ASKSAGE: Sage Q&A Forum - RSS feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Tue, 08 Feb 2022 09:10:41 +0100How to find the spanning elementary subgraphs of a given graphhttps://ask.sagemath.org/question/43581/how-to-find-the-spanning-elementary-subgraphs-of-a-given-graph/consider the following definition: A subgraph $H$
of a graph
$G$
is called an elementary subgraph if each component of
$H$
is either an edge (
$K_{2}$
) or a
cycle of length at least
$3$. A spanning elementary subgraph is a subgraph having all components either path(i.e. $K_{2}$) or cycles and verex set is same as those of $G$. for example consider the graph $C_{4}$ with $V(G)$={1,2,3,4}. then it has 3 spanning elementary subgraphs two edge components namely {12,34};{14,23} and the whole cycle itself.The cycle is named in anticlockwise direction.
Now my problem is:
Consider the following code:
G=graphs.EmptyGraph()
G.add_edges([(1,2),(2,3),(3,4),(4,5),(5,1),(6,5),(6,8),(8,9),(7,9),(7,6),(7,10),(10,11),(10,12),(11,12)])
G.show()
Can we have a sage code that gives all possible spanning subgraphs of this graph.Tue, 04 Sep 2018 15:02:42 +0200https://ask.sagemath.org/question/43581/how-to-find-the-spanning-elementary-subgraphs-of-a-given-graph/Answer by David Coudert for <p>consider the following definition: A subgraph $H$
of a graph
$G$
is called an elementary subgraph if each component of
$H$
is either an edge (
$K_{2}$
) or a
cycle of length at least
$3$. A spanning elementary subgraph is a subgraph having all components either path(i.e. $K_{2}$) or cycles and verex set is same as those of $G$. for example consider the graph $C_{4}$ with $V(G)$={1,2,3,4}. then it has 3 spanning elementary subgraphs two edge components namely {12,34};{14,23} and the whole cycle itself.The cycle is named in anticlockwise direction.
Now my problem is:
Consider the following code:</p>
<pre><code>G=graphs.EmptyGraph()
G.add_edges([(1,2),(2,3),(3,4),(4,5),(5,1),(6,5),(6,8),(8,9),(7,9),(7,6),(7,10),(10,11),(10,12),(11,12)])
G.show()
</code></pre>
<p>Can we have a sage code that gives all possible spanning subgraphs of this graph.</p>
https://ask.sagemath.org/question/43581/how-to-find-the-spanning-elementary-subgraphs-of-a-given-graph/?answer=43592#post-id-43592A perfect matching (if the graph has one) is a *spanning elementary subgraph* according your definition. You can get all the perfect matchings (only 1 in your graph) using
sage: list(G.perfect_matchings())
[[(7, 10), (1, 2), (11, 12), (3, 4), (5, 6), (8, 9)]]
Now if you want all *spanning elementary subgraphs*, you have to design a specific algorithm.
**EDIT:**
A solution is to enumerate all elementary subgraphs and to prune subgraphs without enough vertices.
def elementary_subgraphs(G, nmin=0):
r"""
Iterator over the elementary subgraphs of `G`.
A subgraph `H` of a graph `G` is *elementary* if each of its connected
components is either an edge or a cycle.
INPUT:
- ``G`` -- a Graph
- ``nmin`` -- integer (default: ``0``); lower bound on the number of
vertices involved in the elementary subgraphs of any returned
solution. When set to ``G.order()``, the subgraphs must be spanning.
