Sparsification Upper and Lower Bounds for Graph Problems and Not-All-Equal SAT

We present several sparsification lower and upper bounds for classic problems in graph theory and logic. For the problems 4-Coloring, (Directed) Hamiltonian Cycle, and (Connected) Dominating Set, we prove that there is no polynomial-time algorithm that reduces any n-vertex input to an equivalent instance, of an arbitrary problem, with bitsize O(n^{2-e}) for e>0, unless NP is in coNP/poly and the polynomial-time hierarchy collapses. These results imply that existing linear-vertex kernels for k-Nonblocker and k-Max Leaf Spanning Tree (the parametric duals of (Connected) Dominating Set) cannot be improved to have O(k^{2-e}) edges, unless NP is in coNP/poly. We also present a positive result and exhibit a non-trivial sparsification algorithm for d-Not-All-Equal SAT. We give an algorithm that reduces an n-variable input with clauses of size at most d to an equivalent input with O(n^{d-1}) clauses, for any fixed d. Our algorithm is based on a linear-algebraic proof of Lovasz that bounds the number of hyperedges in critically 3-chromatic d-uniform n-vertex hypergraphs by n choose d-1. We show that our kernel is tight under the assumption that NP is not a subset of coNP/poly.


Introduction
Background.Sparsification refers to the method of reducing an object such as a graph or CNF-formula to an equivalent object that is less dense, that is, an object in which the ratio of edges to vertices (or clauses to variables) is smaller.The notion is fruitful in theoretical [16] and practical (cf.[10]) settings when working with (hyper)graphs and formulas.The theory of kernelization, originating from the field of parameterized complexity theory, can be used to analyze the limits of polynomial-time sparsification.Using tools developed in the last five years, it has become possible to address questions such as: "Is there a polynomial-time algorithm that reduces an n-vertex instance of my favorite graph problem to an equivalent instance with a subquadratic number of edges?" The impetus for this line of analysis was given by an influential paper by Dell and van Melkebeek [8] (conference version in 2010).One of their main results states that if there is an ε > 0 and a polynomial-time algorithm that reduces any n-vertex instance of Vertex Cover to an equivalent instance, of an arbitrary problem, that can be encoded in O(n 2−ε ) bits, then NP ⊆ coNP/poly and the polynomial-time hierarchy collapses.Since any nontrivial input (G, k) of Vertex Cover has k ≤ n = |V (G)|, their result implies that the number of edges in the 2k-vertex kernel for k-Vertex Cover [21] cannot be improved to O(k 2−ε ) unless NP ⊆ coNP/poly.
Using related techniques, Dell and van Melkebeek also proved important lower bounds for d-cnf-sat problems: testing the satisfiability of a propositional formula in CNF form, where each clause has at most d literals.They proved that for every fixed integer d ≥ 3, the existence of a polynomial-time algorithm that reduces any n-variable instance of d-cnf-sat to an equivalent instance, of an arbitrary problem, with O(n d−ε ) bits, for some ε > 0 implies NP ⊆ coNP/poly.Their lower bound is tight: there are O(n d ) possible clauses of size d over n variables, allowing an instance to be represented by a vector of O(n d ) bits that specifies for each clause whether or not it is present.
Our results.We continue this line of investigation and analyze sparsification for several classic problems in graph theory and logic.We obtain several sparsification lower bounds that imply that the quadratic number of edges in existing linear-vertex kernels is likely to be unavoidable.When it comes to problems from logic, we give the-to the best of our knowledge-first example of a problem that does admit nontrivial sparsification: d-Not-All-Equal-SAT.We also provide a matching lower bound.
The first problem we consider is 4-Coloring, which asks whether the input graph has a proper vertex coloring with 4 colors.Using several new gadgets, we give a cross-composition [3] to show that the problem has no compression of size O(n 2−ε ) unless NP ⊆ coNP/poly.To obtain the lower bound, we give a polynomial-time construction that embeds the logical or of a series of t size-n inputs of an NP-hard problem into a graph G with O( ) vertices, such that G has a proper 4-coloring if and only if there is a yes-instance among the inputs.The main structure of the reduction follows the approach of Dell and Marx [7]: we create a table with two rows and O( √ t) columns and O(n O (1) ) vertices in each cell.For each way of picking one cell from each row, we aim to embed one instance into the edge set between the corresponding groups of vertices.When the NP-hard starting problem is chosen such that the t inputs each decompose into two induced subgraphs with a simple structure, one can create the vertex groups and their connections such that for each pair of cells (i, j), the subgraph they induce represents the i • √ t + j-th input.If there is a yes-instance among the inputs, this leads to a pair of cells that can be properly colored in a structured way.The challenging part of the reduction is to ensure that the edges in the graph corresponding to no-inputs do not give conflicts when extending this partial coloring to the entire graph.
The next problem we attack is Hamiltonian Cycle.We rule out compressions of size O(n 2−ε ) for the directed and undirected variant of the problem, assuming NP coNP/poly.The construction is inspired by kernelization lower bounds for Directed Hamiltonian Cycle parameterized by the vertex-deletion distance to a directed graph whose underlying undirected graph is a path [2].
By combining gadgets from kernelization lower bounds for two different parameterizations of Red Blue Dominating Set, we prove that there is no compression of size O(n 2−ε ) for Dominating Set unless NP ⊆ coNP/poly.The same construction rules out subquadratic compressions for Connected Dominating Set.These lower bounds have implications for the kernelization complexity of the parametric duals Nonblocker and Max Leaf Spanning Tree of (Connected) Dominating Set.For both Nonblocker and Max Leaf there are kernels with O(k) vertices [6,11] that have Θ(k 2 ) edges.Our lower bounds imply that the number of edges in these kernels cannot be improved to O(k 2−ε ), unless NP ⊆ coNP/poly.
The final family of problems we consider is d-Not-All-Equal-SAT for fixed d ≥ 4. The input consists of a formula in CNF form with at most d literals per clause.The question is whether there is an assignment to the variables such that each clause contains both a variable that evaluates to true and one that evaluates to false.There is a simple linear-parameter transformation from d-cnf-sat to (d + 1)-nae-sat that consists of adding one variable that occurs as a positive literal in all clauses.By the results of Dell and van Melkebeek discussed above, this implies that d-nae-sat does not admit compressions of size O(n d−1−ε ) unless NP ⊆ coNP/poly.We prove the surprising result that this lower bound is tight!A linearalgebraic result due to Lovász [20], concerning the size of critically 3-chromatic d-uniform hypergraphs, can be used to give a kernel for d-nae-sat with O(n d−1 ) clauses for every fixed d.The kernel is obtained by computing the basis of an associated matrix and removing the clauses that can be expressed as a linear combination of the basis clauses.

