Recently, interesting connections between quantum physics and graph theory have been discovered by Krenn et al. [Krenn2017QuantumEA, Quantum_graphs_2, Quantum_graphs_3]. They have formulated certain questions from photonic quantum physical experiments in graph-theoretic language involving perfect matchings and graph colourings [krenn2019questions]. Krenn expects that answers to these graph-theoretic questions can give exciting new insights into the potential of quantum inference. These questions relate to the design of quantum experiments for producing high-dimensional and multi-partite entangled quantum states using state-of-the-art photonic technology [RevModPhys.84.777]: More precisely, the generation of Greenberger-Horne-Zeilinger (GHZ) states [GHZstates], and their high-dimensional generalizations [Ryu2013GreenbergerHorneZeilingerTF, Ryu2014MultisettingGT, Lawrence2014RotationalCA, Erhard2018ExperimentalGE]. In this work, we focus on the central question(creftype 1) posed by Mario Krenn [krenn_website]. For more details about how this graph-theoretic question correspond to quantum experiments, we refer the reader to [krenn2019questions].
We first define some commonly used graph-theoretic terms. For a graph , let denote the set of vertices and edges, respectively. The minimum number of edges, , whose deletion from the graph disconnects is called the edge connectivity of . The minimum number of vertices, , whose deletion from the graph leaves behind a disconnected or trivial graph, is the vertex connectivity of . denote the maximum and minimum degree of , respectively. For , denotes the induced subgraph of on . denote the set of natural and complex numbers, respectively. The cardinality of a set is denoted by . denotes the set .
We consider the edge of a graph to be an ordered pair vertices. Note that we are only considering undirected graphs and therefore ordered pairs and refers to the same edge. However, we need to consider an edge as an ordered pair instead of unordered pair to define the mixed edge colouring. A mixed edge colouring , associates an ordered pair of colours to each edge. For an edge , if a mixed edge colouring associates the colour pair then, we may visualize that half of the edge nearer to is coloured and the other half nearer to is coloured . We will use the notation to denote and to denote in this case. When , we call the edge to be normally edge coloured or briefly a normal edge and represent its colour by . If , then is called a mixed edge. Thus, with respect to a mixed edge colouring an edge is either a normal edge or a mixed edge.
A graph along with a mixed edge-colouring gives a mixed edge coloured graph . For a colouring , a mixed edge , is said to be contained in a mixed colour class , if or . For a colouring , a normal edge is said to be contained in a normal colour class , if . If all the edges in a perfect matching of belong to the same normal colour class, then is said to be monochromatic. In particular note that a monochromatic perfect matching does not contain a mixed edge. If all the perfect matchings of are monochromatic, we call a perfectly monochromatic edge coloured graph and is called a perfectly monochromatic colouring of . For a perfectly monochromatic colouring of , the number of normal colour classes containing at least one perfect matching is represented by . If colouring of is not perfectly monochromatic, then is undefined. We call an edge to be redundant if it is not contained in any perfect matching.
Any mixed edge in a perfectly monochromatic graph is redundant.
Towards a contradiction, let there be a perfect matching containing a mixed edge. It is easy to see that is non-monochromatic. However, must be monochromatic as is perfectly monochromatic. Contradiction. ∎
A graph is matching covered if every edge of it is part of at least one perfect matching. By removing all redundant edges from the given graph , we get its unique maximum matching covered sub-graph . Note that a colouring of induces a colouring for every subgraph of . When there is no scope for confusion, we use itself to denote this induced colouring. It is easy to see that if is a perfectly monochromatic colouring of , it is also a perfectly monochromatic colouring of and . We also note that all edges in are normal edges from creftype 1.1.
The matching index of is defined as , where is the set of all perfectly monochromatic colourings of . Note that is non-empty, as a mixed edge-colouring in which all edges are normal and belong to the same colour class is always perfectly monochromatic. By definition, if has no perfect matching, is zero. If has a perfect matching, it is easy to see that for a monochromatic edge-colouring , is perfectly monochromatic, since in this case all edges belong to the same normal colour class . It follows that and hence . We note that .
The following theorem is due to Bogdanov and has been described by Krenn et al. in [Krenn2017QuantumEA].
If is isomorphic to , . Otherwise, .
An edge weight function associates a non-zero complex number with each edge. A graph along with a mixed edge colouring and an edge weight function , gives a weighted edge coloured graph . For , denotes the weight of the edge .
