1 Background and New Challenges
1.1 Combinatorial Problems and NLCompleteness
Given a combinatorial problem, it is desirable, for practical reason, to seek for good algorithms that consume fewer computational resources in order to solve the problem, and therefore it is of great importance for us to identify the smallest amount of computational resources required to execute such algorithms. Of various resources, we are focused in this exposition on the smallest “memory space” used by an algorithm that runs within certain reasonable “time span.” The study on the minimal memory space has attracted significant attention in reallife circumstances at which we need to manage vast data sets for most network users who operate memorylimited computing devices. It is therefore useful in general to concentrate on the study of spacebounded computability within reasonable execution time. In the past literature, special attention has been paid to polynomialtime algorithms using logarithmic memory space and two corresponding spacebounded complexity classes: (deterministic logarithmic space) and (nondeterministic logarithmic space).
In association with and , various combinatorial problems have been discussed by, for instance, Jones, Lien, and Laaser [8], Cook and McKenzie [3], Àlvarez and Greenlaw [1], and Jenner [7]. Many graph properties, in particular, can be algorithmically checked using small memory space. Using only logarithmic space^{3}^{3}3Those problems were proven to be in by Reif [10] and complete by Àlvarez and Greenlaw [1], where is the symmetric version of . Since by Reingold [11], nonetheless, all the problems are in . (cf. [1, 10]), for instance, we can easily solve the problems of determining whether or not a given graph is a bipartite graph, a computability graph, a chordal graph, an interval graph, and a split graph. On the contrary, the directed  connectivity problem (DSTCON) and the 2CNF Boolean formula satisfiability problem (2SAT) are known to be complete [8] (together with the result of [6, 12]) and seem to be unsolvable using logarithmic space. To understand the nature of better, it is greatly beneficial to study more interesting problems that fall into the category of complete problems.
1.2 Parameterization of Problems and the Linear Space Hypothesis
Given a target combinatorial problem, it is quite useful from a practical viewpoint to “parameterize” the problem by introducing an adequate “size parameter” as a unit basis of measuring the total amount of computational resources, such as runtime and memory space needed to solve the problem. As quick examples of size parameters, given a graph instance , and respectively denote the total number of the vertices of and the total number of the edges in . For a CNF Boolean formula , in addition, and respectively express the total number of distinct variables in and the total number of clauses in . Decision problems coupled with appropriately chosen size parameters are generally referred to as parameterized decision problems. In this exposition, we are particularly interested in problems parameterized by logspace (computable) size parameters of input . Those parameterized decision problems are succinctly denoted , in particular, to emphasize the size parameter . Their precise definition will be given in Section 2.2.
Among all parameterized decision problems with logspace size parameters , we are focused on combinatorial problems that can be solvable by appropriately designed polynomialtime algorithms using only sublinear space, where the informal term “sublinear” means for an appropriately chosen constant independent of . All such parameterized decision problems form the complexity class [13] (see Section 2.2 for more details). It is natural to ask if all problems parameterized by logspace size parameters (or briefly, parameterized problems) are solvable in polynomial time using sublinear space. To tackle this important question, we zero in on the most difficult (or “complete”) parameterized problems. As a typical example, let us consider the bounded 2SAT, denoted , for which every variable in a 2CNF Boolean formula appears at most 3 times in the form of literals, parameterized by (succinctly, ). It was proven in [14] that is complete for the class of all parameterized problems.
Lately, a practical working hypothesis, known as the linear space hypothesis [13], was proposed in connection to the computational hardness of parameterized problems. This linear space hypothesis (LSH) asserts that cannot be solved by any polynomialtime algorithm using sublinear space. From the completeness of , LSH immediately derives longawaited complexityclass separations, including and , where and are respectively the logspace manyone closure of (deterministic contextfree language class) and (contextfree language class) [13]. Moreover, under the assumption of LSH, it follows that 2way nondeterministic finite automata are simulated by “narrow” alternating finite automata [16].
Notice that the completeness notion requires “reductions” between two problems. The standard completeness notion uses logarithmicspace (or logspace) reductions. Those standard reductions, however, seem to be too powerful to use in many reallife circumstances. Furthermore, is not even known to be close under the standard logspace reductions. Therefore, much weaker reductions may be more suitable to discuss the computational hardness of various reallife problems. A weaker notion, called “short” logspace reductions, was in fact invented and studied extensively in [14, 13]. The importance of such reductions is exemplified by the fact that is indeed closed under short logspace reductions.
1.3 New Challenges in This Exposition
The key question of whether LSH is true may hinge at the intensive study of parameterized “complete” problems. It is however unfortunate that a very few parameterized decision problems have been proven to be equivalent in computational complexity to by short logspace reductions, and this fact drives us to seek out new parameterized decision problems in this exposition in hope that we will eventually come to the point of proving the validity of LSH. This exposition is therefore devoted to proposing a new set of problems and proving their equivalence to by appropriate short logspace reductions.
To replenish the existing short list of parameterized complete problems, after reviewing fundamental notions and notation in Section 2, we will propose three new decision problems in , which are obtained by placing “natural” restrictions on instances of the following three typical complete combinatorial problems: the vertex cover problem (VC), the exact cover by 3sets problem (3XC), and the 3dimensional matching problem (3DM) (refer to, e.g., [5] for their properties). We will then set up their corresponding natural logspace size parameters to form the desired parameterized decision problems.