Related work. Dell and Marx introduced the table structure for compression lower bounds [7]
in their study of compression for packing problems.Hermelin and Wu [15] analyzed similar problems.Other papers about polynomial kernelization and sparsification lower bounds include [5] and [17].The algorithm is a kernel for

Preliminaries
Since a polynomial-time reduction to an equivalent sparse instance yields a generalized kernel, we will use the concept of generalized kernels in the remainder of this paper to prove the non-existence of such sparsification algorithms.We employ the cross-composition framework by Bodlaender et al. [3], which builds on earlier work by several authors [1,8,13].

Definition 1 (Polynomial equivalence relation
).An equivalence relation R on Σ * is called a polynomial equivalence relation if the following conditions hold.

1.
There is an algorithm that, given two strings x, y ∈ Σ * , decides whether x and y belong to the same equivalence class in time polynomial in |x| + |y|. 2. For any finite set S ⊆ Σ * the equivalence relation R partitions the elements of S into a number of classes that is polynomially bounded in the size of the largest element of S.
Definition 2 (Cross-composition).Let L ⊆ Σ * be a language, let R be a polynomial equivalence relation on Σ * , let Q ⊆ Σ * × N be a parameterized problem, and let f : N → N be a function.An or-cross-composition of L into Q (with respect to R) of cost f (t) is an algorithm that, given t instances x 1 , x 2 , . . ., x t ∈ Σ * of L belonging to the same equivalence class of R, takes time polynomial in

(y
Theorem 3 ([3]).Let L ⊆ Σ * be a language, let Q ⊆ Σ * ×N be a parameterized problem, and let d, ε be positive reals.If L is NP-hard under Karp reductions, has an or-cross-composition into Q with cost f (t) = t 1/d+o (1) , where t denotes the number of instances, and Q has a polynomial (generalized) kernelization with size bound O(k d−ε ), then NP ⊆ coNP/poly.
For r ∈ N we will refer to an or-cross-composition of cost f (t) = t 1/r log(t) as a degreer cross-composition.By Theorem 2, a degree-r cross-composition can be used to rule out generalized kernels of size O(k r−ε ).We frequently use the fact that a polynomialtime linear-parameter transformation from problem Q to Q implies that any generalized kernelization lower bound for Q, also holds for Q (cf.[3,4]).Let [r] be defined as

4-Coloring
In this section we analyze the 4-Coloring problem, which asks whether it is possible to assign each vertex of the input graph one out of 4 possible colors, such that there is no edge whose endpoints share the same color.We show that 4-Coloring does not have a generalized kernel of size O(n 2−ε ), by giving a degree-2 cross-composition from a tailor-made problem that will be introduced below.Before giving the construction, we first present and analyze some of the gadgets that will be needed.

Definition 4.
A treegadget is the graph obtained from a complete binary tree by replacing each vertex v by a triangle on vertices r v , x v and y v .Let r v be connected to the parent of v and let x v and y v be connected to the left and right subtree of v.An example of a treegadget with 8 leaves is shown in Figure 1.If vertex v is the root of the tree, then r v is named the root of the treegadget.If v does not have a left subtree, then x v is a leaf of this gadget, similarly, if v does not have a right subtree then we refer to y v as a leaf of the gadget.Let the height of a treegadget be equal to the height of its corresponding binary tree.
It is easy to see that a treegadget is 3-colorable.The important property of this gadget is that if there is a color that does not appear on any leaf in a proper 3-coloring, then this must be the color of the root.See Figure 1a for an illustration.Lemma 5. Let T be a treegadget with root r and let c : Proof.This will be proven using induction on the structure of a treegadget.For a single triangle, the result is obvious.Suppose we are given a treegadget of height h and that the statement holds for all treegadgets of smaller height.Consider the top triangle r, x, y where r is the root.Then, by the induction hypothesis, the roots of the left and right subtree are colored using k.Hence x and y do not use color k.Since x, y, r is a triangle, r has color k in the 3-coloring.
The following lemma will be used in the correctness proof of the cross-composition to argue that the existence of a single yes-input is sufficient for 4-colorability of the entire graph.Lemma 6.Let T be a treegadget with leaves L ⊆ V (T ) and root r.Any 3-coloring c : L → {1, 2, 3} that is proper on T [L] can be extended to a proper 3-coloring of T .If there is a leaf v ∈ L such that c (v) = i, then such an extension exists with c(r) = i.
Proof.We will prove this by induction on the height of the treegadget.For a single triangle, the result is obvious.Suppose the lemma is true for all treegadgets up to height h − 1 and we are given a treegadget of height h with root triangle r, x, y and with coloring of the leaves c .Let one of the leaves be colored using i.Without loss of generality assume this leaf is in the left subtree, which is connected to x.By the induction hypothesis, we can extend the coloring restricted to the leaves of the left subtree to a proper 3-coloring of the left subtree such that c(r 1 ) = i.We assign color i to x.Since c restricted to the leaves in the right subtree is a proper 3-coloring of the leaves in the right subtree, by induction we can extend that coloring to a proper 3-coloring of the right subtree.Suppose the root of this subtree gets color j ∈ {1, 2, 3}.We now color y with a color k ∈ {1, 2, 3} \ {i, j}, which must exist.Finally, choose c(r) ∈ {1, 2, 3} \ {i, k}.By definition, the vertices r, y, and x are now assigned a different color.Both x and y have a different color than the root of their corresponding subtree, thereby c is a proper coloring.We obtain that the defined coloring c is a proper coloring extending c with c(r) = i.Definition 7. A triangular gadget is a graph on 12 vertices depicted in Figure 1c.Vertices u, v, and w are the corners of the gadget, all other vertices are referred to as inner vertices.
It is easy to see that a triangular gadget is always 3-colorable in such a way that every corner gets a different color.Moreover, we make the following observation.
Observation 8. Let G be a triangular gadget with corners u,v and w and let c : V (G) → {1, 2, 3} be a proper 3-coloring of G. Then c(v) = c(u) = c(w) = c(v).Furthermore, every partial coloring that assigns distinct colors to the three corners of a triangular gadget can be extended to a proper 3-coloring of the entire gadget.
Having presented all the gadgets we use in our construction, we now define the source problem for the cross-composition.It is a variant of the problem that was used to prove kernel lower bounds for Chromatic Number parameterized by vertex cover [3].

