1 Introduction

Many algorithmic questions present themselves in the following form: Given the description of a system state and the description of a state we would prefer the system to be in, is it possible to transform the system from its current state into the more desired one without “breaking” certain properties of the system in the process? Such questions, with some generalizations and specializations, have received a substantial amount of attention under the so-called combinatorial reconfiguration framework [1,2,3].

Historically, the study of reconfiguration questions predates the field of computer science, as many classic one-player games can be formulated as reachability questions [4, 5], e.g., the 15-puzzle and Rubik’s cube. More recently, reconfiguration problems have emerged from computational problems in different areas such as graph theory [6,7,8], constraint satisfaction [9, 10], computational geometry [11], and even quantum complexity theory [12]. We refer the reader to the surveys by van den Heuvel [2] and Nishimura [13] for extensive background on combinatorial reconfiguration.

Independent Set Reconfiguration. In this work, we focus on the reconfiguration of independent sets. Given a simple undirected graph G, a set of vertices \(I \subseteq V(G)\) is an independent set if the vertices of this set are pairwise non-adjacent. Finding an independent set of size k, i.e., the Independent Set problem, is known to be NP-hard, but also W[1]-hardFootnote 1 parameterized by solution size k and not approximable within \(O(n^{1-\epsilon })\), for any \(\epsilon > 0\), unless \(\textsf {P}= \textsf {NP}\) [14]. Moreover, Independent Set remains W[1]-hard on graphs excluding \(C_4\) (the cycle on four vertices) as an induced subgraph [15].

We view an independent set as a collection of tokens placed on the vertices of a graph such that no two tokens are placed on adjacent vertices. This gives rise to two natural adjacency relations between independent sets (or token configurations), also called reconfiguration steps. These reconfiguration steps, in turn, give rise to two combinatorial reconfiguration problems.

In the Token Sliding problem, introduced by Hearn and Demaine [16], two independent sets are adjacent if one can be obtained from the other by removing a token from a vertex u and immediately placing it on another vertex v with the requirement that \(\{u, v\}\) must be an edge of the graph. The token is then said to slide from vertex u to vertex v along the edge \(\{u, v\}\). Generally speaking, in the Token Sliding problem, we are given a graph G and two independent sets \(I_s\) and \(I_t\) of G. The goal is to decide whether there exists a sequence of slides (a reconfiguration sequence) that transforms \(I_s\) to \(I_t\). The problem has been extensively studied under the combinatorial reconfiguration framework [17,18,19,20,21,22,23]. It is known that the problem is PSPACE-complete, even on restricted graph classes such as graphs of bounded bandwidth (and hence pathwidth) [24], planar graphs [16], split graphs [25], and bipartite graphs [26]. However, Token Sliding can be decided in polynomial time on trees [19], interval graphs [17], bipartite permutation and bipartite distance-hereditary graphs [20], and line graphs [7].

In the Token Jumping problem, introduced by Kamiński et al. [22], we drop the restriction that the token should move along an edge of G and instead we allow it to move to any vertex of G provided it does not break the independence of the set of tokens. That is, a single reconfiguration step consists of first removing a token on some vertex u and then immediately adding it back on any other vertex v, as long as no two tokens become adjacent. The token is said to jump from vertex u to vertex v. Token Jumping is also PSPACE-complete on graphs of bounded bandwidth [24] and planar graphs [16]. Lokshtanov and Mouawad [26] showed that, unlike Token Sliding, which is PSPACE-complete on bipartite graphs, the Token Jumping problem becomes NP-complete on bipartite graphs. On the positive side, it is “easy” to show that Token Jumping can be decided in polynomial-time on trees (and even on split/chordal graphs) since we can simply jump tokens to leaves (resp. vertices that only appear in the bag of a leaf in the clique tree) to transform one independent set into another.

In this paper we focus on the parameterized complexity of the Token Sliding problem on graphs where cycles with prescribed lengths are forbidden. Given an NP-hard problem, parameterized complexity permits to refine the notion of hardness; does the hardness come from the whole instance or from a small parameter? A problem \(\Pi \) is FPT (fixed-parameter tractable) parameterized by k if one can solve it in time \(f(k) \cdot poly(n)\), for some computable function f. In other words, the combinatorial explosion can be restricted to the parameter k. In the rest of the paper, our parameter k will be the size of the independent set (i.e. the number of tokens). Token Sliding is known to be W[1]-hard parameterized by k on general [23] and bipartite [27] graphs. It remains W[1]-hard on \(\{C_4, \dots , C_p\}\)-free graphs for any \(p \in \mathbb {N}\) [27] and becomes FPT parameterized by k on bipartite \(C_4\)-free graphs. The Token Jumping problem is W[1]-Hard on general graphs [21] and is FPT when parameterized by k on graphs of girth five or more [27]. For graphs of girth four, it was shown that Token Jumping being FPT would imply that Gap-ETH, an unproven computational hardness hypothesis, is false [28]. Both Token Jumping and Token Sliding were recently shown to be XL-complete [29].

Our Result. The complexity of the Token Jumping problem parameterized by k is settled with regard to the girth of the graph, i.e., the problem is unlikely to be FPT for graphs of girth four or less and FPT for graphs of girth five or more. For Token Sliding, it was only known that the problem is W[1]-hard for graphs of girth four or less and the authors in [27] posed the question of whether there exists a constant p such that the problem becomes fixed-parameter tractable on graphs of girth at least p. We answer their question positively and prove that the problem is indeed FPT for graphs of girth five or more, which establishes a full classification of the tractability of Token Sliding parameterized by the number of tokens based on the girth of the input graph.

