Token Sliding Reconfiguration on DAGs
Abstract
Given a graph and two independent sets of same size, the Independent Set Reconfiguration Problem under token sliding asks whether one can, in a step by step manner, transform the first independent set into the second one. In each step we must preserve the condition of independence. Further, referring to solution vertices as tokens, we are only permitted to slide a token along an edge. Until the recent work of Ito et al. [Ito et al. MFCS 2022] this problem was only considered on undirected graphs. In this work, we study reconfiguration under token sliding focusing on DAGs.
We present a complete dichotomy of intractability in regard to the depth of the DAG, by proving that this problem is -complete for DAGs of depth 3 and -hard for depth 4 when parameterized by the number of tokens , and that these bounds are tight. Further, we prove that it is fixed parameter tractable on DAGs parameterized by the combination of treewidth and . We show that this result applies to undirected graphs, when the number of times a token can visit a vertex is restricted.
Keywords and phrases:
Graph theory, FPT algorithms, Reconfiguration, Independent SetsCopyright and License:
2012 ACM Subject Classification:
Mathematics of computing Graph algorithms ; Theory of computation Fixed parameter tractabilityFunding:
This work benefited from state aid managed by the ANR under France 2030 referenced ANR-23-IACL-0006.Editors:
Paweł Gawrychowski, Filip Mazowiecki, and Michał SkrzypczakSeries and Publisher:
Leibniz International Proceedings in Informatics, Schloss Dagstuhl – Leibniz-Zentrum für Informatik
1 Introduction
A reconfiguration problem asks, given two solutions to a combinatorial problem, whether one can change the first solution in a step-by-step manner in order to reach the second one; while ensuring that the transformed set remains a solution throughout the reconfiguration procedure. The research interest in these topics dates back to the 19th century, as many one player games can be considered in these terms [12]. For a more recent introduction to the topic, we refer to [16].
The complexity of reconfiguration problems is often -complete. Specific attention has been given to independent set and (connected) dominating set reconfiguration. For these problems, there are two main reconfiguration paradigms: token sliding, and token jumping. In both paradigms, we view the set to be reconfigured as several tokens placed on the vertices of the graph. Token jumping allows, in each step to move one token to be placed to any other vertex of the graph, while token sliding only allows to move a token to an adjacent vertex (to slide along an edge of the graph).
For both paradigms, independent set reconfiguration is -complete, and remains intractable when parameterized by the size of the set to be reconfigured, or the length of reconfiguration sequence [4, 14]. In order to find efficient algorithms, it is therefore interesting to look at structural restrictions on the graphs over which the sets are being reconfigured. While these problems remain -hard on planar graphs and graphs of bounded bandwidth, they become polynomial on cographs, forest and interval graphs [10, 13, 18]. We however note that the two paradigms differ in complexity over bipartite graphs, where independent set reconfiguration for token sliding (ISR-TS) is -complete, independent set reconfiguration for token jumping (ISR-TJ) is only -complete [13].
The two paradigms differ even more when we look at parameterized complexity. For token jumping, we know that reconfiguration for independent set over general sparse graphs classes, such as bounded degeneracy and nowhere dense, becomes when parameterized by the size of the sets [6, 13, 17]. These classes are generalizations of many well known classes such as planar graph, graph of bounded treewidth, classes excluding a (topological) minor etc.
For token sliding however, the parameterized complexity of independent set reconfiguration is much less understood. On one hand it is known to be on trees [8], graphs of bounded degree [3], and graphs of girth 5 [2], but hard on split graphs [5]. On the other hand, it is still open whether it is tractable on graph of bounded treewidth, or minor-free graphs [6, Question 5]. We refer the readers to the survey [6] for an extensive overview.
We also mention that for both TS and TJ the reconfiguration of every problem that is expressible in is when parameterized by treedepth and the number of tokens [9, Theorem 4.1] and when parameterized by treewidth, the length of the reconfiguration sequence, and the formula describing the considered problem [15, Theorem 7]
Directed graphs.
In this paper, we consider directed graphs and allow tokens to only slide in the direction of the edges. Note, that token jumping is not well suited for directed graphs, as arbitrary jumps stay the same in the directed setting. So far very little is known for directed graph’s reconfiguration. Recently, Ito et al. [11, Theorem 2] proved that Independent Set Reconfiguration for Directed Token Sliding (ISR-DTS) is -hard on DAGs and [1]-hard when parameterized by the number of tokens. On the tractable side of things, they show that it is linear time solvable on orientation of trees [11, Theorem 3]. The hardness extends to -completeness for oriented planar graphs, split graphs, bipartite graphs and most noticeably bounded treewidth graphs [1].