2-3-Coloring with Triangle Split Decomposition
for all x ∈ X?We will refer to such a coloring as a 2-3-coloring of G.

Lemma 9. 2-3-Coloring with Triangle Split Decomposition is NP-complete.
Proof.It is easy to verify the problem is in NP.We will show that it is NP-hard by giving a reduction from 3-nae-sat, which is known to be NP-complete [14].Suppose we are given formula way.For every variable x ∈ U , construct a gadget as depicted in Figure 2a.For every clause C i , construct a gadget as depicted in Figure 2b.
It is easy to verify that G has a triangle split decomposition.In Figure 2, triangles are shown with white vertices and the independent set is shown in black.
Suppose G is 2-3-colorable with color function c : V (G) → {1, 2, 3} and let c(v) ∈ {1, 2} for all v in the independent set.Note that in each of the pairs {x, ¬x}, {b 1 , b 2 }, and {u 1 , u 2 } the two vertices have distinct colors in any proper 2-3-coloring of G. To satisfy F , let x = true if and only if c(x) = 2.To show that this results in a satisfying assignment, consider any clause and c(b 1 ), c(b 2 ) ∈ {1, 2} we obtain c(r 0 ) = 3.Therefore, v 1 and u 0 are colored using colors 1 and 2.
Suppose c(v 1 ) = 1.Thereby, c( 1 ) = 2, implying the first literal of C i is set to true.By c(u 0 ) = 2, we know c(u 1 ) = 1 and c(u 2 ) = 2. Thereby, c(r When c(v 1 ) = 2, we can use the same argument with the colors 1 and 2 swapped, to show that 1 is f alse in C i and 2 or 3 is true, which implies that C i is NAE-satisfied.
Suppose F is a yes-instance, with satisfying truth assignment S. Define color function c : V (G) → {1, 2, 3} as c(x) := 1 and c(¬x) := 2 if x is set to false in S, define c(x) := 2 and c(¬x) := 1 otherwise.Color the remainder of the variable gadgets consistently.We now need to show how to color the clause gadgets.Consider any clause C i = ( 1 ∨ 2 ∨ 3 ).At least one of the literals is true and one is set to false, by symmetry we only consider four cases.The corresponding colorings are depicted in Figure 3, where red corresponds to 1, green corresponds to 2 and blue corresponds to color 3.It is easy to verify that this leads to a proper 3-coloring that only uses colors 1 and 2 on vertices in the independent set.Note that if the roles of 2 and 3 are exactly reversed, you can just exchange colors between their parents to get a proper coloring for that situation.
vertices.For ease of presentation, we will actually give a cross-composition into the 4-List Coloring problem, whose input consists of a graph G and a list function that assigns every vertex v ∈ V (G) a list L(v) ⊆ [4] of allowed colors.The question is whether there is a proper coloring of the graph in which every vertex is assigned a color from its list.The 4-List Coloring reduces to the ordinary 4-Coloring by a simple transformation that adds a 4-clique to enforce the color lists, which will prove the theorem.For now, we focus on giving a cross-composition into 4-List Coloring.
We start by defining a polynomial equivalence relation on inputs of 2-3-Coloring with Triangle Split Decomposition.Let two instances of 2-3-Coloring with Triangle Split Decomposition be equivalent under equivalence relation R when they have the same number of triangles and the independent sets also have the same size.It is easy to see that R is a polynomial equivalence relation.By duplicating one of the inputs, we can ensure that the number of inputs to the cross-composition is an even power of two; this does not change the value of or, and increases the total input size by at most a factor four.We will therefore assume that the input consists of t instances of 2-3-Coloring with Triangle Split Decomposition such that t = 2 2i for some integer i, implying that √ t and log √ t are integers.Let t := √ t.Enumerate the instances as X i,j for 1 ≤ i, j ≤ t .Each input X i,j consists of a graph G i,j and a partition of its vertex set into sets U and V , such that U is an independent set of size m and G i,j [V ] consists of n vertex-disjoint triangles.Enumerate the vertices in U and V as u 1 , . . ., u m and v 1 , . . ., v 3n , such that vertices v 3 −2 , v 3 −1 and v 3 form a triangle, for ∈ [n].We will create an instance G of the 4-List-Coloring problem, which consists of a graph G and a list function L that assigns each vertex a subset of the color palette {x, y, z, a}.Refer to Figure 4 for a sketch of G .

Initialize
Proof.(⇒) Suppose we are given a 4-list coloring c for G .By definition, c(r S ) = a.From Lemma 5 it follows that there is a leaf v of G S such that c(v) = a.This leaf is connected to all vertices in some S i * , which implies that none of the vertices in S i * are colored using a.Therefore all vertices in S i * are colored using x and y.Similarly the gadget G T has at least one leaf v such that c(v) = a, note that this must be a leaf with an odd index.Therefore there exists T j * where all vertices are colored using x,y or z.Thereby in S i * ∪ T j * only three colors are used, such that S i * is colored using only two colors.Using Observation 8 and the fact that G [S i * ∪ T j * ] is isomorphic to the graph obtained from G i * ,j * by replacing triangles by triangular gadgets, we conclude that , furthermore let c (t j ) := z for j = j * and ∈ [3m].For triangular gadgets in T j * the coloring c defines all corners to have distinct colors; by Observation 8 we can color the inner vertices consistently using {x, y, z}.For T j with j ∈ [t ] and j = j * , the corners of triangular gadgets have color z and we can now consistently color the inner vertices using {x, y, a}.
The leaf of gadget G S that is connected to S i * can be colored using a.Every other leaf can use both x and y, so we can properly 3-color the leaves such that one leaf has color a.From Lemma 6 it follows that we can consistently 3-color G S such that the root r S does not receive color a, as required by L(r S ).Similarly, in triangular gadgets in T j * the inner vertices do not have color a.As such, leaf 2j * − 1 of G T can be colored using a and we color leaf 2j * with y.For j ∈ [t ] with j = j * color leaf 2j − 1 with z and leaf 2j using y.Now the leaves of G T are properly 3-colored and one is colored a.It follows from Observation 8 that we can color G T such that the root is not colored a.This completes the 4-list coloring of G .
The claim shows that the construction serves as a cross-composition into 4-List Coloring.
To prove the theorem, we add four new vertices to simulate the list function.Add a clique on 4 vertices {x, y, z, a}.If for any vertex v in G , some color is not contained in L(v), connect v to the vertex corresponding to this color.As proper colorings of the resulting graph correspond to proper list colorings of G , the resulting graph is 4-colorable if and only if there is a yes-instance among the inputs.It remains to bound the parameter of the problem, i.e., the number of vertices.Observe that a treegadget has at least as many leaves as its corresponding binary tree, therefore the graph G has at most 12mt + nt Theorem 10 now follows from Theorem 3 and Lemma 9.

