Boxicity and Poset Dimension
Abstract
Let be a simple, undirected, finite graph with vertex set and edge set . A dimensional box is a Cartesian product of closed intervals . The boxicity of , is the minimum integer such that can be represented as the intersection graph of dimensional boxes, i.e. each vertex is mapped to a dimensional box and two vertices are adjacent in if and only if their corresponding boxes intersect. Let be a poset where is the ground set and is a reflexive, antisymmetric and transitive binary relation on . The dimension of , is the minimum integer such that can be expressed as the intersection of total orders.
Let be the underlying comparability graph of , i.e. is the vertex set and two vertices are adjacent if and only if they are comparable in . It is a wellknown fact that posets with the same underlying comparability graph have the same dimension. The first result of this paper links the dimension of a poset to the boxicity of its underlying comparability graph. In particular, we show that for any poset , , where is the chromatic number of and . It immediately follows that if is a height poset, then since the underlying comparability graph of a height poset is a bipartite graph.
The second result of the paper relates the boxicity of a graph with a natural partial order associated with the extended double cover of , denoted as : Note that is a bipartite graph with partite sets and which are copies of such that corresponding to every , there are two vertices and and is an edge in if and only if either or is adjacent to in . Let be the natural height poset associated with by making the set of minimal elements and the set of maximal elements. We show that .
These results have some immediate and significant consequences. The upper bound allows us to derive hitherto unknown upper bounds for poset dimension such as , since boxicity of any graph is known to be at most its . In the other direction, using the already known bounds for partial order dimension we get the following: (1) The boxicity of any graph with maximum degree is which is an improvement over the best known upper bound of . (2) There exist graphs with boxicity . This disproves a conjecture that the boxicity of a graph is . (3) There exists no polynomialtime algorithm to approximate the boxicity of a bipartite graph on vertices with a factor of for any , unless .
Keywords: Boxicity, partial order, poset dimension, comparability graph, extended double cover.
1 Introduction
1.1 Boxicity
A box is a Cartesian product of closed intervals . A box representation of a graph is a mapping of the vertices of to boxes in the dimensional Euclidean space such that two vertices in are adjacent if and only if their corresponding boxes have a nonempty intersection. The boxicity of a graph denoted , is the minimum integer such that has a box representation. Boxicity was introduced by Roberts [24]. Cozzens [9] showed that computing the boxicity of a graph is NPhard. This was later strengthened by Yannakakis [31] and finally by Kratochvìl [22] who showed that determining whether boxicity of a graph is at most two itself is NPcomplete.
It is easy to see that a graph has boxicity at most 1 if and only if it is an interval graph, i.e. each vertex of the graph can be associated with a closed interval on the real line such that two intervals intersect if and only if the corresponding vertices are adjacent. By definition, boxicity of a complete graph is . Let be any graph and , be graphs on the same vertex set as such that . Then we say that is the intersection graph of s for and denote it as . Boxicity can be stated in terms of intersection of interval graphs as follows:
Lemma 1
Roberts [24]: The boxicity of a noncomplete graph is the minimum positive integer such that can be represented as the intersection of interval graphs. Moreover, if for some graphs then .
Roberts, in his seminal work [24] proved that the boxicity of a complete partite graph is . Chandran and Sivadasan [6] showed that . Chandran, Francis and Sivadasan [5] proved that where, is the chromatic number of . In [14] Esperet proved that , where is the maximum degree of . Scheinerman [25] showed that the boxicity of outer planar graphs is at most 2. Thomassen [26] proved that the boxicity of planar graphs is at most 3. In [11], Cozzens and Roberts studied the boxicity of split graphs.