"""
G._scream_if_not_simple()
def rec(H, low):
if not H.size():
yield {'cycles': [], 'edges': []}, 0
return
if H.order() < low:
# no solution
return
# We select an edge e = (u, v) of H and remove it
u, v = next(H.edge_iterator(labels=False))
H.delete_edge((u, v))
# Case 1: return solutions without e
for sol, n in rec(H, low):
if n >= low:
yield sol, n
# Case 2: select e as an isolated edge
I = H.edges_incident([u, v])
H.delete_vertices([u, v])
for sol, n in rec(H, low - 2):
if n + 2 >= low:
sol['edges'].append((u, v))
yield sol, n + 2
H.add_vertices([u, v])
H.add_edges(I)
# Case 3: select e as part of a cycle
for P in H.all_paths(u, v):
C = [(P[i], P[i + 1]) for i in range(len(P) - 1)]
C.append((u, v))
K = H.copy()
K.delete_vertices(P)
nP = len(P)
for sol, n in rec(K, low - nP):
if n + nP >= low:
sol['cycles'].append(C)
yield sol, n + nP
# Finally restore edge e
H.add_edge(u, v)
for sol, n in rec(G.copy(), nmin):
if n >= nmin:
yield sol
The idea behind the algorithm is, given any edge `e=(u,v)`, to
1. either search for solutions without edge `e`. We return all elementary subgraphs of `G-e`.
2. or `e` is in the solution and is a connected component. We remove the end vertices of `e` from the graph, search for all elementary subgraphs of the remaining graph, and add `e` to each found subgraph before returning it
3. or `e` is part of a cycle. We enumerate all paths between the end vertices of `e`. Each path combined with `e` forms a cycle. For each of these cycles, we search for all elementary subgraphs of the graph `G` without the vertices of the cycle.
We get the spanning elementary subgraphs as follows:
sage: G = Graph([(1,2),(2,3),(3,4),(4,5),(5,1),(6,5),(6,8),(8,9),(7,9),(7,6),(7,10),(10,11),(10,12),(11,12)])
sage: for s in elementary_subgraphs(G, G.order()):
....: print(s)
{'cycles': [], 'edges': [(8, 9), (7, 10), (5, 6), (3, 4), (1, 2), (11, 12)]}
{'cycles': [[(1, 5), (5, 4), (4, 3), (3, 2), (1, 2)], [(10, 11), (11, 12), (10, 12)]], 'edges': [(7, 9), (6, 8)]}
{'cycles': [[(1, 5), (5, 4), (4, 3), (3, 2), (1, 2)], [(10, 11), (11, 12), (10, 12)]], 'edges': [(8, 9), (6, 7)]}
{'cycles': [[(6, 8), (8, 9), (9, 7), (6, 7)], [(1, 5), (5, 4), (4, 3), (3, 2), (1, 2)], [(10, 11), (11, 12), (10, 12)]], 'edges': []}
We can also get the elementary subgraphs with at least `G.order() - 1` vertices:
sage: len(list(elementary_subgraphs(G, G.order()-1)))
32
or all elementary subgraphs (including the empty graph):
sage: len(list(elementary_subgraphs(G, 0)))
647
This method is faster for this graph than the (nice) solution using linear programming:
sage: %timeit list(spanning_elementary_subgraphs(G))
16.2 s ± 3.41 s per loop (mean ± std. dev. of 7 runs, 1 loop each)
sage: %timeit list(elementary_subgraphs(G, G.order()))
14.3 ms ± 459 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
Wed, 05 Sep 2018 17:48:02 +0200https://ask.sagemath.org/question/43581/how-to-find-the-spanning-elementary-subgraphs-of-a-given-graph/?answer=43592#post-id-43592Comment by David Coudert for <p>A perfect matching (if the graph has one) is a <em>spanning elementary subgraph</em> according your definition. You can get all the perfect matchings (only 1 in your graph) using</p>
<pre><code>sage: list(G.perfect_matchings())
[[(7, 10), (1, 2), (11, 12), (3, 4), (5, 6), (8, 9)]]
</code></pre>
<p>Now if you want all <em>spanning elementary subgraphs</em>, you have to design a specific algorithm. </p>
<p><strong>EDIT:</strong></p>
<p>A solution is to enumerate all elementary subgraphs and to prune subgraphs without enough vertices.</p>
<pre><code>def elementary_subgraphs(G, nmin=0):
r"""
Iterator over the elementary subgraphs of `G`.
