private final long k1;

  /**
   * @param c the number of compression rounds (must be positive)
   * @param d the number of finalization rounds (must be positive)
   * @param k0 the first half of the key
   * @param k1 the second half of the key
   */
  SipHashFunction(int c, int d, long k0, long k1) {
    checkArgument(
        c > 0, "The number of SipRound iterations (c=%s) during Compression must be positive.", c);

MinIO protects data against hardware failures and silent data corruption using erasure code and checksums. With the highest level of redundancy, you may lose up to half (N/2) of the total drives and still be able to recover the data.

## What is Erasure Code?

      return entry.value;
    }
    entry = cacheEntry2;
    if (entry != null && entry.key == key) {
      // Promote second cache entry to first so the access pattern
      // [K1, K2, K1, K3, K1, K4...] still hits the cache half the time.
      addToCache(entry);
      return entry.value;
    }
    return null;
  }

  @Override
  void clearCache() {
    super.clearCache();
    cacheEntry1 = null;
    cacheEntry2 = null;
  }

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  }

  private static long random() {
    return randomSource.nextLong();
  }

  // A random value that cannot be 0 and that is unsigned-less-than or equal
  // to the given dividend, so that we don't have half of our divisions being
  // trivial because the divisor is bigger than the dividend.
  // Using remainder here does not give us a uniform distribution but it should
  // not have a big impact on the measurement.

            return roundFloorAsDouble;
          } else if (diff > 0) { // closer to ceiling
            return roundCeilingAsDouble;
          }
          // halfway between the representable values; do the half-whatever logic
          switch (mode) {
            case HALF_EVEN:
              // roundFloorAsDouble and roundCeilingAsDouble are neighbors, so precisely
              // one of them should have an even long representation

      }
      return BigInteger.valueOf(result);
    }

    /*
     * We want each multiplication to have both sides with approximately the same number of digits.
     * Currently, we just divide the range in half.
     */
    int mid = (n1 + n2) >>> 1;
    return oldSlowFactorial(n1, mid).multiply(oldSlowFactorial(mid, n2));
  }

  @Benchmark
  int slowFactorial(int reps) {
    int tmp = 0;

      }
      return BigInteger.valueOf(result);
    }

    /*
     * We want each multiplication to have both sides with approximately the same number of digits.
     * Currently, we just divide the range in half.
     */
    int mid = (n1 + n2) >>> 1;
    return oldSlowFactorial(n1, mid).multiply(oldSlowFactorial(mid, n2));
  }

  @Benchmark
  int slowFactorial(int reps) {
    int tmp = 0;

  static <R, C, V> RegularImmutableTable<R, C, V> forOrderedComponents(
      ImmutableList<Cell<R, C, V>> cellList,
      ImmutableSet<R> rowSpace,
      ImmutableSet<C> columnSpace) {
    // use a dense table if more than half of the cells have values
    // TODO(gak): tune this condition based on empirical evidence
    return (cellList.size() > (((long) rowSpace.size() * columnSpace.size()) / 2))

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