Our Methods. Our result extends and builds on the recent galactic reconfiguration framework introduced by Bartier et al. [30] to show that Token Sliding is FPT on graphs of bounded degree, chordal graphs of bounded clique number, and planar graphs. Let us briefly describe the intuition behind the framework and how we adapt it for our use case. One of the main reasons why the Token Sliding problem is believed to be “harder” than the Token Jumping problem is due to what the authors in [30] call the bottleneck effect. Indeed, if we consider Token Sliding on trees, there might be a lot of empty leaves/subtrees in the tree but there might be a bottleneck in the graph that prevents any other tokens from reaching these vertices. For instance, if we consider a star with one long subdivided branch, then one cannot move any tokens from the leaves of the star to the long branch while there are at least two tokens on leaves. That being said, if the long branch of the star is “long enough” with respect to k then it should be possible to reduce parts of it; as some part would be irrelevant. In fact, this observation can be generalized to many other cases. For instance, when we have a large grid minor, then whenever a token slides into the structure it should then be able to slide freely within the structure (while avoiding conflicts with any other tokens in that structure). However, proving that a structure can be reduced in the context of reconfiguration is usually a daunting task due to the many moving parts. To overcome this problem, the authors in [30] introduce a new type of vertices called black holes, which can simulate the behavior of a large grid minor by being able to absorb as many tokens as they see fit; and then project them back as needed.

Since we need to maintain the girth property,Footnote 2 we do not use the notion of black holes and instead show that when restricted to graphs of girth five or more we can efficiently find structures that behave like large grid minors (from the discussion above) and replace them with subgraphs of size bounded by a function of k that can absorb/project tokens in a similar fashion (and do not decrease the girth of the graph). We note that our strategy for reducing such structures is not limited to graphs of high girth and could in principle apply to any graph.

At a high level, our FPT algorithm can then be summarized as follows. We let \((G, k, I_s, I_t)\) denote an instance of the problem, where G has girth five or more. In a first stage, we show that we can always find a reconfiguration sequence from \(I_s\) to \(I_s'\) and from \(I_t\) to \(I_t'\) such that each vertex \(v \in I_s' \cup I_t'\) has degree bounded by some function of k. This immediately implies that we can bound the size of \(L_1 \cup L_2\), where \(L_1 = I_s' \cup I_t'\) and \(L_2 = N_G(I_s' \cup I_t')\). In a second stage, we show that every connected component C of \(L_3 = V(G) {\setminus } (L_1 \cup L_2)\) can be classified as either a degree-safe component, a diameter-safe component, a bad component, or a bounded component. The remainder of the proof consists in showing that degree-safe and diameter-safe components behave like large grid minors and can be replaced by bounded-size gadgets. We then show that bounded components and bad components will eventually have bounded size and we then conclude the algorithm by showing how to bound the total number of components in \(L_3\).

Finally, we note that many interesting questions remain open. In particular, it remains open whether Token Sliding admits a (polynomial) kernel on graphs of girth five or more and whether the problem remains tractable if we forbid cycles of length \(p \mod q\), for every pair of fixed integers p and q, or if we exclude odd cycles.

2 Preliminaries

We denote the set of natural numbers by \(\mathbb {N}\). For \(n \in \mathbb {N}\) we let \([n] = \{1, 2, \dots , n\}\).

Graphs. We assume that each graph G is finite, simple, and undirected. We let V(G) and E(G) denote the vertex set and edge set of G, respectively. The open neighborhood of a vertex v is denoted by \(N_G(v) = \{u \mid \{u,v\} \in E(G)\}\) and the closed neighborhood by \(N_G[v] = N_G(v) \cup \{v\}\). For a set of vertices \(Q \subseteq V(G)\), we define \(N_G(Q) = \{v \not \in Q \mid \{u,v\} \in E(G), u \in Q\}\) and \(N_G[Q] = N_G(Q) \cup Q\). The subgraph of G induced by Q is denoted by G[Q], where G[Q] has vertex set Q and edge set \(\{\{u,v\} \in E(G) \mid u,v \in Q\}\). We let \(G - Q = G[V(G) \setminus Q]\). We use \(N_Q(v)\) and \(N_Q[v]\) instead of \(N_{G[Q \cup \{v\}]}(v)\) and \(N_{G[Q \cup \{v\}]}[v]\) to denote the open and closed neighborhoods of v in the graph induced by \(Q \cup \{v\}\), respectively.

A walk of length \(\ell \) from \(v_0\) to \(v_\ell \) in G is a vertex sequence \(v_0, \ldots , v_\ell \), such that for all \(i \in \{0, \ldots , \ell -1\}\), \(\{v_i,v_{i + 1}\} \in E(G)\). It is a path if all vertices are distinct. It is a cycle if \(\ell \ge 3\), \(v_0 = v_\ell \), and \(v_0, \ldots , v_{\ell - 1}\) is a path. A path from vertex u to vertex v is also called a uv-path. For a pair of vertices u and v in V(G), by \(\textsf {dist}_G(u,v)\) we denote the distance or length of a shortest uv-path in G (measured in number of edges and set to \(\infty \) if u and v belong to different connected components). The eccentricity of a vertex \(v \in V(G)\), \(\textsf {ecc}(v)\), is equal to \(\max _{u \in V(G)}(\textsf {dist}_G(u,v))\). The diameter of G, \(\textsf {diam}(G)\), is equal to \(\max _{v \in V(G)}(\textsf {ecc}(v))\). The girth of G, \(\textsf {girth}(G)\), is the length of a shortest cycle contained in G. If the graph does not contain any cycles (that is, it is a forest), its girth is defined to be infinity.

