The Power of Data Reduction for Matching
Finding maximum-cardinality matchings in undirected graphs is arguably one of the most central graph primitives. For -edge and -vertex graphs, it is well-known to be solvable in time; however, for several applications this running time is still too slow. We investigate how linear-time (and almost linear-time) data reduction (used as preprocessing) can alleviate the situation. More specifically, we focus on (almost) linear-time kernelization. We start a deeper and systematic study both for general graphs and for bipartite graphs. Our data reduction algorithms easily comply (in form of preprocessing) with every solution strategy (exact, approximate, heuristic), thus making them attractive in various settings.
“Matching is a powerful piece of algorithmic magic” . In Matching, given a graph, one has to compute a maximum-cardinality set of nonoverlapping edges. Matching is arguably among the most fundamental graph-algorithmic primitives allowing for a polynomial-time algorithm. More specifically, on an -vertex and -edge graph a maximum matching can be found in time . Improving this upper time bound, even for bipartite graphs, resisted decades of research. Recently, however,  presented a linear-time algorithm that computes a -approximate maximum-weight matching, where the running time dependency on is . For the unweighted case, the algorithm of  implies a linear-time -approximation, where in this case the running time dependency on is . We take a different route: First, we do not give up the quest for optimal solutions. Second, we focus on efficient data reduction rules—not solving an instance but significantly shrinking its size before actually solving the problem
The spirit behind our approach is thus closer to the identification of efficiently (i.e. linearly) solvable special cases of Matching. There is quite some body of work in this direction. For instance, since an augmenting path can be found in linear time , the standard augmenting path-based algorithm runs in time, where is the number of edges in the maximum matching.  developed an -time algorithm, where is the difference between maximum and minimum vertex degree of the input graph. Moreover, there are linear-time algorithms for computing maximum matchings in special graph classes, including convex bipartite , strongly chordal , and chordal bipartite graphs .
All this and the more general spirit of “parameterization for polynomial-time solvable problems”  (also referred to as “FPT in P” or “FPTP” for short) forms the starting point of our research. Remarkably,  recently developed an algorithm to compute a maximum matching in graphs of treewidth in randomized time.
Following the paradigm of kernelization, that is, provably effective and efficient data reduction, we provide a systematic exploration of the power of polynomial-time data reduction for Matching. Thus, our aim (fitting within FPTP) is to devise problem kernels that are computable in (almost) linear time. A particular motivation for this is that with such very efficient kernelization algorithms it is possible to transform multiplicative () into additive () “(almost) linear-time FPTP” algorithms. Furthermore, kernelization algorithms (typically based on data reduction rules) can be used as a preprocessing to heuristics or approximation algorithms with the goal of getting larger matchings.
As kernelization is usually defined for decision problems, we use in the remainder of the paper the decision version of Matching. In a nutshell, a kernelization of a decision problem instance is an algorithm that produces an equivalent instance whose size can solely be upper-bounded by a function in the parameter (preferably a polynomial). The focus on decision problems is justified by the fact that all our results, although formulated for the decision version, in a straightforward way extend to the corresponding optimization version.
|An undirected graph and a nonnegative integer .|
|Is there a size subset of nonoverlapping (i.e. disjoint) edges?|
Since solving the given instance and returning a trivial yes- or no-instance always produces a constant-size kernel in polynomial time, we are looking for kernelization algorithms that are faster than the algorithms solving the problem. For NP-hard problems, each kernelization algorithm, since running in polynomial time, is (presumably) faster than any solution algorithm. This is, of course, no longer true when applying kernelization to a polynomial-time solvable problem like Matching. While the focus of classical kernelization for NP-hard problems is mostly on improving the size of the kernel, we particularly emphasize that for polynomially solvable problems it now becomes crucial to also focus on the running time of the kernelization algorithm. Moreover, the parameterized complexity analysis framework can also be applied to the kernelization algorithm itself. For example, a kernelization algorithm running in time ( is the “problem specific” parameter) might be preferable to another one running in time. In this paper, we present kernelization algorithms for Matching which run in linear time (see ) or in almost linear time (i.e. in time, see ).
In this paper we present three efficiently computable kernels for Matching(see for an overview).
|Parameter||running time||kernel size|
|Feedback edge number||time||vertices and edges||()|
|Feedback vertex number||time||vertices and edges||()|
|Distance to chain graphs||time||vertices||()|
All our parameterizations can be categorized as “distance to triviality” . They are motivated as follows. First, note that maximum-cardinality matchings be can trivially found in linear time on trees (or forests). So we consider the corresponding edge deletion distance (feedback edge number) and vertex deletion distance (feedback vertex number). Notably, there is a trivial linear-time algorithm for computing the feedback edge number and there is a linear-time factor-4 approximation algorithm for the feedback vertex number . We mention in passing that the parameter vertex cover number, which is lower-bounded by the feedback vertex number, has been frequently studied for kernelization . In particular,  (implicitly) provided a quadratic-size kernel for Matching with respect to the parameter vertex cover number. Coming to bipartite graphs, note that our parameterization by vertex deletion distance to chain graphs is motivated as follows. First, chain graphs form one the most obvious easy cases for bipartite graphs where Matching can be solved in linear time . Second, we show that the vertex deletion distance of any bipartite graph to a chain graph can be 2-approximated in linear time. Moreover, vertex deletion distance to chain graphs lower-bounds the vertex cover number of a bipartite graph.
