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# cython: binding=True r""" Products of graphs
This module gathers everything related to graph products. At the moment it contains an implementation of a recognition algorithm for graphs that can be written as a Cartesian product of smaller ones.
References:
.. [HIK11] Handbook of Product Graphs, R. Hammack, W. Imrich, S. Klavzar, CRC press, 2011
Author:
- Nathann Cohen (May 2012 -- coded while watching the election of Francois Hollande on TV)
Cartesian product of graphs -- the recognition problem ------------------------------------------------------
First, a definition:
**Definition** The Cartesian product of two graphs `G` and `H`, denoted `G\square H`, is a graph defined on the pairs `(g, h)\in V(G)\times V(H)`.
Two elements `(g, h),(g', h')\in V(G\square H)` are adjacent in `G\square H` if and only if :
- `g=g'` and `hh'\in H`; or - `h=h'` and `gg'\in G`
Two remarks follow :
#. The Cartesian product is commutative
#. Any edge `uv` of a graph `G_1 \square \cdots \square G_k` can be given a color `i` corresponding to the unique index `i` such that `u_i` and `v_i` differ.
The problem that is of interest to us in the present module is the following:
**Recognition problem** Given a graph `G`, can we guess whether there exist graphs `G_1, ..., G_k` such that `G=G_1\square \cdots \square G_k` ?
This problem can actually be solved, and the resulting factorization is unique. What is explained below can be found in the book *Handbook of Product Graphs* [HIK11]_.
Everything is actually based on simple observations. Given a graph `G`, finding out whether `G` can be written as the product of several graphs can be attempted by trying to color its edges according to some rules. Indeed, if we are to color the edges of `G` in such a way that each color class represents a factor of `G`, we must ensure several things.
**Remark 1** In any cycle of `G` no color can appear exactly once.
Indeed, if only one edge `uv` of a cycle were labelled with color `i`, it would mean that:
#. The only difference between `u` and `v` lies in their `i` th coordinate
#. It is possible to go from `u` to `v` by changing only coordinates different from the `i` th
A contradiction indeed.
.. image:: ../../../media/cycle.png
That means that, for instance, the edges of a triangle necessarily have the same color.
**Remark 2** If two consecutive edges `u_1u_2` and `u_2u_3` have different colors, there necessarily exists a unique vertex `u_4` different from `u_2` and incident to both `u_1` and `u_3`.
In this situation, opposed edges necessarily have the same colors because of the previous remark.
.. image:: ../../../media/square.png
**1st criterion** : As a corollary, we know that:
#. If two vertices `u,v` have a *unique* common neighbor `x`, then `ux` and `xv` have the same color.
#. If two vertices `u, v` have more that two common neighbors `x_1, ..., x_k` then all edges between the `x_i` and the vertices of `u,v` have the same color. This is also a consequence of the first remark.
**2nd criterion** : if two edges `uv` and `u'v'` of the product graph `G\square H` are such that `d(u,u')+d(v,v')\neq d(u,v') + d(v,u')` then the two edges `uv` and `u'v'` necessarily have the same color.
This is a consequence of the fact that for any two vertices `u,v` of `G\square H` (where `u=(u_G,u_H)` and `v=(v_G,v_H)`), we have `d(u,v) = d_G(u_G,v_G)+d_H(u_H,v_H)`. Indeed, a shortest path from `u` to `v` in `G\square H` contains the information of a shortest path from `u_G` to `v_G` in `G`, and a shortest path from `u_H` to `v_H` in `H`.
The algorithm ^^^^^^^^^^^^^
The previous remarks tell us that some edges are in some way equivalent to some others, i.e. that their colors are equal. In order to compute the coloring we are looking for, we therefore build a graph on the *edges* of a graph `G`, linking two edges whenever they are found to be equivalent according to the previous remarks.
All that is left to do is to compute the connected components of this new graph, as each of them representing the edges of a factor. Of course, only one connected component indicates that the graph has no factorization.
Then again, please refer to [HIK11]_ for any technical question.
To Do ^^^^^
This implementation is made at Python level, and some parts of the algorithm could be rewritten in Cython to save time. Especially when enumerating all pairs of edges and computing their distances. This can easily be done in C with the functions from the :mod:`sage.graphs.distances_all_pairs` module.
Methods ------- """
#****************************************************************************** # Copyright (C) 2012 Nathann Cohen <nathann.cohen@gmail.com> * # * # Distributed under the terms of the GNU General Public License (GPL)* # http://www.gnu.org/licenses/ * #******************************************************************************
from copy import copy
def is_cartesian_product(g, certificate = False, relabeling = False): r""" Tests whether the graph is a Cartesian product.