Reconfiguration. In the Token Sliding problem we are given a graph \(G = (V,E)\) and two independent sets \(I_s\) and \(I_t\) of G, each of size \(k \ge 1\). The goal is to determine whether there exists a sequence \(\langle I_0, I_1, \ldots , I_\ell \rangle \) of independent sets of size k such that \(I_s = I_0\), \(I_\ell = I_t\), and \(I_j \Delta I_{j+1} = \{u, v\} \in E(G)\) for all \(j \in \{0,\ldots , \ell -1\}\). In other words, if we view each independent set as a collection of tokens placed on a subset of the vertices of G, then the problem asks for a sequence of independent sets which transforms \(I_s\) to \(I_t\) by individual token slides along edges of G which maintain the independence of the sets. Note that Token Sliding can be expressed in terms of a reconfiguration graph \(\mathcal {R}(G,k)\); \(\mathcal {R}(G,k)\) contains a node for each independent set of G of size exactly k and we add an edge between two nodes whenever the independent set corresponding to one node can be obtained from the other by a single reconfiguration step. That is, a single token slide corresponds to an edge in \(\mathcal {R}(G,k)\). The Token Sliding problem asks whether \(I_s, I_t \in V(\mathcal {R}(G,k))\) belong to the same connected component of \(\mathcal {R}(G,k)\).

3 Reducing the Graph

Let \((G, k, I_s, I_t)\) be an instance of Token Sliding, where G has girth five or more. The aim of this section is to bound the size of the graph by a function of k. We start with a very simple reduction rule that allows us to get rid of most twin vertices in the graph. Two vertices \(u,v \in V(G)\) are said to be twins if u and v have the same set of neighbours, that is, if \(N(u) = N(v)\).

Lemma 1

Assume \(u, v \in V(G) {\setminus } (I_s \cup I_t)\) and \(N(u) = N(v)\). Then \((G, k, I_s, I_t)\) is a yes-instance if and only if \((G - \{v\}, k, I_s, I_t)\) is a yes-instance.

Proof

Since \(u, v \in V(G) {\setminus } (I_s \cup I_t)\) and \(G - \{v\}\) is an induced subgraph of G, it follows that if there exists a reconfiguration sequence \(\mathcal {S} = \langle I_0,I_1,\ldots ,I_{\ell -1},I_\ell \rangle \) from \(I_s\) to \(I_t\) in \(G - \{v\}\), then the same sequence remains valid in G.

Now assume that there exists a sequence \(\mathcal {S} = \langle I_0,I_1,\ldots ,I_{\ell -1},I_\ell \rangle \) from \(I_s\) to \(I_t\) in G. Since \(u, v \in V(G) \setminus (I_s \cup I_t)\), in \(I_s\) there are no tokens on u and v and the same holds for \(I_t\). Hence, if there exists \(I_j\), \(1 \le j \le \ell -1\) such that \(v \in I_j\), then \(u \not \in I_j\). The reason is that a token can be moved to u only via N(u). By assumption \(N(u) = N(v)\) and N(v) is blocked by the token on v. This implies that we can always choose to slide the token to u instead of v, as needed. \(\square \)

Note that in a graph of girth at least five twins can have degree at most one.

Given Lemma 1, we assume in what follows that twins have been reduced. In other words, we let \((G, k, I_s, I_t)\) be an instance of Token Sliding where G has girth five or more and twins not in \(I_s \cup I_t\) have been removed. We now partition our graph into three sets \(L_1 = I_s \cup I_t\), \(L_2 = N_G(L_1)\), and \(L_3 = V(G) {\setminus } (L_1 \cup L_2)\).

Lemma 2

If \(u \in L_2 \cup L_3\), then u has at most \(\vert L_1\vert \le 2k\) neighbors in \(L_1 \cup L_2\), i.e., \(\vert N_{L_1 \cup L_2}(u)\vert \le 2k\).

Proof

Assume \(u_1\) is a vertex in \(L_2\) and \(u_2 \in N_{L_2}(u_1)\) is a neighbor of \(u_1\) in \(L_2\). If \(u_1\) and \(u_2\) have a common neighbor \(u_3 \in L_1\), then this would imply the existence of a triangle in G, a contradiction.

Now assume \(u_1 \in L_3\) and assume \(u_2, u_3 \in N_{L_2}(u_1)\) are two neighbors of \(u_1\) in \(L_2\). If \(u_2\) and \(u_3\) have a common neighbor \(u_4 \in L_1\) this would imply the existence of a \(C_4\) in G, a contradiction.

Hence, for any vertex \(u \in L_2 \cup L_3\) we have \(N_{L_1}(v) \cap N_{L_1}(w) = \emptyset \) for all distinct \(v, w \in N_{L_2}[u]\). Since each vertex in \(L_2\) has at least one neighbor in \(L_1\) by definition, each vertex \(u \in L_2 \cup L_3\) can have at most one neighbor in \(L_2\) for each of its non-neighbors in \(L_1\), for a total of \(\vert L_1\vert \le 2k\) neighbors in \(L_1 \cup L_2\). \(\square \)

3.1 Safe, Bounded, and Bad Components

Given G and the partition \(L_1 = I_s \cup I_t\), \(L_2 = N_G(L_1)\), and \(L_3 = V(G) \setminus (L_1 \cup L_2)\) we now classify components of \(G[L_3]\) into four different types.