An overview of our main results is given in Table 1. We study kernelization for Matching parameterized by the feedback vertex number, that is, the vertex deletion distance to a forest (see ). As a warm-up we first show that a subset of our data reduction rules for the “feedback vertex set kernel” also yields a linear-time computable linear-size kernel for the typically much larger parameter feedback edge number (see ). As for Bipartite Matching no faster algorithm is known than on general graphs, we kernelize Bipartite Matching with respect to the vertex deletion distance to chain graphs (see ).
Seen from a high level, our two main results employ the same algorithmic strategy, namely upper-bounding (as a function of the parameter) the number of neighbors in the appropriate vertex deletion set ; that is, in the feedback vertex set or in the deletion set to chain graphs, respectively. To achieve this we develop new “irrelevant edge techniques” tailored to these two kernelization problems. More specifically, whenever a vertex of the deletion set has large degree, we efficiently detect edges incident to whose removal does not change the size of the maximum matching. Then the remaining graph can be further shrunk by scenario-specific data reduction rules. While this approach of removing irrelevant edges is natural, the technical details and the proofs of correctness can become quite technical and combinatorially challenging. In particular, for the case of feedback vertex number we could only upper-bound the number of neighbors of each vertex in by .
As a technical side remark, we emphasize that in order to achieve an (almost) linear-time kernelization algorithm, we often need to use suitable data structures and to carefully design the appropriate data reduction rules to be exhaustively applicable in linear time, making this form of “algorithm engineering” much more relevant than in the classical setting of mere polynomial-time data reduction rules.
Notation and Observations.
We use standard notation from graph theory. In particular all paths we consider are simple paths. Two paths in a graph are called internally vertex-disjoint if they are either completely vertex-disjoint or they overlap only in their endpoints. A matching in a graph is a set of pairwise disjoint edges. Let be a graph and let be a matching in . The degree of a vertex is denoted by . A vertex is called matched with respect to if there is an edge in containing , otherwise is called free with respect to . If the matching is clear from the context, then we omit “with respect to ”. An alternating path with respect to is a path in such that every second edge of the path is in . An augmenting path is an alternating path whose endpoints are free. It is well known that a matching is maximum if and only if there is no augmenting path for it. Let and be two matchings in . We denote by the graph containing only the edges in the symmetric difference of and , that is, . Observe that every vertex in has degree at most two.
For a matching for we denote by a maximum matching in with the largest possible overlap (in number of edges) with . That is, is a maximum matching in such that for each maximum matching for it holds that . Observe that, if is a maximum matching for , then . Furthermore observe that consists of only odd-length paths and isolated vertices, and each of these paths is an augmenting path for . Moreover the paths in are as short as possible:
Assume that . Then is a shorter path which is also an augmenting path for in . The corresponding maximum matching satisfies , a contradiction to the definition of .
A parameterized problem is a set of instances where for a finite alphabet , and is the parameter. We say that two instances and of parameterized problems and are equivalent if is a yes-instance for if and only if is a yes-instance for . A kernelization is an algorithm that, given an instance of a parameterized problem , computes in polynomial time an equivalent instance of (the kernel) such that for some computable function . We say that measures the size of the kernel, and if , we say that admits a polynomial kernel. Often, a kernel is achieved by applying polynomial-time executable data reduction rules. We call a data reduction rule correct if the new instance that results from applying to is equivalent to . An instance is called reduced with respect to some data reduction rule if further application of this rule has no effect on the instance.
2Kernelization for Matching on General Graphs
In this section, we investigate the possibility of efficient and effective preprocessing for Matching. As a warm-up, we first present in a simple, linear-size kernel for Matching with respect to the parameter “feedback edge set”. Exploiting the data reduction rules and ideas used for this kernel, we then present in the main result of this section: an exponential-size kernel for the smaller parameter “feedback vertex number”.
2.1Warm-up: Parameter feedback edge number
We provide a linear-time computable linear-size kernel for Matching parameterized by the feedback edge number, that is, the size of a minimum feedback edge set. Observe that a minimum feedback edge set can be computed in linear time via a simple depth-first search or breadth-first search. The kernel is based on the next two simple data reduction rules due to . They deal with vertices of degree at most two.
The correctness was stated by . For completeness, we give a proof.
If has degree zero, then clearly cannot be in any matching and we can remove .
If has degree one, then let be its single neighbor. Let be a maximum matching of size at least for . Then is matched in since otherwise adding the edge would increase the size of the matching. Thus, a maximum matching in is of size at least . Conversely, a maximum matching of size in can easily be extended by the edge to a maximum matching of size in .
If has degree two, then let and be its two neighbors. Let a maximum matching of size at least . If is not matched in , then and are matched since otherwise adding the edge resp. would increase the size of the matching. Thus, deleting and merging and decreases the size of by one ( looses either the edge incident to or one of the edges incident to and ). Hence, the resulting graph has a maximum matching of size at least . Conversely, let be a matching of size at least for . If the merged vertex is free, then is a matching of size in . Otherwise, is matched to some vertex . Then matching in with either or (at least one of the two vertices is a neighbor of ) and matching with the other vertex yields a matching of size for .