In Sections 3–5, we will prove that those new parameterized decision problems are equivalent in computational complexity to by constructing appropriate short logspace reductions. Since is parameterized complete, so are all the three new problems. This completeness result immediately implies that, under the assumption of LSH, those problems cannot be solved in polynomial time using only sublinear space.
2 Fundamental Notions and Notation
We briefly describe basic notions and notation necessary to read through the rest of this exposition.
2.1 Numbers, Sets, Graphs, Languages, and Machines
We denote by the set of all natural numbers including , and denote by the set of all integers. Let . Given two numbers with , the notation expresses the integer interval . We further abbreviate as whenever . All polynomials are assumed to take nonnegative coefficients and all logarithms are taken to the base . The informal notion refers to an arbitrary polynomial in
. Given a (column) vector
(where “” is a transpose operator) and a number , the notation indicates the th entry of . For two (column) vectors and of dimension , the notation means that the inequality holds for every index . A set refers to a set that consists of exactly distinct elements.An alphabet is a finite nonempty set of “symbols” or “letters”. Given an alphabet , a string over is a finite sequence of symbols in . The length of a string , denoted , is the total number of symbols in . The notation denotes the set of all strings over . A language over is a subset of .
In this exposition, we will consider directed and undirected graphs and each graph is expressed as with a set of vertices and a set of edges. An edge between two vertices and in a directed graph is denoted by , whereas the same edge in an undirected graph is denoted by . Two vertices are called adjacent if there is an edge between them. When there is a path from vertex to vertex , we succinctly write . An edge of is said to be a grip if its both endpoints have degree at most . Given a graph , we set and . The following property is immediate.
Lemma 2.1
For any connected graph whose degree is at most , it follows that and .
If a Boolean formula in the conjunctive normal form (CNF) contains variables and clauses, then we set and as two natural size parameters. For later convenience, we call a literal in removable if no clause in contains (i.e., the negation of ). We say that is in a clean shape if each clause of consists of literals whose variables are all different. An exact 2CNF Boolean formula has exactly two literals in each clause.
As a model of computation, we use the notions of
multitape deterministic and nondeterministic Turing machines
(or DTMs and NTMs, for short), each of which is equipped with one readonly input tape, multiple rewritable work tapes, and (possibly) a writeonce^{4}^{4}4A tape is writeonce if its tape head never moves to the left and it must move to the right whenever it writes a nonblank symbol. output tape such that, initially, each input is written on the input tape surrounded by two endmarkers, (left endmarker) and (right endmarker), and all the tape heads are stationed on the designated “start cells.” Given two alphabets and , a function from to (resp., from to ) is computable in time using space if there exists a DTM such that, on each input , produces (resp., ) on the output tape before it halts within steps with accesses to at most worktape cells (not counting the inputtape cells as well as the outputtape cells). We freely identify a decision problem with its corresponding language. A decision problem is defined to be computable within time using at most space in a similar manner.2.2 Parameterized Decision Problems and Short Reductions
Throughout this exposition, we target decision problems (equivalently, languages) that are parameterized by size parameters, which specify “sizes” (i.e., positive integers) of instances given to target problems and those sizes are used as a basis to measuring computational complexities (such as execution time and memory usage) of the problems. More formally, for any input alphabet , a size parameter is a map from to . The information on the instance size is frequently used in solving problems, and thus it is natural to assume the easy “computability” of the size. A size parameter is said to be logspace computable if it is computable using space, where is a symbolic input. A parameterized decision problem is a pair with a language over a certain alphabet and a size parameter mapping to .
For any parameterized decision problem , we say that is computable in polynomial time using sublinear space if there exists a DTM that solves in time polynomial in using space, where is a certain fixed constant in the real interval . A parameterized decision problem with logspace size parameter is in if is computable in polynomial time using sublinear space.
To discuss sublinearspace computability, however, the standard logspace manyone reductions (or mreductions, for short) are no longer useful. For instance, it is unknown that all NLcomplete problems parameterized by natural logspace size parameters are equally difficult to solve in polynomial time using sublinear space. This is because is not yet known to be closed under standard mreductions. Fortunately, is proven to be closed under slightly weaker reductions, called “short” reductions [13, 14].
The short mreducibility between two parameterized decision problems and is given as follows: is short Lmreducible to , denoted by , if there is a polynomialtime, logarithmicspace computable function (which is called a reduction function) and two constants such that, for any input string , (i) iff and (ii) . Instead of using , if we use a polynomialtime logarithmicspace oracle Turing machine to reduce to with the extra requirement of for any query word made by on input for oracle , then is said to be short Treducible to , denoted by .
For any reduction in , we say that two parameterized decision problems and are interreducible (to each other) by reductions if both and hold; in this case, we briefly write .
Lemma 2.2
[13] Let and be two arbitrary parameterized decision problems. (1) If , then . (2) If and , then .
2.3 The Linear Space Hypothesis or LSH
One of the first problems that were proven to be complete in the past literature is the CNF Boolean formula satisfiability problem (3SAT), which asks whether or not a given 3CNF Boolean formula is satisfiable [2]. In sharp comparison, its natural variant, called the CNF Boolean formula satisfiability problem (), was proven to be complete [8] (together with the results of [6, 12]). Let us further consider its natural restriction introduced in [13]. Let .
Bounded CNF Boolean Formula Satisfiability Problem (SAT):

