Ejemplo n.º 1
0
	    /// @brief Makes the inverse relation, mapping codim_to entities
	    /// to their codim_from neighbours.
	    ///
	    /// Implementation note: The algorithm has been changed
	    /// to a three-pass O(n) algorithm.
	    /// @param inv  The OrientedEntityTable 
	    void makeInverseRelation(OrientedEntityTable<codim_to, codim_from>& inv) const
	    {
		// Find the maximum index used. This will give (one less than) the size
		// of the table to be created.
		int maxind = -1;
		for (int i = 0; i < size(); ++i) {
		    EntityRep<codim_from> from_ent(i, true);
		    row_type r = operator[](from_ent);
		    for (int j = 0; j < r.size(); ++j) {
			EntityRep<codim_to> to_ent = r[j];
			int ind = to_ent.index();
			maxind = std::max(ind, maxind);
		    }
		}
		// Build the new_sizes vector and compute datacount.
		std::vector<int> new_sizes(maxind + 1);
		int datacount = 0;
		for (int i = 0; i < size(); ++i) {
		    EntityRep<codim_from> from_ent(i, true);
		    row_type r = operator[](from_ent);
		    datacount += r.size();
		    for (int j = 0; j < r.size(); ++j) {
			EntityRep<codim_to> to_ent = r[j];
			int ind = to_ent.index();
			++new_sizes[ind];
		    }
		}
		// Compute the cumulative sizes.
		std::vector<int> cumul_sizes(new_sizes.size() + 1);
		cumul_sizes[0] = 0;
		std::partial_sum(new_sizes.begin(), new_sizes.end(), cumul_sizes.begin() + 1);
		// Using the cumulative sizes array as indices, we populate new_data.
		// Note that cumul_sizes[ind] is not kept constant, but incremented so that
		// it always gives the correct index for new data corresponding to index ind.
		std::vector<int> new_data(datacount);
		for (int i = 0; i < size(); ++i) {
		    EntityRep<codim_from> from_ent(i, true);
		    row_type r = operator[](from_ent);
		    for (int j = 0; j < r.size(); ++j) {
			EntityRep<codim_to> to_ent = r[j];
			int ind = to_ent.index();
			int data_ind = cumul_sizes[ind];
			new_data[data_ind] = to_ent.orientation() ? i : ~i;
			++cumul_sizes[ind];
		    }
		}
		inv = OrientedEntityTable<codim_to, codim_from>(new_data.begin(),
								new_data.end(),
								new_sizes.begin(),
								new_sizes.end());
	    }
Ejemplo n.º 2
0
 	    /** @brief Prints the relation matrix corresponding to the table.

 	     Let the entities of codimensions f and t be given by
 	     the sets \f$E^f = { e^f_i } \f$ and \f$E^t = { e^t_j }\f$.
 	     A relation matrix R is defined by
	     \f{eqnarray*}{
	       R_{ij} &=& 0  \mbox{ if } e^f_i \mbox{ and } e^t_j \mbox{ are not neighbours }\\
               R_{ij} &=& 1  \mbox{ if they are neighbours with same orientation }\\
	       R_{ij} &=& -1 \mbox{ if they are neighbours with opposite orientation.}
	     \f}
	     @param os   The output stream.
	    */
	    void printRelationMatrix(std::ostream& os) const
	    {
		int columns = numberOfColumns();
		for (int i = 0; i < size(); ++i) {
		    FromType from_ent(i);
		    row_type r  = operator[](from_ent);
		    int cur_col = 0;
		    int next_ent = 0;
		    ToType to_ent = r[next_ent];
		    int next_print = to_ent.index();
		    while (cur_col < columns) {
			if (cur_col == next_print) {
			    if (to_ent.orientation()) {
				os << "  1";
			    } else {
				os << " -1";
			    }
			    ++next_ent;
			    if (next_ent >= r.size()) {
				next_print = columns;
			    } else {
				to_ent = r[next_ent];
				next_print = to_ent.index();
			    }
			} else {
			    os << "  0";
			}
			++cur_col;
		    }
		    os << '\n';
		}
	    }
Ejemplo n.º 3
0
	    int numberOfColumns() const
	    {
		int maxind = 0;
		for (int i = 0; i < size(); ++i) {
		    FromType from_ent(i);
		    row_type r  = operator[](from_ent);
		    for (int j = 0; j < r.size(); ++j) {
			maxind = std::max(maxind, r[j].index());
		    }
		}
		return maxind + 1;
	    }
Ejemplo n.º 4
0
		int size() const
		{
			return left.size() + right.size();
		}