Although are correct, it is not clear whether can be exhaustively applied in linear time. However, for our purpose it suffices to consider the following restricted version which we can exhaustively apply in linear time.
We give an algorithm which exhaustively applies in linear time. First, using bucket sort, sort the vertices by degree and keep three lists containing all degree-zero/one/two vertices. Then one applies in a straightforward way. When a neighbor of a vertex is deleted, then check if the vertex has now degree zero, one, or two. If yes, then add the vertex to the corresponding list.
We next show that this algorithm runs in linear time. First, observe that the deletion of each each degree-zero vertex can be done in constant time as no further vertices are affected. Second, consider a degree-one vertex with a neighbor and observe that deleting and can be done time since one needs to update the degrees of all neighbors of . Furthermore, decreasing by one can be done in constant time for each deleted degree-one vertex. Finally, consider a degree-two vertex with two neighbors and , each of degree at most two. Deleting takes constant time. To merge and iterate over all neighbors of and add them to the neighborhood of . If a neighbor of is already a neighbor of , then decrease the degree of by one. Then, relabel to be the new contracted vertex .
Overall, the worst-case running time to apply exhaustively can be upper-bounded by .
Apply exhaustively in linear time (see ). We claim that the reduced graph has less than vertices and edges. Denote with a feedback edge set for , . Furthermore, denote with , , and the vertices that have degree one, two, and more than two in the . Thus, as each leaf in has to be incident to an edge in . Next, since is a forest (or tree), we have and thus . Finally, each degree-two vertex in needs at least one neighbor of degree at least three since is reduced with respect to . Thus, the vertices in are either incident to an edge in or adjacent to one of the at most vertices in that have degree at least three. Since the sum over all degrees of vertices in is at most , it follows that . Thus, the number of vertices in is . Since is a forest, it follows that has at most edges.
Applying the -time algorithm for Matching  on the kernel yields:
2.2Parameter feedback vertex number
We next provide for Matching a kernel of size computable in time where is the “feedback vertex number”. Using a known linear-time factor 4-approximation algorithm , we can approximate feedback vertex set and use it in our kernelization algorithm.
Roughly speaking, our kernelization algorithm extends the linear-time computable kernel with respect to the parameter “feedback edge set”. Thus, play an important role in the kernelization. Compared to the other kernels presented in this paper, the kernel presented here comes at the price of higher running time and bigger kernel size (exponential size). It remains open whether Matching parameterized by the “feedback vertex number” admits a linear-time computable kernel (possibly of exponential size), and whether it admits a polynomial kernel computable in time.
Subsequently, we describe our kernelization algorithm which keeps in the kernel all vertices in the given feedback vertex set and shrinks the size of . Before doing so, we need some further notation. In this section, we assume that each tree is rooted at some arbitrary (but fixed) vertex such that we can refer to the parent and children of a vertex. A leaf in is called a bottommost leaf either if it has no siblings or if all its siblings are also leaves. (Here, bottommost refers to the subtree with the root being the parent of the considered leaf.) The outline of the algorithm is as follows (we assume throughout that since otherwise the input instance is already a kernel of size ):
Reduce wrt. .
Compute a maximum matching in .
Modify in linear time such that only the leaves of are free ().
Bound the number of free leaves in by ().
Bound the number of bottommost leaves in by ().
Bound the degree of each vertex in by . Then, use to provide the kernel of size ().
Whenever we reduce the graph at some step, we also show that the reduction is correct. That is, the given instance is a yes-instance if and only if the reduced one is a yes-instance. The correctness of our kernelization algorithm then follows by the correctness of each step. We discuss in the following some details of each step.
By we can perform in linear time. By this step is correct.
A maximum matching in can be computed by repeatedly matching a free leaf to its neighbor and by removing both vertices from the graph (thus effectively applying to ). By , this can be done in linear time.
can be done in time by traversing each tree in in a BFS manner starting from the root: If a visited inner vertex is free, then observe that all children are matched since is maximum. Pick an arbitrary child of and match it with . The vertex that was previously matched to is now free and since it is a child of , it will be visited in the future. Observe that do not change the graph but only the auxiliary matching , and thus these steps are correct.
Recall that our goal is to upper-bound the number edges between vertices of and , since we can then use a simple analysis as for the parameter “feedback edge set”. Furthermore, recall that by the size of any maximum matching in is at most plus the size of . Now, the crucial observation is Observe that if a vertex has at least neighbors in that are free wrt. , then there exists a maximum matching where is matched to one of these vertices since at most can be “blocked” by other matching edges. This means we can delete all other edges incident to . Formalizing this idea, we obtain the following reduction rule.
