
  }

  /**

     *
     * The key idea is that based on the number of leading zeros (equivalently, floor(log2(x))), we
     * can narrow the possible floor(log10(x)) values to two. For example, if floor(log2(x)) is 6,
     * then 64 <= x < 128, so floor(log10(x)) is either 1 or 2.
     */
    int y = maxLog10ForLeadingZeros[Integer.numberOfLeadingZeros(x)];
    /*

     *
     * The key idea is that based on the number of leading zeros (equivalently, floor(log2(x))), we
     * can narrow the possible floor(log10(x)) values to two. For example, if floor(log2(x)) is 6,
     * then 64 <= x < 128, so floor(log10(x)) is either 1 or 2.
     */
    int y = maxLog10ForLeadingZeros[Long.numberOfLeadingZeros(x)];
    /*

    // This is an extremely fast implementation of BigInteger.doubleValue(). JDK patch pending.
    BigInteger absX = x.abs();
    int exponent = absX.bitLength() - 1;
    // exponent == floor(log2(abs(x)))
    if (exponent < Long.SIZE - 1) {
      return x.longValue();
    } else if (exponent > MAX_EXPONENT) {
      return x.signum() * POSITIVE_INFINITY;
    }

    /*

    }
  }

  // Relies on the correctness of log2(BigInteger, {HALF_UP,HALF_DOWN}).
  public void testLog2HalfEven() {
    for (BigInteger x : POSITIVE_BIGINTEGER_CANDIDATES) {
      int halfEven = BigIntegerMath.log2(x, HALF_EVEN);
      // Now figure out what rounding mode we should behave like (it depends if FLOOR was
      // odd/even).
      boolean floorWasEven = (BigIntegerMath.log2(x, FLOOR) & 1) == 0;