A subgraph `H` of a graph `G` is *elementary* if each of its connected
components is either an edge or a cycle.
INPUT:
- ``G`` -- a Graph
- ``nmin`` -- integer (default: ``0``); lower bound on the number of
vertices involved in the elementary subgraphs of any returned
solution. When set to ``G.order()``, the subgraphs must be spanning.
"""
G._scream_if_not_simple()
def rec(H, low):
if not H.size():
yield {'cycles': [], 'edges': []}, 0
return
if H.order() < low:
# no solution
return
# We select an edge e = (u, v) of H and remove it
u, v = next(H.edge_iterator(labels=False))
H.delete_edge((u, v))
# Case 1: return solutions without e
for sol, n in rec(H, low):
if n >= low:
yield sol, n
# Case 2: select e as an isolated edge
I = H.edges_incident([u, v])
H.delete_vertices([u, v])
for sol, n in rec(H, low - 2):
if n + 2 >= low:
sol['edges'].append((u, v))
yield sol, n + 2
H.add_vertices([u, v])
H.add_edges(I)
# Case 3: select e as part of a cycle
for P in H.all_paths(u, v):
C = [(P[i], P[i + 1]) for i in range(len(P) - 1)]
C.append((u, v))
K = H.copy()
K.delete_vertices(P)
nP = len(P)
for sol, n in rec(K, low - nP):
if n + nP >= low:
sol['cycles'].append(C)
yield sol, n + nP
# Finally restore edge e
H.add_edge(u, v)
for sol, n in rec(G.copy(), nmin):
if n >= nmin:
yield sol
</code></pre>
<p>The idea behind the algorithm is, given any edge <code>e=(u,v)</code>, to</p>
<ol>
<li>either search for solutions without edge <code>e</code>. We return all elementary subgraphs of <code>G-e</code>.</li>
<li>or <code>e</code> is in the solution and is a connected component. We remove the end vertices of <code>e</code> from the graph, search for all elementary subgraphs of the remaining graph, and add <code>e</code> to each found subgraph before returning it</li>
<li>or <code>e</code> is part of a cycle. We enumerate all paths between the end vertices of <code>e</code>. Each path combined with <code>e</code> forms a cycle. For each of these cycles, we search for all elementary subgraphs of the graph <code>G</code> without the vertices of the cycle. </li>
</ol>
<p>We get the spanning elementary subgraphs as follows:</p>
<pre><code>sage: G = Graph([(1,2),(2,3),(3,4),(4,5),(5,1),(6,5),(6,8),(8,9),(7,9),(7,6),(7,10),(10,11),(10,12),(11,12)])
sage: for s in elementary_subgraphs(G, G.order()):
....: print(s)
{'cycles': [], 'edges': [(8, 9), (7, 10), (5, 6), (3, 4), (1, 2), (11, 12)]}
{'cycles': [[(1, 5), (5, 4), (4, 3), (3, 2), (1, 2)], [(10, 11), (11, 12), (10, 12)]], 'edges': [(7, 9), (6, 8)]}
{'cycles': [[(1, 5), (5, 4), (4, 3), (3, 2), (1, 2)], [(10, 11), (11, 12), (10, 12)]], 'edges': [(8, 9), (6, 7)]}
{'cycles': [[(6, 8), (8, 9), (9, 7), (6, 7)], [(1, 5), (5, 4), (4, 3), (3, 2), (1, 2)], [(10, 11), (11, 12), (10, 12)]], 'edges': []}
</code></pre>
<p>We can also get the elementary subgraphs with at least <code>G.order() - 1</code> vertices:</p>
<pre><code>sage: len(list(elementary_subgraphs(G, G.order()-1)))
32
</code></pre>
<p>or all elementary subgraphs (including the empty graph):</p>
<pre><code>sage: len(list(elementary_subgraphs(G, 0)))
647
</code></pre>
<p>This method is faster for this graph than the (nice) solution using linear programming:</p>
<pre><code>sage: %timeit list(spanning_elementary_subgraphs(G))