We first discuss the correctness and then the running time. Denote by the size of a maximum matching in the input graph and by the size of a maximum matching in the new graph , where some edges incident to are deleted. We need to show that . Since any matching in is also a matching in , we easily obtain . It remains to show . To this end, let be a maximum matching for with the maximum overlap with (see ). If is free wrt. or if matched to a vertex that is also in a neighbor of , then is also a matching in () and thus we have in this case . Hence, consider the remaining case where is matched to some vertex such that , that is, the edge was deleted by . Hence, has neighbors in such that each of these neighbors is free wrt. and none of the edges , was deleted. Observe that by the choice of , the graph (the graph over vertex set and the edges that are either in or in , see ) contains exactly paths (we do not consider isolated vertices as paths). Each of these paths is an augmenting path for . By , we have . Observe that is an edge in one of these augmenting paths; denote this path with . Thus, there are at most paths that do not contain . Also, each of these paths contains exactly two vertices that are free wrt. : the endpoints of the path. This means that no vertex in is an inner vertex on such a path. Furthermore, since is a maximum matching, it follows that for each path at most one of these two endpoints is in . Hence, at most vertices of are contained in the paths of except . Therefore, one of these vertices, say , is free wrt. and can be matched with . Thus, by reversing the augmentation along and adding the edge we obtain another matching of size . Observe that is a matching for and for and thus we have . This completes the proof of correctness.
Now we come to the running time. We exhaustively apply the data reduction rule as follows. First, initialize for each vertex a counter with zero. Second, iterate over all free vertices in in an arbitrary order. For each free vertex iterate over its neighbors in . For each neighbor do the following: if the counter is less than , then increase the counter by one and mark the edge (initially all edges are unmarked). Third, iterate over all vertices in . If the counter of the currently considered vertex is , then delete all unmarked edges incident to . This completes the algorithm. Clearly, it only deletes edges incident to a vertex only if has free neighbors in and the edges to these neighbors are kept. The running time is : When iterating over all free vertices in we consider each edge at most once. Furthermore, when iterating over the vertices in , we again consider each edge at most once.
To finish , we exhaustively apply in linear time. Afterwards, there are at most free (wrt. to ) leaves in that have at least one neighbor in since each of the vertices in is adjacent to at most free leaves. Thus, applying we can remove the remaining free leaves that have no neighbor in . However, since for each degree-one vertex also its neighbor is removed, we might create new free leaves and need to again apply and update the matching (see ). This process of alternating application of stops after at most rounds since the neighborhood of each vertex in can be changed by at most once. This shows the running time . We next show how to improve this to and arrive at the final lemma of this subsection.
In the following, we explain which reduces the graph with respect to and updates the matching as described in . The algorithm performs in () . As described in the previous section, this can be done in linear time. Next, is applied in using the approach described in the proof of : For each vertex in a counter is maintained. When iterating over the free leaves in , these counters will be updated. If a counter reaches , then the algorithm knows that has fixed free neighbors and according to the edges to all other vertices can be deleted (see ). Observe that once the counter reaches , the vertex will never be considered again by the algorithm since its only remaining neighbors are free leaves in that already have been popped from stack . The only difference from the description in the proof of is that the algorithm reacts if the degree of some vertex in is decreased to one (see ). If is matched, then simply remove and its matched neighbor from and . Otherwise, add to the list of unmatched degree-one vertices and defer dealing with to a latter stage of the algorithm.
Observe that the matching still satisfies the property that each free vertex in is a leaf since only matched vertex pairs were deleted so far. When deleting unmatched degree-one vertices and their respective neighbor, the maximum matching needs to be updated to satisfy this property. The algorithm does this from : Let be an entry in such that has degree one in , that is, is a free leaf in and has no neighbors in . Then, following , delete and its neighbor and decrease the solution size by one (see ). Let denote the previously matched neighbor of . Since was removed, is now free. If is a leaf in , then we can simply add it to and in this way deal with it later. If is not a leaf, then we need to update since only leaves are allowed to be free. To this end, take an arbitrary alternating path from to a leaf of the subtree with root and augment along (see ). This can be done as follows: Pick an arbitrary child of . Let be the matched neighbor of . Since is the parent of , it follows that is a child of . Now, remove from and add . If is a leaf, then the alternating path is found with and augmented. Otherwise, repeat the above procedure with taking the role of . This completes the algorithm. Its correctness follows from the fact that it only deletes edges and vertices according to .
It remains to show the running time of . To this end, we prove that the algorithm considers each edge in only two times. First, consider the edges incident to a vertex . These edges will be inspected at most twice by the algorithm: Once, when it is marked (see ). The second time is when it is deleted. This bounds the running time in the first part ().
Now consider the remaining edges within . To this end, observe that the algorithm performs two actions on the edges: deleting the edges () and finding and augmenting along an alternating path (). Clearly, after deleting an edge it will no longer be considered, so it remains to show that edge is part of at most one alternating path used in . Assume toward a contradiction that the algorithm augments along an edge twice or more. From all the edges that are augmented twice or more let be one that is closest to the root of the tree is contained in, that is, there is no edge closer to a root. Let and be the endpoints of the first augmenting path containing and and the endpoints of the second augmenting path containing . Observe that for each augmenting path chosen in it holds that one endpoint is a leaf and the other endpoint is an ancestor of this leaf. Assume without loss of generality that and are the leaves and and are their respective ancestors. Let and ( and ) be the vertices deleted in which in turn made () free. Observe that does not contain any of these four vertices since before augmenting () the vertices and ( and ) are deleted. Since is contained in both paths, either is an ancestor of or vice versa: the case cannot happen since for the second augmenting path the endpoint would not be matched to ; a contradiction (see ).