Definition 1

Let C be a maximal connected component in \(G[L_3]\).

  • We call C a diameter-safe component whenever \(\textsf {diam}(G[V(C)]) > k^3\).

  • We call C a degree-safe component whenever G[V(C)] has a vertex u with at least \(k^2 + 1\) neighbors X in C and at least \(k^2\) vertices of X have degree (at least) two in G[V(C)].

  • We call C a bounded component whenever \(\textsf {diam}(G[V(C)]) \le k^3\) and no vertex of C has degree more than \(k^2\) in G[V(C)].

  • We call C a bad component otherwise.

Lemma 3

A bounded component C in \(G[L_3]\) contains at most \(2k^{2k^3}\) vertices, i.e., \(\vert V(C)\vert \le 2k^{2k^3}\).

Proof

Let T be a breadth-first search tree of C and let \(u \in V(C)\) denote the root of T. Each vertex in T has at most \(k^2\) children given the degree bound of C and the height of the tree is at most \(k^3\) given the diameter bound of C. Hence the total number of leaves in T is at most \(k^{2k^3}\) and the total number of vertices in C is at most \(2k^{2k^3}\). \(\square \)

We now describe a crucial property of degree-safe and diameter-safe components, which we call the absorption-projection property. We note that this notion is similar to the notion of black holes introduced in [30]. The key (informal) insight is that for a safe component C we can show the following:

  1. 1.

    If there exists a reconfiguration sequence \(\mathcal {S} = \langle I_0,I_1,\ldots ,I_{\ell -1},I_\ell \rangle \) from \(I_s\) to \(I_t\), then we may assume that \(I_j \cap N_G(V(C)) \le 1\), for \(0 \le j \le \ell \).

  2. 2.

    A safe component can absorb all k tokens, i.e, a safe component contains an independent set of size at least k and whenever a token reaches \(N_G(V(C))\) then we can (but do not have to) absorb it into C (regardless of how many tokens are already in C). Moreover, a safe component can then project the tokens back into its neighborhood as needed.

Let us start by proving the absorption-projection property for degree-safe components. An s-star is a vertex with s pairwise non-adjacent neighbors, which are called the leaves of the s-star. A subdivided s-star is an s-star where each edge is subdivided (replaced by a new vertex of degree two adjacent to the endpoints of the edge) any number of times. We say that each leaf of a subdivided star belongs to a branch of the star. The length of a branch is the length of the path connecting the center of the star to the leaf of the branch.

Lemma 4

Let C be a degree-safe component in \(G[L_3]\). Then C contains an induced subdivided k-star where all k branches have length more than one.

Proof

Since C is a degree-safe component, it must contain a vertex u with at least \(k^2\) neighbors in C and each one of these neighbors must have another neighbor in C. Note that all of these vertices must be distinct, as otherwise we could find a cycle of length three or four.

Let us call the distance-one and distance-two neighbors of u in C the first level and second level. That is, we let \(N_1(u) = N_C(u) {\setminus } \{u\}\) and \(N_2(u) = N_C(N_1(u)) {\setminus } (N_1(u) \cup \{u\})\).

Note that the first level, \(N_1(u)\), is an independent set, since otherwise that would imply the existence of a triangle. Also, vertices in the second level, \(N_2(u)\), cannot be connected to more than one vertex of the first level, since that would imply the existence of a \(C_4\).

As for the second level, it contains at least \(k^2\) vertices and we can have edges between those vertices. We claim that \(G_2 = G[N_2(u)]\) contains an independent set of size k. Assume first that \(G_2\) contains a vertex v of degree k. Then, since \(G_2\) is triangle free, the k neighbors of v form the required independent set. Otherwise, all vertices of \(G_2\) have degree at most \(k-1\). We iteratively add one vertex v to the independent set and remove N[v] from \(G_2\). This can be repeated for k times leading to the required independent set. Therefore, we get an induced subdivided star with at least k branches of length at least two and there is no edge between the different branches. \(\square \)

Fig. 1
figure 1

An illustration of a degree-safe component C

Full size image

Lemma 5

Let C be a degree-safe component in \(G[L_3]\) and let A be an induced subdivided k-star contained in C where all branches have length exactly two. Let \(B = N_G(V(A))\). If \((G, k, I_s, I_t)\) is a yes-instance, then there exists a reconfiguration sequence from \(I_s\) to \(I_t\) in G where we have at most one token on a vertex of B at all times.

Proof

First, note that the existence of A follows from Lemma 4 and that it is indeed the case that \(I_s \cap B = I_t \cap B = \emptyset \). Let r denote the root of the induced subdivided k-star and let \(N_1\) and \(N_2\) denote the first and second levels of subdivided the star, respectively. Let us explain how we can adapt a transformation \(\mathcal {S}\) from \(I_s\) to \(I_t\) into a transformation containing at most one token on a vertex of B at all times and such that, at any step, the number of tokens in \(V(A) \cup B\) in both transformations is the same and the positions of the tokens in \(V(G) {\setminus } (V(A) \cup B)\) are the same.

Assume that, in the transformation \(\mathcal {S}\), a token is about to reach a vertex \(b \in B\), that is, we consider the step right before a token is about to slide into B. We first move all tokens residing in A, if any, to the second level of their branches, i.e, to \(N_2\). This is possible as A is an induced subdivided star and there are no other tokens on B. Note that we can assume that there is no token on r (and hence every token is on a branch and “the branch” of a token is well defined) since we can otherwise slide this token to one of the empty branches while B is still empty of tokens. Then we proceed as follows:

  • If b is a neighbor of the root r of the subdivided star, then b is not a neighbor of any vertex at the second level of A, since otherwise this would create a cycle of length four. Hence, we can slide the token into b and then r and then some empty branch of A (which is possible since we have k branches in A).

