Fractional matching preclusion for butterfly derived networks

The matching preclusion number of a graph is the minimum number of edges whose deletion results in a graph that has neither perfect matchings nor almost perfect matchings. As a generalization, Liu and Liu [17] recently introduced the concept of fractional matching preclusion number. The fractional matching preclusion number (FMP number) of G, denoted by fmp(G), is the minimum number of edges whose deletion leaves the resulting graph without a fractional perfect matching. The fractional strong matching preclusion number (FSMP number) of G, denoted by fsmp(G), is the minimum number of vertices and edges whose deletion leaves the resulting graph without a fractional perfect matching. In this paper, we study the fractional matching preclusion number and the fractional strong matching preclusion number for butterfly network, augmented butterfly network and enhanced butterfly network.


Introduction
Let G be a graph.A matching M in a graph is a set of pairwise non-adjacent edges.A perfect matching in the graph is a set of edges such that every vertex is incident with exactly one edge in this set.An almost-perfect matching in a graph is a set of edges such that every vertex except one is incident with exactly one edge in this set, and the exceptional vertex is incident to none.All graphs considered in this paper are undirected, finite and simple.We refer to the book [2] for graph theoretical notations and terminology not defined here.For a graph G, let V (G), E(G), and [u, v] (uv for short) denote the set of vertices, the set of edges, and the edge whose end vertices are u and v, respectively.We use G − F to denote the subgraph of G obtained by removing all elements of F .A set F of edges in a graph G = (V, E) is called a matching preclusion set if G − F has neither a perfect matching nor an almost-perfect matching.The matching preclusion number of graph G, denoted by mp(G), is the minimum number of edges whose deletion leaves the resulting graph without a perfect matching or an almost-perfect matching.The concept of matching preclusion was introduced by Brigham et al. [1].Matching preclusion has also connections to a number of theoretical topics, including conditional connectivity and extremal graph theory, and was further studied in [3-13, 15, 20], with special attention given to interconnection networks.
In [20], the concept of strong matching preclusion was introduced.The strong matching preclusion number of a graph G, denoted by smp(G), is the minimum number of vertices and edges whose deletion leaves the resulting graph without a perfect matching or an almostperfect matching.According to the definition of mp(G) and smp(G), we have that smp(G) ≤ mp(G).
A fractional matching is a function f that assigns to each edge a number in [0, 1] so that ∑ e∼v f (e) ≤ 1 for each vertex v, where the sum is taken over all edges e incident with v. Clearly, A fractional perfect matching is a fractional matching f satisfying that and a perfect matching is a fractional perfect matching.Many further ideas and results on fractional graph theory can be found in [24].
In organic molecule graphs, perfect matchings correspond to kekulé structures, playing an important role in analysis of the resonance energy and stability of hydrocarbon compounds.Cyvin and Gutman systematically gave [14] detailed enumeration formulas for kekulé structures of various types of benzenoids.Kardos et al. showed [16] that fullerene graphs have exponentially many kekulé structures.For the details of anti-kekulé number can be found in [22].
Recently, Liu and Liu in [17] introduced some natural and nice generalizations of the above concepts.An edge subset F of G is a fractional matching preclusion set (FMP set for short) if G − F has no fractional perfect matchings.The fractional matching preclusion number (FMP number for short) of G, denoted by f mp(G), is the minimum size of FMP sets of G, that is, f mp(G) = min{|F | : F is an FMP set}.We refer the readers to [17] for more details and additional references.
A set F of edges and vertices of G is a fractional strong matching preclusion set (FSMP set for short) if G − F has no fractional perfect matchings.The fractional strong matching preclusion number (FSMP number for short) of G, denoted by f smp(G), is the minimum size of FSMP sets of G, that is, f smp(G) = min{|F | : F is an FSMP set}.
An FMP (FSMP) set of minimum cardinality is called optimal.An FMP set F is trivial if there is a single vertex of G incident to every edge in F , and an FSMP set F is trivial if there is a vertex v such that every vertex in F is a neighbour of v and every edge in F is incident to v. A graph G is fractional super matched (fractional strongly super matched ) if every optimal FMP (FSMP) set is trivial.
We summarize some knowledge which will be needed later.
If a graph G is Hamiltonian, we can assign to each edge which is in the Hamiltonian cycle of G a number 1  2 , and assign to other edges a number 0. It is obvious that we obtain a fractional perfect matching of G. Thus, the following proposition is immediate.
However, if the number of vertices in G is odd, mp(G) and f mp(G) do not have the same inequality relation.Some examples are given in [17].
Proposition 1.4.[24] A graph G has a fractional perfect matching if and only if i(G − S) ≤ |S| for every set S ⊆ V (G), where i(G − S) is the number of isolated vertices of G − S.
The vertex set of the hypercube Q n is the set of all binary strings of length n.Let r ≥ 1 be an integer.The r-dimensional butterfly network denoted by BF (r) has vertex set V = (i, x) : x ∈ V (Q r ), where i is a non-negative integer and 0 ≤ i ≤ r.Where any two vertices (i, x) and (j, y) of BF (r) are adjacent if and only if j = i + 1 and either: (1) x = y or (2) x differs from y in precisely the jth bit.For x = y, the edge is said to be a straight edge.Otherwise, the edge is a cross edge.For fixed i, the vertex (i, x) is a vertex on layer i.The vertices at layer 0 and r are of degree 2 and all other vertices are of degree 4. Clearly, the number of vertices and edges in the butterfly network BF (r) is equal to (r + 1)2 r and r2 r+1 , respectively.Let E S denote the set of all straight edges in BF (r), and E C denote the set of all cross edges in BF (r) such that E S ∪ E C = E(BF (r)).In [18] Manuel et al. proved that perfect matchings do not exists for a butterfly network of even dimension.BF (r) − F ′ is an induced subgraph obtained by deleting F ′ , where F ′ is the set of vertices or edges.For convenience, we give some symbols as follows.When r is odd, BF (r) has perfect matchings M S and M C , where M S is given by all the straight edges namely [(i, and M C is given by all the cross edges namely Butterfly networks have many weaknesses.It is non-Hamiltonian, not pancyclic and its toughness is less than one.But augmented butterfly network retains most of the favorable properties. Let n ≥ 1 be an integer.The vertices of the n-dimensional augmented butterfly network denoted by ABF (n) are the pairs (r, x) where r is a non-negative integer 0 ≤ r ≤ n called the layer, and In particular, when r = 0, the vertex (0, Clearly, the number of vertices and edges in the augmented butterfly network ABF (n) is equal to (n + 1)2 n and 3n × 2 n , respectively [18].The vertices at layer 0 and n are of degree 3 and all other vertices are of degree 6.
The edges between (r, x) and (r, , where 0 < r ≤ n and between (r, x) and (r, where 0 ≤ r < n are called level edges.The edges between (r, x) and (r + 1, x) are called straight edges.While the edges between (r, x) and (r + 1, Let E S denote the set of all straight edges in ABF (n), E C denote the set of all cross edges in ABF (n), and E L denote the set of all level edges in ABF . Thus E(ABF (n)) is partitioned into three categories of edges namely straight, cross and level edges.
Consider the r-dimensional butterfly network BF (r).Place a new vertex in each 4-cycle of BF (r) and join this vertex to the four vertices of the 4-cycle.The resulting graph is called an enhanced butterfly network EBF (r).This network has 2 r−1 (3r + 2) vertices and r × 2 r+2 edges.Additional vertices are given by the labels (x In this paper, we investigate the fractional matching preclusion (FMP) number and fractional strong matching preclusion (FSMP) number of butterfly network, augmented butterfly network and enhanced butterfly network.