16.2 s ± 3.41 s per loop (mean ± std. dev. of 7 runs, 1 loop each)
sage: %timeit list(elementary_subgraphs(G, G.order()))
14.3 ms ± 459 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
</code></pre>
https://ask.sagemath.org/question/43581/how-to-find-the-spanning-elementary-subgraphs-of-a-given-graph/?comment=61008#post-id-61008Another solution for the enumeration of elementary subgraphs: https://ask.sagemath.org/question/60937/spanning-elementary-subgraphs-on-a-given-number-of-vertices/Tue, 08 Feb 2022 09:10:41 +0100https://ask.sagemath.org/question/43581/how-to-find-the-spanning-elementary-subgraphs-of-a-given-graph/?comment=61008#post-id-61008Comment by rewi for <p>A perfect matching (if the graph has one) is a <em>spanning elementary subgraph</em> according your definition. You can get all the perfect matchings (only 1 in your graph) using</p>
<pre><code>sage: list(G.perfect_matchings())
[[(7, 10), (1, 2), (11, 12), (3, 4), (5, 6), (8, 9)]]
</code></pre>
<p>Now if you want all <em>spanning elementary subgraphs</em>, you have to design a specific algorithm. </p>
<p><strong>EDIT:</strong></p>
<p>A solution is to enumerate all elementary subgraphs and to prune subgraphs without enough vertices.</p>
<pre><code>def elementary_subgraphs(G, nmin=0):
r"""
Iterator over the elementary subgraphs of `G`.
A subgraph `H` of a graph `G` is *elementary* if each of its connected
components is either an edge or a cycle.
INPUT:
- ``G`` -- a Graph
- ``nmin`` -- integer (default: ``0``); lower bound on the number of
vertices involved in the elementary subgraphs of any returned
solution. When set to ``G.order()``, the subgraphs must be spanning.
"""
G._scream_if_not_simple()
def rec(H, low):
if not H.size():
yield {'cycles': [], 'edges': []}, 0
return
if H.order() < low:
# no solution
return
# We select an edge e = (u, v) of H and remove it
u, v = next(H.edge_iterator(labels=False))
H.delete_edge((u, v))
# Case 1: return solutions without e
for sol, n in rec(H, low):
if n >= low:
yield sol, n
# Case 2: select e as an isolated edge
I = H.edges_incident([u, v])
H.delete_vertices([u, v])
for sol, n in rec(H, low - 2):
if n + 2 >= low:
sol['edges'].append((u, v))
yield sol, n + 2
H.add_vertices([u, v])
H.add_edges(I)
# Case 3: select e as part of a cycle
for P in H.all_paths(u, v):
C = [(P[i], P[i + 1]) for i in range(len(P) - 1)]
C.append((u, v))
K = H.copy()
K.delete_vertices(P)
nP = len(P)
for sol, n in rec(K, low - nP):
if n + nP >= low:
sol['cycles'].append(C)
yield sol, n + nP
# Finally restore edge e
H.add_edge(u, v)
for sol, n in rec(G.copy(), nmin):
if n >= nmin:
yield sol
</code></pre>
<p>The idea behind the algorithm is, given any edge <code>e=(u,v)</code>, to</p>
<ol>
<li>either search for solutions without edge <code>e</code>. We return all elementary subgraphs of <code>G-e</code>.</li>
<li>or <code>e</code> is in the solution and is a connected component. We remove the end vertices of <code>e</code> from the graph, search for all elementary subgraphs of the remaining graph, and add <code>e</code> to each found subgraph before returning it</li>
<li>or <code>e</code> is part of a cycle. We enumerate all paths between the end vertices of <code>e</code>. Each path combined with <code>e</code> forms a cycle. For each of these cycles, we search for all elementary subgraphs of the graph <code>G</code> without the vertices of the cycle. </li>