  • Otherwise, if b has no neighbors in the first level \(N_1\) of A, we choose a branch that has a neighbor a of b in \(N_2\) (which exists since b is not adjacent to r nor \(N_1\)). Then, if the branch of a already contains a token, we can safely slide the token into another branch by going to the first level, then the root r, then to another empty branch of A. Now we slide all tokens in A to the first level of their branch (ensuring that b has no neighbors in A containing tokens) and finally we slide the initial token to b and then to a.

  • Finally, if b has neighbors in the first level of A, note that it cannot have more than one neighbor in \(N_1\) since that would imply the existence of a cycle of length four. Let a denote the unique neighbor of b in \(N_1\). If the branch of a has a token on it, then we safely slide it into another empty branch. Now we slide all tokens in A to the first level of their branch and finally we slide the initial token to b and then to a.

Note that all of above slides are reversible and we can therefore use a similar strategy to project tokens from V(A) to B. If, in \(\mathcal {S}\), a token is about to leave the vertex \(b \in B\), then we can similarly move a token from V(A) to b and then perform the same move. Finally, if a reconfiguration step in \(\mathcal {S}\) consists of moving tokens in \(V(A) \cup B\) to \(V(A) \cup B\), we ignore that step. And, if it consists of moving a token from \(V(G) {\setminus } (V(A) \cup B)\) to \(V(G) {\setminus } (V(A) \cup B)\) we perform the same step.

It follows from the previous procedure that whenever \((G, k, I_s, I_t)\) is a yes-instance we can find a reconfiguration sequence from \(I_s\) to \(I_t\) in G where we have at most one token in B at all times, as claimed (see figure 1). \(\square \)

Corollary 1

Let C be a degree-safe component in \(G[L_3]\). If \((G, k, I_s, I_t)\) is a yes-instance, then there exists a reconfiguration sequence from \(I_s\) to \(I_t\) in G where we have at most one token in N(C) (\(\subseteq L_2\)) at all times.

Proof

Let A be an induced subdivided k-star contained in C where all branches have length exactly two and let \(B = N_G(V(A))\). Assume a token slides to a vertex \(c \in N(C)\) (for the first time). If \(c \in B\), then the result follows from Lemma 5. Otherwise, we can follow a path P contained in C that leads to the root of the induced k-subdivided star (such a path exists since \(c \in N(C)\) and C is connected) and right before we reach B we then again can apply Lemma 5. Note that, regardless of whether c is in B or not, once the token reaches N(C) we can assume that it is immediately absorbed by the degree-safe component (and later projected as needed). This implies that we can always find a path P to slide along such that N[P] contains no tokens. \(\square \)

We now turn our attention to diameter-safe components and show that they have a similar absorption-projection behavior as degree-safe components. Given a component C we say that a path A in C is a diameter path if A is a longest shortest path in C.

Fig. 2
figure 2

An illustration of a diameter-safe component C

Full size image

Lemma 6

Let C be a diameter-safe component, let A be a diameter path of C, and let \(B = N_G(V(A))\). If \((G, k, I_s, I_t)\) is a yes-instance, then there exists a reconfiguration sequence from \(I_s\) to \(I_t\) in G where we have at most one token on vertices of B at all times.

Proof

As in the proof of Lemma 5, the goal will consist in proving that we can adapt a transformation \(\mathcal {S}\) from \(I_s\) to \(I_t\) into a transformation containing at most one token on a vertex of B at all times and such that, at any step, the number of tokens in \(V(A) \cup B\) in both transformations is the same and the positions of the tokens in \(V(G) {\setminus } (V(A) \cup B)\) are the same. As in the proof of Lemma 5, all the tokens in \(V(A) \cup B\) will be absorbed into V(A) (and later projected back as needed) and it suffices to explain how we can move the tokens on V(A) when a new token wants to enter in B or leave into B.

We know that two non-consecutive vertices in A cannot be adjacent by minimality of the path. Now assume a token t is about to reach a vertex \(b \in B\) (for the first time). Note that neighbors of b in A are pairwise at distance at least three in A, since otherwise that would imply the existence of a cycle of length less than five. We call the sets of vertices in intervals between consecutive neighbor of b (in A) gap interval sets (with respect to b).

If b has more than k neighbors in A, then we can put the already in A tokens (at most \(k - 1\) of them) in the at most \(k-1\) first gap interval sets; the existence of \(k - 1\) gap interval sets implies the existence of an independent set of size \(k-1\) consisting of the middle vertex of each gap interval set. Indeed, since there is no token on B and A is an induced path, we can freely move tokens where we want and, in particular, to the corresponding independent set. Then, we can slide the token t to b, since none of its neighbors in A have a token on them, and then slide it to the next neighbor of b in A since it has more than k neighbors.