We improve on both the side of hardness, as well as the side of tractability. We first look at the depth of the DAGs and provide sharp dichotomy in both the classical and parameterized setting. In the following statements, refers to the number of tokens.
Theorem 1.1.
ISR-DTS is polynomial on DAGs of depth 2 and for DAGs of depth 3 when parameterized by .
Theorem 1.2.
ISR-DTS is -complete on DAGs of depth 3.
Theorem 1.3.
ISR-DTS is -hard on DAGs of depth 4, when parameterized by .
Additionally, we show that a combination of the parameters and treewidth leads to tractability in DAGs. We do so by adapting the recently developed framework of galactic reconfiguration [3] to directed graphs. The following result can be considered our main contribution.
Theorem 1.4.
ISR-DTS is on DAGs when parameterized by .
Finally, we go back to undirected graph, introduce a new parameter (the iteration ) that measures the maximum number of times a specific token can enter a specific vertex.
Theorem 1.5.
ISR-TS is when parameterized by .
These results for graphs of bounded treewidth is a step toward solving one of the most prominent open problem in this line of work (see [6, Question 5]): is ISR-TS when parameterized by and the treewidth of the graph?
The previous best algorithms only work on the more restricted class of bounded treedepth [9], or require to also parameterize by the length of the reconfiguration sequence [15]. In comparison, our new parameter iteration is much smaller than the length of the reconfiguration sequence. In terms of techniques, parameterizing by enabled the authors of [15] to reduce to the model checking problem for formula which is solved using Courcelle’s Theorem. In our case, we do not believe that we can reduce to Courcelle’s theorem. Thus, the presented algorithm is the first to use dynamic programming on graph of bounded treewidth “by hand” for Independent Set Reconfiguration.
Organization.
The rest of the paper is structured as follows. Section 2 provides the required background and fixes notations. Section 3 is devoted to efficient algorithms and the proof of Theorem 1.1. The proof of Theorem 1.2 is provided in Section 4, followed by the proof of Theorem 1.3 in Section 5. In Section 6 we introduce additional notions around galactic reconfiguration for our main algorithm in order to prove Theorem 1.4. Finally, in Section 7 we discuss implications for undirected graphs, and Theorem 1.5.
2 Preliminaries
Graphs.
We use standard graph notations. All graphs are directed and loop-free unless stated otherwise. Unless preferable for readability we abbreviate an edge with . We denote the open out-neighborhood of a vertex with and the open in-neighborhood with . A graph without cycles is called a DAG. The depth of a DAG is the number of vertices on its longest directed path. Note that if a DAG has depth , then its vertices can be grouped in levels with every edge going from a vertex at some level to a vertex at a greater level. One can inductively define the first layer as the vertices of in-degree 0, and the th layer as the vertices whose entire in-neighborhood is in smaller levels.
For a directed graph , the underling undirected graph is obtained by disregarding the edge direction and removing all double edges. That is the graph . A set of vertices of a directed or undirected graph is called independent if the vertices are pairwise nonadjacent.
Parameterized Complexity.
In the parameterized framework, a problem is said to be fixed parameter tractable (), if there exists an algorithm that solves it in time for a computable function and constant . An alternative characterization of is through the existence of a kernel, that is polynomial time computable function that maps each instance to an instance of size for some computable function . The returned instance should be a positive-instance exactly if the inputted instance is. Kernels are commonly presented in the form of data reduction rules, that provide instructions on how to shrink the size of the input.
Under commonly made assumptions, problems that are -hard are considered not to be . Similar to -completeness, we use -reductions to transfer hardness. However, these reductions may use a longer running time but are more restricted in the sense that the parameter of the resulting instance may not depend on the size but only on the parameter of the inputted instance. More background on algorithms and -reduction can be found in the book [7].
Reconfiguration.
A configuration is a function from a set of tokens to the set of vertices .
For two independent sets and of size in a graph , a reconfiguration-sequence for ISR-DTS is a sequence of configurations such that:
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1.
and .
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2.
For every , and for distinct tokens and : .
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3.
For every , and for distinct tokens and : .
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4.
If , then there exists an edge and for every other token it holds that .
We generally assume that is a sequence from 1 to . For a configuration we call all configurations succeeding or a successor, if . If we say that it is a direct successor.