Hamiltonian cycle
In this section we prove a sparsification lower bound for Hamiltonian Cycle and its directed variant by giving a degree-2 cross-composition.The starting problem is Hamiltonian s − t path on bipartite graphs.It is known that Hamiltonian path is NP-complete on bipartite graphs [14] and it is easy to see that is remains NP-complete when fixing a degree 1 start and endpoint.Proof.By a suitable choice of polynomial equivalence relation, and by padding the number of inputs, it suffices to give a cross-composition from the s − t problem on bipartite graphs when the input consists of t instances X i,j for i, j ∈ [ √ t] (i.e., √ t is an integer), such that each instance X i,j encodes a bipartite graph G i,j with partite sets A * i,j and  5a.Observe that if G contains a path gadget as an induced subgraph, while the remainder of the graph only connects to its terminals in 0 and in 1 , then any Hamiltonian cycle in G traverses the path gadget in one of the two ways depicted in Figure 5a.We create an instance G of Directed Hamiltonian Cycle that acts as the logical or of the inputs.Path 0

For every input instance
, add an arc from in 0 of a i k to in 1 of b j and an arc from in 0 of b j to in 1 of a i k .
If some X i,j has a Hamiltonian s − t path, it can be mimicked by the combination of A i and B j , where for each vertex in X i,j we traverse its path gadget in G , following Path 1.The following construction steps are needed to extend such a path to a Hamiltonian cycle in G .This completes the construction of G .A sketch of G is shown in Figure 5b.In order to prove that the created graph G acts as a logical or of the given input instances, we first establish a number of auxiliary lemmas.

Lemma 13. Any Hamiltonian cycle in G traverses any path gadget in G via directed
Path 0 or Path 1, as shown in Figure 5a.
Proof.Any Hamiltonian cycle in G should visit the center vertex of the path gadget.Since in 0 and in 1 are its only two neighbors in G , the only option is to visit them consecutively, Path 0 and Path 1 are the only two options to do this.

Lemma 14. When any Hamiltonian cycle in G enters path gadget a i
1 at in 0 for some i ∈ [ √ t], the cycle then visits the gadgets a i 2 , a i 3 , . .

. , a i m in order without visiting other vertices in between. Similarly, if any Hamiltonian cycle in G enters path gadget b j
1 at in 0 , the cycle then visits the gadgets b j 2 , b j 3 , . . ., b j n in order without visiting other vertices in between.
Proof.Consider a Hamiltonian cycle in G that enters path gadget a i 1 at in 0 .By Lemma 13 the cycle follows Path 0 and continues to the in 1 terminal of the path gadget.Since that terminal has only one out-neighbor outside the gadget, which leads to the in 0 terminal of a i 2 , it follows that the cycle continues to that path gadget.As the adjacency structure around the other path gadgets is similar, the lemma follows by repeating this argument.The proof when entering group B j at the vertex in 0 of b j 1 is equivalent.
Lemma 15.Let C be a directed Hamiltonian cycle in G , such that its first arc is (start, x 1 ).There are indices i * , j * ∈ [ √ t] such that subpath C x1,y2r of the cycle between x 1 and y 2r contains exactly the vertices where A i * contains all vertices of all gadgets in A i for i = i * , and similarly B j * contains all vertices of all gadgets in B j for j = j * .
Proof.We will first show that when the cycle reaches any x i for i ∈ [r], it traverses exactly one group A with ∈ [r + 1] and continues to y j and x j+1 for some j ∈ [r], without visiting other vertices in between.Similarly, when the cycle reaches any x i for r < i ≤ 2r, it traverses exactly one group B with ∈ [r + 1] and continues to y j for some r < j ≤ 2r.For j < 2r, the cycle then continues to x j+1 , for j = 2r the cycle reached y 2r , which is the last vertex of this subpath.
By Step 6 in the construction, all outgoing arcs of any x i for i ∈ [r] lead to a 1 for some ∈ [ √ t].So for any x i in the cycle there must be a unique ∈ [ √ t] such that the arc from x i to the in 0 terminal of a 1 is in C. By Lemma 14 the cycle visits all vertices in A , and no other vertices, before reaching gadget a m , which is traversed by Path 0 to get to in 1 of this gadget.The only neighbors of in 1 of gadget a m lying outside this gadget are of type y j for j ∈ [r].As such, the cycle must visit some y j next, and its only outgoing arc goes to x j+1 .
The proof for i > r is similar.As such, visiting x i for i ∈ [r] results in visiting all vertices of exactly one group in A before continuing via y j to some x j+1 without visiting any vertices in between.Visiting x i for r < i ≤ 2r results in visiting all vertices of exactly one group in B and returning via y j to either the end of the subpath (j = 2r) or some x j+1 .
Every vertex x i for i ∈ [2r] must be visited by C, it remains to show that it is visited in subpath C x1,y2r .Suppose there exists an x i for i ∈ [2r] such that x i is not visited in the subpath from x 1 to y 2r .As we have seen above, visiting some x i results in visiting all vertices in some group in A or B, continued by visiting some y j for j ∈ [2r].Note that no other vertices are visited in between.Hereby, y j is not in subpath C x1,y2r .This implies j = 2r and thus the next vertex in the cycle is x j+1 .So, for x i not in subpath C x1,y2r , one can find a new vertex x j+1 (where j + 1 = i), such that x j+1 is also not in subpath C x1,y2r .Note that we can not create a loop, by visiting a vertex x i seen earlier, as this would not yield a Hamiltonian cycle in G .For example, the vertex start would never be visited.This is however a contradiction since we only have finitely many vertices x i .
Thus in subpath C x1,y2r , exactly r groups of A are visited and exactly r groups of B are visited, and no other vertices than specified.This leaves exactly one group A i * and one group B j * unvisited in C x1,y2r .
In Step 6 we create a selection mechanism that leaves one group in A and one in B unvisited.The following lemma formalizes this idea.Proof.Vertex next is visited directly after y 2r , since it is the only out-neighbor of y 2r .Furthermore, the arc from next to gadget b 1 must be in the cycle for some ∈ [ √ t], since next only has outgoing arcs of this type.By Lemma 15, all gadgets in all B j for j = j * are visited in the path from x 1 to y 2r , and thus should not be visited after vertex next.Therefore, the arc from next to gadget b j * 1 is in the cycle, which also implies that b 1 is entered at terminal in 1 , it is easy to see that all gadgets are traversed using Path 1.We now construct a Hamiltonian path P for instance Using that every gadget is visited exactly once via Path 1 in C, we see that C is a Hamiltonian path.
(⇒) Suppose X i * ,j * has a Hamiltonian s − t path P .Then we create a Hamiltonian cycle C, for each vertex a * from instance X i * ,j * in P we add Path 1 in path gadget a i * to C and for each vertex b * we add Path 1 in path gadget b j * to C. Let P be ordered such that b * 1 is its first vertex.Now if a * k is followed by b * in P , the arc from terminal in 0 of From b j * n the cycle goes to end, then to start and to x 1 .To visit all groups A i for i = i * and B j for j = j * , do the following.
From x i where i < i * , the cycle continues to gadgets a i 1 , then to a i 2 , a i 3 , . . ., a i m following Path 0, and continue to y i , x i+1 .From x i where i * ≤ i ≤ r it goes to a i+1 1 , a i+1 2 , . . ., a i+1 n and continues with y i , x i+1 .