We associate a vertex colouring with a perfect matching , in which each vertex which is contained in an edge gets the colour . Note that is monochromatic if and only if is monochromatic. denotes the set of all perfect matchings whose associated vertex colouring is . With respect to , if , we call to be a feasible vertex colouring of and if , we call to be an infeasible vertex colouring of .
The weight of a perfect matching , is the product of weights of all edges in . . The weight of a vertex colouring , is the sum of weights of all perfect matchings in .
If a vertex colouring is infeasible, that is if , we set . If is the same for all , we call the vertex colouring to be monochromatic. A weighted edge coloured graph is said to be perfectly monochromatic, if all feasible monochromatic vertex colourings have weight and all non-monochromatic vertex colourings have weight .
Note that unlike in the unweighted case, there can be non-monochromatic perfect matchings in a perfectly monochromatic weighted edge coloured graph since multiple non-monochromatic perfect matchings associated to the same vertex colouring can add up to zero. For example see Fig. 2. The edges and are coloured blue and the remaining edges are coloured red. The edge has weight and the remaining edges have weight . It has four perfect matchings: and . The red monochromatic vertex colouring is associated only with and hence has weight . The blue monochromatic vertex colouring is associated only with and hence has weight . There is only one feasible non-monochromatic vertex colouring in which are coloured red and are coloured blue. The matchings are associated to and hence . Therefore, is not perfectly monochromatic but is perfectly monochromatic. Moreover, .
If is perfectly monochromatic, the number of colours with a feasible monochromatic vertex using that colour, is represented by . Note that, if is not perfectly monochromatic, is not defined. The weighted matching index of , , where is set of all tuples for which is perfectly monochromatic. Recall that is always non-empty for a graph . If , is a perfectly monochromatic colouring of , there exists a weight function such that is also perfectly monochromatic, as presented in the proof of creftype 1.3 below due to Krenn. Therefore is also non-empty.
Let be perfectly monochromatic. From creftype 1.1, all edges are normal. When a monochromatic vertex colouring with colour is feasible, let denote the number of monochromatic perfect matchings of colour in . When monochromatic vertex colouring with colour is infeasible, we set . Consider the weight function on . It is easy to see that the weight of each monochromatic matching of colour is . Since there are many monochromatic perfect matchings of colour , the weight of any feasible monochromatic vertex colourings is . As is perfectly monochromatic, for any non-monochromatic vertex colouring , we know that . Therefore, is perfectly monochromatic and . It follows that . ∎
Recall that is the maximum matching covered graph of . If is perfectly monochromatic, then is also perfectly monochromatic and they have same the number of colour classes with a feasible monochromatic vertex colourings. It follows that and hence .
Krenn conjectured the following:
(Krenn’s Conjecture) If is isomorphic to , . Otherwise,
Krenn has been publicizing this conjecture in various ways since [krenn_website, mixon_website]. He has declared a reward for a resolution of this conjecture and even for a best paper award for making progress towards the solution.
If edge weights are real, positive numbers, then creftype 1 is true.
Consider any perfectly monochromatic weighted edge coloured graph . Note that as the edge weights are real, positive numbers, the weight of any perfect matching is positive, and hence the weight of any feasible vertex colouring is positive. If there were a non-monochromatic perfect matching in , there would be a feasible non-monochromatic vertex colouring of positive weight, which is not allowed. Therefore, all perfect matchings are monochromatic. By discarding the weights, we infer that is a perfectly monochromatic graph and hence has at most feasible monochromatic vertex colourings. From Theorem 1.2, the observation follows. ∎
2 Our results
From Theorem 1.2, we know that for any graph , . We know that is isomorphic to if and only if . We classify the graphs not isomorphic to to be of Type , Type and Type depending on the value being or respectively. Type graphs are the class of graphs, which do not have a perfect matching. It follows from the algorithm of Micali and Vazirani [DBLP:conf/focs/MicaliV80] that Type graphs can be identified in time.
We give a structural characterization of the matching covered subgraph of Type graphs in Theorem 3.12, using which we give a time algorithm to identify whether the given graph belongs to Type or Type in Theorem 3.16.
There is a structural characterization of Type graphs, using which we can get a time algorithm to decide whether the given graph is of Type or Type or Type .