Otherwise, b has at most k neighbors in A. Hence there are at most \(k + 1\) gap interval sets in A (with respect to b). The average number of vertices in the gap interval sets (assuming \(k\ge 4\)) is

$$\begin{aligned} \alpha = \frac{\textsf {diam}(C) - \vert N_{V(A)}(b)\vert }{\vert N_{V(A)}(b)\vert + 1} \ge \frac{k^3 - k}{k + 1} \ge 2k. \end{aligned}$$

Hence at least one gap interval set has length at least \(\alpha \) and therefore we can slide all tokens currently in A (at most \(k-1\) of them) into this gap interval set in such a way that no token is on the border of the gap interval set (since the gap interval set contains an independent set of size at least \(k-1\) which does not contain an endpoint of the gap interval set). Now we can simply slide the token t onto b and then onto any of the neighbors of b in A.

Combined with the fact that the above strategy can also be applied to project a token from V(A) to B, it then follows that whenever \((G, k, I_s, I_t)\) is a yes-instance we can find a reconfiguration sequence from \(I_s\) to \(I_t\) in G where we have at most one token in B at all times, as claimed (see Fig. 2). \(\square \)

Corollary 2

Let C be a diameter-safe component. If \((G, k, I_s, I_t)\) is a yes-instance then there exists a reconfiguration sequence from \(I_s\) to \(I_t\) where we have at most one token in N(C) (\(\subseteq L_2\)) at all times.

Proof

We follow the same strategy as for the degree-safe components. Let A be a diameter path of C and let \(B = N_G(V(A))\). Assume a token slides to a vertex \(c \in N(C)\) (for the first time). If \(c \in B\), then the result follows from Lemma 6. Otherwise, we can follow a path P contained in C that leads to a vertex \(a \in N_G(B) \setminus (V(A) \cup B)\) (with a possibly equal to c). Such a path exists since \(c \in N(C)\) and C is connected. Once we reach a, and before we slide to a neighbor of a in B, say \(b \in B\), we again apply Lemma 6. Consequently, once a token reaches N(C) we can assume that it is immediately absorbed by the diameter-safe component (and later projected as needed). This implies that we can always find a path P to slide along such that N[P] contains no tokens. \(\square \)

Putting Corollaries 1 and 2 together, we know that if \((G, k, I_s, I_t)\) is a yes-instance, then there exists a reconfiguration sequence from \(I_s\) to \(I_t\) where we have at most one token in \(N(C) \subseteq L_2\) at all times, where C is either a degree-safe or a diameter-safe component. We now show how to reduce a safe component C by replacing it by another smaller subgraph that we denote by H.

Fig. 3
figure 3

An illustration of the replacement gadget for a safe component C

Full size image

Lemma 7

Let C be a safe component in \(G[L_3]\) and let \(G'\) be the graph obtained from G as follows:

  • Delete all vertices of C (and their incident edges).

  • For each vertex \(v \in N(C) \subseteq L_2\) add two new vertices \(v'\) and \(v''\) and add the edges \(\{v, v'\}\) and \(\{v', v''\}\).

  • Add a path of length 3k consisting of new vertices \(p_1\) to \(p_{3k}\).

  • Add an edge \(\{p_1, v''\}\) for every vertex \(v''\).

Note that this new component has size \(3k + \vert 2N(C)\vert \) (see Fig. 3). We claim that \((G, k, I_s, I_t)\) is a yes-instance if and only if \((G', k, I_s, I_t)\) is a yes-instance.

Proof

First, we note that replacing C with this new component, H, cannot create cycles of length less than five. This follows from the fact that all the vertices at distance one or two from \(p_1\) have distinct neighbors.

Assume \((G, k, I_s, I_t)\) is a yes-instance. Then, by Corollary 1 and Corollary 2, we know that there exists a reconfiguration sequence from \(I_s\) to \(I_t\) in G where we have at most one token in \(N(C) \subseteq L_2\) at all times, where C is either a degree-safe or a diameter-safe component. Hence, we can mimic the reconfiguration sequence from \(I_s\) to \(I_t\) in \(G'\) by simply projecting tokens onto the path of length 3k in each of the safe components that we replaced.

Now assume that \((G', k, I_s, I_t)\) is a yes-instance. By the same arguments, and combined with the fact that a safe component C can absorb/project the same number of tokens as its replacement component H, we can again mimic the reconfiguration sequence of \(G'\) in G. \(\square \)

3.2 Bounding the Size of \(L_2\)

Having classified the components in \(L_3\) and the edges between \(L_2\) and \(L_3\), our next goal is to bound the size of \(L_2\), which until now could be arbitrarily large. We know that vertices in \(L_2\) are the neighbors of vertices in \(L_1\), hence the size of \(L_2\) will grow whenever there are vertices in \(L_1\) with arbitrarily large degrees. Bounding \(L_2\) will therefore be done by first proving the following lemma.

Lemma 8

Assume a vertex u in \(L_1 = I_s \cup I_t\) has degree greater than \(2k^2\). Moreover, assume, without loss of generality, that \(u \in I_s\). Then, there exists \(I_s'\) such that \(I_s \triangle I_s' = \{u,u'\}\), \(u'\) has degree at most \(2k^2\), and the token on u can slide to \(u'\).

Proof

First note that from such a vertex \(u \in I_s\) we can always slide to a vertex in \(L_2\). Indeed, for every v, \(\vert N(u) \cap N(v)\vert \le 1\) by the assumption on the girth of the graph. Thus, since the degree of u is larger than the number of tokens, there exists at least one vertex in \(L_2\) that the token on u can slide to.

If we slide to a vertex \(v \in L_2\) of degree at most \(2k^2\), then we are done (we set \(u' = v\)). Otherwise, by Lemma 2, we know that most of the neighbors of v are in \(L_3\); since v has degree greater than \(2k^2\) and at most 2k of its neighbors are in \(L_1 \cup L_2\). Hence, we are guaranteed at least one neighbor w of v in some component of \(L_3\).

