}

  @Override
  protected Map<String, Integer> makePopulatedMap() {
    Table<Character, String, Integer> table = HashBasedTable.create();
    table.put('a', "one", 2);
    table.put('a', "two", 4);
    table.put('a', "three", 6);
    table.put('b', "four", 8);
    return Tables.transformValues(table, TableCollectionTest.DIVIDE_BY_2).row('a');
  }

        for (RoundingMode mode : ALL_SAFE_ROUNDING_MODES) {
          BigInteger expected =
              new BigDecimal(p).divide(new BigDecimal(q), 0, mode).toBigIntegerExact();
          assertEquals(expected, BigIntegerMath.divide(p, q, mode));
        }
      }
    }
  }

  private static final BigInteger BAD_FOR_ANDROID_P = new BigInteger("-9223372036854775808");

        for (RoundingMode mode : ALL_SAFE_ROUNDING_MODES) {
          BigInteger expected =
              new BigDecimal(p).divide(new BigDecimal(q), 0, mode).toBigIntegerExact();
          assertEquals(expected, BigIntegerMath.divide(p, q, mode));
        }
      }
    }
  }

  private static final BigInteger BAD_FOR_ANDROID_P = new BigInteger("-9223372036854775808");

    a >>= aTwos; // divide out all 2s
    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.

    }

    private double singleQuantileFromSorted(int index, int scale, double[] dataset) {
      long numerator = (long) index * (dataset.length - 1);
      int positionFloor = (int) LongMath.divide(numerator, scale, RoundingMode.DOWN);
      int remainder = (int) (numerator - positionFloor * scale);
      if (remainder == 0) {
        return dataset[positionFloor];
      } else {

    }

    private double singleQuantileFromSorted(int index, int scale, double[] dataset) {
      long numerator = (long) index * (dataset.length - 1);
      int positionFloor = (int) LongMath.divide(numerator, scale, RoundingMode.DOWN);
      int remainder = (int) (numerator - positionFloor * scale);
      if (remainder == 0) {
        return dataset[positionFloor];
      } else {

        new int[] {GREATEST - 1, GREATEST - 2, 1, 2});
  }

  public void testDivide() {
    for (long a : UNSIGNED_INTS) {
      for (long b : UNSIGNED_INTS) {
        try {
          assertThat(UnsignedInts.divide((int) a, (int) b)).isEqualTo((int) (a / b));
          assertThat(b).isNotEqualTo(0);
        } catch (ArithmeticException e) {
          assertThat(b).isEqualTo(0);
        }
      }
    }
  }

   */
  public static <T extends @Nullable Object> Iterable<T> concat(
      Iterable<? extends Iterable<? extends T>> inputs) {
    return FluentIterable.concat(inputs);
  }

  /**
   * Divides an iterable into unmodifiable sublists of the given size (the final iterable may be
   * smaller). For example, partitioning an iterable containing {@code [a, b, c, d, e]} with a

  }

  /**
   * Verifies that the specified expected result is always produced by collecting the specified
   * inputs, regardless of how the elements are divided.
   */
  @SafeVarargs
  @CanIgnoreReturnValue
  @SuppressWarnings("nullness") // TODO(cpovirk): Remove after we fix whatever the bug is.

      // non-negative int, we can do long-arithmetic on index * (dataset.length - 1) / scale to get
      // a rounded ratio and a remainder which can be expressed as ints, without risk of overflow:
      int quotient = (int) LongMath.divide(numerator, scale, RoundingMode.DOWN);
      int remainder = (int) (numerator - (long) quotient * scale);
      selectInPlace(quotient, dataset, 0, dataset.length - 1);
      if (remainder == 0) {