We next consider the case that is an ancestor of (the other case will be handled subsequently). Denote with the neighbor of on . Observe that since is chosen as being closest to the root. We next distinguish the two cases whether or not is initially matched. If is initially free, then is matched after augmenting along . Then, by choice of , is not changed until the augmentation along . This, however, is a contradiction since augmenting along only happens after the matched edge is deleted. Since and is matched all the time until and are deleted, this means that would be matched to two vertices. Thus, consider the case that is initially matched. Then, after augmenting along , is free and is matched to its parent . As a consequence, is not matched to , neither before nor after the augmentation of . Since the algorithm augments along only after it deleted and where is matched to , it follows that the edge is augmented before the algorithm augments along . Denote with the augmenting path containing the edge . Since is apparently not a free leaf, it follows that needs to contain the matched neighbor of , which is . This means that the edge is augmented at least twice (through and ). However, is closer to the root than , a contradiction to the choice of . This completes the case that is an ancestor of .
We now consider the remaining case where is an ancestor of . In this case we have where is the neighbor of on . Observe that is a child of . Furthermore, observe that after the augmentation along the leaf is free and can be reached by an alternating path from . Hence, before and after the augmentation along it holds that can reach exactly one free leaf via an alternating path ( and ). Observe that this is true even if the algorithm removes since then a new free leaf will be created. Thus, before deleting and (right before the augmentation along ), there is an augmenting path in from to and to the free leaf reachable from . This is a contradiction to the fact that the matching is maximum.
We conclude that each edge in will be augmented at most once. Thus, the algorithm considers each edge at most twice (when augmenting it and when deleting it). Hence, the algorithm runs in linear time.
Summarizing, in we apply in order to obtain an instance with at most free vertices in that are all leaves. By this can be done in linear time. Furthermore, also shows that the step is correct.
In this step we reduce the graph in time so that at most bottommost leaves will remain in the forest . We will restrict ourselves to consider leaves that are matched with their parent vertex in and that do not have a sibling. We call these bottommost leaves interesting. Any sibling of a bottommost leaf is by definition also a leaf. Thus, at most one of these leaves (the bottommost leaf or its siblings) is matched with respect to and all other leaves are free. Recall that in the previous step we upper-bounded the number of free leaves with respect to by . Hence there are at most bottommost leaves that are not interesting. Hence there are at most bottommost leaves with siblings.
Our general strategy for this step is to extend the idea behind : We want to keep for each pair of vertices at most different internally vertex-disjoint augmenting paths from to . (For ease of notation we keep paths although keeping is sufficient.) In this step, we only consider augmenting paths of the form where is a bottommost leaf and is ’s parent in . Assume that the parent of is adjacent to some vertex . Observe that in this case any augmenting path starting with the two vertices and has to continue to and end in a neighbor of . Thus, the edge can be only used in augmenting paths of length three. Furthermore, all these length-three augmenting paths are clearly internally vertex-disjoint. If we do not need the edge because we kept augmenting paths from already, then we can delete . Furthermore, if we deleted the last edge from to (or had no neighbors in in the beginning), then is a degree-two vertex in and can be removed by applying . As the child of is a leaf in , it follows that has at most neighbors in . Thus, an application of to remove takes time.
Counting for each pair and one augmenting path gives in a simple worst-case analysis time per edge, which is too slow for our purposes. Instead, we count for each vertex and for each set one augmenting path. In this way, we know that for each there is one augmenting path from to , without iterating through all . We get an exponential factor in the bound of the bottommost leaves since there are subsets of , but we can perform this step in time as stated in the next lemma. We show below () that the application of to remove takes time. As we remove at most vertices, at most time is spent on in this step.
We now show that after a simple preprocessing one application of in the algorithm above can indeed be performed in time.
The preprocessing is to simply create a partial adjacency matrix for with the vertices in in one dimension and in the other dimension. This adjacency matrix has size and can clearly be computed in time.
Now apply to . Deleting takes constant time. To merge and iterate over all neighbors of . If a neighbor of is already a neighbor of , then decrease the degree of by one, otherwise add to the neighborhood of . Then, relabel to be the new merged vertex .
Since is a leaf in and its only neighbor in , namely , is deleted, it follows that all remaining neighbors of are in . Thus, using the above adjacency matrix, one can check in constant time whether is a neighbor of . Hence, the above algorithm runs in time.
The above ideas are used in which we use for this step ().
The algorithm is explained in the proof of the following lemma stating the correctness and the running time of .
We start with describing the basic idea of the algorithm. To this end, let be an edge such that is an interesting bottommost leaf, that is, without siblings and matched to its parent by . Counting for each pair and one augmenting path gives in a simple worst-case analysis time per edge, which is too slow for our purposes. Instead, we count for each pair consisting of a vertex and a set one augmenting path. In this way, we know that for each there is one augmenting path from to without iterating through all . This comes at the price of considering up to such pairs. However, we will show that we can do the computations in time per considered edge in . The main reason for this improved running time is a simple preprocessing that allows for a bottommost vertex to determine in constant time.
The preprocessing is as follows (see ): First, fix an arbitrary bijection between the set of all subsets of to the numbers . This can be done for example by representing a set by a length- binary string (a number) where the position is 1 if and only if . Given a set such a number can be computed in time in a straightforward way. Thus, can be performed in time. Furthermore, since we assume that (otherwise the input instance is already an exponential kernel), we have that for each . Thus, reading and comparing these numbers can be done in constant time. Furthermore, in the algorithm precomputes for each vertex the number corresponding to its neighborhood in .