If , has no perfect matchings and hence . If , we know that is . As has edges, it has at most disjoint perfect matchings. Hence . We also know that from creftype 1.3. Therefore . It is natural to ask whether for any graph , which if true would imply creftype 1 by Theorem 1.2. We give an affirmative answer to this question for Type graphs and resolve creftype 1 for Type graphs. We also prove that need not be equal to for Type graphs in Section 4.3. We do this by providing a Type graph satisfying in Fig. 2.
If , then .
There exists a graph , with and .
While we settled creftype 1 for non-type graphs, there are no known non-trivial upper bounds for for Type graphs. We first observe a simple upper bound in terms of the minimum degree.
Let be such that . Note that in , there are normal colour classes containing a monochromatic perfect matching. Therefore, any vertex of has at least one incident normal edge from each of the colours. Therefore, has degree at least . Therefore, the minimum degree of is at least . ∎
The minimum degree is considered to be the most intuitive connectivity parameter. There are two other parameters, vertex connectivity, and edge connectivity,, which captures the connectivity of in a more meaningful way. Whitney’s inequality [whitney] given below connects these parameters.
For a connected graph ,
2.1 Results for a simplified version of Krenn’s conjecture
For an edge coloured graph , we call a normal edge-colouring of if all edges of are normal. Restricting to such normal edge-colouring gives more structure to the problem. For instance, the weight of a non-monochromatic vertex colouring can be written as a product of the weight of independent parts of a partition based on colour classes like in creftype 5.1. When mixed coloured edges are there, a matching associated with a non-monochromatic vertex colouring can have edges across colour classes and therefore creftype 5.1 will not hold. This makes the conjecture much harder to deal with when mixed edges are present. Motivated by this, we study the simplified version of creftype 1 where all edges are normal, which is formalized in creftype 2.
We define normal weighted matching index of , , where is the set of all tuples for which is perfectly monochromatic and has only normal edges. It is easy to that .
(Simplified Version of Krenn’s Conjecture) If is isomorphic to , . Otherwise,
We now give an upper bound for in terms of vertex connectivity, .
For a connected graph ,
We note that when is disconnected and has a perfect matching, . Also, when is disconnected and has no perfect matching, . For the special class of graphs with vertex connectivity at most , we resolve creftype 2.
If , then
We also get an upper bound with respect to the maximum degree, . This bound is relevant only when . But it becomes useful to settle creftype 2 for the class of graphs with maximum degree at most , as mentioned in Corollary 2.9.
If , then . Otherwise,
If is not isomorphic to and , then .
Using SAT solvers, Krenn et al. [sat_solver_vertices] resolved creftype 2 for graphs with vertices.
If , then
3 Proof of Theorem 2.1: Unweighted case
Throughout this section, we assume that the graph is not isomorphic to and study the unweighted version. We first prove some simple structural properties of a coloured graph based on its value.
If the two disjoint perfect matchings of an even cycle are contained in two different normal colour classes, we call it a cycle with alternating colours.
is perfectly monochromatic.
has a perfect matching.
has a Hamiltonian cycle with alternating colours.
Statement is true by definition.
If creftype 1 is true and creftype 2 is false, as there are no perfect matchings, . If , must be perfectly monochromatic by Statement . Therefore, if there is a perfect matching it must be monochromatic. But that would imply . Therefore, there is no perfect matching in . Therefore, statement is true.
We prove that if creftype 1, creftype 2 are true and creftype 3 is false, then . If creftype 1 and creftype 2 are true then by statement and statement . Towards a contradiction, let . It follows that . Therefore, there are at least two colours, say red and blue, with a monochromatic perfect matching. Consider the union of a red perfect matching and a blue perfect matching. It must be a disjoint union of even cycle(s) with alternating colours. Clearly, each cycle can be decomposed into two perfect matchings of different colours. If it is a disjoint union of at least two cycles, then by selecting the red matching from one cycle and blue matchings from the remaining cycles, we can construct a non-monochromatic perfect matching. This contradicts the assumption that creftype 1 is true. Therefore, the union of a red and a blue perfect matching is a Hamiltonian cycle with edges of alternating colours. But it is given creftype 3 is false. Contradiction.
If creftype 1, creftype 2 and creftype 3 are true, since there are two monochromatic perfect matchings of different colours which can be formed by selecting the edges from the same normal colour class of the Hamiltonian cycle with alternating colours, . As creftype 1 is true, we know that by Theorem 1.2. Therefore, .