R B
Figure 6 A sketch of G, where t = 2, m = 6 , n = 5 and k = 2. Thereby K should be 5 and W (c 1 ,c 2 ) should contain 10 vertices.In this example we show the constructed graph when choosing K = 1 for simplicity.We use the two colors c1 and c2, corresponding to white and black in the figure.Edges from R to B are left out for simplicity.

6.
Then, add log t triangles, with vertices {t 0 , t 1 , t 2 } for ∈ [log t ].Connect t 0 to all vertices in B j if the 'th bit of j equals 0, connect t 1 to all vertices in B j if the 'th bit of j equals 1. Define T to be the union of all these triangles.By choosing exactly one of the vertices t 0 or t 1 in a dominating set for each , all groups B j except one are dominated automatically.The non-dominated one should then be part of a solvable input instance.7. Finally, add the edges {{s, t i } | ∈ [log t ], i ∈ {0, 1}}.This step ensures that every vertex in T that is contained in the dominating set has s as a neighbor in the dominating set, which implies that there is always a minimum dominating set that is connected.
We now make the following observations.Lemma 20.If G has a dominating set D, then it also has a dominating set D of size at most |D| that does not contain any vertices from B.
Proof.Suppose we are given a minimum dominating set D of G, where vertex v ∈ B is present.In any dominating set, s or s must be present.If s is present and s is not, we replace s by vertex s, and still obtain a valid dominating set of the same size.As such, all vertices in R are now dominated by s.Vertices t 0 and t 1 with ∈ [log t ] are dominated by s.Since t 2 only has neighbors t 1 and t 0 , at least one of these three vertices is present in D for every ∈ [log t ], hereby every vertex in T has a neighbor in D.
Since B is an independent set in G, the vertex v does not dominate other vertices in B. Since the polynomial equivalence relation ensures that there are no isolated vertices in B, vertex v has at least one neighbor u in R. We can safely replace v by u to obtain a valid dominating set that has the same size as D and does not contain any vertices from B. Lemma 21.Any dominating set of G of size at most k + 1 + log t contains at least 1 + log t vertices from {s, s } ∪ {t 0 , t 1 , t 2 | ∈ [log t ]} and thus contains at most k vertices from R.
Proof.In a dominating set D of G, at least log t vertices are needed from T , since t 2 only has neighbors t 1 and t 0 , so one of these vertices must be in D for each ∈ [log t ].Furthermore at least one of the vertices s or s must be present, therefore there are 1 + log t vertices in the set that are not from R.

Lemma 22. Any dominating set of G of size at most k + 1 + log t uses exactly one vertex of each color from R.
Proof.Suppose a dominating set of G of size at most k + 1 + log t uses less than k colors from R. If at most k − 2 colors are used, there must be two colors c 1 and c 2 that are not present in the set.However, this implies that all 2K vertices in W (c1,c2) are not dominated by vertices in R and must therefore be in the set.This contradicts the maximum size of the dominating set, since K = k + 2 + log t .So, we are left with the possibility of using k − 1 colors.Consider some color c 1 that was not used.Look at another color c 2 that is used exactly once, such a color exists by Lemma 21.Suppose the vertex of color c 2 in the dominating set was from set R i for some i ∈ [t ].Then for any x ∈ id(R i ) we have that w is not connected to any vertex in the dominating set and therefore must be in the dominating set itself.Since id(R i ) contains K numbers, there are K vertices that are not dominated by R, which contradicts the maximum size of the dominating set.

