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);

    }
  }

  // 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;

    // Strip off 2s from this value.
    int shift = Long.numberOfTrailingZeros(product);
    product >>= shift;

    // Use floor(log2(num)) + 1 to prevent overflow of multiplication.
    int productBits = LongMath.log2(product, FLOOR) + 1;
    int bits = LongMath.log2(startingNumber, FLOOR) + 1;
    // Check for the next power of two boundary, to save us a CLZ operation.
    int nextPowerOfTwo = 1 << (bits - 1);

    }
  }

  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.

     *
     * 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)];
    /*

 * offering expected O(n + k log k) performance (worst case O(n log k)) for n calls to {@link
 * #offer} and a call to {@link #topK}, with O(k) memory. In comparison, quickselect has the same
 * asymptotics but requires O(n) memory, and a {@code PriorityQueue} implementation takes O(n log
 * k). In benchmarks, this implementation performs at least as well as either implementation, and

import static com.google.common.base.Preconditions.checkPositionIndexes;
import static com.google.common.base.Preconditions.checkState;
import static com.google.common.math.IntMath.divide;
import static com.google.common.math.IntMath.log2;
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.math.RoundingMode.CEILING;
import static java.math.RoundingMode.FLOOR;
import static java.math.RoundingMode.UNNECESSARY;

      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

     *
     * <p>If this returns false, then no query can take more than O(log n).
     *
     * <p>Note that for a RegularImmutableSet with elements with truly random hash codes, contains
     * operations take expected O(1) time but with high probability take O(log n) for at least some
     * element. (https://en.wikipedia.org/wiki/Linear_probing#Analysis)
     *