Let be a token and a sequence with and . We say that the token comes from and ends up in . Similarly we say that a token comes from a subgraph and ends up in a subgraph if and . Additionally, if there is a token on we say that this token blocks all .
Galactic Graphs.
Galactic graphs were recently developed to prove that ISR-TS is in for undirected graphs of bounded degree [3]. We restate their definition, slightly adapted for directed graphs. A galactic graph is a graph where the vertex set is partitioned into planets and black holes .
For a (galactic) graph and a subsets of its vertices we say that it collapses to if with and contains an edge (respectively ) if contains an edge (respectively ) for some . We denote this graph with . For the empty set, we define . Intuitively, this operation takes a set of vertices (planets and black holes) and merges them into one black hole.
Note, that this operation takes a graph on the left-hand site and a vertex set on the right-hand site. Thus, multiple applications of this operator are left-associative.
We will assume that the black holes are named (or labeled) in such a way that they are uniquely identified by the planets that collapsed into it in the original graph. Meaning that for example for disjoint it holds , where and are the black holes created when collapsing and respectively. This can be realized by representing planets as sets containing a single element and black holes as the union of the sets representing vertices (planets or black holes) with an additional marker for black holes added.
Galactic Reconfiguration.
For two multisets and of size in a graph , a galactic reconfiguration sequence for ISR-DTS is a sequence of configurations . Where is a function from a set of tokens to , such that:
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1.
and .
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2.
For distinct tokens and : or .
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3.
All are independent sets.
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4.
If , then there exists an edge and for every other token it holds .
Note that in a galactic reconfiguration sequence, several tokens are allowed to be on the same vertex (including for the first and last configuration), as long as this vertex is a black hole. When we collapse an instance for ISR-DTS and the collapsed vertex set contains start and end positions, we generally assume that these positions in their multiplicity get assigned to the created black hole.
Definition 2.1.
Given a graph , a set of vertices , we say that a reconfiguration sequence , collapses to over if can be obtained by first creating an intermediate reconfiguration sequence where every is mapping to (the created black hole) every token that maps to a vertex of but is otherwise identical. And then removing exhaustively all configurations with an identical direct successor from .
Moreover, if a sequence on a graph , collapses to a sequence on a graph , which itself collapses to a sequence on a graph , we also say that collapses to on .
Note that in the above definition, is a valid reconfiguration sequence, i.e. the tokens only slide along edges and are only on neighboring vertices if at least one of them is a black hole. Furthermore, form this definition, it might look like a single reconfiguration sequence can collapse to multiple sequences on the same graph (depending on the chain of collapses). We later show in Lemma 6.1 that it is not the case, and the collapse notion (with several successive collapses) is therefore well-defined.
Tree-decomposition.
A tree decomposition of a undirected graph is a pair where is a rooted tree and is a mapping that assigns to each node its bag such that the following conditions are satisfied:
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For every vertex , the set of nodes satisfying induces a connected subtree of .
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For every edge , there exists a node such that .
We may assume that every internal node of has exactly two descendants. See [7] for more details on tree decompositions in general. Further, let be a tree decomposition of a graph . The adhesion of a node is defined as and the margin of a node is defined as . The cone at a node is defined as and the component at a node is defined as . Here, means that is a descendant of in . The width of a tree decomposition is the maximum number of vertices in each bag (minus one). The treewidth of a graph , noted , is the minimum width among all possible tree decompositions of . For directed graphs, we define these terms over their underlying undirected graphs.
3 Tractability Results for DAGs of Low Depth
This section is devoted to the proof of Theorem 1.1 that we restate here. See 1.1
We prove the first part of the theorem.
Lemma 3.1.
For DAGs of depth 2 ISR-DTS is polynomial solvable.
Proof.
We are given a DAG and two independent sets and both of size . First, we can assume that is the vertex set of the entire graph, as for every other vertex no token can start in , go to and then end in , contradicting that is a DAG of depth 2.
If is a positive instance, there needs to be at least one vertex in that has just a single neighbor in , as otherwise no token can move. We can therefore start the reconfiguration sequence by moving the token from to . We can now remove from and proceed inductively in the same way. If at any point no token can move, we report that the input is a negative instance to ISR-DTS.
After this small warm up proof, let us now move to the parameterized tractability result. Let us first show, that we can assume the input to have a uniform structure.
Lemma 3.2.