If , then it is easy to see that creftype 1, creftype 2 are true. creftype 3 must be false, because if it is true, that would imply as proved above. Similarly, if , then it is easy to see that creftype 1, creftype 2 are true. creftype 3 must be true, because if it is false, that would imply as proved above. ∎
For a graph of even order and with a Hamiltonian cycle , we define some notation with respect to and . Let the vertices be in clockwise direction along the cycle . We interchangeably refer vertex as . We call
as odd vertices andas even vertices. An odd-even edge is an edge between an odd vertex and even vertex. Similarly, an odd-odd (even-even) edge is an edge between two odd (even) vertices. The edges of the cycle are called -edges. An odd-even edge is called illegal if it is not a -edge. The odd-odd edges and even-even edges are called legal edges. Note that the edges of the graph can be classified into legal edges, illegal edges and -edges.
Note that every edge partitions the remaining vertices of cycle into two parts and . The partition weight of is defined to be equal to . Note that both are odd (even) for legal (illegal) edges. If is odd, we define , a matching formed by the alternate -edges in clockwise order from to .
If an edge has one endpoint in and the other endpoint in , then and are said to be crossing each other. We say that a pair of crossing edges form a legal crossing pair if both the edges are legal. We say that a legal crossing pair is a nice crossing pair if one of the edges is odd-odd and the other edge is even-even.
Definition of drum: Let a nice crossing pair be formed by edges and without loss of generality, assume that the vertices appear in clockwise order on . We say that form a drum if as shown in the left of Fig. 3 or as shown in the right of Fig. 3. In other words an end vertex of and an end vertex of must be adjacent on the cycle (that is, there should be a -edge between them) and the remaining end vertices of must also be adjacent on the cycle .
Note that a drum , say with vertices partitions the remaining vertices of the cycle into two parts, and . It follows that are even for a drum by the definition. We refer to the paths along from to through and from to through to be straps of the drum. The length of a strap is the number of edges on the path. We refer to as the partition weight of the drum . Note that a drum is a -cycle with -edges forming a pair of opposite edges and a nice crossing pair forming the remaining two opposite edges.
It is important to note that even though there might be structures similar to drums formed by two illegal crossing edges, which partition the cycle into two odd parts, we do not refer to them as drums. Only nice crossing pairs can form drums. Therefore, if we colour the Hamiltonian cycle (normally) with alternating colours, both the -edges of a drum will get the same colour.
We emphasize that legal edges, illegal edges, -edges, crossing pairs and drums are defined with respect to a graph and a vertex labelling along the clockwise direction of a Hamiltonian cycle . For example, for a graph , the legal edges with respect to might not be legal edges with respect to a different Hamiltonian cycle . See Fig. 4.
We refer to a Hamiltonian cycle with alternating red-blue edges as a red-blue alternating Hamiltonian cycle . For graphs with such a Hamiltonian cycle , whose vertices are labelled in clockwise cyclic order along , we use a colouring convention that the edges of of the form be red and the edges be blue. We call a drum in monochromatic if all edges of the drum are contained in the same normal colour class.
is a matching covered graph with a colouring for which and . From Theorem 3.1, it has at least one red-blue alternating Hamiltonian cycle, say . Label the vertices in clockwise cyclic order with respect to so that colouring convention is satisfied.
We now make some observations about edge coloured graphs that satisfy creftype 4. Note that when creftype 4 is true, from creftype 1.1, there are no mixed edges as it is a matching covered graph. Therefore, we need to consider only normal edge colourings. We now prove some observations for this case, which we will later use to prove the structural characterization of Type graphs.
If creftype 4 is true, has no illegal edges.
Towards a contradiction, let there be an illegal edge. Let it be for some , without loss of generality. Consider the matching , see Section 3.1 for notation. As an illegal edge cannot be a -edge, both are non-empty. Observe that has colour blue and has colour red by the colouring convention of . Therefore, the matching is a non-monochromatic perfect matching. But all matchings of are monochromatic as is perfectly monochromatic. Contradiction. ∎
If creftype 4 is true, all nice crossing pairs of must form monochromatic drums.
First we prove that all nice crossing pairs form a drum. Towards a contradiction, let there be a nice crossing pair and which do not form a drum. Without loss of generality, we assume that are in clockwise direction on the cycle . There is a perfect matching
Note that is contained in red colour class and is contained in blue colour class by the colouring convention of . If either of these two are empty, the crossing pair forms a drum. Therefore both are non-empty, and is a non-monochromatic perfect matching. But all matchings of are monochromatic as is perfectly monochromatic. Contradiction. Therefore, all nice crossing pairs must form drums.
