One of the fundamental properties of the hypercube is that it is bipancyclic as has a cycle of length for every even integer with . We consider the following problem of generalizing this property: For a given integer with , determine all integers for which there exists an -vertex, -regular subgraph of that is both -connected and bipancyclic. The solution to this problem is known for and . In this paper, we solve the problem for . We prove that contains a -regular subgraph on vertices that is both -connected and bipancyclic if and only if or is an even integer satisfying . For general , we establish that every -regular subgraph of has or at least vertices.
A graph on vertices is an -vertex graph. A -vertex graph is bipancyclic
if it has a cycle of length for
all even integers satisfying
The Cartesian
product of two graphs and
is the graph with vertex set in which two vertices
and are adjacent if and only if either
and is adjacent to in or and is adjacent to
in Throughout the paper denote a positive integer.
The -dimensional hypercube
is the Cartesian product of
copies of the complete graph
It is an -regular, -connected, bipartite, and bipancyclic
graph on vertices with diameter
Because of such rich properties,
hypercubes are one of the most widely used interconnection network
topologies [1]. The
connectivity of a network is an important parameter to evaluate the
reliability and fault tolerance of a network [2]. Bipancyclicity is a fundamental property of the
hypercube networks as it allows the embedding of cycles of various
lengths effectively into hypercubes. Cycle networks are used to design
simple algorithms with low communication cost and it has applications in
image processing and signal processing [1,3].
We consider the following problem of generalizing the property of
bipancyclicity of hypercubes to the existence of -vertex, -regular subgraphs for various values of
that are also -connected and bipancyclic. This will
give subgraphs of with less
number of vertices which retain the important properties of such as regularity, high
connectivity, and bipancyclicity.
Problem 1. For a given integer with determine all integers for which there exists an -vertex, -regular subgraph of that is both -connected and bipancyclic.
This problem is also related to the problem of embedding regular
graphs into hypercubes. Cybenko et al. [4] proved that the problem of deciding whether or
not a given graph is embeddable into a hypercube is NP-complete, in
fact, the problem is NP-complete even for trees [5].
Since the hypercube is a
bipartite graph, every regular subgraph of it has even number of
vertices. For Ramras [6] established that every even
integer from 8 to except 10 can
be the number of vertices of a 3-regular subgraph of Borse and Shaikh [7] improved this result by showing
that such a 3-regular subgraph can be bipancyclic also. They solved the
above problem for and in [7] and [8], receptively. They established that, for has a -regular, -connected, bipancyclic subgraph on
vertices if and only if or is an even integer with The problem
remains open for
Besides hypercubes, the special case of the above problem is settled for the class of the
Cartesian product of cycles in [9] and for the class of the Cartesian product of
paths in [10]. Also, Borse et
al. [11] proved the existence
of a factorization of the Cartesian product of cycles, each of length a power of 2,
into isomorphic -regular, -connected and bipancyclic subgraphs
with the number of vertices a power of 2, for Moreover, the number of
vertices of a smallest -regular
subgraph of an -regular graph
is related to the conditional
-edge-connectivity of [12].
In this paper, we settle Problem 1 for the case The following is the main result
of the paper.
Theorem 1 (Main Theorem). For there exists a -regular, -connected and bipancyclic subgraph of
the hypercube on vertices if and only if or is an even integer such that
For general Borse and Shaikh
[8] obtained the following
result about the non-existence of -regular subgraphs of on a certain number of vertices.
Theorem 2 ([8]). For a given integer with every subgraph of with minimum degree at least either is isomorphic to or has at least vertices.
In this paper, we improve this theorem as follows.
Theorem 3. For a given integer with if is a
subgraph of with minimum degree
at least then one of the
following holds:
is isomorphic to
is a spanning subgraph
of the subgraph of induced by
for some cycle
of length six.
has at least
vertices.
Thus, if and
with then does
not have a -regular subgraph on
vertices and hence no -regular, -vertex graph is embeddable into
We provide preliminary results in Section 2 and prove Theorem 3 in
Section 3. The proof of Main Theorem 1 is divided
into the next three sections. A construction of 5-regular subgraphs of
is given in Section 4. The
connectivity and bipancyclicity properties of these subgraphs are dealt
in Section 5 and Section 6, respectively.