</ol>
<p>We get the spanning elementary subgraphs as follows:</p>
<pre><code>sage: G = Graph([(1,2),(2,3),(3,4),(4,5),(5,1),(6,5),(6,8),(8,9),(7,9),(7,6),(7,10),(10,11),(10,12),(11,12)])
sage: for s in elementary_subgraphs(G, G.order()):
....: print(s)
{'cycles': [], 'edges': [(8, 9), (7, 10), (5, 6), (3, 4), (1, 2), (11, 12)]}
{'cycles': [[(1, 5), (5, 4), (4, 3), (3, 2), (1, 2)], [(10, 11), (11, 12), (10, 12)]], 'edges': [(7, 9), (6, 8)]}
{'cycles': [[(1, 5), (5, 4), (4, 3), (3, 2), (1, 2)], [(10, 11), (11, 12), (10, 12)]], 'edges': [(8, 9), (6, 7)]}
{'cycles': [[(6, 8), (8, 9), (9, 7), (6, 7)], [(1, 5), (5, 4), (4, 3), (3, 2), (1, 2)], [(10, 11), (11, 12), (10, 12)]], 'edges': []}
</code></pre>
<p>We can also get the elementary subgraphs with at least <code>G.order() - 1</code> vertices:</p>
<pre><code>sage: len(list(elementary_subgraphs(G, G.order()-1)))
32
</code></pre>
<p>or all elementary subgraphs (including the empty graph):</p>
<pre><code>sage: len(list(elementary_subgraphs(G, 0)))
647
</code></pre>
<p>This method is faster for this graph than the (nice) solution using linear programming:</p>
<pre><code>sage: %timeit list(spanning_elementary_subgraphs(G))
16.2 s ± 3.41 s per loop (mean ± std. dev. of 7 runs, 1 loop each)
sage: %timeit list(elementary_subgraphs(G, G.order()))
14.3 ms ± 459 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
</code></pre>
https://ask.sagemath.org/question/43581/how-to-find-the-spanning-elementary-subgraphs-of-a-given-graph/?comment=43593#post-id-43593Thank you. I am trying. But I have not been able to find the algorithm .Wed, 05 Sep 2018 18:00:59 +0200https://ask.sagemath.org/question/43581/how-to-find-the-spanning-elementary-subgraphs-of-a-given-graph/?comment=43593#post-id-43593Answer by tmonteil for <p>consider the following definition: A subgraph $H$
of a graph
$G$
is called an elementary subgraph if each component of
$H$
is either an edge (
$K_{2}$
) or a
cycle of length at least
$3$. A spanning elementary subgraph is a subgraph having all components either path(i.e. $K_{2}$) or cycles and verex set is same as those of $G$. for example consider the graph $C_{4}$ with $V(G)$={1,2,3,4}. then it has 3 spanning elementary subgraphs two edge components namely {12,34};{14,23} and the whole cycle itself.The cycle is named in anticlockwise direction.
Now my problem is:
Consider the following code:</p>
<pre><code>G=graphs.EmptyGraph()
G.add_edges([(1,2),(2,3),(3,4),(4,5),(5,1),(6,5),(6,8),(8,9),(7,9),(7,6),(7,10),(10,11),(10,12),(11,12)])
G.show()
</code></pre>
<p>Can we have a sage code that gives all possible spanning subgraphs of this graph.</p>
https://ask.sagemath.org/question/43581/how-to-find-the-spanning-elementary-subgraphs-of-a-given-graph/?answer=43690#post-id-43690There is a generic tool for solving such problems without having to think too much: Integer Linear Programming.
For each edge $uv$ of your graph, you define two binary variables (with values in {0, 1} ): one named $c(uv)$ telling that a cycle from you spanning configuration that is passing through the edge $uv$, and one names $e(uv)$ telling whether the edge the edge $uv$ is selected as an edge of the spanning configuration.