For any instance if is a DAG of depth 3 we can in polynomial time find an equivalent instance such that is equal to the first layer of and to the third and where is also a DAG of depth 3.
Proof.
First note, that if there is a vertex from at the third layer, it can reach nothing, and thus we can return a trivial no instance. While a vertex in in the third layer cannot move, blocking its in-neighborhood, so we can safely remove this vertex together with it’s in-neighborhood. By analogous arguments the same holds if there is a vertex from at the first layer. Additionally, we can remove all vertices without predecessors in (as they cannot be reached) and all vertices with no successors in (as a token here cannot reach a destination).
Observe, that after this a vertex from may not lie on the second layer. If it were so, it would have a predecessor, which then would lie in the first layer. Thus, it would be in which means that is not an independent set.
Furthermore, if a vertex is in the second layer, it may not have successors. Thus, we can add a new vertex make it a successor to all predecessors of as well as itself. Then, while keeping in the graph, we remove from and add instead.
Note that all of these basic checks as well as basic graph manipulations can be carried out in polynomial time.
We show that the following reduction rule, when applied exhaustively, yields a kernel:
Rule 3.3.
If a DAG of depth 3 has two vertices with then remove .
Proof of Correctness.
Let be the graph obtained from applying the reduction rule on . Clearly if there is a reconfiguration sequence in one can use the same sequence in . Thus, we only have to show, that if there is a reconfiguration sequence in there is one in .
Let be the reconfiguration. We show that every time places a token on we can place it on instead. As and have the same neighborhood, the independent set condition is not violated and neither is the requirement of moving tokens only along edges. That is, if there is no token at at that same step. However, a token on , blocking all of , would prohibit the placement of a token on .
Lemma 3.4.
ISR-DTS admits a kernal on DAGs of depth 3 when parameterized by .
Proof.
We first use Lemma 3.2 to obtain an instance, where the first layer is equal to and the third layer to . Then we apply ˜3.3. Note, that now every vertex on the second layer has a distinct neighborhood which is a subset of . As there are at most unique subsets, the total number of vertices is bound by .
Together Lemma 3.4 and Lemma 3.1 complete the proof of Theorem 1.1.
4 NP completeness for DAGs of depth 3
In this section we turn to hardness results, improving the lower bounds of [11]. While they prove that ISR-DTS is -hard on DAGs, their reduction yields DAGs of unbounded depth.
Theorem 1.2. [Restated, see original statement.]
ISR-DTS is -complete on DAGs of depth 3.
The proof of Theorem 1.2 is obtained by reducing from 3-SAT. For a 3-SAT sentence we construct the following instance :
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1.
Add a timing gadget, meaning two vertices and as well as an edge .
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2.
For each variable create four vertices , , and as well as edges , , and . Further, connect this subgraph to the timing gadget with the edge .
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3.
For each clause with the literals add five vertices , and . Further, add edges form to and as well as from and to . Additionally, connect this subgraph to the timing gadget with the edge .
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4.
For every literal add two vertices and and edge . For all (note that ) and containing add edges and respectively.
This construction is represented in Figure 1
Lemma 4.1.
If is satisfiable then is a positive instance for ISR-DTS.
Proof.
Let be a satisfying assignment of . We define the following reconfiguration sequence. In no particular order move a token from to if True, to otherwise.
For all literals where it is possible without violating the independent set condition, slide along . I.e. this move is possible for every literal of the form (resp. ) where sets to True (resp. to False).
As each clause has at least one literal that evaluates to true. Move the token from to over . This move is possible as the token in the “literal gadget” moved from to . Now one is able to move the token from to . Next, move the tokens on all or to and finally from all remaining to .
Lemma 4.2.
If is a positive-instance of ISR-DTS then is satisfiable.
Proof.
We first show, that the flow of token is tightly controlled.
Claim 4.3.
The token coming from will only slide along the edge . Further, for all variables , clauses and literals the tokens from , and will end up in , and respectively.
Proof.
By reachability in the only possibility where this is not the case would be if some token coming from ends up in (for ). Implying the token from ends up in ; implying that the token from ends up in ; implying that the token from slides to . However, as blocks , blocks , blocks and locks in , so none of these moves is doable.
Let be a reconfiguration sequence certifying that is a positive-instance. The previous claim implies that the token in ends up in , so let be the first configuration where a token occupies . We define the following set .
Claim 4.4.
The set contains one literal from each clause. Further, if is in , then is not.
Proof.