2. Preliminaries
We can write as for where is the complete graph A -cycle means a cycle of length
We need the following
lemmas.
Lemma 1 ([3]). Let be an -regular and -connected graph for Then the graph is -regular and -connected.
Lemma 2 ([13]). If
and are non-trivial paths and one
of them has an even number of vertices, then the graph is bipancyclic.
Hence is bipancyclic
if is a non-trivial path or a
cycle of length at least three.
Lemma 3 ([8]). For the hypercube has a
Hamiltonian cycle with a chord
such that there is a -cycle in containing and three edges of
Lemma 4 ([8]). Let be an even integer such that Then there exists an
-cycle in containing six vertices
and there are two vertices in such that
is adjacent to
is adjacent to
and are edges of
We obtain a similar result as follows.
Lemma 5. Let be an even integer such that Then there exists an
-cycle in and a vertex in having four pairwise
non-adjacent neighbours in
Proof. Suppose
Clearly, is a required 8-cycle in as shown by bold lines in Figure 1.
Replacing the edge of by a path of length 3 avoiding gives a 10-cycle and replacing the edge of by a path of length 3 or 5 that is
edge-disjoint with produces
required 12-cycle and 14-cycle. Thus the result holds for
Figure 1. 8-Cycle in
Suppose Write
as Then is obtained from four copies of such that is joined to for by a matching between their
corresponding vertices. Vertices of are joined to the corresponding
vertices of
Figure 2. The -cycle
Since by Lemma 3, each
contains a Hamiltonian
cycle with a chord such that there is a 4-cycle in containing and three edges from For simplicity let Label the set of vertices of
by so that and We now construct a cycle of length in as required, from the cycles and the chord of If then where In this case, take as the cycle shown in Figure 2(a). If
then
with In this case, choose to be the cycle shown in Figure 2(b).
Finally, for take as the cycle given in Figure 2(c). In
each case, is a cycle of length
in and further, the vertex is not on but it has four pairwise non-adjacent
neighbours in This completes
the proof.
In this section, we prove Theorem 3. For a graph let denote the induced subgraph of on a vertex subset The minimum degree of
is denoted by If is isomorphic to a graph then we write Since we can split
into two copies and of If is a subgraph of then there is a subgraph of isomorphic to such that the vertex set of is the set of neighbours of in We say that is the subgraph of corresponding to
We need the following result.
Lemma 6 ([8]). For a given integer with if is a
subraph of isomorphic to then every vertex in has at most one neighbour in
Proof of Theorem 3. By Theorem 2, (i) holds or Suppose (i)
does not hold. Then We prove, by induction on that (ii) holds if otherwise (iii)
holds. Suppose Then or If then it follows that is a chordless 6-cycle or a 6-cycle
with a chord and so (ii) holds. If
then (iii) follows. Thus the result is true for
Suppose Assume that
the result holds for the subgraphs of of minimum degree at least Let be an edge of Without loss of generality, we may
assume that the end vertices of
differ in the first coordinate only. Write
where is the set of all edges in
whose end vertices differ in
the first coordinate only. Then Hence intersects both
and . Let for Then We may assume that
By induction hypothesis, or is a
spanning subgraph of for some -cycle
or Hence or or We consider these three cases separately.
Case (i). Suppose
Consider as Therefore (iii)
holds.
Case (ii). Suppose
In this case, is isomorphic
to As each vertex of has a neighbour in Let be the subgraph of corresponding to Then and Let be the
subgraph of induced by Observe that no vertex
of has a neighbour in and by Lemma 6, it has at most
one neighbour in Therefore
By Theorem 2, or In the
later case, (iii) holds as Suppose Then the subgraph of
induced by the vertices of
is isomorphic to
where is a
path on three vertices. Since is a 6-cycle with a chord, (ii) holds.
Case (iii).
is a spanning subgraph of for some -cycle
We consider two subcases depending on whether the cycle has a chord or not.