After the preprocessing, the algorithm uses a table where it counts an augmenting path from a vertex to a set whenever a bottommost leaf has exactly as neighborhood in and the parent of is adjacent to (see ). To do this in time, the algorithm proceeds as follows: First, it computes in the set which contains all parents of interesting bottommost leaves. Clearly, this can be done in linear time. Next, the algorithm processes the vertices in . Observe that further vertices might be added to (see ) during this processing. Let be the currently processed vertex of , let be its child vertex, and let be the neighborhood of in . For each neighbor , the algorithm checks whether there are already augmenting paths between and with a table lookup in (see ). If not, then the table entry is incremented by one (see ) since and provide another augmenting path. If yes, then the edge is deleted in (we show below that this does not change the maximum matching size). If has degree two after processing all neighbors of in , then by applying , we can remove and contract its two neighbors and . It follows from that this application of can be done in time. Hence, runs in time.
Recall that all vertices in that are free wrt. are leaves. Thus, the changes to by applying in are as follows: First, the edge is removed and second the edge is replaced by for some . Hence, the matching after running has still at most free vertices and all of them are leaves.
It remains to prove that (a) the deletion of the edge in results in an equivalent instance and (b) that the resulting instance has at most bottommost leaves. First, we show (a). To this end, assume towards a contradiction that the new graph has a smaller maximum matching than (clearly, cannot have a larger maximum matching). Thus, any maximum matching for has to contain the edge . This implies that the child of in is matched in with one of its neighbors (except ): If is free wrt. , then deleting from and adding yields another maximum matching not containing , a contradiction. Recall that where since is a leaf in . Thus, each maximum matching for contains for some the edge . Observe that deletes only if there are at least other interesting bottommost leaves in such that their respective parent is adjacent to and (see ). Since , it follows by the pigeon hole principle that at least one of these vertices, say , is not matched to any vertex in . Thus, since is an interesting bottommost leaf, it is matched to its only remaining neighbor: its parent in . This implies that there is another maximum matching
a contradiction to the assumption that all maximum matchings for have to contain .
We next show (b) that the resulting instance has at most bottommost leaves. To this end, recall that there are at most bottommost leaves that are not interesting (see discussion at the beginning of this subsection). Hence, it remains to upper-bound the number of interesting bottommost leaves. Observe that each parent of an interesting bottommost leaf has to be adjacent to a vertex in since otherwise would have been deleted in . Furthermore, after running , each vertex is adjacent to at most parents of interesting bottommost leaves (see ). Thus, the number of interesting bottommost leaves is at most . Therefore the number of bottommost leaves is upper-bounded by .
In this subsection, we provide the final step of our kernelization algorithm. Recall that in the previous steps we have upper-bounded the number of bottommost leaves in by , we computed a maximum matching for such that at most vertices are free wrt. and all free vertices are leaves in . Using this, we next show how to reduce to a graph of size . To this end we need some further notation. A leaf in that is not bottommost is called a pendant. We define to be the pendant-free tree (forest) of , that is, the tree (forest) obtained from by removing all pendants. The next observation shows that is not much larger than . This allows us to restrict ourselves in the following on giving an upper bound on the size of .
Observe that is the union of all pendants in and . Thus, it suffices to show that contains at most pendants. To this end, recall that we have a maximum matching for with at most free leaves. Thus, there are at most leaves in that have a sibling which is also a leaf since from two leaves with the same parent at most one can be matched. Hence, all but at most pendants in have pairwise different parent vertices. Since all these parent vertices are in , it follows that the number of pendants in is .
We use the following observation to provide an upper bound on the number of leaves of .
First observe that each bottommost leaf of is a leaf of since we only remove vertices to obtain from . Thus, it remains to show that each leaf in is a bottommost leaf in .
We distinguish two cases of whether or not is a leaf in : First, assume that is not a leaf in . Thus, all of it child vertices have been removed. Since we only remove pendants to obtain from and since each pendant is a leaf, it follows that is in the parent of one or more leaves . Thus, by definition, all these leaves are bottommost leaves, a contradiction to the fact that they were deleted when creating .
Second, assume that is a leaf in . If is a bottommost leaf, then we are done. Thus, assume that is not a bottommost leaf and therefore a pendant. However, since we remove all pendants to obtain from , it follows that is not contained in , a contradiction.
From it follows that the set of bottommost leaves in is exactly the set of leaves in . In the previous step we reduced the graph such that . Thus, has at most vertices of degree one and, since is a tree (a forest), also has at most vertices of degree at least three. Let be the vertices of degree two in and let be the remaining vertices in . From the above it follows that . Hence, it remains to bound the size of . To this end, we will upper-bound the degree of each vertex in by and then use . We will check for each edge with and whether we “need” it. This check will use the idea from the previous subsection where each vertex in needs to reach each subset at most times via an augmenting path. Similarly as in the previous section, we want to keep “enough” of these augmenting paths. However, this time the augmenting paths might be long, while different augmenting paths might overlap. To still use the basic approach, we use the following lemma stating that we can still somehow replace augmenting paths.