By ˜4.3, for all clauses the token coming from is either at some , or is already on and has previously been in one of the . Thus, no token can be at its neighbor (on the literal gadget). By ˜4.3 there is now a token on .
For the second claim, assume towards a contradiction, that in tokens are placed at both and for some literal . This implies that neither for nor for any succeeding configuration a token is at or where is the variable of the literal . As no token is on there is one at and by ˜4.3 it cannot reach a valid end position.
We find the following assignment . Set True, if and otherwise False. Observe, that all literals in evaluate to true. As contains at least one literal per clause all clauses evaluate to true.
This concludes the proof of Theorem 1.2.
5 W[1]-hardness for DAGs of Depth 4
In this section, we prove the second lower bound of this paper. Here, is the number of tokens.
Theorem 1.3. [Restated, see original statement.]
ISR-DTS is -hard on DAGs of depth 4, when parameterized by .
We show this result by providing a -reduction from Independent Set. That is given an instance for independent set (with parameter ), we build an instance of ISR-DTS with parameter , such that the input instance is a positive instance if and only if the resulting one is. Section 5.1 describes the construction while Lemma 5.1 in Section 5.2 proves the correctness of the reduction.
5.1 Construction
The reduction takes as input an undirected graph of vertices and an integer and outputs the graph as described in the following. This graph is also depicted in Figure 2.
Vertices.
In total the DAG consists of 10 main blocs and three helping gadgets. Let us commence with the main blocs. Six of these consist of isolated vertices. Half of the blocs: , and contains only starting positions: to . The other half: , , and only contain target (or destination) vertices: to .
The remaining four components are each made up of copies of the vertex set of , i.e. isolated vertices. The components are named , , , , and, given , , we call (resp. , , ) the copy of in the th copy of in (resp. , , ).
Lastly, we have six specific vertices, controlling the flow of tokens in the graph: , , , , , , where are starting positions, and destinations.
Edges.
The edges of our construction are represented in Figure 2 and are as follows.
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1.
Full connections: For every and in , connect (resp. , ) to (resp. , ). Further, connect (resp. , ) to (resp. , ).
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2.
Matching connections: For every , and in , connect to . I.e. make a matching between and .
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3.
Co-matching connections: For every , and in if and only if , connect to and connect to . I.e. Make a co-matching between and and between and .
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4.
Neighborhood connections: For every , and in if and only if and , connect to .
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5.
Gadget connections: Connect and each vertex in , , and to . Connect to . Further, connect to , and to each vertex in and . Lastly, connect to and to each vertex in , and connect to each vertex in .
We define the start positions as and destination positions as .
We call our resulting DAG . Notice that has many vertices, and start / destination vertices. Furthermore, the DAG has depth 4 has illustrated in Figure 2.
5.2 Correctness
This section is devoted to the proof of the correctness of the reduction, as stated in the following lemma.
Lemma 5.1.
If the input is a positive instance for the independent set problem if and only if is a positive ISR-DTS instance.
We now prove the first implication. Assume that is an independent set of of size . We move the tokens in as follows.
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1.
For every , move the token from to . Similarly, move tokens from and to , and respectively.
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2.
Move the token from to and the token from to .
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3.
For every , move the token from to . And then from to .
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4.
Move the token from to .
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5.
For every , move the token from to and the tokens from to .
The first set of moves is possible because a token on blocks in every with , but is free. The same holds for . Second, the clocks can move to because , , and , are free of token, freeing . Third, is an independent set in , and therefore is not adjacent to any for . Therefore, is not adjacent to any for . Hence, the tokens can move from to and then to . Forth, as is empty of token, can move to , freeing . Last, the tokens in and can freely move to and respectively.
This shows that is a positive ISR-DTS instance, and conclude the first part of the proof of Lemma 5.1. The second implication in the statement of Lemma 5.1 is harder to prove. We show that any reconfiguration sequence must follow some structure, from which we derive the existence of an independent set in . In what follows, we assume that is a positive ISR-DTS instance.
Claim 5.2.
For any reconfiguration sequence, the tokens in , , and end up in , , and respectively. Furthermore, they move in this order.
Proof.
While other tokens can reach , the token on can only reach . Also, the token in (resp. ) is the only one that can reach (resp. ).
Additionally, notice that both and block all of . As the token reaching need to go through , if moves to before moves to no token can ever reach . So must reach before reaches . And since blocks , must slide to first.
Claim 5.3.
For any reconfiguration sequence, no token slides along the edge , if is in , or .