Subcase (i). Suppose is chordless.
Then is -regular and in fact, Let be the subgraph of corresponding to Then If then As (ii) follows. Suppose
Let
be the subgraph of induced by As is triangle-free, it follows from
Lemma 6 that every vertex of has at most three neighbours in This shows that By Theorem 2, Thus
Therefore (iii) holds in this case.
Subcase (ii).
Suppose has a chord.
Then is a spanning subgraph of
and hence is a spanning subgraph of Let be the vertices of the path in order and let the copy of in corresponding to the vertex for Let be the
subgraph of
corresponding to for Then Since the degree of
every member of
is in we have and If
then (ii) holds (see Figure 3(a)). Suppose
is non-empty and let be the
subgraph of induced by this
set. Then, by Lemma 6, Therefore, by Theorem 2, or In the
later case, (iii) holds.
Suppose It
follows from Lemma 6 that every vertex of has a neighbour in for but no neighbour in Suppose Then
(see Figure 3(b)). It follows that where is a 6-cycle whose six vertices
correspond to in order, and so, (ii) holds. Suppose Then
the graph has minimum
degree at least and hence, it
has at least vertices by
Theorem 2. Thus Therefore (iii) holds. This completes the
proof.
Figure 3. The Subgraph of
Corollary 1. Every -regular subgraph of has or at least
vertices.
4. Construction of
-regular Subgraphs of
In this section, we give a construction of an -vertex, -regular subgraph of the hypercube Suppose has a 5-regular subgraph on vertices. Then is an even integer and by Corollary 1, we
have or We prove that for
every such there exists a -regular subgraph of with that is both -connected and bipancyclic. The case
when is a multiple of 4 follows
trivially from the following result.
Theorem 4 ([8]). If and is an even integer
such that or then there exists a
-vertex, -regular subgraph of that is both -connected and bipancyclic.
Lemma 7. If and is a multiple of
such that or then there exists a
-regular, -connected and bipancyclic subgraph of
on vertices.
Proof. We can write where or is an integer such that Express as Since has a -regular, -connected and bipancyclic subgraph
on vertices by Theorem 4. Therefore, by
Lemma 1, the graph is a -regular and -connected subgraph of on vertices. As is
bipancyclic, it has a Hamiltonian cycle and so has a Hamiltonian path.
Hence, by Lemma 2, is bipancyclic.
Suppose is an even integer
such that but
not a multiple of 4. Then
and We have
the following four cases.
with
with
with
with
Case (i). with
In this case, and Write By Lemma 5,
contains a cycle of length and there is a vertex with four
pairwise non-adjacent neighbours in Let so that
and are 4-cycles and
and are edges of Then is a 5-regular subgraph of
with vertices containing a copy of corresponding to the vertex of Let be the vertices of corresponding to the vertices
respectively.
Let be the subgraph of with and We define a
graph from and as follows. Clearly, is a 5-regular subgraph of on vertices.
Now, to construct -regular
subgraphs in Cases (ii), (iii) and (iv), we choose a cycle in of length by Lemma 4. Then there are
six vertices
on and two vertices in such that is adjacent to and is adjacent to and are edges of Let Then is a
5-regular subgraph of on vertices. As in Case (i), let be a copy of corresponding to the vertex of and let be
the vertices of
corresponding to respectively.
Figure 4. The Subgraph on Vertices
Case (ii). Suppose with
Clearly,
Let be the subgraph of with vertex set and edge set Define a subgraph
of from the graphs and as follows. The graph
is depicted in Figure 5. It is
easy to see that is a 5-regular
subgraph of on vertices.
Figure 5. The Subgraph on Vertices
Case (iii). with
Consider the three edge sets:
and
Let be the subgraph of with vertex set
and edge set We now define a subgraph of which is shown in Figure 6 as Clearly, is a 5-regular subgraph of on vertices.
Figure 6. The subgraph on vertices
Case (iv). with
We define the four edge sets as follows.
and Let
be the subgraph of having
vertex set and edge set We define a subgraph of as follows. The
graph is shown in Figure 7.