1.2 Poset Dimension
A partially ordered set or poset consists of a non empty set , called the ground set and a reflexive, antisymmetric and transitive binary relation on . A total order is a partial order in which every two elements are comparable. It essentially corresponds to a permutation of elements of . A height poset is one in which every element is either a minimal element or a maximal element. A linear extension of a partial order is a total order which satisfies . A realizer of a poset is a set of linear extensions of , say which satisfy the following condition: for any two distinct elements and , in if and only if in , . The poset dimension of (sometimes abbreviated as dimension of ) denoted by is the minimum integer such that there exists a realizer of of cardinality . Poset dimension was introduced by Dushnik and Miller [12]. Clearly, a poset is onedimensional if and only if it is a total order. Pnueli et al. [23] gave a polynomial time algorithm to recognize dimension 2 posets. In [31] Yannakakis showed that it is NPcomplete to decide whether the dimension of a poset is at most 3. For more references and survey on dimension theory of posets see Trotter [28, 29]. Recently, Hegde and Jain [21] showed that it is hard to design an approximation algorithm for computing the dimension of a poset.
A simple undirected graph is a comparability graph if and only if there exists some poset , such that is the vertex set of and two vertices are adjacent in if and only if they are comparable in . We will call such a poset an associated poset of . Likewise, we refer to as the underlying comparability graph of . Note that for a height poset, the underlying comparability graph is a bipartite graph with partite sets and , with say corresponding to minimal elements and to maximal elements. For more on comparability graphs see [19]. It is easy to see that there is a unique comparability graph associated with a poset, whereas, there can be several posets with the same underlying comparability graph. However, Trotter, Moore and Sumner [30] proved that posets with the same underlying comparability graph have the same dimension.
2 Our Main Results
The results of this paper are the consequence of our attempts to bring out some connections between boxicity and poset dimension. As early as 1982, Yannakakis had some intuition regarding a possible connection between these problems when he established the NPcompleteness of both poset dimension and boxicity in [31]. But interestingly, no results were discovered in the last 25 years which establish links between these two notions. Perhaps the researchers were misled by some deceptive examples such as the following one: Consider a complete graph where is even and remove a perfect matching from it. The resulting graph is a comparability graph and the dimension of any of its associated posets is 2, while its boxicity is . In this context it may be worth recalling a result from [16] which relates the poset dimension to another parameter namely the dimension of box orders. A poset is said to be a box order in dimensions if there exists a mapping of its elements to dimensional axisparallel boxes such that in if and only if the box of strictly contains the box of . Box order is a particular type of geometrical containment order (See [16, 28]). The result is as follows: the dimension of is at most if and only if it is a box order in dimensions [18, 20]. But note that boxicity is fundamentally different from box orders. As in the case of the above example, we can demonstrate families of posets of constant dimension whose underlying comparability graphs have arbitrarily high boxicity, which is in contrast with the above result on box orders.
First we state an upper bound and a lower bound for the dimension of a poset in terms of the boxicity of its underlying comparability graph.
Theorem 2.1
Let be a poset such that and its underlying comparability graph. Then,
Theorem 2.2
Let be a poset and let be the chromatic number of its underlying comparability graph such that . Then, .
Note that if is a height poset, then is a bipartite graph and therefore . Thus, from the above results we have the following:
Corollary 1
Let be a height poset and its underlying comparability graph. Then, .
The double cover and extended double cover of a graph are popular notions in graph theory. They provide a natural way to associate a bipartite graph to the given graph. In this paper we make use of the latter construction.
Definition 1
The extended double cover of , denoted as is a bipartite graph with partite sets and which are copies of such that corresponding to every , there are two vertices and and is an edge in if and only if either or is adjacent to in .
We prove the following lemma relating the boxicity of and .
Lemma 2
Let be any graph and its extended double cover. Then,
Let be the natural height poset associated with , i.e. the elements in are the minimal elements and the elements in are the maximal elements. Combining Corollary 1 and Lemma 2 we have the following theorem:
Theorem 2.3
Let be a graph and be the natural height poset associated with its extended double cover. Then, and therefore .
2.1 Consequences
2.1.1 New upper bounds for poset dimension:
Our results lead to some hitherto unknown bounds for poset dimension. Some general bounds obtained in this manner are listed below:

In [3] it is proved that , where is the cardinality of the minimum vertex cover of . Therefore, we have .
Some more interesting results can be obtained if we restrict to belong to certain subclasses of graphs. Note that there are several research papers in the partial order literature which study the dimension of posets whose underlying comparability graph has some special structure – interval order, semi order and crown posets are some examples.