Instance: a 2CNF Boolean formula whose variables occur at most times each in the form of literals.

Question: is satisfiable?
As natural logspace size parameters for the decision problem , we use the aforementioned size parameters and .
Unfortunately, not all complete problems are proven to be interreducible to one another by short logspace reductions. An example of complete problems that are known to be interreducible to is a variant of the directed connectivity problem whose instance graphs have only vertices of degree at most () for any number .
Degree Directed  Connectivity Problem (DSTCON):

Instance: a directed graph of degree at most and two vertices

Question: is there any simple path in from to ?
Given a graph with vertices and edges, we set and as natural logspace size parameters.
Lemma 2.3
[13] Let be any integer. (1) is interreducible to and also to by short mreductions. (2) is interreducible to and further to by short mreductions. (3) is interreducible to by short Treductions.
Notice that we do not know whether we can replace short Treductions in Lemma 2.3(3) by short mreductions. This exemplifies a subtle difference between and .
Definition 2.4
The linear space hypothesis (LSH) asserts, as noted in Section 1.2, the insolvability of the specific parameterized decision problem within polynomial time using only sublinear space.
In other words, LSH asserts that . Note that, since is closed under short Treductions by Lemma 2.2(2), if a parameterized decision problem satisfies , we can freely replace in the definition of LSH by . The use of short Treduction can be relaxed to a much weaker notion of short Treduction [14, 13].
2.4 Linear Programming and Linear Equations
As a basis of later completeness proofs, we recall a combinatorial problem of Jones, Lien, and Laaser [8], who studied a problem of determining whether or not there exists a
solution to a given set of linear programs, provided that each linear program (i.e., a linear inequality) has at most
nonzero coefficients. When each variable further has at most nonzero coefficients in all the linear programs, the corresponding problem is called the entry linear programming problem [13], which is formally described as below.Entry Linear Programming Problem (LP):

Instance: a rational matrix , and a rational (column) vector of dimension , where each row of has at most nonzero entries and each column has at most nonzero entries.