Butterfly Network
Theorem 2.1.Let r ≥ 1 be an integer.Then Proof.We first consider the case that r is odd.Since δ(BF (r)) = 2, it follows from Proposition 1.2 that f mp(BF (r)) ≤ 2. In order to obtain our result, we only show that f mp(BF (r)) ≥ 2, that is, BF (r) − {e} has a fractional perfect matching for any edge e of BF (r).If we delete any straight edge from M S , then there exists a perfect matching M C in BF (r).If we delete any cross edge from M C , then there exists a perfect matching M S in BF (r).Then f mp(BF (r)) > 1, and have f mp(BF (r)) = 2. Next, we consider the case that r is even.Let S = {(i, x)}, where i is odd and 1 , it follows from Proposition 1.4 that BF (r) has no fractional perfect matchings, and have f mp(BF (r)) = 0.The proof is now complete.
which implies BF (r) − F does not have a fractional perfect matching from Proposition 1.4.Thus, we have BF (r) is not fractional super matched.
Theorem 2.2.Let r ≥ 1 be an integer.Then Proof.We first consider the case that r is odd.
, it follows from Proposition 1.4 that BF (r) − F has no fractional perfect matchings.It remains for us to show f smp(BF (r)) ≥ 1, that is, BF (r) has a fractional perfect matching.Since there are perfect matchings in BF (r), it follows that f smp(BF (r)) > 0, and have f smp(BF (r)) = 1.Next, we consider the case that r is even.By Theorem 2.1, f mp(BF (r)) = 0 if r is even, which implies BF (r) has no fractional perfect matchings.So f smp(BF (r)) = 0.The proof is now complete.