Bhowmick and Chandran [1] proved that boxicity of ATfree graphs is at most . Hence, , if is ATfree.
2.1.2 Improved upper bound for boxicity based on maximum degree:
Füredi and Kahn [17] proved the following
Lemma 3
Let be a poset and be the maximum degree of . Then, there exists a constant such that .
From Lemma 2 and Corollary 1 we have , where is the extended double cover of . Note that by construction . On applying the above lemma, we have
Theorem 2.4
For any graph having maximum degree there exists a constant such that .
This result is an improvement over the previous upper bound of by Esperet [14].
2.1.3 Counter examples to the conjecture of [5]:
Chandran et al. [5] conjectured that boxicity of a graph is . We use a result by Erdős, Kierstead and Trotter [13] to show that there exist graphs with boxicity , hence disproving the conjecture. Let be the probability space of height posets with minimal elements forming set and maximal elements forming set , where for any and , . They proved the following:
Theorem 2.5
[13] For every , there exists so that if , then, for almost all .
2.1.4 Approximation hardness for the boxicity of bipartite graphs:
Hegde and Jain [21] proved the following
Theorem 2.6
There exists no polynomialtime algorithm to approximate the dimension of an element poset within a factor of for any , unless .
This is achieved by reducing the fractional chromatic number problem on graphs to the poset dimension problem. In addition they observed that a slight modification of their reduction will imply the same result for even height posets. From Corollary 1, it is clear that for any height poset , . Suppose there exists an algorithm to compute the boxicity of bipartite graphs with approximation factor , for some , then, it is clear that the same algorithm can be used to compute the dimension of height posets with approximation factor , a contradiction. Hence,
Theorem 2.7
There exists no polynomialtime algorithm to approximate the boxicity of a bipartite graph on vertices with a factor of for any , unless .
3 Notations
Let denote where is a positive integer. For any graph , let and denote its vertex set and edge set respectively. If is undirected, for any , means is adjacent to and if is directed, means there is a directed edge from to . Whenever we refer to an bipartite (or cobipartite) graph, we imply that its vertex set is partitioned into nonempty sets and where both these sets induce independent sets (cliques respectively).
In a poset , the notations , in and are equivalent and are used interchangeably. denotes the underlying comparability graph of . A subset of is called a chain if each pair of distinct elements is comparable. If each pair of distinct elements is incomparable, then it is called an antichain. Given an bipartite graph , the natural poset associated with with respect to the bipartition is the poset obtained by taking to be the set of minimal elements and to be the set of maximal elements. In particular, if is the extended double cover of , we denote by the natural associated poset of .
Suppose is an interval graph. Let be an interval representation for , i.e. it is a mapping from the vertex set to closed intervals on the real line such that for any two vertices and , if and only if . Let and denote the left and right end points of the interval corresponding to the vertex respectively. In this paper, we will never consider more than one interval representation for an interval graph. Therefore, we will simplify the notations to and . Further, when there is no ambiguity about the graph under consideration and its interval representation, we simply denote the left and right end points as and respectively. Note that for any interval graph there exists an interval representation with all end points distinct. Such a representation is called a distinguishing interval representation. It is an easy exercise to derive such a distinguishing interval representation starting from an arbitrary interval representation of the graph.
4 Proof of Theorem 2.1
Let . Note that since , cannot be a complete graph and therefore . Let be a set of interval graphs such that . Now, corresponding to each we will construct two total orders and such that is a realizer of .
Let and be an interval representation of . We will define two partial orders and as follows: , belongs to and and for every nonadjacent pair of vertices with respect to ,
Partial orders constructed in the above manner from a collection of closed intervals are called interval orders (See [29] for more details). It is easy to see that (the complement of ) is the underlying comparability graph of both and .
Let and be two directed graphs with vertex set and edge set and respectively. Note that from the definition it is obvious that there are no directed loops in and .
Lemma 4
and are acyclic directed graphs.
Proof
We will prove the lemma for – a similar proof holds for . First of all, since is not a complete graph . Suppose is a total order, i.e. if is an antichain, then it is clear that and therefore is acyclic. Henceforth, we will assume that is not a total order.