Question: is there any vector for which ?
For practicality, all entries in are assumed to be expressed in binary using bits. For any instance of the form given to , we use two logspace size parameters defined as and .
It is known that, for any index , the parameterized decision problem is interreducible to and further to by short mreductions [13].
Lemma 2.5
[13] The following parameterized decision problems are all interreducible to one another by short mreductions: , , and for every index .
We can strengthen the requirement of the form in as follows. Consider another variant of , in which we ask whether or not holds for a certain vector for any given matrix and two (column) vectors and .
Bidirectional Entry Linear Programming Problem (LP):

Instance: a rational matrix , and two rational vectors and of dimension , where each row of has at most nonzero entries and each column has at most nonzero entries.

Question: is there any vector for which ?
Proposition 2.6
For any index , .
Proof. The reduction is easy to verify by setting and with for any instance pair and given to . Since the description size of is proportional to that of , the reduction is indeed “short.”
To verify the opposite reducibility , it suffices to prove that since for any by Lemma 2.5. Take an arbitrary instance given to and assume that is an matrix and and are two (column) vectors of dimension . We wish to reduce to an appropriate instance for , where is a matrix and is a dimensional vector. For all index pairs and , let , , and . For all indices , let and . Moreover, for any pair , we set , , and . In addition, we set , , and . Notice that each column of has at most nonzero entries and each row of has at most nonzero entries.
Let denote a vector of dimension for and let denote a vector of dimension for satisfying and for any . It then follows that the inequality is equivalent to . Furthermore, is equivalent to , which is the same as . Therefore, we conclude that iff . In other words, it follows that iff .
As a special case of by restricting its instances on the form with , it is possible to consider the decision problem of asking whether or not holds for an appropriately chosen vector . We call this new problem the entry linear equation problem (). As shown in Lemma 2.7, falls into , and thus this fact signifies how narrow the gap between and is. For the proof of the lemma, we define the exclusiveor clause (or the clause) of two literals and to be the formula . The problem asks whether, for a given collection of clauses, there exists a truth assignment that forces all clauses in to be true. It is known that is in [8].
Lemma 2.7
For any index , belongs to .
Proof. Consider any instance given to . Since , it suffices to reduce to by standard mreductions. Note that the equation is equivalent to for all , where and are nonzero entries of with . Fix an index and consider the first case where . In this case, we translate into a clause if , and otherwise. In the other case of , on the contrary, we translate into two clauses if , and the other values of are similarly treated. Finally, we define to be the collection of all clauses obtained by the aforementioned translations. It then follows that iff is satisfiable. This implies that iff .
3 2Checkered Vertex Covers
The vertex cover problem (VC) is a typical complete problem, which has been used as a basis of the completeness proofs of many other problems, including the clique problem and the independent set problem (see, e.g., [9, 5]). For a given undirected graph , a vertex cover for is a subset of such that, for each edge , at least one of the endpoints and belongs to .
The problem VC remains complete even if its instances are limited to planar graphs. Similarly, the vertex cover problem restricted to graphs of degree at least 3 is also complete; however, the same problem falls into if we require graphs to have degree at most . We wish to seek out a reasonable setting situated between those two special cases. For this purpose, we intend to partition all edges into two categories: grips and nongrips (where “grips” are defined in Section 2.1). Since grips have a simpler structure than nongrips, the grips need to be treated slight differently from the others. In particular, we request an additional condition, called 2checkeredness, which is described as follows. A subset of is called 2checkered exactly when, for any edge , if both endpoints of are in , then must be a grip. The 2checkered vertex cover problem is introduced in the following way.
2Checkered Vertex Cover Problem (2CVC):

Instance: an undirected graph .