Lemma 23. For any dominating set D of G of size at most
Proof.Suppose there exists two vertices u, v ∈ D such that u ∈ R i and v ∈ R j for some i = j.By Lemma 22, u and v have different colors.Suppose u has color c u and v has color Step 5 of the construction, this means that none of the neighbors of vertex w (cu,cv) x are contained in the dominating set.However, this vertex is not in D and therefore D is not a dominating set of G, which is a contradiction.
Using the previous Lemmas, we obtain: Lemma 24. 1.If there is an input X i * ,j * that has a col-RBDS of size k, then G has a connected dominating set of size k + 1 + log t .2. If G has a (not necessarily connected) dominating set of size k + 1 + log t , then some input X i * ,j * has a col-RBDS of size k.
Proof.(1) Let X i * ,j * have a colored RBDS D of size at most k, then we can construct a dominating set D of G in the following way.For any vertex r * p,q in D, add vertex r i p,q to D .Furthermore add the vertex s to D .Then add vertex t 0 to D if the q'th bit of j * is 1, add vertex t 1 otherwise.Now s is dominated and all vertices in R have neighbor s in D .All vertices in B j * are covered by the vertices in the dominating set from R i * , since D was a col-RBDS of X i * ,j * .All vertices in B j for j = j * have neighbor t 0 or t 1 in D for some ∈ [log t ], since the bit representation of j must differ from the one of j * at some position.It now follows from Step 6 of the construction that all vertices in B j are connected to a vertex in the dominating set.It remains to verify that all vertices in W have a neighbor in D .Consider w , then this vertex is connected to all vertices of color c 1 and exactly one of them is contained in D .If x / ∈ id(R i * ), the vertex w (c1,c2) x is connected to all vertices of color c 2 in R i * and again one vertex of this color in R i * is contained in D .So D is a dominating set of G and it is easy to verify that |D | = k + 1 + log t .Furthermore, D is constructed in such a way that it is connected.We can show this by proving that every vertex in D is a neighbor of s, since we chose s in D. Vertices in D ∩ R and D ∩ T are neighbors of s, by Steps 4 and 7 of the construction of G.
The vertex s and vertices from W and B are not contained in D .Thus, D is a connected dominating set.
(2) Let D be a dominating set of G of size at most k + 1 + log t .Using Lemma 20 we modify D such that it chooses no vertices from B, without increasing its size.By Lemma 22 and 23, D contains exactly k vertices from R, all from the same R i * for some i * and all of different color.D has size at most k + 1 + log t of which k are contained in R and one in {s, s }.Combined with the fact that for any ∈ [log t ] vertex t 2 has t 1 and t 0 as its only two neighbors, it follows that exactly one of these three vertices is contained in D for all .Therefore D contains at most one of the vertices t 0 or t 1   Just as the sparsification lower bounds for Vertex Cover that were presented by Dell and van Melkebeek [8] had implications for the parameterization by the solution size k, Theorem 19 has implications for the kernelization complexity of k-Nonblocker and k-Max Leaf.Since the solution size k never exceeds the number of vertices in this problem, a kernel with O(k 2− ) edges would give a nontrivial sparsification, contradicting Theorem 19.Hence our results show that the existing linear-vertex kernels for k-Nonblocker [6] and k-Max Leaf [11] cannot be improved to O(k 2−ε ) edges unless NP ⊆ coNP/poly.

d-Hypergraph 2-Colorability and d-NAE-SAT
The goal of this section is to give a nontrivial sparsification algorithm for nae-sat and prove a matching lower bound.For ease of presentation, we start by analyzing the closely related hypergraph 2-colorability problem.
Proof.Suppose we are given a hypergraph with vertex set V and hyperedges E, where each hyperedge contains at most d vertices.We show how to reduce the number of hyperedges without changing the 2-colorability status.Let E r ⊆ E denote the set of edges in E that contain exactly r vertices.For each E r we construct a set E r ⊆ E r of representative hyperedges.Enumerate the edges in E r as e r 1 , . . ., e r k .We construct a (0, 1)-matrix M r with N := n r−1 rows and k columns.Consider all possible subsets A 1 , . . ., A N of size r − 1 of the set of vertices V .Define the elements m i,j for i ∈ N and j ∈ k of M r as follows.
m i,j := 1 if A i ⊆ e r j ; 0 otherwise.Using Gaussian elimination, compute a basis B of the columns of this matrix, which is a subset of the columns that span the column space of M r .Let E r contain edge e r i if the i'th column of M r is contained in B, and define E := r∈[d] E r , which forms the kernel.Using a lemma due to Lovász [20], we can prove that E preserves the 2-colorability status.
Lemma 26 ([20]).Let H be an r-uniform hypergraph with edges E 1 , . . ., E m .Let α 1 , . . ., α m be real numbers such that for every (r − 1)-element subset A of V (H), Then for every partition {V 1 , V 2 } of V (H) the following holds: (⇐) Now suppose (V, E ) has a proper 2-coloring.We will show that for each r ∈ [d], no edge of E r is monochromatic under this coloring.All hyperedges contained in E r are 2-colored by definition.Suppose there exists r ∈ [d], such that E r contains a monochromatic hyperedge.Let E r = e r 1 , . . ., e r k and let e i * be a hyperedge in E r whose vertices all receive the same color.
By reordering the matrix M r , we may assume that the basis B of M r contains the first columns, thus i * > .Let m i denote the i'th column of M r .Since m i * is not contained in the basis, there exist coefficients α 1 , . . ., α such that For i ∈ [k], define: Let A j be any size (r − 1)-subset of V .Since m i,j = 1 exactly when e i ⊇ A j , and 0 otherwise, we have: By Lemma 26 we obtain that for any partitioning Consider however the partitioning (V 1 , V 2 ) given by the 2-coloring of the vertices.Then every edge e i ∈ E r contains at least one vertex of each color and is thereby not fully contained in V 1 or V 2 .As such, these edges contribute 0 to both sides of the equation.The edge e i * is the only remaining edge with a non-zero coefficient and by assumption, it is contained entirely within one color class.Without loss of generality, let e i * ⊆ V 1 .But then ei⊆V1 β i = −1 while (−1) r ei⊆V2 β i = 0, which contradicts (1).
To bound the size of the kernel, consider the matrix M r for r ∈ [d].Its rank is bounded by the minimum of its number of rows and columns, which is at most When additionally adding n hyperedges {x i , ¬x i } for i ∈ [n], it is easy to see that the resulting hypergraph is 2-colorable if and only if there is a NAE-satisfying assignment to the formula.The maximum size of a hyperedge matches the maximum size of a clause and the number of created vertices is twice the number of variables.We can therefore sparsify an n-variable instance of d-nae-sat in the following way: reduce it to a d-hypergraph with n := 2n vertices and apply the kernelization algorithm of Theorem 25.It is easy to verify that restricting the formula to the representative hyperedges in the kernel gives an equisatisfiable formula containing 2 • (n ) d−1 ∈ O(2 d−1 n d−1 ) clauses, giving a sparsification for nae-sat.As mentioned in the introduction, the existence of a linear-parameter transformation [18] from d-cnf-sat to (d + 1)-nae-sat also implies a sparsification lower bound for d-nae-sat, using the results of Dell and van Melkebeek [8].Hence we obtain the following theorem.