Suppose is not acyclic. Let be a shortest directed cycle in .
First we will show that ( is the length of ). If , then there should be such that . Since is a partial order, and cannot be simultaneously present in . The same holds for . Thus, without loss of generality we can assume that and . But if , then, and are adjacent in and thus adjacent in . Then clearly the intervals of and intersect and therefore , a contradiction.
Now, we claim that two consecutive edges in cannot belong to (or ). Suppose there do exist such edges, say and which belong to (or ) (note that the addition is modulo ). Since (or ) is a partial order, it implies that (or ) and as a result we have a directed cycle of length , a contradiction to the assumption that is a shortest directed cycle. Therefore, the edges of alternate between and . It also follows that .
Without loss of generality we will assume that . We claim that is an induced poset of . First of all and are not comparable in as they are comparable in . If either or are comparable, then we can demonstrate a shorter directed cycle in , a contradiction. Finally we consider the pair . If , then they are not comparable as they are comparable in while if and if they are comparable, then, it would again imply a shorter directed cycle, a contradiction. Hence, is an induced subposet. In the literature such a poset is denoted as where refers to disjoint sum and 2 is a twoelement total order. Fishburn [15] has proved that a poset is an interval order if and only if it does not contain a . This implies that is not an interval order, a contradiction.
We have therefore proved that there cannot be any directed cycles in . In a similar way we can show that is an acyclic directed graph. ∎
Since and are acyclic, we can construct total orders, say and using topological sort on and such that and (For more details on topological sort, see [8] for example).
For each , we create linear extensions and as described above. We claim that is a realizer of . For each , it is clear from construction that . If and are not comparable in , then , and therefore there exists some interval graph such that . Assuming that the interval for occurs before the interval for in the interval representation of , it follows by construction that and and therefore and . Hence proved.
4.1 Tight Example for Theorem 2.1
Consider the crown poset : a height poset with minimal elements and maximal elements and , for , where the addition is modulo . Its underlying comparability graph is the bipartite graph obtained by removing a perfect matching from the complete bipartite graph . The dimension of this poset is (see [27, 29]) while the boxicity of the graph is [3].
5 Proof of Theorem 2.2
We will prove that . Let , and a realizer of . Now we color the vertices of as follows: For a vertex , if is the length of a longest chain in such that is its maximum element, then we assign color to it. This is clearly a proper coloring scheme since if two vertices and are assigned the same color, say and , then it implies that the length of a longest chain in which occurs as the maximum element is at least , a contradiction. Also, this is a minimum coloring because the maximum number that gets assigned to any vertex equals the length of a longest chain in , which corresponds to the clique number of .
Now we construct interval graphs and show that is an intersection graph of these interval graphs. Let be the permutation induced by the total order on , i.e. if and only if . The following construction applies to all graphs in except . Let . We assign the point interval for all vertices colored . For all vertices colored , we assign and for those colored , we assign . The interval assignment for the last interval graph is as follows: for all vertices colored we assign the point interval and for the rest of the vertices we assign the interval . Next, we will show that .
Claim 1
If and are adjacent in , then they are adjacent in all .
Proof
Let be colored and be colored . It is clear that and without loss of generality we will assume that . By the way we have colored, it implies that in and therefore , . Let , and be the interval graph under consideration. There are 5 possible cases which we consider one by one:
Case 1:
By construction in , and are assigned intervals and respectively and therefore and are adjacent in , .
Case 2:
and are assigned intervals and respectively and therefore are adjacent in , .
Case 3:
is assigned interval and is assigned interval . Since , it follows that is adjacent to in , .
Case 4:
If and , then and therefore is assigned and is assigned . If not, then is assigned the point interval and is assigned . In either case, since , the two vertices are adjacent.
Case 5:
Since , it implies that . Therefore, if and , then and are assigned and respectively. If not, then is assigned the point interval and is assigned . Again, since , in either case the two vertices are adjacent. Hence proved.
Claim 2
If and are not adjacent in , then there exists some such that .