Proof.
As the tokes coming from the clocks stay within them, we may argue as if these vertices and their respective tokens were not present.
The vertices in can only be reached from , thus no token may leave a path from some vertex in to some vertex in which prohibits the use of edges between and or and . Consequently and similarly, the only tokens that can end up in are these that start in , prohibiting the use of the edges between and .
Claim 5.4.
There exists a set such that when moves to , there are tokens on vertices , , and for every .
Proof.
When moves to , already moved to , so , , and are empty of tokens. Furthermore, as and did not move yet, , , and therefore are empty of token. So the tokens in moved to . We then define as the vertex of such that moved to . Finally, note that blocks in every with . So the token from must now be in , as these tokens may not move to other blocs by ˜5.3. Similarly, the token from must now be in .
Claim 5.5.
is an independent set of size in .
Proof.
When goes to , all of is blocked, and remains so for the rest of the reconfiguration. So the vertices in are already occupied with tokens. These tokens must be from by ˜5.3. So before goes to each token in must go to (there is only one outgoing edge from ), and then to . During these moves, the tokens in remains there since is still blocked by . If can freely move to it implies that no token on is adjacent to which implies that no (with ) is adjacent to (or equals) in . As this holds for every , we have that is an independent set in . In conclusion, the existence of a reconfiguration sequence implies the existence of an independent set in . This concludes the proof of Lemma 5.1, and therefore of Theorem 1.3.
6 FPT when parameterized by Treewidth and k
Now that we have covered the impact of the depth of a DAG on the tractability of ISR-DTS, we look at other subfamily of DAGs that could enable algorithms. Recently, Ito et al.[11, Theorem 3] proved that ISR-DTS is polynomial time solvable on DAGs whose underlying graph is a tree. We extend this results by looking at graph of bounded treewidth.
Theorem 1.4. [Restated, see original statement.]
ISR-DTS is on DAGs when parameterized by .
We first prove several results on galactic reconfiguration and the impact of collapses that we defined in Section 2. Then, we explain how to perform a dynamic algorithm on a tree decomposition.
6.1 Reconfiguration and Collapses
The first lemma shows that a reconfiguration defined on a graph only has one possible collapse on a collapse of the graph. Meaning that it does not matter in which sequence of collapse the new graph is obtained.
Lemma 6.1.
Let , , , and galactic graphs such that both and are obtained from through collapses, and is obtained from both and through collapses.
Let be reconfiguration sequences over respectively. If collapses to , collapses to and collapses to , then collapses to .
Proof.
Let be the sequence collapses to in . We show that .
Let be the sequence of graphs in the chain of collapses from over to . For each let denote the function that maps vertices (planets and black holes) from to vertices they correspond to. That is, to itself if they where not collapsed and to the created black hole otherwise.
We have to account for the deletion of consecutive equal configurations, when collapsing reconfiguration sequences. Therefore, let be the function that maps indices of the reconfiguration sequence that collapses to in to the corresponding indices in the reconfiguration sequence that collapses to in .
Let and define analogously. Further, define and analogously for the chain of collapses in from to and finally . By definition for each configuration and each token . Similarly, the same holds for with and .
As the resulting graph for both chains of collapses is the same, . Further, as equality classes of configurations only merge when collapsing, we may assume that consecutive equal configurations get resolved only at the end of a chain of collapses. Thus as also . Thus in conclusion, for all token and steps : .
Observe that these lemma can be iterated to obtain the same statement for arbitrary long chains of collapses, essentially proves that the collapsing is transitive.
The next lemma shows that on DAGs, reconfiguration are restricted.
Lemma 6.2.
Let be a reconfiguration sequence over a DAG without back holes. Then for every and every token we have that implies .
Furthermore, if is a collapse of , then for all and token such that and , we have that is a black hole.
Proof.
The first assertion is trivial. A token returning to a vertex would imply a round walk and thus a cycle. Assume that there exists corresponding indices with – contradicting the first statement.
This in turn bound the number of possible reconfiguration sequences that can be on a DAG or its collapses.
Lemma 6.3.
For a DAG without black holes and a graph obtained by collapses from where additionally does not contain neighboring black holes; the number and length of reconfiguration sequences over that are collapses of reconfiguration sequences over is bounded by .
Proof.
No token can return to a planet after leaving it, due to Lemma 6.2. A token might be at a black hole multiple times. However, as there are no repeating consecutive configurations and by assumption no neighboring black holes, a change for a planet has to happen in between consecutive configurations. This bounds the length of every sequence by .