Conclusion
We have added several classic graph problems to a growing list of problems for which non-trivial polynomial-time sparsification is provably impossible under the assumption that NP coNP/poly.Our results for (Connected) Dominating Set proved that the linearvertex kernels with Θ(k 2 ) edges for k-Nonblocker and k-Max Leaf Spanning Tree cannot be improved to O(k 2−ε ) edges unless NP ⊆ coNP/poly.
The graph problems for which we proved sparsification lower bounds can be defined in terms of vertices: the 4-Coloring problem asks for a partition of the vertex set into four independent sets, Dominating Set asks for a dominating subset of vertices, and Hamiltonian Cycle asks for a permutation of the vertices that forms a cycle.In contrast, not much is known concerning sparsification lower bounds for problems whose solution is an edge subset of possibly quadratic size.For example, no sparsification lower bounds are known for well-studied problems such as Max Cut, Cluster Editing, or Feedback Arc Set in Tournaments.Difficulties arise when attempting to mimic our lower bound constructions for such edge-based problems.Our constructions all embed t instances into a 2 × √ t table, using each combination of a cell in the top row and bottom row to embed one input.For problems defined in terms of edge subsets, it becomes difficult to "turn off" the contribution of edges that are incident on vertices that do not belong to the two cells that correspond to a yes-instance among the inputs to the or-construction.This could be interpreted as evidence that edge-based problems such as Max Cut might admit non-trivial polynomial sparsification.We have not been able to answer this question in either direction, and leave it as an open problem.For completeness, we point out that Karp's reduction [19] from Vertex Cover to Feedback Arc Set (which only doubles the number of vertices) implies, using existing bounds for Vertex Cover [8], that Feedback Arc Set does not have a compression of size O(n 2−ε ) unless NP ⊆ coNP/poly.
Another problem whose compression remains elusive is 3-Coloring.In several settings (cf.[12]), the optimal kernel size matches the size of minimal obstructions in a problem-specific partial order.This is the case for d-nae-sat, whose kernel with O(n d−1 ) clauses matches the fact that critically 3-chromatic d-uniform hypergraphs have at most O(n d−1 ) hyperedges.Following this line of reasoning, it is tempting to conjecture that 3-Coloring does not admit subquadratic compressions: there are critically 4-chromatic graphs with Θ(n 2 ) edges [22].
The kernel we have given for d-nae-sat is one of the first examples of non-trivial polynomial-time sparsification for general structures that are not planar or similarly guaranteed to be sparse.Obtaining non-trivial sparsification algorithms for other problems is an interesting challenge for future work.Are there natural problems defined on general graphs that admit subquadratic sparsification?

t 1 .
i=1 |x i | and outputs an instance (y, k) ∈ Σ * × N such that: the parameter k is bounded by O(f (t)•(max i |x i |) c ), where c is some constant independent of t, and leaves root (a) Treegadget with no red leaf.

Figure 1
Figure 1 Used gadgets with example colorings.

Figure 2
Figure 2The gadgets constructed for the clauses and variables of F .

Theorem 10 . 4 -
Coloring parameterized by the number of vertices n does not have a generalized kernel of size O(n 2−ε ) for any ε > 0, unless NP ⊆ coNP/poly.Proof.By Theorem 3 and Lemma 9 it suffices to give a degree-2 cross-composition from the 2-3-coloring problem defined above into 4-Coloring parameterized by the number of

Figure 3
Figure 3 Valid colorings of a clause gadget, depending on the coloring of the literals 1, . . ., 3.Note that if the roles of 2 and 3 are exactly reversed, you can just exchange colors between their parents to get a proper coloring for that situation.

Figure 4
Figure 4The graph G for t = 4, m = 3 and n = 2. Edges between vertices in S and T are left out for simplicity.

4 . 5 .Claim 11 .
Add a treegadget G S with t leaves to G and enumerate these leaves as 1, . . ., t ; recall that t is a power of two.Connect the i'th leaf of G S to every vertex in S i .Let the root of G S be r S and define L(r S ) := {x, y}.For every other vertex v in G S let L(v) := {x, y, a}.Add a treegadget G T with 2t leaves to G and enumerate these leaves as 1, . . ., 2t .For j ∈ [t ], connect every inner vertex of a triangular gadget in group T j to leaf number 2j − 1 of G T .For every leaf v with an even index let L(v) := {y, z} and let the root r T have list L(r T ) := {y, z}.For every other vertex v of gadget G T let L(v) := {y, z, a}.The graph Hamiltonian s − t path on bipartite graphs Input: An undirected bipartite graph G with partite sets A and B such that |B| = n = |A| + 1, together with two distinguished vertices b 1 and b n that have degree 1. Question: Does G have a Hamiltonian path from b 1 to b n ?

Theorem 12 .
(Directed) Hamiltonian Cycle parameterized by the number of vertices n does not have a generalized kernel of size O(n 2−ε ) for any ε > 0, unless NP ⊆ coNP/poly.
The general structure of the graph, when given 4 inputs with n = 3 and m = 4.

Figure 5
Figure 5 Illustrations for the lower bound for Hamiltonian Cycle.

1 4 . 5 . 6 . 8 .
terminal of a i to the in 0 terminal of a i +1 for all ∈ [m − 1] and all i ∈ [ √ t].Similarly add an arc from the in 1 terminal of b i to the in 0 of b i +1 for all ∈ [n − 1] and all i ∈ [ √ t].Add a vertex start and a vertex end and the arc (end, start).Let r := √ t − 1, add 2r tuples of vertices, x i , y i for i ∈ [2r] and connect start to x 1 .Furthermore, add the arcs (y i , x i+1 ) for i ∈ [2r − 1].For i ≤ r we add arcs from x i to the in 0 terminal of the gadgets a j 1 , j ∈ [ √ t].Furthermore we add an arc from in 1 of a j m to y i for all j ∈ [ √ t] and i ∈ [r].When i > r add arcs from x i to the in 0 terminal of b j 1 for j ∈ [ √ t] and connect in 1 of b j n to y i .7. Add a vertex next and the arc (y 2r , next) and an arc from next to the in 1 terminal of all gadgets b j 1 for j ∈ [ √ t].Furthermore, add arcs from in 0 of all gadgets b j n to end for j ∈ [ √ t].So for each B j , exactly one vertex has an outgoing arc to end and one has an incoming arc from next.