Label the vertices in alternating as odd or even with respect to so that no two consecutive vertices have the same label, is odd, and is even. Analogously, label the vertices in , , and as odd and even with respect to , , and respectively so that is always odd. Since all these paths are augmenting, it follows that each edge from an even vertex to its succeeding odd vertex is in the matching and each edge from an odd vertex to its succeeding even vertex is not in the matching. Observe that intersects each of the other paths at least at two consecutive vertices, since every second edge must be an edge in . Since is a forest and all vertices in are free with respect to , it follows that the intersection of two augmenting paths is connected and thus a path. Since intersects the three augmenting paths from , it follows that at least two of these paths, say and , have a “fitting parity”, that is, in the intersections of with and with the even vertices with respect to are either even or odd with respect to both and .
Assume without loss of generality that in the intersections of the paths the vertices have the same label with respect to the three paths (if the labels differ, then revert the ordering of the vertices in , that is, exchange the names of and and change all labels on to its opposite). Denote with and the first and the last vertex in the intersection of and . Analogously, denote with and the first and the last vertex in the intersection of and . Assume without loss of generality that intersects first with and then with . Observe that and are even vertices and and are odd vertices since the intersections have to start and end with edges in (see for an illustration).
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For an arbitrary path and for two arbitrary vertices of , denote by the subpath of from to . Observe that and are two vertex-disjoint augmenting paths.
We now provide the algorithm for (see for a pseudocode). The algorithm uses the same preprocessing (see ) as . Thus, the algorithm can determine whether two vertices have the same neighborhood in in constant time. As in , uses a table which has an entry for each vertex and each set . The table is filled in such a way that the algorithm detected for each at least internally vertex-disjoint augmenting paths from to .
The main part of the algorithm is the boolean function ‘Keep-Edge’ in which makes the decision on whether or not to delete an edge for and . The function works as follows for edge : Starting at the graph will be explored along possible augmenting paths until a “reason” for keeping the edge is found or further exploration is possible.
If the vertex is free wrt. , then is an augmenting path and we keep (see ). Observe that in we upper-bounded the number of free vertices by and all these vertices are leaves. Thus, we keep a bounded number of edges incident to because the corresponding augmenting paths can end at a free leaf. We provide the exact bound below when discussing the size of the graph returned by . In , the algorithm stops exploring the graph and keeps the edge if has degree at least three in . The reason is to keep the graph exploration simple by following only paths in . This ensures that the running time for exploring the graph from does not exceed . Since the number of vertices in with degree at least three is bounded (see discussion after ), it follows that only a bounded number of such edges are kept.
If is not free wrt. , then it is matched with some vertex . If is adjacent to some leaf in that is free wrt. , then the path is an augmenting path. Thus, the algorithm keeps in this case the edge , see . Again, since the number of free leaves is bounded, only a bounded number of edges incident to will be kept. If has degree at least three in , then the algorithm stops the graph exploration here and keeps the edge , see . Again, this is to keep the running time at overall.
Let denote the neighborhood of in . Thus the partial augmenting path can be extended to each vertex in . Thus, if the algorithm did not yet find paths from to vertices whose neighborhood in is also , then the table entry (where encodes the set ) is increased by one and the edge will be kept (see ). (Here we need paths since these paths might be long and intersect with many other augmenting paths, see proof of for the details of why is enough.) (The proof that paths suffice is based on an exchange argument using .) If the algorithm already found “augmenting paths” from to , then the neighborhood of in is irrelevant for and the algorithm continues.
In , all above discussed cases to keep the edge do not apply and the algorithm extends the partial augmenting part by considering the neighbors of except . Since the algorithm dealt with possible extensions to vertices in in and with extensions to free vertices in in , it follows that the next vertex on this path has to be a vertex that is matched wrt. . Furthermore, since we want to extend a partial augmenting path from , we require that is not adjacent to as otherwise would be another, shorter partial augmenting path from to and we do not need the currently stored partial augmenting path.
Statements on .
For each edge with and we denote by the induced subgraph of on the vertices that are explored in the function Keep-Edge when called in with and . More precisely, we initialize . Whenever the algorithm reaches , we add to . Furthermore, whenever the algorithm reaches , we add to .
We next show that is a path or a path with one additional pendant.
We first show that all vertices in except and its neighbor have degree at most two in . Observe that having more vertices than and in requires to reach .
Let be the currently last vertex when continues the graph exploration in . Observe that the algorithm therefore dealt with the case that has degree at least three in in . Thus, is either a pendant leaf in or (that is, has degree at most two in ). In the first case, there is no candidate to continue and the graph exploration stops. In the second case, has degree at most two in .
We next show that any candidate for continuing the graph exploration in is not a leaf in . Assume toward a contradiction, that is a leaf in . Since the parent of is matched with some vertex (this is how is chosen, see ), it follows that is not matched. This implies that the function ’Keep-Edge’ would have returned true in and would not have reached , a contradiction. Thus, the graph exploration follows only vertices in . Furthermore, the above argumentation implies that is not adjacent to a leaf unless this leaf is its predecessor in the graph exploration.