INPUT:
- ``certificate`` (boolean) -- if ``certificate = False`` (default) the method only returns ``True`` or ``False`` answers. If ``certificate = True``, the ``True`` answers are replaced by the list of the factors of the graph.
- ``relabeling`` (boolean) -- if ``relabeling = True`` (implies ``certificate = True``), the method also returns a dictionary associating to each vertex its natural coordinates as a vertex of a product graph. If `g` is not a Cartesian product, ``None`` is returned instead.
This is set to ``False`` by default.
.. SEEALSO::
- :meth:`sage.graphs.generic_graph.GenericGraph.cartesian_product`
- :mod:`~sage.graphs.graph_decompositions.graph_products` -- a module on graph products.
.. NOTE::
This algorithm may run faster whenever the graph's vertices are integers (see :meth:`~sage.graphs.generic_graph.GenericGraph.relabel`). Give it a try if it is too slow !
EXAMPLES:
The Petersen graph is prime::
sage: from sage.graphs.graph_decompositions.graph_products import is_cartesian_product sage: g = graphs.PetersenGraph() sage: is_cartesian_product(g) False
A 2d grid is the product of paths::
sage: g = graphs.Grid2dGraph(5,5) sage: p1, p2 = is_cartesian_product(g, certificate = True) sage: p1.is_isomorphic(graphs.PathGraph(5)) True sage: p2.is_isomorphic(graphs.PathGraph(5)) True
Forgetting the graph's labels, then finding them back::
sage: g.relabel() sage: g.is_cartesian_product(g, relabeling = True) (True, {0: (0, 0), 1: (0, 1), 2: (0, 2), 3: (0, 3), 4: (0, 4), 5: (5, 0), 6: (5, 1), 7: (5, 2), 8: (5, 3), 9: (5, 4), 10: (10, 0), 11: (10, 1), 12: (10, 2), 13: (10, 3), 14: (10, 4), 15: (15, 0), 16: (15, 1), 17: (15, 2), 18: (15, 3), 19: (15, 4), 20: (20, 0), 21: (20, 1), 22: (20, 2), 23: (20, 3), 24: (20, 4)})
And of course, we find the factors back when we build a graph from a product::
sage: g = graphs.PetersenGraph().cartesian_product(graphs.CycleGraph(3)) sage: g1, g2 = is_cartesian_product(g, certificate = True) sage: any( x.is_isomorphic(graphs.PetersenGraph()) for x in [g1,g2]) True sage: any( x.is_isomorphic(graphs.CycleGraph(3)) for x in [g1,g2]) True
TESTS:
Wagner's Graph (:trac:`13599`)::
sage: g = graphs.WagnerGraph() sage: g.is_cartesian_product() False
Empty and one-element graph (:trac:`19546`)::
sage: Graph().is_cartesian_product() False sage: Graph({0:[]}).is_cartesian_product() False """ raise NotImplementedError("recognition of Cartesian product is not implemented for directed graphs")
# Of course the number of vertices of g can not be prime ! raise ValueError("The graph must be connected !")
# As we need the vertices of g to be linearly ordered, we copy the graph and # relabel it
# Reorder the vertices of an edge
# The equivalence graph on the edges of g
# For all pairs of vertices u,v of G, according to their number of common # neighbors... See the module's documentation !
# u and v are different
# List of common neighbors
# If u and v have no neighbors and uv is not an edge then their # distance is at least 3. As we enumerate the vertices in a # breadth-first search, it means that we already checked all the # vertices at distance less than two from u, and we are done with # this loop ! else:
# If uv is an edge
# Only one common neighbor
# Exactly 2 neighbors # More else: h.add_path([r(u,x) for x in intersect] + [r(v,x) for x in intersect])
# Edges uv and u'v' such that d(u,u')+d(v,v') != d(u,v')+d(v,u') are also # equivalent
# Gathering the connected components, relabeling the vertices on-the-fly
#Print the graph, distinguishing the edges according to their color classes # #from sage.plot.colors import rainbow #g.show(edge_colors = dict(zip(rainbow(len(edges)),edges)))
# Only one connected component ?
# Building the list of factors
# Computing the product of these graphs
# Checking that the resulting graph is indeed isomorphic to what we have. raise ValueError("Something weird happened during the algorithm... "+ "Please report the bug and give us the graph instance"+ " that made it fail !!!") else: return True |