*/
  public static <N> Graph<N> transpose(Graph<N> graph) {
    if (!graph.isDirected()) {
      return graph; // the transpose of an undirected graph is an identical graph
    }

    if (graph instanceof TransposedGraph) {
      return ((TransposedGraph<N>) graph).graph;
    }

    return new TransposedGraph<>(graph);
  }

  /**

      Correspondence.tolerance(ALLOWED_ERROR);

  /**
   * A {@link Correspondence} which accepts either finite values within {@link #ALLOWED_ERROR} of
   * each other or identical non-finite values.
   */
  private static final Correspondence<Double, Double> QUANTILE_CORRESPONDENCE =
      Correspondence.from(
          new BinaryPredicate<Double, Double>() {
            @Override

      Correspondence.tolerance(ALLOWED_ERROR);

  /**
   * A {@link Correspondence} which accepts either finite values within {@link #ALLOWED_ERROR} of
   * each other or identical non-finite values.
   */
  private static final Correspondence<Double, Double> QUANTILE_CORRESPONDENCE =
      Correspondence.from(
          new BinaryPredicate<Double, Double>() {
            @Override

 * {@link #putString(CharSequence, Charset)} is generally only useful for cross-language
 * compatibility (otherwise prefer {@link #putUnencodedChars}). However, the character encodings
 * must be identical across languages. Also beware that {@link Charset} definitions may occasionally
 * change between Java releases.
 *
 * <p><b>Warning:</b> Chunks of data that are put into the {@link Hasher} are not delimited. The

         * ours. Therefore, we have to be careful about whether we create our own TypeVariable:
         *
         * 2a. If the resolved types are identical to the original types, then we can return the
         * original, identical JDK TypeVariable. By doing so, we sidestep the problem entirely.
         *
         * 2b. If the resolved types are different from the original types, things are trickier. The

  /**
   * Compares the specified object with this multimap for equality. Two multimaps are equal when
   * their map views, as returned by {@link #asMap}, are also equal.
   *
   * <p>In general, two multimaps with identical key-value mappings may or may not be equal,
   * depending on the implementation. For example, two {@link SetMultimap} instances with the same

    Object[] array = collection.toArray();
    expectArrayContentsAnyOrder(createSamplesArray(), array);
  }

  /**
   * {@link Collection#toArray(Object[])} says: "Note that {@code toArray(new Object[0])} is
   * identical in function to {@code toArray()}."
   *
   * <p>For maximum effect, the collection under test should be created from an element array of a
   * type other than {@code Object[]}.
   */

        Boolean result1 = (Boolean) constantTimeMethod.invoke(null, password1, password2);
        Boolean result2 = (Boolean) constantTimeMethod.invoke(null, password1, password3);

        assertTrue(result1, "Identical passwords should be equal");
        assertFalse(result2, "Different passwords should not be equal");
    }

    /**
     * Measure the average time for equals operations.
     */

        byte[] out14 = new byte[16];
        byte[] out7 = new byte[8];

        // Act
        NtlmUtil.E(key14, data8, out14);
        NtlmUtil.E(key7, data8, out7);

        // Assert: first block identical to single-chunk encryption
        assertArrayEquals(Arrays.copyOfRange(out14, 0, 8), out7);
        // And second block is present and not all zeros

    }
    int[] stripes = new int[result.size()];
    for (int i = 0; i < result.size(); i++) {
      stripes[i] = indexFor(result.get(i));
    }
    Arrays.sort(stripes);
    // optimize for runs of identical stripes
    int previousStripe = stripes[0];
    result.set(0, getAt(previousStripe));
    for (int i = 1; i < result.size(); i++) {
      int currentStripe = stripes[i];
      if (currentStripe == previousStripe) {