We now have two cases: Either is not adjacent to a leaf in or is a leaf and is its matched neighbor. In the first case, has at most one neighbor since . Hence, has degree two in . In the second case, has at most two neighbors and . Thus, has degree at most three.
We set for
to be the union of all induced subgraphs wrt. .
Since is a tree (or forest), is also bipartite. Let and be its two color classes (so ). We define the two parts and as follows: A subgraph is in if the neighbor of in is contained in , otherwise is in .
We show that all subgraphs in and are pairwise vertex-disjoint. To this end, assume toward a contradiction that two graphs share some vertex. (The case is completely analogous.) Let and be the first vertex in and respectively, that is, and are adjacent to in . Observe that . Let be the first vertex that is in and in . By , and are paths or trees with at most one vertex of degree more than two and this vertex has degree three and is the neighbor of or , respectively. This implies together with that either or . Assume without loss of generality that . Since and and is a vertex in , it follows that followed in the graph exploration from in . However, this is a contradiction since the algorithm checks in whether the new vertex in the path is not adjacent to . Thus, all subgraphs in and are pairwise vertex-disjoint.
We next show that if for some and (recall that maps to a number, see ), then there exist at least internally vertex-disjoint augmenting paths from to .
Note that each time is increased by one (see ), the algorithm found a vertex such that there is an alternating path from to and . Furthermore, since the function Keep-Edge returns true in this case, the edge from to its neighbor on is not deleted in . Thus, there exist at least alternating paths from to vertices whose neighborhood in is exactly . By , it follows that at least half of these paths are vertex-disjoint.
The next lemma shows that is correct and runs in time.
We split the proof into three claims, one for the correctness of the algorithm, one for the returned kernel size, and one for the running time.
Proof. Observe that the algorithm changes the input graph only in two lines: . By , applying yields an equivalent instance. Thus, it remains to show that deleting the edges in is correct, that is, it does not change the size of a maximum matching. To this end, observe that deleting edges does not increase the size of a maximum matching. Thus, we need to show that the size of the maximum matching does not decrease. Assume toward a contradiction that it does.
Let be the edge whose deletion decreased the maximum matching size. Redefine to be the graph before the deletion of and to be the graph after the deletion of . Recall that gets as additional input a maximum matching for . Let be a maximum matching for with the largest possible overlap with and let (cf. ). Since and is free wrt. it follows that there is a path in with one endpoint being .
Recall that since is a path in it follows that is an augmenting path for . Since all vertices in are free wrt. , it follows that all vertices in except the endpoints are in . Let be the second endpoint of this path . We call a vertex on an even (odd) vertex if it has an even (odd) distance to on . (So is an even vertex and and are odd vertices). Observe that is the only odd vertex in adjacent to since otherwise there would be another augmenting path from to which only uses vertices from implying the existence of another maximum matching that does not use , a contradiction. Let be the neighbor of in .
Since no odd vertex on except is adjacent to , it follows that the graph exploration in the function Keep-Edge starting from and in either reached or returned true before. If , then in both cases, the function Keep-Edge would have returned true in and would not have deleted , a contradiction. Thus, assume that . Therefore, the function Keep-Edge considered the vertex in but did not keep the edge . Thus, when considering , it holds that , where encodes and .
By , it follows that there are pairwise vertex-disjoint (except ) alternating paths from to vertices with . Thus, there are internally vertex-disjoint paths from to in . If one of the paths does not intersect any path in , then reverting the augmentation along and augmenting along would results in another maximum matching not containing , a contradiction. Thus, assume that each path in intersects at least one path in .
For each two paths that are intersected by the same path in it holds that each further path in can intersect at most one of and : Assume toward a contradiction that does. Since no path in except contains and it follows that all intersections between the paths are within . Since and are vertex-disjoint and and are internally vertex-disjoint, it follows that there is a cycle in , a contradiction to the fact that is a feedback vertex set.
Since , it follows from the pigeon hole principle that there is a path that intersects at least three paths such that no further path in intersects them. We can now apply and obtain two vertex-disjoint augmenting paths and . Thus, reverting the augmentation along and and augment along and yields another maximum matching for which does not contain , a contradiction.
Proof. We first show that each vertex has degree at most in . To this end, we need to count the number of neighbors where the function Keep-Edge returns true in . By , the function Keep-Edge explores the graph along one or two paths (essentially growing from one starting point into two directions). Recall that denotes the subgraphs induced by the graph exploration of Keep-Edge for the neighbors of . By there is a partition of into and such that within each part the subgraphs are pairwise vertex-disjoint. We consider the two parts independently. We start with bounding the number of graphs in where the function ’Keep-Edge’ returned true (the analysis is completely analog for ).
Since all explored subgraphs are disjoint and all free vertices in wrt. are leaves, it follows that returned at most times true in due to being adjacent to a free leaf in . Also, the algorithm returns at most times true in due to being free. Furthmore, the algorithm returns at most times true in . Finally, we show that the algorithm returns at most times true in , respectively. It follows from the discussion below that , the pendent-free tree of , has at most leaves (denoted by ) and vertices of degree at least three (denoted by ). Let be the vertices of . Since is a tree (or forest), it has more vertices than edges and hence
Thus, returns at most times true in due to being a vertex in . Also, returns at most times true in due to being a vertex in .
Summarizing, considering the graph explorations in , returned at most