Now we show that every drum is monochromatic. Towards a contradiction, let there be a non-monochromatic drum. Recall (see the definition of a drum) that non-adjacent -edges of a drum are in the same colour class, say red. Therefore, are part of the red perfect matching formed by all red edges of . Let the crossing pair of the drum be . Consider the perfect matching . Since has at least vertices, has at least one edge contained in red colour class. As it is a non-monochromatic drum, at least one of must be blue. Therefore, is a non-monochromatic perfect matching. But all perfect matchings of are monochromatic as is perfectly monochromatic. Contradiction. Therefore, all nice crossing pairs must form monochromatic drums. ∎
If creftype 4 is true, every legal edge of is contained in exactly one drum.
Let the legal edge be and of colour red, without loss of generality. Recall that can only be part of the drums with non-adjacent cycle edges or by definition. We claim that the later possibility can be ruled out. This is because cannot form a drum with the legal edge , since would form a nice crossing pair and the drum formed by should be monochromatic by creftype 3.3; However, which would form the -edges of that drum are of colour blue by the colouring convention of . Therefore, can be part of at most one drum. A similar argument holds even if was of colour blue or was an even-even edge.
Now we will show that a legal edge has to be part of a drum. Towards a contradiction, let there be at least one legal edge, which is not part of any drum. Among all such legal edges not part of any drum, let have the minimum partition weight without loss of generality. Let be the smaller among and , without loss of generality. must be in a perfect matching as is matching covered. Observe that there are even vertices and odd vertices in . An even vertex in can not be paired with an even vertex in , as that would contradict our assumption that has the minimum partition weight. As there is one more even vertex than the odd vertices, an even vertex in must match with an even vertex in , since illegal edges are not present by creftype 3.2. It follows that this even-even edge crosses , and since is an odd-odd edge, they form a nice crossing pair. Therefore, from creftype 3.3, they must form a drum. Contradiction. ∎
If creftype 4 is true, each -edge of can be part of at most one drum.
Let a -edge be part of two drums, say towards a contradiction. From creftype 3.4, the legal edges in both drums must be disjoint. Let they be in drum and in drum . Let be the edges incident on even end point of and be the edges incident on the odd end point of . Observe that one of and cross each other. It follows that they form a nice crossing pair and hence form a drum by creftype 3.3. However, a legal edge can be part of only one drum by creftype 3.4. Contradiction. ∎
If creftype 4 is true, all legal crossing pairs of are nice.
Towards a contradiction, let there be a legal crossing pair , which is not nice. Without loss of generality, let both of them be odd-odd edges. From creftype 3.4, must form a drum with a even-even edge, say . As cross, they can not be incident on the same vertex. As is odd-odd and is even-even, they can not be incident on the same vertex. Therefore, should be incident on a vertex from and a vertex from . Therefore, cross each other. As they are a nice crossing pair, from creftype 3.3, they must form a drum. Therefore, is part of two drums. But from creftype 3.4, is part of exactly one drum. Contradiction. ∎
If creftype 4 is true, all crossing pairs of form monochromatic drums.
If creftype 4 is true, every vertex of can have at most two red edges and at most two blue edges incident on it. Moreover, if the vertex has two red(blue) edges incident on it, both of those red(blue) edges must be part of a red(blue) drum.
Let the two cycle edges incident on be . Without loss of generality, let be red and be blue. Every -edge can be part of at most one drum from creftype 3.5. Also, all drums are monochromatic from creftype 3.7. Therefore, can be part of at most one red drum and can be part of at most one blue drum.
Recall that other than -edges, only legal edges are incident on from creftype 3.2. Every legal edge incident on vertex must be part of a unique drum from creftype 3.4. But each of these drums must contain a cycle edge incident on . It follows that can be part of at most one red drum and at most one blue drum, and therefore, can have at most two red edges and two blue edges incident on it. ∎
If is a matching covered graph and , then .
If , it is easy to see that . We now assume that . As , there is a colouring such that . It follows from Theorem 3.1 that creftype 4 holds. Therefore, from creftype 3.8, on each vertex of , at most two red edges and at most two blue edges are incident. Therefore, the degree is for any vertex on and hence . ∎
If creftype 4 is true, all Hamiltonian cycles of are red-blue alternating cycles.