As there are only possible configurations the total number of possible reconfiguration sequences is at most .
Our last lemma of this section is a composition lemma that explains how to glue together two reconfiguration sequences defined on distinct collapses of a given graph.
Lemma 6.4 (Composition lemma).
Let be a graph and disjoint sets of vertices such that has no neighbors in and vise versa. Let and be reconfiguration sequences, for and respectively, that collapse to the same sequence in . Then there exists a reconfiguration sequence in that collapses to and over and respectively.
Proof.
Let and defined as above. Let and the black holes in corresponding to the sets and respectively and let be the shared collapsed reconfiguration sequence in .
We build a configuration sequence over by concatenating sequences (for each ).
Broadly speaking, fix and the configuration , and look at the configurations in and that correspond to it. All these configurations in are equivalent when restricted to , therefore tokens only move in (and are all mapped to in ). Similarly, the configuration in corresponding to this only differ in . So in we can perform all these move independently before moving to the configuration .
Formally let the set of index of corresponding to , and define similarly. Let be the minimum index of minus one and let be the minimum index of . Let be the sequence of length such that for every integer and token , we have:
Note that the definition is consistent as for every and every as soon as is not one of the black hole or . Finally, define as the concatenation of every for .
To see that is indeed reconfiguration sequence, note that each time only one token moves (according to the move defined in or ). Furthermore, since and are disjoint and non-adjacent the set of tokens has to be an independent set (when only considering planets). Otherwise, this inconsistency already holds in or .
By construction, and (resp. ) agree on the positioning of tokens except for (resp. ). And therefore collapses to (resp. ) on (resp. ).
6.2 Dynamic Programming
In a nutshell, our algorithm will try every possible reconfiguration sequence for each bag. Starting from the leaf, the information of what are the possible reconfiguration sequences is computed by a bottom up manner. At every node the algorithm tries to glue the possible reconfiguration sequences from the left and from the right children, checking that they agree on the overlapping part. Thanks to Lemma 6.3 we only have to consider small sequences at the bag level, and the gluing can be done thanks to Lemma 6.4. We now turn to formal definitions. We consider a DAG , and it’s tree-decomposition as defined in Section 2.
Let be a node of the tree-decomposition and its children. We define the significant of noted as . Intuitively, everything that is “above” in the tree decomposition has been collapsed into one black hole. The right-significant, noted as and as the graph . I.e. where one (or both) subtrees of have collapsed into a black hole. Finally, we define as where (respectively ) is the black hole obtained from collapsing (resp ). Note that . I.e. it is only composed of element of the adhesion of with two black holes representing everything that is “above”, and “below” in the tree decomposition. Observe that can be collapsed to obtain all , and in a single step. All these notions are represented in Figure 3, and an example of reconfiguration is presented in Figure 4.
Definition 6.5.
Let be a DAG, together with a tree decomposition, let be a node of the tree decomposition. The profile of is the set of reconfiguration sequences over such that there is a reconfiguration over that collapses to over .
Thanks to Lemma 6.3, the profile of each node is bounded by , where tw is the width of the given tree decomposition. Ne now show that we can compute the profile of each node in a bottom up manner.
Proposition 6.6.
Given a tree decomposition of a DAG of width tw, a node of the decomposition, and the profile of its children and , we can compute the profile of in time , for some computable function .
Proof.
For the leaf of the tree-decomposition, it holds trivially, as we can compute all possible sequences in as it is of size at most tw. Thus, we may assume that is not a leaf. For every possible reconfiguration sequence over the algorithm adds it to the profile of if and only if the following holds.
There are reconfiguration sequences , , and over , , and respectively, such that :
-
1.
(resp. ) belongs to the profile of (resp. ), and
-
2.
collapses to (resp. , ) over (resp. , ).
Claim 6.7.
This checks can be performed in time .
Proof.
By Lemma 6.3, there can be only few possible sequences to consider for . We can then compute its collapses in and , and check that the results belongs to the profiles of and that have already been computed. All of this can therefore be computed in time .
Claim 6.8.
If there exists a reconfiguration sequence over that collapses to in then the algorithm adds to the profile of .
Proof.
We first collapse to a reconfiguration in the graph . By Lemma 6.1 we obtain that also collapses to over .
Now, we collapse to a sequence over , and both collapse to a sequence over . Since is in the significant of , is in the profile of . We define and similarly for .