Proof
Again let be colored and be colored . Recall that . If , then by construction it is clear that and are not adjacent in if and when , then they are not adjacent in . Therefore, without loss of generality we will assume that . Since and are not adjacent in , they are incomparable in and therefore, there exists some such that in which in turn implies that . There are 2 possible cases:
Case 1:
Since , in , and are assigned intervals and respectively and therefore, since and are not adjacent in .
Case 2:
Clearly . If , then it is similar to the previous case. If , then, in , and are assigned and respectively. Since , and are not adjacent in .
Hence we have proved Theorem 2.2.
Consider a complete partite graph on vertices where , i.e. is a partition of where . For any two vertices and , if and only if . is a comparability graph and here is one transitive orientation of : for every pair of adjacent vertices and , where and , make if and only if . Let be the resulting poset. It is an easy exercise to show that . The chromatic number of is and Roberts [24] showed that its boxicity is . From Theorem 2.2 it follows that . Therefore, the complete partite graph serves as a tight example for Theorem 2.2.
However, it would be interesting to see if there are posets of higher dimension for which Theorem 2.2 is tight.
6 Boxicity of the extended double cover
In this section, we will prove Lemma 2. But first, we will need some definitions and lemmas.
Definition 2
Let be an bipartite graph. The associated cobipartite graph of , denoted by is the graph obtained by making the sets and cliques, but keeping the set of edges between vertices of and identical to that of , i.e. , if and only if .
The associated cobipartite graph is not to be confused with the complement of (i.e. ) which is also a cobipartite graph.
Definition 3
(Canonical interval representation of a cobipartite interval graph:) Let be an cobipartite interval graph. A canonical interval representation of satisfies: , and , where the points and are the leftmost and rightmost points respectively of the interval representation.
We claim that such a representation exists for every cobipartite interval graph. Note that if is a complete graph, the claim is trivially true. Therefore we take to be noncomplete. Consider any interval representation of . Since is a clique there exists a point, say which is contained in all intervals corresponding to vertices in . Similarly, let be a point in the intersection of intervals corresponding to vertices of . Since is noncomplete, it is clear that . By definition of and we have and . Without loss of generality we can assume that and as a result and for all vertices . This means no interval ends before the point and no interval starts after the point . Hence, it follows that for any interval containing , we can make its left end point and for an interval containing , we can make its right end point. Therefore, we have a canonical interval representation of .
The following lemma is easy to verify.
Lemma 5
Consider two closed intervals on the real line with left end points , and right end points , . Then, the two intervals intersect if and only if and . In other words, the two intervals do not intersect if and only if or .
Lemma 6
Let be an bipartite graph and its associated cobipartite graph. If is a noninterval graph, then
If is an interval graph, then .
Proof
We first show that . Let and , where are interval graphs. Note that since is a supergraph of a cobipartite graph, it is a cobipartite interval graph. Let us consider a canonical interval representation for each and further assume that the right end points of all vertices in and left end points of all vertices in are distinct. Let be the interval graph obtained by making and keeping the rest of the intervals unchanged. Similarly, let be the interval graph obtained by making . Due to our assumption of distinct end points it is clear that and are independent sets in and respectively. Suppose and . For :
(by construction of from )  
From this, we immediately see that .
Now suppose , i.e. is an interval graph. Then we set and proceed as in the previous case. Hence, . Note that this inequality is tight: take for example , the cycle of length 4. is and therefore an interval graph, but is not.
Now we show that . Let and , where are interval graphs. For each , we create two interval graphs and as follows: Consider an interval representation of . Let and , the leftmost and rightmost points in the interval representation respectively. and are defined as follows:
Now we show that . From the definitions it is clear that in each , and are cliques– for example, the interval corresponding to every vertex in in contains . Therefore we will assume that and .
In , and and in , and . Therefore and are adjacent in both and . Now suppose
In the interval representation of , if , then, by definition and hence, . If , then, and therefore, . Hence proved. ∎
6.1 Proof of Lemma 2
:
Let and where s are interval graphs. For each , we construct interval graphs with vertex set as follows: Consider an interval representation for . For every vertex in , we assign the interval of to and in