Any Hamiltonian cycle can be partitioned into two perfect matchings. Since is perfectly monochromatic, both these matchings must be monochromatic. Therefore, the Hamiltonian cycle is either a red-blue alternating cycle or a monochromatic cycle.
There is a red-blue alternating Hamiltonian cycle from Theorem 3.1. Towards a contradiction, let there be a monochromatic cycle , say of colour red, without loss of generality. Therefore, every vertex must have at least two red edges incident on it. From creftype 3.8, every vertex can have at most two red edges incident on it. It follows that all red edges of the graph form a Hamiltonian cycle . From creftype 3.8, since the degree of all vertices is , each vertex must be part of a drum with respect to . Therefore, the monochromatic red Hamiltonian cycle must contain a red monochromatic drum, which is a -cycle. This is possible only when is a -cycle. But from creftype 4. Contradiction. It follows that there are no monochromatic red Hamiltonian cycles. Using a similar argument, it can be proved that there are no monochromatic blue Hamiltonian cycles. Therefore, all Hamiltonian cycles of are red-blue alternating cycles. ∎
If creftype 4 is true and is not a cycle, a drum exists such that all vertices in have degree .
Among all drums, let be a drum having smallest partition weight equal to . Let be formed by the crossing edges . If , the observation trivially holds. Therefore, we assume that has at least vertex. Towards a contradiction, assume that there is a vertex in with degree greater than . Therefore, there exists some legal edge , incident on .
From creftype 3.4, must be part of some drum along with a crossing edge . As are part of exactly one drum , or cannot be in , as in that case there would another drum involving or other than . From creftype 3.7, as all crossing pairs are drums, neither nor can cross or . Therefore, both the edges of must be on the strap of along . Therefore, . But this contradicts the assumption that has the smallest partition weight. ∎
3.3 Structural characterization of Type graphs
Let be the family of even-order matching-covered Hamiltonian graphs with at least vertices such that all graphs satisfy the following: For all Hamiltonian cycles of , creftype 1, creftype 2 and creftype 3 are satisfied with respect to the vertex labelling obtained from a clockwise cyclic ordering of .
Let be the family of even-order matching-covered Hamiltonian graphs with at least vertices such that all graphs satisfy the following: There exists a Hamiltonian cycle of such that creftype 1, creftype 2 and creftype 3 are satisfied with respect to the vertex labelling obtained from a clockwise cyclic ordering of .
All edges of are legal or -edges.
Every crossing pair of forms a drum.
Each legal edge of is part of exactly one drum.
An example graph is shown in Fig. 4 which satisfies creftype 1, creftype 2 and creftype 3 with respect to the cyclic orderings along two different Hamiltonian cycles. As the properties are satisfied with respect to at least one cycle, . To verify that , we have to consider the vertex labellings with respect to all the Hamiltonian cycles of one by one. Though it may look a little surprising, the following theorem states that creftype 1, creftype 2, creftype 3 are satisfied with respect to all the Hamiltonian cycles of since they are seen to be satisfied with respect to some Hamiltonian cycle of !
. For any graph with , if and only if .
We now proceed to show that . It is easy to see that by definition. Consider any matching covered graph . From Lemma 3.14, . As , from Lemma 3.13, . It follows that and hence . It now suffices to prove Lemma 3.13 and Lemma 3.14.
For any graph with , if , then
Let . If , then . It follows that there exists a perfectly monochromatic colouring such that . From Theorem 3.1, has a perfect matching implying that the order of is even; Moreover has at least one alternating red-blue Hamiltonian cycle. Therefore creftype 4 holds for . From creftype 3.10, all Hamiltonian cycles of are alternating red-blue cycles. Therefore, with respect to any Hamiltonian cycle of creftype 4 holds. It follows that creftype 1, creftype 2 and creftype 3 hold due to creftype 3.2, creftype 3.7 and creftype 3.4 respectively. Therefore, . ∎
If , then .
Let . We first show that there is a normal edge colouring for any graph for which is .
By definition of , there exists an even order Hamiltonian cycle, with respect to which creftype 1, creftype 2 and creftype 3 are satisfied. Colour with alternating red and blue colours. Notice that all uncoloured edges are legal due to creftype 1. Consider a legal edge , due to creftype 3, it is part of a unique drum having another legal edge , where form a crossing pair. By the definition of a drum, both the non-adjacent -edges of must be in the same normal colour class with respect to colouring . Colour with the same colour as the non-adjacent -edges of . Do this for all legal edges. Due to creftype 3, as each legal edge is part of exactly one drum, each legal edge gets coloured exactly once. Thus, we now have a colouring for the entire graph. We now prove that this normal colouring is perfectly monochromatic.