If we reach a bounded component C, i.e., if w belongs to a bounded component, then all vertices of C (including w) have at most \(k^2\) neighbors in C and have at most 2k neighbors in \(L_2\) (by Lemma 2) and thus we can set \(u' = w\).

If we reach a bad component C, then we know that \(\textsf {diam}(G[V(C)]) \le k^3\) (since C is not diameter-safe) and therefore C must have a vertex b with at least \(k^2 + 1\) neighbors in C (as otherwise C is a bounded component). Moreover, since C is not degree-safe, at most \(k^2 - 1\) of the neighbors of b have other neighbors in C. Let z denote a vertex in the neighborhood of b that does not have other neighbors in C. By Lemma 2, z will have degree at most \(2k + 1\) and we can therefore let \(u' = z\).

Finally, if we reach a safe component, then after our replacement such components contain a lot of vertices of degree exactly two and we can therefore slide to any such vertex, which completes the proof. \(\square \)

After exhaustively applying Lemma 8, each time relabeling vertices in \(L_1\), \(L_2\) and \(L_3\) and replacing safe components as described in Lemma 7, we get an equivalent instance where the maximum degree in \(L_1\) is at most \(2k^2\) and hence we get a bound on the size of \(L_2\). We conclude this section with the following lemma.

Lemma 9

Let \((G, k, I_s, I_t)\) be an instance of Token Sliding, where G has girth at least five. Then we can compute an equivalent instance \((G', k, I_s', I_t')\), where \(G'\) has girth at least five, \(\vert L_1 \cup L_2\vert \le 2k + 4k^3 = O(k^3)\), and each safe component of G is replaced in \(G'\) by a component with at most \(3k + 8k^3 = O(k^3)\) vertices.

3.3 Bounding the Size of \(L_3\)

We have proved that the number of vertices in \(L_1\) and \(L_2\) is bounded by a function of k, namely \(\vert L_1 \cup L_2\vert = O(k^3)\). We have also shown that every safe or bounded component in \(L_3\) has a bounded number of vertices, namely safe components have \(O(k^3)\) vertices and bounded components have at most \(2k^{2k^3}\) vertices. We still need to show that \(L_3\) is bounded. We start by showing that bad components become bounded after bounding \(L_2\):

Lemma 10

Let \((G, k, I_s, I_t)\) be an instance where G has girth at least five, \(\vert L_1 \cup L_2\vert \le 2k + 4k^3 = O(k^3)\), and each safe component has at most \(3k + 8k^3 = O(k^3)\) vertices. Then, every bad component in that instance has at most \(k^{O(k^3)}\) vertices.

Proof

Let C be a bad component, hence \(\textsf {diam}(C) \le k^3\) since C is not diameter-safe. Let \(v \in V(C)\) be a vertex in C whose degree is \(d > k^2\). Since C is not a degree-safe component v can have at most \(k^2 - 1\) neighbors in C that have other neighbors in C. Hence, at least \(d - (k^2 - 1) = d - k^2 + 1\) neighbors of v will have only v as a neighbor in C and all their other neighbors must be in \(L_2\). Since, by Lemma 1, we can assume that \(L_3\) contains no twin vertices, \(d - k^2\) of the neighbors of v in C must have at least one neighbor in \(L_2\). But we know that \(L_2\) has size \(O(k^3)\) and if two neighbors of v had a common neighbor in \(L_2\), this would imply the existence of a cycle of length four. Therefore, d must be at most \(O(k^3)\). Having bounded the degree and diameter of bad components, we can now apply the same argument as in the proof of Lemma 3. \(\square \)

Since bounded and bad components now have the same asymptotic number of vertices, in what follows we refer to both of them as bounded components. What remains to show is that the number of safe and bounded components is also bounded by a function of k and hence \(L_3\) and the whole graph will have size bounded by a function of k.

Definition 2

Let \(C_1\) and \(C_2\) be two components in \(G[L_3]\) and \(B_1\) and \(B_2\) be their respective neighborhoods in \(L_2\). We say \(C_1\) and \(C_2\) are equivalent whenever \(B_1 = B_2 = B\) and \(G[V(C_1) \cup B]\) is isomorphic to \(G[V(C_2) \cup B]\) by an isomorphism that fixes B point-wise. We let \(\beta (G)\) denote the number of equivalence classes of bounded components and we let \(\sigma (G)\) denote the number of equivalence classes of safe components.

We are now ready to prove a crucial result for bounding \(L_3\).

Lemma 11

Let \(S_1\) and \(S_2\) be equivalent safe components and let \(B_1\), \(\ldots \), \(B_{k + 1}\) be equivalent bounded components. Then, \((G, k, I_s, I_t)\), \((G - V(S_2), k, I_s, I_t)\) and \((G - V(B_{k + 1}), k, I_s, I_t)\) are equivalent instances.

Proof

Removing vertices from the graph preserves no-instances. As for yes-instances, we will prove equivalence for safe and bounded components separately.

Assume a token reaches the neighborhood of \(S_1\) and \(S_2\) (they have the same neighborhood). Whether the token slides to either of them is irrelevant because both can hold all the tokens together and have the same behavior regarding entering from \(L_2\) and leaving to \(L_2\). Hence, from Corollary 1 and Corollary 2, we can always choose to slide to \(S_1\) and never to \(S_2\) and therefore removing \(S_2\) will preserve yes-instances.