Then, you add the following constraint: for each vertex $v$, there is either exactly one edge adjacent to it that is selected to be an edge of the spanning configuration, or there are exactly two edges adjacent to it that are selected to be part of a cycle of the spanning configuration.
This constraints translates into the following linear equalities: $\forall v \in V(G), \sum_{uv \in E(G)} 2*e(uv) + c(uv) =1$ (E)
Hence, the set of solutions you want can be seen as the set of integer points of the polytope defined by the equation (E) together with the equations $0\leq e(uv)\leq1$ and $0\leq c(uv)\leq1$ for all edge $uv \in E(G)$.
The way to define this polytope in Sage is done by using the `polyhedron` method of the `MixedIntegerLinearProgram` class.
Here is the code of the function (some explanations will follow):
def spanning_elementary_subgraphs(G):
p = MixedIntegerLinearProgram(solver='ppl')
e = p.new_variable(binary=True)
c = p.new_variable(binary=True)
# those are useless constraints, only to control the order in which the variables are created, see discussion below
for edge in G.edges(labels=False):
p.add_constraint(e[edge] >= 0)
for edge in G.edges(labels=False):
p.add_constraint(c[edge] >= 0)
for v in G.vertices():
p.add_constraint(sum(2*e[edge] + c[edge] for edge in G.edges_incident(v, labels=False)) == 2)
P = p.polyhedron(backend='normaliz')
for i,L in enumerate(P.integral_points()):
edges = []
cycle_edges = []
for j,edge in enumerate(G.edges(labels=False)):
if L[j] == 1:
edges.append(edge)
if L[j+G.num_edges()] == 1:
cycle_edges.append(edge)
H = Graph(cycle_edges, format='list_of_edges')
yield {'edges': edges, 'cycles': H.connected_components()}
This function returns an iterator over all solutions. At each iteration, you get a dictionary
Here is it working on your example:
sage: G = graphs.EmptyGraph()
sage: G.add_edges([(1,2),(2,3),(3,4),(4,5),(5,1),(6,5),(6,8),(8,9),(7,9),(7,6),(7,10),(10,11),(10,12),(11,12)])
sage: s = spanning_elementary_subgraphs(G)
sage: list(s)
[{'cycles': [[1, 2, 3, 4, 5], [6, 7, 8, 9], [10, 11, 12]], 'edges': []},
{'cycles': [[1, 2, 3, 4, 5], [10, 11, 12]], 'edges': [(6, 8), (7, 9)]},
{'cycles': [[1, 2, 3, 4, 5], [10, 11, 12]], 'edges': [(6, 7), (8, 9)]},
{'cycles': [], 'edges': [(1, 2), (3, 4), (5, 6), (7, 10), (8, 9), (11, 12)]}]
So, you can see that there are exactly 4 possible spanning configurations.
Now, some more explanations about the code.
First, when we construct the polyhedron, we have to use the `normaliz` backend or the computation will be too slow. To be able to use it, you have to install the `pynormaliz` optional package, by typing from a terminal:
sage -i pynormaliz
Then, there is a total of $2*|E(G)|$ variables: for each edge $uv$, there is $e(uv)$ and $c(uv)$. Each variable will be an element of the basis of the space where the polyhedron lives.
The problem is that, in the current Sage user interface, there is no easy way to see order of the variables that are used to define that basis.
The variables are created on the fly when used for the first time. Since there is a loop over the vertices $v$ and then a loop over the edges $uv$ adjacent to the current vertex $v$ and then a constraint involving both $e(uv)$ and $c(uv)$, it is hard to guess the order of the basis.
Hence, we added some useless constraints in two flat loops in whch the edges appear in a simple-to-handle order.
Note that it trick has nothing to do with interesting algorithmics or mathematics, it is only an annoyance of the current Sage user interface and should be solved by Sage developers (a ticket will follow).Mon, 17 Sep 2018 19:56:17 +0200https://ask.sagemath.org/question/43581/how-to-find-the-spanning-elementary-subgraphs-of-a-given-graph/?answer=43690#post-id-43690