Enhanced Butterfly Network
Lemma 4.1.[21] Let G be the enhanced butterfly network EBF (r), r ≥ 2. Then G has a perfect matching.
3 ) for i = 0, 1, 2, 3, then there exist 8 vertex-disjoint cycles with five vertices such that v is a vertex of a cycle with five vertices from Figure 3 (a).Note that the additional vertex (x 1 x 2 x 3 , x 1 x 2 x 3 ) of L ′ 2 is adjacent to either the vertex (1, x 1 x 2 x 3 ) or the vertex (2, x 1 x 2 x 3 ), so it is easy to find two different paths with six vertices whose end-vertex is (x 1 x 2 x 3 , x 1 x 2 x 3 ); see Figure 3 (b).So EBF (3) − {v} can be decomposed into a path with four vertex, 4 paths with six vertices and 3 cycles with five vertices.Thus, EBF (3) − {v} has a fractional perfect matching.If v ∈ L ′ 2 , then EBF (3) − {v} can be decomposed into 3 paths with six vertices, and 5 cycles with five vertices.Thus, EBF (3) − {v} has a fractional perfect matching.Next, we consider the case that r ) for i = 0, 1, 3, 4, then there exist 16 vertex-disjoint cycles with five vertices such that v is a vertex of a cycle with five vertices; see Figure 4.Note that, there exists a perfect matching M of vertices in L 2 , L ′ 2 and L ′ 3 ; see Figure 4.It follows that EBF (4) − {v} has a fractional perfect matching.
, then v ′ is adjacent to a vertex of a cycle with five vertices, where vv ′ ∈ M .It is easy to find a path with six vertices whose end-vertex is v ′ ; see Figure 5. Thus EBF (4) − {v} has a fractional perfect matching.Assume that the argument is true for EBF (r − 2).We need to show the argument is true for EBF (r).Let H be a subgraph of EBF (r) induced by r−1 adjacent to either the vertex (r − 2, x 1 x 2 . . .x r ) or the vertex (r − 1, x 1 x 2 . . .x r ).Since the subgraph induced by L r ∪ L ′ r ∪ L r−1 can be decomposed into 2 r−1 vertex-disjoint cycles with five vertices, H can be decomposed into 2 r−1  vertex-disjoint paths with six vertices whose end-vertex is (x 1 x 2 . . .x r , x 1 x 2 . . .x r−2 x r−1 x r ).Clearly, EBF (r) can be decomposed into a subgraph H and 4 subgraphs that is isomorphic to EBF (r − 2), say G i for i = 1, 2, 3, 4. If v is a vertex of G i , then G i − {v} has fractional perfect matchings by induction hypothesis, where i ∈ {1, 2, 3, 4}.Moreover, G j has fractional perfect matchings for j ̸ = i.Hence EBF (r)−{v} has a fractional perfect matching.We now consider the case that v is a vertex of H.If v ∈ L ′ r−1 , then H − {v} can be decomposed into a cycle with five vertices and 2 r−1 − 1 paths with six vertices.Hence EBF (r) − {v} has a fractional perfect matching.If v ∈ (L r ∪ L ′ r ∪ L r−1 ), then there exists a path six vertices such that v is a vertex of this path.Let u be a end-vertex of this path such that u ∈ L ′ r−1 .Note that u is adjacent to a vertex in G i , say u ′ , where i ∈ {1, 2, 3, 4}.It is not difficult to see that H − {v, u} can be decomposed into a path with four vertices and 2 r−1 − 1 paths with six vertices.Since G i − {u ′ } has a fractional perfect matching by induction hypothesis, where i ∈ {1, 2, 3, 4}, EBF (r) − {v} has a fractional perfect matching.Hence f smp(EBF (r)) > 1.So f smp(EBF (r)) = 2.The proof is now complete.

, x 1 x 2 )
]}.If e ∈ E S or e ∈ E C , then there exists a perfect matching M L in ABF (n).If e ∈ E L , then there exists a perfect matching M SL or M LC in ABF (n).Thus, we have f mp(ABF (2)) = 2. Next, we consider the case that n ≥ 3. Since δ(ABF (n)) = 3, it follows from Proposition 1.2 that f mp(ABF (n)) ≤ 3. It suffices to prove that f mp(ABF (n)) ≥ 3. Since mp 1 (ABF (n)) = 3 from Lemma 3.1, the resulting graph by deleting any two edges of ABF (n) still has perfect matchings.So f mp(ABF (n)) > 2. Hence f mp(ABF (n)) = 3.

Theorem 3 . 3 .
does not have a fractional perfect matching from Proposition 1.4.So ABF (n) is not fractional super matched.Let n ≥ 1 be an integer.Then f smp(ABF (n)) = 2.
Some cycles with five vertices.(b) Some paths with six vertices.

Figure 3 :
Figure 3: (a) Some cycles with five vertices, (b) Some paths with six vertices.

Figure 4 :
Figure 4: Some cycles with five vertices and a perfect matching M .

Figure 5 :
Figure 5: Some paths with six vertices.