Lemma 16 .
Let C be a Hamiltonian cycle in G , such that its first arc is (start, x 1 ).Let i * and j * satisfy the conditions of Lemma 15.Then cycle C visits b j * 1 before b j * n .Moreover, the subpath of the cycle C b j * 1 ,b j * n between terminal in 1 of b j * 1 and in 0 of b j * n (inclusive) contains all vertices of the gadgets in A i * and B j * and no others.
Furthermore, G has a connected dominating set of size k + 1 + log n if and only if one of the input instances has a col-RBDS of size k.Therefore we have given a degree-2 cross-composition to (Connected) Dominating Set.Using Theorem 3 it follows that Dominating Set and Connected Dominating Set do not have a generalized kernel of size O(n 2−ε ) for any ε > 0, unless NP ⊆ coNP/poly.
Now we can prove the correctness of the presented kernel.Lemma 27. (V, E) has a proper 2-coloring ⇔ (V, E ) has a proper 2-coloring.Proof.(⇒) Clearly, if (V, E) has a proper 2-coloring, then the same coloring is proper for the subhypergraph (V, E ) since E ⊆ E.
at most 2n d−1 hyperedges.Since a hyperedge consists of at most d vertices, the kernel can be encoded in O(n d−1 • d • log n) bits.By a folklore reduction, Theorem 25 gives a sparsification for nae-sat.Consider an instance of d-nae-sat, which is a conjunction of clauses of size at most d over variables x 1 , . . ., x n .The formula gives rise to a hypergraph on vertex set {x i , ¬x i | i ∈ [n]} containing one hyperedge per clause, whose vertices correspond to the literals in the clause.

Theorem 28 .
For every fixed d ≥ 4, the d-nae-sat problem parameterized by the number of variables n has a kernel with O(n d−1 ) clauses that can be encoded in O(n d−1 • log n) bits, but admits no generalized kernel of size O(n d−1−ε ) for ε > 0 unless NP ⊆ coNP/poly.
G as the graph containing t sets of m vertices each, called S i for i ∈ [t ].Label the vertices in each of these sets as s Add t sets of n triangular gadgets each, labeled T j for j ∈ [t ].Label the corner vertices in T j as t j for ∈ [3n], such that vertices t j 3 −2 , t j 3 −1 and t j 3 are the corner vertices of one of the gadgets for ∈ [n].Let L(t j ) := {x, y, z} and for any inner vertex v of a triangular gadget, let L(v) := {x, y, z, a}.3.Connect vertex si k to vertex t j if in graph G i,j vertex u k is connected to v , for k ∈ [m] and ∈ [3n].By this construction, the subgraph of G induced by S i ∪ T j is isomorphic to the graph obtained from G i,j by replacing each triangle with a triangular gadget.
for some m ∈ N.For each instance, label all elements in A * i,j as a * 1 , . . ., a * m and all elements in B * i,j as b * 1 , . . ., b * n such that b * 1 and b * n have degree 1.The construction makes extensive use of the path gadget depicted in Figure

1 .
First of all construct √ t groups of m path gadgets each.Refer to these groups as A i , for i ∈ [ √ t], and label the gadgets within group A i as a i 1 , . . ., a i m .Let the union of all created sets A i be named A. Similarly, construct √ t groups of n path gadgets each.Refer to these groups as B j , for j ∈ [ √ t], and label the gadgets within group B j as b j 1 , . . ., b j n .Let B be the union of all B j for j ∈ [ √ t].
1 is visited before b j * n .It is easy to see that (end, start) is the last arc in C. By considering the incoming arcs of end it follows that some arc from terminal in 0 of b n to end for ∈ [ √ t] is in the cycle.Since the vertices in gadgets b n for = j * are already visited in C x1,y2r by Lemma 15, it follows that (b j * n , end) is in C. By Lemma 15, none of the terminals of gadgets in A i * and B j * are visited in the subpath C x1,y2r or equivalently in the subpath C start,next .Since C is a Hamiltonian cycle these vertices must therefore be visited in C next,start , which is equivalent to saying that C b j * 1 ,b j * n must contain all vertices in A i * ∪ B j * .It is easy to see that this subpath cannot contain any other vertices, as all other vertices are present in C start,next or C end,start .Using the lemmas above, we can now prove that G has a Hamiltonian cycle if and only if one of the instances has a Hamiltonian path.Graph G has a directed Hamiltonian cycle if and only if at least one of the instances X i,j has a Hamiltonian s − t-path.
* Proof.(⇐) Suppose G has a Hamiltonian cycle C.By Lemma 16 there exist i * , j * ∈ [ √ t] such that the subpath of C from gadget b j * 1 to b j * n visits exactly the gadgets in A i * ∪B j * .Since gadget b j * for every ∈ [log t ].We can now define x ∈ {0, 1} for ∈ [log t ], such that t x / ∈ D for all ∈ [log t ].Consider the index j * ∈ [t] given by the binary representation [x 1 x 2 . . .x log t ] 2 .It follows from the bit representation of j * that the vertices in B j * are not connected to any of the vertices in D ∩ T .Since vertices in B j * are only adjacent to vertices in R and vertices of T , it follows that every vertex in B j * has a neighbor in R that is in D .This implies that every vertex in B j * has a neighbor in D ∩ R i * .Since G[R i * ∪ B j * ] is isomorphic to the graph of instance X i * ,j * , it follows that X i * ,j * has a col-RBDS of size at most k, which are exactly the vertices in D ∩ R i * .Given t instances, the graph G constructed above has n•t +m•t +2+3•log t +2 k 2 •2K = O( √ t max |X i,j | 2 ) vertices.It is straightforward to construct G in polynomial time.It follows from Lemma 24 that G has a dominating set of size k + 1 + log t , if and only if one of the input instances has a col-RBDS of size k.
Recall that a hypergraph consists of a vertex set V and a set E of hyperedges; each hyperedge e ∈ E is a subset of V .A 2-coloring of a hypergraph is a function c : V → {1, 2}; such a coloring is proper if there is no hyperedge whose vertices all obtain the same color.We will use d-Hypergraph 2-Colorability to refer to the setting where hyperedges have size at most d.The corresponding decision problem asks, given a hypergraph, whether it is 2-colorable.
Theorem 25. d-Hypergraph 2-Colorability parameterized by the number of vertices n has a kernel with 2 • n d−1 hyperedges that can be encoded in O