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);

     * 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

    }
  }

  // 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;

    }
  }

  public void testLog2NegativeAlwaysThrows() {
    for (int x : NEGATIVE_INTEGER_CANDIDATES) {
      for (RoundingMode mode : ALL_ROUNDING_MODES) {
        assertThrows(IllegalArgumentException.class, () -> IntMath.log2(x, mode));
      }
    }
  }

  // Relies on the correctness of BigIntegerMath.log2 for all modes except UNNECESSARY.

    }
  }

  public void testLog2NegativeAlwaysThrows() {
    for (long x : NEGATIVE_LONG_CANDIDATES) {
      for (RoundingMode mode : ALL_ROUNDING_MODES) {
        assertThrows(IllegalArgumentException.class, () -> LongMath.log2(x, mode));
      }
    }
  }

  /* Relies on the correctness of BigIntegerMath.log2 for all modes except UNNECESSARY. */

     *
     * 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)];
    /*

    int minThresholdPosition = 0;
    // The leftmost position at which the greatest of the k lower elements
    // -- the new value of threshold -- might be found.

    int iterations = 0;
    int maxIterations = IntMath.log2(right - left, RoundingMode.CEILING) * 3;
    while (left < right) {
      int pivotIndex = (left + right + 1) >>> 1;

      int pivotNewIndex = partition(left, right, pivotIndex);

      if (pivotNewIndex > k) {

      return size;
    }
  }

  /** A bit mask were all bits are set. */
  private static final int ALL_SET = ~0;

  private static int ceilToPowerOfTwo(int x) {
    return 1 << IntMath.log2(x, RoundingMode.CEILING);
  }

  /*
   * This method was written by Doug Lea with assistance from members of JCP JSR-166 Expert Group
   * and released to the public domain, as explained at

     * if fewer consecutive positions are filled; see that method for details.)
     */
    static int maxRunBeforeFallback(int tableSize) {
      return MAX_RUN_MULTIPLIER * IntMath.log2(tableSize, RoundingMode.UNNECESSARY);
    }
  }

  /**
   * SetBuilderImpl version that uses a JDK HashSet, which has built in hash flooding protection.
   */