drives we get a total of 128 possible sets, with 4 drives we get a total of 256 possible sets. So algorithm automatically chooses 64 sets, which is *16* 64 = 1024* drives in total.

- *If total number of nodes are of odd number then GCD algorithm provides affinity towards odd number erasure sets to provide for uniform distribution across nodes*. This is to ensure that same number of drives are pariticipating in any erasure set. For example if you have 2 nodes with 180 drives then GCD is 15 but...

	"net/http"
	"time"
)

// Supported amz date formats.
var amzDateFormats = []string{
	// Do not change this order, x-amz-date format is usually in
	// iso8601Format rest are meant for relaxed handling of other
	// odd SDKs that might be out there.
	"20060102T150405Z",
	time.RFC1123,
	time.RFC1123Z,
	// Add new AMZ date formats here.
}

// ErrMalformedDate always returned for dates that cannot be parsed.

    int bTwos = Long.numberOfTrailingZeros(b);
    b >>= bTwos; // divide out all 2s
    while (a != b) { // both a, b are odd
      // The key to the binary GCD algorithm is as follows:
      // Both a and b are odd. Assume a > b; then gcd(a - b, b) = gcd(a, b).
      // But in gcd(a - b, b), a - b is even and b is odd, so we can divide out powers of two.

      // We bend over backwards to avoid branching, adapting a technique from

    final int s1 = min + (1 * range) / 4;
    final int s2 = min + (2 * range) / 4;
    final int s3 = min + (3 * range) / 4;
    final char[] dst = new char[12];

    // Put surrogate pairs at odd indices so they can be split easily
    dst[0] = 'x';
    Character.toChars(min, dst, 1);
    Character.toChars(s1, dst, 3);
    Character.toChars(s2, dst, 5);
    Character.toChars(s3, dst, 7);

     * We round up if either the fractional part of signif is strictly greater than 0.5 (which is
     * true if the 0.5 bit is set and any lower bit is set), or if the fractional part of signif is
     * >= 0.5 and signifFloor is odd (which is true if both the 0.5 bit and the 1 bit are set).
     */
    boolean increment =
        (twiceSignifFloor & 1) != 0 && ((signifFloor & 1) != 0 || absX.getLowestSetBit() < shift);

      return (UnicodeEscaper) escaper;
    } else if (escaper instanceof CharEscaper) {
      return wrap((CharEscaper) escaper);
    }
    // In practice this shouldn't happen because it would be very odd not to
    // extend either CharEscaper or UnicodeEscaper for non-trivial cases.
    throw new IllegalArgumentException(
        "Cannot create a UnicodeEscaper from: " + escaper.getClass().getName());
  }

  /**

# "pip tests" run a similar suite of tests the "nonpip" tests, but do something
# odd to attempt to validate the quality of the pip package. The wheel is
# installed into a virtual environment, and then that venv is used to run all
# bazel tests with a special flag "--define=no_tensorflow_py_deps=true", which

	}
	if isSysErrHandleInvalid(err) {
		return errFileNotFound
	}
	if isSysErrIO(err) {
		return errFaultyDisk
	}
	if isSysErrInvalidArg(err) {
		storageLogIf(context.Background(), err)
		// For some odd calls with O_DIRECT reads
		// filesystems can return EINVAL, handle
		// these as FileNotFound instead.
		return errFileNotFound
	}
	if isSysErrNoSpace(err) {
		return errDiskFull
	}
	return err

    final int s1 = min + (1 * range) / 4;
    final int s2 = min + (2 * range) / 4;
    final int s3 = min + (3 * range) / 4;
    final char[] dst = new char[12];

    // Put surrogate pairs at odd indices so they can be split easily
    dst[0] = 'x';
    Character.toChars(min, dst, 1);
    Character.toChars(s1, dst, 3);
    Character.toChars(s2, dst, 5);
    Character.toChars(s3, dst, 7);

  public void testEviction_maxWeight_zero() {
    CountingRemovalListener<Integer, Integer> removalListener = countingRemovalListener();
    IdentityLoader<Integer> loader = identityLoader();

    // Even numbers are free, odd are too expensive
    Weigher<Integer, Integer> evensOnly =
        new Weigher<Integer, Integer>() {
          @Override
          public int weigh(Integer k, Integer v) {
            return k % 2;
          }