Clearly, it is a 5-regular subgraph of on vertices.
Figure 7. The subgraph on vertices
Thus we have constructed -regular subgraphs and of the hypercube on vertices in each of the four cases. In
the next two sections, we prove that these subgraphs are -connected and bipancyclic.
5. Connectivity
In this section, we prove that all the four subgraphs and of that are constructed in Section 4 and
shown in Figures 4, 5, 6, and 7, respectively are -connected.
If is the matching in consisting of edges having one end
vertex in the lower side (L) and the other end vertex in the upper side
(R) in the figure of then
is an edge-cut of The graph has two components and further,
the components are 4-connected and isomorphic to each other. We prove
that these components together with the matching give a 5-connected graph. We first
prove the following observations for general graphs.
Lemma 8. Let be a simple -connected graph and let be distinct vertices
of Let be a new graph obtained from
by adding a new vertex and edges for Then is -connected.
Proof. The graph is shown in Figure 8(a). Suppose
with If then and so is connected. Therefore is connected as the vertex
has a neighbour in Suppose Then contains at most vertices of and hence is connected. Thus is -connected.
Lemma 9. Let be a simple -connected graph with at least vertices and an independent set where Suppose is a simple graph obtained from
by adding a new vertex and new edges each having one end vertex
including the edges Let
be the graph obtained from the graph by deleting the matching consisting of edges between the two copies of corresponding to the vertices
Then the
graph is -connected.
Proof. The graph is
shown in Figure 8(b). The graph is obtained from and a copy of by adding edges between their
corresponding vertices. Let and
be the vertices of corresponding to and for respectively and let
Then
Since has at least vertices, there are at least edges in between and
Suppose with
It is sufficient to
prove that is connected.
Since is -connected, by Lemma 8, the graph is -connected. Suppose intersects both and Then both and are connected and they
are joined to each other by an edge. Hence is connected. Suppose The degree of in is at least and so it is at least in for any If has a component with vertex
set a subset of the independent set then that component has just one vertex
and so a contradiction.
Thus every component of
has a neighbour in the connected graph in This shows that is connected.
Figure 8. The Graphs of Lemma 8 and Lemma 9
Let and be the 5-regular subgraphs of constructed in Section 4. We now
prove that these graphs are 5-connected.
Lemma 10. The graph is 5-connected.
Proof. The subgraphs of induced by
and by are isomorphic to for some -cycle Hence, by Lemma 1, these two
subgraphs are 4-connected. Now, by Lemma 9, is 5-connected.
Lemma 11. The graph is 5-connected.
Proof. The subgraphs of induced by and by are isomorphic to and so, by Lemma 1, they are 3-connected. Hence, by
Lemma 9, the upper half subgraph of that is induced by is 4-connected. If then is
isomorphic to Thus, by Lemma 9,
is 5-connected.
Lemma 12. The graph is 5-connected.
Proof. By Lemmas 1 and 9,
the subgraphs and of are 3-connected. By similar arguments
of the proof of Lemma 11, we see that the upper half subgraph of induced by is 4-connected. If then is isomorphic to Now, the result follows from Lemma 9.
Lemma 13. The graph is 5-connected.
Proof. Let and let By Lemma 9, the subgraphs
and of are 3-connected. Therefore the graph
is 4-connected.
Hence the graph is also 4-connected. The subgraph of induced by is isomorphic to and so, it is 4-connected.
We now prove that is
5-connected. Suppose with It
is sufficient to prove that
is connected. If intersects both
and then and are connected and they are joined to
each other by an edge leaving connected. Suppose The set of vertices of that do not have any neighbour in is Assume that
has a component such that No member of is isolated in as each has degree five in Hence Observe that the
subgraph of induced by is the forest as shown in Figure 9.
Therefore is a tree containing at
least two pendant vertices. So and every pendant vertex of is adjacent to all the four members of
This gives as a subgraph of and so of a contradiction. Hence every
component of has a neighbour in
the connected graph Therefore
is connected. Similarly,
is connected if
Figure 9. Subgraph of Induced by
6. Bipancyclicity
In this section, we prove that the 5-regular graphs and constructed in Section 4 are all
bipancyclic.