Assume a token reaches the neighborhood of all \(B_i\)’s (they have the same neighborhood). The components not being empty implies that each one can hold at least one token if it can, and hence we can always choose to slide the tokens to one of the first k components since it will be enough to hold all tokens. Therefore removing \(B_{k+1}\) will preserve yes-instances. \(\square \)

After exhaustively removing equivalent components as described in Lemma 11 we obtain the following corollary.

Corollary 3

There are at most \(k\beta (G)\) bounded components and \(\sigma (G)\) safe components.

This leads to the final lemma.

Lemma 12

We have \(\beta (G) = 2^{k^{O(k^3)}}\), \(\sigma (G) = 2^{O(k^6)}\), \(\vert L_3\vert \le k^{O(k^3)}2^{k^{O(k^3)}} + k^3 2^{O(k^6)} = 2^{k^{O(k^3)}}\), and \(\vert V(G)\vert = \vert L_1\vert + \vert L_2\vert + \vert L_3\vert = 2^{k^{O(k^3)}}\).

Proof

Since \(L_2\) and safe components have \(O(k^3)\) size (from Lemma 9) then safe components along with their neighbors in \(L_2\) have size \(O(k^3)\). Hence there are \(2^{O(k^6)}\) equivalence classes of safe components.

Since bounded components have size \(k^{O(k^3)}\) (from Lemma 3) the bounded components along with their neighbors in \(L_2\) have size \(k^{O(k^3)}\) and hence there are \(2^{k^{O(k^3)}}\) equivalence classes of bounded components.

Finally, using the fact that there are \(2^{n \atopwithdelims ()2}\) graphs with n vertices combined with Corollary 3, we get the desired bound on \(L_3\), which implies the desired bound on the size of V(G). \(\square \)

4 The Algorithm

4.1 Outline

Now that we have bounded the size of G by \(f(k) = 2^{k^{O(k^3)}}\) we describe below the complete algorithm for solving an instance \((G, k, I_s, I_t)\) of the Token Sliding problem, where G has girth five or more.

  1. 1.

    Bound the graph size;

    1. (a)

      Remove twin vertices as described in Lemma 1;

    2. (b)

      Repeat the following while \(L_1\) has a vertex of degree greater than \(2k^2\) or there exists an unbounded safe component in \(L_3\):

      • Find safe components as described in Definition 1;

      • Replace safe components as described in Lemma 7;

      • Find a vertex \(u \in L_1\) with degree greater than \(2k^2\);

      • Slide the token to a vertex of degree at most \(2k^2\) (Lemma 8);

    3. (c)

      Test all pairs of \(L_3\) components for equivalence (Lemma 2);

    4. (d)

      Partition the components into equivalence classes;

      • For classes containing a safe component, keep one component and remove the others from the graph (Lemma 11);

      • For each other class, keep k components and remove the others from the graph. If there are already less than k components then do nothing (Lemma 11);

  2. 2.

    Build the graph \(\mathcal {R}(G, k)\);

    • \(\mathcal {R}(G, k)\) will have a node for each independent set of G of size k;

    • Two nodes \(I, J \in \mathcal {R}(G, k)\) will be connected by an edge if the corresponding independent sets are adjacent with respect to the token slide definition, namely \(I \Delta J = \{u, v\} \in E(G)\);

  3. 3.

    Run a breadth-first search (BFS) traversal on \(\mathcal {R}(G, k)\) with source \(I_s\) and destination \(I_t\). Return true if the two are in the same component and false otherwise;

4.2 Analysis

Complexity of step (1). Step (a), removing twin vertices, can be naively implemented to run in \(O(n^3)\)-time. Going to step (b), finding degree-safe components will take O(n)-time by simply checking the degrees of all vertices in a component. As for diameter-safe components, we can find them in \(O(n^2)\)-time by finding for each vertex u in a component C the vertex v furthest away from u in C using a BFS. Replacing a component can be done in O(n)-time. Finding \(u \in L_1\) such that the degree of u is greater than \(2k^2\) and replacing it via slides can be done in O(k)-time. This procedure will be repeated at most 2k times and hence step (b) requires \(O(k^2 + kn^2)\)-time. Going to step (c), we can test isomorphism of components using any exponential-time algorithm. Since the size of the individual components is now bounded by \(k^{O(k^3)}\) and the algorithm will run on all pairs of components, step (c) will require \(2^{k^{O(k^3)}}\)-time in the worst case. Finally, step (d) consists only of removing components and can be done in O(n). Therefore step (1) will take \(O(kn^3 + 2^{k^{O(k^3)}})\)-time.

Complexity of step (2). Building the graph \(\mathcal {R}(G, k)\) will take \(O(\vert V(\mathcal {R}(G, k))\vert + k^2 \vert V(\mathcal {R}(G, k))\vert ^2) = O(k^2{f(k) \atopwithdelims ()k}^2)\)-time since we can check naively for each pair of nodes if they are connected via one slide.

Complexity of step (3). The breadth-first search traversal will take \(O(\vert V(\mathcal {R}(G, k))\vert + \vert E(\mathcal {R}(G, k))\vert ) = O({f(k) \atopwithdelims ()k}^2)\)-time.

Putting it all Together. Therefore, the total running time of the algorithm is

$$\begin{aligned} O(kn^3) + 2^{k^{O(k^3)}} + O(k^2{f(k) \atopwithdelims ()k}^2) \end{aligned}$$

and hence we get the desired result.

Theorem 1

Token Sliding is fixed-parameter tractable when parameterized by k on graphs of girth five or more.