Note that this triple (, , and ) must be considered by the algorithm. Finally, by Lemma 6.1, also collapses to and . So this triple of reconfiguration sequences passes the checks, and the algorithm selects to the profile of .
Claim 6.9.
If the algorithm adds to the profile of , then there exists a reconfiguration sequence in that collapses to in
Proof.
Let be the guessed sequence in and be the sequences that collapses to in , respectively, where and are the left and right children of . As and are in the profile of and respectively, by definition of profile, there exist and in and that collapse to and in and respectively.
Now we use the Composition Lemma 6.4 on and to first get a reconfiguration sequence in . This is possible because they both collapse to according to the check of the algorithms, and because by construction the vertices in are only adjacent to the vertices of (which are in ) and not to the other vertices of .
As collapses to , and the algorithm checked that collapses to , by Lemma 6.1, also collapses to . We can finally use again our Composition Lemma 6.4, because both and collapse to over . And as before, the vertices of that are not in (i.e. the vertices of ) are only adjacent to the vertices of , which are in , and not to the other vertices of . This yields a reconfiguration sequence over . With several uses of Lemma 6.1, we obtain that which collapses to also collapses to , and therefore also collapses to .
In conclusion, this sequence is indeed in the profile of .
This therefore concludes the proof of Proposition 6.6.
With all of that in mind, we can also conclude the proof of Theorem 1.4. Thanks to Proposition 6.6, we compute the profile of every node in time linear in the number of nodes in the tree decomposition (with constant factors depending on and tw). As the significant of the root is the entire graph, the non-emptiness of the profile of the root certifies the existence of a reconfiguration sequence in the entire graph. Such a reconfiguration sequence can be computed in a top-down traversal of the tree decomposition, selecting at each level a sequence from the profile that is compatible with the one selected by the parent.
7 Back to undirected graphs
One of the biggest open problem in token sliding reconfiguration is whether ISR-TS parameterized by is on graphs of bounded treewidth (see [6, Question 5]). Our study of directed graphs and Theorem 1.4 can be seen as a step towards solving this open problem. In fact the techniques that we develop in Section 6 have consequences for undirected graphs.
For example, note that the only place in the proof of Theorem 1.4 where we use that the given graph is directed and is a DAG is in Lemma 6.2. The DAG property is only used to show that a token never goes twice to the same vertex. This in turn bounds the number of possible reconfiguration sequences in collapses of our original DAG (Lemma 6.3).
Our proof then directly provide the following result: Given an (undirected) instance for ISR-TS, deciding whether there is a reconfiguration sequence where no token goes twice to the same vertex can be decided in , parameterized by .
To generalize this idea, we introduce another parameter.
Definition 7.1.
Given a reconfiguration sequence , a token and a vertex , the iteration of is the number of time the token enters (plus one if starts in ).
The iteration of a reconfiguration sequence is the maximum iteration of all pair :
.
For galactic graphs, the iteration is only defined on planets, with no constrains on the number of time a token enters a black holes. Observe then that when taking the collapse of a reconfiguration sequence, its iteration does not increase.
Lemma 7.2 (Analogue of Lemma 6.3).
Let be a (undirected) galactic graph with no neighboring black holes; the number and length of reconfiguration sequences with tokens and of iteration over is bounded by for some computable function .
Proof.
First observe that with the absence of neighboring black holes, each step introduces a change for a planet (losing or receiving a token). As, a planet only witness many changes, this bounds the length of every sequence by , where .
As there are possible configurations, the total number of possible reconfiguration sequences of iteration is at most .
We then call the -profile of the set of reconfiguration of iteration over such that there is a reconfiguration over that collapses to over .
Proposition 7.3 (Analogue of Proposition 6.6).
Given a tree decomposition of a graph of width tw, a node of the decomposition, and the -profile of its children and , we can compute the -profile of in time .
Which in turn yields the proof for Theorem 1.5.
8 Conclusion
We used galactic graphs, to show that ISR-DTS in DAGs is tractable, when parameterized by treewidth and the number of tokens . Otherwise the problem becomes hard, unless either the depth is at most 2, or at most 3 and small. Noticeable, the treewidth result applies to undirected graphs, if the iteration is restricted.
Thus, it might be interesting, to study this parameters behavior in graph classes other then DAGs. Interesting further results might also arise from applying the used techniques to other problems (Dominating Set Reconfiguration might be a good candidate) or to generalize to other directed graph classes that resemble DAGs.
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