A ladder on vertices
has two edges at its two ends. If we identify one of these two edges
with an edge of a -cycle, then it
follows that the resulting graph has cycles of every even length from
to The following lemma is based on this fact.
Lemma 14. Let be even integers and
be an -cycle conatining an edge
and be a ladder on vertices containing an end edge Then the graph has
cycles of all even lengths from
to (see Figure 10).
Figure 10. The Graph
Lemma 15. The graph is bipancyclic.
Proof. Recall that
contains the eight copies of a -cycle Let be the set of edges of with one end vertex in the cycle
and the other in the cycle
Consider the subgraph of Then is isomorphic to where is an 8-cycle. By Lemma 2,
is bipancyclic. Therefore and so
contains a cycle of every even
length from 4 to Let
for such that is
the label to the neighbor of
Then the cycle shown in Figure 11 is a Hamiltonian cycle of Thus the graph is bipancyclic.
Figure 11. Hamiltonian Cycle in
Recall that we have written as where the vertices of
are labeled by so that is a
Hamiltonian cycle of with
chords and
Furthermore, is a -cycle in and is its copy in corresponding to vertex of in the construction of the graphs
and Let be the matching between cycles
and in corresponding to the edge of For each label the vertices of by so that and are the neighbours of the vertex and are the neighbours of
the vertex for some with and Hence the neighbours of on are relabeled as respectively.
Similarly, the vertices are relabeled as respectively.
Lemma 16. The graph is bipancyclic.
Proof. The graph
has vertices. Consider
the subgraph of where Then is isomorphic to where is an 8-cycle. By Lemma 2, it contains cycles of every even
length from 4 to A cycle of length in is shown in Figure 12. Furthermore,
is a cycle of length while is a cycle of length in Thus is bipancyclic.
Figure 12. -cycle in Figure 13. -cycles in for
.5cm
Lemma 17. The graph is bipancyclic.
Proof. The graph is
on vertices. Consider
the subgraph of on vertices defined as Then is
isomorphic to where
is a 6-cycle. Hence contain cycles of all even lengths from
to Therefore these cycles are also
contained in Let be the ladder on vertices defined as Note that has a Hamiltonian cycle Then the subgraph of has vertices. By Lemma 14, it has cycles
of every even length from
to Let and be the cycles of length and as shown in Figure 13(a) and
13(b), respectively. Then is a cycle of length whereas is a -cycle
in Finally, is a cycle of length in Thus is bipancyclic.
Lemma 18. The graph is bipancyclic.
Proof. Recall that
has vertices. We prove
that contain cycles of every
even length from to The subgraph of
is bipancyclic as it is isomorphic to where is a
path on four vertices. This implies that contain cycles of every even length
from to Consider the Hamiltonian cycle of where . We get two paths and each of length from and respectively as follows.
Then is a ladder on vertices. Hence, by Lemma 14,
is a subgraph of on vertices contains cycle of length
for every even integer with
Let and be the cycles in of length and as shown in Figures 14(a) and
14(b), respectively. If and then and are cycles of lengths
and respectively. Finally, and are cycles of lengths and
respectively. Hence is
bipancyclic.
Figure 14. -cycles in for
Remark 1. The problem of the existence of a
-regular subgraph of the hypercube
of a given order is completely
solved for To solve the
problem for the general value of
one needs to identify the even integers in between and that cannot be the order of any -regular subgraph of . By Theorem 3, we have two
intervals and
with the property that no even integer belonging to any of these
intervals is the order of a -regular subgraph of . There is no further gap for However, there seems a
further gap for The next
interval of the gaps can be found by characterizing the -regular subgraph of on vertices. Thus,
Problem 1 remains open for
Acknowledgment
The authors are thankful to the referee for fruitful suggestions to
improve the presentation of the paper. The first author is financially
supported by DST-SERB, Government of India through the project
MTR/2018/000447.
Conflict of
Interest
The authors declare no conflict of interest
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