We first prove that, for a drum with a crossing pair and a perfect matching , either or . Towards a contradiction, let there be a perfect matching , which contains but not . As is legal, and are odd. Since can not match the vertices of within itself and every vertex must be covered by , there must be an edge crossing . But every crossing pair forms a drum due to creftype 2 and is part of only one drum due to creftype 3. Therefore must be . Thus . Contradiction.
We now prove that the colouring is perfectly monochromatic. Towards a contradiction, let there be a non-monochromatic perfect matching . If contains legal edges, they come in pairs such that each pair belongs to a drum. Clearly, all these pairs of legal edges that belong to are disjoint from creftype 3. Replace each such crossing pair in with the associated pair of -edges of the corresponding drum to form a new perfect matching . Observe that since all drums are monochromatic, we are replacing a pair of crossing edges with -edges of the same colour. It follows that since is a non-monochromatic perfect matching, must also be a non-monochromatic perfect matching. However, is a perfect matching from Hamiltonian cycle , which is alternately coloured. Thus must be monochromatic. Contradiction.
Therefore, has no non-monochromatic perfect matchings. Therefore, is perfectly monochromatic. Since there are two perfect matchings of different colours in , . Therefore, . From Theorem 1.2, . It follows that and hence . ∎
3.4 Algorithm to find
We first prove Lemma 3.15, which is useful to prove the correctness of our algorithm.
Let be a matching-covered Type graph. Let and denote the set of all perfect matchings of containing the edge . There exists two edges incident on such that is a Hamiltonian cycle of , for any and any
Let be a perfectly monochromatic colouring of such that . From Theorem 3.1, is perfectly monochromatic and has a red-blue alternating Hamiltonian cycle . For a vertex , let two of its incident edges from be . Without loss of generality, let be red and be blue. For those , we prove that is a Hamiltonian cycle of , for any and .
Since is red, and is perfectly monochromatic, all perfect matchings in must be monochromatic red matchings by definition. Similarly, all perfect matchings in must be monochromatic blue matchings. Therefore, and must be a disjoint union of even cycle(s) of alternating colours. If it is a disjoint union of at least two even cycles, then by selecting the red matching from one cycle and blue matchings from the remaining cycles, we can construct a non-monochromatic perfect matching. However, this is not possible as is perfectly monochromatic. Therefore, is a Hamiltonian cycle. ∎
For any graph , there exists an algorithm to find in time.
From Theorem 3.1, if isomorphic to , . From Theorem 3.1, is a Type graph if and only if there is no perfect matching in . The absence of a perfect matching can be checked using an algorithm of Micali and Vazirani [DBLP:conf/focs/MicaliV80] in time.
We now know that the graph must be either Type or Type . We first find the maximum matching covered subgraph in time using a deterministic algorithm due to Carvalho and Cheriyan [Carvalho2005AnOA]. It can also be computed in time using a randomized algorithm due to Rabin and Vazirani [DBLP:journals/jal/RabinV89]. As, , it is now sufficient to find whether is Type or Type as described in Algorithm 1. The overview of Algorithm 1 is described below, along with its correctness.
If is not Hamiltonian, is Type from Theorem 3.1. If is Hamiltonian and we know a Hamiltonian cycle of , it is easy to check if satisfies creftype 1,creftype 2 and creftype 3 with respect to in time. Due to Theorem 3.12, if all three properties are satisfied with respect to , then and hence Type : If any of the property is not satisfied with respect to , and hence Type . However, finding Hamiltonian cycles in general graphs is an -hard problem. Fortunately, since we are only interested in Type graphs, we can give a simple efficient algorithm that returns a Hamiltonian cycle of , if it is a Type graph using Lemma 3.15.
We pick any vertex , arbitrarily. Note that the maximum degree is at most for Type graphs from creftype 3.9. Therefore, we try to find an Hamiltonian cycle only when . There are at most pairs of edges incident on . Since is a matching covered graph, for each pair , we can find arbitrary perfect matchings containing respectively, using any maximum matching algorithm [DBLP:conf/focs/MicaliV80]. If their union is not a Hamiltonian cycle for all pairs, it is a Type graph from Lemma 3.15. If their union is a Hamiltonian cycle for at least one pair, then we check if