Question: is there a 2checkered vertex cover for ?
Associated with the decision problem , we set up the logspace size parameters: and , which respectively express the total number of the vertices of and that of the edges of .
Given an instance of graph to , if we further demand that every vertex in should have degree at most for any fixed constant , then we denote by (Degree 2CVC) the problem obtained from . There exists a close connection between the parameterizations of and .
Theorem 3.1
.
Proof. Firstly, it is not difficult to show that by Lemma 2.1.
Next, we intend to prove that . Let be any instance to made up of a set of variables and a set of 2CNF Boolean clauses. For convenience, we write for the set of negated variables and define . In the case where a clause contains any removable literal , it is possible to delete all clauses that contain , because we can freely assign (true) to . Without loss of generality, we assume that there is no removable literal in . We further assume that is an exact 2CNF formula in a clean shape (explained in Section 2.1). Since every clause has exactly two literals, each clause is assumed to have the form for any index , where and are treated as “labels” that represent two literals in the clause .
Let us construct an undirected graph as follows. We define and we set to be by writing for and for . We further set as the union of and . Since each clause contains exactly two literals, it follows that . Thus, the edge for each index is a grip. Moreover, since each variable appears at most times in the form of literals (because of the condition of ), . Since no removable literal exists in , we obtain . It follows by the definition that since .
Here, we want to verify that iff . Assume that . Let be any truth assignment that makes satisfiable. We naturally extend to a map from to by setting to be the opposite of . Its corresponding vertex cover is defined in two steps. Initially, contains all elements satisfying . For each index , let . Notice that . If for a certain , then we append to the vertex ; however, if , then we append to the two vertices and instead.
To illustrate our construction, let us consider a simple example: with , , , and . The corresponding graph is drawn in Fig. 1. Take the truth assignment that satisfies . We then obtain , , and . Therefore, the resulting 2checkered vertex cover is the set .
By the definition of ’s, we conclude that belongs to . Conversely, we assume that . Consider any truth assignment for and construct as before. By the construction of , if is a 2checkered vertex cover, then should force to be true, a contradiction. Hence, follows. Overall, it follows that iff . Therefore, we obtain .
Conversely, we need to prove that . Given an undirected graph , we want to define a 2CNF Boolean formula to which reduces. Let and for certain numbers .
Hereafter, we use the following abbreviation: for , for , and for . Notice that, as the notation itself suggests, is logically equivalent to the negation of .
We first define a set of variables to be . For each edge , we define as follows. If one of and has degree more than , then we set to be ; otherwise, we set to be . Finally, we define to be the set of all clauses, namely, . Let denote the 2CNF Boolean formula made up of all clauses in .
Next, we intend to verify that has a 2checkered vertex cover iff is satisfiable. Assume that has a 2checkered vertex cover, say, . Consider obtained from . We define a truth assignment by setting iff . Take any edge . If one of and has degree more than , then either ( and ) or ( and hold, and thus forces to be true. Otherwise, since either or , forces to be true. This concludes that is satisfiable. On the contrary, we assume that is satisfiable by a certain truth assignment, say, ; that is, for any edge , forces to be true. We define a subset of as . Let be any edge. If has the form for , then either or should belong to . If forces in to be true, then either ( and ) or ( and ) hold. Hence, is a 2checkered vertex cover.
The completeness of follows from Theorem 3.1 since (and also ) is complete by standard mreductions [8] (based on the fact [6, 12]).
As an immediate corollary of Theorem 3.1, we obtain the following hardness result regarding the computational complexity of under the assumption of LSH.
Corollary 3.2
Under LSH, letting be any constant in , there is no polynomialtime algorithm that solves using space, where is a symbolic input.
Proof. Assume that LSH is true. If is solvable in polynomial time using space for a certain constant , since is a “natural” subproblem of , Theorem 3.1 implies the existence of a polynomialtime algorithm that solves using space as well. This implies that LSH is false, a contradiction.
4 Exact Covers with Exemption
The exact cover by 3sets problem (3XC) was shown to be complete [9]. Fixing a universe , let us choose a collection of subsets of . We say that is a set cover for if every element in is contained in a certain set in . Furthermore, given a subset , is said to be an exact cover for exempt from if (i) every element in is contained in a unique member of and (ii) every element in appears in at most one member of . When , we say that is an exact cover for . Notice that any exact cover with exemption is a special case of a set cover.
To obtain a decision problem in , we need one more restriction. Given a collection , we introduce a measure, called “overlapping cost,” of an element of any set in as follows. For any element , the overlapping cost of with respect to (w..r.t.) is the cardinality . With the use of this special measure, we define the notion of overlappingness for any as follows. We say that is overlapping if the overlapping cost of every element in w.r.t. is at most .
2Overlapping Exact Cover by Sets with Exemption Problem (XCE):

Instance: a finite set , a subset of , and a overlapping collection of subsets of such that each set in has at most elements.

Question: does contain an exact cover for exempt from ?
The use of an exemption set in the above definition is crucial. If we are given a 2overlapping family of subsets of as an input and then ask for the existence of an exact cover for , then the corresponding problem is rather easy to solve in log space [1].
The size parameter for satisfies , provided that all elements of are expressed in binary symbols. Obviously, is a logspace size parameter. In what follows, we consider parameterized by , , and prove its interreducibility to ..
Theorem 4.1
.